z [see Eq. (4.6)], 17(XJ) 17 (X2) 0 [see Eq. (4.9)] and hence no variation is permitted in
dS2 (4.105) v v ~ Thus, according to the principle of minimum potential energy, the displacement field (x, y. z) that minimizes U and satisfies all the boundary conditions is the one that satisfies the equilibrium equations. In the principle of minimum potential energy, we minimize the functional U, and the resulting equations denote the equilibrium equations while the compatibility conditions are satisfied identically. U(u,
v, w)
u
4.10.2
Principle of Minimum Complementary
Energy
The complementary energy of an elastic body (Ue) is defined as Ue = complementary strain energy in terms of stresses (if)
- work done by the applied loads during stress changes
(W p)
The principle of the minimum complementary energy can be stated as follows: Of all possible stress states that satisfy the equilibrium equations and the stress boundary conditions, the state that satisfies the compatibility conditions will make the complementary energy assume a minimum value. By expressing the complementary energy
Derivation of Equations: Variational Approach Uc in teons of the stresses for compatibility, 8Uc(axx, ayy,""
azx)
aij,
the principle of minimum complementary energy gives,
= 8Jf«axx,
ayy,""
():x) - 8W p(axx, ()yy, ... , azx)
=0 (4.106)
where the variation is taken with respect to the stress components in Eq. (4.106) while the displacements are assumed constant. The complementary strain energy of a linear elastic body can be expressed as (4.107) In the presence of known initial strains Jf =
eo, the complementary
!JJJ aT ([C]a v
strain energy becomes
+ 2eo) dV
(4.108)
where the strain-stress relations are assumed to be of the foon e = [C]a
(4.109)
The work done by the applied loads during stress change, also known as complementary work, is given by Wp
=
ff
u
(¢>x +¢>y'v+¢>zw)d51
=
~
ff J7fr
(4.110)
d51
~
where 51 is the part of the surface of the body on which the values of displacements are prescribed as
Thus, the complementary energy of the body can be expressed. using Eqs. (4.108) and (4.110), as Uc(axx, ayy, ... , azx)
= !JJJ a([C]a + 2£0) dV v
-
Ir ¢Tfr d5 ~
1
(4.111)
In the principle of minimum complementary energy, the functional Uc is minimized and the resulting equations denote the compatibility equations, while the equilibrium equations are satisfied identically. Principle of Stationary Reissner Energy In the principle of minimum potential energy, the potential energy U is expressed in terms of displacements, and variations of u, v, and w are permitted. Similarly, in the case of the principle of minimum complementary energy, the complementary energy Uc is expressed in terms of stresses, and variations ofaxx, ayy ..... azx are pemlitted. In the present case, the Reissner energy Ur is expressed in terms of both displacements and
1
I
I'r
4.10
':.107
Variational Methods in Solid Mechanics
stresses, and variations with respect to both displacements and stresses are pennitted. The Reissner energy for a linearly elastic body is defined as Ur =
III
(internal stresses x strains expressed in tenns of displacements
v
- complementary strain energy in tenns of stresses) dV - work done by applied forces =
III {[ - III -II III
CTxx::
+ CT yy :~ + ... + CTzx(~:
v
(4)xu
+ 4)yv + 4)zw)
dV -
v
II
if} d
+ ~:) ] -
V
(cI>xu + cI>yv + cI>~w) dSz
~
[(u - U)cI>x + (v - V)cI>y
+ (w
- W}cI>z) dSI
51
=
(uTi - !uT[C]u
-
4/ u) dV - JJ uT4>dSZ 52
V
-
JJ(u -
fr)Tq,dSI
51
(4.112) When the variation of Ur is set equal to zero by considering variations in both displacements and stresses, we obtain oUr oUr = "~ -OCTij .. OCTiJ' IJ
+ (OUr -ou
AU
oUr OUr) + --:.-ov + -~-ow uV.
uw
=0
(4.113)
where the subscripts i and j are used to include all the components of stress CTxx,CTyy, ... , CTzx. The first tenn on the right-hand side of Eq. (4.113) gives the stress-displacement relations, and the second tenn gives the equilibrium equations and boundary conditions. The principle of stationary Reissner energy can be stated in words as follows: Of all possible stress and displacement states a body can have, the particular set that makes the Reissner energy stationary gives the correct stress-displacement and equilibrium equations, along with the boundary conditions.
4.10.4 Hamilton's Principle The variational principle that can be used for dynamic problems is called Hamilton's principle. According to this principle, variation of the functional is taken with respect to time. The functional used in Hamilton's principle, similar to U, Uc, and Ur, is called the Lagrangian (L) and is defined as L
=T-
U
= kinetic energy -
potential energy
(4.114)
Development of Hamilton's principle for discrete as well as continuous systems is presented in the following sections. ,,*'
Derivation of Equations: Variational Approach
Hamilton's Principle for Discrete Systems Let a discrete system (system with a finite number of degrees of freedom) be composed of n masses or particles. First, we consider a single particle of mass m, at the position vector r, subjected to a force j(r). The position of the particle, at any time t is given by Newton's second law of motion:
r,
(4.115) If the true path of the particle is r(1), we define a varied path as r + or, where or denotes the variation of the path at any fixed time t. We assume that the true path and the varied path are same at two distinct times tl and t2,so that (4.116) By taking the dot product of Eq. (4.115) with from tl to t2 yields
1
[dm-r
12
... ]
2
2 - j(r) dt
11
or
and integrating with respect to time
. ordt
=0
(4.117)
The first term of the integral in Eg. (4.117) can be integrated by parts as 2
d2rfl2 m-2 ·ordt=-. 11 dt
t [
. 11
dr m-·dt
d(or) .. dr -dt+m-.or dt dt
12
1
11
(4.118)
In view of Eg. (4.116), the second term on the right side of Eg. (4.118) will be zero and Eq. (4.117) becomes
1
12[
I)
dr m-· dt
d -or+ dt
f(r)
'OY
]
dt =0
(4.119)
The kinetic energy of the particle (T) is given by 1 dr dr T = -'m-.2 dt dt
(4.120)
dr dr dr dor oT = m-·o= m-.dt dt dt dt
(4.121)
and hence
Using Eg. (4.121) in Eg. (4.119), the general form of Hamilton's principle for a single mass (particle) can be expressed as 12
1
(oT + j.or) dt = 0
11
(4.122)
l' ,~ •...• ,~,.".,' •.•..... , •.._ •..•~,."i" .•....
I
~."_"
.••~-.,"t.,"1!""
-~·~-I',·~-·..,.=,--.-·-.·-··_",-, . .,...-.-~,-, ...-.~'-""'"""'---"~~'.""~_'"'l:""'~""",'"" ..~"..,."'!'.,." ...••••_-,...,.;'..
'
.. ~_~,"'"'l1~~~
•••'"_,._,
..'. "'-,~~~
__ ~.1'~_
.••. ~.~~J'~~~~~._,-)::l-~.-.
4.10 Variational Methods in Solid Mectmnics . ·Hll)
Conservative Systems For a conservative system, the sum of the potential and kinetic energies is a constant, and the force can be derived from the potential energy U as
1
(4.123)
f = -VU where V denotes the gradient operator. Noting that
_ au
VU.8r = -8x
ax
\',;k"
au
au
ay
az
+ -8y + -,-8z
= 8U(x, y,:)
(4.124)
Hamilton's principle for a single mass in a conservative system is given by 12
8/ = 8
1
L dt = 0
(4.125)
II
where (4.126)
L=T-U
is called the Lagrangian function. Thus, Hamilton's principle for a particle acted by a conservative force can be stated as follows: Of all possible paths that the particle could take from its position at time t1 to its position at time t2, the trUe path will be the one that extremizes the integral --
. 1
12
/ =
(4.127)
L dt
11
r,
Use of Generalized Coordinates If the position of the path at any time t, is expressed in terms of the generalized coordinates q1, q2, and q3 (instead of x, y, and z), the Lagrangian L can be expressed as (4.128)
where qi = dqi! dt (i = 1, 2, 3) denotes the ith generalized velocity. Then the necessary condition for the extremization of / can be written as
If qi are linearly independent, with no constraints among qi, all 8qi are independent, and hence Eq. (4.129) leads to
aL dtd (aL)_o aqi -
aqi -
,
i = 1, 2, 3
(4.130)
Equations (4.130) denote the Euler-Lagrange equations that correspond to the extremization of I and are often called the Lagrange equations of motion.
~.
110
Derivation of Equations: Variational Approach
Nonconservative Systems If the forces are not conservative, the general form of Hamilton's principle, given by Eq. (4.122), can be rewritten as
/2
1
o
T dt
+
1/2
/1
oWnc dt = 0
(4.131)
/1
where oWnc = j.or
denotes the virtual work done by the nonconservative force j. In this case, a functional I does not exist for extremization. If the virtual work oWnc is expressed in terms of generalized coordinates (ql,q2,q3) and generalized forces (Ql,Q2,Q3) as 3
oWnc =
L Qioqi
(4.132)
i=1
where Oqi is the virtual generalized displacement, Eq. (4.131) can be expressed as
1 L [aT--- d (aT)-. /2 3
/1'
aqi
1=1
dt
+Qi
] Oqidt=O
(4.133)
aqi
Thus, the Euler-Lagrange equations corresponding to Eq. (4.133) are given by aqi
dt
i = 1, 2, 3
aqi
(4.134)
System of Masses If a system of n mass particles or rigid bodies with masses mi and position vectors are considered, Hamilton's principle can be expressed as follows. For conservative forces,
ri
/2
o
1
L(ql, q2, ... , 'It, Q2, ... ) dt = 0
(4.135)
II
which is a generalization of Eq. (4.125). For nonconservative forces, o
/2Tdt+ 1/2
1 II
oWncdt=O
(4.136)
I
which is a generalization of Eq. (4.131). The kinetic energy and the virtual work in Eqs. (4.135) and (4.136) are given by (4.137) II
oW(oWnc) =
L };'Ori
(4.]38)
i=l
As can be seen from Eqs. (4.125) and (4.131), Hamilton's principle reduces the problems of dynamics to the study of a scalar integral that does not depend on the coordinates used. Note that Hamilton's principle yields merely the equations of motion of the system but not the solution of the dynamics problem.
..""~~~""",,.-_~~.,
--
~
-,
~
~-"":"";'lS:~~~';!:V..p4""~~
~
,~~"'_'--:or. ~_
-_.~,~~~'-.,.l,-,,-?l,,-.-
J .••.•. , .•
;1!4-_~~:o<"""-,~"1J~.~'Z~~"I;...,.
••~~~"':"f~~~"'T'!~",-~~-~._
.-.~.!-,o .•-~,.;~-"..._.,_~~~~~~
Variational Methods in Solid Mechanics -Ill
4.10
Hamilton's Principle for Continuous Systems energy of the body, T, can be expressed as T =
For a continuous system, the kinetic
t fffv p~T~dV
(4.139)
where p is the density of the material and ~ is the vector of velocity components at any point in the body:
Thus, the Lagrangian can be written as L = ~
III
+ 2uT4;)dV +
(p~T~ - gT[Dli
V
II
T dS ¢ 2
il
(4.140)
~
Hamilton's principle can be stated in words as follows: Of all possible time histories of displacement states that satisfy the compatibility equations and the constraints or the kinematic boundary conditions and that also satisfy the conditions at initial and final times t1 and t2, the history corresponding to the actual solution makes the Lagrangian functional a minimum. Hamilton's principle can thus be expressed as
o
12
1
(4.141)
L dt = 0
11
Generalized Hamilton's Principle For an elastic body in motion, the equations of dynamic equilibrium for an element of the body can be written, using Cartesian tensor notation, as (4.142) i = 1,2,3 where p is the density of the material,
(7ij
= [
(711
(712
(713]
(721
(722
(723
(731
(732
(733
[ (7xx
==
(7xy
(7xy
(7yy
(7xz
(7yz
(4.143)
and (4.144) with Xl = X, x2 = y, X3 = z and U1 = u, U2 = u, U3 = w. The solid body is assumed to have a volume V with a bounding surface 5. The bounding surface 5 is assumed to be composed of two parts, 51 and 52, where the displacements Ui are prescribed on 51 and surface forces (tractions) are prescribed on 52. Consider a set of virtual displacements OUi of the vibrating body which vanishes
,....
--------------------------------------------
112
Derivation of Equations: Variational Approach
over the boundary surface Slo where values of displacements are prescribed, but are arbitrary over the rest of the boundary surface S2, where surface tractions are prescribed. The virtual work done by the body and surface forces is given by
III
¢iOUidV
+
II
(4.145)
s
v
where
= GijVj
L GijVj,
==
(4.146)
i=I,2,3
j=I
where v = {VI V2 V3}T is the unit vector along the outward normal of the surface S with VI, v2, and V3 as its components along the Xl, x2, and X3 directions, respectively. By substituting Eg. (4.146), the second term on the right-hand side of Eq. (4.145) can be written as
II
(4.147)
GijOUiVj dS
s Using Gauss's theorem [7], expression (4.147) can be rewritten in terms of the volume integral as
II
II III
Gjj9UiVjdS=
s
111(GijOUi),jdV
s
=
v
Gij,jOUi dV
+
v
III
GijOUi,j dV
(4.148)
v
Because of the symmetry of the stress tensor, the last term in Eq. (4.148) can be written as
III
GijOUi,j dV =
III
v where
Gij [!(OUi,j
+ oUj.i)]dV
=
v eij
fff
GijOeij dV
(4.149)
v
denotes the strain tensor:
eij
=
r::: ::~ :::] r::: ::: :.::] ==
l e31 e32 e33
lex:
ey:
(4.150)
e::
In view of the equations of dynamic equilibrium, Eq. (4.142), the first integral on the right hand side of Eq. (4.148), can be expressed as (4.151)
....... ~..."...~~""'l'~~~.~'I'
,
__~.""""7.,.....~.,.""~;~;!l1':,....~~'1.~'.~~l"':!'.,~"'~
•.•••••. I'I!"'.~~"'l"
4.10
•..-'~~"".'~"'''''''''''''''
.' ....•......•.. , --.".-...,.~".~'"~:
Variationa1'Methods in Solid Mechanics
113
Thus, the second tenn of expression (4.145) can be written as (4.152)
This gives the variational equation of motion (4.153)
This equation can be stated more concisely by introducing different levels of. restrictions. If the body is perfectly elastic, Eq. (4.153) can be stated in tenns of the strain energy density Jro as (4.154)
or (4.155)
If the variations ~Ui are identified with the actual displacements (audat)dt during a small time interval dt, Eq. (4.155) states that in an arbitrary time interval. the sum of the energy of defonnation and the kinetic energy increases by an amount that is equal to the work done by the external forces during the same time interval. Treating the virtual displacements ~Ui as functions of time and space not identified with the actual displacements, the variational equation of motion. Eq. (4.154), can be integrated between two arbitrary instants of time t1 and t2 and we obtain
(4.156)
Denoting the last tenn in Eq. (4.156) as A, inverting the order of integration, and integrating by parts leads to A =
Iff v
au· ~u. dVlt2 p_1 at "1
If! l
t2
-
dV
,(
au· (a~u. _, p __ ' at at
ap + -~u· at
) dt
(4.157)
I
V
In most problems, the time rate of change of the density of the material, ap/at, can be neglected. Also, we consider ~Ui to be zero at all points of the body at initial and final times t1 and t2, so that (4.158)
114
Derivation of Equations: Variational Approach
In view of Eq. (4.158), Eq. (4.157) can be rewritten as A
=-
12
[
II
Iff Iff v
= _ [12
=-
au· a8u· dVdt p-'--' at at
8
11
v
11
1 au· au· dVdt -p-'-' 2 at at
where T
=
1
2:
[12 Iff
=-
Iff P7itat
aUj aUj d
v
au· p-'8-' at
au· dVdt at
[12 8Tdt 11
(4.159)
(4.160)
V
v is the kinetic energy of the vibrating body. Thus, Eq. (4.156) can be expressed as
/12
8(rr - T) dt =
/'2 III
cf>i8UjdV dt
+
v
/12 II
ct>j8ujd5 dt
(4.161)
52
where rr denotes the total strain energy of the solid body:
III
rr =
rro dV
(4.162)
v If the external forces acting on the body are such that the sum of the integrals on the right-hand side of Eq. (4.161) denotes the variation of a single function W (known as the potential energy of loading), we have
III
+
cf>i8uid~
v
-..
II
ct>j8ujd5 = -8W
(4.163)
52
Then Eq. (4.161) can be expressed as 8[12 Ldt I]
=[12
(rr _ T
+ W)dt = 0
(4.164)
I]
where L=rr-T+W
(4.165)
is called the Lagrangian function and Eq. (4.164) is known as Hamilton's principle. Note that a negative sign is included, as indicated in Eq. (4.163), for the potential energy of loading (W). Hamilton's principle can be stated in words as follows: The time integral of the Lagrangian function between the initial time t1 and the final time t2 is an extremum for the actual displacements (motion) with respect to all admissible virtual displacements that vanish throughout the entire time interval: first, at all points of the body at the instants tl and t2, and second, over the surface 51, where the displacements are prescribed. Hamilton's principle can be interpreted in another way by considering the displacements Uj(xj, X2, X3, t), i = 1, 2, 3, to constitute a dynanlic path in space. Then Hamilton's principle states: Among all admissible dynamic paths that satisfy the prescribed geometric boundary conditions on 51 at all times and the prescribed conditions at two arbitrary instants of time t1 and t2 at every point of the body, the actual dynamic path (solution) makes the Lagrangian function an extremum.
"r"""'-'-~-~.'--"-'''''' .---,
I
..;~!t:...""~.~' _""",~;_~,."".,.""N."1';·,~""_'_"'-:."r.k_,
..7., ~"_':"""'I""'~~l __ .·~-,:r.:>~-,,·c·.·"-"".-"
'o::,.,.....~••..•. ""-~".~""'~
.."'!'_-'"
•••. ~~III':~--,..,,\"".'"
,..-.,-.-. ..,~,;"..., .••.~
4.11.1
__
Applications of Hamilton's PtincrpleUS
4.11
4.11
.."..,:r'." .,~'"
••.~~~_~~~~J'i:1""f"7~~~.~~'~"\f"l--.\'.'.
APPLICATIONS OF HAMILTON'S PRINCIPLE Equation of Motion for Torsional Vibration of a Shaft (Free Vibration) Strain Energy
To derive a general expression for the strain energy of a shaft, consider the shaft to be of variable cross section under a torsional load as shown in Fig. 4.5. If O(x, t) denotes the angular displacement of the cross section at x, the angular displacement of the cross section at x + dx can be denoted as B(x, t) + [oO(x, t)jox]dx, due to the distributed torsional load m,(x, t). The shear strain at a radial distance r is given by y = r(aOjox). The corresponding shear stress can be represented as r Gy Gr(aO jox), where G is the shear modulus. The strain energy density Jro can be represented as Jro ~ry iGr2 (aOjax)2. The total strain energy of the shaft can be determined as
=
=
Jr =
=
fff
L
JrodV = i
=
ff
v
(:~y
2
lGr
L
= li
dAdx
GJ
(:~y dx
(4.166)
A
where V is the volume, L is the length, A is the cross-sectional moment of inertia (for a uniform circular shaft) of the shaft. The kinetic energy
Kinetic Energy expressed as
T
1
area, and J = I p polar
of a shaft with variable cross section can be
{L lo(x) (ao(xat' t»)2
='2 10
dx
(4.167)
where lo(x) = plp(x) is the mass moment of inertia per unit length of the shaft and p is the density. By using Eqs. (4.166) and (4.167), Hamilton's principle can be used to obtain
;;1
'2
(T -Jr)dt
=;; {L
10
'I
[110
{L 10 (00)2
dx
at
2
_1 10
{L GJ (aO)2 dX] dt = 0
ax
2
(4.168) By carrying out the variation operation, the various terms in Eq. (4.168) can be rewritten, noting that;; and ajat as well as ;; and ajax are commutative, as
;;
1 [IlL '2
'I
2
0
10 (aO)2] dx at
11
L
'2
dt = -
'I
a
0
2 a 0 10-280dxdt at
C
I
,_,_,l,-!,-.-,_._. J~,_•.x I I
I I I
x
·1 Figure 4.5
Torsional vibration of a shaft.
(4.169)
116
Derivation of Equations: Variational Approach
e is prescribed
assuming that
o
1 [IlL2
t1 and t2 so that oe = 0 at t1 and t2. Similarly,
GJ (ae)2] dx dt ax
12
II
at
0
=
1',)2 [ GJ-oeaxae
IL 0
lL -ox·0 (af))GJ- ax
OfJdx] dt
0
(4.170) Thus, Eq. (4.168) becomes
1 jl [O -ax ( GJ-axae) 2
2 at
L
ae 1(2) dt = 0 oedx - GJ-oe (4.171) ax Assuming that 8e = 0 atx = 0 and x = L, and 8e is arbitrary in 0 < x < L, Eq. (4.171) /
11
a e]2 10-
0
11
requires that
a ( GJ oe) ox· ax -
10
( G J :~)
2e aat2
O
= 0,
x = 0 and x = L
at
of) = 0
(4.172) (4.173)
Equation (4.172) denotes the equation of motion of the shaft, and Eq. (4.173) indicates the boundary conditions. The boundary conditions require that either G J (oe lax) = 0 (stress is zero) or 8e 0 (e is specified) at x 0 and x L.
=
4.11.2
=
=
Transverse Vibration of a Thin Beam Consider an element of a thin beam in bending as shown in Fig. 4.6. If w denotes the deflection of the beam at any point x along the length of the beam, the slope of the deflected centerline is given by aw lox. Since a plane section of the beam remains plane after deformation according to simple (thin) beam theory, the axial displacement of a fiber located at a distance Z from the neutral axis u due to the transverse displacement w can be expressed as (point A moves to A')
ow ax
u = -z-
(4.174)
Thus, the axial strain can be expressed as
au
ex = -
ax
=
a2w
-z-2 ax
(4.175)
and the axial stress as
. ax = Eex = -Ez
a2w
ax2
(4.176)
The strain energy density of the beam element (]To) is given by (4.177)
I t._~~~ril.A.*:~;i.~.~itllifMf'ti)wL-W;""tai.!V-r'~~~"~'ip:'i'-'.'''~;~'~~~~,~~~~~~,Q,~;~ati',,;,;
.••• \~~~~
4.11
Applications of Hamilton' s'Pri11ciple
H7
z. w
~
I I I I .-.-.-
o
_._~
x
x
Figure 4.6
Beam in bending.
and hence the strain energy of the beam (;rr) can be expressed as rr:
=
Iff
;rrodV =
10[L 21E
v
If
. [dA(x)Z2]
2 (a w)2 ax2
dx
1 [L = 210
EI(x)
2 (a w)2 ax2
dx
A (4.178)
where 1(x) denotes the area moment of inertia of the cross section of the beam at x: I(x) =
ff
dA(x)Z2
(4.179)
A
The kinetic energy of the beam can be expressed as T =~ 2
[L m(x) [aw(x, t)]2 dx
10
(4.180)
at
where m(x) = pA(x) is the mass per unit length of the beam and p is the density of the beam. The virtual work of the applied distributed force, f(x, t), is given by
1
L
8W(t) =
f(x, t)8w(x, t) dx
(4.181)
Noting that the order of integrations with respect to t and x can be interchanged and the operators 8 and d/ dx or 8 and d/ dt are commutative, the variations of;rr and T .can be written as L 2 2 2 L 1 (a-.-2 w)2 dx dt = w) dx dt o;rr = 0 -El EI-a20w (a-2 '1 0 2 ax '1 0 ax ax
=
1'21 1 '2[
'1
1'21
2
L
awl EI-a w 0-ax2 ax
0
1a( L
-
0
ax
2 aw EI-a w) o-dx ax2 ax
] dt
(4.182)
118
Derivation of Equations: Variational Approach
Since
L 2 '21 -a ( EI-a2 w) 0 (aw) 1 ax ax -ax .
'I
dxdt
0
=
'2[ -a ( EI-02w2) ow JL -
1 'I
ax
ax
l
2
a 2 (aEI-w) 2 owdx ] dt -ax ax
L
0
0
2
(4.183)
Eq. (4.182) becomes
L 2 a w (aw)I -ax ax 0
2
IL 1 ax ax + l axLa2 ( E1--aaxw) 8wdx ] dt L oT = 0 '21 -m(x) (aw)2 dx dt = 1'21 m(x)-oaw 1 at at = L [( m--aw) ow 1'2-1'2 -a (aw) m- owdt ] dx l at at at on =
EI-20
'2[
'I
- -a
( £1a w)2
oW
0
2
L
o
-2
0
'I
1 2
'I
o
0
(aw) - dx dt at
'1'1
_ {L o
because
(4.184)
2
[1'2 iat (m aw) owdt] dx at
(4.185)
'I
ow is zero at t = t1 and t =t2. Thus,
Hamilton's principle can be stated as
'2
o
1
.
(T - n + W) dt = 0
'I
or
L 2 l aw 1 m-owdx at l ax EI-ax '2[
-
2
0
'I
-
a2 (
L -2
o
02W) 2
L 2 2 a w2 (aw) - EI-o - IL + -a ( E1-a w)I ow ax ax 0 ox ox2 0 owdx +
l
L
0
fowdt
]
dt
=0
(4.186)
Equation (4.186) leads to the following equations: O
(4.187)
(4.]88)
(4.189)
•...
..1:19
4.12 .Recent Connibutions
Equation (4.187) denotes the equation of motion for the transverse vibration of the beam, and Eqs. (4.188) and (4.189) represent the boundary conditions. It can be seen that Eq. (4.188) requires that either E/~:~
=0
or
=0
0 (~:)
at x = 0 and x = L
(4.190)
while Eq. (4.189) requires that either
a ( a w) ax EI-ax2 2
-
=0
or
ow
=0
at x = 0 and x = L
(4.191)
Thus, Eqs. (4.188) and (4.189) can be satisfied by the following common boundary conditions: 1. Fixed or clamped end:
w
= transverse deflection = 0,
aw ax
-
.
= bendmg slope = 0
(4.192)
2. Pinned or hinged end: w = transverse deflection = 0,
a2w EI-2 = bending moment ax
=0 (4.193)
3. Free end: El
aax2w = bending moment = 0, 2
-a
ax
w) ax 2 = shear force = 0 2
( EIa
(4.194)
4.12 RECENT CONTRmUTIONS Nagem et al. [11] observed that the Hamiltonian formulation of the damped oscillator can be used to model dissipation in quantum mechanics, to analyze low-temperature thermal fluctuations in RLC circuits, and to establish Pontryagin control theory for damped systems. Sato examined the governing equations used for the vibration and stability of a Timoshenko beam from the point of view of Hamilton's principle [12]. He derived the governing equations using an extended Hamilton's principle by considering the deviation of the external force following the deflection of the beam at its tip in terms of the angle s measured from the x axis (which is taken to be along the length of the beam). The variational finite difference method was presented for the vibration of sector plates by Singh and Dev [13]. Conventional finite difference techniques are normally applied to discretize the differential formulation either by approximating the field variable directly or by replacing the differentials by appropriate difference quotients. In general, the boundary conditions pose difficulties, particularly in problems with complex geometric configurations. The difficulties of conventional finite difference analysis can be overcome by using an integral-based finite difference approach in which the
120
Derivation of Equations: Variational Approach
principle of virtual work or minimum potential energy is used. Reference [13] demonstrates the application of the variational finite difference method to vibration problems. Gladwell and Zimmermann [14] presented the variational formulations of the equations governing the harmonic vibration of structural and acoustic systems. Two formulations, one involving displacements only and the other involving forcelike quantities only, were presented along with a discussion of the dual relationship. The principles were applied to the vibration of membranes and plates, to coupled air-membrane and air-plate vibrations, and to the vibration of isotropic elastic solid.
REFERENCES 1. S. S. Rao, Theory of vibration: variational methods, in Encyclopedia of Vibration, S. Braun, D. Ewins, and S. S. Rao, Eds., Academic Press, London, 2002, Vol. 2, pp. 1344-1360. 2. 1. S. Rao, Advanced Theory of Vibration, Wiley Eastern, New Delhi, India, 1992. 3. K. M. Liew, C. M. Wang, Y. Xiang, and S. Kitipornchai, Vibration of Mindlin Plates: Programming the p-Version Ritz Method, Elsevier Science, Oxford, 1998. 4. R. D. Mindlin, Influence of rotatory inertia and shear on flexural motions of isotropic, elastic plates, Journal of Applied Mechanics, Vol. 18, No.1, pp. 31-38, 1951. 5. R. Weinstock, Calculus of Variations: With Applications to Physics and Engineering, McGraw-Hill, New York, 1952. 6. 7. 8. 9. 10. 11.
R. R. Schechter, The Variational Method in Engineering, McGraw-Hill, New York, 1967. Y. C. Fung, Foundations of Solid Mechanics, Prentice-Hall, Englewood Cliffs, NJ, 1965. 1. N. Reddy, Energy and Variational Methods in Applied Mechanics, Wiley, New York, 1984. S. S. Rao, Engineering Optimization: Theory and Practice, 3rd ed., Wiley, New York, 1996. K. Washizu, Variational Methods in Elasticity and Plasticity, Pergamon Press, Oxford, 1968. R. Nagem, B. A. Rhodes, and G. V. H. Sandri, Hamiltonian mechanics of the damped harmonic oscillator, Journal of Sound and Vibration, Vol. 144, No.3, pp. 536-538, 1991. 12. K. Sato, On the governing equations for vibration and stability of a Timoshenko beam: Hamilton's principle," Journal of Sound and Vibration, Vol. 145, No.2, pp. 338-340, 1991. 13. 1. P. Singh and S. S. Dev, Variational finite difference method for the vibration of sector plates, Journal of Sound and Vibration, Vol. 136, No.1, pp. 91-104, 1990. 14. G. M. L. Gladwell and G. Zimmermann, On energy and complementary energy formulations of acoustic and structural vibration problems, Journal of SOllnd and Vibration, Vol. 3, No. 3, pp. 233-241, 1966.
PROBLEMS 4.hFormulate the problem of finding a plane curve of smallest arc length y(x) that connects points (XI, YI) and IX:.
y:).
.u Solve the
problem formulated in Problem 4.1 and show that the shortesl distance between points (XI, YI) and l.t2. Y2) is a straight line.
4.3 A plane curve y(x) is used to connect points (XI, YI) and (X2, Y2) with XI < X2. The curve y(x) is rotated about the X axis to generate a surface of revolution in the range XI =s X =s X2 (Fig. 4.7). Formulate the problem of finding the curve y(x) that corresponds to minimum area of the surface of revolution in the xy plane.
-Problems
·121
y(x)
o
.---.---.---.---.
Figure 4.7 4.4 Solve the problem formulated in Problem 4.3.
=
=
4.5 Given two points A (Xl, Yl) and B (X2, Y2)in the xy plane, consider an arc defined by Y = y(x) > 0,' XI :::: X :::: X2, that passes through A and B whose rotation about the X axis generates a surface of revolution. Find the arc Y y(x) such that the area included in Xl :::: X :::: X2 is a minimum.
=
4.6 Consider the Lagrangian functional L, given by L
=
1/-
pA
o
+
2
l
(au)2 --at
dx -
1/ 0
AE
-
(au)2 ---
ax
2
(a) Find the first variation of the functional L with au(O, t) au(x, t.) au(x, t2) = O. (b) Derive the Euler-Lagrange equations by setting the coefficients of au in (0, I) and at X 1 in the result of part (a) to zero separately.
=
=
4.7 Consider a solid body of revolution obtained by rotating a curve y y(x) in the xy plane passing through the origin (0,0), about the X axis as shown in Fig. 4.8. When this body of revolution moves in the -x direction at a velocity v in a fluid of density p, the normal pressure acting on an element of the surface of the
=
where () is the angle between the direction of the velocity of the fluid and the tangent to the surface. The drag force on the body. P, can be found by integrating the X component of the force acting on the surface of a slice of the body shown in Fig. 4.8(b){9]:
P
= 4rrpv2
10{L
(dy)3 dx
ydx
(4.2)
=
This functional corresponds to the axial vibration of a bar where u(x, t) denotes the axial displacement.
=
(4.1)
dx
+ Fu(l)
fudx
body is given by
Find the curve y y(x) that minimizes the drag on the body of revolution, given by Eq. (4.2), subject to the condition that y(x) satisfies the end conditions y(x = 0) 0 and y(x L) R.
=
=
=
4.8 Consider the functional I (w) that arises in the transverse bending of a thin rectangular plate resting on an elastic foundation:
122
Derivation of Equations: Variational Approach
--x
W Figure 4.8
W Solid body of revolution moving in a fluid.
where w (x, y) denotes the transverse deflection, D the bending rigidity, v the Poisson ratio, k the foundation modulus, and 10 the transverse distributed load acting on the plate with w = 0 on the boundary of the plate.
load
I (x,
y), gives rise to the functional
(a) Find the first variation of the functional I (w) with respect to w. (b) Derive the Euler-Lagrange equation corresponding to the functional I (w ).
Derive the governing differential equation and the ,boundary conditions by minimizing the functional I (w).
(c) Identify the natural and forced boundary conditions of the problem.
4.1L The potential energy of a thin beam, I (w), lying along the x axis subjected to a distributed transverse load I(x) per unit length, a bending moment M] and a shear force VI at the end x = 0, and a bending moment M2 and a shear force V2 at the end x = I is given by
4.9 Consider the problem of minimizing the functional I (y) given by I(y)
= [:
y(x)dx,
yea) = A,
y(b) = B I(w)
subject to the constraint
[bx=a
/1
EI
2 x=O
+
(ddx)2 y
=
I
Derive the Euler-Lagrange equations of the problem using Lagrange multipliers. 4.10 The transverse deflection of a membrane of area A (in the xy plane), subjected to a distributed transverse
•
= -I [1 +M!
(d2W)2 - -2 dx
(dW) dx
- V]wl.r=o
dx -
[1
Iwdx
x=O -M2-
x=O
(dW) dx
x=l
+ V2wlx=1
where w(x) denotes the transverse deflection and EI the bending stiffness of the beam. Derive the governing differential equation and the boundary conditions of the beam by minimizing the potential energy .
Derivation of Equations: Integral Equation Approach 5.1
INTRODUCTION In this chapter we describe the integral formulation of the equations of motion governing the vibration of continuous systems. An integral equation is an equation in which the unknown function appears under one or more signs of integration. The general form of an integral equation is given by
l
b
K(t,~)¢(g)dg
+ao(t)¢(t)
(5.1)
= f(t)
where K(t,~) is a known function of the variables t and ~ and is called the kemal or nucleus, ¢(~) is an unknown function, ao(t) and f(t) are known functions, and a and b are known limits of integration. The function ¢(t) which satisfies Eq. (5.1) is called the solution of the integral Eq. (5.1). Physically, Eq. (5.1) relates the present value of the function ¢(t) to the sum or integral of what had happened to all its previous values, ¢(~), from the previous state, a, to the present state, b. The first and second terms on the left-hand side of Eq. (5.1) are called the regular and exceptional parts of the equation, respectively, while the term on the right-hand side, f(t), is called the disturbance function. In some cases, the integral equation may contain the derivatives of the unknown function ¢(~) as
l
b
K (t, ~)¢(g) d~
+ ao(t)¢(t) + at (t)¢(t) (t) + ... + an(t)¢(n) (t)
where at (t), ... , an(t) are known functions of t and ¢(i)(t) Equation (5.2) is called an integrodifferential equation.
5.2
CLASSIFICATION
= di¢/
= f(t)
dti, i
= 1,2,
(5.2) ... , n.
OF INTEGRAL EQUATIONS
Integral equations can be classified in a variety of ways, as indicated below. 5.2.1
Classification Based on the Nonlinear Appearance of ~(t) If the unknown function (jJ (t) appears nonlinearly in the regular and/or exceptional parts, the equation is said to be a nonlinear integral equation. For example, the equation 123
124
Derivation of Equations: Integral Equation Approach
-l
b
G[¢>(t)]
H[t,~,
¢>(~)]d~ = f(t)
(5.3)
where G and/or H are nonlinear functions of ¢>(t), is called a nonlinear integral equation. On the other hand, if both G and H in Eq. (5.3) are linear in terms of ¢>(t), the equation is said to be a linear integral equation. Thus, Eq. (5.1) is called a linear integral equation, while Eg. (5.2) is said to be a linear integrodifferential equation. 5.2.2
Classification Based on the Location of Unknown Function tfJ(t) Based on the location of the unknown function, the integral equations are said to be of the first, second, or third kind. For example, if the unknown function appears under the integral sign only, the equation is said to be of the first kind. If the unknown function appears both under the integral sign and outside the integral, the equation is considered to be of the second or third kind. In the second kind of integral equation, the unknown function, appearing outside the integral sign, appears alone, whereas in the third kind, it appears in the form of a product ao(t)¢>(t), where ao(1) is a known function of t. According to this classification, Eq. (5.1) is an integral equation of the third kind. The corresponding equations of the second and first kinds can be expressed, respectively, as
l
b
¢>(t) - A
K(t, ~)¢>(~)d~ = f(t)
(5.4)
and
l
b
a
If
f(t)
K(t, n¢>(~)d~ = f(t) __
(5.5)
= 0 in Eq. (5.4), we obtain ¢>(t)
=A
l
b
K(t,~)¢>(~)d~
(5.6)
which is called a homogeneous integral equation. Note that the A in Eqs. (5.4) and (5.6) denotes a constant and can be incorporated into the kernel K (t, O. However, in many applications, this constant represents a significant parameter that may assume several values. Hence, it is included as a separate parameter in these equations. 5.2.3
'\.
Classification Based on the Limits of Integration Based on the type of integral in the regular part, the integral equations are classified as Fredholm- or Volterra-type equations. If the integral is over finite limits with fixed endpoints (definite integral), the equation is said to be of Fredholm type. On the other hand, if the integration limits are variable (indefinite integral), the integral equation is said to be of Volterra type. It can be seen that in Eqs.(5.1) to (5.6), the regular parts involve definite integrals and hence they are considered to be of Fredholm type. If K(t,~) 0 for ~ > t, the regular parts of Eqs. (5.1) to (5.6) can be expressed as indefinite integrals as
=
i ttti~-aIi~:f~~~t;'~~~~Miliwa.;.;u~I~·\:
5.3
Derivation of Integr.il'Equations't25
and hence the resulting equations will be of Volterra type Thus. Volterra-type integral equations of the third, second, and first kind can be expressed. in sequence, as
1
t
+ ao(t)>(t) = f(t)
(5.7)
K(t, ;)>(;)d; = f(t)
(5.8)
K (t, ;)>(;) d;
-1
t
>(t)
1
t
Similar to Eq. (5.6), the Volterra-type
;~-.
K(t,;)>(;)d;
homogeneous
= f(t)
(5.9)
integral equation can be written as
(5.10)
5.2.4
Classification Based on the Proper Nature of an Integral If the regular part of the integral equation contains a singular integral, the equation is called a singular integral equation. Otherwise, the equation is called a normal integral equation. The singularity in the integral may be due to either an infinite range of integration or a nonintegrable or unbounded kernel which causes the integrand to become infinite at some point in the range of integration. Thus, the following equations are examples of singular integral equations:
1
00
-Ai:
>(t)
5.3 5.3.1
K(t, ;)>(;)d; = f(t)
(5.11)
= f(t)
(5.12)
K(t,;)>(;)d;
DERIVATION OF INTEGRAL EQUATIONS Direct Method The direct method of deriving example.
integral equations
is illustrated
through the following
Example 5.1 Load Distribution on a String
Consider the problem of finding the load distribution on a tightly stretched string, which results in a specified deflection shape of the string. Let a string of length L be under tension P. When a concentrated load F is applied to the string at point ;, the string will deflect as shown in Fig. 5.1. Let the transverse displacement of the string at; due to F be 8. Then the displacement w(x) at any other point x can be expressed as
w(x) =
I
8x L ;-'x
8-L -;'
(E5.1.1)
126
Derivation of Equations: Integral Equation Approach w(x) .4.
F
I
I I o
r_J.i)_._J_._I/J. _._
I
x
I·
~
~
Figure 5.1
x
L
)~L-~~
Tightly stretched string subject to a force F.
For small displacements lJ, the conditions for the equilibrium of horizontal and vertical forces can be written as P cose = P cos¢>
(E5.1.2)
F=Psine+Psin¢>
+ tan¢»
~ P(tane =P
LlJ ~(L - ~)
(E5.1.3)
Equation (E5.1.3) can be solved for lJ, which upon substitution in Eq. (E5.1.3) results in F P
w(x) = -g(x,~)
(E5.1.4)
where g (x , ~) is the impulse response junction, also known as Green's junction, given by x(L - ~)
L g(x,~)=
{
' (E5.1.5)
HL-x) L
'
If the external load applied to the string is distributed with a magnitude of f(~) unit length, the transverse displacement of the string can be expressed as w(x) =
~lL
g(x, Of(~)d~
per
(E5.1.6)
If the displacement variation w(x) is specified, Eq. (E5.1.6) becomes an integral equation of the first kind for the unknown force distribution f(x). For free vibration, the force per unit length, due to inertia, is given by f(x,
t) = -p(x)
a2w(x,t) 2 at
(E5.1.7)
"1
_=-< .....-r-~..... "'.,.,....-""_"': ......,....".,......'""""_~=~
I
.•=.,.,."...oJ~·_...,~-:'~~'l
..~~·.""".,.
5.3
·,·••··~.,.."t"<"!~._.~h
•.. ~·,~·-.""'·,~
>
Derivation of Integmi'Equatiens
.•
~~
.. ,.'~~''''
117
where p(x) is the mass density (mass per unit length) of the string at x. Using Eq. (E5.1.7), Eq. (E5.l.6) can be written as
1 [L
a2w(~, t)
10
w(x, t) = - p
g(x, ~)p(~)
ot2
d~
(E5.1.8)
When p(x) is known, Eq. (E5.1.8) denotes the governing integrodifferential equation for the displacement w (x, t). Assuming a simple harmonic solution with frequency w, (E5.1.9)
w(x, t) = W(x) sinwt where W(x) denotes the amplitude
of displacement
of the string at x, Eq. (E5.1.8)
becomes
w21L
W(x) = -
p
g(x,~)p(~)W(~)d~ of the second kind for W (x).
which can be seen to be an integral equation
5.3.2
(E5.1.10)
0
Derivation from the Differential Equation of Motion The equation
of motion for the free vibration
of a string can be expressed
as (see
Eq. (8.9) (5.13) where 2
P
(5.14)
c =p
p is the mass density of the string 'per unit length and P is the tension. If the string is fixed at both ends, the boundary
conditions
are given by
w(O, t)
=0
w(L, t)
== 0
If the string is given an initial displacement
f(x)
(5.15) and initial velocity g(x), we have
w(x,O) = f(x)
oW -at (x, 0) = g(x) Using the separation of variables technique,
(5.16)
w(x. t) can be expressed as
w(x. t) = X(x)T(t)
(5.17)
where X is a function of x and T is a function of t. Using Eq. (5.17), Eq. (5.13) can be rewritten as (5.18)
•• ~.
~.,.",
128
Derivation of Equations: Integral Equation Approach 2
where -'Ac
is a constant. Equation (5.18) yields two ordinary differential d2T
+'Ac2T
-2
dt
d2X dx2 Since T(t) ~ 0, the boundary
+ 'AX
conditions
equations:
=0
(5.19)
=0
(5.20)
can be expressed
as
X (0) = 0
=0
X(L) Integration
of Eq. (5.20) gives
r
dX = _'A
10
dx
where
CI
is a constant. Integration
l
x
C2
+ CI
X d~
(5.22)
of Eq. (5.22) leads to x
= -Al
X where
(5.21)
+ CIX + C2
X d~
dTJ
is a constant. Changing the order of integration, x
X(x) = -'Al
X(~) d~
lX
(5.23)
Eq. (5.23) can be rewritten as
dTJ+ CIX + C2
x
= -'Al
(x - ~)X(~) d~
The use of the boundary conditions,
The differential X (x) as
l
(5.24)
Eq. (5.21), results in
C2
=0
c}
= ~L Jr\L o
- ~)X(~)d~
Eq. (5.13) can thus be expressed
X(x)+'A
+ CIX + C2
as an equivalent
l
(5.25) integral equation in
L
x
o
X
(x-~)X(~)d~-'A-
L
. (L-OX(~)d~=O
0
(5.26)
Introducing K(x,O
=
I~(L-
x),
.\ (L - ~),
Eq. (5.26) can be rewritten
(5.27)
as L
X(X)='A1
K(x,~)X(~)d~
(5.28)
•..
~
.-,~...
...
'T
",.,.,..,
~~."",,",
.. "
.~"""":","",~
'I I I ... _.;r""'"'-''''~':'''"'-i''"t~'~'':'1"':;'"'·?''":'~';-='--n~--:-'7,'"'''',''''''''''~r:-,~~''''~'''''~!'f'''''''"'''·~,,,,,,,"~,"~~I,
5.3
'~j"
.~
••.••••
,,~.~~'~"""'~":'ll""<"''1'''''~''''''=~r.:-:r~.
Derivation oflntegral Equation:;
'.:t29
This equation can be seen to be of the same form as Eq. (E5.1.10) (Volterra-type homogeneous integral equation). Equation (5.28) can be solved using the procedure outlined in the following example. Example 5.2
Find the solution of Eq. (5.28).
SOLUTION
We rewrite Eq. (S.28) as X(x)
+
A1
x
where d =
(x - ~)X(~)d~
IlL
- Axd
=0
(E5.2.1)
(E5.2.2)
(L - ~)X(~) d~
Taking the Laplace transform of Eq. (E5.2.1), we obtain _ X(s)
1-
+ A,"2X(s) s
1 - A"2 d = 0
(E5.2.3)
s
where X(s) is the Laplace transform of X(x). Equation (E5.2.3) yields _ Ad '. X (s) = s2 + A
(E5.2.4)
and the inverse transform of Eq. (E5.2.4) gives X(x) =
hd
(E5.2.5)
sinhx
Substitution of Eq. (E5.2.5) into Eq. (E5.2.2) gives
d l\L
-~)hsinh~d~
=
dL
(E5.2.6)
which can be satisfied when d = 0 or
l\L
-~)hsinh~d~
=
Since d 0 leads to the trivial solution X(x) be satisfied. Equation (E5.2.7) yields
= 0 and
=L w(x, t)
L _ sin.J>::L =L
.J>::
(E5 .2.7)
= 0, Eq. (E5.2.7) must (E5.2.8)
or
sinhL
=0
(E5.2.9)
or n = 1,2, ...
(E5.2.1O)
130
Derivation of Equations: Integral Equation Approach Equations (E5.2.5) and (E5.2.1O) lead to .
nrrx
X(x) = asm--
(E5.2.11)
L
where a is a constant.
5.4 GENERAL FORMULATION OF THE EIGENVALUE PROBLEM 5.4.1
One-Dimensional Systems For a one-dimensional
continuous
system, the displacement t)
w(x,
=
l
w (x, t) can"be expressed as
L
a(x, ~)f(~,
t) d~
(5.29)
where a(x,~) is the flexibility influence function that satisfies the boundary conditions of the system and f (~, t) is the distributed load at point ~ at time t. For a system undergoing free vibration, the load represents the inertia force, so that f() x, t = -m (x )
a2w(x,
where m (x) is the mass per unit length. Assuming during free vibration,
Eq. (5.30) can be expressed
t) (5.30)
a harmonic motion, of frequency w
w(x, t)= W(x)coswt
(5.31)
= w m(x)W(x)
(5.32)
as
f(x, t) Substituting
at2
2
cos wt
Eqs. (5.31) and (5.32) into Eq. (5.29) results in
W(x) =
w21L a(x, ~)m(~)W(~)d~
(5.33)
It can be seen that Eq. (5.33) is a homogeneous integral equation of the second kind and represents the eigenvalue problem of the system in integral form.
Example 5.3 Free Transverse Vwration of a Membrane
Consider a membrane of area A whose equilibrium shape lies in the xy plane. Let the membrane be fixed at its boundary, S, and subjected to a uniform tension P (force per unit length). Let the transverse displacement of point Q (x, y) due to the transverse load f (~, 1]) d ~ d 1] applied at the point R (~, 1]) be w( Q). By considering the equilibrium of a small element of area d.x dy of the membrane, the differential equation can be derived as
(E5.3.1)
5.4
General Formulation of the Eigenvalue Problem
. 131
The Green's function of the membrane, K(x, y;~, 17), is given by [4] K(Q,
R)
= K(x,
y;~,
17)
1
= log -r
- h(Q, R)
(E5.3.2)
where r denotes the distance between two points Q and R in the domain of the membrane: (E5.3.3)
and h (Q, R) is a hannonic function whose values on the boundary of the membrane, S, are the same as those of log(l / r) so that K (Q, R) will be zero on S. For example, if the membrane is circular with center at (0,0) and radius a, the variation of the function K(Q, R) will be as shown in Fig. 5.2. Since the membrane is fixed along its boundary S, the transverse displacement of point Q can be expressed as w(Q) = _1_ 2rr P
If
K(Q, R)f(R)dA
(E5.3.4)
A
K(Q. R)
-'Boundary, S
Figure 5.2
Variation of the Green's function for a circular membrane. (From Ref. [4]).
132
Derivation of Equations: Integral Equation Approach
From this static relation, the free vibration relation can be obtained by substituting _p(R)[02w(R)/ot2] for f(R) in Eq. (E5.3.4) so that
Jf
w(Q) = --2 17r
P
K(Q, R)P(R)-2 02w (R)dA
ot
(E5.3.5)
A
Assuming harmonic motion with frequency w, we have
= W(Q)eiwt
w(Q)
(E5.3.6)
where W(Q) denotes the amplitude of vibration at point Q. Substitution ofEq. (E5.3.6) in Eq. (£5.3.5) yields the relation
Jf
W(Q) = ~
27rP
K(Q, R)p(R)W(R)dA
(E5.3.7)
A
5.4.2 General Continuous Systems The general form of Eq. (5.33), valid for any continuous system, can be expressed as W(x) =
Ai
g(x, ~)m(~)W(~)dV(~)
(5.34)
where W(x) and W(O denote the displacements at points x and ~, respectively. Depending on the dimensionality of the problem, points x and ~ may be defined by one, two, or three spatial coordinates. The general flexibility influence function g(x, ~), also known as the Green'sfunction, is symmetric in x and~, [i.e., g(x,~) = g(~, x)] for a self-adjoint problem. Note that the kernel, g(x, ~)m(~), in Eq. (5.34) is not symmetric unless m(~) is a constant. However, the kernel can be made symmetric by noting the fact that m(~) > 0 and introducing the function ¢(x): ¢(x)
= Jm(x)W(x)
(5.35)
By multiplying both sides of Eg. (5.34) by' .Jm(x) and using Eg. (5.35), we obtain ¢(x) = A
i
K(x, ~)¢(~)dV(~)
(5.36)
where the kernel (5.37) can be seen to be symmetric. An advantage of the transformation above is that a symmetric kernel usually possesses an infinite number of eigenvalues, A, for which Eq. (5.36) will have nonzero solutions. On the other hand, a nonsymmetric kernel mayor may not have eigenvalues [1]. For any specific eigenvalue Ai, Eq. (5.36) has a nontrivial solution ¢i(X), which is related to Wi(x) by Eq. (5.35). The function Wi(x) represents the eigenfunction corresponding to the eigenvalue Ai of the system.
••••.
)0\."'.-
----
...
~
... ',...
~
5.5
5.4.3
-'133
Solution of Integral Eqtiations
Orthogonality of Eigenfunctions It can be shown that the eigenfunctions 4>i (x) are orthogonal in the usual sense. while the functions Wi(x) are orthogonal with respect to the functions m(x). For this. consider Eq. (5.36), corresponding to two distinct eigenvalues Ai and A j:
Multiply obtain
¢i(X)
= Ai
¢j(x)
= Aj
Eq. (5.38) by ¢j(x),
Iv Iv
integrate
K(x, ~)¢i(~)d V(~)
(5.38)
K(x, ~)¢j(~) dV(~)
(5.39)
over the domain
V, and use Eq. (5.39) to
(5.40)
which yields
(5.41)
Since Ai and Aj are distinct, Ai :/; Aj, Eq. (5.41) leads to the orthogonality
relation
(5.42) When Eq. (5.35) is used in Eq. (5.42), eigenfunctions Wi (x) as
we obtain the orthogonality
relation for the
for Ai :/; A j for Ai Aj
(5.43)
=
5.5
.. , .....•..•.....
SOLUTION OF INTEGRAL EQUATIONS Several methods, both exact and approximate methods, can be used to find the solutions of integral equations [1,4-6]. The method of undetermined coefficients and the Rayleigh-Ritz, Galerkin, collocation, and numerical integration methods are considered in this section.
134
5.5.1
Derivation of Equations: Integral Equation Approach
Method of Undetermined Coefficients In this method the unknown function is assumed to be in the form of a power series of a finite number of terms. The assumed function is then substituted into the integral equation and the regular part is integrated. This results in a set of simultaneous equations in terms of the unknown coefficients. Solution of these simultaneous equations yields the solution of the integral equation. Example 5.4
Find the solution of the integral equation
21) SOLUTION
(1- ~+xO¢(~)d~
= -x
+1
(E5.4.I)
Assume the solution of ¢(x) in a power series of two terms as (E5.4.2)
where c) and C2 are constants to be determined. Substitute Eq. (E5.4.2) into Eq. (E5.4.1) and carry out the integration to obtain
21)
(1 - ~ + x~)(c)
+ C2~)
d~ = -x
+1
(E5.4.3)
Upon integration, Eq. (E5.4.3) becomes
(C) + ~C2)
+x
(C) + ~C2)
= -x
+1
(E5.4.4)
Equating similar terms on both sides of Eq. (E5.4.4), we obtain C)
C)
Equations (E5.4.5) yield Eq. (E5.4.1) is given by
C)
= -3
+ ~C2 + ~C2
and ¢ (x)
5.5.2
C2
= 1 = -1
= 3.
(E5.4.5)
Thus, the solution of the integral
= - 3 + 3x
(E5.4.6)
Iterative Method An iterative method similar to the matrix iteration method for the solution of a matrix eigenvalue problem can be used for the solution of the integral Eq. (5.34). The iteration method assumes that the eigenvalues are distinct and well separated such that A) < A2 < A3 .... In addition, the iteration method is based on the expansion theorem related to the eigenfunctions Wj(x). Similar to the expansion theorem of the matrix eigenvalue problem, the expansion theorem related to the integral forrimlation of the eigenvalue problem can be stated as 00
W(x)
= LCiWi(X) j=)
(5.44)
~ I
5.5
where the coefficients
Solution of Integral Equations
135
are determined as
Cj
Cj
=
Iv
(5.45)
m(x)W(x)Wj(x)dV(x)
Equation (5.44) indicates that any function W(x) that satisfies the boundary conditions of the system can be represented as a linear combination of the eigenfunctions Wi (x) of the system. First Eigenfunction The iteration method starts with the selection of a trial function wil) (x) as an approximation to the first eigenfunction or mode shape, WI (x). Substituting Wil)(x) for W(x) on the right-hand side of Eq. (5.34) and evaluating the integral, the next (improved) approximation to the eigenfunction WI (x) can be obtained: (5.46) Using Eq. (5.44), Eq. (5.46) can be expressed as 00
=
W?)(x)
L
Cj
=
f
g(x, ~)m(~)Wj(~)dV(~)
v
j=1 00
"Cj j=1
Wi (x) A'I
(5.47)
The definition of the eigenvalue problem, Eq. (5.34), yields Wj(x)
= Aj
Iv
g(x, ~)m(~)Wj(~) dV(~)
(5.48)
Using W}2)(x) as the trial function on the right-hand side of Eq. (5.48), we obtain the new approximation, W?)(x), as
00
_
"
Cj
Wj(x)
-~
(5.49)
2
A;
i=1
The continuation of the process leads to 00
(n)
WI
"
(x) = ~ j=1
Cj
Wj(x) n-I'
n = 2,3, ...
(5.50)
Ai
Since the eigenvalues are assumed to satisfy the relation A\ < )..2" " the first term on the right-hand side of Eq. (5.50) becomes large compared to the other terms and as
136
Derivation of Equations: Integral Equation Approach n ~
00,
Eq. (5.50) yields
= CI WI (x)
lim W(n-I\x) n-+oo
(5.51)
An-Z
I
I
lim W(n)(x) = n-+oo
and the converged eigenvector
(5.52)
An-I I
Equations (5.51) and (5.52) yield the converged
eigenvalue
.
wt-I)(x)
n-+oo
wt)(x)
A] = hm
WI (x)
CI
I
AI as
(5.53)
can be taken as W,(x)
= n_oo lim
wt)(x)
(5.54)
Higher Eigenfunctions
To determine the second eigenfunction, the trial function Wil) (x) used must be made completely free of the first eigenfunction, W, (x). For this we use any arbitrary trial function Wil) (x) to generate wil) (x) as (5.55)
a,
where is a constant that can be determined eigenfunctions:
Iv
m(x)Wi')
from the orthogonality
condition of the
(x)W, (x) dV(x)
= Iv m(x)Wi')
(x)W, (x) dV(x) - a]
Iv
m(x)[W] (x)f
dV(x)
=0
(5.56)
or w] (x) dV(x) a, = Iv Ivm(x)Wil)(x) m(x)[W, (x)]2 dV(x) When
w] (x)
is normalized
(5.57)
according to Eg. (5.43),
Iv
m(x)[WI (x)f
dV(x)
= I
(5.58)
Eq. (5.57) becomes a, =
Iv
m(x)Wi')(x)W,(x)dV(x)
(5.59)
Once Gj is determined, we substitute Eg. (5.55) for W(x) on the right-hand side of Eq. (5.34), evaluate the integral, and denote the result as WiZ)(x), the next (improved) approximation to the true eigenfunction Wz(x): (5.60)
5.5 For the next iteration,
Solution of hllegral Equations
1J7
we generate W?) (x) that is free of WI (x) as (5.61)
where
02
can be found using an equation .similar to Eq.(5.59) as (5.62)
when WI (x) is normalized according to Eq. (5.43). When the iterative process is continued, we obtain,
as n -+
00,
the converged
result as )..2 =
.
w(n-l)
lim n~oo
W2(X) = lim n~oo
(x)
_2
wt) (x)
_
win>(x)
(5.63) (5.64)
To find the third eigenfunction of the system, we start with any arbitrary trial function WJI)(x) and generate the function WJI) (x) that is completely free of the first and second eigenfunctions
WI (x) and W2(X) as (5.65)
where WI (x) to find to find
the constants al and a2 can be found by making Wjl)(x) orthogonal to both and W2(X). The procedure used in finding the second eigenfunction can be used the converged solution for )..3 and W3(X). In fact, a similar process can be used all other higher eigenvalues and eigenfunctions.
Example 5.5 Find the first eigenvalue and the corresponding
eigenfunction of a tightly stretched string under tension using the iterative method with the trial function w(l)( ) _ I X -
x(L - x) £2
SOLUTION Let the mass of the string be m per unit length and the tension in the string be P. The Green's function or the flexibility influence function, g(x, ~), can be derived by applying a unit load at point ~ and finding the resulting deflection at point x as shown in Fig. 5.3. For vertical force equilibrium, we have
a a P-+P--=l ~
L-~
(ES.5.1)
138
Derivation of Equations: Integral Equation Approach
(a)
)r-L-X~ ---)
r----~;---i-L-; x
I .-.-.-.- .•.-.-.-.-.-+-.-.-.-. Y
I I I
a
(b)
Figure 5.3
which yields a=
~(L - ~)
(E5.5.2)
PL
Thus, the Green's function is given by ax g(x,~)
~>x
T'
=
(E5.5.3)
a(L _ x)
{
L -~
'
~
which can be expressed as
g(x,~)
=
{
x(L - ~) PL ' ~(L _ x)
PL
'
~>x (E5.5.4) ~< x
Using the trial function "'1(I)(X)
= x(L - x) L2
(E5.5.5)
.~
5.5
Solution of Integral Equations
1.39'
for WI (x) on the right-hand side of Eq. (5.34), we obtain the new trial function wf)(x) as W?)(x) =
fL g(x, ~)m(~)W:I)(~) d~ 11;=0
= i:o =
fX
g(x. ~)m(~)W:I)(s) ds
g(x. s)m(~)W?)(s)d~
s(L - x) (m) s(L - ~) d~
11;=0
+
+ i:x
U
PL
fL
s) (m) ~(L ~ s) d~
x(L -
11;=x
PL
(E5.5.6)
L
Equation (E5.5.6) can be simplified as _ 2Lx3 +x4)
W(2)(x) = _m_(L3x 1 12P L2
(E5.5.7)
Using Eqs. (ES.5.5) and (E5.5.7), we obtain (E5.5.8) or 12P
2
(E5.5.9)
Wl~--
mU
or WI ~
3.4641)
P
mL
2
(E5.5.10)
This approximate solution can be seen to be quite good compared to the exact value of the first natural frequency, WI = ;rJp/mU. 5.5.3 Rayleigh-Ritz Method In the Rayleigh-Ritz method, also known as the assumed modes method, the solution of the free vibration problem is approximated by a linear combination of n admissible functions, Ui (x), as n
w(x, t) =
L
Ui (X)T/i (t)
(5.66)
i=1
where 'Ji (t) are time-dependent generalized coordinates. The kinetic energy of the system, T(t), can be expressed as
!
T(t) = !x:.om(x)[w(x, =
t)]2 dx
!I:?=I I:j=1 mijr,i(t)i1j(t)
(5.67)
140
Derivation of Equations: Integral Equation Approach where m(x) denotes the mass per unit length, a dot above a symbol represents a time derivative, and (5.68) The potential energy of the system, U (t), can be expressed in terms of the flexibility influence function g(x, 0 and the distributed load f(x, t) as
= ~fx:o
U(t)
f(x,
t)
[foL g(x,
;)f(;,
t)d;]
dx
(5.69)
Assuming that t) = ulm(x)W(x,
f(x,
t)
(5.70)
for free vibration, Eq. (5.69) can be expressed as U(t) = Substitution
~4iL
m(x)W(x,
t)
[iL
g(x, ;)m(;)W(;,
t) d;]
dx
(5.71)
of Eq. (5.66) into (5.71) leads to
1::7=1 1::J=1 kijT/i(t)T/j(t)
U(t) = ~):2
(5.72)
where k;j = kj;
=
i
L
[i
L
m(x)u;(x)
g(x, ;)m(;)Uj(;)d;]
dx
(5.73)
and 5:. denotes an approximation of w2. Equations (5.66), (5.67), and (5.72) essentially approximate the continuous system by an n-degree-of-freedom system. Lagrange's equations for an n-degree-of-freedom conservative system are given by
d (aT) dt a~k
aT au - aTJk + al]k
k
= 0,
= 1,2,
... , n
(5.74)
where l]k is the generalized displacement and ~k is the generalized velocity. Substitution of Eqs. (5.67) and (5.72) into (5.74) yields the following equations of motion: n
n
+ P Lkk;l];
Lmk;ii; ;=]
For harmonic variation of
= 0,
k=1,2,
... ,n
(5.75)
;=] 1];
(t),
(5.76) and Eqs. (5.75) lead to the matrix eigenvalue problem
5:.[k]ij = [m]ij
=
=
(5.77)
=
where [k] [k;j] and [m] [m;j] are symmetric matrices and 5:. w2. The problem of Eq. (5.77) can be solved readily to find the eigenvalues 5:. and the corresponding
I ;
I,
, j
i, i~~~~~_w~~~:llI,*"".li,/i-'~lJl~~~~~t1;'
""""""" i
55 --solutionof Integral Equations
141
eigenvectors ij. The eigenfunctions of the continuous system can then be determined using Eq. (5.66). Example
5.6 Find the natural frequencies of a tightly stretched string under tension
using the Rayleigh-Ritz method. ~,. .
SOLUTION Let the mass of the string be m per uoit length and the tension in the string be P. The Green's function or flexibility influence function of the string can be expressed as (see Example 5.1) ~(L - x) LP , g(x,~)=g(~,x)=
~
x(L-~)
{
- LP
~>x
,
We assume a two-term solution for the deflection of the string as = Ul
W(x, t)
where the admissible functions
and
Ul(X)
+ U2(X)T/2(t)
(X)T/l (t)
--
are chosen as
U2(X)
U\(x) = ~
L
(E5.6.2)
(1 - ~) L
(E5.6.3)
2
U2(X)
= -x
(
L2
1 - -X
(E5.6.4)
)
L
and 171 (t) and 172(t) are the time-dependent generalized coordinates to be determined. The elements of the matrix (m] can be determined as (E5.6.5) Equation (E5.6.5) gives
mIl
m12
m22
=m
i
L
uf(x)
x=O
= m21 = m =m
dx
i
i
L
u~(x)dx
i
L
2 (
x
=
m L4
1-
x=O
m = "3" L
Ul(X)U2(X)dx
x=o
L
x=O
m = "2 L
i
L
x=O
4 (
x
1-
X)2
L
i
3
dx
mL = 30L2 = mL 30 4
L
3 (
x
X)2
1- -
L
x=O
LX)2
5
dx
=
mL 105L4
dx
mL mL = --3 = -60 60L
= mL 105
The elements of the matrix (k] can be found from (E5.6.6)
142
Derivation of Equations: Integral Equation Approach
In Eq. (E5.6.6) the inside integral for
i
k))
can be evaluated as follows:
L
g(x,;)m(;)u)(;)d;
= m
=
r ;(L - x) i (1_i) d; 10 LP L L .
m UP
(
L
3
-"6x +
1
12x
4
+m
i
L
(I _i)
x(L -;)~ LP L
x
d;
L
L3)
+ 12x
(E5.6.7)
Thus, Eq. (E5.6.6) yields 2
kll =
m (~_ L
[L
10
i
m2
=
L
-4-
L Pol
X ) ~ U UP
(_!::..x3 6
3
+ ~x4 + L 12
x) dx
12
Lx _x2 2 (-2Lx3+x4+L3x)dx
17m2L3
=
(E5.6.8)
5040P Similarly, we can obtain from Eq.(E5.6.6) 2
kj',=k21=
[L m(X
•
2
=
m L
10 3
--p-
(
L2
_X3)~(_!::..x3+x4 L3 L2 P 6
12
+L3x)dX 12
17 ) 10,080·
(E5.6.9)
The inside integral in Eq. (E5.6.6) for
k22
can be evaluated as
(E5.6.1O)
(E5.6.1l) Thus, the eigenvalue problem can be expressed as mL [14 7] 420 7 4
X = m2L3).., [170 50,400
85] 85 44
X
(E5.6.12)
I
i~~f;'bd!~~~~~~~_{~~W~.i~fi.M.i~jiW~;ats!at}~~"~~~;1i~~
t
t
5.5
Solution of Integral Equations
143
or 14 7] [ 7 4
X _ X [170
85] 85 44
-
X
(E5.6.13)
where _ mL2}. }.=-120P
(E5.6.14) (E5.6.15)
2
}. =W
is the eigenvalue and X is the eigenvector (mode shape). The solution of Eq. (E5.6.13) is given by
XI
X2
= 0.0824,
= 0.3333
(E5.6.16)
with
X(l) = {1.OOOO} .
5.5.4
0.0
X(2)
= \-0.4472}
'
(E5.6.17)
0.8944
Galerkin's Method In the Galerkin method, the function ¢J is approximated by a linear combination of n comparison functions, Ui(X), as n
¢J(x, t)
=
L ui(x)T1i
(5.78)
i=1
where T1iare coefficients (or generalized coordinates) to be determined. Consider the eigenvalue problem of the continuous system in the integral form w(x)
=}. [g(X,
~)m(~)w(~)
dV(~)
(5.79)
where w(x) is the displacement at point x and g(x,~) is the symmetric Green's function or flexibility influence function. By introducing ¢J(x) = Jm(x)w(x)
(5.80)
and multiplying both sides of Eq. (5.79) by ..,fm(x), we obtain ¢J(x)
=}. [
K(x, ~)¢J(~) dV(~)
(5.81)
where K (x, ~) denotes the symmetric kernel: K(x,~)
= g(x, ~)Jm(x)Jm(~)
(5.82)
144
Derivation of Equations: Integral Equation Approach When the approximate solution, Eq. (5.78), is substituted into Eq. (5.81), the equality will not hold; hence an error function, €(x), also known as the residual, can be defined as
=
€(X)
¢(x)
Iv
-).
K(x,
s)¢(s) dV(s)
(5.83)
where). denotes an approximate value of A and V indicates the domain of the system. To determine the coefficients TIt. the weighted integral of the error function over the domain of the system is set equal to zero. Using the functions U) (x), U2(X), ... , un(x) as the weighting functions, n equations can be derived:
Iv Substituting
= 0,
S(X)Uk(X) dV(x)
k = 1, 2, ... , n
(5.84)
Eg. (5.83) into (5.84), we obtain n
~
17i
-).
Iv ~
Uk(X)U j(x)
dV(x)
Iv
[Iv
17i
Tlk(X)
K(x,
S)Ui(s)dV(s)]
dV(x)
=0
(5.85)
Defining Uj(X) kjk
= ../m(x)uj(x),
= kkj = Iv
i = 1,2, ... , n
Uk(X)
= Iv m(x)uk(X) mik
= mki =
Iv
[Iv [Iv
K(x,
g(x, s)m(S)ui(S)
Uk(X)Ui(X)
= Iv m(x)uk(X)Ui(X)
S)Ui(S)dV(s)]
(5.86) dV(x)
dV(S)]
dV(x)
(5.87)
dV(x)
(5.88)
dV(x)
Eq. (5.85) can be expressed as )'[kJij = [mJij
(5.89)
which can be seen to be similar to Eq. (5.77).
5.5.5 Collocation Method Consider the eigenvalue
problem of a continuous
system in integral form: (5.90)
55
Sil!ution of Integral Equations
The solution of the free vibration problem is approximated
. n comparison
functions.
Ui(X),
14S
by a linear combination
of
as n
(5.91)
w(x) = I>i(X)l7i i=1
where 1/; are coefficients or generalized coordinates to be determined. When Eq. (5.91) is substituted into Eq. (5.90), the equality will not hold; hence an error function or residual c (x) can be defined as (5.92)
By substituting
Eq. (5.91) into (5.90), the error function can be expressed as c(X) = w(x) - >..
Iv
g(x, ~)m(~)w(~)
n
=
L
n
1/iUi(X) -
5:.
i=1
L
1/i
dV(~)
1
g(x, ~)m(;)Ui(~)
dV(;)
(5.93)
v
i=1
To determine the coefficients T/k, the error function is set equal to zero at n distinct points. By setting the error, Eq. (5.93), equal to zero at the points xk(k = 1,2, ... , n), we obtain k = 1, 2, ... , n Equations
(5.93) and (5.94) lead to the eigenvalue
problem
n
L
(mki - hki)r/i
= 0,
(5.94)
k = 1, 2, ... , n
(5.95)
i=1
which can be expressed
in matrix form as (5.96)
[m]ij = 5:.[k]ij where the elements
of the matrices [m] and [k] are given by mki
=
(5.97)
Ui(Xk)
kki = Iv g(Xk, ;)m(;)ui(;)
dV(;)
(5.98)
It is to be noted that the matrices [m] and [k] are, in general, not symmetric. The solution of the eigenvalue problem with nonsymmetric matrices [m] and [k] is more complex than the one with symmetric matrices [3].
146
5.5.6
Derivation of Equations: Integral Equation Approach Numerical
Integration
Method
In the numerical integration method, the regular part of the integral equation is decomposed into the form of a sum, and the equation is then reduced to a set of simultaneous linear equations with the values of the unknown function at some points in the domain of integration treated as the unknown quantities. The procedure is illustrated through the following example.
Example 5.7
Find the solution of the integral equation
¢(x)
+
1
1
o
(1 + x;)¢(;)
== x2
d; = f(x)
-
23 -x 24
4 +_ 3
(E5.7.1)
numerically and compare the result with the exact solution ¢(x) = x2
2x
-
+1
(E5.7.2)
SOLUTION We use the Gauss integration method for the numerical solution of Eg. (E5.7.1). In Gauss integration, the integral is evaluated by using the formula 1
{get)
n
dt =
-I
L wig(ti) i=1
(£5.7.3)
where n is called the number of Gauss points, Wi are called weights, and ti are the specified values of t in the range of integration. For any specified n, the values of Wi and ti are chosen so that the formula will be exact for polynomials up to and including degree 2n - 1. Since the range of integration in Eq. (E5.7.3) for x is -1 to + 1, the formula can be made applicable to a general range of integration using a transformation of the variable. Thus, an integral of the form f(x) dx can be evaluated, using the Gauss integration method, as
f:
l
b
b
f(x) dx =
L w;f(x;) n
; a
i=1
a
(E5.7.4)
where the coordinate transformation x=
+a +b
(b - a)t
(E5.7.5)
2
is used so that
Xi
=
+a +b
(b - a)ti
2
(E5.7.6)
.......-
'
'
5,6
Recent Contributions
147
Using n = 4, the corresponding values of Wi and ti are given by (2] WI
=
W4
= 0.347854845147454
(E5.7.7)
W2 = W3 = 0.652145154862546
The values of the variable
tl
= -0.861136311594053
t2
= -0.339981043584856
Xi
(E5.7.8)
given by Eq. (E5.7.6) for a
= 0 and b = 1 are
XI = 0.06943184 X2 = 0.33000946 X3 = 0.66999054 X4 = 0.93056816 Treating the values of
= xl -
~~Xi
(E5.7.9)
Xi
as unknowns, Eq. (ES.7.1) can
+ Xi~I)
where
+ 0.18516506
(E5.7.11)
0.18516506
+ 1.324S4112
= 1.30749607
The solution of Eq. (E5.7.1l) is given by
5.6 RECENT CONTRIBUTIONS Strings and Bars Laura and Gutierrez determined the fundamental frequency coefficient of vibrating systems using Rayleigh's optimization concept when solving integral
~
148
Derivation of Equations: Integral Equation Approach
equations by means of the Ritz method [13]. The authors considered the transverse vibration of a string of variable density, the longitudinal vibration of a rod with a nonuniform cross section, and the transverse vibration of a beam with ends elastically restrained against rotation as illustrative examples. Beams An integral equation approach was used by many investigators for the solution of a vibrating beam [13]. In Ref. [9], Bergman and McFarland used Green's functions to study the free vibrations of an Euler-Bernoulli beam with homogeneous boundary conditions, supported in its interior by arbitrarily located pin supports and translational and rotational springs. A method of determining the dynamic response of prismatic damped Euler-Bernoulli beams subjected to distributed and concentrated loads using dynamic Green's functions was presented by Abu-Hilal [8]. The method gives exact solutions in closed form and can be used for single- and multi span beams, single- and multiloaded beams, and statically determinate or indeterminate beams. The responses of a statically indeterminate cantilevered beam and a cantilevered beam with elastic support are considered as example problems. The use of Green's functions in the frequency analysis of Timoshenko beams with oscillators was considered by Kukla [7]. Membranes Spence and Horgan [IO] derived bounds on the natural frequencies of composite circular membranes using an integral equation method. The membrane was assumed to have a stepped radial density. Although such problems, involving discontinuous coefficients in the differential equation, can be treated using classical variational methods, it was shown that an eigenvalue estimation technique based on an integral formulation is more efficient. Gutierrez and Laura [II] analyzed the transverse vibrations of composite membranes using the integral equation method and Rayleigh's optimization suggestion. Specifically, the fundamental frequency of vibration of membranes of nonuniform density was determined. Plates Bickford and Wu [12] considered the problem of finding upper and lower bounds on the natural frequencies of free vibration of a circular plate with stepped radial density. The problem, which involves discontinuous coefficients in the governing differential equation, has been formatted with an integral equation by using a Green's function and the basic theory of linear integral equations.
REFERENCES 1. J. Kondo, Integral Equations, Kodansha, Tokyo, and Clarendon Press, Oxford, 1991. 2. S. S. Rao, Applied Numerical Methods for Engineers and Scientists, Prentice Hall, Upper Saddle River, NJ, 2002. 3. S. S. Rao, Mechanical Vibrations, 4th ed., Prentice Hall, Upper Saddle River, NJ, 2004. 4. F. G. Tricomi, Integral Equations, Interscience, New York, 1957. 5. J. A. Cochran, The Analysis of Linear Integral Equations, McGraw-Hill. New York, 1972. 6. L. Meirovitch. Analytical Methods ill Vibrations, Macmillan, New York, 1967. 7. S. Kukla, Application of Green functions in frequency analysis of Tirnoshenko beams with oscillators, Journal of Sound and Vibration, Vol. 205, No.3, pp. 355-363, 1997.
Problems
,1:49
8. M. Abu-Hilal. Forced vibration of Euler-Bernoulli beams by means of dynamic Green functions. Journal of Sound and Vibration. Vol. 267, No.2, pp. 191-207,2003. 9. L. A. Bergman and D. M. McFarland. On the vibration of a point- supported linear distributed system. Journal of Vibration, Acoustics, Stress, and Reliability in Design. Vol. 110, No.4, pp. 485-492, 1988. 10. J. P. Spence and C. O. Horgan. Bounds on natural frequencies of composite circular membranes: integral equation methods. Journal of Sound and Vibration, Vol. 87, PP'·71-8I, 1983. 11. R. H. Gutierrez and P. A. A. Laura, Analysis of transverse vibrations of composite membranes using the integral equation method and Rayleigh's optimization suggestion, Journal of Sound and Vibration. Vol. 147, No.3, pp. 515-518, 1991. . 12. W. B. Bickford and S. Y. Wu, Bounds on natural frequencies of composite circular plates: integral equation methods. Journal of Sound and Vibration, Vol. 126, No. I, pp. 19-36, 1988. 13. P. A. A. Laura and R. H. Gutierrez, Rayleigh's optimization approach for the approximate solution of integral equations for vibration problems, Journal of Sound and Vibration, Vol. 139. No. I, pp. 63-70,1990.
PROBLEMS 5.1
and the function
Classify the following integral equations:
(a) ¢(x)
+ A Jot(x + y)¢2(y)
dy
=0
(5.2)
(b) ¢(x)-AJ~eX-Y¢(y)dY=f(x) (c) z(x, t) = J; G(x, y) [p(y) -IL(Y)~]]
Show that the function ¢(t) given by Eq. (5.2) is a solution of Eq. (5.1).
dy,
O~x~l 5.2
5.5
Classify the equation ¢(x) = x
+ LX
(~ -
x)¢(~) d~
(5.1)
Show that the function ¢(x) = sin x is a solution of •._Eq. (5.1). 5.3
Consider the integral equation
o~x
¢(x)
5.6
Determine the condition(s) under which the function = sinwx satisfies Eq. (5.1).
~j;:.
Consider the integral equation [1] ¢(t) -
[00
e-('-~)(t - ~)¢(~)
= 2c +CtX
+1
Consider the integral equation
(5.2)
. ¢J(x)
5.4
using the method of undetermined coefficients by assuming the solution ¢(x) to be of the form
~ 1 (5.1)
where - ~), K( x,s!:) _- {x(l ~(l-x),
Solve the equation
= te'
(5.1)
Solve this equation using the method of undetermined coefficients by assuming the solution as
150 5.7
Derivation of Equations: Integral Equation Approach Find the solution of the equation
5.9 Find the solution of the following equation using a numerical method [1]: 1 4>(t) - -
1
404
by assuming a solution of the form
2
71 /
t~4>(nd~
= sint
- -t
Compare your solution with the exact solution, 4>(t) = sin t. Use the method of undetermined coefficients. 5.8
5.10 Find the solution of the following integral equation (eigenvalue problem) using the Gaussian integration method:
Solve the integral equation
using the method of undetermined coefficients. Assume the solution 4>(x) as 4>(x)
= + C2X + C3x2 Cl
Table 5.1
(a) with two Gaussian points; (b) with four Gaussian points; and (c) with six Gaussian points. [Hint: The locations and weights corresponding to different number of Gauss points are given in Table 5.1.]:
Data for Problem 5.10
Number of points, n
Locations,
Weights,
Xi
Wi
1
0.000000ססoo 00000
2
±0.57735 02691 89626
1. 000000000000000
3
±0.774596669241483 0.0000000000 00000
0.555555555555555 0.88888 88888 88889
4
±0.86113 63115 94053 ±0.339981043584856
0.347854845147454 0.652145154862546
5
±0.90617 98459 38664 ±0.538469310105683 0.000000ססoo00000
0.236926885056189 0.4786286704 99366 0.56888 88888 88889
6
±0.93246 95142 03152 ±0.66120 93864 66265 ±0.238619186083l97
0.171324492379170 0.360761573048139 0.467913934572691
Source: Ref. [2].
2.000000000000000
..,'"
"'
Solution Procedure: Eigenvalue and Modal Analysis Approach 5.4
6.1
INTRODUCTION The equations of motion of many continuous systems are in the form of nonhomogeneous linear partial differential equations of order 2 or higher subject to boundary and initial conditions. The boundary conditions may be homogeneous or nonhomogeneous. The initial conditions are usually stated in terms of the values of the field variable and its time derivative at time zero. The solution procedure basically involves two steps. In the first step, the nonhomogeneous part of the equation of motion is neglected and the homogeneous equation is solved using the separation-of-variables teclmique. This leads to an eigenvalue problem whose solution yields an infinite set of eigenvalues and the corresponding eigenfunctions. The eigenfunctions are orthogonal and form a complete set in the sense that any function j (JO that satisfies the boundary conditions of the problem can be represented by a linear combination of the eigenfunctions. This property constitutes what is known as the expansion theorem. In the second step, the solution of the nonhomogeneous equation is assumed to be a sum of the products of the eigenfunctions and time-dependent generalized coordinates using the expansion theorem. This process leads to a set of second-order ordinary differential equations in terms of the generalized coordinates. These equations are solved using the initial conditions of the problem. Once the generalized coordinates are known, complete solution of the problem can be determined from the expansion theorem.
6.2
GENERAL PROBLEM The equation of motion of an undamped continuous sys.tem is in the form of a partial differential equation which can be expressed as
- a
M(X)
2 w(X t)
at
2'
+ L[w(X, -
t)]
=
f(X,
-" t)
s
+~
Fj(t)1>(X
-- - Xj),
X E V
(6.1)
j=l
where X is a typical point in the domain of the system (V), M(X) is the mass distribution, w(X, t) is the field variable or displacement of the system that depends on the spatial variables (X) and time (t), L[w(X, t)] is the stiffness distribution of the" system, f(X, t) is the distributed force acting on the system, Fj(t) is the concentrated 151
152
Solution Procedure: Eigenvalue and Modal Analysis Approach
force acting at the point X = X j of the system, s is the number of concentrated forces acting on the system, and 8 (X - X j) is the Dirac delta function, defined as = 0,
8(X - Xj) [8(X
(6.2)
=
- Xj)dV
Note that the vector X will be identical to x for one-dimensional systems, includes x and y for two-dimensional systems, and consists of x, y, and z for three-dimensional systems. In Eq. (6.1), L and M are linear homogeneous differential operators involving derivatives with respect to the spatial variables X (but not with respect to time, t) up to the orders 2p and 2q, respectively, where p and q are integers with p > q. For example, for a two-dimensional problem in a Cartesian coordinate system, the operator L can be expressed for p = 1 as L[w] = c] w
ow
ow
02w
02w
ox
oy
ox
oy2
+ C2- + C3- + C4- 2 + Cs-
02W
+ C6--
oxoy
(6.3)
and the linearity of L implies that (6.4)
where C],C2,... , C6 are constants. In the case of the transverse vibration of a string having a mass distribution of p (x) per unit length and subjected to a constant tension P, the operators M and L are given by [see Eq. (8.8)] M=p(x)
(6.5)
02
L=P8x2
(6.6)
In the case of the torsional vibration of a shaft, the operators M and L are given by [see Eq. (10.19)] M = Io(x)
L
=~ ox
(6.7)
(Glp~)
(6.8)
ox
where Io(x) is the mass polar moment of inertia of the shaft per unit length, G is the shear modulus, and Ip (x) is the polar moment of inertia of the cross section of the shaft. Similarly, in the case of the transverse vibration of a uniform plate in bending, the operators M and L are given by [see Eq. (14.8)] M=ph
(6.9)
L=D(~+2~+~) ox4
ox20y2
oy4
(6.10)
where p is the density, h is the thickness, and D is the flexural rigidity of the plate.
6.3
Solution of Homogeneous Equations: Separation-of- Variables Technique
i53
The governing differential equation (6.1) is subject to p boundary conditions at every point of boundary S of domain V of the system. The boundary conditions can be expressed as Aj[w]
(6.11)
i = I, 2, ... , p
= ABj[w],
where Ai and Bi are linear homogeneous differential operators involving derivatives of w, with respect to the normal and tangential directions of the boundary, up to the order 2p - 1, and A is a parameter known as the eigenvalue of the system. In some problems, the boundary conditions do not involve the eigenvalue A, in which case Eq. (6.11) reduces to (6.12)
i = 1,2, ... , p
= 0,
Ai[W]
We shall consider mostly boundary conditions of the type given by Eq. (6.12) in further discussions. In the case of free vibration, I and all Fj will be zero, and Eq. (6.1) reduces to the homogeneous
form 2
~
~ 0 w(X, t) M(X) 2
ot
~
+ L[w(X,
XeV
t)] = 0,
(6.13)
6.3 SOLUTION OF HOMOGENEOUS £QUATIONS:
SEPARATION-OF -VARIABLES TECHNIQUE The separation-of-variables technique is applicable to the solution of homogeneous second- and higher-order linear partial differential equations with constant coefficients subject to homogeneous boundary conditions. The partial differential equations may represent initial or boundary value problems. To illustrate the method of separation of variables, we consider a homogeneous hyperbolic equation of the form p(x)
o2w(x,t)
ot 2
+ L[w(x,
t)]
x
= 0,
e G,
t >0
(6.14)
where G denotes a bounded region such as [0, I], x is the spatial variable, t is time, p(x) is positive and independent of t, L is a linear differential operator, and w(x, t) is an unknown function to be determined. The homogeneous boundary conditions can be stated as
The initial conditions
ow
Al w(O, t)
+ BI-(O, ox
A2W(I, t)
+ B2-(I, ox
ow
t)
=0
(6.15)
t)
=0
(6.16)
for Eq. (6.14) can be expressed w(x,O) = I(x),
ow
-ar(x,O)
= g(x),
as
xeG
(6.17)
x eG
(6.18)
154
Solution Procedure: Eigenvalue and Modal Analysis Approach
The separation-of-variables technique replaces the partial differential equation in two parameters, x and t, Eq. (6.14), by two ordinary differential equations. The solution of Eq. (6.14) is assumed to be a product of two functions, each depending on only one of parameters x and t as (6.19) w(x, t) = W(x)T(t) where W(x) is required to satisfy the boundary conditions, Eqs. (6.15) and (6.16), and T(t) is required to satisfy the initial conditions, Eqs. (6.17) and (6.18). Substituting Eq. (6.19) into Eq. (6.14) and dividing throughout by p(x)W(x)T(t) yields T"(t)
= _ L[W(x)j
T(t)
(6.20)
p(x)W(x)
=
where T"(t) d2T(t)/ dt2, Since the left and right sides of Eq. (6.20) depend on different variables, they cannot be functions of their respective variables. Thus, each side of Eq. (6.20) must equal a constant. By denoting this constant by -A, we obtain the following equations for W(x) and T(t) (see Problem 6.1): (6.21) L[W(x)] = Ap(X)W(x) T"(t)
+ AT(t)
(6.22)
=0
The ordinary differential equations (6.21) and (6.22) can be solved by satisfying the boundary and initial conditions to find W(x) and T(t), respectively. Consequently, w(x, t) W(x)T(t) will satisfy both the boundary and initial conditions. Since both the equation and the boundary conditions for W (x) are homogeneous, W (x) 0 will be a solution of the problem (called the trivial solution). However, we require nonzero solutions for W (x) in the boundary··value problem, and such solutions exist only for certain values of the parameter A in Eq. (6.21). The problem of determining the nonzero W (x) and the corresponding value of the parameter A is known as an eigenvalue problem. Here A is called an eigenvalue and W (x) is called an eigenfunction. The eigenvalue problem is known as the Sturm-Liouville problem in the mathematical literature and is discussed in the following section.
=
6.4
=
STURM-LIOUVILLE PROBLEM The mathematical models for the vibration of some continuous systems are in the form of a certain type of two-point boundary value problem known as the Sturm-Liouville problem. The Sturm-Liouville problem is a one-dimensional eigenvalue problem whose governing equation is of the general form
d [ p(x)- dW]
_ dx
dx
or (pW/)/
+ [q(x) + Ar(x)]w(x) + (q + Ar)w
= 0,
= 0, a
a
(6.23)
with boundary conditions in the form A]w(a) A2w(b)
+ Btw/(a) = 0 + B2w/(b) = 0
(6.24) (6.25)
6.4
Sturm-Liouville Problem
155
where a prime denotes a derivative with respect to x, and p~:c), q(x), and r(x) are continuous functions defined in the closed interval a :::x :::b. with p(x) > 0, r(x) > and Ai ~ 0, Bi ~ with Ai + Bi > for i = 1, 2. The boundary conditions of Eqs. (6.24) and (6.25) are said to be homogeneoils because a linear combination of w (x) and w' (x) at x a and x b are both equal to zero.
°
°
=
6.4.1
°
=
Classification of Sturm-Liouville Problems Based on the nature of the boundary conditions and the behavior of p(x) at the boundaries, Sturm-Liouville problems can be classified as regular, periodic, or singular. The problem defined by Eqs. (6.23)-(6.25) is called a regular Swrm-Liouville problem. Note that p(x) > and is continuous in the interval a :::x ::::b with constants Ai and Bi not equal to zero simultaneously in the ith boundary condition, i 1,2 [Eqs. (6.24) and (6.25)]. In this case the problem involves finding constant values of A corresponding to each of which a nontrivial solution w(x) can be found for Eq. (6.23) while satisfying the boundary conditions of Eqs. (6.24) and (6.25). If the function p(x) and the boundary conditions involving w(x) and w'(x) are periodic over the interval a :::x :::b, the problem is called a periodic Sturm-Liouville problem. In this case, the problem involves finding constant values of A corresponding to each of which a nontrivial solution can be found for Eq. (6.23) while satisfying the periodic boundary conditions given by
°
=
p(a) = p(b),\
w(a) = w(b),
w'(a) = w'(b)
(6.26)
If the functions p(x) or r(x) or both are zero at anyone or both of the boundary points a and b, the problem is called a singular Sturm-Liouville problem. In this case the problem involves finding constant values of A corresponding to each of which a nontrivial solution w(x) can be found to satisfy Eq. (6.23) and the boundary conditions of Eqs. (6.24) and (6.25). For example, if the singular point is located at either x = a or x = b so that either p(a) = or p(b) = 0, the boundary condition that is often imposed at the singular point basically requires the solution w(x) to be bounded at that point. Note that Eq. (6.23) always has the solution w(x) = 0, called the trivial solution. For nontrivial solutions (solutions that are not identically zero) that satisfy the specified boundary conditions at x = a and x = b, the parameter A cannot be arbitrary. Thus, the problem involves finding constant values of A for which nontrivial solutions exist that satisfy the specified boundary conditions. Each value of A for which a nontrivial solution can be found is called an eigenvalue of the problem and the corresponding solution w(x) is called an eigenfunction of the problem. Because the Sturm-Liouville problem is homogeneous, it follows that the eigenfunctions are not unique. The eigenfunction corresponding to any eigenvalue can be multiplied by any constant factor, and the resulting function remains an eigenfunction to the same eigenvalue A.
°
Example 6.1 Regular Sturm-Liouville problem
Problem
Find the solution to the eigenvalue
(E6.1.1 )
I .~
156
Solution Procedure: Eigenvalue and Modal Analysis Approach
with the boundary conditions dW W(O) - -(0) = 0 dx
dW
+ -(1)
W(1)
dx
(E6.1.2) (E6.1.3)
=0
SOLUTION Equation (E6.1.l) can be identified as a Sturm-Liouville problem, Eg. (6.23), with p(x) 1, q(x) 0, and r(x) = 1. Theoretically, we can consider three cases: ). < 0, ). = 0, and ). > O. When).. < 0, we set).. = _a2 and write Eq. (E6.1.l) as
=
= 2
d W(x) dx2
=0
_ a2W(x)
(E6.1.4)
which has the general solution (E6.1.5) The boundary conditions, Eqs. (E6.1.2) and (E6.1.3), become C}
C}
(1 - a)
(1 + a)ea
+ c2(1 + a)
+ c2(1 -
a)e-a
=0
(E6.1.6)
=0
(E6.1.7)
It .can be shown, (see Problem 6.2) that for a> 0, C} . and C2 do not have nonzero solution; the only solution is the trivial solution, c} 0 and C2 O. Hence, the problem has no negative eigenvalue. When).. = 0, Eg. (E6.1.1) reduces to
=
2 d W(x)
=
=0
(E6.1.8)
+ C2X
(E6.1.9)
dx2
which has the general solution W(x)
=C}
The boundary conditions, Eqs. (E6.1.2) and (E6.1.3), become C2
=0
(E6.1.10)
+ 2C2
=0
(E6.1.11 )
Cl Cl
The only solution to Eqs. (E6.1.l0) and (E6.1.1l) is the trivial solution C2 O. Hence, A 0 is not an eigenvalue of the problem When).. > 0, we set). = a2 and write Eq. (E6.1.1) as
=
=
d2W(x)
dx2
+ a..2W(x) ,
=0
Cl
= 0 and
(E6.1.12)
which has the general solution W(x)
= Cl cosax + C2 sin ax
(E6.1.13)
6.4
Stuml-Liouville Problem
·1S7
The boundary conditions, Eqs. (E6.1.2) and (E6.1.3), become
=0
(E6.1.14)
+ C2a cosa = 0
(E6.1.15)
Cl -
Cl cosa
+ C2 sina
- cia sina
aC2
The solution of Eqs. (E6.1.l4) and (E6.1.15) is given by Cl
c2[(1 ~ (2)
sin a
+2acosa]
= aC2
(E6.1.16)
=0
(E6.1.17)
If C2 is zero in Eq. (E6.1.l7), Cl will also be zero from Eq. (E6.1.16). This will be a trivial solution. Hence, for a nontrivial solution, we should have
+ 2a cosa
(1 - (2) sina
=0
(E6.1.18)
Equation (E6.1.18) is a transcendental equation whose roots are given by tanai
2ai = -2--' a -1
i = 1,2, ...
(E6.1.l9)
i
Equation (E6.1.19) can be solved numerically to find ai, hence the eigenvalues are given by Ai = and the corresponding eigenfunctions are given by
a;
Wj(X).~
c2(ai cosaix
Example 6.2 Periodic Sturm-Liouville value problem d2W(x)
dx2
+ AW(X)
+ sinaix)
Problem
= 0,
(E6.1.20)
Find the solution of the boundary
0
(E6.2.l)
subject to the boundary conditions W(O) = W(l)
(E6.2.2)
dW dW - -(0) = -(1) dx dx
(E6.2.3)
SOLUTION Equation (E6.2.l) can be identified as the Sturm-Liouville problem of Eq. (6.23) with p(x) 1, q(x) 0, and r(x) 1. We can consider three cases: A < 0, A = 0, and A > O. When A < 0, we set A _a2 and write Eq. (E6.2.l) as
=
=
=
=
2
d W(x)
dx2 which has the general solution
_ a2W(x)
=0
(E6.2.4)
158
Solution Procedure: Eigenvalue and Modal Analysis Approach
The boundary conditions, Eqs. (E6.2.2) and (E6.2.3), become (1 -
CI
+ c2(1
ea)
a
- e- ) a
ea) - c2a(1 - e- )
c)a(l -
=0 =0
(E6.2.6) (E6.2.7)
Equation (E6.2.6) gives (E6.2.8) Substitution of Eq.(E6.2.8) in Eq. (E6.2.7) yields (E6.2.9)
2c)a(l - ea) = 0
=
Since a > 0, Eq. (E6.2.9) gives CJ 0 and hence C2 = 0 [from Eq. (E6.2.8)]. This is a trivial solution and hence the problem has no negative eigenvalue. When A = 0, Eq. (E6.2.l) reduces to
=0
2
d W(x)
(E6.2.l0)
dx2 which has the general solution
= +
W(x)
c)
(E6.2.11)
C2X
The boundary conditions, Eqs. (E9.2.2) and (E6.2.3), become
c)= C2
Equations (E6.2.l2) and (E6.2.l3) Eq. (E6.2.11), reduces to
c)
(E6.2.l2)
+ C2
(E6.2.l3)
= C2
imply that
W(x)
C2
=0
and hence the solution, (E6.2.l4)
= c)
where c) is any nonzero constant. This shows that A = 0 is an eigenvalue of the problem with the corresponding eigenfunction given by Eq. (E6.2.l4). When A > 0, we set A = a2 and write Eq. (E6.2.1) as 2
d W(x) dx2
+ a2W(x)
=0
(E6.2.l5)
which has the gerieral solution W(x)
= c) cosax + C2 sin ax
(E6.2.l6)
The boundary conditions, Eqs. (E6.2.2) and (E6.2.3), become
+ C2 sin a C2a = "':"c)a sin a + C2a cosa c) = c) cosa
(E6.2.l7) (E6.2.18)
6.4
Sturm-Liouville Problem
159
The solution of Eqs. (E6.2.17) and (E6.2.18) yields C2
= C,
1 - cosa
(E6.2.19)
.
sina
and
sina 1 - cos a
(E6.2.20)
C2 = -C1----
which imply that either
Cl
=
°
or (E6.2.21)
cos a = 1
Equation (E6.2.21) implies that a is zero or an integer multiple of 21l', so that m
am = ±m ·21l',
(E6.2.22)
= 0, 1,2, ...
Thus, the eigenvalues of the problem are given by ,
= am2 = 4m 2 1l' 2 ,
m
m
= 0, 1,2, ...
(E6.2.23)
with the corresponding eigenfunctions given by Eq. (E6.2.16): Wm(X) =
Cl
cos2m.1l'x
+ C2 sin2m1l'x,
m = 0, 1,2, ...
(E6.2.24)
Example 6.3 Singular Sturm;'Liouville Problem The free transverse vibration of a circular membrane of radius a, clamped around the edge, is governed by the equation r2
d2W(r) 2 dr
+ r--dW(r) + (w2 dr
m2)W(r) = 0,
0::: r :::a
(E6.3.1)
subject to the requirement of a bounded solution with the condition (E6.3.2)
W(a) = 0
where r denotes the radial direction, W(r) is the transverse displacement, w is the natural frequency (w2 is called the eigenvalue), and m is an integer [see Eq. (13.126)]. Find the solution of the problem . . .i
SOLUTION Equation (E6.3.1) can be identified as Bessel's differential equation of order m with the parameter w. The equation can be rewritten in the form of a Sturm-Liouville equation: -d (dW) r-
dr
dr
m2
+ (22 w r )-
-
r
W
= 0,
(E6.3.3)
=
which can be compared to Eq. (6.23) with the notations p(x) = r, q(x) -m2/r,r(x) =r2, and A=W2. It can be seen that Eq. (E6.3.3) denotes a singular Sturm-Liouville problem because p(O) = O. The solution of Eq. (E6.3.3) is given by
(E6.3.4) where B, and B2 are constants and 1m and Ym are Bessel functions of the first and second kind, respectively. The solution is required to be bounded, but W(r) approaches
160
Solution Procedure: Eigenvalue and Modal Analysis Approach infinity at r = O. Thus, B2 must be set equal to zero to ensure a bounded solution. Thus, Eq. (E6.3.4) reduces to (E6.3.5) The use of the boundary condition
of Eq. (E6.32) in Eq. (E6.3.5) yields (E6.3.6)
Equation (E6.3.6) has infinite roots Wia, i = 1,2, .... Thus, the ith eigenvalue membrane is given by Ai = and the corresponding eigenfunction by
wf,
of the
(E6.3.7) where the constant B I can be selected arbitrarily.
6.4.2
Properties of Eigenvalues and Eigenfunctions The fundamental properties of eigenvalues and eigenfunctions lems are given below.
of Sturm-Liouville
prob-
1. Regular and periodic Sturm-Liouville problems have an infinite number -distinct real eigenvalues 1..1,1..2, ... which can be arranged as
of
AI < 1..2 < ...
AI is finite and the largest one is infinity:
The smallest eigenvalue
lim An =
n -..
00
00
2. A unique eigenfunction exists, except for an arbitrary multiplicative constant, for each eigenvalue of a regular Sturm-Liouville problem. 3. The infinite sequence of eigenfunctions WI (x), W2(X), ... defined over the interval a :::::x :::::b are said to be orthogonal with respect to a weighting function r (x) ?: 0 if
l
b
r(x)wm(x)wn(x)dx
= 0,
When m = n, Eq. (6.27) defines the norm of wn(x),
(6.27) denoted Ilwn(x)ll,
as (6.28)
By normalizing the function
Wm
_
(x) -
W m
-
(x) as Wm (x)
---
IIwm(x)II'
m=1,2,
...
(6.29)
6.4 Sturm-Liouville Problem we obtain the orthonormal
functions
l l
wm(x)
-1(il
with the properties
b
r(x)wm(x)wn(x)
dx
= 0,
r(x)wm(x)wn(x)
dx
= 1,
(6.30)
b
m =n
(6.31)
4. Expansion theorem. The orthogonality of the eigenfunctions WI (x), wz(x), ... over the interval a :::::x :::::b with respect to a weighting function r(x) permits them to be used to represent any function j (x) over the same interval as a linear combination
of
wm(x)
as 00
= LCmWm(X)
j(x)
=CIWI(X)+CZwz(x)+",
(6.32)
m=1.
where CI, Cz, ... are constants known as coefficients of the expansion. Equation (6.32) denotes the eigenfunction expansion of j (x) and is known as the expansion theorem. To determine the coefficients Cm, we multiply Eq. (6.32) by r(x)wn(x) and integrate the result with respect to x from a to b:
l'
=
r(x)j(x)w.(x)~X
f [1'
c.r(x)w.(X)W.(X)dX]
(6.33)
When Eqs. (6.27) and (6.28) are used, Eq. (6.33) reduces to
l
l
b
b
r(x)j(x)wn(x)
dx
=
Cn
Equation (6.34) gives the coefficients
r(x)[wn(x)]z
Cn
= cnllwn(x)llz
(6.34)
as
fa b r(x)f(x)w.n(X)
Cn =
dx
dx
Ilwn(x)IIZ
n '
= 1,2, ...
(6.35)
5. Orthogonality of eigenfunctions. If the functions p(x), q(x), r(x), and r(x) are real valued and continuous with r(x) > on the interval a :::::x :::::b, the eigenfunctions wm(x) and wn(x) corresponding to different eigenvalues Am and An, respectively, are orthogonal with respect to the weighting function r(x). This property can be proved as follows. Since the eigenfunctions satisfy the Sturm-Liouville equation (6.23), we have
°
(pW~)' (pW~)'
+ (q + Amr)wm
= 0,
a
(6.36)
+ (q + Anr)wn
= 0,
a
(6.37)
Multiply Eq. (6.36) by wn(x) and Eq. (6.37) by wm(x) and subtract the resulting equations one from the other to obtain (6.38)
'1
,
I
162
Solution Procedure: Eigenvalue and Modal Analysis Approach
Integration of Eq. (6.38) with respect to x from a to b results in
l
b
(Am - An)
rwm(x)wn(x)
dx = [p(w~(x)Wm(X)
- w~(x)Wn(X)]~
= p(b)[w~(b)wm(b)
- w~(b)wn(b)]
- p(a)[w~(a)wm(a)
Based on whether p(x) is zero or not at x following cases:
=a
or x
= b,
- w~(a)Wn(a)]
(6.39)
we need to consider the
(a) p(a) = 0 and p(b) = 0: In this case, the expression on the right-hand side of Eq. (6.39) is zero. Since Am and An are distinct, we have
l
b
r(x)wm(x)wn(x)dx
= 0,
m
¥= n
(6.40)
Note that Eq. (6.40) is valid irrespective of the boundary conditions of Eqs. (6.24) and (6.25). (b) p(b) = 0 and p(a) ¥= 0: In this case, the expression on the right-hand side of Eq. (6.39) reduces to -p(a)[w~(a)wm(a)
- w~(a)wn(a)]
(6.41)
The boundary condition at x = a can be written as . Alwm(a) Alwn(a)
+ Blw~(a) + Blw~(a)
=0
(6.42)
=0
(6.43)
Multiply Eq. (6.42) by wn(a) and Eq. (6.43) by wm(a) and subtract the resulting equations one from the other to obtain (6.44)
Assuming that BI ¥= 0, the expression in brackets in Eq. (6.44) must be zero. This means that the expression in (6.41) is zero. Hence, the orthogonality condition given in Eq. (6.40) is valid. Note that if BI = 0, then Al ¥= 0 by assumption, and a similar argument proves the orthogonality condition in Eq. (6.40). (c) p(a) = 0 and p(b) ¥= 0: By using a procedure similar to that of case (b) with the boundary condition of Eq. (6.25), the orthogonality condition in Eq. (6.40) can be proved. (d) p(a) ¥= 0 and p(b) ¥= 0: In this case we need to use both the procedures of cases (b) and (c) to establish the validity of Eq. (6.40). (e) p(a) = p(b): In this case the right-hand side of Eq. (6.39) can be written as p(b)[w~(b)wm(b)
- w~,(b)wn(b) - w~(a)wm(a)
+ w~(a)wn(a)]
(6.45)
By using the boundary condition of Eq. (6.24) as before, we can prove that the expression in brackets in Eq. (6.45) is zero. This proves the orthogonality condition given in Eq. (6.40).
6.5
General EigenvaiueProblem
'.:163
6.5 GENERAL EIGENVALUE PROBLEM The eigenvalue problem considered in Section 6.4, also known as the Sturm-Liouville is valid only for one-dimensional systems. A general eigenvalue problem applicable to one-, two-, and three-dimensional systems is discussed in this section. In the case of free vibration, f and all Fj will be zero and Eq. (6.1) reduces to the homogeneous form
problem,
2
-
-aw(X,t) M(X) 2
at
\>-".
+ L[w(X,-
t)] = 0,
For the natural frequencies of vibration, we assume the displacement harmonic function as
",
(6.46)
XEV w(X,
t) to be a
(6.47) where W(X) denotes the mode shape (also called the eigenfunction or normal mode) and w indicates the natural frequency of vibration. Using Eq. (6.47), Eq. (6.46) can be represented as L[W] = AM[W]
(6.48)
where A = w2 is also called"the eigenvalue of the system. Equation (6.48), along with the boundary conditions of Eq. (6.11) or (6.12), defines the eigenvalue problem of the system. The solution of the eigenvalue problem yields an infinite number of eigenvalues AI, A2, ... and the corresponding eigenfunctions WI (X), W2(X), .... The eigenvalue problem is said to be homogeneous, and the amplitudes of the eigenfunctions Wi(X), i = 1,2, ... , are arbitrary. Thus, only the shapes of the eigenfunctions can be determined uniquely.
6.5.1
Self-Adjoint Eigenvalue Problem Before defining a problem known as the self-adjoint eigenvalue problem, two types of functions, called admissible and comparison functions, are introduced. These functions are used in certain approximate methods of solving the eigenvalue problem. As seen in Chapter 4, the boundary conditions of Eq. (6.12) are composed of geometric (or forced) and natural (or free) boundary conditions. A function u(X) is said to be an admissible function if it is p times differentiable over the domain V and satisfies only the geometric boundary conditions of the eigenvalue problem. Note that an admissible function does not satisfy the natural boundary conditions as well as the governing differential equation of the eigenvalue problem. A function u(X) is said to be a comparison function if it is 2p times differentiable over the domain V and satisfies all the boundary conditions (both geometric and natural) of the eigenvalue problem. Note that a comparison function does not satisfy the governing differential equation of the eigenvalue problem. On the other hand, the eigenfunctions Wi (X), i = 1,2, ... satisfy the governing differential equation as well as all the boundary conditions of the eigenvalue problem.
164
Solution Procedure: Eigenvalue and Modal Analysis Approach Definition The eigenvalue problem defined by Eqs. (6.48) and (6.12) is said to be selfadjoint if for any two arbitrary comparison functions u i (X) and U2 (X), the following
relations are valid:
Iv Iv
Ui (X)L[U2(X)]
dV
Ui (X)M[U2(X)]
dV
Iv = Iv =
u2(X)L[Ui(X)]dV
(6.49)
u2(X)M[Ui(X)]dV
(6.50)
Positive Definite Problem An eigenvalue problem, defined by Eqs. (6.48) and (6.12), is said to be positive definite if the operators L and M are both positive definite. The operator L is considered positive if for any comparison function u(X), the following
relation is valid:
Iv
u(X)L[u(X)]
dV
~
0
(6.51)
The operator L is considered positive definite if the integral in Eq. (6.51) is zero only when u(X) is identically equal to zero. Similar definitions are valid for the operator M. The eigenvalue problem is said to be semidefinite if the operator L is only positive and the operator M is positive definite. It is to be noted that the eigenvalue problems corresponding to most continuous systems considered in subsequent discussions are self-adjoint, as implied by Eqs. (6.49) and (6.50). In most cases, the operator M(X) denotes the distributed mass of the system, and hence the positive definiteness of M(X) is ensured. If the system or the eigenvalue problem is positive definite, all the eigenvalues Ai will be positive. If the system is semidefinite, some Ai will be zero. It can be seen that these properties are similar to that of a discrete system. 6.4 The free axial vibration of a uniform bar fixed at both the ends x and x = L is governed by the equation
Example
EA
a2u(x,t) a2u(x,t) ax2 +m at2
=0
0 =
(E6.4.I)
where E is Young's modulus, A is the cross sectional, area, m is the mass per unit length, and u(x, 1) is the axial displacement of the bar. Show that the eigenvalue problem, obtained with u(x,1) = U(x) cos wt
(E6.4.2)
in Eq. (E6.4.I), is self-adjoint. Consider the following comparison functions: (E6.4.3) SOLUTION
The eigenvalue problem corresponding to Eq. (E6.4.1) is given by d2U(x)
EA
di2
=
AmU(x)
(E6.4.4)
6.5
il(tB
General Eigenvalue Problem
where A = (Ji is the eigenvalue. Comparing Eq. (E6.4.4) with Eq. (6.48), we identify the operators L and M as
aZ
L
= EAaxz'
(E6.4.5)
M=m
The eigenvalue problem will be self-adjoint if the following conditions hold true:
1£ 1£
UI (x)L[Uz(x)]
dx
=
dx =
UI (x)M[Uz(x)]
1£ 1£
Uz(x)L[UI
Uz(x)M[UI
(E6.4.6)
(x)] dx
(E6.4.7)
(x)] dx
The boundary conditions of the bar can be expressed as (E6.4.8)
U(L) = 0
U(O) = 0,
The comparison functions given by Eq. (E6.4.3) can be seen to satisfy the boundary conditions, Eq. (E6.4.8). Using UI (x) and Uz(x), we find that
{L
10
UI(x)L[Uz(x)]
dx
{L
= 10
Clx(L
dZ
- x)EA dxZ
(
Cz sin
rr x )
T
dx
4CICzEAL
L 1 o
Uz(X)L[Ul (x)] dx
=
1L 0
rrx Cz sin -EA-z
L
4CICzEAL
(E6.4.9) Z
d Z (ClxL - CIX ) dx dx (E6.4.1O)
It can be seen that Eq. (E6.4.6) is satisfied. Similarly, Eq. (E6.4.7) can also be shown to be satisfied. Thus, the eigenvalue problem is self-adjoint.
6.5.2
Orthogonality of Eigenfunctions The orthogonality property, proved for the Sturm-Liouville problem in Section 6.4.2, can also be established for a general eigenv}lue problem. For ~s, let Ai and Aj denote two distinct eigenvalues, with Wi = Wi(X) and Wj = Wj(X) indicating the corresponding eigenfunctions. Then L[Wd
= AiM[Wd
(6.52)
L[Wj]
= AjM[Wj]
(6.53)
Multiply Eq. (6.52) by Wj and Eq. (6.53) by Wi and subtract the resulting equations from each other:
166
Solution Procedure: Eigenvalue and Modal Analysis Approach
V of the system to obtain
Integrate both sides of Eq. (6.54) over the domain Iv (WjL[W;] If the eigenvalue
- WiL[WjDdV
= IvO'i
WjM[WiD
- Aj WiM[Wj]dV
problem is assumed to be self-adjoint,
(6.55)
then
Iv WjL[W;]dV
= Iv WiL[Wj]dV
(6.56)
Iv WjM[W;]dV
= Iv WiM[Wj]dV
(6.57)
and
In view of Eqs. (6.56) and (6.57), Eq. (6.55) reduces to
=0
(Ai - Aj) Iv WiM[Wj]dV
(6.58)
Since Ai and A j are distinct, Eq. (6.58) yields Iv WiM[Wj]dV
=0
(6.59)
Equations (6.59) and (6.53) can be used to obtain Iv WiL[Wj]dV
=0
for Ai
=I
Aj
(6.60)
Equations (6.59) and (6.60) are known as the generalized orthogonality conditions and the eigenfunctions Wi(X) and Wj(X) are considered to be orthogonal in a generalized sense. The eigenfunctions Wi(X) can_be normalized with respect to M[W;] by setting
i
6.5.3
= 1,2,
...
(6.61)
Expansion Theorem As in the case of the Sturm-Liouv~l~problem, the eigenfunctions constitute a complete set in the sense that any function f (X) that satisfies the homogeneous boundary conditions _of the problem can be represented by a linear combination of the eigenfunctions Wm(X) of the problem as 00
J(X) = L m=!
Cm Wm(X)
(6.63)
6.6
where the coefficients
Cm
Solution of Nonhomogeneous Equations
can be determined as in the case of Eq. (6.35) as
Iv!(X)M[Wm(X)]dV m
C
~167
m = 1,2, ...
IIWm(X)1I2
=
(6.64)
'
Equation (6.63), also known as the expansion theorem, plays an important role in vibration analysis and is commonly used to find the forced vibration response of a system by modal analysis.
6.6
SOLUTION OF NONHOMOGENEOUS EQUATIONS The equation of motion of a continuous system subjected to external forces leads to a nonhomogeneous partial differential equation given by Eq. (6.1):
- a 2w(X-2'
M(X)
at
s t)
+ L[w(X,-
t)] = f(X,- t)
+ '"'- Fj(t)l3(X -- .1 J=
Xj),
XeV (6.65)
subject to the boundary conditions indicated in Eq. (6.12): i
= 1,2, ...
,p
(6.66)
and initial conditions similar to those given by Eqs. (6.17) and (6.18): ",.w(X,
0) = f(X)
(6.67)
-
(6.68)
aw -
= g(X)
a;(X,O)
To find the solution or response of the system, w(X, t), we use a procedure known as modal analysis. This procedure involves the following steps: 1. Solve the eigenvalue problem associated with Eqs. (6.65) and (6.66). The eigenvalue problem consists of the differential equation = AM[W(X)],
L[W(X)]
XeV
(6.69)
with the boundary conditions = 0,
A;[W(X)]
i
= 1,2, ...
,p
(6.70)
where A = w2 is the eigenvalue, W (X) is the eigenfunction, and w is the natural frequency of the system. The solution of Eqs. (6.69) and (6.70) yields an infinite set of eigenvalues A1, A2, .. ' and the corresponding eigenfunctions, also known as mode shapes, WI (X), W2(X), .... The eigenfunctions are orthogonal, so that [M(X)Wm(X)Wn(X)dV
= l3mn
(6.71)
2. Normalize the eigenfunctions so that {
-
-
lv Wm(X)L[Wn(X)]dV
2
= wml3mn
(6.72)
--
168
Solution Procedure: Eigenvalue and Modal Analysis Approach 3. Express the forced response of the system [i.e., the solution of the problem in Eqs. (6.65)-(6.68)] using the expansion theorem as 00
w(X,
t) =
L Wm(X)lIm(t)
(6.73)
m=l
where the lIm(t) are known as the time-dependent generalized coordinates. In Eq. (6.73), the eigenfunctions Wm(X) are known from steps 1 and 2, while the generalized coordinates lIm(t) are unknown and to be determined by satisfying the equation of motion and the initial conditions of Eqs. (6.67) and (6.68). To determine 11m(t), we substitute Eq. (6.73) in Eq. (6.65) to obtain
s
= f(X,
t)
+ L Fj(t)8(X
- Xj)
(6.74)
j=l
which can be rewritten as 00
00
L ~m(t)M(X)Wm(X)
+ L 11m(t)L[Wm
m=l
(X)]
m=l
s
+ L Fj(t)8(X
= f(X,'t)
- Xj)
(6.75)
.-j=l
=
2
where rym(t) d 11m(t)/dt2• By multiplying result over the domain V, we obtain
E Iv
Eq. (6.75) by Wn(X)
and integrating
the
00
rym(t)
E
Wn(X)M(X)Wm(X)dV
00
+ =
J
TJm(t)
Iv
Wn(X)L[Wm(X)]dV s
Wn(X)f(X,
t) dV
+L j=l
V
J
Wn(X)Fj(t)8(X
- Xj) dV
(6.76)
v
Using the property of Dirac delta function given by Eq. (6.2), the last term ofEq. can be simplified as
(6.76)
s
L Wn(Xj)Fj(t)
(6.77)
j=l
In view of Eqs. (6.71), (6.72), and (6.77), Eq. (6.76) can be rewritten as rym(t)
+ W~lIm(t) = Qm(t),
m = 1,2 ....
(6.78)
6.7 where Qm(t)
Forced Response of Viscously Damped Systems
is called the mth generalized
force.
'1>69
given by
(6.79)
Equation (6.78) denotes an infinite set of uncoupled second-order ordinary differential equations. A typical equation in (6.78) can be seen to be similar to the equation of a single-degree-of-freedom system [see Eq. (2.107)]. The solution of Eq. (6.78) can be expressed as [see Eq. (2.109)] 1Jm(t)
= --1
CUm
it
QmCr) sin CUm(t - r) dr
0
.
sin cumt
+ 1Jm(0) cos CUmt + 1Jm(0) --,
m
CUm
= 1,2, ...
(6.80)
where 1Jm(0) and T7m(0) are the initial values of the generalized coordinate (generalized displacement) 1Jm(t) and the time derivative of the generalized coordinate (generalized velocity) T7m(t) = d1Jm(t)/ dt. Using the initial conditions of Eqs. (6.67) and (6.68), the values of 1Jm(O) and T7m(O) can be determined as 1Jm(9).~
0) dV
[M(X)Wm(X)W(X,
=[
(6.81)
M(X)Wm(X)f(X)dV
T7m(O) = [M(X)Wm(X)W(X,
0) dV
= [M(X)Wm(X)g(X)dV
(6.82)
Finally, the solution of the problem (i.e., the forced response of the system) can be found using Eqs. (6.80) and (6.73).
6.7
FORCED RESPONSE OF VISCOUSLY DAMPED SYSTEMS Consider the vibration of a viscously damped continuous system. We assume the damping force, Fd, resisting the motion of the system to be proportional to the velocity and opposite to the direction of the velocity, similar to the case of a discrete system: Fd(X, t)
= -C-a-taw(X,t) -
= - ata C[w(X, -
t)]
(6.83)
where C is a linear homogeneous differential operator, simil~ to the operator L, composed of derivatives with respect to the spatial coordinates X (but not with respect to time t) of order up to 2p. Thus, the equation of motion of the viscously damped system
170
Solution Procedure: Eigenvalue and Modal Analysis Approach can be expressed,
similar to Eq. (6.65), as [3, 4]
- a 2 w(X-2'
M(X)
t)
at
a + -C[w(X,
at
t)]
+ L[w(X, -
t)]
s
= f(X,
t)
+ L Fj(t)o(X
-
X j),
XeV
(6.84)
j=l
subject to the homogeneous
boundary conditions i = 1, 2, ... , p
(6.85)
and the initial conditions w(X,O)
ow at
= f(X)
(6.86)
-
(6.87)
0) = g(X)
-(X,
For the undameed system, we find the eigenvalues Am and the corresponding functions W m (X) by solving the eigenvalue problem
= AM[W(X)],
L[W(X)]
eigen-
(6.88)
XeV
subject to the boundary conditions
= 0,
Ai[W(X)]
i
= 1,2, ...
,p
(6.89)
The orthogonal eigenfunctions are assumed to be normalized according to Eqs. (6.71) and (6.72). As in the case of an undamped system, the damped response of the system is assumed to be a sum of the products of eigenfunctions and time-dependent generalized coordinates TJm(t), using the expansion theorem, as 00
w(X,
t)
= L Wm(X)TJm(t)
(6.90)
m=l
Substitution
of Eq. (6.90) into Eq. (6.84) leads to
00
00
L ~m(t)M(X)Wm(X)
00
+ L Ijm(t)C[Wm(X)]
m=1
+ L TJm(t)L[Wm(X)]
m=1
m=1
s
= f(X,
t)
+ L Fj(t)o(X
- Xj),
(6.91)
XeV
j=1 By multiplying Eg. (6.91) by W,,(X) and integrating the result over the domain of the system V, and using Eqs. (6.71) and (6.72), we obtain 00
~m(t)
+ L cm"ljm(t) + W;'TJm(t)
= Qm(t),
m= 1,2, ...
(6.92)
11=1
where
Cm",
known as the viscous damping coefficients, Cm"
= ( Wm(.X)C[W"C¥)]dV
1v
are given by (6.93)
6.8 Recent Contributions and Qm(t), called the generalizedforees. Qm(t) =
1
171
are given by s
Wm(X)f(X.
t) dV
v
+ L Wm(Xj)Fj(t)
(6,94)
j=l
In many practical situations. the viscous damping operator C is not known. To simplify the analysis. the operator C is assumed to be a linear combination of the operator L and the mass function M: (6.95) where
a,
and
az
can be expressed
are constants. With this assumption,
the viscous damping coefficients
as (6.96)
where ~m is called the damping ratio. Introducing Eq. (6.96) into Eq. (6.92). we obtain a set of uncoupled second-order ordinary differential equations: m
= 1.2 •...
(6.97)
Equations (6.97) are similar to those of a viscously damped single-degree-of-freedom system [see Eq. (2.119)]. The solution of Eq. (6.97) is given by [see Eq. (2.120)] 1}m(t) =
l'
Qm(-r)h(t - r) dr
+ g(t)1}m(O) + h(t)~m(O)
(6.98)
where h(t) = _l_e-~mwm'
(6.99)
sinwdmt
Wdm
~mwm.)smwdmt g(t) = e _~m wm , ( COSwdmt + -Wdm and Wdm is the mth frequency of damped vibration Wdm =
(6.100)
given by
J 1 - ~Ji
Wm
(6.101)
where the system is assumed to be underdamped. Once the 1}m(t) are known. solution of the original equation (6.84) can be found from Eq. (6.90).
the
6.8 RECENT CONTRIBUTIONS A discussion of the various methods of physical modeling and a brief survey of the direct solution techniques for a class of linear vibration systems. including discrete as well as distributed parameter systems. have been presented by Chen [5]. The discussion of discrete systems includes close-coupled. far-coupled. and branched systems. The discussion of continuous systems includes one-dimensional problems of vibrating strings and beams and two-dimensional problems of vibrating membranes and plates. Anderson and Thomas discussed three methods for solving boundary value problems that have
172
Solution Procedure: Eigenvalue and Modal Analysis Approach
both time derivatives of the dependent variable and known time-dependent functions in the boundary conditions [6]. In these methods, the time dependence is eliminated from the boundary conditions by decomposing the solution into a quasistatic part and a dynamic part. The boundary conditions containing the time derivatives of the dependent variable are satisfied identically by imposing special requirements on the quasistatic portion of the complete solution. The method is illustrated with a problem that deals with the forced thickness-stretch vibrations of an elastic plate. Anderson [7] investigated the forced vibrations of two elastic bodies having a surface contact within the framework of the classical linear theory of elasticity. The generalized orthogonality condition and a simple form of the generalized forces are derived. The procedure is illustrated by considering the example of the forced thickness-stretch vibration of a two-layer plate system. As indicated earlier, the modal analysis, based on eigenfunction expansion, is a commonly used technique for the transient analysis of continuous systems. However, the conventional modal expansion is not directly applicable to non-self-adjoint systems whose eigenfunctions are nonorthogonal. In Ref. [8], an exact closed-form solution method was presented for transient analysis of general one-dimensional distributed systems that have non-self-adjoint operators, and eigenvalue-dependent boundary conditions are subject to arbitrary external, initial, and boundary disturbances. In this reference, an eigenfunction series solution is derived through introduction of augmented spatial operators and through application of the modal expansion theorem given in Ref. [9]. The method is demonstrated by considering a cantilever beam with end mass, viscous damper, and spring. Structural intensity can be used to describe the transfer of vibration energy. The spatial distribution of structural intensity within a structure offers information on energy transmission paths and positions of sources and sinks of mechanical energy. Gavric et al. [10] presented a method for the measurement of structural intensity using a normal mode approach. The method is tested on an assembly of two plates.
REFERENCES 1. E. Kreyszig, Advanced Engineering Mathematics, 8th ed., Wiley, New York, 1999. 2. E. Zauderer, Partial Differential Equations of Applied Mathematics, 2nd ed., Wiley, New York, 1989. 3. L. Meirovitch, Analytical Methods in Vibrations, Macmillan, New York, 1967. 4. R. Courant and D. Hilbert, Methods of Mathematical Physics, Interscience, New York. Yol. I, 1937, Yol. II, 1962. 5. F. Y. Chen, On modeling and direct solution of certain free vibration systems, Jou17lalof Sound and Vibration, Yol. 14, No.1, pp. 57-79, 1971. 6. G. L. Anderson and C. R. Thomas, A forced vibration problem involving time derivatives in the boundary conditions, Journal of Sound and Vibration, Yol. 14, No.2, pp. 193-214. 1971. 7. G. L. Anderson, On the forced vibrations of elastic bodies in contact, Journal of Sound and Vibration, Yol. 16, No.4, pp. 533-549, 1971. 8. B. Yang and X. Wu, Transient response of one-dimensional distributed systems: a closed form eigenfunction expansion realization, Journal of Sound and Vibration, Yol. 208, No.5, pp. 763-776,1997.
Problems
173
9. B. Yang, Integral formulas for non-self-adjoint distributed dynamic systems, AIAA Journal, Vol. 34, No. 10, pp. 2132-2139, 1996. 10. L. Gavric. U. Carlsson, and L. Feng, Measurement of structural intensity using a normal mode approach. Journal of Sound and Vibration, Vol. 206, No.1, pp. 87-10 1, 1997.
PROBLEMS 6.1 Prove that the left and right sides of Eq. (6.20) must be equal to a negative constant. 6.2 Show that for a > 0, Eqs. (E6.1.6) and (E6.1.7) do not have a nonzero solution for CI and C2.
6.9 Consider the differential equation corresponding to the transverse vibration of a string fixed at x = 0 and x = l: 2 d W(x)
- x2)
n2]w
-
=0
6.4 Find the eigenvalues and eigenfunctions of the equation d2w
+ AW = 0,
-2
dx
= 0,
w(21f)
6.7 Determine whether equation is self-adjoint: 2
d2w
·6.10 The eigenvalue problem corresponding to the transverse vibration of a uniform beam is given by
E/
i = 1,2, ... the following
+ (x2 -m2)w
differential
=0
the folIowing
0
= csin(1fx/l) = cx(x -1) = cx(2x -1)
d4W(x)
. i1fx Wj(x) = SIn -l-'
d2w dw x2 __ +x2 dx dx
(b) W(x)
=0
6.5 Determine whether the folIowing functions are orthogonal in the interval 0 ::: x :::1:
6.6 Determine whether equation is self-adjoint:
(a) W(x) (c) W(x)
w(O)
(x=,
where a2 = w2 P / P, p is the mass per unit length and P is the tension in the string. Determine whether each of the following functions is an admissible. comparison, or eigenfunction:
22d2w 2 dw (1 - x ) dx2 - 2x(1 - x ) dx
+ (A(1
2W)
~+a
6.3 Convert the folIowing differential equation to Sturm-Liouville form. Eq. (6.23):
differential
dx
4
= AmW(x)
where E/ is the bending stiffness, m is the mass per unit length, W (x) is the transverse displacement (eigenfunction or mode shape), and A w2 is the eigenvalue. Assuming the beam to be simply supported at both ends x 0 and x L. show that the problem is self-adjoint by considering the following comparison functions:
=
=
=
dw
(1 - x ) dx2 - 2x dx - AW = 0 6.8
Consider the Sturm-Liouville equation (xw')'
+ ( w2x
6.11 A uniform shaft with torsional rigidity GJ is fixed at x 0 and carries a rigid disk of mass polar moment of inertia /0 at x L. State the boundary conditions of the shaft in torsional vibration at x 0 and x Land establish that one of the boundary conditions depends on the natural frequency of vibration of the shaft.
=
- ~) w = 0,
0::: x ::: 1
Determine the bounded solution of the equation subject to the condition w(1) = O.
=
=
=
7 Solution Procedure: Integral Transform Methods 7.1 INTRODUCTION Integral transforms are considered to be operational methods or operational calculus methods that are developed for the efficient solution of differential and integral equations. In these methods, the operations of differentiation and integration are symbolized by algebraic operators. Oliver Heaviside (1850-1925) was the first person to develop and use the operational methods for solution of the telegraph equation and the second-order hyperbolic partial differential equations with constant coefficients in 1892 [1]. However, his operational methods were based mostly on intuition and lacked mathematical rigor. Although subsequently, the operational methods have developed into one of most useful mathematical methods, contemporary mathematicians hardly recognized Heaviside's work on operational methods, due to its lack of mathematical rigor. Subsequently, many mathematicians tried to interpret and justify Heaviside' s work. For example, Bromwich and Wagner tried to justify Heaviside's work on the basis of contour integration [2, 3]. Carson attempted to derive the operational method using an infinite integral of the Laplace type [4]. Van der Pol and other mathematicians tried to derive the operational method by employing complex variable theory [5]. All these attempts proved successful in establishing the mathematical validity of the operational method in the early part of the twentieth century. As such, the modern concept of the operational method has a rigorous mathematical foundation and is based on the functional transformation provided by Laplace and Fourier integrals. In general, if a function f(t), defined in terms of the independent variable t, is governed by a differential equation with certain initial or boundary conditions, the integral transforms convert f (t) into F (s) defined by 12
F(s) =
1
f(t)K(s,
t) dt
(7.1)
I]
where s is a parameter, K (s, t) is called the kernal of the transformation, and t] and t2 are the limits of integration. The transform is said to be finite if t1 and t2 are finite. Equation (7.1) is called the integral transformation of f(t). It converts a differential equation into an algebraic equation in terms of the new, transformed function F(s). The initial or boundary conditions will be accounted for automatically in the process of conversion to an algebraic equation. The resulting algebraic equation can be solved 174"
7.2 'FourierTransfotms175
for F(s) without much difficulty. Once F(s) is known, the original function f(t) can be found by using the inverse integral transformation. If a function f, defined in terms of two independent variables, is governed by a partial differential equation, the integral transformation reduces the number of independent variables by one. Thus, instead of a partial differential equation, we need to solve only an ordinary differential equation, which is much simpler. A major task in using the integral transform method involves carrying out the inverse transformation. The transform and its inverse are called the transform pair. The most commonly used integral transforms are the Fourier and Laplace transforms. The application of both these transforms for the solution of vibration problems is considered in this chapter.
7.2 FOURIER TRANSFORMS 7.2.1 Fourier Series In Section 1.10 we saw that the Fourier series expansion of a function f(t) that is periodic with period t" and contains only a finite number of discontinuities is given by
"2:- L ao
f(t)
=
00
(
2mc, t an
cos
-t"-
2mct )
(7.2)
n = 1,2, ...
(7.3)
+ bn sin -r-
n=!
where the coefficients
an
and bn are given by
21'1:/2
ao = t"
f(t)dt
-'1:/2
21'1:/2 2mct an = f(t)cos--dt, t"
r
-'1:/2
21'1:/2 2nrrt bn = f(t) sin -dt, t"
-'1:/2
n = 1,2, ....
t"
Using the identities 2rrt ei(2rrt/'I:) + e-i(27rt/'I:) cos = --------
2
r
. 2rrt ei(2trt/'I:) _ e-i(27rt/r) sm -t"- = ----2-i ---
(7.4)
Eq. (7.2) can be expressed as ao
f(t)
=
00
"2 + L
ei(n.27rt/r)
[
an
+ e-i(n.2rrt/'I:) 2
ei(n.2rrt/'I:) _ e-i(n,27rt/'I:)]
+ bn
2i
n=1
(7.5)
176
Solution Procedure: Integral Transfonn Methods where bo = O. By defining the complex Fourier coefficients
Cn
and
C-n
as
(7.6) Eq. (7.5) can be expressed as
L cnein.21ff/T 00
=
. f(t)
(7.7)
n=-oo
where the Fourier coefficients Cn
=
can be determined
Cn
using Eqs. (7.3) as
2
IjT/2 =f(t) • -T/2
(n'2Jrt cos ---
n'217:t) - i sin ---
dt
•••
= -IjT/2
_._
f(t)e-zn(2rrt/T)
dt
-T/2 (7.8)
Using Eq. (7.8), Eq. (7.7) can be written as ein.21ff/T jT/2
L 00
f(t)
=
n=-oo
7.2.2
•
f(t)e-in(21ff/T)
dt
(7.9)
-T/2
Fourier Transforms When the period of the periodic function f(t) in Eq. (7.9) is extended to infinity, the expansion will be applicable to nonperiodic functions as well. For this, let Wn n· 217:/. and b.wn nwo - (n - I)WO 217:/ •. As • ~ 00, b.wn ~ dw ~ 0 and the subscript n need not be used since the discrete value of Wn becomes continuous. By using the relations nwo = wand dw 217:/. as • ~ 00, Eq. (7.9) becomes
=
=
=
=
f(t)
=
lim r~oo
L 00
jT/2
n=-oo •
= - I JOO 27r
I
_ein(2rrt/T)
. JOO
e-/wt
-00
dt
f(t)e/wt
.
dt dw
(7.10)
-00
Equation (7.10), called the Fourier following Fourier transform pair: F(w)
f(t)e-in(2rrt/T)
-T/2
=
integral,
i:
is often expressed
iwf
f(t)e
1
dt
in the form of the
(7.11 )
00
f(t)
= -21 7r
iwt dw F(w)e .....
(7.12)
-00
where F(w) is called the Fourier transform of f(t) and f(t) is called the inverse Fourier transform of F(w). In Eg. (7.12), F(w) dw can be considered as the harmonic contribution of the function f(t) in the frequency range w to w + dw. This also denotes
7.2
Fourier Transforms"
17i
the limiting value of Cn as'l" -+ 00, as indicated by Eq. (7.8). Thus, Eq. (7.12) denotes an infinite sum of harmonic oscillations in which all frequencies from -00 to 00 are represented. Notes 1. By rewriting Eq. (7.10) as
~l
1
00
OO
f(t)
=
v 2Jr
[
-00
~
V
2Jr
iwt
f(t)e-iw'
dt] e
dw
(7.13)
-00
the Fourier transform pair can be defined in a symmetric form as
1 1
00
=
F(w)
1 r;c:
2Jr
IW1 ,
dt
(7.14)
dw
(7.15)
f(t)e-
-00 00
= r;c: 1 2Jr
f(t)
iwt
F(w)e
-00
It is also possible to define the Fourier transform pair as
1 1
00
F(w) =
f(t)
,
r;c: 1 2Jr
iwt
f(t)e
(7.16)
dt
-00 00
1
= . r;c: 2Jr
IW1.
F(w)e-
(7.17)
dw
-00
2. The Fourier transform pair corresponding to an even function f(t) defined as follows: F(w)
=
1 21
can be
00
f(t)
(7.18)
coswt dt
00
f(t)
=Jr
(7.19)
F(w) coswt dw
0
The Fourier sine transform pair corresponding to an odd function f(t) can be defined as F(w) f(t)
=
1
00
f(t)
21
= -Jr
(7.20)
sinwt dt
00
(7.21)
F(w) sinwt dw
0
3. The Fourier transform pair is applicable only to functions f(t) that satisfy Dirichlet's conditions in the range (-00,00). A function f(t) is said to satisfy Dirichlet's conditions in the interval (a, b) if (a) f(t) has only a finite number of maxima and minima in (a, b) and (b) f(t) has only a finite number of finite discontinuities with no infinite discontinuity in (a, b). As an example, the function f(t) 1/(1 + t2) satisfies Dirichlet's conditions in the interval (-00,00), whereas the function f(t) = 1/(1 - t) does not satisfy Dirichlet's conditions in any interval containing the point t = 1 because f(t) has an infinite discontinuity at t = 1.
=
178 7.2.3
Solution Procedure: Integral Transform Methods
Fourier Transform of Derivatives of Functions Let the Fourier transform of the jth derivative of the function FU)(w). Then, by using the definition of Eq. (7.11), FU)(w)
=
1
i d f~t) eiwt dt dtJ
00
-00
1
dtJ-
_
i W FU-I)(w)
FU-I)(w)
W
of order
(7.23)
1, 2, ... , j - 1 are zero as
FU)(w) = (-i w)i F(w) where F(w) is the complex Fourier transform
7.2.4
(7.22)
-00
is zero as t -r ±oo, Eq. (7.22) reduces to
FU)(w) = -i Again assuming that all derivatives Eq. (7.23) yields
be denoted as
00
= eiwt di-l/(t)1
Assuming that the (j - 1)st derivative of f(t)
f(t)
of f(t)
t
-r ±oo,
(7.24) given by Eq. (7.11).
Finite Sine and Cosine Fourier Transforms The Fourier series expansion [using Eq. (1.32)]
of a function f(t)
00
ao
2
1(
1(
= - +-
f(t)
in the interval 0 ::::t ::::1( is given by
Lan
cosnt
(7.25)
n=l
where an
= i:lr
f(t)cosntdt
(7.26)
Using Egs. (7.25) and (7.26), the finite cosine Fourier transform pair is defined as F(n)
=i:lr
f(t)
= -- + -
f(t)cosntdt 2
F(O)
1(
1(
(7.27)
00
LF(n)cosnt
(7.28)
n=l
A similar procedure can be used to define the finite sine Fourier transforms. Starting with the Fourier sine series expansion of a function f(t) defined in the interval 0 :::: t ::::1( [using Eg. (7.32)], we obtain
2 f(t)
=-
1(
00
Lb
sinnt
(7.29)
sin nt dt
(7.30)
n
n=l
where bn
= i:lr
f(t)
7.2
Fourier Transforms
179
the finite sine Fourier transform pair is defined as
l
1f
F(n) =
sinnt dt
(7.31)
L F(n) sinnt rr
(7.32)
2
f(t) 00
=-
f(t)
n=l
When the independent variable t is defined in the range (O,a) instead of (O,rr), the finite cosine transform is defined as
l
a
=
F(n)
where ~ is yet unspecified. Defining a new dy = (rrja)dt, Eq. (7.33) can be rewritten as a F(n) = -;
(7.33)
f(t)cos~tdt
10r f(y)
y as y
variable
(~-;-ya)
cos
= rrtja
so that
(7.34)
dy
where
_f(y) If ~ajrr
=n
or ~
= nrrja,
(7.35)
then
a- f(y) rr Returning
(ya) -;-
=f
1 = -F(O)
2
L F(n) cosny rr 00
+-
rr
n=l
to the original variable t, we define the finite cosine transform pair as F(n) =
f(t)
l
nrr
a
f(t)
o
cos -
a
t
(7.37)
dt
2~ = -F(O) + ~F(n)cosa a
nrrt
the finite sine Fourier transform pair is defined as F(n)
=
l
a
f(t)
o
2
f(t)
Example 7.1
nrrt sin dt a
(7.39)
nrrt sin -
(7.40)
= -a L F(n) 00
n=l
a
Find the Fourier transform of the function
f(x)
____________________
(7.38)
a
n=l
Similarly,
(7.36)
= {a,
0,
II!!!!!!!!!!!!!'!!!!!!!'!!!!!!!'!!l!II!!!!!!!!!!!!
0
(E7.1.1)
I!!!!!!!I!!!!!!!~~~.,~,._~
__ ,_~_.~,~!":'..,-~._~_.~,,~,!!!!_~~
180
Solution Procedure: Integral Transform Methods
SOLUTION
The Fourier transform of f(x),
~r~:J f(x)e-iWX
F(w) =
=
as defined by Eq. (7.14), is given by
2n
1-
I
fa
dx
DC
..j2iC 10
-iwx ae
a dx
(e-iWX)a
=..j2iC
-iw
0
a(1 _e-iwa)
(E7.1.2)
iw..j2iC
Example
7.2
where
and
Cl
SOLUTION
Find the Fourier transform of the function
C2
are constants.
The Fourier transform of f(x)
1 ~ .1 ..j2iC
can be found using Eq. (7.14) as
00
F(w)
1 = r;:;-::: 2n
f(x)e-IWX. dx
-00
00
= --
. fl (x)e-IWX dx
-00
.= cIFl (w)
1 ..j2iC
00
+ --
~
. !2(x)e-IWX dx
-00
+ C2F2(W)
(E7.2.1)
This shows that Fourier transform is a linear operation; that is, the Fourier transform of a linear sum of a set of functions is equal to the linear sum of the Fourier transforms of the individual functions. Example 7.3 constant.
Find the Fourier transform of the function f(ax),
SOLUTION
The Fourier transform of f(ax) --.
1
,j2ii
By introducing a new variable t as can be rewritten as --
1
is given by [Eq. (7.14)]
foo f(ax)e-IWX.
dx
(E7.3.1)
-00
= ax
t
so that dt = adx, the expression (E7.3.1)
[00 f(t)e-1wr./a
a,j2ii .
Thus, the Fourier transform of f(ax)
where a is a positive
dt
(E7.3.2)
-00
is given by
~F (;)
,
a>O
(E7.3.3)
7.3
Free Vibrationof a Finite String
181
7.3 FREE VIBRATION OF A FINITE STRING Consider a string of length 1 under tension P and fixed at the two endpoints x = 0 and x = I. The equation of motion governing the transverse vibration of the string is given by 2
a2w(x,
c ---
t)
ax2
a2w(x,
t)
= ---'O
at '
(7.41)
-
-
By redefining the spatial coordinate x in terms of p as xrr p=1
(7.42)
Eq. (7.41) can be rewritten as rr2c2 a2w(p, t)
a2w(p,
---= [2 ax2
t)
(7.43)
ot
2
We now take finite sine transform of Eq. (7.43). According to Eq. (7.31), we multiply Eq. (7.43) by sin np and integrate with respect to p from 0 to rr:
a2w = 1" -2 0 at
2 rr2c21" -2-aa w2 sinnpdp
lop
where
" --2-' a2w(p t)-sinnpdp ap
=
1o
(OW -sinnp
) I" _n2
-nwcosnp
ap
(7.44)
sinnpdp
0
1"
wsinnpdp
0
(7.45) Since the string is fixed at p Eq. (7.45) vanishes, so that
= 0 and p = rr,
" a2w(p t) 2' sinnpdp ap 1o
the first term on the right-hand
= _n2
1"
side of
wsinnpdp
(7.46)
wsinnpdp
(7.47)
0
Thus, Eq. (7.44) becomes
n2rr2c21" --2wsinnpdp 1 0
a21"
=
-2
at
0
Defining the finite Fourier sine transform of w(p, t) as [see Eq. (7.31)J
1"
W(n, t) =
w(p, t) sinnpdp
Eq. (7.47) can be expressed as an ordinary differential
d2W(n, t) dt2
+
equation as
rr2c2n2 12
W(n,t)=O
The solution of Eq. (7.49) is given by W(n, t) =
C1ei("cnj/)t
(7.48)
+ C2e-i("cn/I)t
(7.49)
182
Solution Procedure: Integral Transform Methods
or W(n, t)
Jrent
--.+ I
= CI cos
.
Jrent
C2sm --
(7.50)
I
where the constants (:1 and (:2 or CI and C2 can be determined conditions of the string. Let the initial conditions of the string be given by
from the known initial
= 0) = wo(x) . t = 0) = wo(x)
(7.51)
t
w(x,
aw
-(x,
at
In terms of the finite Fourier sine transformW(n, and (7.52) can be expressed as W(n,
t)defined
dW -(n, t
dt
by Eq. (7.48), Eqs. (7.51)
Wo(n)
(7.53)
. = 0) = Wo(n)
(7.54)
= 0)=
t
(7.52)
where Wo(n)
= l1r wo(p)
or '.
Jr
l' /
Wo(n) = - ..
1.01
Wo(n)
=l1r
or .
Wo(n) Equations
= Jr-
(7.55)
sinnpdp
nJr;
wo(;) sin -
(7.56)
d;
(7.57)
wo(p) sin np d p
1/ ..
101
nJr;
wo(;) sm -
d;
(7.58)
(7.53), (7.54), and (7.50) lead to (7.59)
CI = Wo(n) I
.
(7.60)
C2 =-Wo(n) neJr
Thus, the solution, Eq. (7.50), becomes
W(n, t)
= Wo(n)cos-- nJret I
The inverse finite Fourier sine transform
2 w(p, t)
=Jr
I. +-Wo(n) nJre
of W(n,t)
nJret
sin--
I
(7.61)
is given by [see Eq. (7.32)]
00
L n=1
W(n, t) sin np
(7.62)
7.4 Substituting
Forced Vibrationof a Finite String~U3
Eq. (7.61) into (7.62), we obtain
21
2 00 mret . w(p, t) = - '" Wo(n)cos -smnp rr~ 1
1 .. mret - Wo(n) sm-sinnp
00
+ -z'" rre~n
(7.63)
1
n=l
n=1
Using Eqs. (7.56) and (7.58), Eq. (7.63) can be expressed in terms of x and t as
2
w(x, t) = I
L sin --nrr x cos --nrr et oo
1
n=1
1 . nrrx
2 ~ +-~ rre
1
it 0
. nrret
-sm--sm-n 1
n=l
nrr~ wo(~) sin -d~ 1
1
it ..
nrr~ wo(~)sm-d~
0
(7.64)
I
7.4 FORCED VffiRATION OF A FINITE STRING
=
Consider a string of length 1 under tension P, fixed at the two endpoints x 0 and 1, and subjected to a distributed transverse force j(x, t). The equation of motion of the string is given by [see Eq. (8.7)]
x
=
P
aZw(x, t) axz
+
1-( x,t)=p
a2w(x, t) atZ
(7.65)
or (7.66) where
I( x,t )_j(x,t) - --p
(7.67)
As in Section 7.3 we change the spatial variable x to p as xrr p=-
1
(7.68)
so that Eq. (7.66) can be written as
z rr aZw(p, 1Z
apz
t) + I (lPrr ' t)
=
2- aZwZ
(7.69)
e2 at
By proceeding as in the case offree vibration (Section 7.3), Eq. (7.69) can be expressed as an ordinary differential equation: dZW(n, t) dtZ
+
where Wen, t) =
F(n,
t) =
rrzeznz 1Z Wen, t)
in in C:, t)
=e
Z F(n, t)
(7.70)
w(p, t) sinnpdp
(7.71 )
I
(7.72)
sinnpdp
184
Solution Procedure: Integral Transfonn Methods
Assuming the iI)itial conditions of the string. to be zero, the steady-state solution of Eq. (7.70) can be expressed as cl W(n, t) = 1'Cn
l'
n1'CC T) sin -(t - T) dT
F(n,
The inverse finite Fourier sine transform of W(n,
2 w(p,
t)
(7.73)
1
0
t)
is given by [see Eq. (7.32)J
00
L Wn(n,
=1'C
1) sinnp
n=l
2cl ~ 1 = '2" L - sinnp 1'C
n=l
n
l'
or 1
2cl ~ w(x, t) = '2" L
n1'CX
- sin -n 1
1'C n=l
n1'CC F(n, T) sin --(t - T) dT
l' 0
(7.74)
1
0
n1'CC
F(n, T) sin -(t 1·
(7.75)
- T) dT
=
=
7.4 Find the response of a string oflength I, fixed at x 0 and x 1, under the action of the harmonic force f(x, t) = Jo(x)eiw', where w is the forcing frequency. Assume the initial displacement and velocity of the string to be zero.
Example
SOLUTION harmonic as
Since the force is harmonic, the response of the string is assumed to be w(x, t) = u(x)eiw'
(E7.4.1)
and the equation of motion, Eq. (7.66), becomes zdzu(x) C
--;[;2 + w
z u(x)
+
=0
fo(x)
(E7.4.2)
where fo(x)
= fo(x)
(E7.4.3)
p
By introducing the new spatial variable p = 1'Cxjl Eq. (E7.4.2) can be written as 2 2 1'C C
d
2
u(p)
J2 ~
+ WZu(~) + fo
(lP) -;
[as defined In Eq. (7.42)],
=0
(E7.4.4)
We now take the finite Fourier sine transform of Eq. (E7.4.4). According to Eq. (7.31), we multiply Eq. (E7.4.4) by sin np and integrate with respect to p from 0 to 1'C:
(E7.4.5)
7.5
Free Vibration of a Beam
185
where
r d2~ sinnpdp 10 dp~
rr
I0
= (dU sinnp _ nu cosnp) dp
-
r u sinnpdp
n2
(E7.4.6)
10
The first term on the right-hand side of Eq. (E7.4.6) will be zero because u(O) = u(rr) = 0 (since the string is fixed at p = 0 and p = rr) and sin 0 = sinrl7r = O. We define the finite Fourier sine transforms of u(p) and /o(lp/rr) as U(n) and Fo(n):
1 1 C:) rr
U(n) =
(E7.4.7)
u(p)sinnpdp
rr
Fo(rt) =
/0
sin np d p
(E7.4.8)
The inverse finite Fourier sine transforms of Eqs. (E7.4.7) and (E7.4.8) yield
2
00
L U(n) sinnp
u(p) = -
rr
/0
(lP)rr
(E7.4.9)
n=l
f:
= ~ rr
(E7.4.10)
Fo(n) sinnp
n=l
Thus, Eq. (E7.4.5) can be rewritten as rr2c2n2
'....
+ It}U(n) + Fo(n)= 0
--[2-U(n)
or (E7.4.11) where 2 Wn
=
rr2c?n2
(E7.4.12)
-[2-
denotes the natural frequency of the string. Finally, by taking the inverse finite Fourier sine transform of Eq. (E7.4.9) using Eqs. (E7.4.9) and (E7.4.10), we obtain the steadystate forced response of the string as 2eiwt
w(x, t) = --
P[
L ------'-"-------sin(nrrx/ 1) J~.fo(y) sin(nrry/ l)y dy oo
n=l
(n2rr2/
[2)(1 - w2/(2)
(E7.4.13)
n
7.S FREE VmRATION OF A BEAM
=
= [.
Consider a uniform beam oflength [ simply supported at x 0 and x The equation of motion governing the transverse vibration of the beam is given by [see Eq. (3.19)]
a4w 1 a2w ax4 + c2 at2
=0
(7.76)
.~
I
!
I
i
I
.J
186
Solution Procedure: Integral Transform Methods where EI
2
c =-
(7.77)
pA
The boundary conditions
can be expressed as
= 0, x = I
(7.78)
at x = 0, x = I
(7.79)
at x
W(x,t)=O
a
2W
ax2(x,t)=0
We take finite Fourier sine transform of Eq. (7.76). For this, we multiply Eq. (7.76) by sin(mr x / I) and integrate with respect to x from 0 to I:
f a4w --
i o ax4
nrrx
sin --
I
1
dx + -2
a2w nnx sin -dx = 0 at I
if
c
-- 2
0
(7.80)
Here
f a4w . nnx --sm--dx 1o ax4 I
a3w . nrrx If
= --sm-ax3 I
If
nn)21f - (-
a2w . nnx --sm--dx 2
I
In view of the boundary conditions
ax
(7.81)
I
= - (nn - )21f I
0
a2w . nrrx --sm--dx ax2 I
by parts, the integral on the right-hand
aw . nnx II
= -sm- ax I
0
0 -
=1
side of Eq. (7.82) can
11 nn aw nnx --cos--dx 0 I ax / nn -/-
nnxll
w cos -1-
(nn)2 -1-
0 -
Jot w sin nnx -/- dx (7.83)
of Eq. (7.78). Thus, Eq. (7.80) can be expressed as
nnx
wsin --dx 0
(7.82)
nnx Jot w sin -1dx
nrr)2 ( -1-
in view of the boundary conditions
nn)41 (-
-
aw . nrrxll a:; sm -/-
=
I
0
0
of Eq. (7.79), Eq. (7.81) reduces, to
f a4w . nrrx --sm--dx I 1o ax4
f a2w . nnx --sm--dx I 1o ax2
0
a3w . nnx ax3 I
= --sm--
Again using integration be expressed as
f 1 nn a3w nnx ---cos--dx 3 0 I ax I nn a2w nrrx If - ---cos-2 I ax I 0
-
/
I a211 + ?-2
c-
at
0
nnx
wsin --dx /
=0
(7.84 )
7.5
nrr x w(x, t) sin -dx () l
1
1
Eq (7.84) reduces to the ordinary differential
d2 W (n
t)
dt2'
~
of w(x, t) as [see Eq. (7.39)]
Defining the finite Fourier sine transform W(n, t) =
Free Vibration of aBeam
equation
c2n4rr4
+ -l4-W(n,
The solution of Eq. (7.86) can be expressed
(7.85)
t) = 0
(7.86)
as (7.87)
or (7.88) Assuming
the initial conditions
of the beam as
= 0) = wo(x) . t = 0) = wo(x)
(7.89)
w(x, t
--.dw
dt
the finite Fourier sine transforms
(x,
(7.90)
of Eqs. (7.89) and (7.90) yield
= 0) = Wo(n)
W(n, t
dW
(7.91)
.
(7.92)
t = 0) = Wo(n)
-(n,
dt
where
1 . 1
nrrx wo(X) sin dx
1
Wo(n) =
o
(7.93)
I
1
Wo(n) = Using initial conditions
o
nrrx wo(x) sin -dx
(7.94)
l
of Eqs. (7.93) and (7.94), Eq. (7.88) can be expressed as
Finally, the transverse displacement of the beam, w(x, t), can be determined by using the finite inverse Fourier sine transfoml of Eq. (7.95) as
2
w(n, t) =
nrrx
T L W(n, t) sm -l00
•
(7.96)
n=!
_______________________
!I!!!!!!__
!!!!!B
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!.~_!!".
__!!"._~c.~!'!'!._!!"_!!!!'!!!!!!!!!!!!!!!!!!!IJ
188
Solution Procedure: Integral Transform Methods which can be rewritten,
2
using Eqs. (7.93) and (7.94), as 2
2
mr x cn :rr L sin -cos- -t 1 1 1
w(x, t) = -
00
2
n=1
1/
n:rr~
wo(;) sin --
~=o
2l ~ 1 . n:rrx cn2:rr2 ~ "2 sm -sin -2-t c:rr n 1 1
+ -2
1/
n=1
7.6
d~
1
"=0
n:rr~ 1
wo(;) sin --
'
d;
(7.97)
LAPLACE TRANSFORMS The Laplace transform technique is an operational method that can be used conveniently for solving linear ordinary differential equations with constant coefficients. The method can also be used for the solution of linear partial differential equations that govern the response of continuous systems. Its advantage lies in the fact that differentiation of the time function corresponds to multiplication of the transform by a complex variable s. This reduces a differential equation in time t to an algebraic equation in s. Thus, the solution of the differential equation can be obtained by using either a Laplace transform table or the partial fraction expansion method. An added advantage of the Laplace transform method is that during the solution process, the initial conditions of the differential equation are taken care of automatically, so that both the homogeneous (complementary) solution and the particular solution can be obtained simultaneously. The Laplace transformation of a time-dependent function, f(t), denoted as F(s), is defined as
1
00
£[f(t)]
= F(s) =
f(t)e-st
dt
(7.98)
where L is an operational symbol denoting that the quantity upon which it operates is to be transformed by the Laplace integral (7.99) The inverse or reverse process of finding the function f(t) from the Laplace transform F(s), known as the inverse Laplace transform, is donated as
L -1[F(s)]
= f(t)
I
=--. 2:rrl
fC+iOO
F(s)eSI
ds,
t
>0
(7.100)
c-ioo
Certain conditions are to be satisfied for the existence of the Laplace transform of the function f(t). One condition is that the absolute value of f(t) must be bounded as
If(t)1 ~ Ceat for some constants C and a. This means that if the values of the constants can be found such that le-sl f(t)1
~ Ce(a-s)t
(7.101) C and a
(7.1 02)
7.6
then L[f(t)]
=
00
e-SI f(t) dt
1o
=:: C
100
e(a-sll dt
0
--189
Laplace Tansfonns
= --C
(7.103)
s-a
Another condition is that the function f(t) must be piecewise continuous. This means that in a given interval, the function f(t) has a finite number of finite discontinuities and no infinite discontinuity.
7.6.1
Properties of Laplace Transfonns Some of the important properties of Laplace transforms are indicated below. 1. Linearity property. If CI and C2 are any constant and fl (t) and h(t) functions of t with Laplace transforms Fl (s) and F2(S), respectively, then L[ctfl
(t)
+ c2!2(t)]
+ czL[h(t)] + C2F2(S)
= Cl L[ft (t)] = cIFI(S)
are
(7.104)
The validity of Eq. (7.104) can be seen from the definition of the Laplace transform. Because of this property, the operator L can be seen to be a linear operator. 2. First translation or shifting property. If L[f(t)] = F(s) for s > fi, then L[eat f(t)]
(7.105)
= F(s - a)
where s - a > f3 and a may be a real or complex number. To see the validity of Eq. (7.105), we use the definition of the Laplace transform L[eat f(t)]
=
100
eat e-st f(t) dt =
100
e-(s-a)t f(t) dt = F(s - a)
(7.106)
Equation (7.105) shows that the effect of multiplying f(t) by eat in the real domain is to shift the transform of f(t) by an amount a in the s-domain. 3. Second translation L[f(t)]
\:,
or shifting property.
= F(s)
and
g(t)
If
= { of(t,
- a),
t >a t < a
then L[g(t)]
= e-as F(s)
4. Laplace transformation
of derivatives.
L[f'(t)]
= L [d~~t)]
(7.107)
If L[f(t)]
= F(s), then
= sF(s) - f(O)
(7.108)
To see the validity of Eq. (7.108), we use the definition of Laplace transform as L [df(t)] dt
=
[00 e-s,df(t)
10
dt
dt
(7.109)
190
Solution Procedure: Integral Transform Methods
Integrating the right-hand side of Eq. (7.109) by parts, we obtain
-1
00
e-st f(t)lo
= - f(O)
(-se-st)f(t)dt
+ sF(s)
The property of Eq. (7.108) can be extended to the nth derivative of f(t)
(7.110) to obtain
= L[f(n)(t)]
L [d(n) f(t)]
dtn
= _ f(lI-l)(O)
- sf(n-2) (0) - s2 f(n-3) (0) - ... - s(n-l) f(O) -
sfn)
F(sl
(7.111) where (7.112) 5. Convolution theorem. Let the Laplace transforms of the functions f(t) g(t) be given by F(s) and G(s), respectively. Then
* g)(t)]
L[(f
where F * G is called the convolution can be expressed equivalently as
[1
* G(s)
= F(s)
or the faltung
and
(7.113)
of F and G. Equation (7.113)
t
L
(7.114)
= F(s)G(s)
- r) dr]
f(r!.g(t
or conversely, L-1[F(s)G(s)]
=
1
t f(r)g(t
(7.115)
- r)dr
To prove the validity of Eqs. (7.113) to (7.115), consider the definition of the Laplace transform and the convolution operation as L[(f
1 [1 00
* g)(t)] =
t
st
e-
f(r)g(t
- r)dr]
(7.116)
dt
From the region of integration shown in Fig. 7.1, the integral in Eq. (7.116) can be rewritten, by interchanging the order of integration, as
1 [1 00
L[(f
* g)(t)]
=
00
f(r)
st
e-
g(t - r)dt]
(7.117)
dr
By using the second property, the inner integral can be written as e-st G(s). so that Eq. (7.117) can be expressed as L[(f
* g)(t)] =
1
00
G(s)e-sr
= G(s)F(s)
f(r)
dr
= G(s)
1
00
e-ST f(r)
dr (7.118)
7.6
Laplace Transforms
191
r
Figure 7.1
Region of integration in Eq. (7.116).
The converse result can be stated as L -I[F(s)G(s)]
=
it
= f(t)
f(-c)g(t
- -c) d-c
* g(t)
(7.119)
7.6.2 Partial Fraction Method In the Laplace transform method, sometimes
we need to find the inverse transformation
of the function P(s) F(s) = Q(s)
(7.120)
where P(s) and Q(s) are polynomials in s with the degree of P(s) less than that of Q (s). Let the polynomial Q (s) be of order n with roots a 1. az, a3•... , an, SO that (7.121) First, let us consider the case in which all the n roots aI, az, a3•... , an are distinct, so that Eq. (7.120) can be expressed as F(s) where
Cj
= -P(s) Q(s)
are coefficients.
Cl
= --
s-al
Cz
+ -s-az
C3
Cn
+ -+ ... + -s-a3 s-an
(7.122)
The points aI, az, a3, ... , an are called simple poles of F(s).
1 I
192
Solution Procedure: Integral Transfonn Methods
The poles denote points at which the function F(s) becomes infinite. The coefficients Cj in Eq. (7.122) can be found as
=
Cj
. hm
[(s - aj)F(s)]
s ••.•aj
I
P(s) = --Q'(s)
(7.123)
s=ai
where Q'(s) is the derivative of Q(s) with respect to s. Using the result (7.124) the inverse transform of Eq. (7.122) can be found as f(t)
n
=C1ea1t +C2ea2t +...+cneant =L cjeajl
=L-1[F(s)]
j=1 n
=
L lim
S']
[(s - a;)F(s)e
j=1 s-+aj
=
L n
P(s)
j=1
Q'(s)
est
I
(7.125)
s=ai
Next. let us consider the case in which Q(s) has a multiple root of order k, so that (7.126) In this case, Eq. (7.120) can be expressed as F) (s
P(s) CII =--=--+---+ Q(s)
s - al
...+- Clk __
Cl2
(s - a2)2
(s - al)k
Cn-k
C2
C3
s - a2
s - a3
+--+--+
...+---
(7.127)
s - an-k
Note that the coefficients 'Clj can be determined as Clj
=
1
dk-
while the coefficients
Cj,
k
j
(k _ j)! dsk-j
i
[(s - ad
= 2,3,
F(s)]ls=al'
j
= 1. 2. 3, ...
,k
(7.128)
.... n - k, can be found as in Eq. (7.125). Since (7.129)
the inverse of Eq. (7.127) can be expressed as . kI
f(t)
=[ CII +C12t +CI3-t2 +...+Clk--- t 2!
(k-l)!
]
ea11
(7.130)
"7.6 Laplace Transforms-193
7.6.3
Inverse Transformation The inverse Laplace transformation, denoted as L -I [F(s)], is also defined by the complex integration formula
l
a ioo
L -I[F(s)]
= f(t)
= -.1 2rrl
+
est
F(s) ds
(7.131)
a-ioo
where a is a suitable real constant, in Eq. (7.131), the path of the integration is a line parallel to the imaginary axis that crosses the real axis at Re s = a and extends from -00 to +00. We assume that F(s) is an analytic function of the complex variable s in the right half-plane Re s > a and all the poles lie to the left of the line x a. This condition is usually satisfied for all physical problems possessing stability since the poles to the right of the imaginary axis denote instability. The details of evaluation of Eq. (7.131) depend on the nature of the singularities of F(s). The path of the integration is the straight line L as shown in Fig. 7.2 in the complex s plane, with equation s == a + i R, -00 < R < +00 and Re s = a is chosen so that all the singularities of the integrand of Eq. (7.131) lie to the left of the line L. The Cauchy-residue theorem is used to evaluate the contour integral as
=
lest
F(s) ds ==
!r
est F(s) ds
+ lest
F(s) ds
== 2rri[sum of the residues of est F(s) at the poles inside C] (7.132) Ims
a +iR
L
{)
a
Res
I r
i
I a -iR
Figure 7.2
Contour of integration.
1 I
194
Solution Procedure: Integral Transform Methods
r tends
where R -+ 00 and the integral over reduces to the form lim -. 1 R-+oo 27r1
l
a i
+
R
eSI
to zero in most cases. Thus, Eq. (7.131)
F(s) ds
a-iR
= sum of the residues
of
F(s) at the poles of F(s)
est
(7.133)
The following example illustrates the procedure of contour integration. Example 7.5
SOLUTION
Find the inverse Laplace transform of the function s F (s) = s2 + c2
(E7.5.1)
The inverse transform is given by f(t)
l l
a ioo
= -.1
2m
+
est
F(s)ds
a-ioo a ioo
= _1 27ri
+
s
eSI
a-ioo
s2
ds
+ c2
(E7.5.2)
The integrand in Eq. (E7.5.2) has two simple poles at s = ±ic, and residues at these poles are given by
= -ic se 1· = limes + ic) = _e-rct 2 s2 + c 2 R2 = residue of F(s) at s = ic
Rl
= residue
of
est
F(s) at s SI
(E7.5.3)
s-+-ic
eSI
= lim
sest (s - ic) S
S-++IC
Hence, f(t)
= -.1
27r1
7.7
l
a ioo
+
est
2
F (s) ds = Rl
a-ioo
1·
e,cl + c = -2 2
+ R2 =
1·
-2(erCI
+ e-lct.) = cos ct
(E7.5.4)
(E7.5.5)
FREE VIBRATION OF A STRING OF FINITE LENGTH In this case the equation of motion is c2 (l2w
(l2w
(lx2
(lt2
If the string is fixed at x = 0 and x
=0
= I, the boundary
(7.134) conditions· are
W(O, t) = 0
(7.135)
w(l, t) = 0
(7.136)
7.7 Free Vibration of a String of Firlite-i.engtb
Let the initial conditions of the string be given by w(X, t
aw at
-(x,
t
~95 iI
I
= 0) = wo(x)
(7.1;37)
= 0) = wo(x)
(7. Ji38)
I I
Applying Laplace transforms to Eq. (7.134), we obtain c2
d2W(x,
s) 2
dx
2 -
S
+ swo(x) + wo(X)
W(x, s)
=0
(7.139)
where
1
00
W(x, s) =
w(x, t)e-sr dt
(7.140)
Taking finite Fourier sine transform of Eq. (7.139), we obtain (s2
+ c2p;)W(Pn,
s) = sWO(Pn)
+ WO(Pn)
(7.141)
where =
W(Pn,s)
Wo(;n\=
.
WO(Pn)
=
1/ 1/
W(x,
s) sin Pnx
wo(x) sinpnx
dx
(7.142)
dx
(7.143)
t wo(x)sinpnxdx
10
(7.144)
with mr Pn=-
(7.145)
1
Equation (7.141) gives (7.146) I
Performing the inverse finite Fourier sine transform of Eq. (7.146) yields 2 W(x,s)=-L
I
00.
n=1
smpnX
s
2
2
2
+ c Pn
1/
2
I L sin PnX 00
wo(~)
W(x, s)
.j
J
in Eq. (7.147), we obtain
I
i
[cmrt
cos- -
n=1
t . 10
(7.147)
0
Finally, by taking the inverse Laplace transform of w(x,t) =
I
" [swo(~)+wo(~)]smpn~d~
sin Pn~
I
d~
+
sin(cmrt/I) cnrr / I
t. 10
wo(~)
sin Pn~
d~
]
1 (7.148)
1 I
....1 ;
i r i
i
:
196
Solution Procedure: Integral Transform Methods
Figure 7.3
Axial stress at the end of a bar.
Example 7.6 A uniform bar is fixed at x = 0 and subjected to an axial stress ao at x = 1 as shown in Fig. 7.3. Assuming the bar to be at rest initially, determine the axial vibration response of the bar. SOLUTION The equation governing the longitudinal vibration of a bar is giving by [see Eq. (9.15)1: c
2a2U(X,t)
=
ax2
a2u(x,t)
at2
(E7.6.1)
The boundary conditions are given by u(O, t)
au ax
-(1, x)
=0 = ex(l, t) = -aoE
(E7.6.2) (E7.6.3)
and the initial conditions by (E7.6.4)
-u(x,O) = 0
au
(E7.6.5)
-ar(x,O) = 0
By taking Laplace transform of Eq. (E7.6.1) with respect to t, we obtain 2 au 2d2U(x, s) s U(x, s) - su(x, 0) - -ar(x, 0) = c. dx2
(E7.6.6)
which in view of the initial conditions of Eqs. (E7.6.4) and (E7.6.5), reduces to d2 U s2 - -U
dx2
c2
=0
(E7.6.7)
where U(x, s)
Noting that L(l) written as
= lis,
= L[u(x,
t)]
(E7.6.8)
the Laplace transforms of Eqs. (E7.6.2) and (E7.6.3) can be
U(O,s)=O
dU ao -(1,s) =dx Es
(E7.6.9) (E7 .6. ] 0)
7.8
Free Vibration of a Beam of Finite Length
197
The solution of Eq. (E7.6.7) is given by U(x, s) =
sx
+ C2 sinh-sx
cosh -
CI
C
where the constants
and
Cl
C2
can be found using Eqs. (E7.6.9) and (E7.6.1O) as
Cl
s C2 C
(E7.6.1l)
C
=0
sl
(E7.6.12)
0'0
cosh - = C Es
or
= -----
C2
CO"o
(E7.6.13)
Es2 cosh(slfc)
Thus, the solution, U (x, s), becomes CO"o
U(x, s)
=-
sinh(sxfc)
(E7.6.14)
2
E s cosh(slfc)
By taking the inverse Laplace transform of Eq. (E7.6.14), we obtain the axial displacement of the bar as [8] u(x, t)
=
0"0
81 ~
[
E
x
(_l)n
+ Jr2 L-
(2n _
.
(2n - l)Jrx
1)2 sm
(2n -1)Jrct]
21
cos
(E7.6.15)
21
n=1
I r
j
, 7.8
FREE VmRATION OF A BEAM OF FINITE LENGTH
I
The equation of motion for the transverse vibration of a beam is given by 204w(X,t)
02w(x,t)
c---+· 4
ox
ot2
=0
(7.149)
where 2 EI c =-
(7.150)
pA
For free vibration, w(x, t) is assumed to be harmonic with frequency
= w(x)eiwt
w(x, t)
w:
(7.151)
so that Eq. (7.149) reduces to an ordinary differential equation: 4 d w(x) dx4
_ {34w(x)
=0
(7.152)
where {34 -
2
-w
- c2
-
w2 pA __
(7.153)
EI
-
By taking Laplace transforms of Eq. (7.152), we obtain s4W(s)
- s3w(0) - s2w'(0) - sw"(O)
- w"'(O)
- {34W(s)
=0
(7.154)
or W(s)
=
1 4
4 [s3w(0)
s - {3
+s2w'(0)
+sw"(O)
+w"'(O)]
(7.155)
I I
I
198
Solution Procedure: Integral Transfonn Methods
where w(O), w'(O), w"(O), and w"'(O) denote the deflection and its first, second, and third derivative, respectively, at x O. By noting that
=
L-1 L4 ~ /34] = ~(COSh/3X+ cos/3x) L -1
L4 ~ /34] = 2~ (sinh /3x
(7.156)
+ sin /3x)
L-1 L4 ~ /34] = 2~2(COSh/3X-
(7.157)
cosf3x)
(7.158)
L -1 L4 ~ /34] = 2~3 (sinh/3x - sin/3x)
(7.159)
the inverse Laplace transfonn of Eq.(7.155) gives w(x)
= ~(cosh/3x + cos/3x)w(O) + ~(sinh/3x + sin/3x)w'(O) 2 2/3 + ~(COSh/3X
-
2/3
7.9
+ ~(Sinh/3x
cos/3x)w"(O)
- sin/3x)w"'(O)
2/3
(7.160)
FORCED VIBRATION OF A BEAM OF FINITE LENGTH The governing equation is given by 84w(x, t) 8x4
E/
...
+ pA
82w
8t2 = f(x,
t)
(7.161)
where f(x, t) denotes the time-varying distributed force. Let the initial deflection and velocity be given by wo(x) and wo(x), respectively. The Laplace transfonn of Eq. (7.161), with respect to t with s as the subsidiary variable, yields d4W(x, s) dx 4
+ pA E/s
2
W(x,
s)
=
pA E/ [swo(x)
+ wo(x)] + E/1
F(x,
s)
(7.162)
or d4W(x, s) dx4
-
/3
4·
.
W(x, s)
= G(x,
(7.163)
s)
where /34
= _ pAs2
(7.164)
E/ G(x, s)
pA
= -/ E
[swo(x)
1
+ wo(x)] + ·-F(x, E/
s)
(7.165)
Again, by taking Laplace transfonn of Eq. (7.163) with respect to x with p as the subsidiary variable, we obtain (p4 _ /34)W(p,
s)
= G(p,
s)
+ p3W(O,
s)
2-' -II +p Wx(O,s)+pWx(O,s)+
-",
Wx (O,s)
7.9
Forced Vibration of a Beam of Finite Length
199
or W
-
)_ (p, S -
G(p,S) 4 f34
P -
3-
+
2-'
p W(O,S)+p
-ff
-fIf
Wt(O,S)+pWx(O.S)+Wx(O,S) 4 f34
.
P -
(7.166) -
-/~,
_1/1
where W(O. s), Wx(O, s), Wx(O, s), and Wx (0, s) denote the Laplace transforms with respect to f of w(x, f), (aw/ax)(x, f), (a2w/ax2)(x, t) and (a3w/ax3)(x, f) respectively, at x = 0. Next, we perform the inverse Laplace transform of Eq. (7.166) with respect to x. For this, we use Eqs. (7.156)-(7.159) and express the inverse transform of Eq. (7.166) as W(x, s)
=~
2f3
10r G(1], s)[sinhf3(x
1-
+ 2 W(O,
-1]) - sinf3(x
-1])]dYJ
1-,
s)(coshf3x
+ cosf3x) + 2f3 Wx(O,
1 ~,
1
+ 2f32 Wx(O, s)(coshf3x
-fIf
+ 2f33 Wx
- cosf3x)
+ sinf3x)
s)(sinhf3x •
(0, S)(Slnhf3X - sinf3x)
(7.167) Finally, we perform the inverse Laplace transform of Eq. (7.167) with respect to t to find the desired solution, w(x, t) ..The procedure is illustrated in the following example. Example 7.7
A uniform beam of length I is subjected to a concentrated harmonic force at x = g, 0 < g < i. Assuming the end conditions of the beam to be simple supports and the initial conditions to be zero, determine the response of the beam.
fo sin wf
SOLUTION
The boundary conditions of the beam can be expressed as w(O, t)
a
2w
ax2
a ax2
(E7.7.1)
° =°
(E7.7.3)
(I, t) = 0
(E7.7.4)
(E7.7.2)
(0, t) =
w(I, t) 2w
=0
Taking Laplace transforms, Eqs. (E7.7.1)-(E7.7.4) can be written as
°
(E7.7.5)
W;(O, s) = 0
(E7.7.6)
I
(E7.7.7)
!
W(O, s) =
W(I,s)=O
W;(l,
s)
=0
(E7.7.8)
I
I,
The applied concentrated force can be expressed as r f(x,
t)
=
foo(x
- g)h(t)
== fo sinwto(x
-~)
(E7.7.9) f
)
200
Solution Procedure: Integral Transform Methods
The Laplace transform of Eq. (E7.7.9) gives (E7.7.10)
F(x, s) = loo(x - ;)H(s)
where
= L[h(t)] = L[sinwt] =
H(s)
2 S
w 2 +w
(E7.7.11)
Since the initial conditions are zero, wo(x) = wo(x)
=0
(E7.7.12)
and hence Eq. (7.165) yields G(x, s) =
F(x, s) EI
10 = Elo(x
- ~)H(s)
(E7.7.13)
Using the boundary conditions at x = 0 [Eqs. (E7.7.5) and (E7.7.6)] and Eq. (E7.7.13), Eq. (7.167) can be expressed as W(x, s) = ~
r 10
Elo 0(71 - ;)H(s)[sinh,fj(x
2,fj I + Cl sinh,fjx + C2sin,fjx
- 71) - sin,fj(x - 71)]d71
(E7.7 .14)
where
1:-,
111/
Cl = 2,fjWx(O, s) + 2,fj3Wx (0, s) 1 Willi/ C2= 2,fj x(O, s) - 2,fj3Wx (0, s)
(E7.7.15) (E7.7.16)
By differentiating W(x, s) given by Eq. (E7.7.14) with respect to x and using the conditions of Eqs. (E7.7.7) and (E7.7.8), we obtain W(l, s) = 0 =
10H(s) 3
2,fj E I
1/
..
0(71 - ;)[Slnh,fj(l - 71) - s1O,fj(l- 1])] d1]
0
+clsinh,fjI+C2sin,fj1 I
W ' (l, s) = 0 = x
10H(s) 3
1/
..
0(1] - ;)[Slnh ,fj(l
2,fj E I 0 +clsinh,fjl-c2sin,fj1
(E7.7.17) -1])
+ s1O,fj(l- 71)] d1] (E7.7.18)
The solution of Eqs. (E7.7.17) and (E7.7.18) gives Cl
=-
C2
=
10H(s) sinh,fj(l-;)
(E7.7.19)
2,fj3E 1 sinh ,fjl
10H(s) sin,fj(l -;) 2,fj3E 1 sin ,fjl
(E7.7.20)
7.10
Recent Contributions
201
Thus, Eq. (E7.7.18) becomes
.. ior o(1]-~)[stnhP(x-1])-smp(x-1])]d1]
/oB(s) 2{J3EI
W(x,s)=
/oB(s)
sinhP(1 -~)
2p3 E I /oB(s)
sinhpx sin pl- sinp(l-~) (sinh pi sin PI)
sinpx sinh pi
..
= 2p3 E I [smh P(x -~)
- sm P(x - ~)]o(x - ~)
sinh P(l - ~) sinh px sin pi - sin P(lsin pi
+ -----------~--------sinh pi
~) sin px sinh pi
(E7.7.21)
which results in the solution w(x, t) =
/0 sinwt 3
{
2p EI
[sinP(x -~)
sinhp(l-~)
+ The value of
p4
.
- SlnhP(X - ~)]O(x -~)
sinh,8x sin pi - sin,8(I-~) sinh,81 sin,81
sinpx Sinh,8I}
can be obtained from Eq. (7.164), by substituting s pAw2 ,84 = -EI
(E7.7.22)
= iw as (E7.7.23) r
I
7.10 RECENT CONTRIBUTIONS
]
Fast Fourier Transforms' The fast Fourier transform algorithm and the associated programming considerations in the calculation of sine, cosine, and Laplace transforms was presented by Cooley et al. [13]. The problem of establishing the correspondence between discrete and continuous functions is described.
f
I I I
Beams Cobble and Fang [14] considered the finite transform solution of the damped cantilever beam equation with distributed load, elastic support, and the wall edge elastically restrained against rotation. The solution is based on the properties of a Hermitian operator and its orthogonal basis vectors.
I
I
j
Membranes The general solution of the vibrating annular membrane with arbitrary loading, initial conditions, and time-dependent boundary conditions was given by Sharp [15]. Hankel Transform The solution of the scalar wave equation of an annular membrane, in which the motion is symmetrical about the origin, for arbitrary initial and boundary conditions was given in Ref. [16]. The solution is obtained by using a finite Hankel transform. An example is given to illustrate the procedure and the solution is compared to the one given by the method of separation of variables.
J
I
j
j
I
I
I I
202
Solution Procedure: Integral Transform Methods
Plates
A method of determining a finite integral transform that will remove the presence of one of the independent variables in a fourth-order partial differential equation is applied to the equation of motion of classical plate theory for complete and annular circular plates subjected to various boundary conditions by Anderson [17]. The method is expected to be particularly useful for the solution of plate vibration problems with time-dependent boundary conditions. Forced torsional vibration of thin, elastic, spherical, and hemispherical shells subjected to either a free or a restrained edge was considered by Anderson in Ref. [18].
z Transform
Application of the z-transform method to the solution of the wave equation was presented by Tsai et aI. [19]. In the conventional method of solution using the Laplace transformation, the conversion, directly from the s domain to the t domain to find the time function, sometimes proves to be very difficult and yields a solution in the form of an infinite series. However, if the s domain solution is first transformed to the z domain and then converted to the time domain, the process of inverse transformation is simplified and a closed-form solution may be obtained.
REFERENCES 1. O. Heaviside, Electromagnetic Theory, 1899; reprint by Dover, New York, 1950. 2. T. Bromwich, Normal coordinates in dynamical systems, Proceedings of the London Mathematical Society, Ser. 2, Vol. 15,'pp, 401-448, 1916. 3. K. W. Wagner, Uber eine Forme1 Von Heaviside zur Berechnung von Einscha1tvorgangen, Archiv fuver Elektrotechnik, Vol. 4, pp. 159-193, 1916. 4. 1. R. Carson, On a general expansion theorem for the transient oscillations of a connected system, Physics Review, Ser. 2, Vol. 10, pp. 217-225,1917. 5. B. Van der Pol, A simple proof and extension of Heaviside's operational calculus for invariable systems, Philosophical Magazine, Ser. 7, pp. 1153-1162, 1929. 6. L. Debnath, Integral Transforms and Their Applications, CRC Press, Boca Raton, FL, 1995. 7. I. N. Sneddon, Fourier Transforms, McGraw-Hill, New York, 1951. 8. M. R. Spiegel, Theory and Problems of Laplace Transforms, Schaum, New York, 1965. 9. W. T. Thomson, Laplace Transformation, 2nd ed., Prentice-Hall, Englewood Cliffs, NJ. 1960. 10. C. J. Tranter, Integral Transforms in Mathematical Physics, Methuen, London, 1959. 11. E. C. Titchmarsh, Introduction to the Theory of Fourier Integrals, Oxford University Press, New York, 1948. 12. W. Nowacki, Dynamics of Elastic Systems, translated by H. Zorski, Wiley, New York, 1963. 13. J. W. Cooley, P. A. W. Lewis, and P. D. Welch, The fast Fourier transform algorithm: progranuning considerations in the calculation of sine, cosine and Laplace transforms. Journal of Sound and Vibration, Vol. 12, No.3, pp. 315-337,1970. 14. M. H. Cobble and P. C. Fang, Finite transform solution of the damped cantilever beam equation having distributed load, elastic support, and the wall edge elastically restrained against rotation, Journal of Sound and Vibration, Vol. 6, No.2. pp. 187-198, 1967. 15. G. R. Sharp, Finite transform solution of the vibrating annular membrane, Journal of Sound and Vibration, Vol. 6, No.1, pp. 117-128, 1967.
Problems
203
16. G. R. Sharp, Finite transform solution of the symmetrically vibrating annular membrane, Journal of Sound and Vibration, Vol. 5, No.1, pp. 1-8, 1967. 17. G. L. Anderson, On the determination of finite integral transforms for forced vibrations of circular plates, Journal of Sound and Vibration, Vol. 9. No. I, pp. 126-144, 1969. 18. G. L. Anderson, On Gegenbauer transforms and forced torsional vibrations of thin spherical shells, Journal of Sound and Vibration, Vol. 12, No.3. pp. 265-275, 1970. 19. S. C. Tsai, E. C. Ong, B. P. Tan, and P. H. Wong, Application of the z-transform method to the solution of the wave equation, Journal of Sound and Vibration, Vol. 19, No. t, pp. 17-20, 1971.
PROBLEMS 7.1 Find the Fourier transforms of the following functions: (a) f(x)
=
(b) f(x)
= 8(x _ a)e-
(b) f(x)
C2X2
= {~:
= e-x
(b) f(x)
= {~:
O
= {1-X2, 0,
f(t)
= {sinx, 0,
= {1-X2, 0,
O~x=?l
otherwIse
0~ x ~ n otherwise O~x
=?
=t = eat
a)
= {0:I
t>a
t
7.9 A single-degree-of-freedom spring-mass-damper system is subjected to a displacement Xo and velocity io at time t O. Determine the resulting motion of the mass (m) using Laplace transforms. Assume the spring and damping forces to be kx and ci, where k is the spring constant and i dxl dt is the velocity of the mass.
=
7.10 Derive an expression for the response of a uniform beam of length t fixed at both ends when subjected to a concentrated force fo (t) at x g, 0 < g < t. Assume the initial conditions of the beam to be zero. Use Fourier transforms.
=
1
otherwise
7.6 Find the Laplace transforms of the following functions: (a) f(t)
= U(t -
=
0~ x ~ n otherwIse
7.5 Find the Fourier sine transforms of the following functions:
(b) f(t)
(a) f(t) == {cos(t - 7'( 14), t > 7'(14 0, t<7'(14 (b) Heaviside's unit step function:
O
(a) f(x) = {sinx, 0,
n < t < 2n
7.8 Find the Laplace transforms of the following functions:
7.4 Find the Fourier cosine transforms of the following functions:
(b) f(x)
= {~: ~: ~ < 2 = {sint, 0 < t < n 0,
(a) f(x)
(a) f(x)
(a) f(t) (b) f(t)
= e-x
7.3 Find the Fourier sine transforms of the following functions:
(b) f(x)
(d) f(t)
= sinat = cosat
7.7 Find the Laplace transforms of the following functions:
x2~a2
7.2 Find the Fourier cosine transforms of the following functions: (a) f(x)
(c) f(t)
7.11 Find the response of a uniform beam of length t fixed at both the ends when subjected to an impulse G at x g, 0 < g < t. Assume that the beam is in equilibrium before the impulse is applied. Use Fourier transforms.
=
I
,
j
J
I
204
Solution Procedure: Integral Transform Methods
7.12 Find the free transverse displacement of a semiinfinite string using Fourier transforms. The governing equation is 2
cPw
cPw
ox
ot
with w(O) = 1 and dw/ dt(O)
= -1.
7.17 The longitudinal vibration of a uniform bar of length 1 is governed by the equation
c--=-2 2 and the initial conditions are
= wo(x),
w(x,O)
ow -(x, ot
0)
. = Wo
Assume that the string is fixed at x = 0 and stretched along the positive x axis under tension P.
=
7.13 Consider a finite string of length 1 fixed at x 0 and x 1, subjected to tension P. Find the transverse displacement of a string that is initially at rest and subjected to an impulse F at point x a, 0 < a < 1 using the Fourier transform method.
=
=
7.14 Find the steady-state transverse vibration response of a string of length 1 fixed at both ends, subjected to the force f(x,
t)
= fo sin nt
using the Fourier transform method.
where c2 = E / p with E and p denoting Young's modulus and the mass density of the bar respectively. The bar is fixed at x = 0 and free at x = 1. Find the free vibration response of the bar subject to the initial conditions u(x,O)
= uo(x),
au at
-(x,O)
. (x) = Uo
using Laplace transforms. 7.18 Find the longitudinal vibration response of a uniform bar fixed at x = 0 and subjected to an axial force f(t) at x 1, using Laplace transforms. The equation of motion is given in Problem 7.17.
=
7.19 A uniform bar fixed at x = 0, is subjected to a sudden displacement of magnitude Uo at x = 1. Find the ensuing axial motion of the bar using Laplace trans-
7.15 Find the Laplace transforms of the following func- .. forms. The governing equation of the bar· is given in tions:,Problem 7.17. (a) f(t)
= {O,
(b) f(t)
= = 2e-2t sin 3t
(c) f(t)
7.20 A taut string of length 1, fixed at x = 0 and x = 1 is subjected to tension P. If the string is given an initial displacement
t <0 at, t ~ 0 eat, ex is real
7.16 Find the solution of the following differential equation using Laplace transforms: d2w dw dx2 +4 dx + 3w
= sin2t
w(X, 0)
= wo(x) =
2x 0 < x < 0.5 { 20-'x), 0.5 < x < 1
and released with zero velocity, determine the ensuing motion of the string.
Transverse Vibration of Strings 8.1
INTRODUCTION It is well known that most important musical instruments, including the violin and the guitar, involve strings whose natural frequencies and mode shapes play a significant role in their performance. The characteristics of many engineering systems, such as guy wires, electric transmission lines, ropes and belts used in machinery, and thread manufacture, can be derived from a study of the dynamics of taut strings. The free and forced transverse vibration of strings is considered in this chapter. As will be seen in subsequent chapters, the equation governing the transverse vibration of strings will have the same form as the equations of motion of longitudinal vibration of bars and torsional vibration of shafts.
8.2 EQUATION OF MOTION 8.2.1 Equilibrium Approach Figure 8.1 shows a tightly stretched elastic string or cable of length I subjected to a distributed transverse force f(x,t) per unit length. The string is assumed to be supported at the ends on elastic springs of stiffness k1 and k2. By assuming the transverse displacement of the string w(x,t) to be small, Newton's second law of motion can be applied for the motion of an element of the string in the z direction as net force acting on an element = inertia force acting on the element
(8.1)
e
If P is the tension, p is the mass per unit length, and is the angle made by the deflected string with the x axis, Eq. (8.1) can be rewritten, for an element of length dx, as (P
+ dP)
sin(e
+ de) + f dx
= pdx-2a w 2
- P sine
at
(8.2)
Noting that
ap
dP = -dx
(8.3)
ax
sine ~ tane = -
(8.4)
ax
+ de)
~ tan(O
I I f t
OW
sin(e
I
aw
o2w
ax
ax
+ dO) = - + --2
dx
(8.5) 205
j
I I
l
I
206
Transve~se Vibration of Strings z,w
t
f(x,/)
I I I I I
I_ds-: I I I I I
I I I I I
...•...- ...•.. -.-.-.-.-.-.-.-.-
.-.-.-.-
x
.
.-._x
x+dx
(a)
z,w
~
I I I I I I I I I I I
p
(b)
Figure 8.1
(a) Vibrating string; (b) differential element.
Eq. (8.2) can be expressed as a
'::I
uX
[p aw(x,t)] ax
2
+ f(x,t)
= p(x) a w(x ,t) at 2
(8.6)
If the string is uniform and the tension is constant, Eq. (8.6) takes the form p a2w(x,t) ax2
For free vibration,
+
2
f() x,t
=p
a w(x,t) at2
(8.7)
f (x , t) = 0 and Eq. (8.7) reduces to p a2w(x,t)
ax2
a2w(x,t)
_ -
p
at2
(8.8)
8.2
;:l07
Equation of Motion
which can be rewritten as
(8.9) where
(8.10) Equation (8.9) is called the one-dimensional wave equation.
8.2.2 Variational Approach Strain Energy
There are three sources of strain energy for the taut string shown in Fig. 8.1. The first is due to the defonnation of the string over 0::: x ::: l, where the tension P (x) tries to restore the deflected string to the equilibrium position; the second is due to the defonnation of the spring at x = 0; and the third is due to the defonnation of the spring at x = l. The length of a differential element dx in the defonned position, ds, can be expressed as
ds=
2] 1/2 [
[ (dx)2+(~:dX)
=
l+(~:)2] 1/2
dx~
[
2] l+~(~:)
dx
(8.Il) by assuming the slope of the deflected string, awjax, to be smaIl. The strain energy due to the defonnations of the springs at x 0 and x l is given by ~kl w2(O,t) and k2W2(l,t), and the strain energy associated with the defonnation of the string is given by the work done by the tensile force P (x) while moving through the distance ds -dx:
=
t
tIt o P(x)[ds(x,t)
- dx]
= 210
=
P(x)
[aw(x t)]2 ax' dx
(8.12)
Thus, the total strain energy, 7r, is given by
7r
Kinetic Energy
III
=2
P(x)
[aw(x,t)]2 ---
dx
ax
0
I I W2 + -2kIW2(O,t) + -k2 (l,t) 2
(8.13)
The kinetic energy of the string is given by I
T
t p(x)
= 210
[aw(x,t)]2 at
dx
(8.14)
where p(x) is the mass of the string per unit length.
Work Done by External Forces load acting on the string, f(x,t),
The work done by the nonconservative can be expressed as
1
distributed
1
W =
f(X,t)w(x,t)dx
l I
! (8.15)
I
J I
I
208
Transverse Vibration of Strings
Hamilton's principle gives 12
8
1
(T - n
+ W)
=
dt
°
(8.16)
I]
or
~ f [H'
p(x) (~~)'
1/
+
- ~k2W2(l,t)
H'
dx -
e~)'
P(x)
dx - ~klwZ(O.t)
°
dX] dt =
f(x,t)w
(8.17)
The variations of the individual terms appearing in Eq. (8.17) can be carried out as follows:
12 oT dt = 112 dt 1 p(x)-o-aw 1
1
I]
at
0
I)
=
aw dx at
112 1/
aw a(ow) p--dx at at
dt
0
I]
(8.18)
using the interchangeability of the variation and differentiation processes. Equation (8.18) can be evaluated by using integration by parts with respect to time:
t [1 Jo
12
I]
p aw a(ow) dt] dx = [/{ (p aw OWl:;) at at at
Jo
Using the fact that ow = Eq. (8.19) yields
1
/2
°
- 1/ [[/2 at t
oT dt
==
=
I)
:t (p aa~) OWdt]}
tl and t= t2 and assuming p(x)
-1/
2
I)
(
t p a2~ ow dX)
Jo
at
dx
(8.19)
to be constant,
dt
(8.20)
The second term of Eq. (8.16) can be written as 2
1
1
1 Jot 12
on dt =
II
[
I]
P(x) aw ~(ow) ax ax
+k2W(l,t)OW(l,t)]
dx
+ kl w(O,t)ow(O,t) (8.21)
dt
By using integration by parts with respect to x, Eq. (8.21) can be expressed as
1
/2on
I]
dt
=
1 [p 12
II
aw owlh _ ax
+ k2W(l,t)OW(l,t)]
t Y- (p
Jo
ax
aw) ax
ow dx
+ k]w(O,t)ow(O,t) (8.22)
dt
The third term of Eq.(8.16) can be written as
[12 oW dt = ll2 [1/ f(x,t)oW(X,t)
dX] dt
(8.23)
8.3
Initial and Boundary Conditions
209
By inserting Eqs. (8.20), (8.22), and (8.23) into Eq. (8.16) and collecting the terms, we obtain
1"{10t [-p
a2~ at
tl
-
+ ~ (p aw) + ax
ax
+ k2W) OW1X=I}
(p ~;
fJ
+
ow dx
(p awax -
dt = 0
kl w). owl
x=o (8.24)
ow
Since the variation over the interval 0 < x < I is arbitrary, Eq. (8.24) can be satisfied only when the individual terms of Eq. (8.24) are equal to zero:
2 = p a w, (Jt2
+f
~ (p aw) ax ax
(p ~; - ow = (p ~; + 2W) ow =
0
(8.25)
k1 W )
0,
x
=0
(8.26)
k
0,
x =I
(8.27)
Equation (8.25) denotes the equation of motion while Eqs. (8.26) and (8.27) represent the boundary conditions. Equation (8.26) can be satisfied when w(O,t) = 0 or when p[awjax](O,t) - k1w(0,t) = O. Since the displacement w cannot be zero for all time at x 0, Eq. (8.26) can only be satisfied by setting
=
aw ax
P-
- kl W
=0
at x
=0
(8.28)
Similarly, Eq. (8.27) leads to the condition
aw ax
P-+k2W=0
atx=l
(8.29)
I I I
Thus, the differential equation of motion of the string is given by Eq. (8.25) and the corresponding boundary conditions by Eqs. (8.28) and (8.29).
8.3 INITIAL AND BOUNDARY CONDITIONS The equation of motion, Eq. (8.6), or its special forms (8.7) and (8.8) or (8.9), is a partial differential equation of order 2 in x as well as t. Thus, two boundary conditions and two initial conditions are required to find the solution, w(x,t). If the string is given an initial deflection wo(x) and an initial velocity wo(x), the initial conditions can be stated as w(x, t 0) = wo(x)
=
I
I j i
aw (x, t = 0) = wo(x)
at
= O,t)
!
~ i
If the string is fixed at x = 0, the displacement tions will be w(x
(8.30)
= 0,
is zero and hence the boundary condi-
t~O
(8.31)
j 1 .1
j
,
Al.i"~
i
210
Transverse Vibration of Strings If the string is connected to a pin that is free to move in a transverse direction, the end will not be able to support any transverse force, and hence the boundary condition will be (8.32) P(x)-(x,t) 0 Ox
ow
=
If the axial force is constant and the end x
ow ax
-(O,t)
= 0 is free,
= 0,
If the end x = 0 of the string is connected boundary condition will be P(x)
ow (X,t)1 ax
(8.33)
t2:0 to an elastic
= kW(X,t)\
x=o
Eq. (8.32) becomes
spring of stiffness
k, the
(8.34)
'
x=o
Some of the possible boundary conditions
of a string are summarized
in Table 8.1.
8.4 FREE vmRATION OF AN INFINITE STRING Consider a string of infinite length. The free vibration equation of the string, Eq. (8.9), is solved using three different approaches in this section.
8.4.1
Traveling- Wave Solution The solution of the wave equation (8.9) can be expressed
as (8.35)
where Fl and F2 are arbitrary functions of (x - et) and (x + et), respectively. The solution given by Eq. (8.35) is known as D'Alembert's solution. The validity of Eq. (8.35) can be established by differentiating Eq. (8.35) as (8.36) (8.37) where a prime indicates a derivative with respect to the respective argument. By substituting Eqs. (8.36) and (8.37) into Eq. (8.9), we find that Eq. (8.9) is satisfied. The functions FJ(x - et) and F2(x + et) denote waves that propagate in the positive and negative directions of the x axis, respectively, with a velocity c. The functions Fl and F2 can be determined from the known initial conditions of the string. Using the initial conditions of Eq. (8.30), Eq. (8.35) yields Fl (x) -eF{(x)
+ F2(X) = wo(x) + eF~(x) = wo(x)
(8.38) (8.39)
8.4 Table 8.1
Free Vibration of an Infinite String';zU
Boundary Conditions of a String Boundary conditions to be satisfied
Support conditions of the string 1. Both ends fixed w(x.t)
.4
I I I o
=0
w(O,t) w(I,t)
=0
·-··-·-·--....x
a
2. Both ends free w(X,t)
P
t
I
aw ax (O,t) aw
P
I:~._._._._._._._._._.
ax
o
(l,t)
=0
=0
3. Both ends attached with masses aZw mi at2 (O,t)
P
.
m
oL'-'-'_'-'-'-'-'-'-'-'I'~'-'-'~x
aZw -m,-(l 2
- at '
aw
= P ax (O,t)
aw t) = P-(I,t)
ax
4. Both ends attached with springs w(x,t) p
•••
p
I
k2
o
'-'-'-'-'-'-'-'-'-'-
I
-.-.-.-.~
kiw(O,t)
aw = P-(O,t) ax
-kzw(l,t)
aw = P-(l,t) ax
x
5. Both ends attached with dampers w(x.t)
aw at
•••
I
cll-l._._._._._.
P
Ci-(O,t)
._L-;J c:_._._.~ I
x
-C2
aw at
(I,t)
aw = P-(O,t) ax =P
aw ax
(/,t)
212
Transverse Vibration of Strings
where a prime in Eq. (8.39) denotes a derivative with respect to the respective argument at t = 0 (i.e., with respect to x). By integrating Eq. (8.39) with respect to x, we obtain (8.40) where Xo is a constant. Equations (8.38) and (8.40) can be solved to find Fl (x) and F2(X) as (8.41) Fl(x) = ~ [wo(X) - ~ wo(X)dX] 2 c Xo
IX
+
= ~ [wo(X)
F2(X)
2
~Ix c
(8.42)
wo(X) dX]
xo
By replacing x by x - ct and x + ct, respectively, in Eqs. (8.41) and (8.42), we can express the wave solution of the string, w(x,t), as w(x,t)
= Fl(X
- ct)
= -1 [ wo(x 2
+ F2(X + ct)
- ct)
+ wo(x + ct) + -1
I
x ct
2c
+
wo(X) dx ]
(8.43)
x-ct
The solution given by Eq. (8.:43) can be rewritten as w(x,t)
=
Wd(X,t)
+ wv(x,t)
(8.44)
where
Wd(X,t) represents a wave propagating due to a known initial displacement with zero initial velocity, andwv(x,t) indicates a wave moving due to the initial velocity wo(x) with zero initial displacement. A typical wave traveling due to initial displacement (introduced by pulling the string slightly in the transverse direction with zero velocity) is shown in Fig. 8:2. wo(x)
I I I
!\
/\/\ A Figure 8.2
AAtt=t3>t2
Propagation of a wave caused by an initial displacement.
8.4
8.4.2
Free Vibration of an Infinite String
213
Fourier Transform-Based Solution To find the free vibration response of an infinite string (-00 < x < 00) under the initial conditions of Eq. (8.30), we take the Fourier transform of Eq. (8.9). For this, we multiply Eq. (8.9) by eiax and integrate from x -00 to x 00:
=
1
1
. 1 02 elaXdx=Z-2 C ot
02w(x.t) ,. ox-
00
-00
Integration
l
-00
00
.
w(x,t)e,axdx
(8.45)
-00
of the left-hand side of Eq. (8.45) by parts results in
02W(X,t)
°0
=
iaxd'
---e 2 ox
oo
x=
ow iax/ . -e -la ox. -00
. = -ow_e1ax ox
1
00
-00
ow iaxd -e x ox
00
00
-
1
[
. 1 iawelaX
-00
21
00
- (ia) -00
welaX dx
.]
-00
oo
= (ow
ox
eiax/
_ iaw)
_ a21°O
weiax dx
(8.46)
-00-00
Assuming that both w and ow/ox tend to zero as Ixl ~ 00, the first term on the righthand side of Eq. (8.46) vanishes; Using Eq. (7.16), the Fourier transform of w(x,t) is defined as
1
00
W(a,t)
= ~1 2Jl'
w(x,t)eiax
dx
(8.47)
-00
and Eq. (8.45) can be rewritten in the form
(8.48) Note that the use of the Fourier transform reduced the partial differential equation (8.9) into the ordinary differentiation equation (8.48). The solution of Eq. (8.48) can be expressed as (8.49) where the constants C I and C2 can be evaluated using the initial conditions, Eqs. (8.30). By taking the Fourier transforms of the initial displacement (w wo(x)] and initial velocity [ow/ot = wo(x)], we obtain
=
2Jl'
dW -(a,t dt
=0) =
~1 v 2Jl'
1 1
wo(x)e iax d x = Wo(a)
(8.50)
1 1 I
(8.51 )
f
-00
00
-00
j j
00
W(a, t = 0)=~ 1
j
• ( . Wo x)e iax d x = Wo(a)
1 ! i
214
Transverse Vibration of Strings The use of Eqs. (8.50) and (8.51) in (8.49) leads to Wo(a)
= Cl
Wo(a)
=
+ C2
iac(C}
(8.53)
- C2)
whose solution gives
1.
Wo
Cl
= -2 + --Wo 2iac
C2
= -2 -
(8.54)
1.
Wo
Thus, Eq. (8.49) can be expressed
(8.52)
(8.55)
--Wo 2iac
as
.. 1· .. = -21 Wo(a) (ewct + e-lact) + -. _Wo(a)(e1act 2lac
W(a,t)
_ e-1act)
,
(8.56)
By using the inverse Fourier transform of Eq. (8.47), we obtain
1 -J2ii
00
w(x,t)
W(a,t)e-WX .
= --1
(8.57)
da
-00
which, in view of Eq. (8.56), becomes
{_1_1 -J2ii + ~ {_1_1 -Jiii OO
w(x,t)
=~
+ e-ia(x+ct)]
Wo(a)[e-ia(x-ct)
2
da}
-00
OO
2c
W~(a) [e-ia(X-Cf)
-00
Note that the inverse Fourier transforms can be obtained as
_ e-ia(X+ct)]da}
(8.58)
za of Wo(a) and Wo(a), Eqs. (8.50) and (8.51),
1 1
00
wo(x)
1 = r-;= •••.71:
wo(~)
1 = ~
WX .
Wo(a)e-
da
(8.59)
da
(8.60)
-00 00
271:
..
Wo(a)e-W
~
-00
so that wo(x =F ct)
1 -J2ii
00
= --1
Cf) . Wo(a)e-W(."I:
(8.61 )
da
-00
By integrating Eq. (8.60) with respect to ~ from x - ct to x
l
cf
+
X
X-Cf
wo(~)d~
1 __ -Jiii
= -- 1
00 ~
-00
(a) ~_[e-ia(X-ct) Ia
+ ct,
we obtain
_ e-ia(x+ct)]da
(8.62)
8.4 Free Vibration of an Infinite String ':215
When Eqs. (8.61) and (8.62) are substituted into Eq. (8.58), we obtain w(x,t)
1 = -[wo(x 2
+ ct) + wo(x
- ct)]
+ -I
2c
l
x c1
+
wo(~) d~
(8.63)
x-cr
which can be seen to be identical to Eq. (8.43).
8.4.3
Laplace Transform-Based Solution The Laplace transforms of the terms in the governing equation (8.9) lead to (8.64) (8.65) where
i
oo
W(x, s) =
e-srw(x,t)
dt
(8.66)
Using Eqs. (8.64) and (8.65) along with the initial conditions of Eq. (8.30), Eq. (8.9) can be expressed as d2W c2-_2
dx
= s2W - swo(x) - wo(x)
(8.67)
Now, we take the Fourier transform of Eq. (8.67). For this, we multiply Eq. (8.67) by eipx and integrate with respect to x from -00 to +00, to obtain c2
1
d2 W --2
00
. dx = s2 e'Px
dx
-00
1"'" We'Px dx -':)0
- s
100 wo(x)e1pX. dx - 100 wo(x)e1pX. dx -00 -00 (8.68)
The integral on the left-hand side of Eq. (8.68) can be evaluated by parts: 00
1
-00
d2W . dW . --2 e1PXdx = _e1PX dx dx dW
= _e'Px dx
.
1+00- 100 -.dW-ipe'Px. dx -00 -00 dx 1+00- ipWe'Px. +00- p2 1+00 Weipx dx -00
-00
Assuming that the deflection, W(x, s), and the slope, dW(x, s)/dx, x ~ ±oo, Eq. (8.69) reduces to +00 d2W 1+00 --2 eipx dx = - p2 Weipx dx -00 dx -00
1
(8.69)
-00
tend to be zero as
(8.70)
216
Transverse Vibration of Strings and hence Eq. (8.68) can be rewritten as
or
or -
sWo+Wo c2p2 + s2
W=
(8.71)
where (8.72)
27r
-00
(8.73)
27r ...:.... Wo(p) =
1+
-00 00
1
r;:;-::: 27r
.
wo(x)e'pX dx of W (p, s) to obtain
Now we first take the inverse Fourier transform
1 ./iii'.
W()X,s
= --
00
1
,-00
sWo(p) + Wo(p) -ipx d e x c2 p2 + s2
and next we take the inverse Laplace transform
(8.75)
of W (x, s) to obtain
= L -1[W(x,
w(x,t)
(8.74)
-00
s)]
(8.76)
Noting that L -I
[ 2
c p
2S
] +s 2 = cos pet
(8.77)
and L
_] [
2
1
c p
2
+ s2
]
1, pc
(8.78)
= - sm pet
Eqs. (8.76) and (8.75) yield w(x,t)=
r::;=
1
~7r
-00
1
00 [-
1 ...:.... Wo(p)cospct+-Wo(p)sinpct
where wo(x)=
wo(x)
].
pc
1 r;:;-::: 27r
= ~27r
1 1
e-lpxdp
00 -
. Wo(p)e-lpXdp
(8.79)
(8.80)
-00 00
ipX
Wo(p)e-
(8.81 )
dp
-00
••• "••••".~~~~':!c
__
~"',""',!:;!",·'\._~iiIl"~liII-~ •.
8.5
of a"String of Finite 'Length
Free Vibration
'217
From Eqs. (8.80) and (8.81), we can write wo(x
./Ii J-oo
l
= _1_
x+ct wo(~)d~
the following
[00
./Ii Loo
x-ct
In addition,
r~
± et) = _1_
Wo(p)e-ip(x±ct)
dp
W~(P)[e-iP(X-ct)
(8.82)
_ eiPP"+Cl)]dp
IP
(8.83)
identities are valid:
I"
" + e-lpct )
cos pet = _(e,pct 2
1" " 2i (e,pct - e-lpct)
=
sin pet
(8.84) (8.85)
Thus Eq. (8.79) can be rewritten as w(x,t)
1 = -[wo(x 2
+ et) + wo(x
- et)]
+ -1
2e
l
x ct
+
x-ct
wo(~) d~
(8.86)
"
which can be seen to be the same as the solution given by Eqs. (8.43) and (8.63). Note that Fourier transforms were used in addition to Laplace transfonns in the current approach.
8.5 FREE VmRATION OF A STRING OF FINITE LENGTH The solution of the free vibration equation, Eq. (8.9), can be found using the method of separation of variables. In this method, the solution is written as
=
w(x,t)
W(x)T(t)
where W(x) is a function of x only and T(t) Eq. (8.87) into Eq. (8.9), we obtain
e2 d2W W dx2
(8.87)
is a function of t only. By substituting
I d2T =
T dt2
(8.88)
Noting that the left-hand side of Eq. (8.88) depends only on x while the right-hand side depends only on t, their common value must be a constant, a, and hence e2 d2W W dx2 Equation
=
I d2T T dt2 = a
(8.89)
(8.89) can be written as two separate equations: d2W
a
---W=O dx2 e2
" (8.90)
,
I I
d2T
--aT=O
dx2
(8.91)
]
I r
218
Transverse Vibration of Strings The constant a is usually negative! (8.91) can be rewritten as
and hence, by setting a = _w2, Eqs. (8.90) and
d2 W
w2
-+-W=O dx2 c2 d2r 2 -+w dt2
r
(8.92) (8.93)
=0
The solution of Eqs. (8.92) and (8.93) can be expressed wx
W (x) = A cos -
c T(t) = C coswt
as
. wx -
+ B sm
(8.94)
c
+ D sinwt
(8.95)
where w is the frequency of vibration, the constants A and B can be evaluated from the boundary conditions, and the constants C and D can be determined from the initial conditions of the string.
8.5.1
Free Vibration of a String with Both Ends Fixed If the string is fixed at both ends, the boundary conditions w(O,t) = w(l,t) = 0,
are given by (8.96)
t2:0
Equations (8.96) and (8.94) yield W (0) = 0
(8.97)
W (I) = 0
(8.98)
The condition of Eq. (8.97) requires that
A=O
(8.99)
in Eq. (8.94). Using Eqs. (8.98) and (8.99) in Eq. (8.94), we obtain Bsin-
wI
c
(8.100)
=0
ITo show that a is usually a negative quantity, multiply Eq. (8.90) by W(x) and integrate with respect to x from 0 to I to obtain l d2W(x) a l · W(x)-d-,-dx=2 W2(x)dx (a) o xc 0
i
i
Equation (a) indicates that the sign of a will be same as the sign of the integral on the left-hand side. The left-hand side of Eq. (a) can be integrated by parts to obtain
II
i
d2W(x) - dW(x) W(x)-dx = Wv) -o dx2 dx l
i
l
0
0
[dW(X)]2 -dx dx
(b)
The first term on the right-hand side of Eq. (b) can be seen to be zero or negative for a string with any combination of fixed end (W 0), free end (dWjdx 0), or elastically supponedend (P dWjdx -kW), where k is the spring constant of the elastic support. Thus, the integral on the left-hand side of Eq. (a), and hence the sign of a is negative.
=
=
=
8.5
Free Vibration of a String of Finite Lenglh ':219
For a nontrivial solution, B cannot be zero and hence . wl
sm - = 0 c
(8.10 1)
Equation (8.10 l) is called the frequency or characteristic equation, and the values of W that satisfy Eq. (8.101) are called the eigenvalues (or characteristic values or natural frequencies) of the string. The nth root of Eq. (8.101) is given by wnl
-- = n7!, c
n = 1,2, ...
(8.102)
and hence the nth natural frequency of vibration of the string is given by nC7!
= --, l
Wn
n = 1,2, ...
(8.103)
The transverse displacement of the string, corresponding to wn, known as the nth mode or nth harmonic or nth normal mode of the string is given by
of vibration
wn(x,t)
n7!x = sin -l-
= Wn(x)Tn(t)
(
nC7!t
Cn cos -l-
nC7!t) + Dn sin -l-
(8.104)
In the nth mode, each point of the string vibrates with an amplitude proportional to the value of Wn at that point with a circular frequency Wn• The first four modes of vibration are shown in Fig. ~t3.The mode corresponding to n = 1 is called the fundamental mode, WI is called the fundamental frequency, and 1:1
21r 2l = -.= WI C
(8.105)
is called the fundamental period. The points at which Wn = 0 for t ::: 0 are called nodes. It can be seen that the fundamental mode has two nodes (at x 0 and x l), the second mode has three nodes (at x = 0, x = l/2, and x = l), and so on.
=
=
The free vibration of the string, which satisfies the boundary conditions of Eqs. (8.97) and (8.98), can be found by superposing all the natural modes wn(x) as w(x,t)
=
L wn(x,t) = L sin -l-n1rX ( Cn cos -l-nC7!t +. Dn sin -l-nC7!t) 00
00
(8.106)
n=I
n=1
This equation represents the general solution of Eq. (8.9) and vibrations of the string. The particular vibration that occurs is by the initial conditions specified. The initial conditions provide constants Cn and Dn in Eq. (8.106). For the initial conditions Eq. (8.106) yields
includes all possible uniquely determined unique values of the stated in Eq. (8.30),
00
'"' Cn sin -ln7!X
= wo(x)
(8.107)
. = wo(x)
(8.108)
n=1 00
'"' n=1
nC7! -l-Dn
n7!x
sin -l-
j !I I
1
!
!
220
Transverse Vibration of Strings
·-·-·-·-x I
·-·-·-·-x I
·-·-·-·-x I
Figure 8.3 Mode shapes of a string. Noting that Eqs. (8.107) and (8.108) denote Fourier sine series expansions of wo(x) and wo(x) in the interval 0 ~ x ~ 1, the values of Cn and Dn can be determined by multiplying Eqs. (8.107) and (8.108) by sin and integrating with respect to x from 0 to 1. This gives the constants Cn and Dn as
n7x
21
1
Cn = 1 Dn
0
2 = -nC1r
wo(x)
1/ ..
. nrr x sm --
wo(x)
0
dx
(8.109)
mrx dx 1
(8.110)
1
sm --
Note that the solution given by Eq. (8.106) represents the method of mode superposition since the response is expressed as a superposition of the nonnal modes. As indicated earlier. the procedure is applicable in finding not only the free vibration response but also the forced vibration response of any continuous system. Example 8.1 Find the free vibration response of a fixed-fixed string whose middle point is pulled out by a distance h and then let it go at time t = 0 as shown in Fig. 8.4.
8.5
FreeNibration of a String of Finite Length
221
wo(x)
t
hl--------I I I I o·
_
._._.--1._._._.L._._.--1._._._
..
_._._~.x
1
l
Figure 8.4
Initial deflection of the string.
424
~
SOLUTION The free vibration solution if the string is given by Eq. (8.106) with CII and Dn given by Eqs. (8.109) and (8.110), respectively. In the present case, the initial displacement can be represented as [
for 0 < x < -
-
-2 (E8.1.1)
[
.
for - < x < [
2-
-
and the initial velocity is zero: wo(x) = 0
(E8.1.2)
Equations (E8.1.2) and (8.106) yield Dn =0
(E8.1.3)
Thus, the free vibration response becomes 00
w(x,t)
=
~ . nrrx £- Cn sm -[-cos
ncrrt -[-
(E8.1.4)
11=1
The constant CII can be evaluated using Eq. (8.109) as l
Cn
= -21 [ 0 2 =-
. nrrx wo(x) sm -- dx [
[ll/2
[
0
2hx nrrx -sin--dx+ [ I
jl l/2
2h nrrx ~(l-x)sin-dx [
]
I
[
I
= {
8h . nrr -sm2 rr n2 2
for n = 1,3,5, ...
o
for n = 2, 4, 6, ...
j (E8.1.5)
I
I
~ ~I
222
Transverse Vibration of Strings
Noting the relation n = 1,3,5, ...
(E8.1.6)
the free vibration response of the string can be expressed as 8h ( Jr x Jr ct 1 3Jr x 3Jr ct w(x t) = sin - cos - - sin -- cos -, Jr2 I I 9 I I
+ ...)
(E8.1.7)
The solution given by Eq. (E8.1.7), using different number of terms, is shown in Fig. 8.5. The fast convergence of the series of Eq. (E8.1.7) can be seen from the figure.
.... .•......... .•..•..
""
.
" .•..•.. .•..•.. -...:
v
o
~
-------_._.~-----_._-_.-
.-.-.-.+- x
I
2
o
.-.-.-.+-
I
x
I
2 .•.........
~
.•..•.. ..
o ....--.---.-.-.~----.-.-.-.~
-
-
~
.-.-.-.+- x
I 2
--.-.-.+- x
o 2
Figure 8.5
A string of length I fixed at both ends is struck at t = 0 such that the initial displacement distribution is zero and the initial velocity distribution is given by
Example 8.2
8.5
Free Vibration of a String of Finite,bmgtfl
223
wo(~)
A
I
a
L
_
I I I I o
_____ --1
_
-----
1
__L 31
-
4
2
Figure 8.6
l
-4
1
~x
1
Initial velocity of the string
(Fig. 8.6):
4ax -1-'
- -4
4a(~-x) 2
=
wo(x)
1
O
1
'
1 1 -
0,
2-
(E8.2.1)
-
Find the resulting free vibration response of the string. SOLUTION
Since the initial displacement
of the string is zero,
Wo(x) = 0
(E8.2.2)
and hence Eq. (8.106) gives
Cn =0
(E8.2.3)
Thus, the free vibration solution becomes
. nJrx L Dnsm--sm--1 00
w(x,t) =
n=l
The constant Dn can be evaluated
2 Dn = -nCJr =~ nCJr 8a = --2 2 Jr n
1
1
wo(x)sm
0
[t/104 (
2sin-
•
4
(E8.2.4)
using Eqs. (8.110) and (E8.2.1) as
nJr x --dx 1
4ax sin nJrx dx 1 1 Jrn
. nCJrt 1
. Jrn)
-sm-
2
+ [1/2 1/4
4a (~ _ x) sin nJrx dX] 1 2 1 (E8.2.5) -,
I I I
224
Transverse Vibration of Strings
Thus, the free vibration response of the string is given by 8al [~ . nx . nct 1. 2nx . 2nct (..•.... 2 -l)smsm + - sm-sm-n3c I I 4 I I
w(x,t) = -
. 3n x + ./i + 1 sm--sm--
. 3n ct I
I
27
,J2 + 1 125
. 5n x . 5n ct sm--sm--+"· I I
]
(E8.2.6)
It is to be noted that the modes involving n = 4, 8, 12, ... are absent in Eq. (E8.2.6). The solution given by Eq. (E8.2.6), using a different number of terms is shown in Fig. 8.7. The fast convergence of the series of Eq. (E8.2.6) can be seen from the figure.
/ / / / /
o
/
/'
,, ,
,,
_._._._._.~. __
._._._
1
._._._·_·~x 1
2 First term
, "-
"o-'-'-'-'-'~ 1 2 First two terms
7 / /
o
/
"._._._._._.~. I
._._._._.~ 1
x
_._._._.~ 1
x
2 First three terms
o 2 First four terms
Figure 8.7
Example 8.3 Find the natural frequencies and mode shapes of a taut wire supported at the ends by springs as shown in Fig. 8.8.
8.5 Free'~Vibration of a String of Finite Length
225
z,w A
I
·_·_·_~x
o Figure 8.8
SOLUTION
Taut wire supported by springs at the two ends.
The nth mode of vibration of the wire can be represented as
(E8.3.1) At the two ends of the wire, the spring force must be in equilibrium with the z component of the tensile force P in the wire. Thus, the boundary conditions can be expressed as
aw
P-(O,t)
ax
aw
P-(l,t)
ax
= k1w(0,t)
(E8.3.2)
= -k2w(l,t)
(E8.3.3)
Substituting Eq. (E8.3.1) into Eqs. (E8.3.2) and (E8.3.3), we obtain P dWn (0) dx
= k1 Wn (0)
(E8.3.4)
or (E8.3.5) (E8.3.6) or Anwn . wnl P ( --sm c c
Bnwn + -c
wnl)
cos -
c
= -k2
(
An
wnl
CDS -
c
. Wnl) + Bn sm c
(E8.3.7)
Equations (E8.3.5) and (E8.3.7) can be rewritten as (E8.3.8) (E8.3.9)
J
226
Transverse
Vibration of Strings
For a nontrivial solution of the constants An and Bn' the determinant of their coefficient matrix must be zero: wnl
k] ( k2sine
PWn + --cos-.
wnl)
e
e
PWn + --
(
e
wnl
k2COS-
-
e
PWn
--sine
wnl)
e
=0 (E8.3.1O)
Defining IV
wnl
__
n-
(E8.3.11)
e plw2 n
R.
n-
(E8.3.12)
k] plw~
(E8.3.13)
Yn = --
k2
the frequency equation, Eq. (E8.3.1O), can be expressed as (1-~)
tan an -
13nYn
(.2.-13n +~) an = 0 Yn
(E8.3.14)
Using the relation k]e
= An-an 13n
Bn = An-P Wn
(E8.3.15)
From Eq. (E8.3.8), the modal function Wn(x) can be written as Wn(x)
=C
n
WnX ( cos-- e
an. + -sm-
wnx)
e
13n
=C
n
(anx
cos-
an + -sin-
l
13n
anX) l
(E8.3.16)
Notes 1. If k] and k2 are both large, k] ~ Eq. (E8.3.1O), reduces to
00
and k2 ~
00
and the frequency equation,
. wnl 0 sm-=
(E8.3.17)
e
Equation (E8.3.17) corresponds to the frequency equation of a wire with both ends fixed. 2. If k] and k2 are both small, 1/ 13n~ 0 and I/Yn ~ 0 and Eq. (E8.3.14) gives the frequencies as tanan = 0
or
an = n:rc
or
Wp!
= n:rce --l
(E8.3.] 8)
and Eq. (£8.3.16) yields the modal functions as wllX
Wn(x) = Cn cos-
I
(£8.3.]9)
It can be observed that this solution corresponds to that of a wire with both ends free.
8.6
3. If k1 is large and k2 is small, kl Wnz
cos -
c
-+ 00
=0
or
and k2 Wnz C
Forced Vibration
227
0, and Eq. (E8.3.IO) yields
-+
(2n - 1)1l'
-
2
or
=
Wn
(2n - 1)1l'c 2Z
(E8.3.20)
and Eq. (E8.3.16) gives the modal functions as . O!nX . WnX Wn(x) = Cn sm -Z- = Cn sm-;;-
(E8.3.21)
This solution corresponds to that of a wire which is fixed at x = 0 and free at = Z.
x
8.6 FORCED VIBRATION The equation of motion governing the forced vibration of a uniform string subjected to a distributed load f(x,t) per unit length is given by P
w
02 (x,t) ot2
- P
02w(x,t) ox2
f
=
)
(8.111)
(x,t
Let the string be fixed at both ends so that the boundary conditions become w(O, t) = 0
(8.112)
w(l,t) = 0
(8.113)
The solution of the homogeneous equation [with f(x,t) = 0 in Eq. (8.111)], which represents free vibration, can be expressed as [see Eq. (8.106)] w(x,t)
00
= LSin
nrr x ( ncrrt -ZCncos -Z-
n=l
nC1l't)
+ Dn sin-Z-
(8.114)
The solution of the nonhomogeneous equation [with f(x ,t) in Eq. (8.111)], which also satisfies the boundary conditions of Eqs. (8.112) and (8.113), can be assumed to be of the form 00
w(x,t)
~
nrr x
(8.115)
= L..-sin -Z-TJn(t) n=l
where TJn(t) denotes the generalized coordinates. By substituting Eq. (8.115) into Eq. (8.111), we obtain 00
z-
PLsinn=l
nrr x d2TJn(t) 00 dt2 +PL(-z
n1l')2
n1l'x sin-z-TJn(t)
=
f(x,t)
(8.116)
n=l
j
228
Transverse Vibration of Strings Multiplication of Eq. (8.116) by sin(mrx/i) the use of the orthogonality of the functions
and integration from 0 to i, along with sin(irr x/i), leads to
(8.117) where
=
Qn(t)
1 o
The solution of Eq. (8.117), including integral, can be expressed as T}n(t) = Cn COS
ncrrt -. -
i
+
nrr x sin dx
1
f(x,t)
. ncrrt Dn sm -i
(8.118)
i
the homogeneous
+ --
1t
2
ncrrp
solution
and the particular
. ncrr(t - T) Qn(T) sm dT i
0
(8.119)
Thus, in view of Eq. (8.115), the forced vibration response of the string is given by
L
=
00
w(x,t)
n=l
(
ncrr t Cncos-i-+Dnsin- i
ncrr t )
t
2 ~ 1 . nrrx +-~ -:sm -crr p n i
1
. ncrr(t - T) Qn(T) sm dT i
0
n=l
where the constants Cn and Dn are determined
nrr x sin- i
(8.120)
from the initial conditions of the string.
Example 8.4
Find the forced vibration response of a uniform taut string fixed at both ends when a uniformly distributed force fo per unit length is applied. Assume the initial displacement and the initial velocity of the string to be zero. SOLUTION
For a uniformly
1 . nrrx fosm-dx
1
Qn(t) =
o
distributed
2ifo
=-,
i
= fo,
force f(x,t)
Eq. (8.118) gives
n = 1,3,5, ...
nrr
(E8.4.I)
and hence t
. nrrc(t - T) d 2/101t. Q" (T) sm T = / mro 1o 2/fo = ----
sm I
1l1T n1TC
nrrc(t - T)
1
0
n1T~C
l-cos--
dT
.
sm ydy
nrrct/I
2
21 fo (
= -~-
i
.•
nrrct)
i
,
n = 1,3,5, ...
(E8.4.2)
8.6 Forced Vibration
.219
Thus, the forced response of the string becomes [Eq. (8.120)]: =
w(x,t)
2:::: 00
(
ncrrt
+ Dn sin -/-
ncrrt. )
nrrx sm -z-
1. nrex( -sm-n3 /
l-cos-- lIcret)
Cn cos -/-
n=1 00
2::::
2
4/ fo +-c2re3p
n= 1.3.5 •...
(E8.4.3)
/
Use of the initial conditions W(x,
(E8.4.4)
0) =0 =0
aw(x,O)
at
(E8.4.5)
Cnsin-/-nrr x =0
(E8.4.6)
in Eq. (E8.4.3) yields 00
'" n=1 00
ncre
.
2:::: -Dnsm--=O /
117'(
X
(E8.4.7)
/
n=1
Equations (E8.4.6) and (E8.4.7) result in " Cn
= Dn = 0
(E8.4.8)
and hence the forced vibration response of the string is given by [see Eq. (E8.4.3)]: 00
2
w(x,t)
= --4/ fo c2rr3p
2::::
nrex ( 1 - cos--ncret) -1 sin --
n=1.3.5 •...
n3
/
(E8.4.9)
/
Example 8.5 Find the steady-state forced vibration response of a fixed-fixed string subjected to a concentrated force F(t) = Fo at x = Xo. SOLUTION
The applied force can be represented as = F(t)8(x
f(x,t)
(E8.5.l)
- xo)
The steady-state forced vibration response of the string can be expressed, using Eq. (8.120), as W(X,t)
~ 1 . nrex k..J - sm -crr p n /
=-
2
n=l
where
Qn(t)
it 0
is given by Eq. (8.118): Qn(t)
=
ti o
f(x,t)
nrex sin dx =
nrr Xo = F(t)sin-I
l
.
Qn(r) sm
it 0
ncrr(t - r)
F(t)8(x
l
(E8.5.2)
dr
nrex - xo) sin dx I
(E8.5.3)
230
Transverse Vibration of Strings
The function F(t) = Fo can be denoted as
= FoH(t)
F(t)
(E8.5.4)
where H(t) is the Heaviside function, defined by
t:::O
0, { 1,
H(t) =
(E8.5.5)
t:::O
Substitution of Egs. (E8.5.4) and (E8.5.3) into (E8.5.2) results in
1/ .
2Fo ~ 1 . mrx w(x,t) = ~ - sm -crrp n=l n 1 2Fol
00
= -.-" 2 2
rr c p ~
n=l
mrxo
H(r) sm -l-
sin
mrc(t - r) I
0
1 nrrxo. -sin--sm-2 n 1
nrrx
()
1
l-cos--
nrrct
1
dr
(E8.5.6)
Example 8.6 Find the steady-state response of a fixed-fixed string subjected to a load moving at a constant velocity v given by f(x,t)
o ::: vt
F(t)8(x - vt),
={ 0,
::: 1
vt > 1
(E8.6.1)
where F(t) is a suddenly applied force Fo. SOLUTION
The steady-state response of the string is given by Eg. (8.120) as
1/
2 ~ 1 . nrrx w(x,t) = ~ - sm -crr p n=l n 1
0
nrrc(t - r) Qn(r) sin ----dr I
(E8.6.2)
where Qn(t) is given by Eg. (8.118): Qn(t) =
i
t
o
f(x,t)
nrr x
sin -
I
dx
(E8.6.3)
Using Eg. (E8.6.1), Qn(t) can be evaluated as Qn(t) =
i
t
o
mrx nrrvt F(t)8(x - vt) sin dx = F(t) sin-I I
Thus, Eg. (E8.6.2) becomes w(x,t)
= -crr2 p ~~ -nI sm. --nrrx I n=l
it 0
nrrvt nrrc(t - r) F(r) sin -sin ----dr I I
(E8.6.4 )
(E8.6.5)
Using F(t) = FoR(t)
(E8.6.6)
I ,.•••. ~"~,~~n'~t~;,-~~~~,
.••,'~I}!,"~~
,"'j.'•"iy::,~.~~~-,"'P':.:\,'~"'C:~:Im:'''\'''~t,_,:,,~!J!~~'''fi~
••••
'._:~'~',,!1f .'''';",''\,~'9,,~ii~'''~~'''"~'~~':,,
. 'Y"'"'l!~-'~~",'W;-~_t'.!'~"J~-~,.,
I
Recent ContributIons --131
8.7
Eq. (E8.6.5) can be evaluated as W(X,t)
2Fo =-
crr P
I:
oo
n=1
=:
2Fo ~ = crrp
1 nJfX - sin -n
Z
l' 0
nrrvt . nrrc(t - r) sm ----dr
H(r) sin --
Z
Z
sin(nrrxfl) (nrrv . nrrc nrrc. nrrv) n (n2rr2v2/12 _ n2rr2c2/Z2) -t- sm -z-t - -t- sm -t-t (E8.6.7)
8.7
RECENT CONTRmUTIONS The D'Alembert's solution ofEq. (8.8), as given by Eq. (8.35), is obtained by assuming that the increase in tension due to stretching is negligible. If this assumption is not made, Eq. (8.9) becomes [4] (8.121) where (8.122) with Po denoting the density of the string. Here CI denotes the speed of compressional longitudinal wave through the string. An approximate solution of Eq. (8.121) was presented by Bolwell [4]. The dynamics of cables, chains, taut inclined cables, and hanging cables was considered by Triantafyllou [5, 6]. In particular, the problem of linear transverse vibration of an elastic string hanging freely under its own weight presents a paradox, in that a solution can be obtained only when the lower end is free. An explanation of the paradox was given by Triantafyllou [6], who also showed that the paradox can be removed by including bending stiffness using singular perturbations. A mathematical model of the excitation of a vibrating system by a plucking action was studied by Griffel [7]. The mechanism is of the type used in musical instruments [8]. The effectiveness of the mechanism is computed over a range of the relevant parameters. In Ref. [9], Simpson derived the equations of in-plane motion of an elastic catenary translating uniformly between its end supports in an Eulerian frame of reference. The approximate analytical solution of these equations is given for a shallow catenary in which the tension is dominated by the cable section modulus. Although the mathematical description of a vibrating string is given by the wave equation, a quantum model of information theory was used by Barrett to obtain a one-degree-of-freedom mechanical system governed by a second-order differential equation [10]. The vibration of a sectionally uniform string from an initial state was considered by Beddoe [11]. The problem was formulated in terms of reflections and transmissions of progressive waves and solved using the Laplace transform method without incorporating the orthogonality relationships. The exact equations of motion of a string were
232
Transverse Vibration of Strings
fonnulated by Narasimha [12], and a systematic procedure was described for obtaining approximations to the equations to any order, making only the assumption that the strain in the material of the string is small. It was shown that the lowest-order equations in the scheme, which were nonlinear, were used to describe the response of the string near resonance. Electrodischarge machining (EDM) is a noncontact process of electrically removing (cutting) material from conductive workpieces. In this process, a high potential difference is generated between a wire and a workpiece by charging them positively and negatively, respectively. The potential difference causes sparks between the wires and the workpiece. By moving the wire forward and sideways, the contour desired can be cut on the workpiece. In Ref. [13], Shahruz developed a mathematical model for the transverse vibration of the moving wire used in the EDM process in the form of a nonlinear partial differential equation. The equation was solved, and it was shown that the transverse vibration of the wire is stable and decays to zero for wire axial speeds below a critical value. A comprehensive view of cable structures was presented by Irvine [14]. The natural frequencies and mode shapes of cables with attached masses have been determined by Sergev and Iwan [15]. The linear theory of free vibrations of a suspended cable has been outlined by Irvine and Caughey [16]. Yu presented explicit vibration solutions of a cable under complicated loads [17]. A theoretical and experimental analysis of free and forced vibration of sagged cable/mass suspension has been presented by Cheng and Perkins [18]. The linear dynamics of a translating string on an elastic foundation was considered by Perkins [19].
REFERENCES 1. W. Nowacki, Dynamics of Elastic Systems, translated by H. Zorski, Wiley, New York, 1963. 2. N. W. McLachlan, Theory of Vibrations, Dover, New York, 1951. 3. S. Timoshenko, D. H. Young, and W. Weaver, Jr., Vibration Problems in Engineering, 4th ed., Wiley, New York, 1974. 4. J. E. Bolwell, The flexible string's neglected term, Journal of Sound and Vibration, Vol. 206, No.4, pp. 618-623, 1997. 5. M. S. Triantafyllou, Dynamics of cables and chains, Shock and Vibration Digest, Vol. 19, pp. 3-5 1987. 6. M. S. Triantafyllou and G. S. Triantafyllou, The paradox of the hanging string: an explanation using singular perturbations, Journal of Sound and Vibration, Vol. 148, No.2, pp. 343-351, 1991. 7. D. H. Griffel, The dynamics of plucking, Journal of Sound and Vibration, Vol. 175, No.3, pp. 289-297, 1994. 8. N. H. Fletcher and T. D. Rossing, The Physics of Musical Instruments, Springer-Verlag. New York, 1991. 9. A. Simpson, On the oscillatory motions of translating elastic cables, Journal of Sound alld Vibration, Vol. 20, No.2, pp. 177-189, 1972. 10. T. W. Barrett, On vibrating strings and information theory, Journal of Sound and Vibratioll. Vol. 20, No.3, pp. 407-412, 1972.
Problems
233
II. B. Beddoe, Vibration of a sectionally uniform string from an initial state. Journal of Sound and Vibration, Vol. 4, No.2, pp. 215-223, 1966. 12. R. Narasimha, Non-linear vibration of an elastic string, Journal of Sound and Vibration, Vol. 8, No. I, pp. 134-146, 1968. 13. S. M. Shahruz, Vibration of wires used in electro-discharge machining, Journal of Sound and Vibration, Vol. 266, No.5, pp. 1109-1116, 2003. 14. H. M. Irvine, Cable Structures, MIT Press, Cambridge, MA, 1981. 15. S. S. Sergev and W. D. Iwan, The natural frequencies and mode shapes of cables with attached masses, Journal of Energy Resources Technology, Vol. 103, pp. 237-242. 1981. 16. H. M. Irvine and T. K. Caughey, The linear theory of free vibrations of a suspended cable, Proceedings of the Royal Society, London, Vol. A-341, pp. 299-315, 1974. 17. P. Yu, Explicit vibration solutions of a cable under complicated loads, Journal of Applied Mechanics, Vol. 64, No.4, pp. 957-964, 1997. 18. S.-P. Cheng and N. C. Perkins, Theoretical and experimental analysis of the forced response of sagged cable/mass suspension, Journal of Applied Mechanics, Vol. 61, No.4, pp. 944-948, 1994. 19. N. C. Perkins, Linear dynamics of a translating string on an elastic foundation, Journal of Vibration and Acoustics, Vol. 112, No. I, pp. 2-7, 1990.
PROBLEMS 8.1 Find the free vibration response of a fixed-fixed string of length l which is given an initial displacement . 2itx l
wo(x) = h SIn--
8.7 Derive the boundary conditions corresponding to support conditions 3, 4, and 5 of Table 8.1 from equilibrium considerations. 8.8 The transverse vibration of a string of length l is governed by the equation
and initial velocity wo(x) = O.
iz
8.2 A steel wire of diameter in. and length 3 ft is fixed at both ends and is subjected to a tension of 200 lb. Find the first four natural frequencies and the corresponding mode shapes of the wire. 8.3 Determine the stress that needs to be applied to the wire of Problem 8.2 to reduce its fundamental natural frequency of vibration by 50 % of the value found in Problem 8.2.
=
8.4 A string of length l is fixed at x 0 and subjected to a transverse force f (t) = fo cos wt at x = l. Find the resulting vibration of the string. 8.5 Find the forced vibration response of a fixed-fixed string of length l that is subjected to the distributed transverse force f(x,t) = F(x)eiwt• 8.6 A uniform string of length l is fixed at both ends and is subjected to the following initial conditions: w(x,O)
. 21Z'x l
=XOSIn--,
.(
W
x,
0
)
. 21Z'x = -yO SID -l
o2w
16- = ox2
=2
o2w
-ot2
The boundary and initial conditions of the string are given by w(O,t) = 0, w(X, 0)
= 0.lx(2
w(2,t) = 0 - x),
ow
7it(x,O)
=0
Find the displacement of the string, w(x,t).
=
8.9 A semi-infinite string has one end at x 0 and the other end at x 00. It is initially at rest on the x axis and the end x 0 is made to oscillate with a transverse displacement of w(O,t) = c sin nt. Find the transverse displacement of the string, w(x,t).
= =
8.10 Find the natural frequencies of transverse vibration of a taut string of length l resting on linear springs of stiffnesses k, and k2 at the ends x = 0 and x = [, respectively. Assume the following data: P 1000 N, p = 0.1 kg/m; and k, = k2 = 5000 N/m.
=
j
9 Longitudinal Vibration of Bars 9.1
INTRODUCTION A straight elastic bar can undergo longitudinal, torsional, and lateral vibration. Among these, the longitudinal vibration is the simplest to analyze. If x denotes the longitudinal (centroidal) axis and y and z represent the principal directions of the cross section, the longitudinal vibrations take place in the x direction, torsional vibrations occur about the x axis, and lateral vibrations involve motion in either the xy plane or the xz plane. These vibrations may be coupled in some cases. For example, if the cross section is not symmetric about the y or z axis, the torsional and lateral vibrations are coupled. If the bar is pretwisted along the x direction, the lateral vibrations in the xy and xz planes are coupled. We consider first the longitudinal vibration of a bar using a simple theory.
9.2
EQUATION OF MOTION USING SIMPLE THEORY We consider a simple theory for the longitudinal vibration of bars based on the following assumptions: 1. The cross sections of the bar originally plane remain plane during deformation. 2. The displacement components in the bar (except for the component parallel to the bar's longitudinal axis) are negligible. These assumptions permit the specification of the displacement as a function of the single space coordinate denoting location along the length of the bar. Although lateral displacement components exist in any cross section, the second assumption can be shown to be valid for vibrations involving wavelengths that are long compared to the cross-sectional dimensions of the bar. We shall derive the equation of motion using two different approaches: by applying Newton's second law of motion and from Hamilton's principle. Using New1on's Second Law of Motion For an elastic bar of length I, Young's modulus E, and mass density p with varying cross-sectional area A(x) as shown in Fig. 9.l(a), the equation of motion has been derived, using Newton's second law of motion, in Section 3.4 as o [OU(X,t)] ox E(x)A(x)
23.:f° ..,.
ox
+ f(x,
t)
= p(x)A(x)
02u(x.t) ot2
(9.1)
E:quation of Motion Using Simple Theory ..:,235
9.2
(a)
c a
I I
..•-
: a'
I
I
:
P :
._._~-_._
I
I
I
: I I I
Ib'
l .•.•.--
P+dP
I
'-r'-'-'
b
,
_----Ic
__
..
- -----Itf d
.t-:-
dx -\ (b)
Figure 9.1
Longitudinal vibration of a bar.
For a uniform bar, Eq. (9.1) reduces to 2 EA a u(x, t) ax2
9.2.2
+ f(
t) _ x,
-p
2 A a u(x, t) at2
(9.2)
Using Hamilton's Principle During longitudinal vibration, the cross section of the bar located at a distance x from the origin undergoes an axial displacement of u while the cross section located at a distance x + dx undergoes an axial displacement of u + du u + (au/ax) dx, as shown in Fig. 9.1(b). Since the deformation of the cross section in the y and z directions (v and w) is assumed to be negligible, the displacement field can be expressed as
=
u
= u(x,
t),
w=o
v =0,
(9.3)
The strains in the cross section at x are given by
au Exx
= ax'
Eyy
= Ezz = 0,
Exy
=
Eyz
= Ezx = 0
(9.4)
Note that the displacements v and w, and the strains Eyy and Ezz, will not be zero, due to Poisson's effect in a real bar; they are assumed to be zero in the simple theory. The
236
Longitudinal Vibration of Bars stresses acting in the cross section at x, corresponding are
au
O'xx
= E ax
'
= O'zz = 0,
O'yy
to the strains given by Eq. (9.4),
O'xy
=
O'yZ
=
O'zx
=0
(9.5)
The strain and kinetic energies of the bar can be found as
1r
=~ 2
T = ~
t
Jo
= ~
Adx
O'xxexx
2
t pA
(au)2
2 Jo
dx
dx
at
The work done by the external force
tEA (au)2 ax
Jo
f (x,
(9.6)
(9.7)
t) is given by
1
1
W = The generalized Hamilton's
f(x,
t)u dx
(9.8)
principle can be stated as
a
t2
1
(T -
7r
+ W)dt
=0
(9.9)
t}
The substitution of Eqs. (9.6)-(9.8) into Eq. (9.9) yields the equation of motion and the associated boundary conditions as (see Problem 9.8)
a ( ... au) ax EA ax
+f
a2u = pA at2
(9.10)
EA au au 11 = 0
ax
(9.11)
0
Note that Eq. (9.11) will be satisfied for a free boundary where O'xx
or when the displacement boundary condition is
au ax
=EA- =0
is specified at the boundary with
(9.12)
au
= 0; for a fixed end, the
u=O
(9.13)
9.3'" FREE VffiRATION SOLUTION AND NATURAL FREQUENCIES The equation governing the free vibration of bars can be obtained by setting Eqs. (9.1) and (9.2). For nonuniform bars:
2 ~ [EA(X) au(x, t)] = pA(x) a u(x, t) ax ax at2
f =0
in
(9.14)
I ••~l:':;"'!.
"".
p····'<;,·".".t.,"";:;;·"~~~.1:,~~"'''·,.--.·~'';:,_'!l·._f''iV ••',,.~"~",,,..,..l"_"')'''':;''~·:'·'·'·''_··'''''''':"
~"~"'.~~"f",\t14".~~~'1""Ji'~';.,.~~~!,,,,:;'~""-'~"'";.~~.~~.,
9.3
Free Vibration Solution and Natural Frequencies
237
For uniform bars:
or (9.15) where
[E
C=y;
(9.16)
The solution of Eq. (9.15) can be obtained using either the wave solution approach or the method of separation of variables. The wave solution of Eq. (9.15) can be expressed, as in the case of vibration of strings, as u(x, t) = 11(x - ct)
+ hex + ct)
(9.17)
Although this solution [Eq. (9.17)] is useful in the study of certain impact and wave propagation problems involving impulses of very short duration, it is not very useful in the study of vibration problems. The method of separation of variables followed by the eigenvalue and modal analyses is more useful in the study of vibrations.
9.3.1
Solution Using Separation of Variables To develop the solution using the method
of separation
of variables, the solution of
Eq. (9.15) is written as (9.18)
Vex, t) = V(x)T(t) where V and T depend on only x and t, respectively.
Substitution
of Eq. (9.18) into
Eq. (9.15) leads to
c2 d2V 1 d2T V dx2 = T dt2
(9.19)
Since the left-hand side of Eq. (9.19) depends only on x and the right-hand side depends only on t, their common value must be a constant, which can be shown to be a negative number (see Problem 9.7), denoted as _w2• Thus. Eq. (9.19) can be written as two separate equations: (9.20) (9.21)
•.·.T·''''.~~'<'·''''''.~;!!~"T,_4"
238
Longitudinal Vibration of Bars
The solution of Egs. (9.20) and (9.21) can be represented as wx
+ B sin-wx
Vex) = A cos -
C T(t)
=
(9.22)
C
+ Dsinwt
Ccoswt
(9.23)
where w denotes the frequency of vibration, the function V (x) represents the nonnal mode, the constants A and B can be evaluated from the boundary conditions, the function T(t) indicates hannonic motion, and the constants C and D can be determined from the initial conditions of the bar. The complete solution of Eg. (9.15) becomes u(x, t)
=
V(x)T(t)
WX)
wx
= ( A cos 7"+ B sin 7
(C coswt
+D
sinwt)
(9.24)
The common boundary conditions of the bar are as follows. For the fixed end:
u=o For the free end:
(9.25)
au ax
-=0
(9.26)
Some possible boundary conditions of a bar are shown in Table 9.1. The application of the boundary conditions in Eq. (9.22) leads to the frequency equation whose solution yields the eigenvalues. The substitution of any specific eigenvalue in Eg. (9.22) gives the corresponding eigenfunction. If the axial displacement and the axial velocity of the bar are specified as uo(x) and uo(x), respectively, at time t 0, the initial conditions can be stated as
=
u(x, t
au -(x, at
t
= 0) = uo(x)
(9.27)
= 0) = uo(x)
(9.28)
The following examples illustrate the formulation of boundary conditions, the determination of natural frequencies for specified boundary conditions of the bar, and the method of finding the free vibration solution of the bar in longitudinal vibration. Example 9.1 The ends of a uniform bar are connected to masses, springs, and viscous dampers as shown in Fig. 9.2(a). State the boundary conditions of the bar in axial vibration. SOLUTION If the axial displacement, velocity, and acceleration of the bar at x = 0 are assumed to be positive with values u(O, t), ou/ot(O, t), and o2u/ot2(0, t), respectively, the spring force kju(O, t), the damping force Cj[ou/ot](O, t), and the inertia force m I [a2u/at2](0, t) act toward the left as shown in the free-body diagram of Fig. 9.2(b). The boundary condition at x = can be expressed as
°
[force (reaction) in bar at x = 0]
= (sum of spring, damper,
and inertia forces at
x
= 0)
,..;
'-'I l:::
:::
--
II
II
-
N•
II
~I--..: II II
a
a
a
ON
>0(
-
l::: ..-.
+N i:::
N
'-'
.S
II
'"
u"
o II
11 '" '"o ()
=
.2
OJ...
II
.J:J
:> "OJ
:e=
a '00
=
o
....l
o II
0
II
...... e:~ :: :: .-.. ..-.
.S ~
lJ:l
'.g'=-" o
'"
:e
=o
U
~
"0
=:o:l
lJ:l
239
v '"
'uc
c
v 0
•••
::l .~
ff~ ~:9 ;> "a
3
~I-..; II II a
•••••
0
z~
...;
N
ON
'"I- .... ~ ;::0
II
II
a ;::
;::
...;
ON
ON
~I-..; II II a
~/-..;
~I'"
~ I '"
II
II
;::
;::
a
,-.. '"
'" .~ § v
0..
"'..E o
::E
II
0
c
'-"
0 ::l
~
0':;:
0"
fJ. v
II
~ 0 ~
~I~ I M~
'"0
II
,-... '" ~ G' ':5
§
is>'c
~I-
~I '" .5
..c c ~"a
,-... ~
,-... '"0 II
(,)
G'
'-"
':5
~
':5
"0
0::l
C .~ ::l"O
EI~~I-
I'tl.
o c lXl
8
II II
,-... •.. 0 '-"
::l
"l*f
24()'
~
-
II
II
'tl.
~
':5
~
0
~ I
0()
I II
II
a.5I'"
II
~ 0 u ~
~ ~ c
'"
-
~I~
-o:r::
-
II 0()
Q
::lIN •••• Q C'eC'C;:::" ::l
M
~
-
'-"
'-"
::ll~
G'
'-"
'tl.
(,)
0
(,)
G'
~
~r:l
'"0 ,-... ~
(,)
~ I
0
0
0
II
II
II
,-.. •.. 0 '-"
,-... •..
-I ~
,-... •..
,-... •..
'-"
8.
-
::ll ~ ::ll ~
'-"
• ::l
::ll~
~
'-"
N::lIN •.•
~ I
0
II
~ 8.
,-... •.. •.. ,-.. :::::.
• ::l
-
I ~ '-':::
::ll ~:"I:
I
9.3
Free Vibration Solution and Natural Frequencies2..J!
I$=Q_' __
'_-_'~_X
----"~
(a)
au
P-:X AE-
I m.,u- \
--+-.t=1
ku
1.lt=1
x=~ ~C1U\
-
Free-body diagram of mass ml
+X,+U,+
.r=l
Free-body diagram of mass ml
au
a;
(b)
Figure 9.2 Bar with masses. springs,and dampers at ends.
or
au
..' ,au
AE-(O.
t)
ax
= klU(O,
t)
+ Cl-(O. at
a2u
t)
+ m1-a
t
2 (0, t)
(E9.1.1)
In a similar manner. the boundary condition at x = 1 can be expressed as (E9.1.2)
Example 9.2
A uniform bar oflength 1. cross-sectional area A, density P. and Young's modulus E. is fixed at x = 0 and a rigid mass M is attached at x = 1 [Fig.9.3(a)]. Determine the natural frequencies and mode shapes of longitudinal vibration of the bar. SOLUTION
The solution for the free longitudinal vibration of a bar is given by
Eq. (9.24): U(x, 1)
wx
= ( A cos -
C
) + B sin wx -
(C coswt
. + D S1Dwt)
(E9.2.1)
C
The boundary condition at the fixed end. x = 0, is given by u(O, t) = 0
(E9.2.2)
242
Longitudinal Vibration of Bars
~_~:E~P I,. (a)
EJ --~ (b)
Figure 9.3 Longitudinal vibration of a bar, fixed at x = 0 and mass attached at x = l: (a) bar with end mass M; (b) free body diagram of mass M.
The boundary condition at x mass shown in Fig. 9.3(b) as
= I can be expressed from the free-body diagram
au reaction force = P(/, t) = Aa(l, t) = AE-(/, ax
t)
a2u (l,t)
= -inertia force = -M-
atz
Equations (E9.2.2) and (E9.2.1) give
of the
(E9.2.3)
A=O
(E9.2.4)
and hence Eg. (E9.2.1) becomes .
wx
u(x, t) = ~ B sm-(c cOswt c Equation (E9.2.5) gives
au
w
wx 2
-z at = -Bw ~
(E9.2.5)
.
-ax = ~ B -c cos - c (C cos wt azu
. + D smwt)
+ D sm wt)
wx sin -(C COswt + D sinwt) c
(E9.2.6) (E9.2.7)
Using Egs. (E9.2.6) and (E9.2.7), Eg. (E9.2.3) can be expressed as AE~B c- cos wi c (C coswt
+ D sinwt)
= Mw2 -B sin wi c (C coswt
+ D sinwt)
or
wi tan-=_ c By introducing the mass of the bar, m, as
AE
Mwc
m = pAL Eg. (E9.2.8) can be rewritten as CI
tan ex = f3
(E9.2.8)
(E9.2.9)
(E9.2.1O)
9.3
Free Vibration Solution and Natural Frequencies - :243
where wi a= -
(E9.2.11)
c
{3 _ pAL -
M
-
!!!..
(E9.2.12)
M
Equation (E9 .2.10) is the frequency equation in the form of a transcendental equation which has an infinite number of roots. For the nth root, Eq. (E9.2.10) can be written as (E9.2.13)
n = 1,2, ...
an tan an = {3, with a n---
wnl
or
wn---
anc
C
The mode shapes corresponding to the natural frequency Un ( x)
. = -Bnsm-, c WnX
(E9.2.14)
I Wn
can be expressed as (E9.2.15)
n = 1,2, ...
The first 10 roots of Eq. (E9.2.13) for different values of the mass ratio {3are given in Table 9.2.
Table 9.2
Roots of Eq. (E9.2.13) Value of an for:
n 1 2 3 4 5 6 7 8 9 10
fJ = 0
fJ=lO
fJ=1
fJ=fo
fJ -\00 - \
0 3.1416 6.2832 9.4248 12.5664 15.7080 18.8496 21.9911 25.1327 28.2743
1.4289 4.3058 7.2281 10.2003 13.2142 16.2594 19.3270 22.4108 25.5064 28.6106
0.8603 3.4256 6.4373 9.5293 12.6453 15.7713 18.9024 22.0365 25.1724 28.3096
0.3111 3.1731 6.2991 9.4354 12.5743 15.7143 18.8549 21.9957 25.1367 28.2779
0.0998 3.1448 6.2848 9.4258 12.5672 15.7086 18.8501 21.9916 25.1331 28.2747
Example 9.3 A uniform bar oflength I, cross-sectional area A, density p, and Young's modulus E is free at x = 0 and attached to a spring of stiffness K at x = I, as shown in Fig. 9.4(a). Determine the natural frequencies and the mode shapes of longitudinal vibration of the bar.
I
244
Longitudinal Vibration of Bars
(a)
KuU,t)
t=JPU.t)
(b)
Figure 9.4
SOLUTION Eq. (9.24):
Bar free at x = 0 and attached to a spring at x = 1.
The solution for the free longitudinal vibration of a bar is given by wx ( Acos-+Bsm-
u(x.t)=
-
c
.
WX)
-
. (C coswt+Dsmwt)
c
(E9.3.1)
Since the end x = 0 is free. we have
au
AE
ax (0. t)
=0
au
ax (0. t)
or
=0
(E9.3.2)
Equations (E9.3.1) and (E9.3.2) yielq B=O
(E9.3.3)
Thus. Eq. (E9.3.1) reduces to . + D smwt)
wx u(x. t ) = Acos-(Ccoswt c
(E9.3.4)
The boundary condition at x = I can be expressed as [Fig. 9.4(b)] reaction force = -spring force that is.
au ax
AE-(I.
t) = -Ku(I, t)
(E9.3.5)
Equations (E9.3.4) and (E9.3.5) lead to ~E - A --
C
~
.
~
sin - (C cas wt + D sin wt) = - K A cas - (C cas wf + D sin Wf) c
-
c
or AEw
-cK
= cot-
wI c
(E9.3.6)
.~\~"""""",,"''''h"
.."P'''"''''''''''''''''f,'I'<;I'''~'''\!.'''''''''''''.'.,,,.,,,.,,~,,,,,,,,,,,,"''''''''Y'>.~',n., .,.,,, .•..,,•.•,"""~,.,,,~~~
.•,..~r,~."""".,, •.C.•..,"'''r:c· '~~-'j
\
I I I
9.3
Free Vibration Solution and Natural Frequencies
'US
Introducing the mass of the bar, m, as m = pAL
(E9.3.7)
a cota = {3
(E9.3.8)
Eq. (E9.3.6) can be rewritten as
where wl a=-
(E9.3.9)
k
(E9.3.1O)
c
f3
=
K
AE k=1
(E9.3.1l)
denotes the stiffness of the bar. Equation (E9.3.8) denotes the frequency equation in the form of a transcendental equation with an infinite number of roots. For the nth root, Eq. (E9.3.8) can be expressed as (E9.3.12)
n = 1,2, .. , with
(E9.3.13) The mode shape corresponding to the natural frequency
Wn
can be expressed as (E9.3.14)
n = 1.2, .. ,
The first 10 roots of Eq. (E9.3.12) for different values of the stiffness ratio f3 = k/ K are given in Table 9.3. Table 9.3
Roots of Eq. (E9.3.12) Value of an for:
n 1 2 3 4 5 6 7 8 9 10
P=1
p_! -5
3.145 4.495 6.285 7.725 9.425 10.905 12.565 14.065 15.705 17.225
1.435 3.145 4.665 6.285 7.825 9.425 10.975 12.565 14.125 15.705
P = to 1.505 3.145 4.695 6.285 7.845 9.425 10.985 12.565 14.135 15.705
p--50\ 1.555 3.145 4.705 6.285 7.855 9.425 10.995 12.565 14.135 15.705
p--\00\ 1.565 3.145 4.715 6.285 7.855 9.425 10.995 12.565 14.135 15.705
~__ ----------t
246
Longitudinal Vibration of Bars
Example 9.4
Find the natural frequencies
of vibration
and the mode shapes of a bar
with free ends.
SOLUTION
The boundary
conditions
au
=
of a free-free
dU dx (0)
bar can be expressed
as
= 0,
t~O
(E9.4.I)
dU -(i, t) = -(i) = 0, ax dx
t~O
(E9.4.2)
ax (0, t)
au
In the solution wx U (x) = A cos -
+ B sin
C
wx _
-
dU w wx -(x) = -A- sin dx - C C
(E9.4.3)
C
w
+ -B-C cos-
wx C
(E9.4.4)
use of the condition, Eq. (E9.4.1), gives
B =0
(E9.4.5)
The condition of Eq. (E9.4.2) leads to the frequency equal to zero for a nontrivial solution):
equation (noting that
6 cannot
be
. wi
sm- =0
(E9.4.6)
C
which yields
wi - = nJr,
n = 1,2, ...
C
(E9.4.7)
As different values of n give different frequencies of the various modes of vibration, the nth frequency and the corresponding mode shape can be expressed as nJrc
wn=-,
n = 1,2, '"
i
wnx nJrx Un(x) = Acos = Acos --, -
C
i
-
The first three frequencies and the corresponding and (E9.4.9), are shown in Table 9.4.
9.3.2
n = 1,2, ...
(E9.4.8) (E9.4.9)
mode shapes, given by Eqs. (E9.4.8)
Orthogonality of Eigenfunctions The differential equation governing the free longitudinal Eq. (9.1) with f 0, can be written in general form as
=
L[u(x, t»)
+M
vibration of a prismatic
bar,
a2u
at2 (x, t) = 0
(9.29)
9.3 Table 9.4
Free Vibration Solution and Natural Frequenci.es
247
First Three Mode Shapes of a Free-Free bar Natural frequency,
Mode number, n
Mode shape, Un(X)
Wn
rrc
2rrc 2
x
o
3rrc 3
x
I
where 82
==
L = -R 8x2 R
82
M = pA
-EA 8x2'
(9.30)
= EA denotes
the axial rigidity and pA indicates the mass per unit length of the bar. For free vibration (with harmonic motion) in the ith natural mode, we can write Uj(x, t)
Substituting
= Uj(X)(Ci
COSWjt
+ Dj sinwjt)
(9.31)
Eq. (9.31) into Eq. (9.29), we obtain RUr(x)
+ Mw;Uj(X)
=0
(9.32)
where a prime denotes a derivative with respect to x. Equation (9.32) can be rewritten as an eigenvalue
problem (9.33)
where Uj (x) is the eigenfunction conditions
or normal function
determined
from the boundary
and
Mw~
Aj
(Wj)2
= --j- = - 7
(9.34)
248
Longitudinal Vibration of Bars
is the eigenvalue with
(9.35) Let Ui(x) and Uj(x) denote the eigenfunctions corresponding to the natural frequencies and Wj, respectively, so that
Wi
= AiUi
(9.36)
Uj' = AjUj
(9.37)
U!'
Multiply Eq. (9.36) by Uj and Eq. (9.37) by Ui and integrate the resulting equations from 0 to I to obtain
1
1
1/
U!'Uj dx
= Aj
UjUi dx = Aj
1/ 1/
UjUj dx
(9.38)
UjUj dx
(9.39)
Integrate the left-hand sides of Eqs. (9.38) and (9.39) by parts: U;Uj 1&
-1/
U;Uj dx = Aj
iI&-l/
UjU
UjUjdx
=Ajl/
1/
UjUj dx
(9.40)
UjUjdx
(9.41)
The first terms on the left-hand sides of Eqs. (9.40) and (9.41) are zero if the end of the bar is either fixed or free. Subtract Eq. (9.41) from (9.40) to obtain (Aj - Aj)
When the eigenvalues are distinct Aj for normal functions:
1/
1/
=1= Aj,
UjUj dx
=0
(9.42)
Eq. (9.42) gives the orthogonality principle
UiUj dx = 0,
(9.43)
In view of Eq. (9.43). Eqs. (9.40) and (9.39) yield
JtU~U'dx=O o I
1/
)
U!'Uj dx
,
= 0,
(9.44)
(9.45)
Equations (9.43)-(9.45) indicate that the orthogonality is valid not only among the eigenfunctions, but also among the derivatives of the eigenfunctions.
~"""'""!o.'_~
..
_.~"".~"",.~~.,""~l;,,,,--.",,l"~~":"'.;l';;,
..~'~7
"".~'''~:''~'''''~-'',.'
.1·~'.'j:·':""-;·~~""f,.·~;'I',,;o·"~'d~~ .:,.••. ,-<:,,..~':'!t~..~__ ~,
..:1l'.J!T'":."',...~.
,~-'"~~
.,,,,,,._,.,,:,.~;:'I./""4-~.,).";r'f"<."''' ..,,,,,,,,,,,,,,,:,~''-'''''~-''.~-,.r'·I-"'~-"~-!~"
·7..;7,<,\~·~1~·~~~~"1"·-"'-~;"
I
9.3
Free Vibration Solution and Natural Frequencies
249
Note The orthogonality relationships for a bar with a mass 11 attached at the end x = I [as in Fig. 9.3(a)] can be developed as follows. Rewrite Eq. (9.32) corresponding i and j as
to two distinct eigenvalues
(9.46)
J' RU"J' = -mw,2U J
=
=
with m == M pA. To include the boundary condition at x I in the orthogonality relation, we write the boundary condition for eigenvalues i and j as (9.47) Using a procedure similar to the one used in deriving Eq. (9.43), we obtain the orthogonality condition as (see Problem 9.21) (9.48)
9.3.3
Free Vibration Response due to Initial Excitation The response or displacement of the bar during longitudinal vibration can be expressed in terms of the normal functions Ui(X), using the expansion theorem, as 00
""u(x,'t)
=L
(9.49)
Ui(x)r/i(t)
i=l
Substitution
of Eq. (9.49) into Eq. (9.29) results in 00
+ Miji (t)Ui
L[R1]i(t)Uf'(x)
(x)]
=0
(9.50)
i=l
Multiplication
of Eq. (9.50) by Uj(x)
~ [Rn,lt) In view of the orthogonality
f.'
and integration
U!,Ujdx +Mij,
relationships,
f.'
from 0 to I yields
U,UjdX] = 0
(9.51)
Eqs. (9.43) and (9.45), Eq. (9.51) reduces
to
where Mi and Ri denote the generalized respectively,
i = 1,2, ...
(9.52)
mass and generalized
stiffness (or rigidity),
in mode i: Mi = M Ri
=R
1/ u? 1/
(9.53)
dx
Uf'Ui dx
= -R
1/
(U[)2dx
= WrMi
(9.54)
250
Longitudinal Vibration of Bars If the eigenfunctions
are nonnalized
with respect to the mass distribution
i
as
l
Mi = M Eq. (9.54) gives Ri =
w; and Eq. (9.52)
U? dx = 1
(9.55)
becomes
i
= 1,2, ...
(9.56)
The solution of Eq. (9.56) is given by
=
TJi(t)
~i(O) + --
TIi(O)coswit
.
smwit
(9.57)
Wi
=
=
where TJi(0) TJiOand ~i (0) TJiOare the initial values of TJi(t) and TJi(t), which can be detennined from the initial values of the displacement and velocity given by Eqs. (9.27) and (9.28). For this, first we express uo(x) and uo(x) using Eq. (9.49) as 00
=
uo(x)
L
Ui(X)TJiO
(9.58)
Ui(X)TJiO
(9.59)
i=1 00
=
uo(x)
L i=!
Multiplication
of Eqs. (9.58) and (9.59) by Uj(x)
r 10
r 10
100
uo(x)Uj(x)
dx
=L
r 10
I
r 10
00
= L~iO
uo(x)Uj(x)dx
i=1
from 0 to I result in
1
TJiO
i=1
and integration
Ui (x)Uj
(x) dx
= TJjO
(9.60)
= TJjO
(9.61)
1
Ui(x)Uj(x)dx
in view of the orthogonality of the nonnal modes. Thus, the jth generalized coordinate can be detennined from Eq. (9.57). The total response of the bar can be expressed as [Eq. (9.49)]
u(x. t)
=
00
EUi(x.
t)
=
E 00
(
Ui(x)
')
TJiOCOSWit
Example 9.5
Find the free vibration response of a unifonn initial displacement and velocity.
SOLUTION The free vibratory motion of the free-free expressed, using Eq. (9.24), as un(x, t)
= Un(x)Tn(t) = cos --n7rX I
(n7rc C cos n
--t
I
+ :~ sin Wit
(9.62)
bar with free ends due to
bar in the nth mode can be
+ DII sin-tn7rC I
) (E9.5.1)
9.3
Free Vibration Solution and Natural Frequencies
251
where Cn and Dn are constants. By superposing the solutions given by Eq. (E9.5.1), we can represent any longitudinal vibration of the bar in the form 00
u(x,
""
t) = ~
nrrx -z-
cos
(nrrc Cn COS
nrrc ) -z-t + Dn sin -z-t
(E9.5.2)
n=1
where the constants Cn and Dn can be determined from the initial conditions specified. If the initial displacement and initial velocity of the bar are specified as u(x,O)
ou ot
= uo(x),
0)
-(x,
. = uo(x)
(E9.5.3)
-z-
(E9.5.4)
then Eq. (E9.5.2) gives 00
U(x, 0)
= uo(x) = ~
"" Cn COS nrr x
n=1
ou -(x,O)
.
ot
nrrc nrrx L: -Z-Dn cosn=1 I 00
= uo(x)
=
(E9.5.5)
To determine the constant Cn in Eq. (E9.5.4), we multiply both sides of Eq. (E9.5.4) by the mth mode shape, cos(mrrx/ I), and integrate from 0 to I:
t
mrrx uo(x) cos -1- dx
o
nrrx mrrx = 10t ~ ~ Cn COS -z- cos -1- dx
.0
n=1
Noting that 1
nrrx
1
mrrx
cos--cos-o I
I
=
I
0, m =1= n I - m=n 2'
(E9.5.6)
(E9.5.7)
Eq. (E9.5.6) can be simplified to obtain Cn =
T2 10t uo(X)
nrr x cos -1- dx
(E9.5.8)
Using a similar procedure, the constant Dn in Eq. (E9.5.5) can be determined as Dn
= -2
nrrc
1
1
• nrrx uo(X) cos -1- dx
(E9.5.9)
0
Example 9.6 Consider a free-free bar of uniform cross-sectional area. It is subjected to an axial compressive force at each end. Find the free vibration response of the bar when the forces are suddenly removed. SOLUTION We assume that the middle of the bar remains stationary. The displacement of the bar just before the forces are removed (one-half of the initial displacement .at each end) is given by Uo
sol 2
= U(x, 0) = - -
sox
(E9.6.l)
,
I
I
j 1
j
252
Longitudinal Vibration of Bars
and the initial velocity by
.
Uo
au
= -at (x,
0)
=0
(E9.6.2)
where £0 denotes the compressive strain at the ends at time t = O. Using Eqs. (E9.5.8) and (E9.5.9) and Eqs. (E9.6.1) and (E9.6.2), the constants Cn and Dn can be evaluated as Cn
=~
1/
(£~1 -
sox) cos n~x
£011/
=cos--mfX d (nnx) -nrr 0 1 1 0,
dx -2£0-- 1
n2rr2
1/ 0
nrrx nrrx (nnx) --cos--d-1
1
1
n is even
4£01 - { --
n2rr2 '
(E9.6.3)
n is odd
Dn = 0
(E9.6.4)
Thus, the general solution for the longitudinal vibration of the free-free bar can be expressed as [see Eq. (E9.5.2)] u(x t)
,
4Eol = -n2
L 00
n=1.3.5 •...
-n12
nrr x nn ex cos -cos-1
1
(E9.6.5)
Example 9.7 A bar of uniform cross-sectional area A, length 1, modulus of elasticity E, and density p is fixed at both ends. It is subjected to an axial force Fo at the middle [Fig. 9.5(a)] and is suddenly removed at t = O. Find the resulting vibration of the bar.
(a)
~
I-
t
-I-
f
-I
(b)
Figure 9.5 tion. uo(x).
(a)
Bar subjected to axial force
Fo
at the middle;
(b)
initial displacement distribu-
9.3
SOLUTION
Free Vibration Solution and Natural Frequencies
253
The tensile strain induced in the left half of the bar is given by Fo e=--
(E9.7.1)
2EA
which is equal in magnitude to the compressive strain in the right half of the bar. Thus, the initial displacement of the bar can be expressed as [see Fig. 9.5(b)] Fox
u(x, 0) = uo(x) =
-
(E9.7.2)
Fo(l - x) e(l - x) = '2EA '
\
/ - 2
O
ex = -2EA'
Z
-
2-
-
Since the initial velocity is zero, we have
au at
.
(E9.7.3)
0) = uo(x) = 0,
-(x,
To find the general solution of the bar, we note the boundary conditions u(O, t)
= 0,
(E9.7.4)
u(l, t)
= 0,
(E9.7.5)
The use of Eq. (E9.7.4) in Eq. (9.24) gives !\ = 0, and the use of Eq. (E9.7.5) gives the frequency equation: . wZ c
Bsm- =0 -
or
. wZ
S1O-=
0
(E9.7.6)
C
The natural frequencies are given by wnl - = n1f,n = 1, 2, ... C
or n1fC
Wn
= -Z-,
n = 1,2, ...
(E9.7.7)
and the corresponding mode shapes by n1fx Un(x) = §n sin -t-'
n
= 1,2, ...
(E9.7.8)
The general free vibration solution of the bar can be expressed using the mode super~ position approach as u(x,t)
00
00
n=l
n=l
= L:>n(X,t) = Lsin-l-
n1fX (
n1fct . n1fct) Cncos-Z+Dnsm-Z-
(E9.7.9)
254
Longitudinal Vibration of Bars
Using the initial velocity condition, Eq. (E9.7.3), in Eq. (E9.7.9) gives Dn = O. The use of the initial displacement condition, Eq. (E9.7.2), in Eq. (E9.7.9) yields 12 /
21 Cn = 1
Fox. nrrx --sm--dx+2EA
0
2
12 /
Fo 1 = EAI --n2rr2
1
1
II . II
Fo 1 ---2 2 EAI n rr
1/2
1
(nrrx) 1
1
nrrx . nrrx d (nrrx) --sm--_ 1
2Fol (_l)(n-l)/2 -A-E-rr-2 n2 {
EA
(nrrx)
1
1/2
2
Fo(l- x) . nrrx ---sm--dx 1/2
1
nrrx sm--d-_
EA nrr
=
1
nrrx . nrrx --sm--d--
0
1
Fol + --
2fl
1
1
1
..
If n IS odd (E9.7.1O)
o
if n is even
Thus, the free vibration solution of the bar becomes 2Fol u(x, t) = AErr2
~ L.,
(_l)(n-l)/2
nrrx
nrrct
1
1
n
n=I,3,5, ...
9.4
.
----sm--cos2
(E9.7.1l)
FORCED VffiRATION The equation of motion for the longitudinal vibration of a prismatic bar subjected to a distributed force f(x, t) per unit length can be expressed in a general form as -Ru"(x, t)
+ Mii(x, t) = f(x,
t)
(9.63)
or
-c2u"(x, t)
+ ii(x, t)
= lex, t)
(9.64)
where R and M are given by Eq. (9.30), c by Eq. (9.35), and
f=.£=1M pA
(9.65)
-
In modal analysis, the forced vibration response is assumed to be given by the sum of products of normal modes and generalized coordinates as indicated by Eq. (9.49). By substituting Eq. (9.49) in Eq. (9.64) for u(x, t), multiplying by Uj(x), and integrating from 0 to I, we obtain
1
1
2
1)-C 00
;=1
=
1];
U!'(x)Uj(x)dx
0
Jto Uj(x)f(x, -
+;;; _
t)dx
1
1
U;(x)Uj(x)dx]
0
(9.66)
,.,-."'"","""~",.~"~."."""",""",",,,,,,p
"M' '"W"';"'_"""l"
".".",~""""""",!~~"'"""",,","""·"~"'~~""""~"'"""~""".,~,,
'," ,-,,,.,1"',,,",,,,,"''''
••
9.4
Forced Vibration
~"••-\
2SS
In view of the orthogonality relations, Eqs. (9.43) and (9.45), Eq. (9.66) becomes (for i = j) iji
=
+WrT/i
10t Ui(x)f(x,t)dx -
(9.67)
Equation (9.67) represents a second-order ordinary differential equation for the gener~ alized coordinate T/i(t). The solution of Eq. (9.67) can be obtained, using a Duhamel integral, as (9.68)
Thus, the total steady-state forced longitudinal vibration response of the bar is given by (ignoring the effect of initial conditions) u(x, t)
= L U·(x) 00
_1_
it it Ui(X)
Wi
i=l
0
r) sin Wi(t - t) dr dx
f(x,
(9.69)
0 -
=
Note If the bar is subjected to an axial concentrated force Fm(t) at x xm, there is no need for integration over the length of the bar, and Eq. (9.69) takes the form u (x, t)
it -
oo Ui(X )Ui(x = xm) =L . '-", Wi 1=1
Fm(t).
0
'"
pA
sm Wi
(t
-
t
)d
r
(9.70)
Example 9.8
Consider a prismatic bar fixed at both ends. Find the steady-state response of the bar if the following loads are applied suddenly at the same time (see Fig. 9.6): a uniformly distributed longitudinal force of magnitude fo per unit length, and an axial concentrated force Fo at the middle point of the bar, x 1/2.
=
~:- - - - ...:'!:~ ~~ - - 0F'-'-'~x --Fo
I
Figure 9.6
~I
~
~
-I .
-\
Bar subjected to distributed and concentrated loads.
SOLUTION We can find the steady-state response of the bar by superposing the responses due to the two loads. To find the response due to the uniformly distributed load, we use fo f(x, r) = -A p in Eq. (9.69) to obtain ~ Ui(X) u(x, t) = ~ -i=l
Wi
it 0
Ui(X)-
fo
pA
it .
smWi(t - r) dr dx
(E9.8.!)
(E9.8.2)
0
------------------------------,
;
256
Longitudinal Vibration of Bars
where from the free vibration analysis [Eq. (E9.7.8)] we have . ir{X
= Bj sm-
Uj(x)
(E9.8.3)
/
irrc Wj=-
(E9.8.4)
/
and Bj is a constant (i = 1,2, ... ). When Uj(x) is normalized as
1/
=I
Ul(x)dx
or
(E9.8.S)
/
irrx sin2 dx =1
1
B~
10/
(E9.8.6)
we obtain
(E9:8.7) and hence irrx = ~. -sm-
Uj(x)
/
(E9.8.8)
/
Thus, Eq. (E9.8.2) becomes u(x, t) =
~ ~ --- / L.-
/ nrrc
n=l =
. -nrr x sm
/~.
0
4/0/2 1 ---'-sm-L rr n=1,3,S,... c pA n 3 2
nrr x _. -/ sm ----
1"
/
.
3
/ mfX
/
(
it .
/0
smwn(t _ r:) dr: dx
pA
I-cos--
T=O
nrrct)
(E9.8.9)
/
To find the response of the bar due to the concentrated load, we use F m (r:) Eq. (9.70), so that u(x, t) =
=
o o~Sin(nrr
L n=l
xl /)-J2i7 sin(nrr 12) Fo
------------/
nrrcl/
pA
= Fo
in
it° .
smwn(t _ r:)dr:
2
"" 2 F.20/ 12sin ~(_1)(n-l)/2 L.rr c pA n I n=I.3.S....
(1 - cos~)
/
(E9.8.l0)
Thus, the total response of the bar is given by the sum of the two responses given by Eqs. (E9.8.9) and (E9.8.1O): u(x,t)
=
2/ 2 2 rr c pA
L sm-. nrrx n=1,3.S,...
I
( I-cos-
nrrct) /
0/
[2/ __
rrn3
Fo + _(-l)(n-I n2
)/
2J
(E9.8.II)
9.5
9.5
Response of a Bar Subjected to Longitudinal S~ppon Motion
257
RESPONSE OF A BAR SUBJECTED TO LONGITUDINAL SUPPORT MOTION
=
Let a prismatic bar be subjected to a support or base motion, Ub(t) p(t) in the axial direction as shown in Fig. 9.7. The equation of motion for the longitudinal vibration of the bar can be obtained as
a2u(xat '
t)
a2
- EA ax2 [u(x, t) - Ub(t)] = 0
(9.71)
By defining a new variable v(x, t) that denotes the displacement
of any point in the
pA
2
bar relative to the base as v(x, t)
(9.72)
= u(x, t) - Ub(t)
we can write
a
2v a2U at2 = at2
+
a2 p at2
(9.73)
Using Eqs. (9.72) and (9.73), Eq. (9.71) can be rewritten as a2v(x, t) EA a2v(~, t) pA at2 ax2
=
2 A a p(t) -p ~
(9.74)
A comparison of Eq. (9.74) with Eq. (9.10) shows that the term on the right-hand side of Eq. (9.74) denotes equivalent distributed loading induced by the base motion. By dividing Eq. (9.74) by pA, we obtain a2v at2 -
2
c
2 a2v a p ax2 = - at2
(9.75)
Since Eq. (9.75) is similar to Eq. (9.64), we can find the equation for the ith generalized coordinate
TJi(t) in Eq. (9.49) as 2 a2TJi 2 a p at2 +WiTJi = - at2
10t Ui(x)dx,
The solution of Eq. (9.76) can be expressed, TJi(t). = -- 1 Wi
_---.
",,(t)
1 0
= 1,2, ...
1a 0
2
p nwi(t 7T(r)si lJt
= p(t) Figure 9.7
(9.76)
using a Duhamel integral, as
t
1
Ui(x)dx
i
Bar with support motion.
- r)dr
(9.77)
258
Longitudinal Vibration of Bars for v(x, t) can be obtained
The total solution responses as
=-
v(x, t)
u.(x) L.....!..-. 00
j=1
Finally, the longitudinal
Wj
1/
Uj(x)dx
l'
0
by superposing
82p -a-r(r) sin
(t - r)dr
(9.78)
t
0
vibrational
Wj
all the nonnal-mode
motion of the bar can be found from Eq. (9.72) as
U(x, t) = Ub(t) + v(x, t)
(9.79)
9.6 RAYLEIGH THEORY 9.6.1
Equation of Motion In this theory, the inertia of the lateral motions by which the cross sections are extended or contracted in their own planes is considered. But the contribution of shear stiffness to the strain energy is neglected. An element in the cross section of the bar, located at the coordinates y and z, undergoes the lateral displacements -vy(8uj8x) and -vz(auj8x), respectively, along the y and z directions, with v denoting Poisson's ratio [2, 3, 6]. Thus, the displacement field is given by U
8u(x,t) v = -vy- - 8x
= u(x, t),
w
8u(x, t) = -vz---
(9.80)
8x
The strain energy of the bar and the. work done by the external forces are given by Eqs. (9.6) and (9.8), while the kinetic energy of the bar can be obtained as
T=2Jo1
t dx JfAo
= -211/0
pdA
pA (au)2 at
11/ pA (8u)2 at
=2
at
(8u)2
dx
11/ l 11/ (
2
2
pdA
A
+-
dx+-
0
1 t fA +2Jo dx Jo
dx
0
0
pdA
2 -vy-- a u)2 ax8t
[(
0
pv2/p
at
[(av)2
+ (8W)2] at + ( -vz-- a2u )2] ax8t
2u )2 dx 8x8t
--8
(9.81)
where / p is the polar moment of the inertia of the cross section, defined by
(9.82) The application
of extended Hamilton's
1" l' [1
o - dt '1
0
- pA (au)2 2 8t
1
principle
+ -2 pv2 / p
gives
2 ( 8xat a u ). I (au)2 -2EA ax --
2
-
.
+ f u]
dx
=0 (9.83)
9.6
Rayleigh Theory
-Z;;9
yielding the equation of motion and the boundary conditions as __ o
ox
(2
u pv / --o3 )
OU ( EAox
(OU) EA-
- -0
P oxot2
oxot2
ox
+ pA-o2u
=f
II = 0
u
o3 ) + pv 2 /P--2
(9.85)
OU
oxot
(9.84)
0
Note that Eq. (9.85) is satisfied if the bar is either fixed or free at the ends x = 0 and
x = l. At a fixed end,
U
= 0 and
hence
OU
= 0,
while
OU 2 o3u E/ ox +pv /poxot2
(9.86)
=0
at a free end.
9.6.2
Natural Frequencies and Mode Shapes For the free axial vibration of a uniform bar, we set
f
= 0 and Eqs. (9.84) and (9.85)
reduce to (9.87)
(9.88)
The natural frequencies
of the bar can be determined
using a harmonic solution (9.89)
u(x, t) = U(x) coswt Using Eq. (9.89), Eq. (9.87) can be expressed as (_pv2/pw2 The solution of the second-order
d2U
+ EA) dx2 + pAw2U = 0
ordinary differential
(9.90)
equation, Eq. (9.90) can be writ-
ten as
U(x) = Cl
COS
px
+ C2 sin px
(9.91)
where p= and C I and C2 are constants to be determined
__ ----------t
(9.92)
from the boundary conditions of the bar.
260
Longitudinal Vibration of Bars
Bar with Both Ends Fixed
For a bar fixed at both ends,
U (x
= 0) = 0
= I) = 0
U (x
and
(9.93)
Equations (9.91) and (9.93) lead to
Equation (9.95) gives the frequencies pi =
nn,
C1 = 0
(9.94)
sin pi = 0
(9.95)
of vibration:
n = 1,2, ...
or
n = 1,2, ... The mode shape corresponding
to the frequency
Un(x) = sinnnx,
Wn
(9.96)
is given by
n = 1,2, '"
(9.97)
It can be seen that the mode shapes [Eq. (9.97)J are identical to those given by the simple theory, whereas the natural frequencies [Eq. (9.96)J are reduced by the factor
( 1+
·2/ v pn 2n 2) 1/2 A[2
compared to those given by the simple theory.
9.7 BISHOP'S THEORY 9.7.1
Equation of Motion This theory considers the effect not only of the inertia of the lateral motions but also of the shear stiffness [1, 3, 6J. The displacement field is given by Eq. (9.80). The strains in the cross section can be obtained as
exx
exy
=
au ax'
eyy
av = -ay =
-v
2u .(au av) a = ay + ax = -vy ax2' ezx
au ax
-,
ezz
eyZ
au = -aw az = -v-, ax
aw) = 0, = (avaz + a;
= (au + aw) = -vz a2u az
ax
ax2
(9.98)
9.7
The stresses induced in the cross section of the bar can be determined. three-dimensional Hooke's law. as axx
1- v
v
v v
I- v
0 0 0
0 0 0
an' a:: axy ay: a:x
=
Substitution
E (l
+ v)(l
- 2v)
v v I-v 0 0 0
v
l-Zv -Z-
0 0 0 0 0
0
l-Zv -Z-
0 0 0 0
0 0 0 1-2v -2-
0 0
26J
Bishop's Theory
using the
(9.99)
of Eq. (9.98) in Eq. (9.99) results in
(9.100)
=
The strain energy of the bar can be computed
as
(9.101) The kinetic energy of the bar and the work done by the external forces are given by Eqs. (9.81) and (9.8), respectively. The extended Hamilton's principle can be expressed as t2
8
1
(9.102)
(T-rr+W)dt=O
tl
By substituting Eqs. (9.81), (9.101), and (9.8) for T, rr, and W, respectively, in Eq. (9.102) and simplifying results in the following equation of motion and the associated boundary conditions,
2 ( a2u) vZGI _. axZ p axZ
_a
( v plp--a3u) ax axatZ
- -a
2
- -a
ax
( EA-au) ax
+pA-aZu Z =
at
f
(9.103)
it can be seen that if an end is rigidly fixed,
o in
Eq. (9.104).
u
=
au/ax
= 0 and hence
8u
= 8(au/ax)
=
262 9.7.2
Longitudinal Vibration of Bars
Natural Frequencies and Mode Shapes For a unifonn bar undergoing free vibration written as 4 a u a4u a2u 2 2 v G!p ax4 - pv !p ax2at2 - EA ax2 3
EA(au
ax
+ V2p!p-a a au 2
-
x t
(f = 0), Eqs. (9.103) and (9.104) can be a2u 8t2 = 0
+ pA
3 2 v G!p_ a -3u) 8u II ax 0
+(
(9.105)
2 v2G!P-2a u) 8 (au)/I ax ax
0
=0
(9.106)
The natural frequencies of the bar can be found using a harmonic solution: u(x, t) = U(x)coswt
(9.107)
Substitution of Eq. (9.107) into (9.105) leads to 4
d U 2 v G!p dx4
+ (pv2!pw2
d2 U - EA) dx2 - pAw2U
=0
(9.108)
By assuming the solution of Eq. (9.108) as U(x)
= CePx
(9.109)
where C and P are constants, the auxiliary equation can be obtained as 2 v G!pp4
+ (pv2!pw2
- EA)p2 - pAw2 = 0
(9.110)
Equation (9.110) is a quadratic equation in p2 whose roots are given by 2 (EA - pv !p(2)
2
± J(EA
P =
- pv2!p(2)2 2 2v G!p
+ 4v2G!ppAw2 = a ±b
(9.111)
where
a=
EA - pv2! w2 p
(9.112)
2v2Glp
b = J(EA
- pv2!p(2)2 + 4v2G!ppAw2 2v2G!p
(9.113)
Since b > a, the roots can be expressed as PI = -P2
=
Sj
=.Ja + b,
(9.114)
Thus, the general solution of Eq. (9.110) can be written as U(x) = CleS1X
+ C2e-S1X + C3eis2 + C4e-is2
(9.115)
where the constants Cj, C2, C3, and C4 are to be determined from the boundary conditions of the bar. Noting that sinh x t(eX - e-X), cosh x = t(eX + e-X), sinx iX ix iX Oj2i)(e - e- ), and cosx = t(e + e-ix), Eq. (9.115) can be rewritten as
=
U(x) = Cj coshsjx
+
C2 sinhslx
where CI, C2, C3, and C4 are constants.
=
+
C3 COSS2X + C4 sins2x
(9.116)
, """~~,
~".,.,"'rI:~'!'1"1""'"~ l_I'1'_>"?";;l,~."i,""~\;,:'rl,"':':':,!",,,"'l~~~t'.,~.".';l~r.'v,;_' .,~;',I,r~~~'-rt'C,f-~",,-,,~,~-,
•• ~~,~~~~~~,"",,-'1q~';';"4>-t"~,,,-,..,
.
.
1
..,,~.... ~.-<.l~i'-"";:l'-~}·~-·~""~·i,~_·l.~'r~.'·"'-'·'"'-'-~·""'-_~·':'·'f!!~, •.• '-,<,~,'~'\J''':~'~ '7"~~~'_1'-""'_'-."'-,,-. "''''.•..•,,-,.,.~-~,11_.
I,
9.7 Bishop's Theory~§3
Bar Fixed Loosely at Both Ends If the bar is fixed loosely at both ends, the axial displacem~nt and shear strain will be zero at each end, so that
=0 U (1) = 0
(9.117)
dZU dxz (0) = 0
(9.119)
dZU
(9.120)
U (0)
-z dx
(1)
(9.118)
=0
Equations (9.116)-(9.120) lead to CI
+ Czsr SinhSll -
(9.121)
=0
(9.122)
Cisr - c3si = 0
(9.123)
c3si cosszl - c4si sinszl = 0
(9.124)
+ Cz sinhsll + C3cos szl + C4 sinszl
CI coshsil
Cisr coshsll
+ C3 = 0
Equations (9.121) and (9.123) give Cl = C3
(9.125)
=0
and Eqs. (9.122) and (9.124) reduce to
+ C4 sinszl = 0
(9.126)
Czsr sinhsll - c4si sinszl = 0
(9.127)
Cz sinhsll
The condition for a nontrivial solution of Cz and C4 in Eqs. (9.126) and (9.127) is
1-
sinhsll sinszl 0 sr sinh sll -si sin szl \ or (9.128) Since sinhsll
#- 0 for nonzero
values of sll, Eq. (9.128) leads to the frequency equation sinszl = 0
(9.129)
The natural frequencies are given by szl = nrr,
n
= 1,2, ...
(9.130)
Using Eqs. (9.114), (9.112), and (9.113) in (9.130), we can express the natural frequencies as (see Problem 9.6) (J)z
n
nZrrzE (AElZ = -plz AElz
+ vZGl nZrrZ) p + vZElpn2rr2
(9.131)
264
Longitudinal Vibration of Bars
The mode shape corresponding to the frequency Un(x) = sinmrx,
Wn
is given by
n = 1,2, ...
(9.132)
It can be observed that the mode shapes [Eq. (9.132)J are identical to those given by the simple theory, whereas the natural frequencies [Eq. (9.131)J are reduced by the factor
compared to those given by the simple theory.
9.7.3
Forced Vibration Using Modal Analysis The equation of motion of a prismatic bar in longitudinal vibration, Eq. (9.105), can be expressed as Mu
+ Lu
=
f
(9.133)
where
(9.134) (9.135) In modal analysis, the solution is expressed as the sum of natural modes as 00
u(x, t) =
L Ui (X)71i (t)
(9.136)
i=l
so that the equation of motion for the ith normal mode becomes
+ (L[Ui(X)])71i(t)
(M[Ui(X)])~i(t)
= f(x,
t)
(9.137)
By multiplying Eq. (9.137) by Uj(x) and integrating from 0 to l, we obtain
l'
(M[U, (x)])Uj (x)., (I) dx
+
l'
(L[ U, (x)])Uj (x)", (ll dx =
l'
f (x. t}Uj
(x) dx
(9.138) In view of the orthogonality relations among natural modes, Eq. (9.138) reduces to Mi~i
+ Ki71i = j;,
i = 1,2, ...
(9.139)
9.7
Bishop's Theory
where Mj is the generalized mass, Kj is the generalized stiffness, and alized force in the ith mode, given by
II = II = II
Ij
·.uS
is the gener-
Mj =
(M[Uj(x)])Uj(x)dx
(9.140)
Kj
(L[Uj (x)])Uj (x) dx
(9.141)
I(x,
(9.142)
Ii
t)Ui(X) dx
The solution of Eq. (9.139) can be expressed, using a Duhamel integral, as i/i(O) T/i(t) = T/oCOSWjt + Wi
where
Wi
.
Sill
Wit
it .
+ -- 1
Mi Wi
Hi
Ii
(1:)
SlfiWi(t - r)d1:,
0
i = 1,2, ...
(9.143)
is the ith natural frequency given by
,-
" •. -
__ I
Mj
i
,
= 1,2, ...
(9.144)
and T/i(O) and ~i(O) are the initial values of the generalized displacement T/i(t) and generalized velocity 1Ji(t). If uo(x) = u(x, 0) and uo(x) = u(x, 0) are the initial values specified for longitudinal displacement and velocity, we can express 00
uo(x) =
L Ui (X)1/i (0)
(9.145)
i=l 00
uo(x)
=
L Ui(X)~i(O)
(9.146)
i=l
Multiplying Eqs. (9.145) and (9.146) by M[Uj(x)]
and integrating from 0 to I results
lfi
{lOOt
10
uo(x)M[Uj(x)]dx
toot
10
uo(x)M[Uj(x)]dx
= {;
10
T/i(O)Ui(x)M[Uj(x)]dx
= {;
10
~j(O)Ui(x)M[U/x)]dx
(9.147)
(9.148)
When the property of orthogonality of normal modes is used, Eqs. (9.147) and (9.148) yield
i i
l
1/i(O) = -1
Mi
uo(X)M[Ui(X)]dx,
i = 1,2, ...
(9.149)
uO(X)M[Ui(x)ldx,
i
= 1,2, ...
(9.150)
0 l
~j(O) = -1 Mj
0
266
Longitudinal Vibration of Bars
Finally, the total axial motion (displacement) of the bar can be expressed as 00
u(x, t) = '"
[
17j(O) COSWjt
l'
+ ~ (0) sinwjt + -- 1 •.
M·w· I I
W·I
;=1
fj(T)
sinwj(t
- T) dT
]
Uj(x)
0
(9.151) Example 9.9 Determine the steady-state response of a prismatic bar fixed loosely at both ends when an axial force Fo is suddenly applied at the middle as shown in Fig. 9.8.
Figure 9.8
Bar supported loosely at ends.
SOLUTION The natural frequencies and normal modes of the bar are given by Eqs. (9.131),and (9.132): i2rr2 E AEl2 + v2G I i2rr2 w2 = -p p[2 AEl2+ v2Elpi2n2
(E9.9.1)
I
Ui(X) = sinirrx
(E9.9.2)
The generalized mass Mi and the generalized stiffness Ki in mode i can be determined as
M,
=
l
(M[U, (x)])U, (x) dx
=
l [(PA -
= (pA K, =
l
(L[U, (x)])U, (x) dx =
v' pIp
+ v2plpi2rr2)~
a~') sin i" x ]
dx
(E9.9.3)
2
l [(v'C Ip a::- a~') E1
2 = (v Glpi4rr4
sin in
+ EIi2rr2)_
l 2
sin in ] sin i" x dx (E9.9.4)
The applied axial force can be represented as
f(x,
t)
= FoH(t)8
(x - ~)
(E9.9.5)
9.8
Recent Contributions
267
where H(t) is the Heaviside unit step function and 0 is the Dirac delta function. The generalized force in the i th normal mode can be computed as
=
fi(t)
1
1
I(x,
t)Uj(x)dx
= FoH(t)Uj Thus, the steady-state
=
1
1
FoH(t)o (x -~)
Uj(x)dx (E9.9.6)
(x = ~) = FoH(t) sin i~l
solution of the ith generalized
coordinate
is given by the solution
of Eq. (9.139) as T/j(t) = _1_ MjWj
t fi(r)
10
i = 1,2, ...
sinwj(t - r) dr,
(E9.9.7)
which can be written as
1
I1j(t) = -Mjwj
=
1
t
• . d sm -irrl smwj(t - r) r 2
17
roH(r)
0
Fo sin(irrlf2)
M~
1
t
H(r) sinwj(t - r)dr =
2
M~
0
Thus, the total steady-state
(1- COSWjt) (E9.9.8)
response of the bar is given by
u(x, t)
~
= £...J j=l
9.8
Fo sin(irrlf2)
Fosin(irrlf2) 2
MjWj
(1 - cos Wjt)
(E9.9.9)
RECENT CONTRIBUTIONS Additional problems of longitudinal vibration, including the determination of the natural frequencies of nonuniform bars, and free and forced vibration of uniform viscoelastic and viscoelastic ally coated bars, are discussed in detail by Rao [3]. A comparative evaluation of the approximate solutions given by discretization methods such as the finite element and finite difference methods for the free axial vibration of uniform rods was made by Ramesh and Itku [14]. The solution of the wave equation, which describes the axial free vibration of uniform rods in terms of eigenvalues and eigenfunctions, was used as a basis for comparison of the approximate solutions. It was observed that the frequencies given by the discretization methods were influenced significantly and the mode shapes were relatively insensitive to the choice of mass lumping scheme. Kukla et al. [15] considered the problem of longitudinal vibration of two rods coupled by many translational springs using the Green's function method. The frequencies of longitudinal vibration of a uniform rod with a tip mass or spring was considered by Kohoutek [8]. Raj and Sujith [9] developed closed-form solutions for the free longitudinal vibration of inhomogeneous rods. The longitudinal impulsive response analysis of variable-cross-section bars was presented by Matsuda et al. [10]. Exact analytical solutions for the longitudinal vibration of bars with a nonuniform cross section were presented by Li [11] and Kumar and Sujith [12].
\
"
268
Longitudinal Vibration of Bars
The solutions are found in terms of special functions such as the Bessel and Neumann as well as trigonometric functions. Simple expressions are given to predict the natural frequencies of nonuniform bars with various boundary conditions. The equation of motion of a vibrating Timoshenko column is discussed by Kounadis in Ref. [13].
REFERENCES 1. R. E. D. Bishop, Longitudinal waves in beams, Aeronautical Quarterly, Vol. 3, No.2, pp. 280-293, 1952. 2. J. W. S. Rayleigh, The Theory of Sound, Dover, New York, 1945. 3. J. S. Rao, Advanced Theory of Vibration, Wiley Eastern, New Delhi, India, 1992. 4. A. E. H. Love, A Treatize on the Mathematical Theory of Elasticity, 4th ed., Dover, New York, 1944. 5. S. Timoshenko, D. H. Young, and W. Weaven, Jr., Vibration Problems in Engineering, 4th ed., Wiley, New York, 1974. 6. E. Volterra and E. C. Zachmanoglou, Dynamics of Vibrations, Charles E. Merrill, Columbus, OH, 1965. 7. S. K. Clark, Dynamics of Continuous Elements, Prentice-Hall, Englewood Cliffs, NJ, 1972. 8. R. Kohoutek, Natural longitudinal frequencies of a uniform rod with a tip mass or spring, Journal of Sound and Vibration, Vol. 77, No.1, pp. 147-148, 1981. 9. A. Raj and R. I. Sujith, Closed-form solutions for the free longitudinal vibration of inhomogeneous rods, Journal of Sound and Vibration, Vol. 283, No. 3-5, pp. 1015-1030,2005. 10. H. Matsuda, T. Sakiyama, C. Morita, and M. Kawakami, Longitudinal impulsive response analysis of variable cross-section bars, Journal of Sound and Vibration, Vol. 181, No.3, pp. 541-551,1995. 11. Q. S. Li, Exact solutions for free longitudinal vibration of non-uniform rods, Journal of Sound and Vibration, Vol. 234, No.1, pp. 1-19,2000. 12. B. M. Kumar and R. I. Sujith, Exact solutions for the longitudinal vibration of non-uniform rods, Journal of Sound and Vibration, Vol. 207, No.5, pp. 721-729, 1997. 13. N. Kounadis, On the derivation of equation of motion for a vibrating Timoshenko column, Journal of Sound and Vibration, Vol. 73, No.2, pp. 174-184, 1980. 14. A. V. Ramesh and S. Utku, A comparison of approximate solutions for the axial free vibrations of uniform rods, Journal of Sound and Vibration, Vol. 139, No.3, pp. 407 -424, 1990. 15. S. Kukla, J. Przybylski, and L. Tomski, Longitudinal vibration of rods coupled by translational springs, Journal of Sound and Vibration, Vol. 185, No.4. pp. 717-722, 1995.
PROBLEMS 9.1 Derive the frequency equation for the longitudinal vibration of the bar shown in Fig. 9.9. 9.2 Derive the equation of motion for the longitudinal vibration of a bar by including the damping force that is proportional to the longitudinal velocity. 9.3 A unifonn bar is fixed at one end and free at the other end. Find the longitudinal vibration response of
the bar subject to the initial conditions u(x, 0) and u(x, 0) = O.
= Uox2
9.4 Consider a unifonn bar fixed at one end and carrying a mass M at the other end. Find the longitudinal vibration response of the bar when its fixed end is subjected to a harmonic axial motion as shown in Fig. 9.10.
Problems
269
Figure 9.9
... u(t)
= Uosin
~t
~--~~.:, I~ Figure 9.10
9.5 Specialize Eq. (E9.2.10) to the case where the mass of the bar is negligible compared to the mass attac~ed. Solve the resulting equation to find the fundamental frequency of vibration of the bar.
9.11 Derive Eqs. (9.84) ton's principle.
9.6
9.13 Consider a uniform free-free bar. If the ends x 0 and x 1 are subjected to the displacements u(O, t) = UleiP.1 and u(t, t) U2eml. determine the axial motion of the bar, u(x, t), 0 < x < I, t > O.
Derive Eq. (9.131).
9.7 Show that the expressions on either side of the equality sign in Eq. (9.19) is equal to a negative quantity. 9.8
Derive Eqs. (9.10) and (9.11) from Eq. (9.9).
9.9 Derive the frequency equation for the longitudinal vibration of a uniform bar fixed at x 0 and attached to a mass M and spring of stiffness k at x 1 (case 5 of Table 9.1).
=
=
9.10 Derive the frequency equation for the longitudinal vibration of a uniform bar free at x 0 and attached to a mass M at x = 1 (case 7 of Table 9.1).
=
h-.--'--'--X
and
(9.85) from Hamil-
9.12 Derive Eqs. (9.103) and (9.104) from Hamilton's principle.
=
=
=
9.14 A uniform bar fixed at x = 0 and free at x = 1 is subjected to a distributed axial force f (x, t) x2 sin 2t. Determine the resulting axial motion of the bar.
=
9.15 The ends of a uniform bar are connected to two springs as shown in Fig. 9.11. Derive the frequency equation corresponding to the axial vibration of the bar.
A,E.p
_~~\_1( -.~ Figure 9.11
270
Longitudinal Vibration of Bars
=
9.16 A uniform bar is fixed at x 0 and is subjected to a sudden axial force fo (shown in Fig. 9.12) at x 1. Find the ensuing axial motion of the bar at x = I.
=
9.17 A uniform bar of length I, cross-sectional area A, Young's modulus E, and mass density p strikes a spring of stiffness k with a velocity V as shown in Fig. 9.13. Find the resulting axial motion of the bar, u(x, t), measured from the instant the bar strikes the spring. 9.18 A uniform bar of length I, cross-sectional area A, Young's modulus E, and mass density p is fixed at x 0 and carries a mass M at x I. The end x I is subjected to an axial force F(t) Fo sin Qt as shown in Fig. 9.14. Determine the steady-state response, u(x, t), of the bar.
=
=
=
=
9.19 Find the longitudinal vibration response of a uniform bar of length I, fixed at x = 0 and free at x = I, when the end x = 0 is subjected to an axial harmonic displacement, Ub (t) = c sin wt where c is a constant and w is the frequency. 9.20 Find the steady state axial motion of a prismatic bar of length I, fixed at x 0, when an axial force F(t) = Fo acts at the end x = I using the Laplace transform approach.
=
9.21 Derive the orthogonality relationships for a bar, fixed at x 0, carrying a mass ¥ at x I.
=
=
/(t)
fa
o Figure 9.12
k
Figure 9.13
I:-IA~'~
0~=FoSinnt
x
)I
/Figure 9.14
1'0 Torsional Vibration of Shafts 10.1 INTRODUCTION Many rotating shafts and axles used for power transmission experience torsional vibration, particularly when the prime mover is a reciprocating engine. The shafts used in high-speed machinery, especially those carrying heavy wheels, are subjected to dynamic torsional forces and vibration. A solid or hollow cylindrical rod of circular section undergoes torsional displacement or twisting such that each transverse section remains in its own plane when a torsional moment is applied. In this case the cross sections of the rod do not experience any motion parallel to the axis of the rod. However, if the cross section of the rod is not circular, the effect of a twist will be more involved. In this case the twist will be accompanied by a warping of normal cross sections. The torsional vibrations of uniform and nonuniform rods with circular cross section and rods with noncirctilar section are considered in this chapter. For noncircular sections, the equations of motion are derived using both the Saint-Venant and Timoshenko-Gere theories. The methods of determining the torsional rigidity of noncircular rods is presented using the Prandtl stress function and the Prandtl membrane analogy .
10.2 ELEMENTARY THEORY: EQUATION OF MOTION 10.2.1
Equilibrium Approach Consider an element of a nonuniform circular shaft between two cross sections at x and x + dx, as shown in Fig. 1O.l(a). Let Mt(x, t) denote the torque induced in the shaft at x and time t and Mt(x, t) + dMt(x, t) the torque induced in the shaft at x + dx and at the same time t. If the angular displacement of the cross section at x is denoted as O(x, t), the angular displacement of the cross section at x + dx can be represented as B(x, t) + dB(x, t). Let the external torque acting on the shaft per unit length be denoted mt(x, t). The inertia torque acting on the element of the shaft is given by 10dx(azB /atz), where 10 is the mass polar moment of inertia of the shaft per unit length. Noting that dMt (a Mrlax) dx and dB (aO/ax)dx, Newton's second law of motion can be applied to the element of the shaft to obtain the equation of motion as
=
a Mt ( Mt + --dx ax
=
)
- Mt +mtdx
aZe
= lodx-z
at
(l0.1)
271
~
I
272
Torsional Vibration of Shafts
From strength of .materials, the relationship between the torque in the shaft and the angular displacement is given by [l]
,
i,
0(J
Mt = Glpox
(10.2)
where G is the shear modulus and Ip = J is the polar moment of inertia of the cross section of the shaft. Using Eq. (10.2), the equation of motion, Eq. (10.1), can be expressed as o ox
10.2.2
(GI
p
o(J(x, t») ox +mt(x,t
)
=
l 02(J(x, t) ot2 0
(10.3)
Variational Approach The equation of motion of a nonuniform shaft, using the variational approach, has been derived in Section 4.11.1. In this section the variational approach is used to derive the equation of motion and the boundary conditions for a nonuniform shaft with torsional springs (with stiffnesses ktl and kd and masses (with mass moments of inertia ho and /zo) attached at each end as shown in Fig. 10.1. The cross sections of the shaft are assumed to remain plane before and after angular deformation. Since the cross section of the shaft at x undergoes an angular displacement
I
I
I I
I I
.-.-.-.-.>-.-1-.-.-. x-Idx!(a)
t
x M,(x,t) + dM,(x,t)
9(x.t) + d9(x,t) (b)
Figure 10.1
Torsional vibration of a nonuniform shaft.
10.2 Elementary Theory: Equation of Motten
273
z
t
.-.-y
Figure 10.2
Rotation of a point in the cross section of a shaft.
e(x, t) about the center of twist, the shape of the cross section does not change. The cross section simply rotates about the x axis. A typical point P rotates around the x axis by a small angle as shown in Fig. 10.2. The displacements of point P parallel to the y and z axes are given by the projections of the displacement PP' on oy and oz:
e
v(y, z) = OF' cosa - OPcos(a
- e)
= OP' cosa - OPcosacose w(y, z) = OP' sina - OPsin(a
= Op'sina Since
e is
- OPsina
sine
(10.4)
- e)
- OPsinacose
+ OPcosasine
small, we can write
e,
sine ~
cose~l
OPcosa
~ OP' cosa = y
OPsina
~ 0 P' sin a = z
(10.5)
so that v(y, z) Thus, the displacement can be expressed as
= -ze,
components
w(y, z) = ye
of the shaft parallel to the three coordinate
(10.6) axes
U(x, t) = 0
vex, t) = -ze(x, t) w(x, t) = yO(x, t)
(10.7)
274
Torsional Vibration of Shafts The strains in the shaft are assumed to be
and the corresponding
OU
OV
Exy
= -oy + -ox = -zox
Exz
OU =Oz
+ -ox
Exx
= Eyy = Ezz = Eyz
ow
00
00 = yox
(10.8)
=0
stresses are given by
axy = -Gz-
00
ox
00 axz = Gy-
(10.9)
ox
axx
= ayy = aZZ = ayz = 0
The strain energy of the shaft and the torsional springs is given by
7r
=~
fff
(axxExx
+ ayyEyy + azzEzz + axyExy + axzExz + aYZEyZ)
dV
v
(10.10) where Ip =
ff
(y2
+ z2)dA.
The kinetic energy of the shaft can be expressed as
A
+ =
1 [ '2IIO
1 t '210 pIp
(00at (0, t) )2 + '2ho 1 (00at(l, t) )2]
(at 00) 2 dx + [ '2IIO 1 (00ar(O, t) ) 2 + '21 (00ar(/, t) ) 2] 120
The work done by the external torque ml (x, t) can be represented
(10.11)
as (10.12)
,
l·~~~~~~~
.••.•. ~.~·~trw.J.\r~:'~'".A""~i·~I~·~~~'M·~~~~~~;~.ialuf~~liilP"i'~-c:,;,,,,:m;.UI!:Iil:,,-_,~,
10.2
Ekmentary Theory: Equation of Motion
275
The application of the generalized Hamilton's principle yields 12
8
1
(Jl' -
T - W) dt = 0
II
or
81 1 ~ Jt 12
II
2 o
GIp
(~e)Z
_~t pI (oe)2 2 Jo at
ax
dx _
p
-1
dx
+ [~klleZ(O, 2
[~IIO 2
t)
+ ~kt2ez(l.
(aeat (0, t))2
2
+ ~Izo 2
t)]
(aoat (I, t))2]
t
mle dX} dt = 0
(10.13)
The variations in Eq. (10.13) can be evaluated using integration by parts to obtain
(10.14)
(10.15)
(10.16)
(10.17)
Note that integration by parts with respect to time, along with the fact that 1>8 = 0 at t = t1 and t = tz, has been used in deriving Eqs. (10.16) and (10.17). By using
______
~~I
276
Torsional Vibration of Shafts
Eqs. (10.14)-(10.17) in Eq. (10.13), we obtain 1
f
'2lGlp
IJ
ao 001
ax
0
+ k,IOoOlo
+ 110 a2~ 00/ +k'200011 at 0
+ ho a2~ 0(11)
dt
at
l
+1'2
{l
[-a:
(GIP:~)+Plp~:~
-mIJOOdx}
dt=O
(10.18)
By setting the two expressions under the braces in each tenn of Eq. (10.18) equal to zero, we obtain the equation of motion for the torsional vibration of the shaft as 2 a 0 a ( ao) 10 at2 = ax Glp ax + m,(x, t) (10.19) where 10 = pIp is the mass moment of inertia of the shaft per unit length, and the boundary conditions as ( -Glp
ao ax
ao ( Glp ax
2 a 0) + ktlO + ho at2 oe = 0
at x = 0
2
+ k,20 + 120 aat20)
00 = 0
at x = 1
(10.20)
Each of the equations in (10.20) can be satisfied in two ways but will be satisfied only one way for any specific end conditions of the shaft. The boundary conditions implied by Eqs. (10.20) are as follows. At x == 0, either 0 is specified (so that 00 = 0) or a20)
ao ( Glpax -ktlO-hoat2 At x
=0
(10.21)
=0
(10.22)
= 1, either 0 is specified (so that 00 = 0) or ao ( Glp ax
a20 )
+ ktlO + 110 at2
In the present case, the second conditions stated in each of Eqs. (10.21) and (10.22) are valid.
10.3
FREE VffiRATION OF UNIFORM SHAFTS For a unifonn shaft, Eq. (10.19) reduces to
a20 Glp ax2 (x, t)
+ mt(x,
t)
=
a20 10
at2 (x, t)
(10.23)
By setting m,(x, t) = 0, we obtain the free vibration equation
a ax
2 0 c2_2(x,
t)
a20 at-
= -, (x, t)
(10.24)
IlU
free Vibration of Uniform Shal"ts
277'
where
c = /Glp
(10.25)
10
It can be observed that Eqs. (10.23)-(10.25) are similar to the equations derived in the cases of transverse vibration of a string and longitudinal vibration of a bar. For a uniform shaft, 10 = pIp and Eq. (10.25) takes the form (10.26) By assuming the solution as (10.27)
O(x, t) = 8(x)T(t)
Eq. (10.24) can be written as two separate equations: d28(x) ox2
(I}
+ ;r8(x)
d2T
-
ot2 + (liT(t)
=0
(10.28)
=0
(10.29)
The solutions of Eqs. (10.28) and (10.29) can be expressed as wx wx 8(x) = Acos - + Bsinc c T(t) = Ccoswt + Dsinwt
(10.30) (10.31)
where A, B, C, and D are constants. If (un denotes the nth frequency of vibration and 8n(O) the corresponding mode shape, the general free vibration solution ofEq. (10.24) is given by 00
O(x, t)
= L 8n (0) Tn (t) n=l
(10.32) The constraints Cn and Dn can be evaluated from the initial conditions, and the constraints An and Bn can be determined (not the absolute values, only their relative values) from the boundary conditions of the shaft. The initial conditions are usually stated in terms of the initial angular displacement and angular velocity distributions of the shaft. 10.3.1
Natural Frequencies of a Shaft with Both Ends Fixed For a uniform circular shaft of length l fixed at both ends, the boundary conditions are given by 0(0, t)
=0
O(l,t)=O
(10.33) (10.34)
278
Torsional Vibration of Shafts The free vibration solution is given by Eq. (10.27):
wx ( A cos"7
==
e(x, t) = e(e)T(t)
+ B sin"7WX) (C coswt + D sinwt)
(10.35)
Equations (10.33) and (10.35) yield (10.36) and the solution can be expressed
as
= sin wx (C' cos wt
e (x, t)
+ D' sin wt)
(10.37)
C
where C' and D' are new constants. quency equation
The use of Eq. (10.34) in (10.37) gives the fre. wI sm - = 0
(10.38)
C
The natural frequencies
of vibration are given by the roots of Eq. (10.38) as wI
- = n7r,
=
n
C
1, 2, ...
or n7rC Wn
The mode shape corresponding en(x)
n = 1,2, ...
= -1-'
to the natural frequency
= ...Bn sm -,wnx
n
C
The free vibration solution of the fixed-fixed its normal modes:
(10.39) Wn
= 1,2,
can be expressed as ...
(10.40)
shaft is given by a linear combination
of
00
e(x, t)
=
L sin wnx (C~ n=!
COS
wnt
+ D~ sinwnt)
(10.41)
C
10.3.2 Natural Frequencies of a Shaft with Both Ends Free Since the torque, M, = G I p (ae / ax), is zero at a free end, the boundary a free-free shaft are given by
ae ax ae -(l,t)=O ax
-(O,t)=O
In view of Eq. (10.27), Eqs. (10.42) and (10.43) can be expressed
de (0) = 0 dx
de
conditions of
(10.42) (10.43) as (10.44)
.
-(I) = 0 dx
(10.45)
~.. ",,~.,,-,.....~.,.
lO.3
Free Vibration of UnifllnllSbafts
-.279
Equation (10.30) gives wx
e(x) = A cos -
de
C
Aw
. wx
+ B SlOc wx c
-(x) = --sindx c
Bw c
+ -cos-
(l0.46) wx c
(10.47)
Equations (10.44) and (10.47) yield (10.48)
B =0 and Eqs. (10.45) and (10.47) result in . wI S10 -
c
(10.49)
=0
The roots of Eq. (10.49) are given by W n --
nJrc - nrrJ% I
-
I
p
,
n = 1,2, ...
(10.50)
= 1,2, ...
(10.51)
The nth normal mode is given by wnx c
en(x) = An cOS-,
n
The free vibration solution of the shaft can be expressed as [see Eq. (10.32)] (10.52) where the constants Cn and Dn can be determined from the initial conditions of the shaft. 10.3.3
Natural Frequencies of a Shaft Fixed at One End and Attached to a Torsional Spring at the Other For a uniform circular shaft fixed at x = 0 and attached to a torsional spring of stiffness kr at x = I as shown in Fig. 10.3, the boundary conditions are given by (10.53)
0(0, t) = 0
ao ax
Mr(l, t) = Glp-(I, t) = -krO(l, t)
o~_rG,p'I'_~, leI Figure 10.3
Shaft fixed at x = 0 and a torsional spring attached at x = l.
(10.54)
280
Torsional Vibration of Shafts
The free vibration solution of a shaft is given by Eq. (10.27): O(x, t) = G(O)T(t)
==
( A cos
wx . WX) 7" + B sm 7"
(C coswt
+ D sinwt)
(10.55)
The use of the boundary condition of Eq. (10.53) in Eq. (10.55) gives A=O
(10.56)
and the solution can be expressed as wx O(x, t) = sin -(C coswt
_
+ D sin wt)
c
(10.57)
The use of the boundary condition of Eq. (10.54) in Eq. (10.57) yields the frequency equation wGlp
wI
--cosC
= -k,sin-
C
wI C
(10.58)
Using Eq. (10.26), Eq. (10.58) can be rewritten as
a tan a =-fJ
(10.59)
where wI
a=-,
(10.60)
C
The roots of the frequency equation (10.60) give the natural frequencies of vibration of the shaft as anc I
Wn=-,
n
= 1,2, ...
(10.61)
and the corresponding mode shapes as Gn(x
)
= Bn sm --,c .
wnX
n = 1,2, ...
(10.62)
Finally, the free vibration solution of the shaft can be expressed as 00
'\"
. wnx O(x, t) = L.."sm -(Cn c
coswnt
. + Dn smwnt)
(10.63)
n=\
Several possible boundary conditions for the torsional vibration of a unifonn shaft are given in Table 10.I along with the corresponding frequency equations and the mode shapes. 10.1 . Detennine the free torsional vibration solution of a unifonn shaft carrying disks at both the ends as shown in Fig. 10.4.
Example
'"0 e;... ·u c:
:::I (;j
Z
0
:::I
0" 0
.z:
~
'-'
-t!
~
N
+J:::
N
("'l
II
II
i
:> <:l
"'"
"'"
0
("'l '"'"
••
- ~I- 0- ~I- ..: ~I- N
("'l
II
II
s:::
i
II
II
s:::
i
'"
.S
i:
II
II
s:::
~
s:::
i
~
>-<
~
c + s:::
o .S:: 0.(.)
~
~
c:
"'..z
.gO; o ::;E
e
0
5
N
~I,-,
~I- ~I.S '"
("'l '"'"
.S
0
rJ
II
'"'"I:
II >-< '-'I:
II >-< '-'I:
0
0
0
II
II
II
C,,)
..-
..->-
.-.
'"'"I:
m
m
m
m
rJ
I:
rJ
.-.>-<
c:
';jj
'"
(.)
'"
g>-c: 0 (1) .~ :::I ~
0":::1
c:
.S:: ~ .c
~ ~
0" 0
11 '-'
;>
11'-'
11'-'
';jj
'"
c:
'"
0 (.)
~I~
<:cl.
11'-' II
II
1:1
1:1
c:
S
.S
1:1
0;
c:
...
.-. ...
~
N~IN ... ~~
0
';jj
~
.S ¢:
~ ~ CIl
e
J2
'2
~
~§
:::I"1:l
o r::Q
c:
8
;:J
0
0
0
II
II
n
-;::-
~
•.•...
0
.-. ...
-;::-
'" c:
.g :.a
II
0 0
..-... .-....
0
II .-. ... .-. .-. ... ... II
~ ~~ ~ ~ ~ 0
'"'" '-' '-' ~ ~I >-<~I>-<
0
I
0
~'':: c: .-
'-' ~I>-<
8: ~
-
8: '-' ~ ~
-
'-' ~I>-< ~ ~
...., c:>
c:
.::'
0
U
~
"0
c: :::I 0
r::Q
c: '"
0
.:::
~
.0 .... :.a~
Q ....
~ Z ~
c:~
8 '" "1:l •.••.• c: 0 t.l.l
I "1:l
0
><
~
...;
""
A
I
] I
~ r-i
t ""
I
~
"0 4J
""
+
>< 4J
><
~
....;
I
~ '" :.a I "0 4J
><
~
.,f
...• <:i
~
""
t I
281
ca .-'"0 _ 0 :; g
ctl ::l
ar .:::
Z
,..,.)
~( ~ ~ II
"
§
M
"N'
~1-..:
•....•
§
"N'
~1-..:II
" "
:::
,..,.)
M
"N'
II
:::
§
~I-..:
" "
:::
:::
~ ~
~
~"I
.••... C eLl .~
"'..2
o
::E
~I-.
~I-.
0..ctl u
..:: c
~ca
C
'Vi
§ 0
c
'-'
c 'Vi
u
tJ
rJ
"
.••... ~ '-' ~
~I-.
'"
~ ~
"I~-
0
u
rJ
II
CD
I
'"
0
.••...
~I~
II
.••... ~ '-' ~
CD
II
'"8
"
rJ
.••...'-"" '-"~
CD
CD
..•.....•... ~
~c c 0 Q.)
',=
J:
0
I
::l ctl C"::l C"
s
~I- -,
1-. '"'d
~
.:.(
I
~~ ctl 0 ''::
C
.-
8
II
II
II
II
~
~
~ 0 u ~
~
co c
..,::t
.t:
I
0
I
II
II
II
.••... •..
-::-
•...
0 .~
482
~
;§
;.ac::: C ctl 0"::
u '" . "0.
J:j
'C
B I
"0 0
><
Ii:
, .,.,
'-" ~
~ '-"I ~
:t:> ~
...,
2:
'-"
~ I~~ ~ I ~~ ~ ..., ~
<:t>
..;
.••... •..
•...
.••... •..
8
-;:-
-
~
co
C
.t:
"
"
II
II
~
~
Cl:l.
-;:-
.••... •..
'-"
8
~I'"' ~I'"'
'"' •.. '"'~ <"C•.. ~~~ ..:;
I
.••... •..
.••... •..
....:
8 '-" ~I ~ '"~ ...,I ~~ ~ ...,'" ~
~
.",-
•...•
N
'"
••
••
t
,.\(
I
0
;.a'" I
~ ~ 0>
~t I
l
c
.8 •... '" B I
0
J:
r-.:
••
!
I I
1
II
~I~
P"@
~I~
s 1-.
....••
.",-
II
1;'~
II
'-" ~I ~~ ~~I...,~~
::;1c::t ••••
~ C .:::
.••... •..
~
'"
-. c'" ...• .g
II Cl:l.
I
c
'-"
II
~
'-'
"@
~ c
II
~I'"'
0
~'"'
~ 0 u ~
'"'~ ~•..
P-
·5
0 ...•
~
-
~ .••...
'"'d
'-' ~ II
.••... •.. 0
~I~~ s' -.
.••... •.. ....:
..; 0
'-" ~
"'1::l
d
II
:::"0
o c Q:l
~.
~ c .::: ~
-;:-
""0
e£e£
::;/ +~ I c::t ..; _
••
+ I
,.\(
;.a'" I
,.\(
is'" oC
- 1
•...•
~I~ '"'
10.3
Free Vibration of UnifOlTIlShafts
283
CD
----I Figure 10.4
Shaft with disks at both ends, under torsional vibration.
SOLUTION The boundary conditions, with the inertial torques exerted by the disks, can be expressed as ae a2e GJ ax (0, t) = It at2 (0, t) ae GJ ax (t, t) =
(EIO.l.l)
a2e
(ElO.1.2)
-h at2 (t, t)
Assuming the solution in the nth mode of vibration as (ElO.1.3)
en (x, t) = en (x)(Cn coswnt + Dn sinwnt)
where Wnx
en (x) = An cos -
c
'.
WnX
+ Bn sm -
(ElO.1.4)
c
the boundary conditions of Eqs. (EIO.l.l) and (EIO.1.2) can be rewritten as den
GJ-(O)
dx
= -Itw~en(O)
or (EIO.1.5) and
or . wnl G J -Wn ( -An sm + Bn cos -wnl) c c c
l
. wn ) = hWn2 ( An cos wnl + Bn sm . (ElO.1.6)
c
c
Equations (EIO.1.5) and (EIO.1.6) represent a system of two homogeneous algebraic equations in the two unknown constants An and Bn, which can be rewritten in matrix form as
w;
It Wn
.
wnl
[ GJ- smC
c
. 2
GJ..-!!.
wc
wnl
+ 12wn cos-
c
Wn
wnl
] 2 .
-GJ-cos+hwnsmc c
wnl
c (ElO.1.7)
284
Torsional Vibration of Shafts The determinant of the coefficient matrix in Eq. (E10.1.7) is set equal to zero for a nontrivial solution of An and Bn to obtain the frequency equation as
-hw
3GJ wnl -cos-+I]hw ne e
Rearranging
4
wnl
.
2
sm--G e
n
the terms, Eq.(ElO.1.8)
2 W~ . wnl w~ J -sm--GJ-hcos-=O
c2
c
c
wnl c (ElO.1.8)
can be expressed as
_ 1) tan an = an ( fha~ fh
(~+~) fJ]
(ElO.1.9)
fJ2
where wnl an=c
(ElO.1.l0)
pJl
10
pJl
10
h
h
fh =- =h h fJ2=-=Thus, the mode shapes or normal in (ElO.1.3), as.
functions
(ElO. 1. 11) (ElO.1.l2)
can be expressed,
using Eq. (ElO.1.5)
Gn(X) = An (cos anx _ an sin anx) .
I
Thus, the complete free vibration
fJl·
•
(ElO.1.13)
I
solution of the shaft with disks is given by
00
B(x, t) =
L Gn(x)(Cn coswnt + Dn sinwnt) n=]
00
= '"
n=]
where 9n and initial conditions
pn
(
anx
cos -
I
an
- -
fh
anx )
sin -
I
denote new constants specified.
(Cn coswnt -
.
+ Dn smwnt) -
whose values can be determined
(ElO.1.14 ) from the
Notes 1. If the mass moments of inertia of the disks I] and h are large compared to the mass moment of inertia of the shaft 10, fJ] and fJ2 will be small and the frequency equation, Eq. (E1O.1.9), can be written as (ElO. 1.1 5) 2. If the mass moments of inertia of the disks II and /2 are small compared to the mass moment of inertia of the shaft 10, [3] and fJ2 will be large and the frequency equation, Eq. (ElO.1.9), can be written as tan all~O
(ElO.1.l6)
1""
'''''''',
~, ,
..•••• ~"...,~ •. -, •...•. ,,~
10.3 Free Vibration of Uniform Shafts .,z85
x=O
Figure 10.5
x=l
Shaft attached to a heavy disk and a torsional spring at each end.
Example 10.2 Derive the frequency equation of a uniform shaft attached to a heavy disk and a torsional spring at each end as shown in Fig. 10.5. SOLUTION
The free vibration of the shaft in the ith mode is given by ei(x, t) = E>i(X)(Ci COSWit + Di sin wit)
(EIO.2.1)
where
= Ai
ei(x)
W-x COS
_1_
e
+ Bi sin W·X
(EI0.2.2)
_I_
e
The boundary conditions, considering the resulting torques of the torsional springs and the inertial torques of the heavy disks, can be stated as ae '. aZe Glp ax (0, t) = It atZ (0, t)
+ ktje(O,
(EIO.2.3)
t)
ae aZe Glp ax (I, t) = -/z atZ (1, t) - kt2e(l, t)
(EI0.2.4)
Using Eq. (EI0.2.1), Eqs. (EI0.2.3) and (EI0.2.4) can be expressed as Glp
dei (0) dx
Glp~
dei (1)
Z
= -Itwiei(O) Z
= /zwi ei(l)
(EIO.2.5)
+kt(ei(O)
(EIO.2.6)
- kt2ei(l)
Equation (EI0.2.2) gives dei(x)
Aiwi.
WjX
= ---sme
dx
e
BjWi + -'-' -cos-
e
WiX
(EIO.2.7)
e
In view of Eqs. (ElO.2.2) and (ElO.2.7), Eqs. (EI0.2.5) and (EI0.2.6) yield Aj(hwj
Z
- kt()
GlpWi Ai [ -e +Bj
G I pWi
+ Bj-wil
sin -
e
e
+ (lzwi Z -
GlpWj wi1 ---coS-+(lzwi
[
e
=0
c
(EI0.2.8)
kt2) COS Z
Wi1]
-
e
. Wi1] -kt2)sm=0
e
(EIO.2.9)
286
Torsional Vibration of Shafts
For a nontrivial solution of Aj and Bj, the detenninant of the coefficient matrix must be equal to zero in Eqs. (EIO.2.8) and (ElO.2.9). This gives the desired frequency equation as /Iwt - k'l Glpwj
c
sin ~
c
+ (hw2,.
c
- k(2) cos ~
-
c
cos ~
Glpwj pWj eel
+ (/2w2
=0 - k, ) sin ~ 2 C
(EIO.2.10) Example 10.3
Derive the orthogonality relationships for a shaft in torsional vibration.
SOLUTION Case (i): Shaft with simple boundary conditions The eigenvalue problem of the shaft, corresponding to two distinct natural frequencies of vibration Wj and W j, can be expressed as [from Eq. (10.28)] lJ.i
e;'(x)
+ -tej(x)
ej(x)
+ -fej(x)
c
=0
(EIO.3.1)
=0
(ElO.3.2)
w~ c
where a prime denotes a derivative with respect to x. Multiply Eq. (EIO.3.1) by ej(x) and Eq. (EI0.3.2) by ej(x) and integrate the resulting equations from 0 to 1 to obtain (EIO.3.3) (EIO.3.4) Integrating the left-hand sides of Eqs. (ElO.3.3) and (ElO.3.4) by parts results in 10 -
e;ej 1
10 -
elej J
1
1/ 1/ e;e'. o
e;e'.dx J
0
J
dx
w21/ + --.!... 2 c
+
w?1/
--.!...
c2
0
=0
(EIO.3.5)
dx = 0
(EIO.3.6)
ejejdx
0
ejej
If the ends of the shaft are either fixed (ej = eJ· = 0) or free (e~= e J = 0), the first tenns of Eqs. (ElO.3.5) and (ElO.3.6) will be zero. By subtracting the resulting Equation (ElO.3.6) from (EIO.3.5), we obtain l
I
(wt -
W])
1/ eje
j
dx
=0
(ElO.3.7)
For distinct eigenvalues, Wj #- Wj, and Eq. (EIO.3.7) gives the orthogonality relation for nonnal modes of the shaft as
1/
ejej
dx = 0, i
#- j
(ElO.3.8)
Free Vibration of Unifurm-Sfiafts ,--287
10.3
In view of Eq. (ElO.3.8), Eqs. (EIO.3.5) and (EIO.3.3) give
1/
e;ej
dx = 0,
(ElO.3.9)
l'
e;'ej
dx = 0,
(ElO.3.10)
Equations (EI0.3.8)-(ElO.3.1O) denote the desired orthogonality relationships for the torsional vibration of a shaft. Case (ii): Shaft with disks at both ends The eigenvalue problem of the shaft corresponding to two distinct natural frequencies of vibration can be expressed as [from Eq. (10.28)]
+ w; loei(x) + w]loej(x)
Glpe;'(x) Glpej(x)
Multiplying Eq. (EIO.3.11) by ej(x) the length of the shaft, we have Glp
e;'ej
dx
'j---
10
(ElO.3.11)
=0
(EIO.3.12)
and Eq. (EIO.3.12) by ei(X) and integrating over
1/
Glp (
=0
e'~ei dx 1
l'
= -low;
= -low~1 10(
eiej
dx
(EI0.3.13)
eiej
dx
(ElO.3.14)
/
The disk located at x = 0 and x = 1 must also be considered in developing the orthogonality relationship and hence the boundary conditions given by Eqs. (EIO.2.5) and (EIO.2.6) (without the torsional springs) are written for modes i and j as Glpe;o = Glpejo
(ElO.3.15)
-!tw;eio
= -!tW]e
(EIO.3.16)
jo
Glpe;1 = hW;ei/
(EIO.3.1?)
= hw]ej/
(EIO.3.18)
Glpej/
where
e''0
= de_'dx
(x
e'
= 0) '
Multiply Eqs. (EIO.3.15)-(EI0.3.19),
'I
de· = _, (x = 1) dx
respectively by ejo'
eio' eft
(ElO.3.19) and
ei/
to obtain
GI p e'io e 10 -
-
-
-liW?e· I
e·10
(ElO.3.20)
Glpejoeio
=
-!tW]eioejo
(EIO.3.21)
G I pe;/
'0
ej/ = hw;eit ej/
Glpej/ei/
= hW]ei/eft
(EIO.3.22) (ElO.3.23)
288
Torsional Vibration of Shafts
Add Eqs. (EI0.3.20) and (E1O.3.21) to (EI0.3.13) and subtract Eqs. (E1O.3.22) and (E1O.3.23) from (E1O.3.14) to produce the combined relationships Glp
1/
e;'ej dx
- -low? -
Glp
I
1/
+ Glpe;oejo + Glpejoeio
10t e·e· dx J
I
- hw2Ie·
ejei dx - Glpe;lejl
- -low?
-
I
1./o e·e· dx -
e·JO -
Ilw~e· e·Jo J '0
(EIO.3.24)
12W2e· e· J 'I JI
(EI0.3.25)
- Glpejleil
12W2e 'I e·JI
J
I
'0
I
-
Carrying out the integrations on the left-hand sides of Eqs. (ElO.3.24) and (E1O.3.25) by parts, we obtain Glpe;lejl
- Glp
-- -low?I
1/ o
1/ e;ej
e·e· I J dx
- -low~J
1/ o
e·e·J dx I
+ Glpejoeio
- I)w?e· I
- Glpejoeio - Glp
-
dx
1/
'0
e·Jo -
eje; dx
-.. 12Ule· I 'I e· JI
I)w~e· J '0 e· Jo
(ElO.3.26)
- Glpe;lejl
-
12W~e· J 'I 8· JI
.
(ElO.3.27)
Subtract Eqs. (EIO.3.20) and (ElO.3.21) from (ElO.3.13) to obtain Glp
1/
e;'ejdx
- Glpe;oejo - Glpejoeio
= -low21/ e·e· dx + I)w2e· e· + I)w~e· e· I
o
.
J
I
I
.0
Jo
J
'0
JO
(E1O.3.28)
Integration on the left-hand side of Eq. (ElO.3.28) by parts results in
(E1O.3.29) Subtraction of Eq. (EIO.3.29) from Eq. (EIO.3.27) results in 2Glp(e;08jo - e;/ej,)
= (w2 I
w~)Io J
10t e·8· dx I
J
- I) (w2 + w?)e· e· J I '0 JO
-
I~(w2 + w~)e· e·JI I J 'I (E1O.3.30)
,..{OA.Free
Vibration Response due to Initial Conditions: Modal Analysis
2S9
By using two times the result obtained by subtracting Eq. (E 10.3.22) from (E 10.3.20) on the left-hand side of Eq. (E1O.3.30), we obtain, after simplification, (wf - w;)(/o
For
Wi
¥=
Wj'
1/
8i8j dx
+ /18io8jo + h8it8it) = 0
(E 10.3.31)
+ h8io8jo + h8it8it
(EIO.3.32)
Eq. (ElO.3.31) gives /0
1/
8i8j dx
=0
Addition of Eqs. (£10.3.13) and (E1O.3.20) and subtraction of Eq. (E1O.3.22) from the result yields
(1
1
Glp
8;'8j dx
= -wf(/o
1/
+ 8;o8jo
8i8j dx
- 8;/8it)
+ h8io8jo + 128it8j/)
(E1O.3.33)
In view of Eq. (E1O.3.32), Eq. (EI0.3.33) gives
(l/
G/p
8;'8jdx+
8;o8jo - 8;t8it)
=0
(EI0.3.34)
Finally, the addition of Eqs. (EI0.3.26) and (EI0.3.27) gives
1
1
-2G/p
8;8jdx
f
= -(:u
+w;) (/0 1/ 8i8jdx
+h8io8jo + 128i/8it) (EI0.3.35)
In view of Eq. (ElO.3.32), Eq. (EI0.3.35) reduces to (EI0.3.36) . Equations (EI0.3.32), (EI0.3.34) and (EI0.3.36) denote the orthogonality relations for torsional vibration of a uniform shaft with heavy disks at both ends.
10.4
FREE VmRATION RESPONSE DUE TO INITIAL CONDITIONS: MODAL ANALYSIS The angular displacement of a shaft in torsional vibration can be expressed in terms of normal modes 8i (x) using the expansion theorem, as 00
9(x, t) =
L 8 (X)TJi i
i=1
(t)
(10.64)
290
Torsional Vibration of Shafts
where 7/;(t) is the ith generalized coordinate. Substituting Eq. (10.64) into Eq. (10.24), we obtain 00
c2
L
00
=
8~'(X)71i(t)
L
i=1
(10.65)
8i(x)17i(t)
i=)
where 8;'(x) = d28i(X)/dx2 and 17i(t) = d271i(t)/dt2. by 8j (x) and integration from 0 to l yields
Multiplication of Eq. (10.65)
(10.66) In view of the orthogonality relationships, Eqs. (EI0.3.8) and (E1O.3.1O),Eq. (10.66) reduces to
or
(1 f 1
-wf
(1
1
8 (X)dX)
=
71i(t)
f
8 (X)dX)
17i(t)
(10.67)
Equation (10.67) yields i = 1,2, ...
(10.68)
The solution of Eq. (10.68) is given by TJi( t )
=
. + -71iosmwit
TJiocoswit
(10.69)
Wi
where 7]io = TJi(t = 0) and ~io = ~i (t = 0) denote the initial values of the generalized coordinate 7]i(t) and the generalized velocity ~i(t), respectively. Initial Conditions
If the initial conditions of the shaft are given by = eo(x)
e(x,O)
ae at
(10.70)
.
= eo(x)
-(x,O)
(10.71)
Eg. (10.64) gives 00
eo(x)
=
L
8i(x)TJio
(10.72)
8i(X)~io
(10.73)
i=) 00
8o(x)
=
L i=)
10.4 Free Vibration Response due to Initial Conditions: Modal Analysis Eqs. (10.72) and (10.73) by 8j(x)
By multiplying
and integrating
291
from 0 to 1, we
obtain /
f
00
=
Oo(x)8j(x)dx
10 10
dx
= L T7io f
8i(x)8j(x)
10
i:t
=
= 17jo'
j
1,2, ...
(10.74)
= T7jo'
j = 1. 2, ...
(10.75)
I
00
eo(x)8j(x)
8i(X)8j(x)dx
10
i:l /
f
/
L17io f
dx
in view of the orthogonality of normal modes [Eq. (E10.3.8)]. Using the initial values of 17j(t) and T7j(t), Eqs. (10.74) and (10.75), the free vibration response of the shaft can be determined from Eqs. (10.69) and (10.64): O(x, t)
=
f
8i(x)
(17iO coswit
+ ~~sin Wit)
i:1
(10.76)
'
Example 10.4
Find the free vibration response of an unrestrained in Fig. 10.6 when it is twisted by an equal and apposite angle
uniform shaft shown
ao at the two ends at
t = 0 and then released.
SOLUTION
The initial displacement
of the shaft can be expressed as
= Oo(x) = ao (27- - 1)
O(x,O)
which gives the angular deflections as -ao at x = 0 and
(EI0A.1)
ao at x = 1. The initial velocity
can be assumed to be zero: (EI0.4.2) The natural frequencies
and the mode shapes of the shaft are given by Eqs. (10.50)
and (10.51):
w,
=
i; f%. i
8i (x) = Ai The mode shapes are normalized
1
1
8f(x)
lUiX COS -,
dx
o
= Af
(EIOA.4)
i = 1, 2, ...
c
as
(EI0A.3)
= 1.2•...
1/ 0
2 W·X -'-
cos
c
dx
=1
(EI0A.5)
rG,p,lp
(~._._._._.~
I~ Figure 10.6
()
x
---_t\ Unrestrained (free-free) shaft.
292
Torsional Vibration of Shafts
which gives (ElOA.6) and hence
i
= 1,2, ...
(ElOA.7)
The initial values of the generalized displacement and generalized velocity can be determined using Eqs. (10.74) and (10.75) as:
1 1
1
1Jio=
Oo(x)Bi (x) dx 1
=
ao (27- -
__ {
(ElOA.8)
1) ji
(27- -
1) cos W~x dx
(ElOA.9)
i = 1,3,5, ...
4../2i"2~2ao , "
(ElOA.lO)
= 2, 4, 6, ...
i
0,
1
1
1
cos w~x dx = aoji
1
~io =
Bo(x)Bi(x)
dx = 0
(ElOA.ll)
Thus, the free vibration response of the shaft is given by Eq. (10.76)
8aoL:00
O(x t) = -- -. ,
1!2 . 1=1,3,5, ...
iJrx
i1!ct
I
I
cos-cos--
(EI0A.12)
10.5 FORCED VIBRATION OF A UNIFORM SHAFT: MODAL ANALYSIS The equation of motion of a uniform shaft subjected to distributed external torque, m,(x, t), is given by Eq. (10.23):
a20 Glp ax2 (x, t)
+ m,(x,
a20
t) = 10 at2 (x, t)
(10.77)
The solution of Eq. (10.77) using modal analysis is expressed as 00
O(x,
t)
= L: Bn (x)1Jn(t)
(10.78)
n=1
w~ere Bn (x) is the nth normalized normal mode and 1Jn(t) is the nth generalized coordinate. The normal modes Bn(x) are determined by solving the eigenvalue problem d2Bn(x)
Glp
dx2
2
+ 10wnBn(x) = 0
(10.79)
10.5
29.~
Forced Vibration of a Unifonn Shaft: Modal Analysis
by applying the boundary conditions of the shaft. By substituting Eq. (10.78) into (10.77), we obtain 00
L Glpe~(x)17n(t)
00
+ m,(x,
t)
= L 108n (x)ijn (t)
n=1
(10.80)
"=1
where (10.81) Using Eq. (10.79), Eq. (10.80) can be rewritten as 00
- L 10w~en(X)17n(t) + m,(x,
00
t) =
n=1
L 108n (x)ijn (t)
(10.82)
n=1
Multiplication of Eq. (10.82) by 8m(x) and integration from 0 to I result in
L 00
- 10W~17n(t) 8n(x)8m(x) n=1
1
dx
+
1/
m,(x, t)em(x) dx
0
1
= loijn(t)
(10.83)
e.n,(x)8m(x)dx
In view of the orthogonality relationships, Eq. (EI0.3.8), Eq. (10.83) reduces to ijn(t)
+ W~17n(t)=
Qn(t),
n
= 1,2, ...
(10.84)
where the normal modes are assumed to satisfy the normalization condition
1
1
e~(x) dx
= I,
n = 1,2, ...
(10.85)
and Qn(t), called the generalized force in nth mode, is given by Qn(t) = ~
(I m,(x,
t)en(x)
dx
(10.86)
1010 The complete solution of Eq. (10.84) can be expressed as (10.87) where the constants An and Bn can be determined from the initial conditions of the shaft. Thus, the forced vibration response of the shaft [Le., the solution of Eq. (10.77)], is given by
294
Torsional Vibration of Shafts
The steady-state response of the shaft, without considering the effect of initial conditions, can be obtained from Eq. (10.88), as B(x, t) =
9 (x) l' L _n_ Qn('r) sinwn(t n 00
n=1
W
- T)dT
(10.89)
0
Note that if the shaft is unrestrained (free at both ends), the rigid-body displacement, (f(t), is to be added to the solution given by Eq. (10.89). If M,(t) denotes the torque applied to the shaft, the rigid-body motion of the shaft, (f(t), can be determined from
the relation d2(f(t)
10--;}t2
= M,(t)
(10.90)
where 10 = plIp denotes the mass moment of inertia of the shaft and d2(f(t)jdt2 indicates the acceleration of rigid-body motion. Example 10.5 Find the steady-state response of a shaft, free at both ends, when subjected to a torque Mt = aot, where ao is a constant at x = I. SOLUTION
The steady-state response of the shaft is given by Eq. (10.89): B(x, t) =
9 (x) l' L _n_ Qn(T) 00
Wn
n=1
0
sin Wn (t - T) dT
(ElO.5.1)
..
where the generalized force Qn(t) is given by Eq. (10.86):
=~. t m,(x,
Qn(t)
t)9n(x)dx
1010
(ElO.5.2)
Since the applied torque is concentrated at x = I, m,(x, t) can be represented as m,(x,
t)
= M,8(x
-I)
= aot8(x
-I)
(E1O.5.3)
= ao -9n(/)t
(ElO.5.4)
where 8 is the Dirac delta function. Thus, Qn(t)
= -1 ~
.1
1
aot8(x -1)9n(x)
dx
0
~
For a free-free shaft, the natural frequencies and mode shapes are given by Eqs. (10.50) and (l0.51): =
W
n
~.= I
nn I
Wnx
9n(x)=Ancos-,
c
Y[G p'
n = 1, 2, ...
n=I,2,
...
(EI0.5.5) (E1O.5.6)
When 9n(x) is normalized as
1
1
9~(x) dx = 1
(E1O.5.7)
10.6 Torsional Vibration of Noncircular Shafts: Saint-Venant's Thool)"
we obtain
If.
An =
(ElO.5.8)
n = I, 2, ...
(i
mr X
8n(x)=YTcos-z-'
·-295
n=I,2
(El0.5.9)
....
The integral in Eq. (E10.5.1) can be evaluated as
r Qn(r) sinwn(t -r)
dr = ao ~
~YT
h
COSnJr
r r sinwn(t - r) dr
(ElO.5.lO)
h
(ElO.5.l1)
=~ ~ cos mr (t - _1 sin wnt) lown Y T Wn Thus, the steady-state response, given by Eq. (E10.5.1), becomes e(x, t)
=
f: 2..Y~T
n=l
L
2a0
00
=
n=2.4....
--2
1I0wn o
_
cos ~
Wn
o
L
n=1.3 •...
(~ ~ cosnrr) [Own T
nrrX COS-
(
-
Z
2ao' --cos-IIow2
1. t - -smwnt
(ElO.5.12)
)
Wn
nrrx (
1
n
_1.. sinwnt) Wn
(t
Y
z
1. t - -smwnt W
)
(ElO.5.13)
n
Since the shaft is unrestrained, the rigid-body motion is to be added to Eq. (ElO.5.13). Using Mt = aot in Eq. (10.90) and integrating it twice with respect to t, we obtain the rigid-body rotation of the shaft as _ ao t3 (ElO.5.l4) e(t) = -10 6 Thus, the complete torsional vibration response of the shaft becomes ao e(t) = -t 6/0
3
2ao
L 00
n=1.3 ....
10.6
L
1 nrr x 2cos-Z10 [ n=2,4.... Wn
+ -1-
00
1 nrrx ( 1. 2cos-t - -smwnt Wn 1 Wn
(
1 t - -sinwnt
)
Wn
)]
.
(El0.5.15)
TORSIONAL VmRATION OF NONCIRCULAR SHAFTS: SAINT·VENANT'S THEORY For a shaft or bar of noncircular cross section subjected to torsion, the cross sections do not simply rotate with respect to one another as in. the case of a circular shaft, but they are deformed, too. The originally plane cross sections of the shaft do not remain plane but warp out of their own planes after twisting as shown in Fig. 10.7. Thus, the
296
Torsional Vibration of Shafts
-.(--
.-.-((0)
-·f--
.-.{(b)
Figure 10.7 Shaft with rectangular cross section under torsion: (a) before deformation; (b) after deformation.
points in the cross section undergo an axial displacement. A function as the warping function, is used to denote the axial displacement as u=
ao
1/f(y, z) ax
1/f(y, z), known (10.91)
where aolax denotes the rate of twist along the shaft, assumed to be a constant. The other components of displacement in the shaft are given by v = -zO(x, t)
w =
(10.92)
yO(x, t)
(10.93)
The strains are given by
au ax
Exx
=-
Eyy
== -
Ezz
=-
Ex
=
au
y
ay
aw az
=0
(since a 0 lax is a constant)
=0 =0
au + av = (a1/f _ z) ae ay ax ay ax
Exz
= -au + -aw- = (a1/f -az + Y ) az ax
Eyz
== -
av aw az +-ay
= -0 +0 = 0
ao ax
_
(10.94)
___________________
_.10.6 Torsional Y.ilir.atiQnGfNoncir~-ular-Shafts:·SaintThe corresponding
stresses can be determined
Uxx
= Uyy
U •
= G (81/f + Y) 8z
x.
= uYZ = 0,
= Uzz
Verumt's Theory.297
as
= G (81/1 _,.) ae a"Y ax '
U
xv.
ae 8x
(10.95)
The strain energy of the shaft is gi yen by 1T
=~
fff v
=
211/x=o
+ Uyyeyy + Uzzezz + Uxyexy + uxzexz + UYZeyz) dV
(uxxexx
If
G
[(81/1
ay -
z
)2 + (81/f a; + Y )2]. (8e)2 ax
(10.96)
d A dx
A
Defining the torsional rigidity (C) of its noncircular
c=
section of the shaft as
II G[(~~ -z)' +(~;+Y)']dA
(10.97)
A
the strain energy of the shaft can be expressed 1T=
~
2
as
t C (8e)2
10
dx
(10.98)
8x
Neglecting the inertia due to axial motion, the kinetic energy of the shaft can be written as in Eq. (10.11). The work done by the applied torque is given by Eq. (10.12). Hamilton's principle can be written as
of /,'{lj
-z)' + (~; + Y )']dA} (:;)' dxdt _01'2 10t ~plp (88)2 dxdt _01'2 10t m,8dxdt 2 ~G [(~;
=0
8t
'1
(10.99)
'1
The first integral of Eq. (10.99) can be evaluated
as
° 1,[ 10t If 2 [(81/fay-Z )2 + (81/1 a;+Y )2] (80)2 8x dAdxdt [(81/1 )2 + (81/f-+y )2] ---dAdxdt 80 a(80) = ['21/ If ---z 8z 8x 8x 1G
'2
A
G
'1
8y
0
A
+
t
2
tlf (aO)2[(81/f
l't 10
G
ax
-z) 8y
8(81/f) +(81/1 +Y) 8(81/1)] dAdxdt 8y
8z
8z
A
(10.100)
B!!!!!!!!!!!!!!!!!!!I!!I!l!!!!I!!I!l!!!!!!!!!!!!I!!
!!!!!!!!!!!!!!!!~~~.
~.<.~ ..~"'~~.~~ ...~!!!! .._~._~.!!!!!!!!!!!!!!!!!!!!!!!
298
Torsional Vibration of Shafts
The first integral term on the right-hand side of Eq. (10.100) can be expressed as [see Eq. (10.97)]
1 1. / 2
1
0
IJ
1 [1/ 12
000(00) GC---dxdt= 8x 8x
11
0
-- 0 (00) GC8x 8x
80] oO+GC-oOI& 8x
dt (10.101)
The second integral term on the right-hand side of Eq. (10.100) is set equal to zero independently:
1 1/ If (88)2 [(81/1) 2
1
II
G
0
A
-
8x
8(01/1) --+ 8y
--z 8y
(81/1 ) -+y-8z.
8(01/1)] dydzdxdt=O oz (10.102)
Integrating Eq. (10.102) by parts, we obtain
f [l
G (:~)'
dX] [-
ij :Y (~~ - Z)NdYdZ + f.(~~- z)
-ij :Z (~~ + Y)8,;dYdZ + f.(~~+ Y)
1,8';
1,8';
d!
d!] dt = 0 (10.103)
where ~ is the bounding curve of the c.ross section and ly(lz) is the cosine of the angle between the normal to the bounding curve and the y(z) direction. Equation (10.103) yields the differential equation for the warping function 1/1 as 021/1 821/1) ( 8y2 + 8Z2 . = and the boundary condition on
1/1
2
"1
1/1
(10.104)
=0
as (10.105)
Physically, Eq. (10.105) represents that the shear stress normal to the boundary must be zero at every point on the boundary of the cross section of the shaft. When Eq. (10.101) is combined with the second and third integrals of Eq. (10.99), it leads to the equation of motion as 2 8 0(x,t) pIp ot2 and the boundary conditions on
o( 80(X,t») = 8x C 8x 0
+ ml(x,
t)
(10.106)
as (10.107)
10.7 Torsional Vibration of Noncircular Shafts. Inchidin£:Axial Inertia
10.7
TORSIONAL VmRATION OF NONCIRCULAR INCLUDING AXIAL INERTIA
299
SHAFTS,
Love included the inertia due to the axial motion caused by the warping of the cross section in deriving the equation of motion of a shaft in torsional vibration [4, 10]. In this case, the kinetic energy of the shaft is given by
T=
H' 11
(:t~S
2
+,2 (~~)' + Y'
+V
A
=
e~)']
dAdx
h +h
(10.108)
where
= -1
h
2
1/ If (
p1/f2 --02e ) otox
0
2
(10.109)
dAdx
A
h=
-21110 pIp (oe)2 ot
(10.110)
dx
Note that h denotes the axial inertia term. The variation associated with ton's principle leads to 8
12
1
=
hdt
~
h in Hamil-
112
8h~t
1 (f""f 1 ~""
=
1
2
1
p
0
1\
)
02() 02(8()) 1/f2dA -----dxdt otox otox
A
(10.111) Denoting (10.112)
18
=
/ 1 o
P
(02())2 -otox
(10.113)
dx
the integrals in Eq. (10.111) can be evaluated to obtain 8
12
1
hdt
11211
=-
1\
0
1\
'21 1
1
=
'\
+
0
1'2 If
0(8()) -0 ( pl",-- 02() ) --dxdt+ ot otox oX
--02 ( pl",-- 02() ) MJdxdt oxot otox
£'2
18 111/f81/fdAdt A
18
11
1
1/f81/fdAdt
A
12
1\
02() ) 8()I&dt -0 ( pl1f;-ot otox (10.114)
300
Torsional Vibration of Shafts
The first, second; and third terms on the right-hand side of Eq. (10.114) contribute to the equation of motion, Eq. (10.106), the boundary conditions on e, Eq. (10.107), and to the differential equation for 1/1, Eq. (10.104), respecti vely, The new equations are given by (10.115) (10.116) (10.117) (10.118) where Ig =
10.8
{I G (oe)2 ox
10
(10.119)
dx
TORSIONAL VffiRATION OF NON CIRCULAR SHAFTS: TIMOSHENKO-GERE THEORY In this theory also, the displacement components of a point in the cross section are assumed to be [4,11, 17]
\
oe
u =1/f(y, z) ox (x, t) v =
-ze(x,
(10.120)
t)
(10.121)
w = ye(x, t) where as
oe lox
(10.122)
is not assumed to be a constant. The components of strains can be obtained
ou
cxx
= ox
o2e
= 1/f(y, z) ox2
ov = -oy = 0 ow Ezz = oz = 0 = ov + ou
(10.123)
Cyy
Cxy
ox
Exz
oy
(10.124) (10.125) =
=:~+~;
(01/1 _ oy
= (:~
z)
00 ox
+
y) :~
ov ow ~ = -oz +-oy = -0 +e = 0
Er
(10.126) (10.127) (10.128)
-l,Q.,S. T~.sional Vibration of Noncircular Shafts: Timoshenko-Gere
Theory
..301
The component of stress are given by I-v
v
v
v
I-v
v
axx ayy azz axy ayz
=
E
v
V
l-v
(1 + v)(1 - 2v)
0 0
0
0 0
azx
0 0
0
0 0 0
0 0 0
1-2v -2-
0 1-2v ---y-
0 0
0
0
0 0 0 0
€.u
€yy €::
€xy By: €:x
0 1-2v -2-
(10.129) that is, E(1 - v)1/1
axx = --------
cPa
(1 + v)(1 - 2v) 8x2
a
-a
axy
=
a
82
~ E1/1-
(10.130)
8x2
a2a
Ev1/1
(10.131)
---------~O zz - (1 + v)(1 - 2v) 8x2
yy -
E 1 - 2v (1 + v)(1 _ 2v) 2
(81/1
ay -
) z
8a 8x
(81/f
= G ay -
z
) 8a 8x
(10.133)
ayz = 0 (j
zx
(10.132)
1 - 2v
E
= (1 + v)(1
- 2v) . 2
az y)
(81/1 +
8a 8x
=
az + )
G (81/f
Y
8a (10.134) 8x
Note that the effect of Poisson's ratio is neglected in Eqs. (10.130) and (10.131). The strain energy of the shaft can be determined as rr
=~
fff
+ ayysyy + azzszz + axysxy + axzsxz + ayZsyz)
(axxsxx
dV
v (10.135)
=it+h where
it =
11/x=o If
2
E
(
1/f
2a)2
8
8x2
dAdx
A
h=H~ff lG[(~~-z) A
The variation of the integral/I
:~r+G[(~~ +Y) :~ndAd.
can be evaluated as
(10.136)
(10.137)
302
Torsional Vibration of Shafts
where
1/12dA
I1/t=!!
(10.139)
A
When the second term on the right-hand side of Eq. (10.138) is integrated by parts, we obtain
'J, =
f L!! E,"NG:~)'
dAdxdt+
A
f [EI.:> G~)I~
10~2 (EI1/t::~)oedx] 1
- o~ (EI1/t::~)oelb+
dt
(10.140)
The variation of the integral h can be evaluated as indicated in Eqs. (10.100), (10.101) and (10.103). The expressions for the.kinetic energy and the work done by the applied torque are given by Eqs. (10.108) and (10.12), respectively, and hence their variations can be evaluated as indicated earlier.}'he application of Hamilton's principle leads to the equation of motion for e(x, t):
2 02e2 0 ( 02e ) 0 ( oe) 02 ( 02e) pIp ot - ot ox . pI1/tox ot - ax C ox + ox2 EI1/tox2
= m,(x,
t)
(10.141) and the boundary conditions
022e)] 8e Ii0 [coeox + PI1/t~ot2 ox _ ~ox (EI1/t ax
=0
02e28 (00) EI1/tox ox Ii0 =0 The differential equation for the warping function
1
1o
G
2 (00)2(021/1 0 1/1) + dx -+ox oy2 OZ2
1/1
(10.142) (10.143)
becomes
.[1 p (0--ox2°)211at dx1
0
0
J
28)2 E (0-ox2 dx 1/1=0 (10.144)
with the boundary condition on
1/1
given by
(~~ - z) iy + (~~ +
Y) iz = 0
(10.145)
10.9 Torsional Rigidity of Noncircular Shafts
10.9
303
TORSIONAL RIGIDITY OF NON CIRCULAR SHAFTS It is necessary to find the torsional rigidity C of the shaft in order to find the solution of the torsional vibration problem, Eq. (10.106). The torsional rigidity can be determined by solving the Laplace equation, Eq. (10.104): 2
0 1/1 = -02t/f + =0 oy2 OZ2
\721/1
subject to the boundary condition,
(10.146)
Eq. (10.105):
(~~ - z) ly + (~~ + y) lz = 0
(10.147)
which is equivalent to (10.148) Since the solution of Eq. (10.146), for the warping function t/f, subject to the boundary condition of Eq. (10.147) or (10.148) is relatively more difficult, we use an alternative procedure which leads to a differential equation similar to Eq. (10.146), and a boundary condition that is much simpler in form than Eq. (10.147) or (10.148). For this we express the stresses O'xy and O'xz in terms of a function (y, z), known as the Prandtl stress function, as [3, 7] 0<1> O'xz = --
0<1> O'xy '=, " -0 z '
The stress field corresponding to Saint-Venant's Eq. (10.149), satisfies the equilibrium equations:
theory,
oO'xx
oO'xy
oO'xz
0
oO'xy
oO'yy
oO'yZ
0
-+-+-= ox oy oz
-+-+-= ox oy oz
oO'xz oO'YZ oO'zz 0 -a;+ ay + az = By equating the corresponding (10.149), we obtain
expressions
G0f) (ot/f ox oy G of) (ot/f ox ozoy
(10.149)
oy
Eq. (10.95),
along
with
(10.150)
of O'xy and O'xz given by Eqs. (10.95) and
-z) = + y) = _
0<1> oz 0<1>
(10.151)
(10.152)
Differentiating Eq. (10.151) with respect to z and Eq. (10.152) with respect to y and subtracting the resulting equations one from the other leads to the Poisson equation: 02<1> 02<1> V2<1> = + oy2 OZ2
= -2Gf3
(10.153)
304
Torsional Vibration of Shafts
where (10.154) is assumed to be a constant. The condition to be satisfied by the stress function ct> on the boundary can be derived by considering a small element of the rod at the boundary as shown in Fig. 10.8. The component of shear stress along the nonna! direction n can be expressed as (10.155) since the boundary is stress-free. In Eq. (10.155), the direction cosines are given by ly
= cosa = dz -,dt
lz
. = sma = --dydt
(10.156)
where t denotes the tangential direction. Using Eq. (10.149), the boundary condition, Eg. (10.155), can be written as act>
act>
az
oy
-ly - -lz = 0
(10.157)
The rate of change of ct> along the tangential direction at the boundary (t) can be expressed as dct>
act> dy
act> dz
act>.
act>
-dt = -ay-- dtaz+ -~ dt = --sma +..-cosa oy oz'
=0 .
(10.158)
using Eqs. (10.156) and (10.157). Equation (10.158) indicates that the stress function ct> is a constant on the boundary of the cross section of the rod. Since the magnitude
n .(
t
A.
/
1~-dy
(a)
Figure 10.8
(b)
Boundary condition on the stresses.
I
10.9 Torsional RigidiD' of Noncircular Shafts
305
of this constant does not affect the stress, which contains only derivatives of <1>,we choose. for convenience, (10.159) to be the boundary condition. Next we derive a relation between the unknown angle ,B(angle of twist per unit length) and the torque (Mt) acting on the rod. For this. consider the cross section of the twisted rod as shown in Fig. 10.9. The moment about the x axis of all the forces acting on a small elemental area dA located at the point (y,z) is given by (10.160) The resulting moment can be found by integrating the expression in Eq. (10.160) over the entire area of cross section of the bar as (10.161)
Each term under the integral sign in Eq. (10.161) can be integrated by parts to obtain (see Fig. 10.9):
=
I (yl~;-l~ dz
=-
I 1~2 dz
=-
II
A
(10.162)
z
t
I
O'xy
L
y
I
IdA
I
(y,z)
L. . .
Figure 10.9
z
.L--·
Forces acting on the cross section of a rod under torsion.
306
Torsional Vibration of Shafts
z
t t b
+ b
--1-
~a-+-a~ Figure 10.10
since
¢
= 0 at the points
If A
a¢ -zdA az
Elliptic cross section of a rod.
PI and P2. Similarly,
If =I =
A
dy (¢z
I l
p4
a¢ -zdydz az
=
dy
P3
I~: -l:4
a¢ -zdz az
4>dZ) = -
II
4>dA
(10.163)
A
Thus, the torque on the cross section (M,) is given by .. M,._.=2
II
¢dA
(10.164)
A
The function 4>satisfies the linear differential (Poisson) equation given by Eq. (10.153) and depends linearly on GfJ, so that Eq. (10.164) produces an equation of the form M, = GJfJ = CfJ, where J is called the torsional constant (J is the polar moment of inertia of the cross section for a circular section) and C is called the torsional rigidity. Thus, Eq. (10.164) can be used to find the torsional rigidity (C). Note There are very few cross-sectional shapes for which Eq. (10.164) can be evaluated in closed form to find an exact solution of the torsion problem. The following example indicates the procedure of finding an exact closed-form solution for the torsion problem for an elliptic cross section. ExampLe 10.6 (Fig. 10.10). SOLUTION
Find the torsional rigidity of a rod with an elliptic cross section The equation of the boundary of the ellipse can be expressed as fey,
Noting that
\72
z)
= 1-
v2
Z2
~2 -
b2
=0
f is a constant, we take the stress function ¢(y, z) = c (1 - ~: - ~~)
(E1O.6.l) ¢(y, z) as
(E1O.6.2)
10.9 Torsional Rigidity of Noncircular Shafts where c is a constant. Using Eq. (ElO.6.2) in Eq. (l0.153), 2
V'2
2
= -aay2
( 1
-2 a
+ -12 ) b
307
we obtain
= -2GfJ
(ElO.6.3)
If we choose
GfJa2b2
c = a2
+ b2
(EIO.6.4)
the function
2GfJa'1 (a2 + b2) z
(E10.6.5)
2GfJb2 = (a2 + b2) Y
(ElO.6.6)
a
(1xy =
az
=
a
O"xz=
-ay
The torque (Mt) can be obtained as
Mt
=
II
II
= a;?b2
(O"xzY- O"xyz)dA
but also the
A
2
Clb
2 2 a ) dA
+z
(ElO.6.7)
A
Noting that
II idA=11 II = II A
ldydz
= Iz = ~rrba3
(E10.6.8)
2 z dzdy
= Iy = ~rrab3
(E10.6.9)
A
z2dA
A
A
Eq. (E1O.6.7) yields (ElO.6.1O) When Mt is expressed
as
Mt Eq. (ElO.6.10)
= GJfJ
= CfJ
(E10.6.1l)
gives the torsional constant J as
rra3b3
(E10.6.12)
J=---
a2
+ b2
and the torsional rigidity C as
C
=
The rate of twist can be expressed
rra3b3 Mt a2 + b2 G =
(E10.6.13)
in terms of the torque as Mt
fJ =
7i
C=
Mt (a2
+ b2)
Grra3b3
(ElO.6.14)
f"
308
Torsional Vibration of Shafts
10.10 PRANDTL'S MEMBRANE ANALOGY Prandtl observed that the differential equation for the stress function, Eq. (10.153), is of the same form as the equation that describes the deflection of a membrane or soap film under transverse pressure [see Eq. (13.1) without the right-hand-side inertia term]. This analogy between the torsion and membrane problems has been used in determining the torsional rigidity of rods with noncircular cross sections experimentally [3, 4]. An actual experiment with a soap bubble would consist of an airtight box with a hole cut on one side (Fig. 10.11). The shape of the hole is the same as the cross section of the rod in torsion. First, a soap film is created over the hole. Then air under pressure (p) is pumped into the box. This causes the soap film to deflect transversely as shown in Fig. 10.11. If P denotes the uniform tension in the soap film, the small transyerse deflection of the soap film (w) is governed by the equation [see Eq. (13.1) without the right-band-side inertia term] (10.165) or (10.166) in the hole region (cross section) and
w=O
(10.167)
on the boundary of the hole (cross section). Note that the differential equation and the boundary condition, Eqs. (10.166) and (10.167), are of precisely the same form as for the stress function 4>,namely, Eqs. (10.153) and (10.159): 2
V 4> = -2Gf3
(10.168)
in the interior, and (10.169) on the boundary. Thus, the soap bubble represents the surface of the stress function with
w piP
4>
(10.170)
= 2Gf3 Deflected soap film
Hole in box with shape similar to the cross section of the rod in torsion
Air pressure
Figure 10.11
l ;,,~"~~,,~~
"M'ii'''~'k~~~~~~11'''"'ii{~il~':z.~~~Aii~~~i:.\o~-..i~~~~~;~~~~,Q:~l.'iWilS!i~.i;;.K~.;.id~~~~~,:>,
Soap film for the membrane analogy.
10.10
Prandtl's Membrane Anal~gy
309
or (10.171)
= Cw
where
9 denotes
a proportionality constant: C
-
= 2Gf3P
(10.172)
P
The analogous quantities in the two cases are given in Table 10.2. The membrane analogy provides more than an experimental technique for the solution of torsion problem. It also serves as the basis for obtaining approximate analytical solutions for rods with narrow cross sections and open thin-walled cross sections. Table 10.3 gives the values of the maximum shear stress and the angle of twist per unit length for some commonly encountered cross-sectional shapes of rods. Example 10.7 Determine the torsional rigidity of a rod with a rectangular cross section as shown in Fig. 10.12. SOLUTION We seek a solution of the membrane equation, (10.166), and use it for the stress function, Eq. (10.171). The governing equation for the deflection of a membrane is
(E1O.7.1)
-a :::y ::: a, -b :::z :::b
Table 10.2
Prandtl's Membrane Analogy
Soap bubble (membrane) problem
W 1
G 2{3
P
P
Torsion problem
aw aw
- az' ay
2 (volume under bubble)
z
-- ...•
--x M,
M,
Figure 10.12
Rod with a rectangular cross section .
. ,I
310
Torsional Vibration of Shafts
Table 10.3
Torsional Properties of Shafts with Various Cross Sections Angle of twist per unit length, ()
Cross section
Maximum shear stress, Tmax
1. Solid circular shaft
'-.1 O
2. Thick-walled tube
fm"" RO
R·I
2T 7r R3
T G
2TRo
= torque = shear modulus
7r(Rri - Ri)
2T 7rG(R6 - Ri)
3. Thin-walled tube
4. Solid elliptic shaft
.j2(a2 + b2)T 47rGa2b2t
T
27rabt
(continued
on next page)
10.10
Prandtl"s Membrane Analogy
311
(continued)
Table 10.3
Maximum shear stress, !max
Angle of twist per unit length, f)
Cross section 6. Solid square shaft
4.808T -a-3-
7.092T
(];;4
T otGab3
7. Solid rectangular shaft u b
1.0 2.0 3.0 5.0 10.0 00
ot
f3
0.141 0.229 0.263 0.291 0.312 0.333
0.208 0.246 0.267 0.292 0.312 0.333
8. Hollow rectangular shaft
(at2 + btl)T 2Gtlt2a2b2
T b
.i
!maxl
=
T 2abtl
!max 2
=
T 2abt2
tl
9. Solid equilateral triangular shaft
26T
Ga4
13T
-;;3
10. Thin-walled tube
S
=
A
=
circumference of the centerline of the tube (midwall perimeters) area enclosed by the midwall perimeters
T 2.4t
312
Torsional Vibration of Shafts and the boundary conditions
are
we-a, w(y,
= w(a,
z)
= 0, b) = 0, z)
= w(y,
-b)
-b
The deflection shape, w(y, z), that satisfies Eq. (EIO.7.2), can be expressed as
:s z
:S b
(EIO.7.2)
-a :S y :S a
(ElO.7.3)
the boundary
conditions
at y
= =fa,
00
w(y, z)
'"' ~
=
ai cos irry 2a Zi(Z)
(EIO.7.4)
i=I,3,5.... where ai is a constant and Zj (z) is a function Eq. (E1O.7.4) into Eq. (ElO.7.I), we obtain
of z to be determined.
Substituting
(ElO.7.5) When the relation
t j=I.3.5 .... is introduced on the right-hand
~(_1)(i-I)/2cos irr
2a
=1
(ElO.7.6)
side of Eq. (ElO.7.5), the equation yields
d2 Z ·2 --j - ~Zi dz2
2
irrz 2a
(ElO.7.7)
P irraj
differential
equation can be expressed
irrz
+ Bj
= Aj cosh -
4 P -_(_1)(i-1)/2
=-
4a2 ...
The solution of this second-order Zi(Z)
irry
+.
sinh - -
2a
I6pa2
as
.
(_1)(1-1)/2
(ElO.7.8)
Pi2rr2ai
where Ai and Bj are constants to be determined from the boundary conditions at = =fb. Using the condition Zj(z = -b) = 0 in Eq. (EIO.7.8) yields Bj = 0, and the condition Zj (z = b) = 0 leads to Z
I6pa2
1
A· - -------I
Thus, the function expressed as
=
Z(z)
w(y, z
)
=
-
Pi2rr2aj
Zj (z) and the deflection 2 I6pa Pi2rr2aj I6pa2 Prr2
(_1)(i-1)/2
(ElO.7.9)
cosh(irrb/2a)
[1-
of the rectangular
cosh(irrz/2a)]
membrane
can be
(ElO.7.1O)
cosh(irrb/2a) ~ . ~
1=1.3.5....
1 i2 (-1)
(i-I)/2 [
coshUrrZ/2a)] 1 - cosh(irrb/2a)
irr)' cos 2a (ElO.7.JJ)
,. l,~~Lt-'~~1o.."'·--'~~$O~~~jl~U"~~~L:~~~~~~~~~~\o;,~.i&~~~...b.>~l~~~;..
10.11
Recent Contributions
313
Equations (ElO.7.11), (10.171), and (10.172) yield the stress function (y, z) as "'(
) _ 32GfJa2 ~ 1 - ---::-L-- -(2 , rr i=I.3.5.... i2
y
Z
1)(i-l)/2 [
cosh(irrZ/2a)] 1 - -----:---:-cosh(irrb/2a)
irry cos 2a
(E1O.7.12)
The torque on the rod, Mt, can be determined as [see Eq. (10.164)] Mt = 2
ff
d A
A
64GfJa2ja jb ~ 1 ( 1 (i-l)/2 [ cosh(irrZ/2a)] = -~- 2 L-- -2 - ) 1- ---'--rr -a -b i=I.3.5.... i cosh(irrb/2a)
= 32GfJ(2a)\2b) rr4
~ -'4 L-- I i=I.3.5....
_
64GfJ(2a)4 rr5
irry cos-dydz 2a
.!..
~ tanh _irr_b L-- i5 2a i=I.3.5....
(ElO.7.!3)
Using the identity (E1O.7.14)
Eq. (EI0.7.l3) can be rewritten as .. 1 3 ( 192a ~ 1 irrb) Mt = 3GfJ(2a) (2b) 1 - rr5b i=~ .... i5 tanh 2a
(ElO.7.l5)
The torsional rigidity of the rectangular cross section (C) can be found as C = Mt = kG(2a)3(2b)
(EI0.7.16)
fJ
where oo
192 a k = -1 ( 1 - -3 rr5 b
L
i=I.3.5....
. -15 tanhlrri
b)
(E1O.7.17)
a
For any given rectangular cross section, the ratio b/a is known and hence the series in Eq. (ElO.7.1?) can be evaluated to find the value of k to any desired accuracy. The values of k for a range of b/a are given in Table 10.4.
10.11
RECENT CONTRIBUTIONS The torsional vibration of tapered rods with rectangular cross section, pre-twisted uniform rods, and pre-twisted tapered rods is presented by Rao [4]. In addition, several refined theories of torsional vibration of rods are also presented in Ref. [4].
314
Torsional Vibration of Shafts Table 10.4 Values of k in Eq. (E1O.7.16) bla
k
bla
k
1.0 1.5 2.0 2.5 3.0
0.141 0.196 0.229 0.249 0.263
4.0 5.0 6.0 10.0
0.281 0.291 0.299 0.312 0.333
00
Torsional Vibration of Bars The torsional vibration of beams with a rectangular cross section is presented by Vet [8). An overview of the vibration problems associated with turbomachinery is given by Vance [9]. The free vibration coupling of bending and torsion of a uniform spinning beam was studied by Filipich and Rosales [16]. The exact solution was presented and a numerical example was presented to point out the influence of whole coupling. Torsional Vibration of Thin- Walled Beams The torsional vibration of beams of thinwalled open section has been studied by Gere [11). The behavior of torsion of bars with warping restraint is studied using Hamilton's principle by Lo and Goulard [12]. Vibration of a Cracked Rotor The coupling between longitudinal, lateral, and torsional vibrations of a cracked rotor was studied by Darpe et al. [13]. In this work, the stiffness matrix of a Timoshenko beam element was modified to account for the effect of a crack and all six degrees of freedom per node were considered. Torsional Vibration Control The torsional vibration control of a shaft through active constrained layer damping treatments has been studied by Shen et al. [14). The equation of motion of the arrangement, consisting of piezoelectric and viscoelastic layers, is derived and its stability and controllability are discussed. Torsional Vibration of Machinery Drives The startup torque in an electrical induction motor can create problems when the motor is connected to mechanical loads such as fans and pumps through shafts. The interrelationship between the electric motor and the mechanical system, which is effectively a multimass oscillatory system, has been examined by Ran et al. [15).
REFERENCES 1. W. F. Riley, L. D. Sturges, and D. H. Morris, Mechanics of Materials, 5th ed., Wiley, New York, 1999. 2. S. Timoshenko, D. H. Young, and W. Weaver, Jr., Vibration Problems in Engineering, 4th ed., Wiley, New York, 1974. 3. W. B. Bickford, Advanced Mechanics of Materials, Addison-Wesley, Reading, MA, 1998. 4. J. S. Rao, Advanced Theory of Vibration, Wiley, New York, 1992.
Problems ..315 5. S. K. Clark, Dynamics of Continuous Elements, Prentice-Hall, Englewood Cliffs, NJ, 1972. 6. S. S. Rao, Mechanical Vibrations, 4th ed., Prentice Hall, Upper Saddle River, NJ, 2004. 7. J. H. Faupel and F. E. Fisher, Engineering Design, 2nd ed., Wiley-Interscience. New York, 1981. 8. M. Vet, Torsional vibration of beams having rectangular cross sections, Journal of tire Acoustical Society of America, Vol. 34, p. 1570, 1962. 9. J. M. Vance, Rotordynamics of Turbomachinery, Wiley, New York, 1988. 10. A. E. H. Love, A Treatise on the Mathematical Theory of Elasticity, 4th Ed., Dover. New York, 1944. 11. J. M. Gere, Torsional vibrations of beams of thin walled open section, Journal of Applied Mechanics, Vol. 21, p. 381, 1954. 12. H. Lo and M. Gou1ard, Torsion with warping restraint from Hamilton's principle, Proceedings of the 2nd Midwestern Conference on Solid Mechanics, 1955, p. 68. 13. A. K. Darpe, K. Gupta, and A. Chawla, Coupled bending, longitudinal and torsional vibration of a cracked rotor, Journal of Sound and Vibration, Vol. 269, No. 1-2, pp. 33-60, 2004. 14. I. Y. Shen, W. Guo, and Y. C. Pao, Torsional vibration control of a shaft through active constrained layer damping treatments, Journal of Vibration and Acoustics, Vol. 119, No.4, pp. 504-511, 1997. 15. L. Ran, R. Yacamini, and K. S. Smith, Torsional vibrations in electrical induction motor drives during start up, Journal of Vibration and Acoustics, Vol. 118, No.2, pp. 242-251, 1996. 16. C. P. Filipich and M. B. Rosales, Free flexural-torsional vibrations of a uniform spinning beam, Journal of Sound and Vibration, Vol. 141, No.3, pp. 375-387, 1990. 17. S. P. Timoshenko, Theory of bending, torsion and buckling of thin-walled member of open cross-section, Journal of the Franklin Institute, Vol. 239, pp. 201, 249, and 343, 1945.
PROBLEMS 10.1 A shaft with a uniform circular cross section of diameter d and length 1 carries a heavy disk of mass moment of inertia h at the center. Find the first three natural frequencies and the corresponding modes of the shaft in torsional vibration. Assume that the shaft is fixed at both the ends. 10.2 A shaft with a uniform circular cross section of diameter d and length 1 carries a heavy disk of mass moment of inertia h at the center. If both ends of the shaft are fixed, determine the free vibration response of the system when the disk is given an initial angular displacement of eo and a zero initial angular velocity.
=
10.3 A uniform shaft supported at x 0 and rotating at an angular velocity n is suddenly stopped at the end x = O. If the end x = 1 is free and the cross section of the shaft is tubular with inner and outer radii rj and ro,
respectively, find the subsequent time variation of the angular displacement of the shaft.
=
10.4 A uniform shaft of length 1 is fixed at x 0 and free at x = 1. Find the forced vibration response of the shaft if a torque Mt (t) = Mto cos nt is applied at the free end. Assume the initial conditions of the shaft to be zero. 10.5 Find the first three natural frequencies of torsional vibration of a shaft of length 1 m and diameter 20 rom for the following end conditions: (a) Both ends are fixed. (b) One end is fixed and the other end is free. (c) Both ends are free. Material of the shaft: steel with p = 7800 kglm3 and G = 0.8 X 10" N/m2.
316
Torsional Vibration of Shafts
10.6 Solve Problem 10.5 by assuming the material of the shaft to be aluminum with p = 2700 kg/m3 and G 0.26 X 1011 N/m2•
=
Consider two shafts each of length 1 with thinwalled tubular sections, one in the form of a circle and the other in the form of a square, as shown in Fig. 10.13.
(b) Cross section: hollow with inner diameter 80 rom and outer diameter 120 mm; material: aluminum with p 2700 kglm3 and G 0.26 x IOJ] N/m2.
=
=
10.7
o Figure 10.13
Assuming the same wall thickness of the tubes and the same total area of the region occupied by the material (material area), compare the fundamental natural frequencies of torsional vibration of the shafts.
10.10 Find the first three natural frequencies of torsional vibration of a shaft fixed at x 0 and a disk of mass moment of inertia II = 20 kg·m2/rad attached at x = 1. Shaft: uniform circular cross section of diameter 20 rom and length 1 m; material of shaft: steel with p = 7800 kg/m3 and G = 0.8 X 1011 N/m2.
=
10.11 Find the fundamental natural frequency of torsional vibration of the shaft described in Problem 10.10 using a single-degree-of-freedom model.
10.12 Find the free torsional vibration response of a uniform shaft of length 1 subjected to an initial angular displacement O(x, 0) f)o(x / 1) and an initial angular velocity O(x, 0) 0 using modal analysis. Assume the shaft to be fixed at x 0 and free at x 1.
=
= =
=
10.13 Find the free torsional vibration response of a uniform shaft of length 1 subjected to an initial angular displacement O(x, 0) 0 and an initial angular velocity 10.8 Solve Problem 10.7 by assuming the tube wall thickness and the interior cavity areas of the tubes to be O(x,O) = Vo8(x -1) using modal analysis. Assume the the same. '. .shaft to be fixed at x = 0 and free at x = 1.
=
10.9 Determine the velocity of propagation of torsional waves in the drive shaft of an automobile for the following data:
(a) Cross section: circular with diameter 100 rom; material: steel with p 7800 kg/m3 and G 0.8 X 1011 N/m2•
=
=
10.14 Derive the frequency equation for the torsional vibration of a uniform shaft with a torsional spring of stiffness kt attached to each end. 10.15 Find the steady-state response of a shaft fixed at both ends when subjected to a torque Mt (x, t) = Mto sin Ot at x 1/4 using modal analysis.
=
,
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Ii Transverse Vibration of Beams 11.1 INTRODUCTION The free and forced transverse vibration of beams is considered in this chapter. The equations of motion of a beam are derived according to the Euler-Bernoulli, Rayleigh, and Timoshenko theories. The Euler-Bernoulli theory neglects the effects of rotary inertia and shear deformation and is applicable to an analysis of thin beams. The Rayleigh theory considers the effect of rotary inertia, and the Timoshenko theory considers the effects of both rotary inertia and shear deformation. The Timoshenko theory can be used for thick beams. The equations of motion for the transverse vibration of beams are in the form of fourth-order partial differential equations with two boundary conditions at each end. The different possible boundary conditions of the beam can involve spatial derivatives up to third order. The responses of beams under moving loads, beams subjected to axial force, rotating beams, continuous beams, and beams on elastic foundation are considered using thin beam (Euler-Bernoulli) theory. The free vibration solution, including the determination of natural frequencies and mode shapes, is considered according to the three theories.
11.2 EQUATION OF MOTION: EULER-BERNOULLI THEORY The governing equation of motion and boundary conditions of a thin beam have been derived by considering an element of the beam shown in Fig. l1.1(b), using Newton's second law of motion (equilibrium approach), in Section 3.5. The equation of motion was derived in Section 4.11.2 using the extended Hamilton's principle (variational approach). We now derive the equation of motion and boundary conditions corresponding to the transverse vibration of a thin beam connected to a mass, a linear spring, and a torsional spring at each end [Fig. 11.1(a), (b)] using the generalized Hamilton's principle. In the Euler-Bernoulli or thin beam theory, the rotation of cross sections of the beam is neglected compared to the translation. In addition, the angular distortion due to shear is considered negligible compared to the bending deformation. The thin beam theory is applicable to beams for which the length is much larger than the depth (at least 10 times) and the deflections are small compared to the depth. When the transverse displacement of the centerline of the beam is w, the displacement components of any point in the cross section, when plane sections remain plane and normal
317
_.<.rl".~"'.
-i
318
Transverse Vibration of Beams
(0)
z
fix,t)
0'
.-----
V(x,t) w(x,t)
I I
,
+ dV(x,t)
I I
,
I I I
I I I
dx
,''''
~I
I
I I I I
I I
x (b)
Figure 11.1(a), (b)
Beam in bending.
to the centerline, are given by [Fig. 11.1(c)] aw(x, t)
u
= -z ax
v
= 0,
w
= w(x,
t)
01.1 )
where u, v, and w denote the components of displacement parallel to x, y, and z directions, respectively. The components of strain and stress corresponding to this
J12
Equation of Motion:.Euler-Bemoulli Theory _319
---
-t-_~~I, Original beam
{---_J_--~,
\
w
(c)
Figure 1l.1(c)
Beam in bending.
displacement field are given by
au
exx
C1xx
a2w
= ax = -z ax2 ' =
(11.2)
a2w
-Ez dX2
= ezz = exy = eyZ = e:x = 0
eyy
'
. C1yy
= C1zz = C1xy = C1yz = C1zx = 0
The strain energy of the system (rr) can be expressed as rr = ~
ff f
(C1xxexx
+ C1yyeyy + C1zzezz + C1xyexy+C1YZeyZ + C1zxe:;.t) d V
v
where the first term on the right-hand side of Eq. (11.3) denotes the strain energy of the beam, the second term, in brackets, indicates the strain energy of the springs, and I denotes the area moment of inertia of the cross section of the beam about the y axis: I = ly =
ff A
Z2 d A
(11.4)
320
Transverse Vibration of Beams The kinetic energy of the system (T) is given by
[1
1 t!!(aw)2 T = 210 P ar
)2 1
dA dx + 2m) (aw ar(O, t) + 2m2 (aw ar(l, t) )2]
A
III pA (aw)2 -at
= 2-
[1
dx + 2-m) (aw -(0, t) at
0
)2 + -m2 1 (awat 2
-(1,
t) )2]
(11.5)
where the first term on the right-hand side of Eq. (11.5) represents the kinetic energy of the beam, and the second term, in brackets, indicates the kinetic energy of the attached masses. The work done by the distributed transverse load f(x, t) is given by
i fwdx l
W = The application of the generalized
Hamilton's
81/
(11.6)
principle
gives
2
or
81'21~10t E/ (aax2w)2 +
Uk,
-H' +
dx
2
2
'1
(n - T - W) dt = 0
'1
w'(O, t l+
pA (~~)'
~k"C:
(0,
k,
1))' + ~
w' (/, t l+
~k"(:: (/,1)']
dx -[ ~m, (~~ (0,1))'
~m,(~~(I,1))} l'
fWdX}
dt
=0
(11.7)
The variations in Eq.(1I.7) can be evaluated using integration by pans, similar to those used in Eqs. (4.181),(4.184) and (4.185), to obtain 8
1 III0 '2-
2
'1
2 (a E/ -2w)2 dxdt= ax
1/
[
II
a28w (aw)11 -_ a ( EI-a2w) 8w 2 ax ax 0 ax ax2 0
2
EI-
/1
2
+ 10t axa 2(E/ a2~) ax 8WdX] dt
8
1 -k) w-(O, t) -k (aw-ax = 1 w(O,t)8w(O, t) + 11
1 (aw t) )21 + 2-k2w 2 (1,t) + 2-k'2 -(1, ax k] ktl aw(O, t) 8 (aw(o, t») ax ax + k2w(l, t)8w(l, t) + k'2 aw(l, t) 8 (aw(l, t»)] dt ax ax
/2[1' 2
+
1 tl 2
(0,
r)
)2]
dt
12 [
II
(11.8)
ll.2
Equation of Motion: Euler-Bernoulli Theory
at dx ] dt 1 [12. 10r' pA (aw)2 2 aw 1(2) dx - l' (1"pA-az w owdt ) dx = pA-ow at 1 at,( ) dt 2atowdx I"(11 pA-aZw o 1 [1 (awat t))Z + -mz1 (awat t))Z] dt a2w(0, t) aZw(l, t) ] ow(O,t) + mz Z ow(l, t) dt at 1 at
.321
'2
0'1
1(
o
0'1
=-
'1
'2
'1
-ml
2
0
-(0,
-(I,
2
'2[
=-
ml
(11.9)
2
'1
Note that integration by parts with respect to time, along with the fact that t = tl and t = tz, is used in deriving Eqs. (11.9).
i'2 (1
fWdX) dt
i'21
= 0 at
1
1
o
ow
=
(11.10)
fowdxdt
In view of Eqs. (11.8)-(11.10), Eq. (11.7) becomes
(aw)11 ax ax 1,[ '21 £I-zoaZw 0
I
aZw aw (aw)11 atZ ow 0 + kzwowl I + k'2a;o ax + m2 aZw atZ ow
+
m1
+
1
'2{
t1
II
-a ( £I-zaZw) ow +klwow +ktl-oaw (aw) ax ax 0 ax ax o
0
III dt
Z 10t [aaxZ ( £1 aZw) ax2 + pA aZw atZ - f ] owdx } dt = 0
(11.11)
By setting the two expressions within the braces in each term of Eq. (11.11) equal to zero, we obtain the differential equation of motion for the transverse vibration of the beam as (11.12) and the boundary conditions as
2 £1-2aZw 0 (aw)11 - -a ( £1-2a w) ow +k1wowlo +k'l-oaw (aw)1 ax ax 0 ax ax 0 ax ax +ml-zaZwowlo+k2wowl ,aw +kt2-0 (aw) II +mz-2aZwowl1 =0 at ax ax at
II
0
(11.13)
322
Transverse Vibration of Beams To satisfy Eq. (11.13), the following
conditions
are to be satisfied:
(_Ela2~ +ktlaw)o(aw)j ax ax ax + kr2 aw) (E 1 a2~ ax ax
[!.(Ela2~) ax ax
+kjw+mj
0
(aw) ax
a2~]owl at
=0
(11.14)
=0
(11.15)
=0
(11.16)
=0
(11.17)
x=o
I
x=l
x=o
[- aa (E 1 a2~) + k2w + m2 a2~] owl x ax at x=l
Each of the equations in (11.14)-(11.17) can be satisfied in two ways but will be satisfied in only one way for any specific support conditions of the beam. The boundary conditions implied by Eqs. (11.14)-(11.17) are as follows, At x = 0,
aw ax = constant
[so that 0
(aw) ax = 0]
or
2 (-EI aaxw2+ ktl aw) ax = 0
or
a2w awl [ EI-ax2 +kr2-ax
(11.18)
and
w
At x
= constant(so
that
ow = 0)
or
= I,
~: = constant
= 0]
[so that 0 (~:)
=0
(11.20)
and
w = constant(so
that
ow = 0)
or
In the present case, the second conditions . valid.
stated in each of Eqs. (11.18)-(11.21
)are
Note The boundary conditions of a beam corresponding to different types of end supports are given in Table 11.1. The boundary conditions for supports, other than the supports shown in the table, can be obtained from Hamilton's principle by including the appropriate energy terms in the formulation of equations as illustrated for the case of attached masses and springs in this section.
11:3 FREE VffiRATION EQUATIONS For free vibration, the external excitation is assumed to be zero:
f (x,
t) = 0
(11.22)
II
---
II
-
0
II
~
0
-::-
0
31'"
~ ~>-:
M
'-' M
•...
t:J ~31 ~>-: ~
M
..•
~ I~ ~
t:J
II
.-. ..
'-'
3
II
-
.-. ..
.-. ..
-
'-'
~31>-:~
'-' M
II
-
31 >-: «:> ~
M
'-'
3
'-'
~31>-:~
"-"
«:>I~
.-. ..
0
.-.
~I~
,..-.....
II
t:J
•...
e
0
II
II
-::-
..
ci '-'
3
31"'<~ «:>
ci '-'
0
ci '-'
II
-:::-
e:3
31'" "'< ~~
M
....• ~
0
II
.-. ..
e: 31"'< ~~
M
t:J
~I~
31 ~ «:>«:>
M~31'" «:> "'<
t:J
~
M
•... "-" «:>I~
II
..
<:l
-3 -•... '-'
..:c,
~ ~ '-' M M31 >-: .~ ~ ~ \'5
"
t:J
'-'
0
II
.. ci.-...
.-.
..•
8
"-"
..:c,
«:>I~
;:>.
.-.
s
,..-.....
~ ~'\l
0
..• --..
~31'"~>-:
•... t:J "-"
II
0
~
....•
t:J
.-. ..
'-'
M
0
~31 ~~
M
31 ~ <0«:>
M
II
M
31 >-: ~~
-
,..-.....
M
II
M
M
•...
t:J
0
0
---
II
8
0
~
0
-::-
II
..
.-.
0
---
--
0 0
II
.-. ..
3 I
'-' M
~31 ~~
M
....• ~
c:
.g :.a c: o
u
....;
323
II
II
-~
~r ....
,...., II
'-'
-
~
~
~
II
---. ..
-
II 0
---. .. ,....,..II ,....,..II N~IN'-: ...: '-' .... '-' ~ '-' ~ ~I" N~IN'-:'-'~ N~IN.. N~IN'-:N~IN'-: '->,
N '-: ...:,....,..
~
0
....
~
"0
c:
1::.
....
....
~
-
~r
~
~
N~IN'-: ~-:::.... N~IN.. ~ '-'
'-'
...,
-:::-
.....,. 0
~
1::,
0
.E
u=~ "u - u~
co .~ •..
<
II
~
II
JJ
~ "0
c:
0
N '-: 0'-'..
~r ....
~
6'
II
e
~
'-'
,....,
,
....•0
II
~
/if I
~
~ ~
II
---. .•.
-:::-
0
~
~IN'-: 0'-' -:::- ,....,..II ~0'-'.. N~IN'-: 0.. N '-' e: .... N~IN.. e: '-'0 ~ ~ .... N~IN ~I" .. ~ ~IN N '-: '->, ~ 1::, N~IN'-:N~IN'-: ....." '-' I II
-:::-
II
r
~
....
I
~
0
....
I
~
,....,
0
....
~ ~
'-'
1::, I
.::::
..!!
<
ITa
~~
324
0 II
~
.,~ 0 II
~
~ E'
!~
'.
H.4
Free .vibrationSoJyt~@ 325
and hence the equation of motion, Eq. (11.13), reduces to
£
[EI(X)
a2w(x,
t)]
ax2
ax2
+ pA(x)
2
a w(x. t) = 0 at2
(11.23)
For a unifonn beam, Eq. (11.23) can be expressed as
a4w
a2w
+ -2
c2-4 (x, t) ax
at
(11.24)
(x, t) = 0
where (11.25)
11.4 FREE VffiRATION SOLUTION The free vibration solution can be found using the method of separation of variables as
= W(x)T(t)
w(x, t)
(11.26)
Using Eq. (11.26) in Eq. (11.24) and rearranging yields d4W(x) W(x) dx4,
2
c2
1 d T(t) --------a-w T(t) dt2 -
2
(11.27)
where a = w2 can be shown to" be a positive constant (see Problem 11.20). Equation (11.27) can be rewritten as two equations: d4W(x)
- ,84W(x) = 0
dx4 d2T(t)
--
+ w2T(t)
dt2
where
w2
=0
pAw2
,84 - - - -- c2 EI The solution of Eq. (11.29) is given by T(t) = A coswt
+ B sinwt
(11.28) (11.29)
(11.30)
(11.31)
where A and B are constants that can be found from the initial conditions. The solution of Eq. (11.28) is assumed to be of exponential form as W(x)
= Cesx
(11.32)
where C and s are constants. Substitution of Eq. (11.32) into Eq. (11.28) results in the auxiliary equation S4 -
(11.33)
,84 = 0
The roots of this equation are given by SI,2
= ±,8,
S3,4
= ±i,8
(11.34)
326
Transverse Vibration of Beams Thus, the solution of Eq. (13.28) can be expressed as W(x)
= C]e/3x
+ Cze-/3x + C3e;/3x + C4e-;/3x
C1,
where Cz, C3, and C4 are constants. conveniently as
= C] cos
W(x)
fJx
Equation
+ Cz sinfJx + C3 cosh
(11.35)
(11.35) can be expressed
fJx
+ C4 sinh fJx
more
(11.36)
or W(x)
+ coshfJx) + Cz(cosfJx - coshfJx) + C3 (sin fJx + sinh fJx) + C4 (sin fJx - sinh fJx)
= C] (cos
fJx
(11.37)
where C], Cz, C3, and are different constants in each case. The natural frequencies of the beam can be determined from Eq. (11.30) as C4,
w
= p2; pA El = (P/)2; E 1 pAl4
(11.38)
The function W (x) is known as the normal mode or characteristic function of the beam and w is called the natural frequency of vibration. For any beam, there will be an infinite number of normal modes with one natural frequency associated with each normal mode. The unknown constants C] to C4 in Eq. (11.36) or (11.37) and the value of fJ in Eq. (11.38) can be determined from the known boundary conditions of the beam. If the ith natural frequency is denoted as W; and the corresponding normal mode as W;(x), the total free vibration'response of the beam can be found by superposing the normal modes as 00
w(x, t)
=
L
W;(x)(A;
cosw;t
+
B; sinw;t)
(11.39)
;=]
where the constants beam.
11.5
A; and B; can be determined
from the initial conditions
of the
FREQUENCIES AND MODE SHAPES OF UNIFORM BEAMS The natural frequencies and mode shapes of beams with a uniform cross section with different boundary conditions are considered in this section.
11.5.1 Beam Simply Supported at Both Ends The transverse displacement and the bending moment are zero at a simply supported (or pinned or hinged) end. Hence, the boundary conditions can be stated as
W (0) = 0 dZW E I dxz (0) = 0
(11.40) or
(1l.41)
l.t:"5. 'Frequencies 'and Mode'Shapes m'Uniform .£eaR~ . :327
(11.42)
W(l) = 0
d2 W
E I dx2 (l) = 0
d2 W
or
dx2 (l)
=0
(11.43)
When used in the solution of Eq. (11.37), Eqs. (11.40) and (1l.4l) yield (11.44)
CI = C2 = 0 When used with Eq. (11.37), Eqs. (11.42) and (11.43) result in C3(sinf3l
+ sinhf3l) + C4(sinf3l
-C3(sin f3l - sinh f3l) - C4(sin f3l
=0
(11.45)
+ sinh f3l) = 0
(11.46)
- sinhf3l)
Equations (11.45) and (11.46) denote a system of two equations in the two unknowns C3 and C4. For a nontrivial solution of C3 and C4, the determinant of the coefficients must be equal to zero. This leads to -(sinf3l
+ sinh,8l)2 + (sinf3l
- sinhf3l)2
=0
or sin,8l sinh f3l
=0
(11.47)
=
It can be observed that sinh f3l is not equal to zero unless f3 = O. The value of f3 0 need not be considered because it implies, according to Eq. (11.38), W = 0, which corresponds to the beam at rest. Thus, the frequency equation becomes (11.48)
sinf3l = 0
The roots of Eq. (11.48), f3nl, are given by (11.49)
n = 1,2, ... and hence the natural frequencies of vibration become Wn
EI )1/2 = (f3nl)2 ( pAl4 = n2rr2
(EI
pAl4
)1/2
'
Substituting Eq. (11.48) into Eq. (11.45), we find that gives the mode shape as mrx Wn(x) = Cn sinf3nx = Cn sin -l-'
= 1,2, ... C3 = C4. Hence, n
n
= 1,2, ...
(11.50) Eq. (11'.37)
(11.51)
The first four natural frequencies of vibration and the corresponding mode shapes are shown in Fig. 11.2. The normal modes of vibration are given by Wn(x, t) = Wn(x)(An coswnt
+ Bn sinwnt),
= 1,2, ...
(11.52)
+ Bn sinwnt)
(11.53)
n
The total (free vibration) solution can be expressed as 00
00
. ~ ~ nrrx w~x, t) = ~ wn(x, t) = ~ sin -l-(An n=1
n=1
coswnt
328
Transverse Vibration of Beams W1(X)
nfftn
_
o
.81/
I"
4 W (X) ~~
= 3.1416
.84/=12.5664
o
/
Figure 11.2 Natural frequencies and mode shapes of a beam simply supported at both ends. (fJnl)2(EI/pAL4)1/2, fJnl mr.
Wn
=
If the initial conditions
=
are given by w(x,O)
Eqs. (11.53)-(11.55)
=
wo(x)
(11.54)
aw( 0 . at x, ) = wo(x)
lead to
00
""
(11.55)
. mrx
AnS10 -1- =
wo(x)
(11.56)
n=l 00
""
mr X wnBn sin -1- = wo(x)
(11.57)
n=l
Multiplying Eqs. (11.56) and (11.57) by sin(mrr x/I)
All
=
2
2 wnl
Bn = -
11.5.2
t
T Jo
and integrating
from Oto 1 yields
. nrrx dx
WO(X)S1O
i
-I-
l
0
.. nrr x wo(x) S10 -dx
1
(11.58)
9
(11.5 )
Beam Fixed at Both Ends At a fixed end, the transverse displacement and the slope of the displacement Hence, the boundary conditions are given by W (0) = 0
dW
(0) dx
=0
are zero.
(11.60) (11.61)
11.5 Frequencies and Mode Shapes of Uniform Beams '62.9 W (i) = 0
dW dx
(i)
(11.62)
=0
(11.63)
When Eq. (11.37) is used, the boundary conditions (11.60) and (11.61) lead to Ct
= C3 = 0
(11.64)
and the boundary conditions (11.62) and (11.63) yield CZ(cos {3i - cosh {3i)
-Cz(sin {3i
+ C4(sin {3i -
+ sinh {3i) + C4(COS{3i
-
= cosh {3i) = 0 sinh {31) 0
(11.65) (11.66)
Equations (11.65) and (11.66) denote a system of two homogeneous algebraic equations with Cz and C4 as unknowns. For a nontrivial solution of Cz and C4, we set the determinant of the coefficients of Cz and C4 in Eqs. (11.65) and (11.66) to zero to obtain
I
cos {3i - cosh {3i -(sin {3i + sinh {3i)
I
sin {3i - sinh {3i =0 cos {3i - cosh {3i
or
=
(cos {3i - cosh {31)z + (sinz {3i - sinhz {31) 0
(11.67)
Equation (11.67) can be simplified to obtain the frequency equation as cos {31cosh {31- 1 = 0
(11.68)
Equation (11.65) gives C4 =
cos {31- cosh {31C sin {31_ sinh {31 z
(11.69)
If {3n1 denotes the nth root of the transcendental equation (11.68), the corresponding mode shape can be obtained by substituting Eqs. (11.64) and (11.69) in Eq. (11.37) as Wn(x)
= Cn [ (cos{3nx
- cosh {3nx) -
cos {3n1- cosh {3nl .. ] . {3I 'nh{3 l (s1O{3nx- Sinh {3nx) sm n - SI n (11.70)
The first four natural frequencies and the. corresponding mode shapes are shown in Fig. 11.3. The nth normal mode of vibration can be expressed as Wn(X, t)
= Wn(x)(An
coswnt
+ Bn sinwnt)
(11.71)
and the free vibration solution as 00
w(x, t) =
L wn(x,
t)
n=\
~ [ =L.., (cos{3nx-coshpnx)-. , n=l
cos {3n1- cosh {3nl 1 'hRl(sin{3nx-sinh{3nx) s1O{3n- sm pn
]
(11.72)
1
,
, ,,,,
"c".·
"~._,,....J
330
Transverse Vibration of Beams
@
W,(x)
_
______
@ N=4.7300
o
1
~
2
W (x)
..
---............ @ fJ21 = 7.8532
~
~b.51
~
o
@
W,«)
1
--------~i' ...
@ P,I.IO.9956
o
1
@~~
W,«)
@P,loI4.1372
o Figure 11.3
1
Natural frequencies and mode shapes of a beam fixed at both ends. + l);rr /2.
(Un
(~nl)2(E I / pAL 4)1/2, ~nl ::::(2n
=
where the constants An and Bn in Eq. (11.72) can be detennined from the known initial conditions as in the case of a beam with simply supported ends.
11.5.3 Beam Free at Both Ends At a free end, the bending moment and shear force are zero. Hence, conditions of the beam can be stated as d2W
El
d3W
El
(0) dx3
d2W
El
By differentiating d2W --2 dx
(x)
dx
3
=0
(11.73)
= 0 or
d3 W (0) dx3
=0
(11.74)
=0
(11.75)
=0
(11.76)
= 0 or
(l) dx3
d2 W (I) dx2 d3 W (I) dx3
Eq. (11.37), we obtain
= fi2[C)
(- cos fix
+ C3( d3 W (x)
d2 W (0) dx2
= 0 or
(l) dx2
d3W
El
or
=0
(0) dx2
the boundary
= fi3[C)
sin fix
(sin fix
+ cosh fix) + C2(-
+ sinh fix) + C4(+ sinh fix) + C2(sin
cos fix - cosh fix)
sin fix - sinh fix)] fix - sinh fix)
(11.77)
~",
331
11.5 -Frequencies and"Mode Shapes of Uniform Beams
."
Equations (11.73) and (11.74) require that (11.79) in Eq. (11.37), and Eqs. (11.75) and (11.76) lead to C I(- cos fJi
C 1 (sin fJI
+ cosh fJi) + C3 (- sin fJI + sinh fJI) = 0 + sinh fJi) + C3 (- cos fJi + cosh fJi) = 0
( 11.80) (11.81)
For a nontrivial solution of the constants CI and C3 in Eqs. (11.80) and (11.81), the determinant formed by their coefficients is set equal to zero:
1
+ cosh fJi + sinh fJI
cos fJi
-
sin fJI
- sin fJI - cos fJi
+ sinh fJi I = 0 + cosh fJi
(11.82)
Equation (11.82) can be simplified to obtain the frequency equation as cos fJI cosh fJi - 1
=0
(11.83)
Note that Eq. (11.83) is the same as Eq. (11.68) obtained for a beam fixed at both ends. The main difference is that a value of fJoi = 0 leads to a rigid-body mode in the case of a free-free beam. By proceeding as in the preceding case, the nth mode shape of the beam can be expressed as Wn(x) = (cosfJnx
+ coshfJnx)
-
cos fJni - cosh fJni .. . fJ I 'nhfJ i (smfJnx
smn-Sl
n
+ slnhfJnX)
(11.84)
The first five natural frequencies given by Eq. (11.83) and the corresponding mode shapes given by Eqs. (11.80) and (11.81) are shown in Fig. 11.4. The nth normal mode and the free vibration solution are given by (11.85) and 00
w(x, t) =
L wn(x,
t)
n=1
~
=~
[
(cosfJnx
+ coshfJnx)
-
cos fJni - cosh fJnI . sinfJnisinhfJnI (smfJnx
+ sinhfJnx)
]
(11.86)
11.5.4
Beam with One End Fixed and the Other Simply Supported At a fixed end, the transverse deflection and slope of deflection are zero, and at a simply supported end, the transverse deflection and bending moment are zero. If the
332
Transverse Vibration of Beams
---------1- ----
--------l------
Rigid-body motion(s)
~ fJol=o
r---
~0.224/
-----.......
N'4.7300
I~
0.776/
o
/
/;~
~~,~
0.132/
0.5/
""=78532
0.868/
o
/
L~·
0.094/
,~
0.35~0.644/
fJ3/=1O.9956 0.9~
o
/ ~
../' 0.073/
L,.. 0.277/
)
~
'---tJ.5/
'~ 0.723/
./-
fJ4/-
14.1372
0.927/
o
I
Figure 11.4 Natural frequencies and mode shapes of a beam with free ends. (Pnl)2(E I / pAL 4)1/2, Pni ::: (2n + 1)11"/2.
beam is fixed at x = 0 and simply supported at x = l, the boundary stated as
W (0) dW dx (0) W (l)
=0
(11.87)
=0 =0
(l) =0
(11.88) (11.89) or
(11.90)
Condition (11.87) leads to
(11.91) . in Eq.(11.36),
dW/
and Eqs. (11.88) and (11.36) give
-dx x=o = /3( -Cj sin /3x
=
conditions can be
d2W EI dx2
{Un
+ C2 cos /3x + C3 sinh /3x + C4 cosh
/3x).r=o
=0
or
(11.92)
-B.S ·Frequenciesand -Mode-Snapes'of UnitQrmBeams
333
Thus, the solution, Eq. (11.36), becomes
+ C2(sin fJx
= CI (cos fJx - cosh fJx)
W(X)
- sinh fJx)
( 11.93)
Applying conditions (11.89) and (11.90) to Eq. (11.93) yields CI(cosfJ1-Cl
=0 C2(sin fJ1 + sinh fJ1) = 0
+ C2(sinfJ1-
coshfJ1)
(cos fJ1 + cosh fJ1) -
sinhfJl)
(11.94) (11.95)
For a nontrivial solution of CI and C2 in Eqs. (11.94) and (11.95), the determinant of their coefficients must be zero:
I
cos fJ1 - cosh fJ1 -(cosfJ1
sin fJ1 - sinh fJ1 -(sinfJ1 + sinhfJ/)
+ coshfJ1)
1-
0
-
(11.96)
Expanding the determinant gives the frequency equation cos fJ1 sinh fJ1 - sin fJI cosh fJ1 = 0 or tan fJ1
= tanh fJ1
(11.97)
The roots of this equation, fJn1,· give the natural frequencies of vibration:
EI )
Wn
= (fJn1)2 ( pAl4
1/2
'
n = 1,2, ...
(11.98)
If the value of C2 corresponding to fJn is denoted as C2n, it can be expressed in terms of Cln from Eq. (11.94): C2n
= -Cln
cos fJnl - cosh fJnl . fJ n 1 - SI'nhfJ n 1 sm
(11.99)
Hence, Eq. (11.93) can be written as Wn(x)
- coshfJnx) -
= Cln' [ (cosfJnx
cos fJnl - cosh fJnl ] sinfJn1- sinhfJn1 (sinfJnx - sinhfJnx) (11.1 00)
The first four natural frequencies and the corresponding mode shapes given by Eqs. (11.98) and (11.100) are shown in Fig. 11.5.
11.5.5 Beam Fixed at One End and Free at the Other
=
If the beam is fixed at x 0 and free at x = 1, the transverse deflection and its slope must be zero at x = 0 and the bending moment and shear force must be zero at x = 1.
334
Transverse Vibration of Beams
~
TTf7:trr
~0.5591
f32/ = 7.0686
/
~
ao
~
~
~0.294/
TTf7:trr
.0.52~651
/
f341= 13.3518
Figure 11.5 Natural frequencies and mode shapes of a fixed simply supported beam. (/3n1)2(EI/pAL4) 1/2, /3n1:::: (4n + 1)1r/4.
Thus, the boundary conditions
W n
=
become
W (0) dW dx{O)
d2W
EI-(l) 2
dx
=0
(11.101)
=0
(11.102)
=
'.
0
or
(11.103)
d3W EI dx3 (l) =0
or
(11.1 04)
When used in the solution of Eq. (11.37), Eqs. (11.10 1) and (11.102) yield (11.1 05) When used with Eq. (11.37) Eqs. (11.103) and (11.104) result in C2(cos fJl C2( - sin fJl
+ cosh fJl) + C4(sin fJl + sinh fJl) + sinh fJI) + C4(cos fJl + cosh fJl)
Equations (11.106) and (11.1 07) lead to the frequency cos fJI cosh fJl
+I
=0 =0
(11.106) (11.1 07)
equation
=0
(11.108)
Using Eq. (11.106), we obtain cos fJl sin fJl
+ cosh fJl + sinh fJl
C2
(11.1 09)
'., ..
-.,
...~.~ .• ~~,",.
Frequencies and Mode Shapes·()f Uniform, Beams
11:5
W\(x)
-------....
~ 0
J3S
=
f311 1.8751
I
W2(x)
~ 0
/
-------
=
f3l1 4.6941
~O.7831
~ W3(x)
~
~.5041
1
0
~ W4(x)
~
=
f331 7.8547
0.868/"-........
,/
~0.3581
0.644~.9061
M=
10.9956
1
0
Figure 11.6 Natural frequencies and mode (/3nl)2(E 1/ pAL 4)1/2. /3n1 ~ (2n - l)n /2.
shapes
of
a fixed- free
beam.
Wn
=
and hence the nth mode shape can be expressed as Wn(x)
= (cos{3nx
. cos {3nl . {3 l sm n
- cosh {3nx) -
+ cosh {3nl . x + SI'nh{3n l (sm{3n
- sinh{3nx)
(11.110)
The first four natural frequencies and the corresponding mode shapes given by Eqs. (11.108) and (11.110) are shown in Fig. 11.6. Example 11.1 Determine the natural frequencies and mode shapes of transverse vibration of a uniform beam fixed at one end and a mass M attached at the other end. SOLUTION If the beam is fixed at x = 0, the transverse deflection and its slope are zero at x = O. At the other end, x l, the bending moment is zero and the shear force is equal to the inertia force due to the attached mass M. Thus, the boundary conditions can be stated as
=
(E11.1.1)
W (0) = 0 dW dx (0)
=0
d2 W E 1 dx2 (l)
=0
a3w(l. E1
ax3
t)
=M
(E11.1.2) d2 W dx2 (l) = 0
or
a2w(l,
ot2
t)
(E11.1.3)
336
Transverse Vibration of Beams
or d3W EI dx3 (I) The boundary conditions (ElI.U)
= -MCllW(l)
(EI1.1.4)
and (El1.1.2) require that (El1.1.S)
in Eq. (I 1.37). Thus, the solution becomes W(X) = C2(cos{3x - cosh{3x) + C4(sin{3x - sinh{3x)
(El1.1.6)
When used in Eq. (EI1.1.6), the conditions (EI1.1.3) and (El1.1.4) lead to C2(COS{31 + cosh {31) + C4(sin {31+ sinh {31) = 0 C2[(-EI{33)(sin{3l-
sinh{3l) - M(J}(cos{3l-
(EI1.1.7)
cosh {31)]
{31 + cosh {31) - M(J}(sin {31- sinh {31)] = 0
+ C4[(EI{33)(cos
(El1.1.8)
For a nontrivial solution of the constants C2 and C4, the determinant formed by their coefficients is set equal to zero: cos {31 + cosh {31 [-EI{33(sin{31 - M
(J)2
sin {31+ sinh {31
- sinh{3l)
[EI{3\coS{3I+coSh{3l)
(cos {31 - cosh {31)]
- M
(J)2
=0
(El1.1.9)
(sin {31 - sinh {31)]
The simplification of Eq. (El1.1.9) leads to the frequency equation 1
I
+ cos {31cos h {31 -
R{31(tan {31- tanh {31) = 0
(EI1.1.l0)
M R=-
pAL
(EI1.UI)
denotes the ratio of the attached mass M to the mass of the beam. Equation (EI1.1. 7) gives cos {31+ cosh {31 . {3I + SInh . {31 C2 sm
(EI1.U2)
and hence the nth mode shape can be expressed as Wn(x)
= C2n [(COS{3nx -
cosh {3nx)
cos fJnl + cosh fJnl ... ] x . fJ 1 . hfJ I (sm{3n smh{3nx) sm n + sm n
(EII.I.13)
where C2n is a constant. The first two natural frequencies given by Eq. (EI1.1.l0) for different values of the mass ratio (R) are given in Table 11.1.2.
,11.5 .. PreqUCtlcies and Mede.Shapes-of Unitoonlk.;.uns
.337
Table 11.1.2 Natural Frequencies of a Beam with One End Fixed and a Mass Attached at the Other R
n
/3nL
0.01
1 2 1 2 1 2
1.852 4.650 1.723 4.399 1.248 4.031
0.1
R 10 100 00
n
/3nL
1 2 1 2 1 2
0.736 3.938 0.416 3.928 0 3.927
Example 11.2
The ends of a beam carry masses and are supported by linear springs and linear viscous dampers as shown in Fig. 11.7(a). State the boundary conditions of the beam using an equilibrium approach.
x=o (a)
d2w
d2w mi -
a~
"'2 -arZ
(O,t)
(l,t)
V(O,t)
aw
Ci-(O,t)
at
aw
C2- (l,t)
at
(b)
Figure 11.7(a), (b)
°
SOLUTION If the transverse displacement, velocity, and acceleration of the beam at 2 2 are assumed to be positive with values w(O, t), 8w(O, t)j8t, and 8 w(O, t)j8t , respectively, the spring force klW(O, t), damping force cl[8w(O, t)j8t], and the inertia force ml[82w(O, t)j8t2] act downward, as shown in Fig. 11.7(b). The positive shear force (V) at x = is equal to the negative of the forces of spring, damper, and inertia
x =
°
338
Transverse Vibration of Beams
Beam, EI
x=o
x= I (c)
x=o
x=l (d)
Figure 1l.7(c), (d)
at x = O. This boundary condition can be expressed as
a [ EI(x) a2w(x,t)] ax
Vex, t) = ax
2
2
aw =- [ k1w(x,t)+clar(x,t)+ml
a w(x, t)] at
In addition, the bending moment must be zero at x E I (x)
a2W(~,
ax-
at
2
t)
=0
x
=0
(EI1.2.1)
= 0: x
at
=0
(EI1.2.2)
In a similar manner. the shear force boundary condition at x = I can be expressed as 2
Vex, t)
= axa [ EI(x) a w(x,0] ax2 = k,w(x, t) -
+ C2 aw(x,
at
t)
2w(x, t) at2
+ m2 a
at
x =I
(EI1.2.3)
In addition, the bending moment must be zero at x = I: 2
EI(x)a W(x,t) ,.;
.:,'
ax2
=0
at
x =I
(EI ].2.4)
11.6 Orthogonality of Normal Modes
339
Note: If the ends of the beam carry mass moments of inertia and are supported by torsional springs and torsional dampers as shown in Fig. 11.7(c), the reaction moments at the ends are shown in Fig. l1.7(d). Thus, the bending moment and shear force boundary conditions at the ends can be expressed as a2W(x,t) ax2
M(x, t) = E/(x)
a2w(x, t) axat -
-Ctl V(x, t)
=
a [ ax E/(x)
a3w(x, t) axat2
/01
2 a w(x,t)] ax2
=0
at at
x =0
x =0
(El1.2.5) (El1.2.6)
a2w(x, t) Ow(x, t) ox2 = kr2 ox
. M(x, t) = E/(x)
o2w(x, t) oxot
+ Ct2 V(x, t)
aw(x,t)
= -ktl-a-x-
= oxo [ E/(x)
+
l 02
02W(X,t)] ox2
o3w(x, t) oxot2
=0
at at
x =1
x=l
(El1.2.7) (El1.2.8)
11.6 ORTHOGONALITY OF NORMAL MODES The eigenvalue problem corresponding to a nonuniform beam can be obtained assuming a harmonic solution with frequency w in Eq. (11.23) as 2 d -2 dx
[ E/(x)
2 d W(X)] 2 dx
= w2pA(x)W(x)
by
(11.111)
wf
W]
To derive the orthogonality relations for beams, consider two eigenvalues and and the corresponding eigen or normal functions Wj(x) and Wj(x), respectively, so that 2
d
- -2
dx
2 [ E/(x) d Wj(X)] 2 dx
= wrpA(x)Wj(x)
(11.112)
= wj2pA(x)
(11.113)
and 2 d2 [ E/(x) d W.(X)] -2 j2 dx dx
Wj (x)
Multiply Eq. (11.112) by Wj(x) and integrate over the length of the beam to obtain
t 10
2 d2W Wj(x) d [E/(X) 2 dx dx
i;X)]
dx
=wr 10t pA(x)Wj(x)Wj(x)dx
(11.114)
Integrating the left-hand side of Eq. (11.114) by parts twice and using any combination of the boundary conditions among fixed, pinned, and free ends of the beam [given by
340
Transverse Vibration of Beams
Eqs. (11.18)-(11.21)], we obtain 2 d W(X)] '2 dx
1 o
1 d2 [ Wj(X)-2 EI(x) dx
2 j d W (x)]/1 dx2
= W.(X)~ [EI(X) J dx
=
o
o
1 EI(x)
d2Wj(X) d2Wj(x) dx
2
dx
Thus, Eq. (11.114) can be written as
1 o
1 EI(x)
o
EI(x/2Wj(X) dx2
_ dWj(x) dx
11 0
1 d2W·(x) d2Wj(x) EI(x)· J 2 dx dx 2 dx
1 1 +
dx
d2W·(x) d2 Wj (x) J2 d 2 dx = dx x
dx
2
(11.115)
w; 1pA(x)Wj(X)Wj(x)dx 1
0
(11.116)
Similarly, by multiplying Eq. (11.113) by Wj (x) and integrating over the length of the beam, we can obtain
t
io
EI(x)
2 d Wj(x) d2W·(x) dx2 d:2 dx =
t
w710
pA(x) Wj (x) Wj (x) dx
(11.117)
Noting that the order of the subs~ripts i and j ~nder the. integrals is immaterial and subtracting Eq. (11.117) from Eq. (11.116) yields (w; -
(7)
1
1
=
pA(x)Wj(x)Wj(x)dx
°
(11.118)
Since the eigenvalues are distinct, Eq. (11.118) gives
1
1 dx = 0,
pA(x)Wj(x)Wj(x)
i, j = 1,2, ... ,
(11.119)
In view of Eq. (11.119), Eq. (11.116) or (11.117) gives
1 o
1 EI(x)
d2Wj(X) d2Wj(x) dx
2
dx
2
dx=O,
i,j=1,2,
...
(J},,/ -'- w~ J
(11.120)
Equation (11.119) is called the orthogonality relation for nonnal functions. Equation (11.120) represents another fonn of the orthogonality condition for the normal modes. In fact, by nonna1izing the ith nonnal mode as
1
1 pA(x)wl(x)
dx = 1,
i
= 1,2, ...
(11.121)
Eqs. (1l.I 19) and (11.121) can be expressed in the following fonn: (11.122)
'-I
11.7 Free Vibration Response
ctueto 1nitial"Conditions
. ..341
where dij is the Kronecker delta: i#j
~.. _ { 0, I}
-
I,
(11.123)
i=j
Using Eq. (11.122) in Eq. (11.114), another fonn of orthogonality relation can be derived as
I
t
o
d2 [ Wj(x)-2 E/(x) dx
d2Wi(X)] 2 dx
2
(11.124)
dX=Widij
According to the expansion theorem, any function W (x) that satisfies the boundary conditions of the beam denotes a possible transverse displacement of the beam and can be expressed as a sum of eigenfunctions as 00
W(x) =
I>i Wi (x)
(11.125)
i=l
where the constants
Cj
Ci
and CjWi2
=
I
t
=
are defined by
I
t pA(x) Wi (x)W(x)
2 Wj(X)-2d o dx
dx,
2 [ E/(x) d W(X)] 2 • dx
(11.126)
i = 1,2, ...
dx,
i
= 1,2, ...
(11.127)
Note that the derivative 2
d
dx2
2
[E / (x) d W (x) ] dx2
is assumed to be continuous in Eq. (11.127).
11.7 FREE VmRATION RESPONSE DUE TO INITIAL CONDITIONS The free vibration response of a beam can be expressed as a linear combination of all the natural or characteristic motions of the beam. The natural or characteristic motions consist of the natural modes multiplied by time-dependent harmonic functions with frequencies equal to the natural frequencies of the beam. To show this, consider the response in the following fonn: 00
w(x, t) =
L Wi(X)17i(t)
(11.128)
i=l
where Wi(x) is the ith natural mode and 17i(t) is a time-dependent function to be determined. By substituting Eq. (11.128) in the equation of motion of free vibration of the beam, Eq. (11.23), we obtain
f; 00
d2 [ dx2 E/(x)
d2Wi(X)] dx2
00
17i(t) + f;PA(X)Wi(X)~
d217j(t)
=0
(11.129)
342
TransverseVibrationof Beams Multiplying Eq. (11.129) by
and integrating over the length of the beam yields
Wj(x)
(11.130) In view of the orthogonality conditions, Eqs. (11.122) and (11.124), Eq. (11.130) gives the following equations, which are known as modal equations: d211i(t) dt2
2
+ Wi 11i(t)
= 0,
i = 1,2, ...
(11.131)
where 11i(t) is called the ith modal displacement (coordinate) and Wi is the ith natural frequency of the beam. Each equation in (11.131) is similar to the equation of motion of a single-degree-of-freedom system whose solution can be expressed as =
11i(t)
Ai COSWit
+ Bi sin Wit,
i = 1,2, ...
(11.132)
where Ai and Bj are constants that can be determined from the initial conditions. If = 0) =
11i (t
11i (0)
and
d 11i (t = 0) = dt
1]i
(0)
(11.133)
are the initial values of modal displacement and modal velocity, Eq. (11.132) can be expressed as l'Ji(t)
= 7Ji(O)COSWit+
1]i (0)
--smwit,
.
i = 1,2, ...
Wi
(11.134)
If the initial displacement and velocity distributions of the beam are specified as w(x, t
=: 0) =:
wo(x),
aw (x, t
at
= 0) =
w(x, 0)
=
wo(x)
(11.135)
the initial values of modal displacement and modal velocity can be determined as follows. Using Eq. (11.135) in Eq. (11.128), we find 00
w(x, t
=: 0) =:
L
Wi (X)7Ji (0)
=:
wo(x)
(11.136)
i=] 00
at
~
(11.137)
i=1
By multiplying Eq. (11.136) by pA(x) Wj (x), integrating over the length of the beam, and using the orthogonality condition of Eq.(11.122), we obtain 7Ji(O)
=:
l'
pA(x)Wi(x)wo(x)dx,
i =: 1,2, ...
(11.138)
1"1:7 FreeVibralion 'Response due to initial Conditions -c>343
A similar procedure with Eq. (11.137) leads to 17j(O)
1/
=
pA(x)Wj(X)WO(X)
i
dx,
= 1,2, ...
(11.139)
Once TJj (0) and 17j(0) are known, the response of the beam under the specified initial conditions can be computed using Eqs. (11.128), (11.134), (11.138), and (11.139). The procedure is illustrated in the following example. Example 11.3 A uniformly distributed load of magnitude to per unit length acts on the entire length of a uniform simply supported beam. Find the vibrations that ensue when the load is suddenly removed. SOLUTION The initial deflection of the beam under the distributed load of intensity fo is given by the static deflection curve [2] wo(x)
=
+ Z3x)
- 2lx3
~(x4
24£1 and the initial velocity of the beam is assumed to be zero: Wo(x)
(EI1.3.1)
(El1.3.2)
=0
For a simply supported beam, the normalized normal modes can be found from Eq. (11.121):
1
1
= 1,
PA(x)~l(x)dx
i
= 1,2, ...
(El1.3.3)
In the present case, the normal modes are given by Eq. (11.51): irrx = C; sm. -Z-,
Wj (x)
i
= 1, 2, ...
(EI1.3.4)
where Cj is a constant. Equations (El1.3.3) and (El1.3.4) lead to pAC;
t sin
10
2
irrx
-Z- dx = 1
(E11.3.5)
or
c- = I
f2
i = 1,2, ...
VPAi'
(EI1.3.6)
Thus, the normalized normal modes are given by Wj(x) =
sin f;I; pAZ
irrx -Z-,
i = 1,2, ...
(El1.3.7)
The response of a beam subject to initial conditions is given by Eqs. (11.128) and (11.134): (EI1.3.8)
344
Transverse Vibration of Beams
where 17i(0) and 17i(0) are given by Eqs. (11.138) and (11.139): ., (0) = (' pA(x) W,(x)wo(x) dx = PAl
2 {' sin in pAL I
Jo
=
2"j2pAllo/4 El1r5i5
1
~(X4
Jo
_
2lx'
+ Z'x) dx
24E/
i = 1,3,5, ...
(EI1.3.9)
1
17;(0)=
dx =0,
pA(x)W;(x)wo(x)
i=I,2,
(EI1.3.1O)
...
Thus, the response of the beam can be expressed as [Eq. (EI1.3.8)]
~ V;Ai /2 sin -1i7rX 2"j2pAllo/ E/7r i
4
w(x, t) = . ~
/4 4/0 =- 5 E/7r
coswit
55
1=1,3•...
L
OO
i=I,3, ...
1. i7rx -sm-cosw·t i5 I
(El1.3.1l)
1
11.8 FORCED VmRATION The equation of motion of a beam under distributed transverse force is given by [see Eq. (11.12)] 2
8
8x2
[
E/(x)
2 8 w(x, t)] 8x2 +pA(x)
82w(x, t) 8t2 =/(x,t)
(11.140)
Using the normal mode approach (modal analysis), the solution of Eq. (11.140) is assumed to be a linear combination of the normal modes of the beam as 00
w(x, t) =
L Wi(X)17;(t)
(11.141)
;=1
where W;(x) are the normal modes found by solving the equation (using the four boundary conditions of the beam) 2 2 d [ d W;(X)] -2 E/(x) 2 dx dx
-
pA(x)w;Wi(x)
=0
(11.142)
and 17;(1) are the generalized coordinates or modal participation coefficients. Using Eq. (11.141), Eq. (11.140) can be expressed as
L 00
2 d [d2W;(X)] dx2 E/(x) dx2
17i(t) + pA(x)
;=1
d217;(t)
L Wi(X)~ 00
= I(x, t)
(11.143)
;=1
Using Eq. (11.142), Eq. (11.143) can be rewritten as 00
pA(x) "" w;Wi (X)17i(t) ~ 1=1
+ pA(x)
00
"" Wi (x) ~ 1=1
d217i(t) dt
2
= I(x,
t)
(11.144)
"11":gforced Vibration
345
Multiplying Eq. (11.144) by Wj(x) and integrating from 0 to [ results in 00
L 7li(t) i=1
/
( pA(x)wfWj(x)Wj(x)
10
d27Ji(t) + L~ 00
dx
1/
1/
0
0
pA(x)Wj(x)Wi(x)dx=
i=l
(11.145)
Wj(x)f(x,t)dx
In view of the orthogonality condition, Eq. (11.122), all terms in each of the summations on the left side of Eq. (11.145) vanish except for the one term for which i = j, leaving i = 1,2, ...
(11.146)
where Qi(t) is the generalized force corresponding to the ith mode given by Qi(t) =
1/
Wi (x)f(x,
(11.147)
i = 1,2, ...
t) dx,
The complete solution of Eg. (11.146) can be expressed as [see Eq.(2.109)] 7Ji(t) = Ai coswit
r
+ Bi sin wit + ~
10
Wi
(11.148)
Qi(r) sin Wi(t - r)dr
Thus, the solution of Eg. (11.140) is given by Eqs.(11.141) and (11.148): w(x, t) =
f:
[Ai coswit
+ Bi sin wit +
i=1
~"1t I
Qj(r) sinwi(t - r) dr]
Wi(x)
0
(11.149) Note that the first two terms inside the brackets denote the free vibration, and the third term indicates the forced vibration of the beam. The constants Ai and Bi in Eq.( 11.149) can be evaluated using the initial conditions of the beam. Example 11.4 Find the response of a uniform simply supported beam subjected to a step-function force Fo at x as shown in Fig. 11.8. Assume the initial conditions of the beam to be zero.
= ~,
SOLUTION The natural frequencies and the normal modes of vibration of a uniform simply supported beam are given by w. I
Wi(x)
___________________
IB!!I!!!!!!!!!!!!!!
= ;',,' / [2 =C
i
sin-
£1
(El1.4.1)
pA irrx
(El1.4.2)
[
!!!'!""" __
--------.-.<
-<........ol-
346
Transverse Vibration of Beams F(t) = Fo
1
I
I
\\!\ \\
~)
.1 I~'" (0) F(t)
o (b)
Figure 11.8
where Cj is a constant. If the normal modes are normalized according to Eg. (11.122), we have
1/
pA(x)W?(x)
dx=
1/
sin2
pAC;
i1rX
-
dx
=1
001
or
C.= I
(2
yPAi
(El1.4.3)
Thus, the normalized mode shapes are given by Wj(x)
=
f£ -
pAl
sin-
i1rX
(El1.4.4)
I
The force acting on the beam can be expressed as f(x,
t)
= Foo(x
-~)
(El1.4.5)
and the generalized force corresponding to the ith mode can be determined using Eq.(11.147) as Q;(t) =
1/
= /2
W;(x)f(x,
t) dx
f'sin in Foo(x _ ,)dx = /2
pAl Jo
I
pAl
Fosin in, I
(El1.4.6)
"'1'1'.8 Forced Vibration --'-347
The generalized
T'Jj(t)
in the ith mode is given by Eq.(11.l48):
coordinate
= Aj COSWit + Bi sin Wit + /
. fl;
= A· cosw·t + B· sinw·t + Fio I
I
I
r
2 Fo sin i7rl~ ~ pAL Wi
I
pAL
SIO
i7r~
--(1 l
10
1
w?
sinwi(t - r) dr
- cosw·t)
(El1.4.7)
I
I
where the constants Aj and Bi can be determined from the initial conditions of the beam. In the present case, the initial conditions are zero and hence T'Ji(t) can be expressed as
T'J·(t)
=
fl;
pA
l4
i7r~
sin -(1-
F.o -----
pAL i4rr4 EI
I
(El1.4.8)
cosw·t)
l
I
Thus, the response of the beam is given by [see Eq.(11.149)]
=
w(x, t)
2Fol3 ~ -4-
7rEli=ll
1 . irrx
L...J 74 sm -
.
i7r~
S10 -l-(1
(El1.4.9)
- COSWit)
l
Example 11.5 A uniform beam is subjected to a concentrated harmonic force Fo sin ru at x = ~ (Fig. 11.9).
F(t)
__
= Fa sin Or 8(x -
~)
1_-_
• x
I:
·1
Figure 11.9
(a) Find an expression
for the response of the beam valid for all support conditions.
(b) Find an expression for the response of a simply supported beam when ~ Assume zero initial conditions. SOLUTION Eq.(11.147): Qj(t)
(a) The generalized
=
Fo
1/
force corresponding
Wi (x) sin !;1to(x - ~)dx
=
= 1/2.
to the ith mode is given by
FoWj(~)sinnt
(El1.5.1)
348
Transverse Vibration of Beams
For zero initial conditions, the generalized coordinate in the ith mode becomes [see Eq.(1l.148)]
'fJi(t)
= -~1
it
Qi ('r) sin Wi (t - r) dr
0
Fo
= zWi(S Wi
1
1:)
2
Po = --..£ ~
it
Wi(~) sin Quin Wi(t _ r) dr
0
( .•...• Q. sm~,t - -smwit Wi
2
1 - Q j wi
)
(El1.S.2)
The response of the beam can be expressed as [Eq.(1l.141)] 00
w(x, t)
'"
= Fo L..i=1
Wi(X)Wi(~) 2 _
( .
Q.)
sm Qt - --:- sm Wit
2
Q
Wi
(E11.S.3)
W.
where Wi(x) and Wi(;) correspond to the normalized normal modes. (b) For a simply supported beam, the normalized normal modes are given by [see Eq.(E11.4.4)]
Wi(x)=
[;f.
inx
-smpAL
(E11.S.4)
I
and hence i = 2, 4, 6, ... i=1,5,9,
...
(EI1.S.S)
i = 3, 7, 11, ... Thus, the response of the beam, Eq. (E11.S.3), becomes
w(x,t)=
2Fo [
~
p
sin(inxjl)
L..-
Al
2_
i=I,5,9, ...
-£
. '=3.7.11, ...
Wi
sin~inXj:) wi -Q
2
Q
( . Q smQt---:-sinwit W,
(Sin Qt -
.£: sin Wit)]
Wi
)
.
(E11.S.6)
where
(E11.S.7)
Example 1];6 Find the dynamic response of a uniform beam simply supported at both ends and subjected to a harmonically varying load: l(x,
t)
= 10sm. --nnx I
sinwt
(E11.6.1)
_______
11.8 Forced Vibration
":349
where 10 is a constant, n is an integer, t is the length of the beam, and w is the frequency of variation of the load. Assume the initial displacement and initial velocity of the beam to be zero. SOLUTION
The equation of motion of the beam is given by [see Eq.(l1.l2)]
EI
4 a w(X, t) ax
4
+pA
~ a-w(x, at
t)
=
2
F.' JO
sm
mrx
.
(E11.6.2)
-t- sm wt
Although the solution of Eq. (El1.6.2) can be obtained using Eq. (11.141), a simpler approach can be used because of the nature of the load, Eq.(E11.6.1). The homogeneous (or free vibration) solution of Eq.(EI1.6.2) can be expressed as 00
•
. -[-( IJrX C i COSWit w(x, t) = "" L.,. sm
+ Di sinwit)
(El1.6.3)
i=l
where
Wi
is the natural frequency of the beam, given by [see Eq.(11.50)] ·2
=I
Wi
The particular integral of Eq.(El1.6.2)
2
Jr
(El
Y pXi4
(E11.6.4)
can be expressed nJrx
w(x, t) = an sin -.
[
as
.
(El1.6.5)
smwt
where the expression for an can be found by substituting
Eq.(El1.6.5)
into Eq.(Ell.6.2)
as 10[4
an = The total solution of Eq. (El1.6.2) the particular
EI(nJr)4
[1 -
(El1.6.6) (wjwn)2]
is given by sum of its homogeneous
solution and
integral: 00
w(x, t) =
L .
•
IJrX
sin -(Ci
coswit
[
+ Di
..
smwit)
+ an sm
mrx
-
[
sinwt
(El1.6.7)
1=1
The initial conditions
of the beam are given by w(X,O)
aw at
-(x,O) Substituting
Eq. (El1.6.7)
=0
(E11.6.8)
=0
(El1.6.9)
into Eqs. (E11.6.8) and (El1.6.9),
we obtain (El1.6.1O)
for all i for i =1= n
(E11.6.1l) for i = n
,
I
IB!!!!!!!!!!!!!~
~~~
=,~~,
••..• "'.__
350
Transverse Vibration of Beams
Thus, the.solution of the beam becomes w(x, t) =
1014
E I (mr)
4[
. mr 2]sm --
1
1 - (w / Wn)
X( smwt .
W sinwnt )
- -
W
n
(El1.6.12)
11:9 RESPONSE OF BEAMS UNDER MOVING LOADS Consider a uniform beam subjected to a concentrated load P that moves at a constant speed Vo along the beam as shown in Fig. 11.10. The boundary conditions of the simply supported beam are given by =0
(11.150)
(0, t) = 0
(11.151)
w(l, t)
=0
(11.152)
a2w - -2 (1, t) ax
=0
(11.153)
w(O, t)
a
2w
. ax2
The beam is assumed to be at rest initially, so that the initial conditions can be written as w(x,O) = 0
(11.154)
aw - (x, 0) = 0 at
(11.155)
Instead of representing the concentrated force using a Dirac delta function, it will be represented using a Fourier series. For this, the concentrated load P acting at x = d is assumed to be distributed uniformly over an elemental length 2~x centered at x d as shown in Fig. 11.1L Now the distributed force, 1(x), can be defined as
=
for 0 < x < d - ~x = { ;~,
f(x)
for d - ~x =:: x =:: d for d + ~x < x < 1
+ ~x
(11.156)
From Fourier series analysis, it is known that if a function 1(x) is defined only over a finite interval (e.g., from Xo to Xo + L), the definition of the function I(x) can be
1
d =
Vol
•. x
I· Figure 11.10
Simply supported beam subjected to a moving concentrated load.
,-
11.9 Response of'Bemns under Mu"'ingLoads
I
~351
p
.-x
1'---:
Figure 11.11
d
~y
Concentrated load assumed to be uniformly distributed over a length 2Llx.
extended for all values of x Fourier series expansion of i(x) in the original interval is defined over the interval terms is given by
and can be considered to be periodic with period L. The the extended periodic function converges to the function from Xo to Xo + L. As a specific case, if the function i(x) 0 to I, its Fourier series expansion in terms of only sine 00
=~ '"'
_ i(x) '.
in
where the coefficients
n=l
are giveri by in =
2
t i(x) sm. -t-nrr
7 10
In the present case, the Fourier coefficients for i(x), as
[1
d aX
in = ~
1/ =- 1
d+ax
P Il:1.x
"in can
x
(11.158)
dx
be computed. using Eq. (11.156)
l:::
x
(0) (sin n~x) dx
-
+
(11.157)
nJf X in sin -t-
+
2~x (sin n~x) dx
(0) (sin nrrX) dX] I
d+ax
d-ax
" nrrx d sm -x t
nrr d sin(nrr l:1.x/l) = -2P. sm ------I I nrrl:1.x/t
Since P is actually a concentrated load acting at x with
= d, we let l:1.x~
. sin(nrr l:1.x/l) hm ----=1
ax--+o (llrr l:1.x/ t)
(11.159) 0 in Eq.(11.159)
(11.160)
to obtain the coefficients 2P
in = -sin-I
nrr d
t
(11.161)
I
i
1
i
i
i I
J
352
Transverse Vibration of Beams
Thus, the Fourier sine series expansion of the concentrated load acting at x = d can be expressed as 2P
nrrd LS10 -100
mrx
•
= -1-
f(x)
sin -1-
(11.162)
n=!
Using d and t as f(x
'
= vot
in Eq. (11.162), the load distribution can be represented in terms of x
2P (. rrx . rrvot . 2rrx . 2rrvot . 3rrx t) = S1O-S1O-- +S1O--sm--+sm--s1O I I I I I I
. 3rrvot ) __ + ' .. I
(11.163) The response of the beam under the nth component of the load represented by Eq. (11.163) can be obtained using Eq. (El1.6.l2) as w(x, t) =
EI(nrr)4
3 2PI [1 - (2rrvO/lwn)2]
nrrx ( 2rrvo sin --t I I
sin --
2rrvo
- -sinwnt wnl
)
(11.164) The total response of the beam considering all components (or harmonics) of the load, given by Eq.(11.163), can be expressed as w(x, t) =
2PI3 ~ 1 1 ~""4 • EIrr n=! n 1 - (2rrvo/lwn)
--4
2 sin
nrrx ( 2rrvo 2rrvo ) - -1- sin --t - --I sinwnt I
W
n
(11.165)
11.10
TRANSVERSE VffiRATION OF BEAMS SUBJECTED TO AXIAL FORCE The problem of transverse vibration of beams subjected to axial force finds application in the study of vibration of cables, guy wires, and turbine blades. Although the vibration of a cable can be studied by modeling it as a taut string, many cables fail due to fatigue caused by alternating flexure induced by vortex shedding in a light wind. In turbines, blade failures are associated with combined transverse loads due to fluids flowing at high velocities and axial loads due to centrifugal action.
11.10.1
Derivation of Equations Consider a beam undergoing transverse vibration under axial tensile force as shown in Fig. 11.12(a). The forces acting on an element of the beam of length dx are shown in Fig. 11.12(b). The change in the length of the beam element is given by
ds - dx =
I
(dx)2
+
aw(x, t) [ ax dX]
2
)
!/2
- dx
(11. 166)
______
11:10 Transverse Vibration of'Beains'SiIbJecteti'to A:'{ht1Force
353
[(x.t)
._
..
·_.+.+._._._. -
.
__
I
. I
L
x
P(X,t)
._.~x
I. --jdxr(a)
w(x.t)
t I I I I
I
I
1._._._.-1._._._.-1._._._._._._ o x x+dx
x
(b) P +
.
dS ._.-.
". /.-
T
p
w(x.t)
ap ax
aJ:, dx
dw=-
r---=:---t---~:~ " ax
_I_._._._._._._._._._._l-.x (c)
Figure 11.12
For small amplitudes of vibration, Eq. (11.166) can be approximated as [Fig. 11.12(c)]
ds - dx ~ ..;dx2
+ dw2
-
(OW)2
dx ~ -1 -
2
ox
dx
(11.167)
The small displacement values w(x, t) are assumed to cause no changes in the axial force P (x, t) and the transverse distributed force f (x, t). The work done by the axial force against the change in the length of the element of the beam can be expressed as
Wp =
1
r' P(x, t) [OW(X,ox
-2: 10
t)]2
dx
(11.168)
I ~~!\!!!!!!!!!!!!!!!!!!!!!!!!!!!'
~'T'~
354
Transverse Vibration of Beams
I (x,
The work done by the transverse load
l'
Wf=
t) is given by
I(x,t)w(x,t)dx
(11.169)
Thus, the total work done (W) is given by W = Wp
+ WI
t)]2 210r' P(x, t) [OW(X, ox
= -~
+
dx
10r' I(x,
t)w(x, t)dx
(11.170)
The strain and kinetic energies of the beam are given by 1 r' [o2W(Xox ' t)]2 . dx = 210 EI(x) 2
7'l
11'
pA(x) [OW(X,t)]2 dx
T =-
2
(11.171) (11.172)
ot
0
The extended Hamilton's principle can be expressed as
o
l2
f
(T -
7'l
+ W) dt
=0
(11.173)
II
The various. terms of Eq. (11.173) can be evaluated as follows: o
f
l' f
l
l2
T dt =
0
II
=
f'2
7'ldt
= fh -
'I
o
1
1/
0
f = f'• + 1/ =
'2
O
'I
2 [
'I
0
f
l
II
dx
2
II
-0 ot
(OW) pA-
ot
o2w owdx dt pA-2
ot
owdt
]
dx
(11.174)
== tl and t = t2.
l'
11
12
1
pA-ow ot
II
o
ow 0 pA--(ow)dt ot ot
1[, ow f 21'
=since ow(x, t) = 0 at t
2
II
(.o2W) EI-2o2w o -2 dx dt. ox ox o2w o2(ow) EI---dx ox2 ox2
I' w)
EI- o2w --- o(ow) ox2 ox
2 a2( EI- a 2ax o ox -2
0
dt
W - -o( EI- 02 ) ow 1/ ax ox2 0 owdx
] dt
(11.175)
_______________________
35.5
11.10 Transverse Vibration of Beams Subjected to Axial Force
o
1/ ] 1/ ]
1" Wdt= 1/2[-P-oawax II
/1
1 1 1 P-owawax 1
=
/2 [
-
0
/1
=
(aw) - dx+ fowdx dt ax 0 aw a P--(ow)dx + fowdx dt ax ax 0
/2 [
1/-ox0 ( P-aw)ax owdx + 1 fowdx 1
1/ +
-
0
/1
0
]
dt
0
(11.176) Substitution of Eqs. (11.174)-(11.176)
into Eq. (11.173) leads to
2
-1" /1
a2w t [PA a 2w + ~ 2 (E1 2) _ ~ (paw) - f]oWdt 10 ot ox ox ox ox
_lhElo2~ t1
ax
o(aw)ll ax
0
dt+1'2[~(Elo2~)_paow]owll ox ox
dt=O
/1
(11.177)
0
X
Since ow is assumed to be an arbitrary (nonzero) variation in 0 < x < 1, the expression under the double integral in Eq. (11.177) is set equal to zero to obtain the differential equation of motion:
2 ~ (E102W) __'°i.(pow) + pA a w = f(x, t) ox2 ox2 ax ox ot2
(11.178)
Setting the individual terms with single integrals in Eq. (11.177) equal to zero, we obtain the boundary conditions as
02W E10 2
(ow) ox
ox
[ -ax0 (02E1-ox2 W)
ow]
-p-
ox
1/
(11.179)
=0
0
8w1/ =0
(11.180)
0
Equation (11.179) indicates that the bending moment, E I (o2w/ ax2), or the slope, ow/ax, must be zero at x =0 as well as at x =1, while Eq. (11.180) denotes that either the total vertical force,
a (02w) ax EI-ox2
-
or the deflection, w, must be zero at x 11.10.2
ow ax
-p-
= 0 and also at x = 1.
Free Vibration of a Uniform Beam For a uniform beam with no transverse force, Eq. (11.178) becomes
o4w a2w a2w EI ox4 + pA ot2 - P ox2 = 0
-_--""""
(11.181)
!'!""" __
-----------
••••••••••••.
,,!!!!I.J
356
Transverse Vibration of Beams
The method of separation of variables is used to find the solution of Eq. (11.181): w(x, t) = W(x)(A
coswt
+ B sin wt)
(11.182)
By substituting Eq. (11.182) into Eq. (11.181), we obtain d4 W d2 W EI-- p-- 2 - pAw2W 4 dx
dx
=0
(11.183)
By assuming the solution of W(x) in the form
= Cesx
W(x)
(11.184)
where C is a constant, Eq. (11.183) gives the auxiliary equation p
4
2
--s EI The roots of Eq. (11.185) are given by S
2
2
s),s2
P
= 2EI
pAw2
---=0 EI
(11.185)
PA(2))/2
(P2
± 4E212 + EI
(11.186)
Thus, the solution of Eq. (11.183) can be expressed as W(x)
= C) coshs)x + C2 sinhs)x + C3 COSS2X + C4 sins2x
(11.187)
wh.ere the constants C) to C4 are to be determined from the boundary conditions of the beam. Example 11.7 Find the natural frequencies of a uniform simply supported beam subjected to an axial force P. SOLUTION
The boundary conditions of the beam are given by W (0) = 0 d2W -2 dx
(Ell. 7.1)
(0) =0
(ElI.7.2)
=0
(E11.7.3)
(l) = 0 dx2
(Ell. 7.4)
W(l) d2W
When Eqs. (E11.7.l) and (El1.7.2) are used in the solution, Eq. (11.187), we obtain C) C3 = O. This leads to
=
(E11.7.5) Equations (Ell. 7.3)-(E11.7.5) yield C2 sinhs)l C2S~ sinh s)l -
+ C4 sins2l
c si 4
=0
(E11.7.6)
sin S2l = 0
(E11.7.7)
t 1.11 ·Vibrution .of. a Rotating .&amJS1
For a nontrivial solution of C2 and C4 in Eqs. (EIl.7.6) and (EIl.7.7). the detenninant of their coefficient matrix is set equal to zero. This leads to the frequency equation (EIl.7.8) Noting that sit ~ O. the roots of Eq. (EIl.7.8) are given by n=0.1.2
(Ell.7.9)
•...
Equations (Ell.7.9) and (11.186) yield the natural frequencies of the beam as (Ell.7.1O) If the axial force is compressive (P is negative). Eq. (El1.7.10) can be rewritten as =
W n
1l'2
t2
(£1)1/2 (n n2~)1/2 4 _
pA
(El1.7.11)
Peri
where (El1.7.12) is the smallest Euler buckling load of a simply supported beam under compressive load. Notes:
1. If P = O. Eq. (E11.7.11) reduces to Eq. (11.50). which gives the natural frequencies of vibration of a simply supported beam. 2. If £1 O. Eq. (E11.7.1O) reduces to Eq. (8.103). which gives the natural frequencies of vibration of a taut string fixed at both ends. 3. If P -+ Peri' the fundamental natural frequency of vibration approaches zero
=
(WI
-+ 0).
4. If P > 0, the values of the natural frequencies increase due to the stiffening of the beam.
11.11
VmRATION OF A ROTATING BEAM Let a unifonn beam rotate about an axis parallel to the z axis at a constant angular velocity n. The radius of the hub r is considered to be negligibly small (Fig. 11.13). The beam is assumed to be fixed at x 0 and free at x t. At any point x along the beam, the centrifugal force induces an axial force. P(x). given by
=
=
(11.188)
______________________~-------J
I
358
Transverse Vibration of Beams z
t
W(X,t)
l--j Figure 11.13
Note that the force induced due to the longitudinal elastic displacement of the beam is neglected in deriving Eq. (11.188). The equation of motion for the transverse vibration of the rotating beam can be obtained using Eq. (11.188) for P in Eq. (11.178):
(11.189) Because of the coordinate system indicated in Fig. 11.13, the axial force, P, given by Eq. (11.188) will be zero at x = Z, and hence the boundary conditions of the beam will be same as those of a nonrotating beam: w =0
and
aw ax
-=0
at x = 0
(11.190)
and
(11.191) For the free vibration of the rotating beam, a harmonic solution of the form w(x, t) = W(x)cos(wt
is assumed. UsingEq. (11.192),Eqs.(11.l89)-(11.l91) 4
- ¢)
(11.192)
can. be expressed as
2
d W - -pAQ-Z 1 ~ 2 -d [( 1- X- ) dWJ = w-pAW ~ £1-dx4 2 dx Z2 dx
(11.193)
11.12
Natural Frequencies of Continuous Beams on Many S~pports
359
or d4W 1 d2W £1--4- _pAn2(l2 - x2)--2 dx 2 dx dW(x)
= --
W(x)
dx
dx
- pACliw = 0 (11.194) (11.195)
x=O
=0,
_ d3W(x) dx3
d2W(x) dx2
+ pAn2/x-dW
_ -,
0
(11.196)
x =/
The exact solution of the problem defined by Eqs. (11.194)-(11.196) is difficult to find. However, approximate solutions can be found using the methods of Chapter 17.
11.12 NATURAL FREQUENCIES MANY SUPPORTS
OF CONTINUOUS BEAMS ON
Consider a continuous beam supported at n points as shown in Fig. 11.14. Fot the vibration analysis of the beam, we consider the span (or section) between each pair of consecutive supports as a separate beam with its origin at the left support of the span. Hence the solution given by Eq. (11.36) or (11.37) is valid for each span of the beam. Thus, the characteristic function or normal mode of span i can be expressed, using Eq. (11.36), as i = 1,2, ... , n - 1 (11.197) where (11.198)
i = 1,2, ... , n - 1
and Pi, Ai, £j, and h denote the values of p, A, E, and I, respectively, for span i. The following conditions are used to evaluate the constants Aj, Bj, Ci, and Dj, i = 1,2, ... , n - 1:
1. The deflection is zero at the origin of each span (except possibly the first span): (11.199)
Wj(O) = 0
Span n - 1
Span 1 1
,A,.
I
2
n-l
~
~x 4-x II
Figure 11.14
Continuous beam on multiple supports.
I
360
Transverse Vibration of Beams
2. At the end of each span, the deflection is zero (except possibly when i = n _ 1) since the deflection is zero at each intermediate support. W;(/;) = 0
(11.200)
3. Since the beam is continuous, the slope and bending moment just to the left and to the right of any intermediate support are the same. Thus, dW;_1 (/;-d
-----
dx
E
dW;(O)
= ---
2
;-1/;-1
d W;_I(l;_I) dx2
(11.201)
dx
=
E ;1;
d2W;(0) dx2
(11.202)
4. At each of the end supports 1 and n, two boundary conditions can be written, depending on the nature of support (such as fixed, simply supported, or free condition). When Eqs. (11.199)-(11.202) are applied for each span of the beam (i = 1,2, 0.0, n - 1), along with the boundary conditions at each end of the beam, we get a total of 4(n - 1) homogeneous algebraic equations in the unknown constants A;, B;, C;, and D;, i = 1,2, .. 0' n - 1. Using the condition for the nontrivial solution of the constants, we can obtain the frequency equation of the system. Example 11.8
Determine the natural frequencies and mode shapes of a beam resting on three simple supports as shown in Fig. 11.15(a). Assume the beam to be uniform with /1 = 12 = 10
Span 1
Span 2 2
3
x
x (0)
(b)
(c)
Figure 11.15
Uniform beam resting on three simple supports.
11.12 SOLUTION expressed as
Natural Frequencies of ContinuousB_eams on Many Supports
The characteristic
functions
36J
in the two spans of the beam can be ','
WI(X) = Al cosf3tx Wz(x) = Az cos .Bzx
+ BI sin,Blx + CI cosh,Blx + DI sinh,Blx + Bz sin ,Bzx + Cz cosh .B2x + Dz sinh .Bzx
(EIl.8.I) (EIl.8.2)
To simplify the computations, the x axis is taken from support I to the right for span I and from support 3 to the left for span 2. The simply supported end conditions at support I are given by (EIl.8.3)
WI(O) = 0 Ellt Equations
(EI1.8.1),
d2Wt(O) dxz
(EI1.8.3),
0
(EIl.8.4)
=
and (EI1.8.4)
give
+ CI = 0 + CI = 0
Al
-AI or Al Thus, Eq. (EI1.8.1)
= CI = 0
(EIl.8.S)
reduces to
= BI sin.Blx
WI(X)
Since the displacement
+ DI
(EIl.8.6)
sinh,Blx
is zero at support 2, WI (II) = 0 and hence Eq. (EI1.8.6) BI sin.BI1I
+ DI
sinh.BIII
gives
=0
or sin ,BIll DI = -Bl SI·nh.B I I 1 Thus, Eq. (EI1.8.6)
(EI1.8.7)
Carl be written as WI(X)
Next, the simply supported
= Bl
( sin.Blx -
end conditions
sin.BIII sinh,BIII
)
sinh.Blx
at support 3 are given by (EIl.8.9)
WZ(O) = 0 E I dZWz(O) = 0 z z dxz Equations
(EI1.8.2),
(EI1.8.9),
and (EIl.8.10) Az -Az
(EIl.8.8)
or
dZ Wz (0) dxz =0 yield
+ Cz = 0 + Cz = 0
(EI1.8.1O)
•.>" k""II'f,<,
•.
362
Transverse Vibration of Beams
or (E11.8.11) Thus, Eq. (E11.8.2) reduces to (E11.8.12) Using the condition that the displacement at support 2 is zero in Eq. (E11.8.12), we obtain
(E11.8.13) and hence Eq. (E11.8.12) becomes
(E11.8.14) The slope is continuous at support 2. This yields dW1 (II> dx
-
dWz(lz)
(E11.8.15)
dx
which can be expressed, using Eqs. (E11.8.8) and (E11.8.14), as,
Finally, the bending moment is continuous at support 2. This leads to
(E11.8.17) which becomes, in view of Eqs. (E11.8.8) and (E11.8.14), BIE]II~?sin~1/1
- BzEzlz~isin~2/2
=0
(El1.8.18)
Equations (E11.8.16) and (E11.8.18) denote two simultaneous homogeneous algebraic equations in the unknown constants BJ and Bz. For a nontrivial solution of these constants, the determinant of'their coefficient matrix is set equal to zero. This gives the frequency equation as
(E11.8.19)
11.12
Natural Frequencies
of Continuous
Beams on Many Supports
= Ez = E,
For a uniform beam with identical spans, Et and II = 12 = /, and Eq. (E11.8.19) reduces to
It
363
= Iz = I, fJt = 132 = 13, (E11.8.20)
(cos 131- sin 131coth 131) sin 131 = 0 Equation (E11.8.20) will be satisfied when sin 131 = 0
(E11.8.21)
tan 131 - tanh 13/ = 0
(E11.8.22)
or
Case (i): When
sin 13/ = 0 This condition gives the natural frequencies as fJnI
= nJr,
n
= 1,2, ...
or
Wn
=n
2
2 Jr
[EI
n = 1,2, ...
y;Ai4'
(EI1.8.23)
These natural frequencies can be seen to be identical to those of a beam of length / simply supported at both ends; When Eq. (E11.8.21) is valid, Eq. (E11.8.16) gives BI
= -B2
(E11.8.24)
and hence the mode shape becomes WIn (x)
. sin fJnI. ) = C2n ( sm fJnx - sinh fJnI sinh fJnx -C2n
. (
for span 1 (E11.8.25)
sinfJnI inhR ) sm pnX - sinh fJnI s pnX R
for span 2
where the constants C2n can be assumed to be I, for simplicity. The mode shape given by Eg. (E11.8.25) denotes an antisymmetric mode with respect to the middle support 2, as shown in Fig. I 1. 15(b). Case (ii): When tan 131 - tanh 13/ 0 This condition can be seen to correspond to a beam of length I fixed at one end and simply supported at the other end. The roots, fJn/, of Eq. (EIl.8.22) are given in Fig. 11.4. When Eq. (EIl.8.22) is valid, Eq. (EIl.8.I8) gives BI = B2 and the mode shape becomes
=
WIn(x)
= W2n(X) = CZn (sin sinfJnx -
fJn/
sinhfJn/ sinhfJnx
)
for spans 1 and 2 (EI1.8.26)
where the constant C2n can be assumed to be 1, for simplicity. Notice that the mode shape given by Eq. (EIl.8.26) indicates a symmetric mode with respect to the middle support 2, as shown in Fig. 11.15(c).
364
Transverse Vibration of Beams
z
V+dV
•I I I I I
W(X,t)
fix,t)
I I 0
I
_
I
---------
Figure 11.16
11.13
x
x
Free-body diagram of a beam on an elastic foundation.
BEAM ON AN ELASTIC FOUNDATION Let a unifonn beam rest on an elastic foundation, such as a rail track on soil, as shown in Fig. 11.16. The continuous elastic foundation is denoted by a large number of closely spaced translational springs. The load per unit length of the beam necessary to cause the foundation to deflect by a unit amount, called the foundation modulus, is assumed to be k f· By considering a small element of the vibrating beam on an elastic foundation, the equation of motion of the beam can be expressed as 2 2 a (a w) ax2 E1ax2 where
11.13.1
f (x,
t)
2 a w +pAat2 +kfw=f(x,t)
(11.203)
denotes the distributed load on the beam.
Free Vibration For the free vibration of a unifonn beam, Eq. (11.203) reduces to
a4w E I ax4
+ pA
a2w at2
+ k fW
=0
(11.204)
,":''''''.-,
LL.L3
The free vibration
Beamon· -an-Elastic Foundation --365
solution of the beam is expressed
as
00
w(x. t)
=
L
cosw;t
W;(x)(C;
+ D;
sinw;t)
(11.205)
;=1
where W; is the i th natural frequency and W; (x) is the corresponding natural mode shape of the beam. Substitution of the ith modal solution into Eq. (11.204) yields
(PAW?
d4W;(x)
kf)
+ -E/"" +
dx4
.
(11.206)
EI
(11.207)
W;(x) = 0
EI
Defining 4
ot;
=(-
kf
EI
+ w?) c2
and
c =
J
pA
Eq. (11.206) can be written as d4W;(x)
dx4
4
- ot· W(x) I
The solution of Eq. (11.208) can be expressed Wi(x)
I
=0
(11.208)
as
= Cli COSotiX+ C2f SinotiX + C3i coshaix
+ C4i sinhot;x
(11.209)
where the constants Cli. C2i. C3i. and C4i can be evaluated from the boundary conditions of the beam. Noting that Eq. (11.209) has the same form as that of a beam without a foundation [see Eq. (11.36)]. the solutions obtained for beams with different end conditions in Section 11.5 are applicable to this case also if /3; is replaced by a;. The natural frequencies of the beam on an elastic foundation are given by Eq. (11.207):
1+~
(11.210)
Elot~ I
Assuming the beam on elastic foundation to be simply supported at the ends. the normal modes can be expressed as Wi(X) = C; sina;x where Ci is a constant. The natural frequencies
ail = irr.
(11.211)
of the beam can be found as
i=I.2.3
•...
or i = 1. 2. 3, ...
(11.212)
The free-end forced response of a simply supported beam on an elastic foundation can be found from the corresponding results of a simply supported beam with no foundation by using Eq. (11.212) for Wi instead of Eq.(11.50).
.. <"
.~
••
~ •••. ~ .• -_.,
366
Transverse Vibration of Beams
11.13.2
Forced Vibration The forced transverse vibration of a uniform beam on elastic foundation is governed by the equation 2
8 w(x, t) 8t2
El 84w(x, t) = f(x, t)
kf
+ pA
w(x, t)
+
pA
8x4
(11.213)
pA
When the normal mode method is used, the solution of Eq. (11.213) can be expressed as 00
w(x, t) =
L W (x)1]n (t)
(11.214)
n
n=)
where Wn(x) is the nth normal mode and 1]n is the corresponding generalized coordinate of the beam. Noting that the normal mode Wn(x) satisfies the relation [see Eq. (11.206)] 4
d Wn(x) = dx4
(PAW; _ E1
kf) ~ X E1 n( )
(11.215)
Eq. (11.213) can be reduced to d21]n(t) dt2
2
+ Wn1]n(t)
where the natural frequency, Qn (t), by
(Un,
= Qn(t),
n = 1,2, ...
(11.216)
is given by Eq.(1l.21O) and the generalized force,
(11.217) The solution of Eq. (11.216) can be expressed as
n
= 1,2,
...
(11.218)
where the initial values of the generalized displacement 1]n (0) and generalized velocity l]n(O) are determined from the initial conditions of w(x, t) [see Eqs. (11.138) and (11.139)]: 1]n(O) =
l]n(O) =
where wo(x)
= w(x,
l' l'
pA(x)Wn(x)wo(x)
dx
(11.219)
pA(x)Wn(x)wo(x)
dx
(11.220)
0) and wo(x) = w(x, 0).
11.13
Beam on an Elastic Foundation
367
• Va f(x,t)
Elastic foundation (foundation modulus, kf)
Figure 11.17
11.13.3
Beam on an elastic foundation subjected to a moving load.
Beam on an Elastic Foundation Subjected to a Moving Load Let an infinitely long uniform beam on an elastic foundation be subjected to a distributed transverse load f (x, t) traveling at a constant speed Vo along the beam as shown in Fig. 11.17. The vibration response of a railroad track under the moving weight of the rail can be determined using the present analysis. Since the load moves at a constant speed along x, the distributed load can be denoted in terms of x and t as f (x - vot). The equation of motion for the'transverse vibration of the beam is given by EI
84w(x,t) 8x4
82w(x,t) 8t2
+ pA
+ kfw(x,
t)
= f(x
(11.221)
- vot)
Using z = x - vot Eq. (11.221) can be rewritten EI
(11.222)
as
d4w(z) dz4
+ pAv5
d2w(z) dz2
+ kfw(Z)
(11.223)
= f(z)
If f(x, t) is a concentrated load Fo moving along the beam with a constant velocity vo, the equation of motion can be written as (11.224) and the concentrated load Fo is incorporated as a known shear force at solution. The solution of Eq. (11.224) is assumed to be w(z) Substitution
Z
= 0 into
= eJ.LZ
the
(11.225)
of Eq. (11.225) into Eq. (11.224) yields the auxiliary equation EI/-L4
+ pAv5/-L2 + kf = 0
(11.226)
368
Transverse Vibration of Beams
The roots of Eq. (11.226) can be expressed as
J"fi
(11.227)
+ J"fi
(11.228)
±i/a -
J,Ll,Z
=
J,L3,4
= ±i / a
where pAv2 0 a = __
(11.229)
2El ~ _
pZA2v4
-
4E212
0
__
kf
(11.230)
El
Thus, the solution of Eq. (11.224) becomes w(z)
= C1ei.ja-.JPz
+ C2e-i.ja-.JPz
+ C3ei.ja+.JPz + C4e-i.ja+.JPz
(11.231)
where the constants C1, C2, C3, and C4 can be determined using the conditions w=O d2w
-=0z dz
dw =0 dz'
at
z=
00
(11.232)
at
z=
00
(11.233)
at
z=o
E1d3w = Fo dz3 2
at
(11.234)
z=o
(11.235)
[Note that the concentrated load Fo at z = 0 causes discontinuity of the shear force. By considering the shear forces immediately on the left- and right-hand sides of Fo, we obtain -EI-3
d3w
3
dz
(z
= 0+) + El- ddzw-3 (z = 0-)
= Fo
(11.236)
Due to symmetry at z = 0, Eq. (11.236) yields Eq.(11.235) as z --+ 0.] To satisfy Eqs. (11.232) and (11.233), C1 and C3 must be zero in Eq. (11.231). This gives (11.237) The use of conditions (11.234) and (11.235) in Eq. (11.237) yields
2/a - J"fi - iC4/a
-iC
iCz(/a - J"fiY + (/a iC4
+ J"fi = 0
+ J"fiY
= ~o
(11.238) (11.239)
The solution of Eqs. (11.238) and (11.239) is given by C2 =
Fo
(11.240)
------=== i .4EI,JpJa -,Jp
Fo C4 = -----::=-;::::==== i .4EI
(11.241)
,JpJa +,Jp
Thus, the solution of Eq. (11.224) can be expressed as w(z)
=_ i .4EI
+
e-i..ja-./Pz
Fo
,JpJa -,Jp
Fo i·
4EI,JpJa
e-i..ja+.jfJz
+JP (11.242)
11.14 RAYLEIGH'S THEORY The inertia due to the axial displacement of the beam is included in Rayleigh's theory. This effect is called rotary (or rotatory) inertia. The reason is that since the cross section remains plane during motion, the axial motion of points located in any cross section undergoes rotary motion about the y axis. Using u = -z(ow/ox) from Eq. (11.1), the axial velocity is given by
au , at -
o2w at ax
..... - - -z--
(11.243)
and hence the kinetic energy associated with the axial motion is given by
(11.244) The term associated with Ta in Hamilton's principle can be evaluated as
1 '21/ 1 2
Ia = 0
t
Tadt
=0
'\
=
~
0
1'2 11/ pI ( -~-ato2w )2 dx dt -2
0
t\
vX
W (o2 --) at ax
o2w pI --0 at ax
(11.245)
dxdt
Using integration by parts with respect to time, Eq. (11.245) gives Ia
=-
1 1/ 2
t
'\
a3w -2--
pI
at ax
0
0 (ow) ax
dxdt
(11.246)
Using integration by parts with respect to x, Eq. (11.246) yields
Ia =
1
3
'2 [
'1
-pI
a w -2-
at ax
ow 1/ 0
+
1/0
3 a ( pI -2a w) ax at ax
owdx ] dt
(11.247)
370
Transverse Vibration of Beams
When - 1a is added to Eq. (11.11), the equation of motion, Eq. (11.12), and the boundary conditions, Eq.(11.13), will be modified as follows (by neglecting the springs and masses at the ends): 2 0 ( 02w) a ( 03W) OX2 E1 OX2 - AX pI ot20x 02W EI-o ox2
(OW) ax
II - [a-ax
W (02 EI-)
0
+ pA
W -pl-- 03 ] . ot20x
ox2
02W ot2 = f(x,
t)
II
ow =0 0
(11.248)
(11.249)
For a uniform beam, the equation of motion and the boundary conditions can be expressed as
(11.250) (11.251)
(11.252)
(11.253) For harmonic oscillations, the solutionis assumed as W(x, t) = W(x) cos(wt - ¢)
(11.254)
Equations (11.253) and (11.254) lead to d4 W EI-- 4 dx
+ p1w2 d__
W dx2 2
-
pAw2W = 0
(11.255)
Using W(x)
=e
Sx
(11.256)
the auxiliary equation corresponding to Eq. (11.255) can be derived as E1s4 + p1w2s2 """
Denoting the roots of Eq. (11.257) as s), can be expressed as
-
pAw2
S2, S3,
and
=0 S4,
(11.257) the solution of Eg. (11.255)
4
W(x)
= L Cies;x
(11.258)
i=l
where the constants Cl, C2. C3, and C4 can be determined from the boundary conditions.
Example 11.9 Determine the natural frequencies of vibration of a simply supported Rayleigh beam. SOLUTION
By introducing the parameters 2 EI a =-
(El1.9.1)
2 I r =A
(El1.9.2)
pA
the equation of motion, Eq. (11.253), can be written for free vibration as a2
a4w ax4
__
2w at2
+ -a
-
a4w ax2at2
r2
=0
(E1193)
..
The solution of Eq. (El1.9.3) can be assumed as w(x, t)
=C
. nrrx
sm -/-
(EI1.9.4)
coswnt
where C is a constant and Wn denotes the nth natural frequency of vibration. Equation (El1.9.4) can be seen to satisfy the boundary conditions of Eqs. (11.251) and (11.252). By substituting Eq. (El1.9.4) into Eq. (El1.9.3), we obtain the frequency equation as
a
2
(n1r)4 2 -/- - wn
(
1+
2 22
n rr r )_ /2 - 0
(El1.9.5)
Equation (El1.9.5) gives the natural frequencies of vibration as W2n -
a2 (nrr j /)4
n=l,
1 + (n2rr2j /2)r2'
2, ...
(El1.9.6)
11.15 TIMOSHENKO'S THEORY 11.15.1
Equations of Motion The effect of shear deformation, in addition to the effect of rotary inertia, is considered in this theory. To include the effect of shear deformation, first consider a beam undergoing only shear deformation as indicated in Fig. 11.18. Here a vertical section, such as PQ, before deformation remains vertical (PI Q') after deformation but moves by a distance w in the z direction. Thus, the components of displacement of a point in the beam are given by u =0,
v =0,
w =
w(x,t)
(11.259)
The components of strain can be found as
au
av
Exx
= -ax = 0,
Eyz
= -az + -8y = 0,
av
Eyy
= -ay = 0,
aw
au
Ezx
Szz
aw
aw
=-
8z
aw
= -az + -ax = -ax
=0
Exy
au = ay
av
+ ax
= 0, (11.260)
372
Transverse Vibration of Beams
o
(0)
C P'~
- ---ow
"- ~-=-"l~: ox D'
Q'
(b)
Figure 11.18
Beam in shear deformation.
=
The shear strain ezx is the same as the rotation {3(x, t) awjax(x, t) experienced by any fiber located parallel to the centerline of the beam, as shown in Fig. 11.18(b). The components of stress corresponding to the strains indicated in Eq. (11.260) are given by O'xx
= O'yy = O'zz = O'xy
= O'yz
= 0,
O'zx
= Gezx
aw ax
= G-
(11.261)
Equation (11.261) states that the shear stress O'zx is the same (uniform) at every point in the cross section of the beam. Since this is not true in reality, Timoshenko used a constant k, known as the shear correction factor, in the expression for O'zx as O'zx
= kG--
aw ax
(11.262)
The total transverse displacement of the centerline of the beam is given by (see Fig. 11.19): w = Ws
+ Wb
(11.263)
and hence the total slope of the deflected centerline of the beam is given by ow aws ax = ax
+
aWb ax
(11.264)
Since the cross section of the beam undergoes rotation due only to bending. the rotation ' of the cross section can be expressed as aWb f/J
= ax
aw
= ax -
aws aw ax = ax - {3
(11.265)
z
t
.
p
~-._._-._-._.f-9-.o~._.+.~ x
Q
1-----
X
-+-j (a)
(b)
(c)
(d)
Figure 11.19 Bending and shear deformations: (a) element with no deformation; (b) Element with only shear deformation; (c) element with only bending deformation; (d) element with total deformation.
=
where f3 aWslax is the shear deformation or shear angle. An element of fiber located at a distance z from the centerline undergoes axial displacement due only to the rotation of the cross section (shear deformation does not cause any axial displacement), and hence the components of displacement can be expressed as
u
-
v
- 0 - w(x,
W
-z (~:
-
f3 ) ==
-z
t)
374
Transverse Vibration of Beams
where the strains corresponding to the displacement field given by Eq. (11.266) are Exx
Eyy
Ezz
Exy
Eyz
Ezx
au a¢ ax = -z- ax av = -=0 ay = aw =0 az au av = -+-=0 ay ax aw av = -+-=0 ay az au aw = -+-=-¢+az ax = -
(11.267)
aw ax
The components of stress corresponding to the strains of Eq. (11.267) are given by
(11.268) Ciyy=
Cizz
= Cixy = CiyZ = 0
The strain energy of the beam can be determined as
j(
III H'ff
=~
(CixxExx
v
=
[E,2
A
= "21
Jor' [(a¢)2 £1 ox
+ CiyyEyy + CizzEzz+CixyExy
+ CiyzEyz + CizxEzx) dV
e:r e:-~r] +kG
dAdx
+ kAG (awax - ¢)2] dx
(11.269)
The kinetic energy of the beam, including rotary inertia, is given by
T
= "21 Jr'o [(oW)2 pA at + pI (o¢)2] at
The work done by the external distributed load
1
f (x,
dx
(11.270)
t) is given by
1
w
=
fw dx
(11.27])
11.l51imoshenko·s Application
of extended
Hamilton's
o
-375
gives
l2
f
(rr - T - W) dt = 0
11
or
ll2
principle
Theory
{1/ [EI :~ 0 (:~)
+ kAG (~: - ~) 0 (~:)
_t [PA aw 8 (aw) 10 at at
+ pla~
at
0
- kAG (~: - ~) O~] dx
(a~)] dx - t f OWdX} dt at 10
=0
(11.272) The integrals in Eq. (11.272) can be evaluated using integration by parts (with respect to x or t) as follows.
21/ a~ (a~) f 0 EI- ax 8 -ax dxdt 1
11
=
l f 2 [ EI- a~ o~II ax 0 I.
1/ -axa (a~)EI- ax
o~dx ] dt
0
(11.273)
--~ ) 0 (aw) dxdt f 21/ 0 kAG (aw ax ax l = f 2 [kAG (aw -~)'owl/ _ t kAG ~ (aw ax 0 10 ax ax 1
11
11
-¢) OWdX] dt (11.274)
tpAaw
1
_f 2
tpAa2~ owdxdt at 1 2 - fl2 11 pI- a~ 0 (a~) - dxdt = f 21 pI-2a ¢ o¢dxdt 0 at at 0 at
10
I.
at
o(aw)dxdt=-fI2 at
11
10
(11.275)
l
11
Substitution
(11.276)
11
of Eqs. (11.273)-(11.276)
into Eq. (11.269) leads to
_~) Owil + EI a~ o¢11 f 21 kAG (aw ax 0 ax 0 l
11
+ t [-~(kAG o ax
i
+ l [ -- a (a¢) EIo
ax
(aw _ ax
¢)) + pA aat22w - f]oWdX
-kAG (aw) --¢ ax ax
Equation (11.277) gives the differential
equations
2 +pI- a 2¢] o¢dx } dt=O at
(11.277)
of motion for w and ¢ as
-aax [kAGa: (~: -¢)]+PAaa2t~ =f(x,t) 2 _~ (Ela¢) -kAG (aw _~) +pl a 2¢ =0 ax ax ax at
(11.278)
(11.279)
376
Transverse Vibration of Beams
and the boundary conditions as
(11.280)
(11.281) 11.15.2
Equations for a Unifonn Beam Equations (11.278) and (11.279) can be combined into a single equation of motion for a uniform beam. For a uniform beam, Eqs. (11.278) and (11.279) reduce to -kAG
a2w ax2
+ kAG
a
(11.282)
or a
a2w
p a2w
f
-=-----+-ax ax2 kG at2 kAG
(11.283)
and a~ aw ax2 - kAG ax
-EI
+ kAGct> + pI
a~ at2
=0
(11.284)
which upon differentiation with respect to x becomes
£2 (act»
-EI
ax
ax
2 _ kAG a w+ kAG a
=0
(11.285)
Substitution of Eq. (11.283) into Eq. (11.285) yields the desired equation: 4
aw EI ax4
2 a w ( E) + pA at2 - pI 1 + kG
EI + kAG
a2 f ax2
_~ kAG
a2 f at2
a4w ax2at2
p2I a4w at4
+ kG
__ 0 f -
(1 1.286)
For free vibration, Eq. (11.286) reduces to 4
aw EI ax4
2 a w ( E) + pA at2 - pI 1 + kG
a4w ax2at2
+
p2I a4w kG at4
=0
(11.287)
The terms in Eq. (11.287) can be identified as follows. The first two terms are the same as those of the Euler-Bernoulli theory. The third term, -pI (a4w/ax2at2), denotes . the effect of rotary inertia. In fact, the first three terms are the same as those of the Rayleigh theory. Finally, the last two terms, involving kG in the denominators, represent the influence of shear deformation. Equations (11.280) and (11.281) will be satisfied by the following common support conditions. At a clamped or fixed end: ct>= 0,
w=o
(1 1.288)
11.1'5--·Timosnenko'-s ;:r-heory377
At a pinned or simply supported end: E I a
ax
= 0,
w
(11.289)
=0
At a free end: a
ax = 0,
(11.290)
EI-
11.15.3 Natural Frequencies of Vibration The natural frequencies of vibration of uniform Timoshenko beams can be found by assuming a harmonic time variation of solution and solving Eq. (11.287) while satisfying the specific boundary conditions of the beam. In some cases, it is more convenient to solve the simultaneous differential equations in
By dividing Eq. (11.287) by pA and defining 2 Ol
r2
EI =-
(11.291)
pA I
(11.292)
=A
Eq. (11.287) can be rewritten as 4 0l2_w
a
ax4
2
aw +_ 2
r2 ( 1 + -E ) kG
at
(11.293)
The boundary conditions can be expressed as w(x,t) a
ax
=0,
x=O,l
=0,
x=O,
(11.294) I
(11.295)
Equation (11.295) can be expressed, using Eq. (11.283) with 2 2 a
°
x
f
= 0, as
°
= ,
I
(11.296)
When harmonic time variations are assumed for
°
X
= 0, I
(11.297)
in view of Eq. (11.294). Thus, the boundary conditions can be stated in terms of w as w (0, t) = 0, d2w -2
dx
(0, t) = 0,
w(/,
t) =
°
d2w -2
dx
(I, t) =
(11.298)
°
( 11.299)
378
Transverse Vibration of Beams
The solution of Eq. (11.293), which also satisfies the boundary conditions of Eqs. (11.298) and (11.299), is assumed as n:rrx
w(x, t) = C sin --
1
coswnt
(11.300)
where C is a constant and Wn is the nth natural frequency of vibration. Substitution of Eq. (11.300) into Eq. (11.293) gives the frequency equation 2
2
2 2 4 pr 2 ( n :rr r n n w kG - w 1 + -1-2 -
!!.-)kG
2 2 2
+
r 12
n :rr
+
a2n4:rr4_
14
-
(11.301)
0
w;
w;
Equation (11.301) is a quadratic equation in and gives two values of for any value of n. The smaller value of corresponds to the bending deformation mode, and the larger value corresponds to the shear deformation mode.
w;
Fixed-Fixed
Beam
The boundary conditions can be expressed as w(O, t) = 0,
¢(O, t)
=0
(11.302)
w(l, t) = 0,
¢(l, t) = 0
(11.303)
Since the expression for ¢ in terms of w is not directly available, Eqs. (11.283) and (11.284) are solved, with f 0, simultaneously. For this, the solution is assumed to be of the form
=
= W(x)
w(x, t)
coswnt
(11.304)
¢(x, t) = eI>(x)cos wnt
(11.305)
Substitution of Eqs. (11.304) and (11.305) into Eqs. (11.283) and (11.284) gives (by setting f = 0) -kAG -EI-
d2e1> dx2
d2W -+kAGdxz dW
-kAG-
dx
del> 2 -pAw W=O dx n Z
+kAGeI> - pIw eI> n
=0
(11.306) (11.307)
The solutions of Eqs. (11.306) and (11.307) are assumed to be W(x)=C1 eI>(x)=Cz
exp(at) exp(at)
(11.308) (11.309)
where a, C), and C2 are constants. Substitution of Eqs. (11.308) and (11.309) into Eqs. (11.306) and (11.307) leads to ( -kAG;:
y)
(-kAG
C]
+ (-EI
- PAW~) C1
;: -kAG
T
+ (kAGT)
Cz = 0
(11.310)
+kAG - pI w~) C2 = 0
(11.311)
l1.tS--"Timoshenko's Theory
'379
For a nontrivial solution of the constants C t and C2, the determinant of their coefficients in Eqs. (11.310) and (11.311) is set equal to zero. This yields the equation a4
+ [ai[2 tI
(E-E + ~)] kG
a2
+ [ai n [4 (W~p2 kGE
_
PA)] EI
=0
(11.312)
The roots of Eq. (11.312) are given by (11.313) where al
=w~[2p
(
~
+ k~ )
w~P ( kG EEl
a2=w~[4p
(11.314)
A )
(11.315)
The four values of a given by Eq. (11.313) can be used to express W(x) and
=
=
= ~.
SOLUTION The natural frequencies of vibration are given according to Timoshenko theory [Eq. (11.301)]: (EILlO.l) According to Rayleigh theory [terms involving k are to be deleted in Eq. (EILlO.l)], _
By Euler-Bernoulli
2 (
wn
2 2
1+
n2rr r ) [2.
2 4
+
a n Jl'4_ [4 -
0
(EILlO.2)
theory [term involving r2 is to be deleted in Eq. (EILlO.2)], _w2n
a2n4rr4
+ --0--=0 [4
=
For the given beam, A = (0.05)(0.15) 0.0075 m2, I 6 4 10- m , and Eqs. (El1.10.1)-(EILl0.3)become 2.1706 x 10-9 w~ - (1 + 76.4754 x 1O-3n2)w~ -(1
+ 18.5062
x 1O-3n2)w~ -w~
(EILlO.3)
= 12(0.05)(0.15)3 = 14.063 X
+ 494.2300 + 494.2300 + 494.2300
x 103n4
=0
(EILlO.4)
4
=0
(EILlO.5)
x 103n
x 103n4 = 0
(Ell.l0.6)
380
TransverseVibrationof Beams Table 11.3
Computationfor Example 11.l0 Natural frequency(radls)
Euler- Bernoulli
Rayleigh
703.0149 2,812.0598 6,327.1348
696.5987 2,713.4221 5,858.0654
Timoshenko Bending Shear
n
1 2 3
677.8909 2,473.3691 4,948.0063
22,259.102 24,402.975 27,446.297
The natural frequencies computed from Eqs. (E11.10.4)-(E11.10.6) Table 11.3.
11.16
are given in
COUPLED BENDING- TORSIONAL VffiRATION OF BEAMS In the transverse vibration of beams considered so far, it is implied that the cross section of the beam has two axes of symmetry (y and z axes). If the cross section of the beam has two axes of symmetry, the centroid and the shear center (or center of flexure) coincide and the bending and torsional vibrations are uncoupled. In all the cases considered so far, the transverse vibration of the beam is assumed to be in a plane of symmetry (xz plane). On the other hand, if the cross section of a beam has only one axis of symmetry, the shear center lies on the axis of symmetry. When the load does not act through the shear center, the beam will undergo twisting in addition to bending. Note that, in general, the' shear center need not lie on a principal axis; it may lie outside the cross section of the beam as shown in Fig. 11.20 for a beam with channel section. In Fig. 11.20, the line GG' represents the centroidal axis, the x axis (00') denotes the shear center axis, the z axis indicates the axis of symmetry, and the y axis represents a direction parallel to the web. For the thin-walled channel section, the locations of the shear center and the centroid from the center of the web are given by [3)
a2b2t c=--
(11.316)
lz
b2 d=--2(b
I, ~ ~a3t
-
3
(11.317)
+ a) + 2 (.2..bt3 + bta2) 12
~
~a3t
3
+ 2a2bt
(11.318)
where 2a, b, and t are the height of web, width of flanges, and thickness of web and flanges as shown in Fig. 11.20. The distance between the centroid and the shear center of the channel section, e, is given by (11.319)
x
/
~
/ y
T --L
t
20
I
o~·
f-c
f-b~ Figure 11.20
11.16.1
Beam with a channel section, GG' centroidal axis; 00' shear center axis.
Equations of Motion Consider the beam with a channel section subjected to a distributed load f (x) acting along the centroidal axis GG' as shown in Fig. 11.21. Since the load does not pass through the shear center axis 00', the beam will undergo both bending and torsional deflections. To study the resulting coupled bending-torsion motion of the beam, the load acting through the centroidal axis is replaced by the same load and a torque of magnitude fe, distributed along the shear center axis 00' (see Fig. 11.21). Then the equation governing the bending deflection of the beam in the xy plane can be written as 04V
E lz
ox4
=
f
(11.320)
where v is the deflection of the beam in the y direction and E lz is the bending rigidity of the cross section about the z axis. For the torsional deflection of the beam, the total torque acting on any cross section of the beam, T(x), is written as the sum of the Saint-Venant torque (Tsy) and the torque arising from the restraint against warping (Tw) [3]: T(x) = Tsy(x)
+ Tw(x)
drp d3rp = GJ dx - EJw dx3
(11.321)
where rp is the angle of twist (rotation), GJ is the torsional rigidity under uniform torsion (in the absence of any warping restraint), E Jw is the warping rigidity, and Jw
382
Transverse Vibration of Beams
(a)
(b)
(c)
Figure 11.21 (a) Load acting through a centroid; (b) load acting through the shear center; (c) torque acting about the shear center.
is the sectional moment of inertia of the cross section with [3] 1
J = -3L,.. "bitt
1
2
3 = -(2bt + 2at3) = -t\a + b) 3 3
a2b3t 4at+ 3bt + 6bt
(11.322) (11.323)
Jw = --3-2at
By differentiating Eq. (11.321) with respect to x and using the relation dT / dx = we obtain
Ie,
(11.324) Note that the solution of Eqs. (11.320) and (11.324) gives the bending deflection, vex), and torsional deflection, ¢(x), of the beam under a static load, I(x). For the free vibration of the beam, the inertia forces acting in the y and ¢ directions are given by
a2 at2
-pA-(v
- e¢)
(11.325)
and (11.326) where v - e¢ denotes the net transverse deflection of the beam in the y direction, p is the density, A is the cross-sectional area, and I G is the polar moment of inertia
of the cross section about its centroidal axis. The equations of motion for the coupled bending-torsional vibration of the beam can be written, by using the term (11.325) in
II .16
place of
Coupled Bending - 'fi.lrsional. Vibration "i.lf-Beams. -- .383
f and including the inertia force in Eq. (11.324), as 04v Elz ox4 o 2 if> GJ-
o 4 if> - EJwox2 ox4
= -pA
oZ
(11.327)
otZ (v - eif» 2
= -pAe-(v ot02
+ plG-
- eif»
o 2 if> ot2
(11.328)
11.16.2 Natural Frequencies of Vibration For free vibration of the beam, the solution is assumed to be in the form v(x, t)
=
V(x)C1
if>(x , t)
=
cos(wt
+ 01) + 02)
(11.329) (11.330)
where V (x) and
d2
= pAwz(X
- e
d4
- EJw-
(11.331)
- e
(11.332)
The normal modes of the beam, V(x) and
The simply supported boundary conditions can be expressed as V(x) d2V -(x) dxz
=I
(EILlLl)
=0
at
x =0,
x
=0
at
x=O,
x =1
(EI Ll 1.2)
=0
at
x =0,
x =1
(EILl1.3)
=0
at
x =0,
X
=I
(EI Ll 1.4)
The following functions can be seen to satisfy the boundary conditions of Eqs. (EI 1.1Ll)-(EI Ll1.4): Vj (x)
.
jrrx
= A j sm -, =
I
. jrrx Bj sm -1-'
j
= 1,2,3,
j=I,2,3,
...
(EILl1.5)
...
(EI Ll 1.6)
384
Transverse Vibration of Beams
where A j and B j are constants. By substituting Eqs. (El1.l1.5) and (El1.l1.6) into Eqs. (11.331) and (11.332), we obtain the following equations for finding the jth natural frequency (Wj): (El1.l1.7) (El1.11.8) Equations (El1.l1.7) and (El1.l1.8) can be rewritten as (p2 - w])Aj (q2w])Aj
+ (w]e)Bj
+ (r2
- w])Bj
=0
(El1.l1.9)
=0
(El1.11.l0)
where E /zj41(4 pAZ4
(El1.l1.11)
2
Ae /G + Ae2
(E11.l1.12 )
2
G J Z2 j2rr2
2
P = q =
r =
pZ4(lG
+ E Jwj4rr4 + Ae2)
(El1.l1.13)
For a nontrivial solution of A j and Bj ,_~e detenninant of their coefficient matrix must be equal to zero. This leads to
=0 or wj(1 - q2e) - W](p2
+ r2) + p2r2
=0
(El1.11.l4)
The solution of Eq. (El1.11.l4) gives 2 _
W·1
p2
+ r2
=F [(p2 - r2)2
+ 4p2r2q2e]I/2
2(1 - eq2)
(El1.l1.l5)
Equation (El1.l1.l5) gives two values of W], corresponding to two possible modes of the coupled bending-torsional vibration of the beam. The mode shapes corresponding to the two natural frequencies W] can be detennined by solving Eqs. (El1.l1.9) and (E11.l1.1 0). To fino the physical significance of the two natural frequencies given by Eq. (El1.l1.l5), consider a beam with symmetric cross section with e = O. For this case, Eq. (E11.l1.l5) gives ?
p-
(El1.11.l6)
11.17
Transform Methods:rree -Vibration of an"Infinite Beam
385
=
From Eqs. (E11.11.16), (E11.l1.1l), and (ElLlLl3), we find that W] p2 corre2 sponds to the flexural vibration mode and W] = r corresponds to the torsional vibration mode. For a beam with a nonsymmetric cross section with e i= 0, one of the natural frequencies given by Eq" (EI1.1 1.15) will be smaller and the other will be larger than the values given by Eq. (Ell.l1.l6).
11.17 TRANSFORM METHODS: FREE VmRATION OF AN INFINITE BEAM As indicated in Chapter 7, Laplace and Fourier transform methods can be used to solve free and forced vibration problems. The applicability of the methods to beam problems is illustrated in this section by considering the free vibration of an infinite beam. The equation of motion for the transverse vibration of a uniform beam is given by (11.333) where 2
EI
(11.334)
c =-
p
Let the initial conditions of the beam be given by (11.335)
t = 0) = wo(x)
w(x,
aw (x, t = 0") = Wo ( x) at
(11.336)
-
By taking the Laplace transform of Eq. (11.333), we obtain C
2d4W(x, dx4
s)
+ s 2 W(x,
. + wo(x)
= swo(x)
s)
(11.337)
Next, we take Fourier transform of Eq. (11.337). For this, we multiply Eq. (11.337) by eipx and integrate the resulting equation with respect to x from -00 to 00: c2
1
[d4W(X,
00
4
dx
-00
s)
s2 + 2'W(x,
s)
J .
1
00
e'Px dx
=
c
[swo(x)
+ uio(x)]
.
e'Px dx
-00
(11.338)
The first term of Eq. (11.338) can be integrated by parts to obtain
1
00d4W(x,
s) ----e 4
-00
dx
ipx d
2
3
W . d W x=(d---IP---P 2 3 dx
+ p4
i:
dx
ipx
We
dx
. 2dW
-
dx
+.lp 3W»)
eiPX!OO -00
(11.339)
386
Transverse Vibration of Beams
Since W(x, s), dW(x, s)/ dx, d2W(x, s)/dx2 Ix I -+ 00, Eq. (11.339) reduces to
1
and d3W(x, s)/dx3
d4W ipx --4 e dx = p4 dx
00
-00
1
tend to zero as
00
Weipx dx
(11.340)
-00
Defining the Fourier transforms of W(x, s), wo(x), and wo(x) as -W(p, s)
=~
-Wo(p) =
1 1 1
00
1
2rr
. W(x, s)e'PXdx
(11.341)
-00
00
~1 2rr
. wo(x)e1pX dx
(11.342)
-00
00
-=1 Wo(p) = ~
2rr
wo(x)e1pX dx .
(11.343)
-00
Eq. (11.338) can be rewritten as
or _ sWo(p) + Wo(p) W( p,s ) C2p4 + s2
(11.344)
The inverse Fourier transform of Eg".(11.344) yields
1 -J2ii
W()X,S
= --
00
1
sWo(p)
c2 p4
-00
+ Wo(p) + s2
e
-ipx d
x
(11.345)
Finally, we take the inverse Laplace transform of W(x, s). Noting that L -1[W(x,
L -I
[
L-1
[
2
c P
s]
4
+s
s)] = w(x, t) 2
(11.346)
= cos P 2ct
1 2] = -21. sm p 2 ct +s pc
2 4
cp
(11.347) (11.348)
Eg. (11.345) gives w(x, t)
= v~2rr
1
00
[Wo(P) cos p2ct
+ Wo(p)
sinp2ct] e-ipx dp
(11.349)
-00
The convolution theorem for the Fourier transforms yields (11.350)
Noting the validity of the relations (11.351) 2iT
2v ct
-00
_1_100 .J2ii
dp =
sin p2cte-ipx
'(cos ~
- sin~)
,
(11.352)
2 - sin -17 ) d17
(11.353)
4ct
can be used to express Eq. (11.349) as
1
2
2 wo(x - 17) (17 cos -
00
w(x, t} =
4ct
4ct
2~
-00
Eqs. (11.350)-(11.352)
_1_
4ct
1 -J2iiCi 2 2iTct
4ct
-00
1
2 uio(x - 17) (17 cos -
+ sin 4ct -17 )
00
+ ~1
2vV 2iTct
4ct
-00
d17
4ct
Introducing 2
)..2=!!.-
(11.354)
4ct
Eq. (11.353) can be rewritten
as
(11.355)
11.18 RECENT CONTRIBUTIONS Higher-Order Theories
The second-order theories account for the deformation of the beam due to shear by considering a second variable in deriving the governing differential equations. Some theories, such as Timoshenko theory, include a shear coefficient to account for the nonuniformity of shear deformation across the cross section and the accompanying cross-sectional warping. Other theories include a third variable to characterize the degree to which bending warping occurs. Ewing [6] presented a model of the latter type for application to constant-cross-section beams undergoing symmetric bending vibration (with no bending-torsion coupling).
Beams on an Elastic Foundation
A power series solution was presented for the free vibration of a simply supported Euler-Bernoulli beam resting on an elastic foundation having quadratic and cubic nonlinearities [7]. The problem was posed as a nonlinear eigenvalue problem by assuming the time dependence to be hannonic.
Multispan Beams
In Ref. [8], Wang determined the natural frequencies of continuous Timoshenko beams. A modal analysis procedure was proposed by Wang [9] to investigate the forced vibration of multispan Timoshenko beams. The effects of span number, rotary inertia, and shear deformation on the maximum moment, maximum deflection, and critical velocity of the beam are determined.
_________________________
!!!!!I!I!
!!!"""~~------.-,
•.-.~,-~_-,..........ol-,-,
388
Transverse Vibration of Beams
Self-Excited Vibration In tall buildings and pylons of suspension bridges, the flexibility and the transverse motion of the structures generates self-excited lift and drag aerodynamic forces, known as galloping, which cause nonlinear dynamic behavior. An approximate solution of this problem is presented by Nayfeh and Abdel-Rohman [10]. Damped Beams Wang et al. [11] studied the free vibration of a transmission-line conductor equipped with a number of Stockbridge dampers by modeling it as a tensioned beam acted on by concentrated frequency-dependent forces. An exact solution is obtained using integral transformation. Beams with Tip Mass Zhou gave the exact analytical solution for the eigenfrequencies and mode shapes of a cantilever beam carrying a heavy tip mass with translational and rotational elastic supports in the context of antenna structures [12]. Beams under Axial Loads The vibrational behavior of initially imperfect simply supported beams subject to axial loading has been considered by Ilanko [13]. The natural frequencies of beams subjected to tensile axial loads are investigated by Bokaian [14]. Moving Loads and Bridge Structures The analysis of the vibrational behavior of . structural elements traversed by moving forces or masses can be used to study the dynamics of a bridge traveled by a car or of rails traveled by a train. The problem was studied by Gutierrez and Laura [15]. Waves An analytical method for finding the vibrational response and the net transmitted power of bending wave fields in-systems consisting of coupled finite beams has been studied by Hugin [16]. Beams with Discontinuities A method of finding the bending moments and shear forces as well as the free vibration characteristics of a Timoshenko beam having discontinuities was presented by Popplewell and Chang [17]. Beams with Variable Properties The eigenfunction method using shear theory was used by Gupta and Sharma [18] to analyze the forced motion of a rectangular beam whose thickness, density, and elastic properties along the length vary in any number of steps. A beam of two steps clamped at both edges and subjected to a constant or half-sine pulse load was considered. Computation of Elastic Properties Larsson [19] discussed accuracy in the computation of Young's modulus and the longitudinal-transverse shear modulus from flexural vibration frequencies. Coupled Vibrations Yaman [20] presented an exact analytical method, based on the wave propagation approach, for the forced vibrations of unifonn open-section channels. Coupled wave .numbers. various frequency response curves, and mode shapes were presented for undamped and structurally damped channels using the Euler-Bernoulli
'~,,~,References •.,..J89
model. Bercin and Tanaka studied the coupled flexural-torsional vibrations of monosymmetric beams [21]. The effects of warping stiffness, shear deformation and rotary inertia are taken into account in the formulations. Frames The vibration of frame structures according to the Tlffioshenko theory was studied by Wang and Kinsman [22]. A portal frame subjected to free and forced vibrations is used to illustrate the method.
REFERENCES
".1l
,~i ,
. .."
'ol!' .
1. L. Fryba. Vibration of Solids and Structures Under Moving Loads, Noordhoff International Publishing, Groningen, The Netherlands, 1972. 2. J. E. Shigley and L. D. Mitchell, Mechanical Engineering Design, 4th ed., McGraw-Hill, New York, 1983. 3. W. B. Bickford, Advanced Mechanics of Materials, Addison-Wesley, Reading, MA, 1998. 4. S. S. Rao, Natural vibrations of systems of elastically connected Timoshenko beams, Journal of the Acoustical Society of America, Vol. 55, pp. 1232-1237,1974. 5. C. F. Garland, Normal modes of vibration of beams having noncollinear elastic and mass axes, Journal of Applied Mechanics, Vol. 62, p. 97, 1940. 6. M. S. Ewing, Another second order beam vibration theory: explicit bending, warping flexibility and restraint, Journal of Sound and Vibration, Vol. 137, No. I, pp. 43-51,1990. 7. M. 1. Qaisi, Normal modes of a continuous system with quadratic and cubic nonlinearities, Journal of Sound and Vibratioll;.Voi. 265, No.2, pp. 329-335, 2003. 8. T. M. Wang, Natural frequencies of continuous Tunoshenko beams, Journal of Sound and Vibration, Vol. 13, No.3, pp. 409-414, 1970. 9. R.-T. Wang, Vibration of multi-span Timoshenko beams to a moving force, Journal of Sound and Vibration, Vol. 207, No.5, pp. 731-742, 1997. 10. A. H. Nayfeh and M. Abdel-Rohman, Analysis of galloping responses in cantilever beams, Journal of Sound and Vibration, Vol. 144, No.1, pp. 87-93, 1991. 11. H. Q. Wang, J. C. Miao, J. H. Luo, F. Huang, and L. G. Wang, The free vibration of long-span transmission line conductors with dampers, Journal of Sound and Vibration, Vol. 208, No.4, pp. 501-516, 1997. 12. D. Zhou, The vibrations of a cantilever beam carrying a heavy tip mass with elastic supports, Journal of Sound and Vibration, Vol. 206, No.2, pp. 275-279, 1997. 13. S. Danko, The vibration behaviour of initially imperfect simply supported beams subject to axial loading, Journal of Sound and Vibration, Vol. 142, No.2, pp. 355-359, 1990. 14. A. Bokaian, Natural frequencies of beams under tensile axial loads, Journal of Sound and Vibration, Vol. 142, No.3, pp. 481-498, 1990. 15. R. H. Gutierrez and P. A. A. Laura, Vibrations of a beam of non-uniform cross-section traversed by a time varying concentrated force, Journal of Sound and Vibration, Vol. 207, No.3, pp. 419-425, 1997. 16.C. T.Hugin,A physical description of the response of coupled beams, Journal of Sound and Vibration, Vol. 203, No.4, pp. 563-580, 1997~ 17. N. Popplewell and D. Chang, Free vibrations of a stepped, spinning Timoshenko beam, Journal of Sound and Vibration, Vol. 203, No.4, pp. 717 - 722, 1997. 18. A. P. Gupta and N. Sharma, Effect of transverse shear and rotatory inertia on the forced motion of a stepped rectangular beam, Journal of Sound and Vibration, Vol. 209, No.5, pp. 811-820, 1998.
390
Transverse Vibration of Beams 19. P.-O. Larsson, Determination of Young's and shear moduli from flexural vibrations of beams, Journal of Sound and Vibration, Vol. 146, No.1, pp. 111-123, 1991. 20. Y. Yaman, Vibrations of open-section channels: a coupled flexural and torsional wave analysis, Journal of Sound and Vibration, Vol. 204, No. I, pp. 131-158, 1997. 21. A. N. Bercin and M. Tanaka, Coupled flexural-torsional vibrations of Timoshenko beams, Journal of Sound and Vibration, Vol. 207, No.1, pp. 47-59, 1997. 22. M. Wang and T. A. Kinsman, Vibrations of frame structures according to the Timoshenko theory, Journal of Sound and Vibration, Vol. 14, No.2, pp. 215-227, 1971. 23. S. S. Rao, Mechanical Vibrations, 4th ed., Prentice Hall, Upper Saddle River, NJ, 2004. 24. W. T. Thomson, Laplace Transformation, Second Edition, Prentice-Hall, Englewood Cliffs, NJ,1960.
PROBLEMS 11.1 Derive the equation of motion and the boundary conditions of a Timoshenko beam resting on an elastic foundation using Newton's second law of motion. Assume that the beam is supported on a linear spring, of stiffness Ko, at x 0 and a rotational spring, of stiffness Klo, at x
= I.
spans have identical values of mass density p, crosssectional area A, and flexural rigidity E 1.
=
11.2 Derive the equation of motion and the boundary conditions of a Rayleigh beam resting on an elastic foundation using a variational approach. Assume that the beam is supported on a linear spring of stiffness K at x = 0 and carries a mass M at x I.
2/
=
11.3 Find the natural frequencies and normal modes of a uniform beam simply supported at x 0 and free at x=l.
=
11.4 Derive the orthogonality· relationships of the transverse vibration of a uniform beam that is fixed at x 0 and carries a mass mo and mass moment of inertia 10 at x I, and is supported on a linear spring of stiffness K and torsional spring of stiffness K, at x = I as shown in Fig. 11.22.
=
=
Figure 11.23
11.6 Compute the first four natural frequencies of transverse vibration of a uniform beam for the following boundary conditions: (a) fixed-fixed; (b) pinned-pinned; (c) fixed-free; (d) free-free; (e) fixed-pinned. Data: E 30 x 106 Ib/in2, A = 2 4 2 in , I = ~ in , p = 732.4 X 10-6 Ib-sec2 lin4, and I = 20 in.
=
11.7 A uniform beam simply supported at both ends is deflected initially by a concentrated force F applied at the middle. Determine the free vibration of the beam when the force F is suddenly removed. 11.8 Determine the free vibration of a uniform beam simply supported at both ends that is subjected. to an initial uniform transverse velocity 1'0 at
Figure 11.22
11.5 Determine the frequency equation for the twospan beam shown in Fig. 11.23. Assume that the two
0< x < I.
11.9 A uniform cantilever beam is deflected initially by a concentrated force F applied at the free end. Determine the free vibration of the beam when the force F is suddenly removed.
Problems
11.10 A unifonn fixed-fixed beam is deflected initially by a concentrated force F applied at the middle. Determine the free vibration of the beam when the force F is suddenly removed.
11.15 Find the first four natural frequencies of vibration of a unifonn beam with a channel section as shown in Fig. 11.24 and simply supported at both ends.
11.11 A railway track can be modeled as an infinite beam resting on an elastic foundation with a soil stiffness of k per unit length. A railway car moving on the railway track can be modeled as a load Fa moving at a constant velocity Va. Derive the equation of motion of the beam.
T
11.12 Compare the natural frequencies of vibration of a unifonn beam, simply supported at both ends, given by the Euler-Bernoulli, Rayleigh, and Timoshenko theories for the following data: (a) £ = 30 x 106psi, p = 732.4 x 1O-61b-sec2/in4, A
= 2in
2, [
k = (b) £
(c) £
(d) £
5
.
6' G = 11.5 X
106pSl
= 8in2, [ = 10~in4, [=
k
5 = 6' G = 11.5
.
106pSl
106psi, p
= 253.6 x
2in2,
~in4, 3
A
=
k
5 = 6' G = 3.8
[=
1O-61b-sec2/in4, 20 in.,
[=
.
X
106pSl
= 10.3 x 106psi, p = 253.6 x
k
20 in.,
3
X
1O-61b-sec2/in4,
= 8in2, [ = 1O~in4, [ = 20 in., 3 5
= 6' G = 3.8 X
Figure 11.24
10-61b-sec2 /in4,
A
A
10 in.
= ~in4, [ = 20 in., 3
= 30 x 106psi, p = 732.4 x
= 10.3 x
391
.
106pSl
11.13 Plot the variations of WRn/WEn and WTn/WEn over the range 0 ::: nr / I ::: 1.0 for £/ kG = 1.0, 2.5 and 5.0, where WEn, WRn, and WTn denote the nth natural frequency of vibration of a unifonn beam given by the Euler-Bernoulli, Rayleigh, and Timoshenko theories, respectively. 11.14 Derive the equation of motion of a transversely vibrating beam subjected to an axial force P using Newton's second law of motion.
,11.16 In the frequency equation of a simply supported Timoshenko beam, Eq. (11.301), the contribution of the first term, w~(pr2 / kG), can be shown to be negligibly small for nr / I « 1. By neglecting the first tenn, the approximate frequency of vibration can be found as w2 n
a2n411'4/ ~
------------1 n211'2r2/
+
[2
+
[4
(n211'2r2/
[2)£ / kG
Find the first three natural frequencies of vibration of the beam considered in Example ILl 0 using the approximate expression given above, and find the error involved compared to the values given by the exact equation (11.301). 11.17 Consider a beam subjected to a base motion such as one experienced during an earthquake shown in Fig. 11.25. Ifthe elastic deflection of the beam, w(x, t), is measured relative to the support motion, wg(t), the total displacement of the beam is given by Wt(X,
t)
= wg(t)
+ w(x,
t)
(a) Derive the equation of motion of the beam, subjected to base motion, considering the inertia force associated with the total acceleration of the beam. (b) Suggest a method of finding the response of a beam subjected to a specified base motion wg(t).
392
Transverse Vibration of Beams
T
._._._._._._._._.~._.~ !~ '-. -----_-------------
.•..•
1-- -Figure 11.25
x
~I Beam subjected to support excitation.
11.18 A uniform simply supported beam is subjected to a step-function force Fo at the midspan (at x 1/2). Assuming zero initial conditions, derive expressions for the following dynamic responses: (a) displacement distribution in the beam; (b) bending moment distribution in the beam;(c) bending stress distribution in the beam.
=
11.19 Determine the response of a simply supported beam subjected to a moving uniformly distributed load, as shown in Fig. 11.26.
11.20 Show that the constant a in Eq. (11.27) is positive. 11.21 Derive the expressions for the constants An and En in Eq. (11.72) in terms of the initial conditions of the beam [Eqs. (11.54) and (11.55)]. 11.22 Using Hamilton's principle, derive the equation of motion, Eq. (11.203), and the boundary conditions of a beam resting on an elastic foundation. 11.23 Derive Eq. (11.216) from Eqs. (11.213) and (11.214) for the forced vibration of a beam on an elastic foundation. 11.24 Derive the characteristic equation for a fixed-fixed Timoshenko beam using the solution of Eqs. (11.308) and (11.309) and the boundary conditions of Eqs. (11.302) and (11.303).
Figure 11.26 Simply supported beam subjected to a traveling distributed load.
11.25 Determine the natural frequencies of vibration of a fixed-fixed beam using Rayleigh's theory.
12 Vibration of Circular Rings and Curved Beams 12.1 INTRODUCTION The problems of vibration of circular rings and curved beams (or rods) find application in several practical problems. The vibration of circular rings is encountered in an investigation of the frequencies and dynamic response of ring-stiffened cylinders such as those encountered in airplane fuselages. circular machine parts such as gears and pulleys. rotating machines. and stators of electrical machines. The vibration of a curved rod can be categorized into four types when the centerline of an undeformed rod is a plane curve and its plane is a principal plane of the rod at each point. In the first type, flexural vibrations take place in the plane of the ring without undergoing any extension of the centerline of the ring. In the second type. flexural vibrations. involving both displacement at right angles to the plane of the ring and twist. take place. In the third type. the curved rod or ring vibrates in modes similar to the torsional vibrations of a straight rod. In the fourth type. the ring possesses modes of vibration similar to the extensional vibration of a straight rod. It is assumed that the undeformed centerline of the ring has a radius R. the cross section of the ring is uniform. and the crosssectional dimensions of the ring are small (for a thin ring) compared to the radius of the centerline of the ring. The vibration of curved beams is important in the study of the dynamic behavior of arches.
12.2 EQUATIONS OF MOTION OF A CIRCULAR RING 12.2.1 Three-Dimensional Vibrations of a Circular Thin Ring Next. we derive the general equations of motion governing the three-dimensional vibrations of a thin rod which in the unstressed state forms a circular ring or a portion of such a ring [Fig. 12.1(a)]. The effects of rotary inertia and shear deformation are neglected. The following assumptions are made in the derivation: 1. The centerline of the ring in an undeformed state forms a full circle or an arc of a circle. 2. The cross section of the ring is constant around the circle. 3. No boundary constraints are introduced on the ring (Le.• the rim is assumed free) . 393
394
Vibration of Circular Rings and Curved Beams F
\
Section I-I
\ \ d()
I
(.
\
p(6, t)
,,,P+d~
\
2
o
~Ml+dMl (a)
(b)
M,
-t""""_._._~ec~ q(6,t)
'\
'\
'\
Section 1-1
M2+dM2
0
M,+dM, (c)
Figure 12.1
(d)
Element of a circular ring.
Consider the free-body diagram of an element of a circular ring abed shown in Fig. 12.1(b), where MI(B, t) is the in-plane bending moment, taken positive when it tends to reduce the radius of curvature of the beam, F (B, t) is the shearing force, taken positive when it acts in a radially inward direction on a positive face, P (B, t) is the tensile force, and f(B, t) and pCB, t) are the external radial and tangential forces, respectively, per unit length. The radial and tangential inertia forces acting on the element of the circular ring are given by and
A(B)R dB (Pw(B, r) P
at2
where p is the density, A is the cross-sectional area, and R is the radius of the centerline of the ring. The equations of motion in the radial and tangential directions can be
12.2 expressed as (by neglecting
Equations of Motion of a Circular Ring <~95
small quantities
of high order):
a2u
aF
-ae + P + fR = pAR- at2
(12.1)
ap a2w - - F+ pR =pARae at2
(12.2)
If v(e, t) denotes the transverse displacement, Q(e, t) the transverse shear force, and q(e, t) the external distributed transverse force in a direction normal to the middle plane of the ring [Fig. 12.1(c) and (d)], the transverse inertia force acting on the element of the ring is 2
p
A(e)R de a v(e, t)
at2
and the equation of motion in the transverse
aQ
direction is given by
a2v
ae +qR
= pAR at2
(12.3)
The moment equation of motion in the middle plane of the ring (about an axis normal to the middle plane of the ring) leads to
aMI
ae
+ FR
=0
(12.4)
Let M2(e, t) be the bending moment in the ring about the radial axis, Mr(e, t) the torsional moment about the tangential axis, and mo(e, t) the distributed external torque acting on the ring. Then the dynamical moment equilibrium equation about the radial axis of the element of the ring is given by
aM2 ae - QR+Mr Similarly, the dynamical expressed
moment equilibrium
=0
(12.5)
equation about the tangential axis can be
as
aMr
--
ao
- M2
+ moR
=0
(12.6)
The six equations(l2.1)-(12.6) are the equations of motion governing the threedimensional vibrations of a ring in 10 unknowns: F, P, Q, Mlo M2, Mr, u, v, w, and n [n and v appear in the expression of Mr, as shown in Eq. (12.18)]. To solve the equations, four more relations are required. These relations include the moment-displacement relations, along with the condition of inextensionality of the centerline of the ring.
12.2.2
Axial Force and Moments in Terms of Displacements To express the forces and moments in the ring in terms of the deformation components, consider a typical element of the ring, ab, located at a distance x from the centroidal axis of the ring, as shown in Fig. -12.2(a). If the axial stress in the element is G, the
396
Vibration of Circular Rings and Curved Beams
(a)
(b)
Figure 12.2 (a) Differential element of a curved beam; (b) Undefonned and defonned positions of the central axis of the curved beam.
axial force and the bending moment can be expressed as
,p=ffadA ,
(12.7)
A
M} =
ffaXdA
(12.8)
A
where a = E£
(12.9)
£ is the axial strain, E is Young's modulus, and A is the cross-sectional area of the ring. The strain in the element can be expressed as [1, 5, 6]
£
=
2. [-u + aw _ ae
R
~~ (w + au)] R ae ae
(12.10)
where u and ware, respectively, the radial and tangential displacements of a typical point A lying on the central axis of the ring as shown in Fig. 12.2(b). By substituting Eqs. (12.10) and (12.9) into Eqs. (12.7) and (12.8) and performing the integrations, we obtain P =
EA ( aw) R -u + M
M - EII~ I R2 ae
(w + aeau)
(12.11) (12.12)
12.2
Equations of MOli01t{)f.aCu-cular-Rillg
397
where /1 denotes the moment of inertia of the cross section about an axis perpendicular to the middle plane of the ring and passing through the centroid:
ff
It =
2
(12.13)
x dA
A
Note that the x contribution of Eq. (12.10) does not enter in Eq. (12.11) since the axis is the centroidal axis. The condition of inextensionality of the centerline of the ring is given by s =0
(12.14)
By neglecting products of small quantities in Eq. (12.10), Eq. (12.14) leads to
aw ao
-
=u
(12.15)
Equation (12.12) denotes the differential equation for the deflection curve of a thin curved bar with a circular centerline. If the inextensionality of the centerline of the ring, Eq. (12.15), is used, Eq. (12.12) takes the form MI
= -Elt R2
(
u)
2
u
+ -aao2
(12.16)
It can be seen that for infinitely large radius, Eq. (12.16) coincides with the equation for the bending moment of a straight beam. Proceeding in a similar manner, two more relations for bending in a normal plane and for twisting of the ring can be derived as [5, 7] (12.17)
GJ (dVdO + R dn) dO
(12.18)
R2
Mt =
2
d V) Rn - -d0 2
M2 = -EIz
(
/if
where E Iz is the flexural rigidity of the ring in a normal plane, GJ is the torsional rigidity of the ring (GJ is denoted as C for noncircular sections), and n is the angular displacement of the cross section of the ring due to torsion. 12.2.3
Summary of Equations and Classification of Vibrations The 10 governing equations in 10 unknowns for the three-dimensional vibrations of a circular ring are given by Eqs. (12.1)-(12.6) and (12.15)-(12.18). It can be seen that these equations fall into two sets: one consisting of Eqs. (12.1), (12.2), (12.4), (12.15), and (12.16), where the variables v and n do not appear, and the other consisting of Eqs. (12.3), (12.5), (12.6), (12.17) and (12.18), where the variables u and w do not appear. In the first set of equations, the motion can be specified by the displacement u or w, and hence they represent flexural vibrations of the ring in its plane. In the second set of equations, the motion can be specified by v or n, and hence they represent flexural vibrations at right angles to the plane of the ring, also called coupled twist-bending vibrations involving both displacement at right angles to the plane of the ring and twist.
398
Vibration of Circular Rings and Curved Beams
In general, it can be shown [5] that the vibrations of a curved rod fall into two such classes whenever the centerline of the undeformed rod is a plane curve and its plane is a principal plane of the rod at each point. In case the centerline is a curve of double curvature such as a helical rod, it is not possible to separate the modes of vibration into two classes, and the problem becomes extremely difficult. In addition to the two types of vibration stated, a curved rod also possesses modes of vibration analogous to the torsional and extensional vibrations of a straight rod. When u w 0 and v is assumed to be small compared to Rn, Eq. (12.6) becomes the primary equation governing the torsional vibrations of a ring. On the other hand, when v = n = 0 and the inextensionality condition of Eq. (12.15) is not used, the ring undergoes extensional vibrations.
= =
12.3 IN-PLANE FLEXURAL VffiRATIONS OF RINGS 12.3.1
Classical Equations of Motion As indicated in Section 12.2.3, the equations of motion for the in-plane flexural vibrations of a thin ring are given by Eqs. (12.1), (12.2), (12.4), (12.15) and (12.16). These equations can be combined into a single equation as ' 6 4 2 4 8w a w aw R (af ) pAR4 82 (a2w ) 6 ao +2 ao4 + ao2 - Eh ao - P + Ell at2 ao2 - w = 0 (12.19) Equation (12.19) represents the equation of motion for the in-plane vibrations of a thin ring in terms of the radial deflection w. Natural Frequencies of Vwration For free vibration, the external forces f and p are assumed to be zero. Then, by assuming the solution to be harmonic, as w(O, t)
= W(O)eiwr
(12.20)
where w is the frequency of vibration, Eq. (12.19) becomes d4W
6
dW d06
+
2
d04
+
2 d W _ pAR4(J)2 (d2W _ d02 Eh d02
w)
=
0
(12.21)
The solution of Eq. (12.21) is assumed as W(O) = C] sin (nO
+ <1»
(12.22)
where C] and are constants. When Eq. (12.22) is substituted, Eq. (12.21) gives the natural frequencies of vibration as 2 (J)n
E I] n6
= pAR 4
2n4
-
n
2
+ n2
+1
n
= 2, 3, ...
(12.23)
Note that for a complete ring, n in Eq. (12.22) must be an integer, for this gives n complete waves of deflection W in the circumference of the ring and W must be a function whose values recur as 0 increases by 2rr. Also, n cannot take the value of ] since it represents pure rigid-body oscillation without any alteration of shape, as shown in Fig. 12.3. Thus, n can only take the values 2, 3, .... The normal modes of the ring
12.3
In-Plant: FlexuralVibr,ati~ns~f.Rings .•.}99
.'}:
---::.-
..... ,;,;
,;
,;
II
"
I"
I I I I I I \ \
- ::-" ..•..
..•...•..
" , "" ,
,;
"
\
\
\
I I
\
\
\ \
,
\ \
\
I
I
I I I
\
\
\ \
I I I I \ \ \
I
,,
-
" ••.•...•.. "
I I I I
/
.
,;
"
..........-=. __
/ ,;
"
"
/
n=l
Figure 12.3
can be expressed
as Wn(O,
where C 1 and
12.3.2
Mode shapes of a ring.
t)
= Cl sin(nO +
(12.24)
from the initial conditions of the ring.
Equations of Motion That Include Effects of Rotary Inertia and Shear Deformation A procedure similar to the one used in the ease of a straight beam by Timoshenko can be used to include the effects of shear deformation and rotatory inertia in the equations of motion of a ring. The slope of the deflection curve (1/1) depends not only on the rotation of cross sections of the ring, but also on the shear. If
1/I=
But 1/1 can be expressed in terms of the displacement components u and W once the geometry of deformation is known. From Fig. 12.4, the total slope of the deflection curve can be expressed as
1 au
W
ao
R
'1/1=--+R
(12.26)
400
Vibration of Circular Rings and Curved Beams F
P
/ P+dP
~F + dF
/
.~
~
/
/
~
~
I
..~.~ ~
/
fJ
·'.t:.._L._ o
Figure 12.4
Composition of the slope of a deflection curve.
and hence Eq. (12.25) yields
fi
(au-ae + w - R¢J )
= -1
R
(12.27)
As for the shear force, the exact manner of distribution of the shear stress over the cross section is not known. Hence, to account for the variation of f3 through the cross section, a numerical factor k that depends on the shape of the cross section is introduced and the shear force is expressed as F = kfiAG
(12.28)
where G is the rigidity modulus of the material of the ring. Since the exact determination of the factor k for rings involves consideration of the theory of elasticity, the value of k of a straight beam can be used in Eq. (12.28) as an approximation. Equations (12.27) and (12.28) can be combined to obtain kAG F=R
(au-+w-R¢J ) ae
(12.29)
The bending moment can be expressed in terms of the displacement components as [Eq. (12.16) is modified in the presence of shear deformation] M
J
=
Ell a¢J
R
ae
(12.30)
The differential equation of rotation of an element abed, shown in Fig. 12.4, can be found as (12.3])
12.3
In-Plane Flexural Vibrations of Rings :-401
The differential equations for the translatory motion of the element in the radial and tangential directions are given by a2u pAR at2
aF
ao + P + f R = ap ao - F + pR = pAR
(12.32)
aw 2
(12.33)
at2
By eliminating P from Eqs. (12.32) and (12.33), we obtain aZF aez
+F
= pAR
(~_ aeat2
aZw) atZ
+ R (p _ af)
(12.34)
ae
The condition for inextensionality of the centerline is given by Eq. (12.15). Thus, the set of equations (12.29), (12.30), (12.31), (12.34), and (12.15) govern the in-plane flexural vibrations of a ring when the effects of rotary inertia and shear deformation are included. From these equations, a single equation of motion can be derived as a6w ae6
a4w + 2 ae4
= (PRZ E
+
+ +
PRZ ( E
4 pR - kEAG
aZw aez
PRZ) kG
+
4 2 _ pAR _ PR ) + Eh kG 4 Z 4 Z 4 aZw (P R ) a w _ (P R ) atZ + kEG at4 kEG
~ ae4atZ
PAR4) Eh
Z a ( af) atZ p - ae
4 a w ae2atZ 6 a w aeZat4
(2PRZ E
4 R ( af) - Ell P - ae
Z
+
R kAG
(
af) P - ae
(12.35)
Equation (12.35) denotes the governing differential equation of motion of a ring that takes into account the effect of rotary inertia and shear deformation. If the effect of shear deformation is neglected, the terms involving k are to be neglected in Eq. (12.35). Similarly, if the effects of rotary inertia and shear deformation are neglected, Eq. (12.35) can be seen to reduce to Eq. (12.19). Finally, R --+ 00 and Eq. (12.35) reduces to the equation of motion of a Timoshenko bearn. Natural Frequencies of Vibration For free vibration, the external forces f and p are assumed to be zero. By assuming a harmonic solution as in Eq. (12.24), the frequency equation can be derived as Z Ki(-nzSiS, - SiS,) + Kz(n4Sz + n4SzS1 - 2Sznz +nzSzS, + n + Sz + 1)
+
(_n6
+ 2n4
_ nZ)
=0
(12.36)
where E Sl = kG' Equation (12.36) is a quadratic in K2 and hence two frequency values are associated with each mode of vibration (Le., for each value of n). The smaller of the two
402
Vibrationof CircularRings and Curved Beams w values corresponds to the flexural mode, and the higher value corresponds to the thickness-shear mode. Similar behavior is exhibited by a Timoshenko beam as well. In Eq. (12.36), n must be an integer with a value greater than 1. The values of the natural frequencies given by Eq. (12.36) for various values of 8) and 82 are given in Tables 12.1. Note that if the terms involving 82 are neglected in Eq. (12.36), we obtain a frequency equation that neglects the effect of shear deformation (but considers the effect of rotary inertia). Similarly, if the terms involving 8) and 82 are neglected, Eq. (12.36) reduces to Eq. (12.23), which neglects both the effects of rotary inertia and shear deformation.
12.4 FLEXURAL VIBRATIONS AT RIGHT ANGLES TO THE PLANE OF A RING 12.4.1
Classical Equations of Motion As stated in Section 12.2.3, the equations of motion for the coupled twist-bending vibrations of a thin ring are given by Eqs. (12.3), (12.5), (12.6), (12.17), and (12.18). All these equations can be combined to obtain a single equation as
a6v a4v a2v pAR4 a4v pAR4 a2v 6 ae + 2 ae4 + ae2 + E h ae2at2 - -C at2 4 2 4 R a q R 3( a2mo 2 - E h ae + Cq + R C + Eh ae2 = 0
1 1)
(12.37)
where C = GJ. Thus, Eq. (12.37) denotes the classical equation for the flexural vibrations involving transverse displacement and twist of a thin ring. The twist n is related to the transverse deflection v by
2 aQ ae2
(E h a4v = Eh + C R ae4 1
C R
3
a2v a2v ae2 7- pAR at2 -
3)
R q
Table 12.1
In-PlaneFlexural Vibrationsof a Ring, NaturalFrequencies,and Valuesof Wn J pA R4 / E h; Effects of Rotary Inertia and Shear Deformation Included E/kG
/J/AR2
n
Flexural mode
1.0
0.02
2 3 2 3 2 3 2 3 2 3
2.543 6.682 2.167 5.017 2.459 6.289 2.380 5.950 1.321 2.682
0.10 2.0
0.02
3.0
0.02 0.25
Thickness-shearmode 52.759 56.792 12.382 15.127 38.587 42.667 32.400 36.800 4.695 6.542
(12.38)
12.4
Flexural Vibrations
at Right Angles to the Plane of a Ring
=
403
=
Natural Frequencies of Vibration For free vibration, q mo 0, andEq. (12.37) gives the natural frequencies of vibration, by assuming a harmonic solution, as w
E Iz n - 2n + n =------n pAR4 nZ + ElzlC 6
2
Note that the value of n of the ring are given by
= 1 corresponds
vn(e, t)
12.4.2
4
Z
n=2,3,oo.
'
to rigid-body motion and the normal modes
+ cp)eiwnt,
= Ct sin(ne
(12.39)
n
= 2, 3,
00.
(12.40)
Equations of Motion That Include Effects of Rotary Inertia and Shear Deformation Using a procedure similar to that of a Timosherko beam, the slope of the transverse deflection curve is expressed as 1
R
av ae
=a
+ f3
(12.41)
where a is the slope of the deflection curve when the shearing force is neglected and f3 is the angle of shear at the neutral axis in the same cross section. The transverse shearing force F is given by Q
= kf3AG
(12.42)
where k is a numerical factor taken to account for the variation of f3 through the cross section and is a constant for any given cross section. Using Eq. (12.41), Q can be expressed as 1 av ( ---a Rae
Q=kAG
)
(12.43)
The moment-displacement relations, with a consideration of the shear deformation effect, can be expressed as M2 =
Eh
M = C t
aa)
(
(12.44)
Ii'" n - ae R
(a + an) ae
(12.45)
The differential equation for the translatory motion of an element of the ring in the transverse direction is given by
aQ
aZv
ae
atZ
- + Rq = pAR-
(12.46)
The equations of motion for the rotation of an element of the ring about the radial and tangential axes can be expressed as
aM2
aZa
ae aMt ae
atZ aZn = pJR- Z at
-+Mt-QR=-phR- Mz +moR
(12.47) (12.48)
;1
404
Vibration of Circular Rings and Curved Beams
In view of Eqs. (12.43)-(12.45), Eqs. (12.46)-(12.48) can be rewritten as
a [kAG ( ---ex 1 av)] -ae R ae·
2 +qR-pAR-=Oaat2v (12.49)
a [Eh (aex)] ae -R n-ae
-kARG
(1---ex av R ae
) +-c (an) a2ex R ex+-ae +phR-=O at2 (12.50)
[C (
a - ex+-an)] -- Eh (aex) a2n -ae R ae R n-ae +moR-pJR-=Oat2 (12.51) Equations (12.49)-(12.51) thus represent the equations of motion of a ring for the coupled twist-bending vibrations of a ring, including the effects of rotary inertia and shear deformation. These three equations can be combined to obtain a single equation in terms of the displacement variable v as [7, 9]
(12.52)
· I
12.4 Flexural Vibrations at R~ght,~n.gles to the Plane of a Ring
405
Various special cases can be derived from Eqs. (12.52) and (12.53) as follows: 1. When the effect of shear deformation is considered without the effect of rotary inertia:
(12.55) 2. When the effect of rotary inertia is considered without the effect of shear deformation:
(12.56)
3. When the effects of both rotary inertia and shear deformation are neglected:
a6v a4v a2v pAR4 a4v pAR4 a2v R4 a2q ae6 + 2 ae4 + ae2 + E h ae2at2 at2 - E h ae2 R4 a2mo 3 (12.58) + Cq + R E h + c ae2 = 0
(1
1)
---c-
(12.59) Further, if R is made equal to infinity and the terms involving the torsional motion are neglected, Eq. (12.52) will reduce to the Timoshenko beam equation.
406
Vibration of Circular Rings and Curved Beams
Natural Frequencies of Vibration harmonic solution and introducing SI
Setting q
= mo = 0
E
for free vibration with the 1 S4 =k'
=-, G
Eh
Ss=C
(12.60) the frequency equation can be derived as _ T2
T3SIS?S4SS S3
+
[n2s1 (SIS
4
T [SIS,S4 + S, +.' 4
+ (2n
-
6
n
-
2 n )
(1
+ Ss + SIS4SS) + SlSls4 + sIsls4ss S3
+~ -
S3
S3
2SIS,S. +
+ S2SS] S3
S,S,) + .·S, (1 + SIS, + ~:)]
=0
(12.61)
This equation is a cubic in T and gives three frequency values for each mode number n with the lowest one corresponding to the flexural mode. The two higher frequencies correspond to the torsional and transverse thickness-shear modes. For a complete ring, n must be an integer with values greater than 1. Corresponding to any Wn given by Eq. (12.61), the normal mode can be expressed as vn(O, t) = C1 sin(nO
+ ¢)eiwnt
(12.62)
In this case, the solution ofEq. (12.53) can be expressed as (by considering the periodic nature of n): nn(O,t)=-(Eh+C)n2
1
[ n E17+n 2CR-pRwn 2(CkG+AR 2)
- n2phRw~
4
2
(1 +k~) +
p2k~ h
W~] Cj sin(nO + ¢)eiwnt (12.63)
Note that if the effect of rotatory inertia only is under consideration, terms involving k in Eqs. (12.61) and (12.63) are to be omitted. The in-plane and normal-to-plane
natural frequencies of vibration of a ring with circular cross section are compared in Table 12.2.
12.5 TORSIONAL VIBRATIONS For the torsional vibration of a circular ring, the in-plane displacement components u and w are assumed to be zero. In addition, the transverse displacement v, perpendicular to the plane of the ring, is assumed to be small compared to Rn, where R is the radius of the undeformed centerline of the ring and n is the angular deformation of the cross section of the ring (Fig. 12.5). In this case, ignoring the terms involving Ct, Eq. (12.51) gives
(12.64)
~_""""''''''''''_~,,,,,,._-~
.••••••••••••• ~.-....~~
•."",~'~''''_'''''-';'~'''''''''''''''''''''''''''''~''''
••<'''''."."".•''''''"'''''.'''''''''''''' .••.li;.'''''''''''',,, •..,,,,.,,~,_ i
••••.•..
12.6 Table 12.2
wnJ mAIr
/ EI according to: Shear deformation considered
Rotary inertia considered
Classical theory
2 3 4 5 6
407
Comparison of In-Plane and Normal-to-Plane Frequencies of a Ring" Value of
Mode shape, n
Extensional Vibrations
NormalIn-plane to-plane
In-plane
Normalto-plane
In-plane
2.606 7.478 14.425 23.399 34.385
2.544 6.414 10.765 15.240 19.706
2.011 4.860 7.875 10.796 13.590
2.011 4.572 7.169 9.709 12.191
2.683 7.589 14.552 23.534 34.524
"Ring with circularcross section, r I R
= 0.5, k = 0.833,
E
NormalNormalto-plane In-plane to-plane 2.259 5.028 7.671 10.192 12.636
1.975 4.446 6.974 9.478 II. 948
1.898 4.339 6.886 9.431 II. 946
I G = 2.6.
I I
_.--n
Both shear and rotary inertia considered
n--.
-r----_C_~~<::-
::--,-'lf~_L .-
I I
Figure 12.5 Torsional vibration of a ring
For a complete circular ring, the free vibrations involving n wavelengths in the circumference and frequency w are assumed to be of the form Q(O, t)
= Cl sin (nO + r!J)eiwr
(12.65)
Substituting Eq. (12.65) into (12.64), we obtain Cn2 + E/z pJR2
2 _ -
wn
(12.66)
12.6 EXTENSIONAL VIBRATIONS For the extensional vibration of a circular ring, we assume v and Q to be zero. In this case, the centerline of the ring extends by (II R)[(8wI80) - u] and tension developed is given by [1, 5] p = EA R
(8W _ u) 80
(12.67)
408
Vibration of Circular Rings and Curved Beams
The equations of motion governing u and w are given by Eqs. (12.1) and (12.2), which can be rewritten, for free vibration, as
aF
"M + P
= pAR
ap
"M -
F = pAR
a2u at2 a2w at2
(12.68) (12.69)
Neglecting F, Eqs. (12.68) and (12.69) can be written as P
ap ao
=
EA R
(aw _ de
u)
2W _ au) ae2 ao
= EA (d R
= PARa2u at2
=. ARa2w p at2
(12.70) (12.71)
During free vibration, the displacements u and w can be assumed to be of the form u(O, t)
= (C}
w(O, t)
= n(C}
sin nO cosne
+ C2cosnO)eiwt -
C2 sinnO)eiw1
(12.72) (12.73)
Using Eqs. (12.72) and (12.73), Eqs. (12.70) and (12.71) yield the frequency of extensional vibrations of the ring as
(12.74)
12.7 12.7.1
VIBRATION OF A CURVED BEAM WITH VARIABLE CURVATURE Thin Curved Beam Consider a thin uniform curved beam with variable curvature as shown in Fig. 12.6. The curved beam is assumed to have a span I and height h and its center line (middle surface) is defined by the equation y = y(x), where the x axis is defined by the line joining the two endpoints (supports) of the curved beam. The curvature of its centerline is defined by p(O) or p(x), where the angle 0 is indicated in Fig. 12.6(a). The radial and tangential displacements of the centerline are denoted u and w, respectively, and the rotation of the cross section of the beam is denoted ¢, with the positive directions of the displacements as indicated in Fig. 12.6(a). of Motion The dynamic equilibrium approach will be used for derivation of the equations of motion by including the effect of rotary inertia [21, 22]. For this we consider the free-body diagram of an element of the curved beam shown in Fig. 12.6(b). By denoting the inertia forces per unit length in the radial and tangential directions, respectively, as Fi and Pi and the inertia moment (rotary inertia) per unit length as Mi, Equations
_________________
12.7
Vibration of a Curved Beam with Variable Curvature
409
Axis of curved beam
u
P+dP
C \ \
P
\ \ \
P
\
\ \ \ \ \ \ \ \ \ \ \ \ \ \
B
__.J_ w
\ \
\ \-dB \\
o
W
Figure 12.6 Curved beam analysis: (a) geometry of the curved beam; (b) free-body diagram of an element of the curved beam.
the equilibrium equations can 'be expressed as follows [22. 23]. Equilibrium of forces in the radial direction: '
aF ae - P + pF
i
=0
(12.75)
=0
(12.76)
Equilibrium of forces in the tangential direction:
ap
-+F+PPi
ae
Equilibrium of moments in the xy plane: aM
ae -
pF - pMi = 0
(12.77)
The tangential force (P) and the moment (M) can be expressed in terms of the displacement components u and w as [5]
P = EAp [dW + w + _1_ (d + u)] de Ap de EI (d M = -pi de + 2
U 2
2
(12.78)
2
U
)
2
U
(12.79)
The rotation of the cross section (» is related to the displacement components u and W as [13. 21] >
=~ p
(dUde -
w)
!!!!!!!I!!BI
(12.80)
----.....'!""""""'~~
•.~
410
Vibration of Circular Rings and Curved Beams
During free vibration, u and w can be expressed as
= U(e)coswt
(12.81)
w(e, t) = W(e)coswt
(12.82)
u(e, t)
where w is the frequency of vibration and U (e) and W (e) denote the time-independent variations of the amplitudes of u(e, t) and w(e, t), respectively. Thus, the inertia forces and the inertia moment during free vibration of the curved beam are given by Pi = mw2U
(12.83)
Pi = mw2W
(12.84)
M. = I
(dUde _ w)
2
mw J
Ap
(12.85)
where m is the mass of the beam per unit length. Equations (12.78), (12.79), (12.77), (12.83)-(12.85), used in Eqs. (12.75) and (12.76) to obtain 4 I (d U { - E1 p3 de4
2 3 d U) 5 dp (d U de2 - p4 de de3
+
1d
2 _ cmw J [~(d2U _dW) A p de2 de
P
+
dU) de
+
2 3J dp (d U
de2
- Ap4 de
p2 de
+U
2 U --- 2 dp (d--+U p4 de de2
3 (d U Ap3 de3
)]
J
- E1
[1
w)]
_
de
+pmw 2U=O
J
2
(d W d()2
_..!.. dP(dU
2 (d-+U U)] Ap2 de2
- -EA [dW -+W+p de 1
dU)
+ de
2p 2 J(d U dW)} - p4 de2 de2 - de
P + 2 [4p5 (dde)2
EA [
(12.81), and (12.82) can be
+
p3
dU) de
3 (d U de3
2 )] --- cmw J (dU --W Ap de
(12.86)
1 dp (dW - p2 de de
+U
)
dU)
+ de
) +pmw
2
W=O
(12.87)
where c is a constant set equal to 0 or 1 if rotary inertia effect (Mi) is excluded or included, respectively. It can be seen that Eqs. (12.86) and (12.87) denote two coupled differential equations in the displacement variables U (e) and W (e). Introducing the nondimensional parameters x
y
~-- I'
1}
p ~-- I'
h h=I'
=
l'
u--UI '
- -
r=~H I J'
-
W
W=-
(12.88)
I
Q. = I
w./ I
mI4 EJ
(12.89)
12.7 Vibration of a Curved Beam with V.aIi;lbleCurvature : ..4)1 where
denotes
Wi
i = 1,2 •...•
the ith frequency;
Eqs. (12.86) and (12.87) can be
expressed as [21, 22]
d4T} _ de4 -
PI
+ d2
W
dez
d3T} de3
+
I
( PI
_ CP4QZ) dT} [4 de I
(PS + P6Q~) U + (1- ':"Q2) P3 dWde [4
d2 T}
= P7 de2
+
2 (" CP3Q2) d Tj + P~ + [4 de2
P3 -
+ C
I
[4
-
(C Qi2 - 1) [4
Q + CP4 t~ ?W
I
d T}
de + psT} + P9
d
W
de
(12.91)
Q~ W
[4
(12.90)
1_
1_
where 5 d~ PI = --
(12.92)
P2=~d2~ _.!(d~)2_2 ~ d()2 ~2 de
(12.93)
P3 =
(12.94)
~ de
_[2~2
P4 = -r ps =
d~
2
(12.95)
~de
2 d2~
f de2
-
8 (d~)2 ~2
2 2
de
- [ ~ - 1
(12.97)
P6 = [4~4 1 d~
P7
(12.96)
(12.98)
= [2~3 de
ps = ~d~
~ de
(1 + _1_)
(12.99)
[2~2
1 d~ P9 = --
(12.100)
~ de
The coefficients PI to P9 can be computed by starting from the equation of the curved beam. y = y(x). Using the nondimensional parameters defined in Eqs. (12.88) and (12.89) for ~. 1/. ~ and [, we can express the equation of the curved beam in nondimensional form as 1/ = 1/(~). Using the relation rr e=--tan 2
_ld1/
d~
(12.101)
we obtain (12.102)
412
Vibration of Circular Rings and Curved Beams
Equations (12.101) and (12.102) can be used to compute the first and second derivatives of ~ as
d~ _ d~ d~ de - d~ de d2~ de2
(12.103)
d (d~) de
d~ de
= d~
The coefficients PI to P9, defined in Eqs. (12.92)-(12.100), Eqs. (12.101)-(12.104).
(12.104) can be computed using
Numerical Solution The boundary conditions of the curved beam or arch can be stated as follows. For a clamped or fixed end: w=o
or
W=o
(12.105)
u=o
or
U=O
(12.106)
au =0 ae
or
dTJ de
w=o
or
w=O
(12.108)
u=o
or
U=O
(12.109)
=0
(12.107)
For a pinned or hinged end:
au ae2=0 2
d2 or
U de-2 =0
(12.110)
The equations of motion, Eqs. (12.90) and (12.91), can be solved numerically to find the frequency parameter Qi and the mode shape defined by f!i(n and Wi(~)' Numerical results are obtained for three types of curved beams: parabolic, sinusoidal, and ellipticshaped beams [22J. For a parabolic-shaped curved beam, the equation of the beam is given by 4hx y(x) = -y(x
-l),
(12.111)
or (12.112) For a sinusoidal-shaped curved beam, shown in Fig. 12.7, the equation of the curved beam is given by _
.
1TX
y=smL
x
= otl + x,
(12.113) y=H-h+y
(12.114)
or 1]
= !J. - d
+ d sin(e~ + eot),
(12.115)
12.7
Vibration of a Curved Beam with Variable Curvature
413
Curved beam
-1-'H
J
-.-.-.- f h
.-
Y
', ,
I
,
'I I
BI'
I'
I I I
,
1/
A
-
I I I
1/~~~ave .
, ,
,
-'-1'-'-'-'- - ~ :i-=:j-'-
I
I
·~/~.I;-
- -.Figure 12.7
\
Sinusoidal-shaped curved beam.
where
d=
h
(12.116)
1 - sinae 1T
(12.117)
e=--·1+2a
For the elliptic-shaped curved beapl shown in Fig. 12.8, the equation of the curved beam is given by (12.118) where
f =
g!! g - (a
(12.119)
+ (2)1!2
(12.120)
g =!(1 +2a)
'-~.':~ /' :
( ~-'-'-'-_'_~~-T.1h
j
._:.
\
al
,
~
A
B:
I
I
,"
'\j
_:.a.
I
I
I
I
I
.•.•..••......
---------------------Figure 12.8
---
I
"
I
~//
..... '--
Elliptic-shaped curved beam .
. j.'
414
Vibration of Circular Rings and Curved Beams Table 12.3 Natural Frequency Parameters of Curved Beams Frequency parameter, i= 1
ni
i=2
i=3
Without rotary inertia, c=o
With rotary inertia, c= 1
Without rotary inertia, c=o
With rotary inertia, c= 1
Without rotary inertia, c=o
With rotary inertia, c=1
Parabolic shape, pinned - pinned: !: = 0.1,[ = 10
11.47
10.94
29.49
28.72
38.69
33.73
Sinusoidal shape, pinned-fixed: ex = 0.5, !: = 0.3, [ = 10
16.83
16.37
23.32
22.88
35.61
31.78
Elliptic shape, fixed- fixed: ex = 0.5, !: = 0.5, [ = 10
16.92
16.69
17.24
16.98
28.49
26.20
Geometry and boundary conditions of the curved beam
Source: Ref. 22.
Frequency parameters corresponding to the three types of curved beams with different boundary conditions are given in Table 12.3.
12.7.2
Curved Beam Analysis, Including the Effect of Shear Deformation Equations of Motion The dynamic equilibrium approach will be used to derive the equations of motion by considering the free-body diagram of an element of the curved beam as shown in Fig. 12.6(b). When the effects of rotary inertia and shear deformation are considered, the equilibrium equations can be obtained, similar to those of Section 12.7.1, as follows [21]. Equilibrium of forces in the radial direction:
aF
-ae - P - pF:- = 0 I
(12.121)
Equilibrium of forces in the tangential direction:
ap
ae + F -
PPi = 0
(12.122)
Equilibrium of moments in the xy plane: 1 aM
p
ae -
F
+ Mi
=0
(12.123)
Rotation of the tangent to the centroidal axis is given by [21]
y = .!.p (auae _ w)
(12.124)
.•
"'#~--'''''''~'ll!>oIl...'''''''''''''''''''~~'","",,-''''''~''''''''''''''''''''''''''~'''''''o_'''''''-·''"~''''''''''''"'''''''-'''''''''''''''''''''''r''',"',.'-'~'~"~
12.7 Vibration of a Curved Beam with Variable Curvature
415
When shear deformation is considered, the rotation angle y can be expressed as (12.125) where t/J' is the rotation angle with no shear deformation and fJ is the angular deformation due to shear. Equations (12.124) and (12.125) yield . fJ = -1 p
(au-aB -
)
pt/J'
W -
(12.126)
When the effects of rotary inertia. shear deformation. and axial defonnation are considered, the bending moment (M), normal force (P). and shear force (F) in the curved beam are given by [19. 20] Eldl/f
(12.127)
M=--p dB
+ EI
P = EA (dW +u) p dB kAG F = kAGfJ = -
p
dl/f
(12.128)
p2 dB
(dU-
-
dB
W -
pt/J'
)
(12.129)
where k is the shear factor. To find the natural frequencies of vibration of the curved beam, all the displacement components are assumed to be harmonic with frequency w, so that w(B. t) = W(B) coswt
(12.130)
u(B. t) = U(B) coswt
(12.131)
l/f(B. t) = w(B) coswt
(12.132)
Thus, the inertia forces are given by
= -mAw2W(B)
(12.133)
Fj(B) = -mAw2U(B)
(12.134)
Mj(B) = -mlw2W(B)
(12.135)
Pj(B)
where m is the mass density of the curved beam. Using a procedure similar to that of Section 12.7.1, the equations of motion can be expressed in nondimensional form as [21]: 2 d l} = ~ d( dl} dB2 ( dB dB
1
d(
+ 2.
(1 _
J.L
- f dB !Y +
(
(2 w~) U [2 -
1)
(+ J.L([2
I
-
dw dB
+
(1 + 2.) d!y J.L dB
(12.136)
416
Vibration of Circular Rings and Curved Beams
(12.137)
(12.138) where kG E
(12.139)
I-L=-
n. = (V.rl _ I
'-
ffff YE
(12.140)
and the other symbols are defined by Eqs. (12.88) and (12.89). Numerical Solution The boundary conditions of the curved beam are as follows. For a clamped or fixed end: u=o
or
U=O
(12.141)
w=O
or
W=O
(12.142)
1/1=0
or
\11=0
(12.143)
u=O
or
U=O
(12.144)
w =0
or
W=O
(12.145)
a1/1= 0
or
d\l1 = 0 dO
(12.146)
For a pinned or hinged end:
ao
The frequency parameters corresponding to three types of curved beams with different boundary conditions are given in Table 12.4.
12.8 RECENT CONTRIBUTIONS Curved Beams and Rings An analytical procedure was proposed by Stavridis and . Michaltsos [12] for evaluation of the eigenfrequencies of a thin-walled beam curved in plan in response to transverse bending and torsion with various boundary conditions. Wasserman [13] derived an exact formula for the lowest natural frequencies and critical loads of elastic circular arches with flexibly supported ends for symmetric vibration in a direction perpendicular to the initial curvature of the arch. The values of frequencies and critical loads were shown to be dependent on the opening angle of the arch, on the stiffness of the flexibly supported ends, and on the ratio of the flexural rigidity to the torsional rigidity of the cross section. Bickford and Maganty [14] obtained the expressions for out-of-plane modal frequencies of a thick ring, which accounts for the variations in curvature across the cross section.
12.8 Table 12.4 Natural Frequencies Effect of Shear Deformation
Recent Contributions
417
of Curved Beams. Including the
Q
n
Frequency
parameter.
i= 1
i -"
i=3
Parabolic shape. Pinned-pinned: f.z = 0.3, ! = 75
21.83
56.00
102.3
Sinusoidal shape, fixed-fixed: a 0.5, f.z O.1,!
56.30
66.14
114.3
35.25
57.11
83.00
Geometry and boundary conditions of the curved beam
=
=
Elliptic shape, fixed - pinned: a 0.5, f.z 0.2, !
=
=
j
= 100 = 50
Source: Ref. 21. aFor shear coefficient, I-' = 0.3.
Vibration of Multispan Curved Beams Culver and Oeste I [15J presented a method of analysis for determining the natural frequencies of multispan horizontally curved girders used in bridge structures. The method was illustrated by deriving the frequency equation of a two-span curved girder. Numerical results and comparison with existing solutions were also given .. Vibration of Helical Springs
The longitudinal and torsional vibrations of helical springs of finite length with small pitch were analyzed by Kagawa [16J on the basis of Love's formulation for a naturally curved thin rod of small deformation. The drivingpoint impedance at one end of the spring while the other end is free or supported was discussed.
Vrbration of Gears Ring gear structural modes of planetary gears used in modem automotive, aerospace, marine, and other industrial drivetrain systems often contribute significantly to the severity of the gear whine problem caused by transmission error excitation. The dynamics and modes of ring gears have been studied utilizing the analytical and computational solutions of smooth rings having nearly the same nominal dimensions but without the explicit presence of the spline and tooth geometries by Tanna and Urn [17]. Vibration of Frames of Electrical Machines The frames of electrical machines such as motors and generators can be modeled as circular arcs with partly built-in ends. The actual frequencies of vibration of these frames are expected to lie within the limits given by those of an arc with hinged ends and of an arc with fixed ends. The vibration of the frames of electrical machines were studied by Erdelyi and Horvay [18]. Vtbration of Arches and Frames General relations between the forces and moments developed in arches and frames as well as the dynamic equilibrium equations have been presented by Borg and Gennaro [19J and Henrych [20J. The natural frequencies
!....
418
Vibration of Circular Rings and Curved Beams
of noncircular arches, considering the effects of rotary inertia and shear deformation, have been investigated by Oh et al. [21).
REFERENCES 1. L. L. Philipson, On the role of extension in the flexural vibrations of rings, Journal of Applied Mechanics, Vol. 23 (Transactions of ASME, Vol. 78), p. 364, 1956. 2. B. S. Seidel and E. A. Erdelyi, On the vibration of a thick ring in its own plane, Journal of Engineering for Industry (Transactions of ASME, Ser. B), Vol. 86, p. 240, 1964. 3. S. Timoshenko, On the correction for shear of the differential equation for transverse vibration of prismatic bars, Philosophical Magazine, Ser. 6, Vol. 41, p. 744, 1921. 4. J. P. Den Hartog, Vibrations of frames of electrical machines, Journal of Applied Mechanics, Transactions of ASME, Vol. 50, APM-50-11, 1928. 5. A. E. H. Love, A Treatise on the Mathematical Theory of Elasticity, 4th ed., Dover, New York, 1944. 6. S. S. Rao and V. Sundararajan, Inplane flexural vibrations of circular rings, Journal of Applied Mechanics, Vol. 36, No.3, pp. 620-625, 1969. 7. S. S. Rao, Effects of transverse shear and rotatory inertia on the coupled twist-bending vibrations of circular rings, Journal of Sound and Vibration, Vol. 16, No.4, pp. 551-566, 1971. 8. S. S. Rao, On the natural vibrations of systems of elastically connected concentric thick rings, Journal of Sound and Vibration. Vol. 32, No.3, pp. 467-479, 1974. 9. S. S. Rao, Three dimensional vibration of a ring Journal of the Royal Aeronautical Society, London, 10. R. E. Peterson, An experimental investigation of study of gear-noise, Transactions o/the American 52, APM-52-1, 1930.
on an elastic foundation, Aeronautical Vol. 75, pp. 417-419, 1971. ring vibration in connection with the Society of Mechanical Engineers, Vol.
11. E. R. Kaiser, Acoustical vibrations of rings, Journal of the Acoustical Society of America, Vol. 25, p. 617,1953. 12. L. T. Stavridis and G. T. Michaltsos, Eigenfrequency analysis of thin-walled girders curved in plan, Journal of Sound and Vibration, Vol. 227, No.2, pp. 383-396, 1999. 13. Y. Wasserman, Spatial symmetrical vibrations and stability of circular arches with flexibly supported ends, Journal of Sound and Vibration, Vol. 59, No.2, pp. 181-194, 1978. 14. W. B. Bickford and S. P. Maganty, On the out-of-p1ane vibrations of thick rings, Journal of Sound and Vibration, Vol. 108, No.3, pp. 503-507, 1986. 15. C. G. Culver and D. 1. Oestel, Natural frequencies of mu1tispan curved beams, Journal of Sound and Vibration, Vol. 10, No.3, pp. 380-389, 1969. 16. Y. Kagawa, On the dynamical properties of helical springs of finite length with small pitch, Journal of Sound and Vibration. Vol. 8, No.1, pp. 1-15, 1968. 17. R. P. Tanna and T. C. Lim, Modal frequency deviations in estimating ring gear modes using smooth ring solutions, Journal of Sound and Vibration, Vol. 269, No. 3-5, pp. 1099-1110, 2004. 18. E. Erdelyi and G. Horvay, Vibration modes of stators of induction motors, Journal of Applied Mechanics, Vol. 24 (Transactions of ASME, Vol. 79), p. 39, 1957. 19. S. F. Borg and J. J. Gennaro, Modern Structural Analysis, Van Nostrand Reinhold, New York, 1969. 20. J. Henrych, The Dynamics of Arches and Frames, Elsevier, Amsterdam, 1981.
Problems
''419
21. S. J. Oh, B. K. Lee, and 1. W. Lee, Natural frequencies of non-circular arches with rotatory inertia and shear deformation, Journal of Sound and Vibration, Vol. 219, No.1, pp. 23-33. 1999. 22. B. K. Lee and J. F. Wilson, Free vibrations of arches with variable curvature, Journal of Sound and Vibration, Vol. 136, No.1, pp. 75-89, 1989. 23. K. F. Graff, Wave Motion in Elastic Solids, Ohio State University Press, Columbus, OH, 1975.
PROBLEMS 12.1 Derive Eqs. (12.17) and (12.18). 12.2 Derive Eq. (12.24) from Eqs. (12.19)-(12.23).
cross section for the following data: radius of the cross section = 1 cm, radius of the centerline of the ring = 15 em, Young's modulus = 207 GPa, shear modulus =
123 Derive Eq. (12.26) from Eqs. (12.19), (12.20),
79.3 GPa, and unit weight
= 76.5 kN/m3.
(12.22), (12.23), and (12.25).
12.11 Find the first five natural frequencies of exten-
12.4 Derive Eq. (12.10) by considering the deformations shown in Fig. 12.2(b).
sional vibrations of a circular ring with circular cross section for the following data: radius of cross section = I cm, radius of the centerline of the ring = 15 cm, Young's modulus = 207 GPa, and unit weight = 76.5 kN/m3•
12.5 Find the first five natural frequencies of inplane flexural vibrations of a circular ring with Itl AR2 0.25 and E I kG 2.0 (a) according to classical theory; (b) by considering the effect of rotary inertia only; (c) by
=
=
12.12 Derive Eqs. (12.131) and (12.132).
considering the effects of both rotary inertia and shear ',12.13 Derive Eqs. (12.136)-(12.138). deformation. 12.14 Derive the following differential equations for 12.6 Derive Eq. (12.50) from Eqs. (12.45)-(12.49). determining the radial and tangential components of 12.7 Derive Eq. (12.61) from Eqs. (12.45), (12.48), displacement (u and v) of an arch: (12.49), (12.59), and (12.60). u
12.8 Derive Eq. (12.78) from Eqs. (12.69)-(12.71) and (12.72)-(12.74). 12.9 Find the first five natural frequencies of coupled twist-bending vibrations of a circular ring with rl R 0.25, k = 0.8333, and EIG = 3.0 (a) according to classical theory; (b) by considering the effect of rotary inertia only; (c) by considering the effects of both rotary inertia and shear deformation.
=
12.10 Find the first five natural frequencies of pure torsional vibrations of a circular ring with a circular
dv
N
M
-+-=---+p ds EA pEA d2u
u
-ds2 + -p2
Me EI
M =-EI
where N is the tangential force, M is the bending moment, E is Young's modulus, A is the cross-sectional area, I is the area moment of inertia of the cross section, p is the radius of curvature of the centroidal axis, e is the distance between the centroidal axis and the neutral axis of the cross section, and s is the tangential coordinate.
Vibration of Membranes 13.1 INTRODUCTION A membrane is a perfectly flexible thin plate or lamina that is subjected to tension. It has negligible resistance to shear or bending forces, and the restoring forces arise exclusively from the in-plane stretching or tensile forces. The drumhead and diaphragms of condenser microphones are examples of membranes.
13.2 EQUATION OF MOTION 13.2.1
Equilibrium Approach Consider a homogeneous and perfectly flexible membrane bounded by a plane curve C in the xy plane in the undeformed state. It is subjected to a pressure loading of intensity f(x, y, t) per unit area in the transverse or z direction and tension of magnitude P per unit length along the edge as in the case of a drumhead. Each point of the membrane is assumed to move only in the z direction, and the displacement, w (x, y, t), is assumed to be very small compared to the dimensions of the membrane. Consider an elemental area of the membrane, dx dy, with tensile forces of magnitude P dx and P dy acting on the sides parallel to the x and y axes, respectively, as shown in Fig. 13.1. After deformation, the net forces acting on the element of the membrane along the z direction due to the forces P dx and P dy will be [see Fig. 13.1(d)]
(p ~:~
dx dY)
and
(p ~:~
dx dY)
The pressure force acting on the element of the membrane in the
z
direction is
f (x,
y)
dx dy. The inertia force on the element is given by
p(x, y)
a2w
at2
dx dy
where p (x, y) is the mass per unit area. The application of Newton's second law of motion yields the equation of motion for the forced transverse vibration of the membrane as (13.1) 420
'.~ _~""',.,--< ,""""";~_ - _-"""""". "",·i"'.~.·,-•.,",,,,,,,,,,.,,,,",,,,,,,,,,"-,,,,,,~~,,
•.•,,,,,,,.,,,,",,,,",.,,,.,,,;,_,.~."~"-'
13.2 Equationef:Motion
When the external force f(x,
421
y)
= 0, the free vibration equation can be expressed as
c2
(a2w + a2w) ax2 ay2
2 aw at2
=
(13.2)
where (13.3) Equation (13.2) is also known as the two-dimensional the wave velocity.
wave equation. with c denoting
Initial and Boundary Conditions Since the equation of motion, Eq. (13.1) or (13.2), involves second-order partial derivatives with respect to each of t. x, and y, we need to specify two initial conditions and four boundary conditions to find a unique solution
y Boundary, C
r
dy
1
I
o
x
--dx(a)
z
t I I I ~ .. I' .
f(x,y,t)
tlJb~+ (
I
I
~ w+awdx
aww
ax
. Pd L._.L._.
o
Pdy
t
£,. aw a a ax +ax
(a~dx
Deflected
lax I
I
I
,
f+ dx~
._._._._._.~
X
Undeflected (b)
Figure 13.1 (a) Undefonned membrane in the xy plane; (b) defonned membrane as seen in the xz plane; (c) defanned membrane as seen in the yz plane; (d) forces acting on an element of the membrane.
422
Vibration of Membranes J,.Y
I
I
Pdx~.
dW + l... -1Q d dY dY'~' Y
.
i . w+dWd
Deflected f(x,y,t)
~I __!.Y.-
I I
.!j'
. b
:::...
_ "E
_
T dy
i
\
i.\ \
dW
L_ _.-.-._._
I
....l
0
Z
(c)
(d)
Figure 13.1
(continued)
of the problem. Usually, the displacement and velocity of the membrane at t = 0 are specified as wo(x) and wo(x), respectively. Thus, the initial conditions are given by w(x, y, 0)
ow
alex,
= wo(x,
y)
(13.4)
y, 0) = wo(x, y)
(13.5)
The boundary conditions of the membrane can be stated as follows: 1. If the membrane is fixed at any point (x),y) must be zero, and hence w(x) , y),
,
t) = 0,
on the boundary, the deflection
t;:::O
(13.6)
.
i ·~hi.W~~;~~_-~_:-:li::i_';;:"_._'·_":":··I_.',",~.~1i.~~-.>~-"·!lloj-"'-It~"~~~~~~,~;:;_,~~_~M:~W:~~~~".~~",~'t'~~~.~ll:7t.,
I
13.2 Equation of MOtion -:423 2. If the membrane is free to deflect transversely (in the z direction) at any point (X2,y2) of the boundary, there cannot be any force at the point in the z direction. Thus, OW
P-(xz,
on
where 0 W / on indicates the derivative to the boundary at the point (XZ,yz).
13.2.2
t::;:O
Yz, t) = 0, of
W
(13.7)
with respect to a direction n normal
Variational Approach To derive the equation of motion of a membrane using the extended Hamilton's principle, the expressions for the strain and kinetic energies as well as the work done by external forces are needed. The strain and kinetic energies of a membrane can be expressed as
,,=
Hf W:)'+ e;)'] p
(13.8)
dA
A
T=~ff p(oo~r dA
(13.9)
A,
The work done by the distributed
pressure loading f(x,
w=
ff
y, t) is given by (13.10)
fw dA
A
The application of Hamilton's
principle
o
gives
12
1
(rr - T - W) dt = 0
(13.11)
II
or
{{ ~ij
+ (~;)']
p [(~:)'
dA - ~
ij
p (~~)'
dA-
ij
fWdA} dt
=0
(13.12) The variations in Eq. (13.12) can be evaluated using integration by parts as follows:
If (OW)Z dAdt 2
1 If
ow a --(ow)dAdt ax ax 1 =p1 [1fc aWOwL."Cdc-If!...(OW)owdA] dt ox ax ox 12
12 P
It = 0
=P
-
II
oX
II
A
A
12
II
A
(13.13)
424
Vibration of Membranes
h= 0
1 If 1 [1.rc ill 12 -p 2
IJ
A
I]
ow owly dC oy
I)
A
(ow) oy
oy
ow 0 --(ow)dAdt oy oy
OWdA] dt 12
(oo;r
dAdt
0
A
dAdt
If 1
12
2
0
I)
A
(13.15)
A
By using integration by parts with respect to time, the integral
h = -p
(13.14)
A
12
h = 01
12
If ~ = ill 1 (~;r
12
= P
1If
=P
(OW)2 dAdt oy
dtdA = p (OW)2 ot
If A
12
-ow [ow ot
1
12
-
1
h can be written as
I)
I)
-0 (OW) owdt ] dA ot ot (13.16)
Since ow vanishes at t1 and t2, Eq. (13.16) reduces to
1 If 1II 12
h = -p
o2w -2owdAdt
A
12
14 =, 0
(13.17)
ot
I)
1 II 12
=
fwdAdt
A
fowdAdt
(13.18)
A
Using Eqs. (13.13), (13.14), (13.17), and (13.18), Eq. (13.12) can be expressed as
1 [-If
o2w (p
12
I)
A
W +po2w +f_p02 )OWdA] 2 ox oy2 ot2
+ 112[tp(~:lx+~;ly)owdC]dt=0
dt
(13.19)
By setting each of the expressions under the brackets in Eq. (13.19) equal to zero, we obtain the differential equation of motion for the transverse vibration of the membrane as 02W P ( ox2
+
02W) oy2
+f
o2w = P ot2
(13.20)
and the boundary condition as
J
Jc
P (owlx ox
+ OWly) oy
owdC
=0
(13.21)
Note that Eq. (13.21) will be satisfied for any combination of boundary conditions for a rectangular membrane. For a fixed edge: w
=0
and hence
0w
=0
(13.22)
For a free edge with x = 0 or x = a, Ly = 0 and Lx = 1:
ow ax
p-=o
(13.23)
With y = 0 or y = b, Lx = 0 and Ly = 1:
ow oy
p-=O
(13.24)
For arbitrary geometries of the membrane, Eq. (13.21) can be expressed as ipow{,wdC=O
(13.25)
Yc an
which will be satisfied when either the edge is fixed with w
=0
and hence
(,w
=0
(13.26)
or the edge is free with
ow an
p-=o
(13.27)
13.3 WAVE SOLUTION The functions y, t) = f(x
- et)
(13.28)
wz(x, y, t) = f(x
+ et)
(13.29)
Wi(X,
W3(X,
y, t) = f(x cos()
+ y sin() -
et)
(13.30)
can be verified to be the solutions of the two-dimensional wave equation, Eq. (13.2). For example, consider the function W3 given by Eq. (13.30). The partial derivatives of W3 with respect to x, y, and t are given by OW3
-
ax
o2W3
ox
2
OW3
-
oy
02W3
= = =
oyZ
=
OW3
=
at
02W3
otZ
I
f cos ()
(13.31)
f"
(13.32)
f
cos
2 ()
I'
SlD ()
f'"
Z () SlD
(13.33)
(13.34)
-ef'
(13.35)
= cZ f"
(13.36)
426
Vibration of Membranes
where a prime denotes a derivative with respect to the argument of the function. When W3 is substituted for w using the relations (13.32), (13.34), and (13.36), Eq. (13.2) can be seen to be satisfied. The solutions given by Eqs. (13.28) and (13.29) are the same as those of a string. Equation (13.28) denotes a wave moving in the positive x direction at velocity c with its crests parallel to the y axis. The shape of the wave is independent of y and the membrane behaves as if it were made up of an infinite number of strips, all parallel to the x axis. Similarly, Eq. (13.29) denotes a wave moving in the negative x direction with crests parallel to the x axis and the shape independent of x. Equation (13.30) denotes a parallel wave moving in a direction at an angle e to the x axis with a velocity c.
13.4
FREE VIBRATION OF RECTANGULAR MEMBRANES The free vibration of a rectangular membrane of sides a and b (Fig. 13.2) can be determined using the method of separation of variables. Thus, the displacement w(x, y, t) is expressed as a product of three functions as w(x, y, t) = W(x, y)T(t)
== X(x)Y(y)T(t)
(13.37)
where W is a function of x and y, and X, Y, and T are functions of x, y, and t, respectively. Substituting Eq. (13.37) into the free vibration equation, Eq.(13.2), and dividing the resulting expression through by X(x)Y(y)T(t), we obtain c2
2 1_d X(x) X(x) dx2
+ _1_d2Y(y)]
[ __
·Y(y)
= _1_d2T(t)
dy2
(13.38)
dt2
T(t)
Since the left-hand side of Eq. (13.38) is a function of x and y only, and the right-hand side is a function of t only, each side must be equal to a constant, say, kI: 2 [
c
2 d X(x) X(x) dx2
I
a
1 d2Y(y)]
+ Y(y)
)I
dy2
.-
=
1 d2T(t) T(t) ~
Figure 13.2
x
=k
Rectangular
= -w
2
(13.39)
membrane.
IThe constant k can be shown to be a negative quantity by proceeding as in the case of free vibration of suings (Problem 13.3). Thus. we can write k _w2, where w is another constant.
=
13.4 Free Vibration of Reclanguhtr-Membrancs
Equation (13.39) can be rewritten as two separate equations: d2X(x) dx2
I X(x)
d2T(t)
--
w2
I
+;;r
=-
+ w2T(t)
dt2
Y(y)
d2y(y) dy2
(13.40)
=0
(13.41)
It can be noted, again, that the left-hand side of Eq. (13.40) is a function of x only and the right-hand side is a function of y only. Hence, Eq. (13.40) can be rewritten as two separate equations: (13.42) (13.43) where a2 and f32 are new constants related to w2 as 2
f3
2
W a =-rc
2
(13.44)
Thus, the problem of solving a partial differential equation involving three variables, Eq. (13.2), has been reduced to the problem of solving three second-order ordinary differential equations, Eqs. (13.41)-(13.43). The solutions of Eqs. (13.41)-(13.43) can be expressed as2 .
+ B sinwt = Cl cosax + C2 sin ax = C3 cos f3y + C4 sin f3y
(13.45)
T(t) = A cos wt X(x) Y(y)
(13.46) (13.47)
where the constants A and B can be determined from the initial conditions and the constants C 1 to C4 can be found from the boundary conditions of the membrane. 2The solution given by Eqs. (13.45)-(13.47) can also be obtained by proceeding as follows. The equation governing the free vibration of a rectangular membrane can be expressed, setting f 0 in Eq. (13.1), as
=
a2w
I a2w
a2w
-+-=-ax2 ay2 c2 at2 By assuming a harmonic solution at frequency
(J)
(a)
as
w(x, y, t)
= W(x,
y)eiwt
(b)
Eq. (a) can be expressed as
a2w(x,
"2 uX
y)
+
a2W(x, "2 uy
y)
+
(J)2W( c2
) _ 0
x, Y -
(c)
By assuming the solution of W(x, y) in the form W(x, y)
= X(x)Y(y)
and proceeding as indicated earlier, the solution shown in Eqs. (13.45)-(13.47) can be obtained.
(d)
428 13.4.1
Vibration of Membranes
Membrane with Clamped Boundaries If a rectangular membrane is clamped or fixed on all the edges, the boundary conditions can be stated as w(O, y, t) 0, (13.48) 0:::: y:::: b, t:::O w(a, w(x, w(x, In view of Eq. (13.37), restated as
= y, t) = 0, 0, t) = 0, b, t) = 0,
the boundary
b,
t:::O
(13.49)
0:::: x:::: a,
t:::O
(13.50)
O::::x:::: a,
t:::O
(13.51)
conditions
of Eqs. (13.48)-(13.51)
can be
x (0)
=0
(13.52)
X(a)
=0
(13.53)
Y (0) = 0
(13.54)
=0
(13.55)
Y(b)
=
y::::
0::::
=
=
The conditions X(O) 0 and Y(O) 0 [Eqs. (13.52) and (13.54)] require that C1 0 in Eq. (13.46) and C3 = 0 in Eq. (13.47). Thus, the functions X(x) and Y(y) become X(x)
= C2 sin ax
(13.56)
Y(y)
= C4 sin}9y
(13.57)
For nontrivial solutions of X (x) and Y (y), [Eqs. (13.53) and (13.55)] require that
the conditions
X (a) = 0 and Y (b) = 0
sinaa = 0
(13.58)
sin}9b = 0
(13.59)
Equations (13.58) and (13.59) together define the eigenvalues of the membrane through Eq. (13.44). The roots of Eqs. (13.58) and (13.59) are given by
m
= 1,2, ...
n=1,2, The natural frequencies
of the membrane,
wmn
= 7r C
[
m
(-;;)
2
(n)2]1/2
+ b
The following observations
...
(13.61)
can be determined
using Eq. (13.44) as
= c2(a2m +.. R2)
w2mn
or
Wmn,
(13.60)
fJn
'
m=1,2,
... ,
n=1,2,
...
(13.62)
can be made from Eq. (13.62):
1. For any given mode of vibration, the natural frequency will decrease if either side of the rectangle
is increased.
13.4
Free Vibration of RectlingutarMembranes'-;429
2. The fundamental natural frequency, WI [, is most influenced by changes in the shorter side of the rectangle. 3. For an elongated rectangular membrane (with b » a), the fundamental natural frequency, W11, is negligibly influenced by variations in the longer side. The eigenfunction or mode shape, Wmn(x, y), of the membrane corresponding to the natural frequency Wmn is given by
. mrrx . = Cmnsm--sm--, a
mry b
m, n
= 1,2, ...
(13.63)
where Cmn = C2mC4n is a constant. Thus, the natural mode of vibration corresponding to Wmn can be expressed as . mrrx . nrry wmn(x, y, t) = sm -sm --(Amn a b
coswmnt
+ B')mn smwmnt
(13.64)
where Amn = CmnA and Bmn = CmnB are new constants. The general solution of Eq. (13.64) is given by the sum of all the natural modes as (13.65)
The constants Amn and Bmn in Eq. (13.65) can be determined using the initial conditions stated in Eqs. (13.4) and (13.5). Substituting Eq. (13.65) into Eqs. (13.4) and (13.5), we obtain 00
00
"" "" Amn sin -amrr x sin nrry b
= wo(x,
y)
(13.66)
m=l n=1 00
00
""
"" Bmnwmnsm-a-Slnb . mrrx m=1n=1
. nrry
. = wo(x,y)
(13.67)
Equations (13.66) and (13.67) denote the double Fourier sine series expansions of the functions wo(x, y) and wo(x, y), respectively. Multiplying Eqs. (13.66) and (13.67) by sin(mrrx/a) sin(nrry/b) and integrating over the area of the membrane leads to the relations
41 l a
Amn
b
=-
ab
0
4
Bmn = -abwmn
0
mrrx nrry wo(x, y) sin -sin -dxdy a b
Lal
b
0
0
mrrx nrry wo(x, y) sin -sin -b dx dy a
(13.68) (13.69)
Example J3.1 Find the free vibration response of a rectangular membrane when it is struck such that the middle point experiences a velocity Vo at t = O.
430
Vibration of Membranes
SOLUTION
the constants Amn
Assuming the initial conditions as
Amn
and
Wo(x, y) = 0
(EI3.1.1)
Wo(x, y) = V08 (x - ~) 8 (y - ~)
(E13.1.2)
given by Eqs. (13.68) and (13.69) can be evaluated as
Bmn
=0
(E13.1.3)
Bmn = _4_ abwmn
r {b Voo (x _~)2
10 10
4Vo
= -abwmn
o(y
_~) 2
sin mrrx sin mfY dxdy a b
. mrr . mf SIO -
2
SIO -
(EI3.1.4)
2
Thus, the free vibration response of the membrane is given by Eg. (13.65): 00
4Vo "" w(x, y, t) = L.... ab m=!
00
1 L.... --Wmn n=l
. mrrx
SIO --
a
. nrry b
SIO --
. mrr . nrr
SIO -
2
SIO -
2
.
smwmnt
(EI3.1.5)
The displacements of the membrane given by Eq. (E13.1.5) using values of each of m and n up to 10 at different instants of time are shown in Fig. 13.3.
13.4.2
Mode Shapes Equation (13.64) describes a possible displacement variation of a membrane clamped at the boundary. Each point of the membrane moves harmonically with circular frequency Wmn and amplitude given by the eigenfunction Wmn of Eq. (13.63). The following observations can be made regarding the characteristics of mode shapes [8]. 1. The fundamental or lowest mode shape of the membrane corresponds to m = n = I. In this modal pattern, the deflected surface of the membrane will consist of one half of a sine wave in each of the x and y directions. The higher values of m and n correspond to mode shapes with m and n half sine waves along the x and y directions, respectively. Thus, for values of m and n larger than I, the deflection (mode) shapes will consist of lines within the membrane along which the deflection is zero. The lines along which the deflection is zero during vibration are called nodal lines. For specificity, the nodal lines corresponding to m, n 1, 2 are shown in Fig. 13.4. For example, for m 2 and n I, the nodal line will be parallel to the y axis at x = a/2, as shown in Fig. 13.5(a). Note that a specific natural frequency is associated with each combination of m and n values.
=
=
=
2. Equation (13.62) indicates that some of the higher natural frequencies (wmn) are integral multiples of the fundamental natural frequency (wpp = pw!!), where p is an integer, whereas some higher frequencies are not integral multiples of WI!· For example, W!2, W2!, W13, and W3! are not integral multiples of WI!.
13.4
Free Vibration of RC<:t;lnguUlr.Membrancs
t=
oSJl
1. 4
Figure 13.3 Deflection of a membrane at different times, initial velocity at the middle. Times given, t, are in 'terms of fractions of the fundamental natural period of vibration. (Source: Ref. [10].)
432
Vibration of Membranes y
+
m= I,n=2
m=2,n=2 (a)
y
y
+
+ I I I I
+
-------+------+
I I I
+
:
L------------'·_.x m=I,n=2
m=2,n=2 (b)
Figure 13.4 (a) modal patterns and nodal lines; (b) schematic illustration of mode shapes (+, positive deflection; -, negative deflection).
y
+
l--
---l
__
X
m= I,n= 1
m = 2, n = I (0)
y
y
+
+r-----r----, +
L-
+ L..-
--'. __X
m = I, n = I
-'-
--' __
X
m = 2, n = I (b)
Figure 13.5
Modal patterns with m = I, n
= 1 and m = 2, n = I.
13.4
Free Vibration of Ra..'tangulM
3. It can be seen that when a2 and b2 are incommensurable, no two pairs of values of m and n can result in the same natural frequency. However, when a2 and b2 are commensurable, two or more values of Wmn may have the same magnitude. If the ratio of sides K == a/b is a rational number, the eigenvalues Wmn and Wij will have the same magnitude if (13.70)
1,
For example, WJ5 == W54, W53 == W46, etc. when K == and WI3.4 == when K == 4. If the membrane is square, a == b and Eq. (13.70) reduces to
i.
WI2.5.
etc.
(13.71) and the magnitudes of Wmn and Wnm will be the same. This means that two different eigenfunctions W mn (x, y) and Wnm (x, y) correspond to the same frequency wmn(== wnm); thus, there will be fewer frequencies than modes. Such cases are called degenerate cases. If the natural frequencies are repeated with Wmn == Wnm' any linear combination of the corresponding natural modes Wmn and Wnm can also be shown to be a natural mode of the membrane. Thus, for these cases a large variety of nodal patterns occur. 5. To find the modal patterns and nodal lines of a square membrane corresponding to repeated frequencies, consider, as an example, the case of Wmn with m == 1 and n == 2, For this case, W12 == W21 == J5rrc/a and the corresponding distinct mode shapes can be expressed as (with a == b)
. rrx . 2rry (
W12(X,
y, t) == sm -
sm --
a
,J"5rrct
Al2
b
cos ---
a
. ,J"5rrct)
+ Bl2 sm --- a
(13.72)
. 2rrx . rry (
W21(X,
y, t) == sm --
a
sm -
,J"5rrct
A21
b
cos ---
a
+ B21
,J"5rrct)
sin ---
a
(13.73) Since the frequencies are the same, it will be of interest to consider a linear combination of the maximum deflection patterns given by Eqs. (13.72) and (13.73) as
. rrx . 2rry
W
== A sm -
a
sm --
b
. 2rrx . rry
+ B sm --a
sm-
(13.74)
b
where A and B are constants. The deflection shapes given by Eq. (13.74) for specific combinations of values of A and B are shown in Fig. 13.6. Figure 13.6(a) to 13.6(d) correspond to values of B == 0, A == 0, A == B, and A == -B, respectively. When B == 0, the deflection shape given by Eq. (13.74) consists of one-half sine wave along the x direction and two half sine waves along the y direction with a nodal line at y == bj2. Similarly, when A == 0, the
~
434
Vibration of Membranes
r
r
i
0
0
r
0'
0,
i
,, ,, ,
,,
----------
"2
u_ 0
x
0
0
0
0
"2 (b)
(0)
r
x
0
I' /
""
""
I
I
T 0
'x
/
•
/
/
I
0
00
I
/
"
0
I I
I I
/ /
I
0
I
I' I'
x
r
0'
I'
"-
0
(c)
y
1
""
'-
t
0
/
,, ,, ,, ,
T
I
"2
I
0
I
"-
/
i
/
0
x
0
W
Figure 13.6 Deflection shapes given by Eq. (13.74): (a) B A = -B; (e) A = B/2; (f) A = 2B.
I
"2
i
."
I
0
"
= 0;
(b) A
"-
0
00
= 0;
(c) A
x
= B;
(d)
nodal line will be at x = a/2. When A = B, Eq. (13.74) becomes W
7rX . = A (.sm-sm-a
27rY . 27rx +sm--smb a
. 7rY) b
. 7rX 7rY (7rX 7r ) = 2A sm - sin cas - + cas - Y a b a b
(13.75)
It can be seen that w = 0 in Eq. (13.75) when . 7rX sm- =0 a
. 7ry
or sm- =0 b
(13.76)
or 7rX 7rY cos - + cos - = 0 a b
(13.77)
The cases in Eq. (13.76) correspond to w = 0 along the edges of the membrane, while the case in Eq. (13.77) gives w = 0 at which 7rX a
-=7r--
7rY a
or
x+y=a
(13.78)
Equation (13.78) indicates that the nodal line is a diagonal of the square as shown in Fig. 13.6(c). Similarly, the case A = -B gives the nodal line along
13.4 Free Vibration of Rectangular Membranes -'.435
the other diagonal of the square as indicated in Fig. 13.6(d). For arbitrary values of A and B, Eq. (13.74) can be written as
.
rrx . 2rry
w = A ( sm - sm -a b
. 2rrx + R sm -a
rrv)
(13.79)
sin -' b
where R = B / A is a constant. Different nodal lines can be obtained based on the value of R. For example, the nodal line [along which w = 0 in Eq. (13.79)] corresponding to K = 2 is shown in Fig: 13.6(e) and (f). The following observations can be made from the discussion above: (a) A large variety of nodal patterns can exist for any repeated frequency in a square or rectangular membrane. Thus, it is not possible to associate a mode shape uniquely with a frequency in a membrane problem. (b) The nodal lines need not be straight lines. It can be shown that all the nodal lines of a square membrane pass through the center, x = y = a/2, which is called a pole. 6. For a square membrane, the modal pattern corresponding to m = n = 1 consists of one-half of a sine wave along each of the x and y directions. For m = n = 2, no other pair of integers i and j give the same natural frequency, W22· In this case the maximum modal deflection can be expressed as .
2rrx
w =sm--sm-a
.
2rry b
. rrx
rrx.
=4sm-cos-sm-cosa a
rry
rry
b
b
(13.80)
The nodal lines corresponding to this mode are determined by the equation
. rrx ~x. sm-cos-sm-cosa a
rry
rry
b
b
(13.81)
=0
Equation (13.81) gives the nodal lines as (in addition to the edges) a
x-- 2'
y=-
a
(13.82)
2
which are shown in Fig. 13.7.
.f!..1-
2
I I I I
..J
_
I I
I I
I
o
a
"2
a'~
Figure 13.7 brane .
Nodal lines corresponding to
W22
of a square mem-
..•
436
Vibration of Membranes
=
=
7. Next, consider the case of m 3 and n 1 for a square membrane. In this case, W31 WI3 .Ji'O(1Tc/a) and the corresponding distinct mode shapes can be expressed as
=
W31(X,
=
. 31TX sin -1TY ( A cos --.Ji'01Tct + 31 a b a
y, t) = sm --
.Ji'01Tct) sin _,....__ a
B3]
(13.83) W13(X,
. -1TX sm . -31TY ( A COS --JT01Tct y, t) = sm 13 a
b
a
+
B13
. JT01Tct) sm _ a
(13.84) Since the frequencies are the same, a linear combination of the maximum deflection.patterns given by Egs. (13.83) and (13.84) can be represented as W
. 31TX . 1Ty . 1TX . 31TY = Asm--sm+Bsm-sm-a a a a
(13.85)
where A and B are constants. The nodal lines corresponding to Eq. (13.85) are defined by w 0, which can be rewritten as
=
sin :x sin 1T:
[A (4 cos
2
1T: _
1) + B (4 cos ? _1)] = 0 2
(13.86)
Neglecting the factor sin(1Tx/a) sin(1Ty/a), which corresponds to nodal lines along the edges, Eq. (13.86) can be expressed as A
(4 cos :X - 1) + B (4 cos ? _1) = 0 2
2
(13.87)
It can be seen from Eq. (13.87) that: (a) When A (b) When B (c) When A
= 0, y = a/3 and 20/3 denote the nodal lines. = 0, x = a/3 and 2a/3 denote the nodal lines. = -B, Eq. (13.87) reduces to 1TX 1TY cos = ± cos a a
(13.88)
or x =y
and
=a-
y
(13.89)
21TX 21TY cos + cos -+ 1= 0 a a
(13.90)
x
denote the nodal lines. (d) When A = B, Eq. (13.87) reduces to 2 1TX 2 1TY 1 cos - + cos - = a a 2
or
which represents a circle. The nodal lines in each of these cases are shown in Fig. 13.8.
j
:l
__
""'__"""'''''''''_"''''"''''''";''''''~''''.'''.'''.''"''''''''''''''''~'''"''''''''''''''"'_i,'__... .,•..•.• ~ .•"""'"'_",,,"""_~~"'"" ..,...
13.4 Free Vibration ofReaangular Membrane:; ~'~37
ty a 1;--
---, I
2a ----------T a ---------"3
---..l .. _
L-
o
a
I I I I I I I
I I I I I I I I I
x
o
a
1 I I
3"
2a
a
T
'x
(b)
(a)
F a;-I
-,
I',
o
"
" ,, ,, ."""" , " ~<" "" ,, " '' "" ', '
(er
a
..•.
-- ...• ,
/
\
( \
'x
I I
,'--'" "
--1."_
1.-
o
a
x
(d)
Figure 13.8 Nodal lines of a square membrane corresponding to B = 0; (e) A -B; (d) A = B.
=
CV31
= (LI\3:
(a) A
= 0;
(b)
8. Whenever, in a vibrating system, including a membrane, certain parts or points remain at rest, they can be assumed to be absolutely fixed and the result may be applicable to another system. For example, at a particular natural frequency w, if the modal pattern of a square membrane consists of a diagonal line as a nodal line, the solution will also be applicable for a membrane whose boundary is an isosceles right triangle. In addition, it can be observed that each possible mode of vibration of the isosceles triangle corresponds to some natural mode of the square. Accordingly, the fundamental natural frequency of vibration of an isosceles right triangle will be equal to the natural frequency of a square with m = 1 and n = 2:
-/5rr c
w=--
(13.91)
a
The second natural frequency of the isosceles right triangle will be equal to the natural frequency of a square plate with m = 3 and n = 1:
w= The mode shapes corresponding (13.92) are shown in Fig. 13.9.
(13.92)
a
to the natural frequencies of Eqs. (13.91) and
I
438
Vibration of Membranes
(0)
(b)
Figure 13.9
13.5 FORCED VIBRATION OF RECTANGULAR 13.5.1
(a) w = .J57rc/a; (b) w =
M7rc/a.
MEMBRANES
Modal Analysis Approach The equation of motion governing the forced vibration of a rectangular membrane is given by Eq. (13.1): 2 a w(x, y, t) -p [ ax2
+
a2w(x, y, t)] ay2
+p
a2w(x, y, t) at2
=
f(x,
y, t)
(13.93)
We can find the solution of Eq. (13.93) using the modal analysis procedure. Accordingly, we assume the solution of Eq. (13.93) as 00
w(x,
y, t)
00
LL
=
Wmn(x, y)17mn(t)
(13.94)
m=l n=l
where Wmn(x, y) are the natural modes of vibration and 17mn(t) are the corresponding generalized coordinates. For specificity we consider a membrane with clamped edges. For this, the eigenfunctions are given by Eq. (13.63): Wmn(x, y)
.
mrrx
= Cmn sm --
.
nrry
sm --,
a
m,n = 1,2, ...
b
(13.95)
The eigenfunctions (or normal modes) can be normalized as (13.96) or
1 1b . Q
o
2 mrrx. --
pC~n sm
0
a
sm2
nrrv --'
b
dx dy
=1
(13.97)
The simplification of Eq. (13.97) yields 2
Cmn = -.J pab
(13.98)
Thus, the normal modes take the form Wmn(x, y)
=
2 . mrrx . nrry r:::=-r:. sm -sm --,
pab
a
b
m, n = 1,2, ...
( 13.99)
13.5
Forced Vibration of· ReL"tangular Memtlranes
--439
Substituting Eq. (13.94) into Eq. (13.93), multiplying the resulting equation throughout by WmnCr), and integrating over the area of the membrane leads to the equation
+ w~n1Jmn(t) = Nmn(t),
~mn(t)
m, n
= 1,2, ...
(13.100)
where (13.101)
=
11
=
2 r::-:::r:.
a
Nmn(t)
b
pab
Wmn(x,
ll
y)f(x,
b
a
0
y, t) dx dy mrrx y, t) sin --
f(x,
a
0
nrry
sin -b
dx dy
(13.102)
The solution of Eq. (13.100) can be written as (see Eq. (2.109)] TJmn(t)
= -W 1
1
tN
0
mn
mn ()'r smWmn ()d t - r
T]mn(O) smwmnt . + TJmn(0) coswmnt + --
r
Wmn
(13.103)
The solution of Eq. (13.93) becomes, in view of Eqs. (13.94) and (13.103), w(x,
y, t) =
ff
/..(ii{ib2
pab
y mrrx sin nrr [TJmn(O) coswmnt b .a 'Wmn
sin
m=ln=l
2 "" + [ --::-'-sin-'
1
mrr x sm-. nrry a b
Wmn
..j pab m=l n=l
1
+ T]mn(O) sinwmnt]]
"
t
x
sin Wmn (t - r) dr]
Nmn(r)
(13.104)
It can be seen that the quantity inside the braces represents the free vibration response (due to the initial conditions) and the quantity in the second set of brackets denotes the forced vibrations of the membrane. Example 13.2 Find the forced vibration response of a rectangular membrane of sides a and b subjected to a harmonic force Fo sin nt at the center of the membrane. Assume all edges of the membrane to be fixed and the initial conditions to be zero. SOLUTION
The applied force can be described as (see Fig. 13.10) y, t) = Fo sin nt<5 (x -
f(x,
i,
y - ~)
(E13.2.1)
where Fo sin nt denotes the time-dependent amplitude of the concentrated force and <5(x - a/2, y - b/2) is a two-dimensional spatial Dirac delta function defined as <5
(x - ~2'
y - ~)
2
11 a
=0
b
<5 (x -
X -I- ~
'
r
i,
y -~)
2
and/or
dx dy = 1
(E13.2.2)
440
Vibration of Membranes
o
a
a
'2
Figure 13.10
Harmonic force at center.
The nonnalized eigenfunctions or nonnal modes of the membrane, with all edges fixed, are given by [see Eq. (13.99)] Wmn(x, y)
=
2 . m1rX . n1rY r;:-;;r: sm -sm -b' v pab a
m, n
= 1,2,
...
(E13.2.3)
and the corresponding natural frequencies by [see Eq. (13.101)] m, n = 1,2, ...
(E13.2.4)
The response or deflection of the membrane can be expressed, using Eq. (13.94), as 00
00
=2: 2:
w(x, y, t)
Wmn(x, Y)77mn(t)
(E13.2.5)
m==l n==l
where the generalized coordinates 77mn(t) can be determined using Eq. (13.103). For zero initial conditions, Eq. (13.103) gives 77mn(t) = - 1 Wmn
it
- r)dr
Nmn(r)sinwmn(t
(E13.2.6)
0
The generalized force Nmn(t) can be found as [see Eq. (13.102)] Nmn(t) =
2 r;:-;;r: v pab
ll
b
a
0
m1r x y, t) sin -sin -n1rY dx dy a b
f(x,
0
(E13.2.7)
Using Eqs. (E13.2.1) and (E13.2.2) in Eq. (E13.2.7), we obtain Nmn(t) =
2Fo. r;:-;;r: sm Qt v pab
ll
2Fo .. = --smQtsm-sm-Jpab
a
0
b
. 0
m1r
m1rX . n1rY sm -sm --0 a b .
a
b) d~ dy 2
x - -, y - -
2
n1r
2
Equations (E13.2.6) and (E13.2.8) yield ( = 77mnt)
(
2Fo. m1r . n1r r;:-;;r: sm sm wmnvpab 2 2
(E13.2.l'n
2
1
1 .
0
,...,
sm ~,r sinwmnCt- r) dr
(EI3.2.l))
13.5 Forced Vibration oLRectangular Membranes
-441
The integral in Eq. (EI3.2.9) can be evaluated using a trigonometric identity for a product as
l'
sinQrsinwmn(t
=~
l'
- r)dr
+ wmn)r
{cos[(Q
- wmnt] - cos[(Q - wmn)r
+ wmnt]}
= ~[sin[(Q +wmn)r - wmnt] _ sin[(Q - wmn)r +wmnt]]' 2 Q + Wmn Q - Wmn
dr
0
1 ( sin Qt sin Qt sin Wmn t sin Wmn t ) =- -------+---+--2 Q + Wmn Q - Wmn Q + Wmn Q - Wmn =
.
1
2
wmn[1 - (Qfwmn)
(El3.2.1O)
(sin Qt - -Q- sin wmnt)
]
Wmn
Equations (EI3.2.9) and (EI3.2.1O) give l1mn(t)
= .JPlib
2Fo 2
pabwmn(1
. m;r . n;r ( . Q. ) sm -2 sm sm Qt - smwmnt Q fwmn) 2 Wmn (El3.2.ll) 2
-
2
Thus, the steady-state response of the membrane can be expressed as [see Eq. (EI3.2.5)] 4Fo ~
w (x, y, t ) = --
pab
sin(m;rf2) sin(n;rf2) sin(m;rxfa)
~
L...., L...., m=l n=l
2
2
sin(n;ryfb)
2
wmn(l - Q fwmn)
-E..... sin W
x (sin Qt -
(EI3.2.12)
Wmn t)
mn
with
Wmn
given by Eq. (EI3.2.4).
13.5.2 Fourier Transform Approach The governing equation can be expressed as [see Eqs. (13.1)-(13.3)] 1 o2w
o2w
c2 ot2 = ox2
+
o2w oy2
I(x,
+
y, t) p
,
o ::: x
(13.105)
::: a,
where w(x, y, t) is the transverse displacement. Let the membrane be fixed along all the edges, x = 0, x = a, y = 0, and y = b. Multiplying Eq. (13.105) by sin(m;rxfa) sin(n;ryjb) and integrating the resulting equation over the area of the membrane yields 1 d2W(m,
n, t)
2
2(m
2 n )
1
(13.106)
442
Vibration of Membranes
where
W(m, n, t) w(x, y, t) and I(x,
and
F(m, n, t) denote y, t), respectively:
F(m,
1 1b 1 1b a
=
W(m, n, t)
o
• m:rrx
w(x, y, t)
sin --
I(x,
sin --
0
a
=
n, t)
the double finite Fourier sine transforms of
o
y, t)
nJry
sin -
a
m:rr x
0
a
n:rry
sin -
dx dy
(13.107)
dx dy
(13.108)
b b
The solution of Eq. (13.106) can be expressed as
W(m, n, t)
=
Wo(m, n) cos:rrCCXmnt
Wo(m, + ---
n)
. smJtccxmnt
1
:rr CCXmn
f
C + iT 2 PCX mn
where Wo(m, n) and Wo(m, values of wand 8w/8t: w(x, y, t
8w alex,
y, t
n)
= 0) =
(13.110)
wo(x, y)
(13.111)
'l i 11 =
b
o
,'"
om,n
)
=
wo(x, y)
p
a
Ti, (
(13.109)
y)
a
Wo(m, n)
F(m,n,r)siniTCCXmn(t-r)dr
are the double finite Fourier sine transforms of the initial
= 0) = wo(x,
,
0
o
b .
wo(x,y
0
)
. -miTX sm . -niTY sm a b
dx dy
. m:rrx . niTY sm--sm--dxdy a b
(13.112) (13.113)
where (13.114) The displacement of the membrane can be found by taking the double inverse finite sine transform of Eq. (13.109). The procedure is illustt:ated for a simple case in the following example. Example 13.3 Find the response of a rectangular membrane subjected to an impulse f; applied at (x = ~,y = 71) in the transverse direction. Assume the initial displacement of the membrane to be zero . • ON
SOLUTION
The initial conditions of the membrane can be expressed as wo(x, y)
=0
wo(x, y)
=
f; 8(x p
(E13.3.l) - ~)8(y
-
71)
(E13.3.2)
13.5 Forced Vibrl.ltioo~·R.ectqngular Membr~nes
,.
443
where p is the mass per unit area of the membrane. The free vibration response of the membrane can be obtained from Eq. (13.109) by setting F 0:
=
W(m, n, t) = Wo(m, n)
Wo(m, n) .
+ ----
COS1l'camnt
l l
= cosrrccxmnt .
b
a
o
Sln1fCCXmnt
rrCCXmn
mrrx nrry wo(x, y, t) sin -sin -- dx dy
b
a
0
Lal
b
sinJl'ccxmnt + ---rrCCXmn
0
mrrx nrry wo(x, y) sin -sin -b dx dy 0 a
(E13.3.3) Taking the inverse transform of Eq. (E13.3.3), we obtain3
2: 2: 00
4 w(x, y, t) = ab
00 . mrr x . nrry sm -sm -b cosrrccxmnt a
m=l n=l
[l l a
X
b
y'=o
x'=O
wo(x', y') sin _m_rr_x_' sin _nrr_y_' dx' dY']
2: 2: .
0000
4 +rrc
a
mrrx
x
. nrry
sm--sm---------
..m=l n=l
lb 1 a
a
smrrccxmnt
b (m2b2
+ n2a2)1/2
mrrx' nrry' -sin -b-dx'
wo(x', y')sin
y'=o
x'=O
b
.
dy'
(E13.3.4)
a
By substituting the initial conditions, Eqs. (E13.3.1) and (E13.3.2), and noting that a
L
lb
mrrx . nrry (; . mrr~ . nrr17 wo(x, y) sin -sm dx dy = - sm -sm -
a
00
b
p
a
(E13.3.5)
b
we find the response of the membrane as w(x, y, t)
=:
4(; 00 00 1 =-",,-----sm 2 2 2 rrpc ~
. mrr~
x sm --
a
+ n a2)1/2
(m b
. nJl'17 . mrrx
sm --
sm --
b
. [
a
(m
n
a
b
2
2
rrc-+2 2
. nrry
sm --
)1/2]
t
(E13.3.6)
b
3If p(x, y) denotes a function of the variables x and y that satisfies the Dirichlet's condition over the region ~ a, 0 ~ y ~ b, its finite double Fourier sine transform, P(m, n), is defined by
o~x
F(m, n)
=
jb
l
a
x=O
mrrx nrry p(x, y) sin -a- sin dx dy
T
(a)
y=O
The inverse transform of F(m, n), given by Eq. (a), can be expressed as p(x, y)
=
4 ab
mrrx
L L P(m, n) sin -a00
00
•••=1.=1
. nrry SID
T dx
dy
(b)
444
13.6 13.6.1
Vibration
of Membranes
FREE VIBRATION OF CIRCULAR MEMBRANES Equation of Motion Noting that the Laplacian operator in rectangular coordinates is defined by 02 + _02 OX2 oy2
V2 = _
(13.115)
the equation of motion of a rectangular membrane, Eq. (13.1), can be expressed as a2w(x,y,t) 2
2
PVw(x,y,t)+!(x,y,t)=p
(13.116)
ot
For a circular membrane, the governing equation of motion can be derived using an equilibrium approach by considering a differential element in the polar coordinates r and e (see Problem 13.1). Alternatively, a coordinate transformation using the relations x = r cos e,
y=rsine
(13.117)
can be used to express the Laplacian operator in polar coordinates as (see Problem 13.2)
(13.118) Thus, the equation of motion for the forced vibration of a circular membrane can be expressed as 02w(r, e, t) or2
P[
~ ow(r, e, t)! 02w(r, e, t)]! ar + r2 oe2
+r
_ 02w(r, e, (r,e,t) - p ot2
+
0
(13.119) For free vibration, Eq. (13.119) reduces to c
2
[02w(r, e, t) or2
+ 1 ower, e, t) +_12 02w(r,2e, t)] = r
or
r
oe
02w(r, e, t) ot2
(13.120)
where c is given by Eq. (13.3). As the displacement, w, is now a function of r,
e, and
t, we use the method of separation of variables and express the solution as
w(r, e, t) = R(r)8(e)1'(t) where R, 8, and l' are functions of only r, Eq. (13.121) into Eq. (13.120), we obtain R8-
2 2 d 1' R 2 (d-81' = c dt2 dr2
1 dR + --81'
r dr
e,
(13.121)
and t, respectively. By substituting 28)
1 d + -R-1' 2
r
de2
(13.122)
1 d28 de2
(13.123)
which, upon division by R8T, becomes 1 1 d2T 1 d2 RId R c2 T dt2 = R dr2 + r R d;
+ r28
________________________
13.6
Free Vibration of Circular
Membranes
445
Noting that each side of Eq. (13.123) must be a constant with a negative value (see Problem 13.3), the constant is taken as _w2, where w is any number, and Eq. (13.123) is rewritten as (13.124) (13.125) Again, we note that each side of Eq. (13.125) must be a constant. Using ex2 as the constant, Eq. (13.125) is rewritten as 2
d R(r) IdR(r) --+---+ dr2 r dr
(2
w
ex2) 0 R(r)= r2
--
e
d2
2 -+ex e=0 2
(13.126) (13.127)
d0
Since the constant ex2 must yield the displacement w as a periodic function of 0 with a period 2rr [i.e., w(r, 0, t) = w(r, 0 + 2rr, t)] ex must be an integer: m=0,1,2,
ex=m,
(13.128)
...
The solutions of Eqs. (13. 124}and (13.127) can be expressed as T(t)
=
At coswt
+A2
e(o) = Clm cosmO
(13.129)
sinwt
+ C2m sinmO,
m = 0, 1, 2, ...
(13.130)
Equation (13.126) can be rewritten as d2R r2--2 dr
dR dr
+ r-
+ (r2w2
2
- m
)R = 0
which can be identified as Bessel's equation of order solution of Eq. (13.131) is given by [9]:
m
(13.131)
with the parameter w. The (13.132)
where 81 and 82 are constants to be determined from the boundary conditions, and 1m and Ym are Bessel functions of first and second kind, respectively. The Bessel functions are in the form of infinite series and are studied extensively and tabulated in the literature [2],[3] because of their importance in the study of problems involving circular geometry. For a circular membrane, w(r, 0, t) must be finite (bounded) everywhere. However, Ym (wr) approaches infinity at the origin (r = 0). Hence, the constant 82 must be zero in Eq. (13.132). Thus, Eq. (13.132) reduces to (13.133) and the complete solution can be expressed as w(r, 0, t)
= Wm(r,
O)(AI coswt
+ A2 sinwt)
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!I!!I!II!I!I!
(13.134)
!!!!!!!!!!!'!~~!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!'!!IJ
446
Vibration of Membranes
where Wm(r, 0) with Clm and 13.6.2
= Jm(wr)(Clm C2m
cosmO
+ C2m
sinmO),
m
= 0, 1, 2, ...
(13.135)
denoting some new constants.
Membrane with a Clamped Boundary If the membrane is clamped or fixed at the boundary can be stated as
r = a, the boundary
m = 0, 1,2, ...
conditions (13.136)
Using Eq. (13.136), Eq. (13.135) can be expressed as Wm(a, 0) = ClmJm(wa) cosmO
+ C2m Jm (wa)
sinmO = 0,
m
= 0, 1,2, ... (13.137)
The Bessel function of the first kind, Jm(wr), is given by [2],[3] 00
"
Jm(wr) =
to'
.
(_1)' i!r(m + i
(wr)m+2i
+ 1) "2
(13.138)
Equation (13.137) has to be satisfied for all values of O. It can be satisfied only if m = 0, 1,2, ...
(13.139)
This is the frequency equation, Which has an infinite number of discrete solutions, wmn' for each value of m. Although wa = 0 is a root of Eq. (13.139) for m 2: 1, this leads to the trivial solutions w = 0, and hence we need to consider roots with wa 2: O. Some of the roots of Eq. (13.139) are given below [2],[3]. For m = 0, Jo(wa) = 0: y
= wa = 2.405,5.520,8.654,11.792,14.931,18.071, 21.212,24.353, ...
For
m
= 1, JI(wa) = 0: y
= wa = 3.832,7.016,
10.173, 13.323, 16.470,
19.616,22.760,25.903, For m
...
= 2, lz(wa) = 0: y = wa = 5.135, 8.417,11.620,14.796,17.960,
21.117, 24.270, 27.421, ... For m = 3, h(wa) = 0: y = wa = 6.379,9.760,13.017,16.224,19.410,
22.583,25.749,28.909,
...
13.6 Free Vibration of Circular Membranes
For m
= 4, l4(wa) = 0: y = wa = 7.586,11.064,14.373,17.616,20.827, 24.018,27.200,30.371,
For
m
447
= 5,
...
= 0:
ls(wa)
y
= wa = 8.780,
12.339, 15.700, 18.982,22.220,
25.431,28.628,31.813,
...
It can be seen from Eqs. (13.134) and (13.135) that the general solution of w(r, e, t) becomes complicated in view of the various combinations of 1m, sinme, cosme, sinwmnt, and coswmnt involved for each value of m = 0,1,2, .... Hence, the solution is usually expressed in terms of two characteristic functions W~~(r, e) and w~~(r, e) as indicated below. If Ymn denotes the nth solution or root of 1m(Y) = 0, the natural frequencies can be expressed as Wmn
= -Ymn a
(13.140)
Two characteristic functions W~~(r, e) and W~~(r, e) can be defined for any Wmn as W~~(r, e) = Clmn1m(wmnr) cosme W~~(r,e)
(13.141)
= CZmn1m(wmnr) sin me
It can be seen that for any given values of m and n (m =1= 0) the two characteristic functions will have the same shape; they differ from one another only by an angular rotation of 90°. Thus, the two natural modes of vibration corresponding to Wmn are given by w~~(r, e, t) = 1m(wmnr) cosme[A~~
coswmnt
w;;~(r, e, t) = 1m(wmnr) sinme[A~~ coswmnt
+ A~~ sinwmnt] + A~~ sinwmnt]
(13.142)
The general solution of Eq. (13.120) can be expressed as 00
w(r, e, t) = L
00
L[w~~(r,
e, t)
+ w;;~(r,
e, t))
(13.143)
m=O n=O
where the constants A2~, ... , A~~ can be determined from the initial conditions. 13.6.3 Mode Shapes It can be seen from Eq. (13.141) that the characteristic functions or normal modes will have the same shape for any given values of m and n, and differ from one another only by an angular rotation of 90°. In the mode shapes given by Eq. (13.141), the value of m determines the number of nodal diameters. The value of n, indicating the order of the root or the zero of the Bessel function, denotes the number of nodal circles in the mode shapes. The nodal diameters and nodal circles corresponding to m, n = 0, 1, and
448
Vibration of Membranes
,-10 '-to ® 1-20--1
I- 20--1 I- 20--1
I- 2a--l
m=O
m=l
m=2
m=3
m = number of nodal diameters (a)
z
z
z
z
~~~~ I
a
r--t
I
O.463a
O.278a ~ O.638a ~
~
i a-I
:0 @I'~J
.1
I
O.204a O.468a O.734a
I
~: ~
a
a
'-" +
1-20--1 1-20--1 I- 20--1 n=O
n=l n=2 n = number of nodal circles
I- 2o-J n=3
(b)
Figure 13.11
Mode shapes of a clamped circular membrane.
2 are shown in Fig. 13.11. When m= 0, there will be no diametrical nodal lines but there will be n circular nodal lines, including the boundary of the membrane. When m 1, there will be one diametrical node and n circular nodes, including the boundary. In general, the mode shape Wmn has m equally spaced diametrical nodes and n circular nodes (including the boundary) of radius ri (wmdwmn)a, i 1,2, ... , n. The mode shapes corresponding to a few combinations of m and n are shown in Fig. 13.11.
=
=
13.7
=
FORCED VIBRATION OF CIRCULAR MEMBRANES The equation of motion governing the forced vibration of a circular membrane is given by Eg. (13.119): 02W p ( ar2
1 aw
1 02W)
+;:a; + r2
ae2
+f
=P
a2w 2
ot
(13.144)
Using modal analysis, the solution of Eq. (13.144) is assumed in the form 00
w(r, e, t)
=L
00
L Wmn(r, e)TJmn(t)
(13.145)
m=O n=O
where Wmn(r, e) are the natural modes of vibration and TJmn(t) are the corresponding generalized coordinates. For specificity we consider a circular membrane clamped at the edge, r = a. For this, the eigenfunctions are given by Eq. (13.141), since two
13.7 Forced Vibration of Circular Membranes
eigenfunctions are used for any Wmn' the modes will be degenerate except when and we rewrite Eq. (13.145) as 00
w(r, e, t) =
00
m
449
= 0,
00
L WJ~)(r. e)'16~(t) + L L W~~(r, e)'1~~(t) n=l
m=ln=l 00
+L
00
L W~~(r, e)'1~~(t)
(13.146)
m=ln=l
where the normal modes WJ~), W~~, and W~~ are given by Eq. (13.141). The normal modes can be normalized as
II
II 2rr
p[W~~(r,e)]2dA=
A
0
a
pCrmnJ~(Wmnr)cos2merdrde
0
Jr
2
= 2:pClmna
2
2
Jm+1 (Wmna) = 1
(13.147)
where A is the area of the circular membrane. Equation (13.147) yields 2
C1mn ._
II
= Jrpa 2J2m+l2 (wmna )
II 2rr
p[W~~(r,e)]2dA::
A
0
(13.148)
a
pc1nnJ~(wmnr)sin2merdrde
0
Jr
2
= 2:pC2mna
2
2
Jm+1 (wmna)
=1
(13.149)
or
2
2
(13.150)
C2mn = Jrpa 2J2m+l ( wmna ) Thus, the normalized normal modes can be expressed as W~~ (2)
Wmn
],J2
cos me sin ne
= -------Jm(wmnr)
.J1fPaJm+l (wmna)
m=O,I,2,
... ,
n=I.2,3,
} •
...
(13.151)
When Eq. (13.145) is used, Eq. (13.144) yields the equations iimn(t)
+ w;n'lmn(t)
= Nmn(t)
(13.152)
where N mn (t) denotes the generalized force given by
1l 21T
Nmn(t)
=
aWmn(r,e)!(r,e,t)rdrde
(13.153)
450
Vibration of Membranes
Neglecting the contribution due to initial conditions, the generalized coordinates can be expressed as [see Eq. (2.109)]
1 1 1 t
1)On(t) = - 1
Nci~\r)sinwon(t - r)dr
Won
(13.154)
0
t
1)~2(t) = --1 Wmn
N~lJ(r)sinwmn(t - r)dr
(13.155)
0
t
1)~2(t) = --1 Wmn
where the natural frequencies by Eq. (13.153).
Wmn
N~J(r)sinwmn(t
- r)dr
(13.156)
0
are given by Eq. (13.140), and the generalized forces
Example 13.4 Find the steady-state response of a circular membrane of radius a subjected to a harmonically varying uniform pressure allover the surface area, as shown in Fig. 13.12. Assume that the membrane is fixed at the boundary, r a.
=
Figure 13.12
SOLUTION
Circular membrane subjected to uniform pressure.
The load acting on the membrane can be described as = focosr2t.
f(r,e,t)
o ::: r
:::a,
0:::
e :::2Jf
(E13.4.I)
The transverse displacement of the membrane can be expressed in terms of the normal modes,Wmn(r,
e),
as 00
w(r, e, t)
00
=-2: 2: Wmn(r,
e)1)mn(t)
(EI3A.2)
m=On=l
,.
where the Wmn(r, e) for a circular membrane clamped at r = a are given by Eq. (13.151). Equation (EIOA.2) can be rewritten in view of Eq. (13.141) as
13.7
Forl:cd Vibration of Cirl:ular Membranes
00
w(r, e, t) =
00
L Wci~)(r,e)l/6~(t)
+L
n=l
451
00
L W~~(r, e)l/~2(t)
m=l n=( 00
00
+L
L W~~(r, e)l/~2
(E13.4.3)
(t)
m=1 n=1
where the generalized coordinates l/6~(t), l/~~(t), Eqs. (13.154)-(13.156). Using, from Eq. (13.151),
and l/~~(t)
are given by
(E13.4.4)
(E13.4.5)
(E13.4.6) the generalized forces can be evaluated as
(E13.4.7)
In view of Eqs. (E13.4.7)-(EI3.4.9), (I)
l/On(t)=
=
21ffo
2
Eqs. (13.154)-(13.156) become
JO(WOna)
"jifPWOn J( (WOna) 21ffoJO(Wona)(cosWOnt "jifPWon(Q2
1
t
.
cos~2rsmwOn(t-r)dr
0
- cos Qt)
- W5n)JI (WOna)
(E13.4.10)
452
Vibration of Membranes 77~~(t)
=0
(E13.4.11)
77;;~(t)
=0
(E13.4.12)
Thus, the response of the membrane can be expressed, using Eqs. (13.145), (13.151), and (E13.4.l0)-(E13.4.12), as w(r,
e, t)
2,J"i!o ~
= ---
pa
~
JO(WOna) JO(WOn r) 2
2
2
n=1 J1 (WOna)WOn(n
(COS
WOnt - cos nt)
(E13.4.13)
- WOn)
13.8 MEMBRANES WITH IRREGULAR SHAPES The known natural frequencies of vibration of rectangular and circular membranes can be used to estimate the natural frequencies of membranes having irregular boundaries. For example, the natural frequencies of a regular polygon are expected to lie in between those of the inscribed and circumscribed circles. Rayleigh presented an analysis to find the effect of a departure from the exact circular shape on the natural frequencies of vibration of uniform membranes. The results of the analysis indicate that for membranes of fairly regular shape, the fundamental or lowest natural frequency of vibration can be approximated as (13.157) where T is the tension, p is the density per unit area, A is the surface area, and Ol is a factor whose values are given in Table 13.1 for several irregular shapes. The factors given in the table 13.1 indicate, for instance, that for the same values of tension, density, and surface area, the fundamental natural frequency of vibration of a square membrane is larger than that of a circular membrane by the factor 4.443/4.261 = 1.043. Table 13.1 Values of the Factor ex in Eq. (13.157) Shape of the membrane
ex
Square Rectangle with b/a 2 Rectangle with b/a = 3 Rectangle with b/a 3/2 Circle Semicircle Quarter circle 60° sector of a circle Equilateral triangle Isosceles right triangle
4.443 4.967 5.736 4.624 4.261 4.803 4.551 4.616 4.774 4.967
= =
Sourr:e: Ref. [8)
_______________________
13.10
o
'-'~r
Figure 13.13
Recent Contributions
453
Circular sector membrane.
13.9 PARTIAL CIRCULAR MEMBRANES Consider a membrane in the form of a circular sector of radius a as shown in Fig. 13.13. Let the membrane be fixed on all three edges. When the zero-displacement conditions along the edges (j = 0 and (j y are used in the general solution of Eq. (13.134), we find that the solution becomes
=
w(r, (j)
= CJ
nrr/y ( Ar)
.
mr(j
sm --
t3
cos(dt
+ ~) Q
(13.158)
where C is a constant and n is an integer. To satisfy the boundary condition along the edge r = a, Eq. (13.158) is set equal to zero at r = a. This leads to the frequency equation Jnrr /y (Aa)
=0
(13.159)
If Aia denotes the ith root of Eq. (13.159), the corresponding natural frequency of vibration can be computed as Aia Wi=-=Ai
a
(13.160)
For example, for a semicircular membrane, y = 7C and for n = 1, Eq. (13.159) becomes Jl(Aa)
=0
(13.161)
whose roots are given by Ala = 3.832, A2a = 7.016, A3a = 10.173, and A4a = 13.324, .... Thus, the natural frequencies of vibration of the semicircular membrane will be WI = 3.832ja, Wz = 7.016ja, W3 = 1O.173ja, ....
13.10
RECENT CONTRIBUTIONS Spence and Horgan [11] derived bounds on the natural frequencies of composite circular membranes using an integral equation method. The membrane was assumed to have a stepped radial density. Although such problems, involving discontinuous coefficients in the differential equation, can be treated using the classical variational methods, it was shown that an eigenvalue estimation technique based on an integral formulation is more efficient. For a comparable amount of effort, the integral equation method is expected to provide more accurate bounds on the natural frequencies. The transient response of hanging curtains clamped at three edges was considered by Yamada et al. [12]. A hanging curtain was replaced by an equivalent membrane
.!'!'!.•• """! .. _!'!'!,.•.!I!'! .•. !.'!!.
_
454
Vibration of Membranes
for deriving the equation of motion. The free vibration of the membrane was analyzed theoretically and its transient response when subjected to a rectangularly varying point force was also studied by using Galerkin's method. The forced vibration response of a uniform membrane of arbitrary shape under an arbitrary distribution of time-dependent excitation with arbitrary initial conditions and time-dependent boundary conditions was found by Oleer [13]. Leissa and Ghamat-Rezaei [14] presented the vibration frequencies and mode shapes of rectangular membranes subjected to shear stresses and/or nonuniform tensile stresses. The solution is found using the Ritz method, with the transverse displacement in the form of a double series of trigonometric functions. The scalar wave equation of an annular membrane in which the motion is symmetrical about the origin was solved for arbitrary initial and boundary conditions by Sharp [15]. The solution was obtained using a finite Hankel transform. A simple example was given and its solution was compared with one given by the method of separation of variables. The vibration of a loaded kettledrum was considered by De [16]. In this work, the author discussed the effect of the applied mass load at a point on the frequency of a vibrating kettledrum. In a method of obtaining approximations of the natural frequencies of membranes was developed by Torvik and Eastep [17], an approximate expression for the radius of the bounding curve is first written as a truncated Fourier series. The deflection, expressed as a superposition of the modes of the circular membrane, is forced to vanish approximately on the approximated boundary. This leads to a system of linear homogeneous equations in terms of the amplitudes of the modes of the circular membrane. By equating the determinant of coefficients to zero, the approximate frequencies are found. Some exact solutions of the vibration of nonhomogeneous membranes were presented by Wang [18], including the exact solutions of a rectangular membrane with a linear density variation and a nonhomogeneous annular membrane with inverse square density distribution.
REFERENCES 1. K. F. Graff, Wave Motion in Elastic Solids, Ohio State University Press, Columbus, OH, 1975. 2. N. W. McLachlan, Bessel Functions for Engineers, Oxford University Press, New York, 1934. 3. G. N. Watson, Theory of Bessel Functions, Cambridge University Press, London, 1922. 4. L. Meirovitch, Analytical Methods in Vibrations, Macmillan, New York, 1967. 5. S. Timoshenko, D. H. Young, and W. Weaver, Jr., Vibration Problems in Engineering, 4th ed.. Wiley, New York, 1974. 6. E. Volterra and E. C. Zachmanoglou, Dynamics of Vibrations, Charles E. Merrill, Columbus,OH, 1965. 7. S. K. Clark, Dynamics of Continuous Elements, Prentice-Hall, Englewood Cliffs, NJ, 1972. 8. Lord Rayleigh, The Theory of Sound. 2nd ed., Vol. 1, Dover, New York, 1945. 9. A. Jeffrey, Advanced Engineering Mathematics, Harcourt/Academic Press, San Diego, CA,2002. 10. P. M. Morse, Vibration and Sound, McGraw-Hill, New York, 1936.
Problems
455
11. 1. P. Spence and C. O. Horgan, Bounds on natural frequencies of composite circular membranes: integral equation methods, Journal of Sound and Vibration, Vol. 87, No. I, pp. 71-81, 1983. 12. G. Yamada, Y. Kobayashi, and H. Hamaya, Transient response of a ~anging curtain, Journal of Sound and Vibration, Vol. 130, No.2, pp. 223-235, 1989. 13. N. Y. Olcer, General solution to the equation of the vibrating membrane, Journal of Sound and Vibration, Vol. 6, No.3, pp. 365-374, 1967. 14. A. W. Leissa and A. Ghamat-Rezaei, Vibrations of rectangular membranes subjected to shear and nonuniform tensile stresses, Journal of the Acoustical Society of America, Vol. 88, No.1, pp. 231-238, 1990. 15. G. R. Sharp, Finite transform solution of the symmetrically vibrating annular membrane, Journal of Sound and Vibration, Vol. 5, No.1, pp. 1-8, 1967. 16. S. De, Vibrations of a loaded kettledrum, Journal of Sound and Vibration, Vol. 20, No. I, pp. 79-92, 1972. 17. P. J. Torvik and F. E. Eastep, A method for improving the estimation of membrane frequencies, Journal of Sound and Vibration, Vol. 21, No.3, pp. 285-294, 1972. 18. C. Y. Wang, Some exact solutions of the vibration of non-homogeneous membranes, Journal of Sound and Vibration, Vol. 210, No.4, pp. 555-558, 1998.
PROBLEMS
=
=
13.1 Starting from the free-body diagram of an element Fo cos nt at the point (x 2a/3, y b/3). Assume the of a membrane in polar coordinates, derive the equation "membrane to be clamped on all four edges. of motion of a vibrating membrane in polar coordinates 13.7 Derive the equation of motion of a membrane in using the equilibrium approach. polar coordinates using a variational approach. 13.2 Derive the expression for the Laplacian operator 13.8 A thin sheet of steel of thickness 0.01 mm is in polar coordinates starting from the relation stretched over a rectangular metal framework of size 25 mm x 50 mm under a tension of 2 kN per unit length of periphery. Determine the first three natural frequencies of vibration and the corresponding mode shapes and using the coordinate transformation relations x = of the sheet. Assume the density of steel sheet to be r cos (}and y r sin (}. 76.5 kN/m3•
=
13.3 Show that each side of Eqs. (13.38) and (13.123) is equal to a negative constant. 13.4 Consider a rectangular membrane with the ratio of sides a and b equal to K = a/ b Find the distinct natural frequencies Wmn and Wij that will have the same magnitude.
= 1.
13.5 Find the forced vibration response of a rectangular membrane of sides a and b subjected to a suddenly applied uniformly distributed pressure fo per unit area. Assume zero initial conditions and the membrane to be fixed around all the edges. 13.6 Find the steady-state response of a rectangular membrane of sides a and b subjected to a harmonic force
13.9 Solve Problem 13.8 by assuming the sheet to be aluminum instead of steel, with a density of 26.5 kN/m3•
13.10 A thin sheet of steel of thickness 0.01 mm is stretched over a circular metal framework of diameter 50 mm under a tension of 2 kN per unit length of periphery. Determine the first three natural frequencies of vibration and the corresponding mode shapes of symmetric vibration of the sheet. Assume the density of the steel sheet to be 76.5 kN/m3• 13.11 Solve Problem 13.10 by assuming the sheet to be aluminum instead of steel, with a density of 26.5 kN/m3.
456
Vibration of Membranes
13.12 Derive the frequency equation of an annular membrane of inner radius r) and outer radius r2 fixed at both edges. 13.13 Find the free vibration response of a rectangular membrane of sides a and b subjected to the following initial conditions: •
w(x, y, 0)
TCX
= Wo sm -;;-
• 7Cy SID
b'
aw
at'
(x, y, 0)
=0
Assume that the membrane is fixed on all the sides. 13.14 Find the free vibration response of a rectangular membrane of sides a and b subjected to the following initial conditions: w(x, y, 0)
= 0,
aw (x, y, 0) = Wo
13.16 Find the free vibration response of a circular membrane of radius a subjected to the following initial conditions: w(r,
aw (r, e, 0) = 0
e, 0) = wor,
-
at
Assume that the membrane is fixed at the outer edge, r =a . 13.17 Find the response of a rectangular membrane of sides a and b fixed on all four sides when subjected to an impulse {; at (x xo, y Yo).
=
=
13.18 Find the steady-state response of a circular membrane of radius a when subjected to the force Fo sin Qt at r = O.
Assume that the membrane is fixed on all sides.
13.19 Derive the frequency equation of an annular membrane of inner radius b and outer radius a assuming a clamped inner edge and free outer edge.
13.15 Find the steady-state response of a rectangular membrane fixed on all sides subjected to the force f(x, y, t) = fo sin Qt.
13.20 Derive the frequency equation of an annular membrane of inner radius b and outer radius a assuming that it is free at both the inner and outer edges.
at
I
Transverse Vibration of Plates 14.1 INTRODUCTION A plate is a solid body bounded by two surfaces. The distance between the two surfaces defines the thickness of the plate, which is assumed to be small compared to the lateral dimensions, such as the length and width in the case of a rectangular plate and the diameter in the case of a circular plate. A plate is usually considered to be thin when the ratio of its thickness to the smaller lateral dimension (such as width in the case of a rectangular plate and diameter in the case of a circular plate) is less than :}o. The vibration of plates is important in the study of practical systems such as bridge decks, hydraulic structures, pressure vessel covers, pavements of highways and airport runways, ship decks, airplanes, missiles, and machine parts. The theory of elastic plates is an approximation of the three-dimensional elasticity theory to two dimensions, which permits a description of the deformation of every point in the plate in terms of only the deformation of the midplane of the plate. The equations of motion of plates are derived using the thin plate theory as well as Mindlin theory, which considers the effects of rotary inertia and shear deformation. Free and forced vibration of rectangular and circular plates are considered. The vibration of plates with variable thickness, of plates on elastic foundation, and of plates subjected to in-plane loads is also outlined.
14.2 EQUATION OF MOTION: CLASSICAL PLATE THEORY 14.2.1
Equilibrium Approach The small deflection theory of thin plates, called classical plate theory or Kirchhoff theory, is based on assumptions similar to those used in thin beam or Euler-Bernoulli beam theory. The following assumptions are made in thin or classical plate theory: 1. The thickness of the plate (h) is small compared to its lateral dimensions. 2. The middle plane of the plate does not undergo in-plane deformation. Thus, the midplane remains as the neutral plane after deformation or bending. 3. The displacement components of the midsurface of the plate are small compared to the thickness of the plate. 4. The influence of transverse shear deformation is neglected. This implies that plane sections normal to the midsurface before deformation remain normal to the midsurface even after deformation or bending. This assumption implies that the transverse shear strains, Bxz and Byz' are negligible, where z denotes the thickness direction. 5. The transverse normal strain Ezz under transverse loading can be neglected. The transverse normal stress au is small and hence can be neglected compared to the other components of stress.
457
458
Transverse Vibration of Plates
The equation of motion for the transverse vibration of a thin plate has been derived using an equilibrium approach in Section 3.6. Some of the important relations and equations are summarized below for a rectangular plate (see Fig. 3.3). Moment resultant-transverse displacement relations: o2W Mx = - D ( ox2
o2W)
o2W ( --+voy2
My=-D
(14.1)
+ v oy2
o2W) ax2
(14.2)
o2w
(14.3)
Mxy = Myx = -(1 - v)D-ox oy in which D represents the flexural rigidity of the plate: Eh3 D = ---12(1 - v2)
(14.4)
where h is the thickness, E is Young's modulus, and v is Poisson's ratio of the plate. Shear force resultants: 2 2 Qx = aMx + oMxy = -D~ (a + a (14.5) ox ay ax ox2 ay2
w
Q = aMy y oy
+ oMxy ox
= -D~
(o2W oy ox2
Equation of motion (force equilibrium in the aQx-
-
ax
aQy
+-
oY·
z
w)
2
+a
w) ay2
(14.6)
direction):
+ f(x,y,t)
a2w = ph- 2 at
(14.7)
or a4w D ( ax4
a4w ay2
+ 2 ax2
+
where p is the density of the plate and the plate per unit area.
o4W) ay4
+ ph
o2w at2 = f(x,y,t)
(14.8)
f is the distributed transverse load acting on
14.2.2 Variational Approach Because of assumptions 4 and 5 in Section 14.2.1, the state of stress in a thin plate can be assumed to be plane stress. Thus, the nonzero stresses induced in a thin plate are given by an. ayy, and axy. The strain energy density (no) of the plate can be expressed as (14.9) The strain components can be expressed in terms of the transverse displacement of the middle surface of the plate, w(x,y), as follows:
au a2w Exx = ax = -z ox2
(14.10)
14.2 Equation of Motion: Classical Plate Theory
av E
yy -
a2W
-
-
ay -
-z-ay2
(14.11) a2w ax ay
au av Ex =-+-=-2z-Y ay ax Ezz
aw
= -az ~ 0,
....,,59
Exz
(14.12)
au
av Eyz = -
aw
= -az + -ax = 0,
aw
oz + -ay
=
°
(14.13)
The stress-strain relations permit stresses to be expressed in terms of the transverse displacement, w(x,y), a
=-
2 Ez (a w 1 _ v2 ax2
+ v aay2w)
(14.14)
+ VExx) = -
2 Ez (a w 1 _ v2 ay2
+ v ax2
o2W)
(14.15)
=
E 1 _ v2 (Exx + VEyy)
=
E 1 _ v2 (Syy
G'xx
G'yy
E
= Gsx y = ---Ex y 2(1 + v)
G'x
2
o2w
Ez
By substituting Eqs. (14.10)-(14.12) and Eqs. (14.14)-(14.16) strain energy density can be written in terms of w as
Jro
= _E_z2_[(_a2_w)2+ 2(1 - v2)
ax2
a2w oy
= -2Gz-- ax ay = ----y 1 + vax
(14.16)
into Eq. (14.9), the
2 (_a _w)2 +2v_a2_w_a2_w+2(1- v) (_O_2W_)2] . ay2 ax2 ay2 ax ay (14.17)
Integrating Eq. (14.17) over the volume of the plate (V), the strain energy of bending can be obtained as
,,=
1[1
"odV 2
fj Chi' dA
2
wa w + 2v-.a -2 -2 ax
=
ay
+2(1- v)
= 2(1 ~ .')
"odz
( o2w -ax oy
)2]
dA
{I [e:~)'+ (~:~)'
lh/2 2 z dz z=-h/2
(14.18)
where dV = d A dz denotes the volume of an infinitesimal element of the plate. Noting that --
E
jh/2
1 - v2 z=-h/2
Z2
Eh3
d z = -----:::12(1 - v2)
is the flexural rigidity of the plate (D), Eq. (14.18) can be rewritten as
(14.19)
460
Transverse Vibration of Plates
Considering only the transverse motion and neglecting the effect of rotary inertia, the kinetic energy of the plate (T) can be expressed as T =
ph T
if (aw)2 at
(14.21)
dxdy
A
If there is a distributed transverse load, f(x, y, t), acting on the plate, the work done by the extemalload (W) is given by W =
II
(14.22)
fwdxdy
A
The generalized Hamilton's principle can be used to derive the equations of motion: 8
1'2
L dt = 8
11
112
. (n - W - T) dt = 0
(14.23)
11
Substituting Eqs. (14.20)-(14.22) into Eq. (14.23), Hamilton's principle can be written as
h -
(::~)']}
dx dy - P2
ij (~~)' ij dxdy
-
dt ~ 0
fWdXdY)
(14.24) where
V2
denotes the harmonic operator with o2w
V2w = - 2 ox
o2w
+ -oy2
(14.25)
Performing the variation of the first integral term in Eq. (14.24), we have II
= 8 JI2
~ II (V2w)2dxdydt
=D
A
JI2 II V2.wV28wdxdydt
(14.26)
A
By using the two-dimensional Green's theorem [40), Eq. (14.26) can be written as
h=D
t {ij
V'wow
dx d y
+ [[ V'w a~:) - ow a<::w) ] dC
} dt
(14.27)
where C denotes the boundary of the plate, n indicates the outward drawn normal to the boundary, and V4 represents the biharmonic operator, so that o4w 4 2 2 V w = V (V w) = ax4
a4w
+ 2 ax2ay2 +
a4w oy4
(14.28)
Note that the integration on the boundary, fe' extends all around the boundary of the plate. Similarly, we can express the variation of the second integral term in
14.2 Equation of Mot~oR:CJ~ssicalelate The~ry
461
Eq. (14.24) as
h=8
-D(1-v)
l2
f 11
If If
2
2
[azw --- 2 a w ax ayZ
(a--w ax ay
)Z]
dxdydt
A
l
= -D(l
_ v) f 2 11
+
[aZw 8 (aZw) ax2 ayZ
aZw 8 (aZw) ayZ axZ
A 2
-2 a w ax ay
8(
aZw)] ax ay l2
=-D(1-v)
f 11
If
dxdydt [aZw aZ(8w) aZwaZ(8w) ---+----2----Z ax ayZ ayZ axZ
2
a2w a (8W)] ax ay ax ay
dxdydt
A
(14.29) Noting that the quantity written as
i. [aZw ax
ayZ
under the sign of the area integral in Eq. (14.29) can be
+ i. [aZw
a(8w) _ aZw a(8W)] ax ax ay ay
axz
ay
a(8w) _ aZw a(8W)] ay ax ay ax
= -ahaxl + -ahz ay
(14.30)
where
aZwa(8w) ayZ ax
aZw a(8w) -----. ax ay ay
(14.31)
aZwa(8w) ax ay
aZw a(8w) ax ay ax
(14.32)
hi = -----.-
hz=--------z Equation (14.29) can be expressed
h
= -D(1-
v)
as l2
f 11
If
(ah! -
ax
+ -ahz) ay
dxdydt
(14.33)
A
Using the relations
ff:~dXdY=
[tCOS(}dC
(14.34)
[tsiU(}dC
(14.35)
A
ff:~dXdY= A
Eq. (14.33) can be rewritten
as I2
h=-D(1-V)f
{(hlcos(}+hzsin(})dC
11
Jc
(14.36)
where () is the angle between the outward drawn normal to the boundary (n) and the and a(ow)jay appearing in hi
x axis as shown in Fig. 14.1. The quantities a(8w)jax
462
Transverse Vibration of Plates y
n
Plate, area A
Boundary, C
x
Figure 14.1
Normal to the boundary of plate.
and h2 [Eqs. (14.31) and (14.32)] can be transformed (n, s) as follows: o(8w) ox
= o(8w) on + o(8w) os
o(8w) oy
=
=
into the new coordinate
system
o(8w) cosO _ o(8w) sinO on os
(14.37)
o(8w) on + o(8w) os = o(8w) sinO + o(8w) cosO on oy os oy on os
(14.38)
on
ox
os
ox
where s is the tangential direction to the boundary. Eqs. (14.31) and (14.32) can be expressed as o2w [O(8W) --cosO oy2 on
O(8W).] - --SInO os
hi = -
In view of Eqs. (14.37) and (14.38),
o2w [O(8W) - ----sinO ox oy on
o(8w) + --cosO os
] (14.39)
h2
[O(8W). = -o2w --on ox2
o(8w) ] o2w [O(8W) O(8W)] SIn 0 + -cos 0 - ---cos 0 - -sin 0 os ox oy on os (14.40)
When Eqs. (14.39) and (14.40) are substituted, 12
= -D(1 -
v)
1'2 {[c J
'1
+
[o8w Jc os
[(o2~ ox
Eq. (14.36) becomes
o8w (Oo2~ cos20 + o2~ sin20 _ 2 o2w sinO COSO) dC on r ox ox oy
_ o2~)cososino+ oy
o2w (Sin20-COs20)] ox oy
dC}
dt (14.41 )
The second integral involving using the relation
1 c
o8w -~-g(x,y) uS
integration with respect to C can be integrated by parts
dC
= g(x,y)8wlc
-
1 c
8w-
og
os
dC
(14.42)
14.2 Equation of Motion: CI;\Ssicai£late~l}f
where in the present case,
·-463
is given by
g(x, y)
(14.43) Since the variation of displacement, OW, on the boundary is zero. Eq. (14.42) reduces to
1 as c
aow -g(x,y)dC
=-
1 c
ag ow-dC
(14.44)
os
Inserting Eq. (14.43) in (14.44) and the result in Eq. (14.41), we obtain 12 = -D(1-
v)
{ 1 1 12
C
II
an
a2w ) 2-- sin B cos B d C ax ax ay 2 2 2 w - -2 a w) cosB sinB + --(sin a w2 ow- a [(a -2 B - cos2B) ] dC } dt c as ox ay ax oy (14.45)
o2W x ( -2 cos2B ay
+
a(ow)
--
a2w 2 + -2 sin B -
1
The variation of the third integral term in Eq. (14.24) can be expressed as
By using integration by parts with respect to time, the integral 1J can be written as
h
= -ph
If 1 If [
aw a(ow) ----dxdydt at at
12
11
A
= -ph
-ow aw at
A
12
1
12
-
1 1 I
Since the variation of the displacement
11
(ow)
-a (aw) at at is zero at
owdt ] dxdy
tl
and
t2,
(14.47)
Eq. (14.47) reduces to (14.48)
where iiJ = 82w/8t2.
The variation of the last integral term in Eq. (14.24) yields
i II 12
14 = -0
A
fwdxdydt
=
_£12 II f owdxdydt A
(14.49)
464
Transverse Vibration of Plates Using Eqs. (14.27), (14.45), (14.48), and (14.49), Hamilton's can be expressed as
(fj + + Dl [\72W -
1,"
(D~4 w
aaw an
phi» - f)8w dx dy
v)
(1-
c
principle of Eq. (14.24)
(a2~ ax
sin2()
_
2aa2w sin()cos() x ay
+
a2~ ay
COS2())]
dC
_ D
1{a\7an2w
.;..(1 _
c
2 (sin2 ax ay
+ aw
() _
cos2
2 2 [(a w 2 _ a w) as ax ay2
v)~
ow dC)
()]}
cos() sin()
dt = 0
(14.50)
To satisfy Eq. (14.50), each of the three terms within the outside parentheses must be zero. Furthermore, since is arbitrary, the expression inside the parentheses under the area integral must be zero. This leads to the relations
ow
D\74w
2 {\7 w _
(1-
v)
(a2~ oX
oaw = 0 on C on 2W _(1_V)~[(02W 0\7 { on as ow = 0 on C
+ phw
sin2()
_
-
f =0
2 2 a w ox oy
in A
sin()cos()
+
(14.51)
a2~ oy
COS2())}
(14.52)
_
ox2
02W)cOS()Sin()+
oy2
o2w (Sin2()-COs2()]} ox oy (14.53)
It can be seen that Eq. (14.51) is the equation of motion for the transverse vibration of a plate and Eqs. (14.52) and (14.53) are the boundary conditions. Note that for a clamped or fixed edge, the deflection and the slope of deflection normal to the edge must be zero (Fig. 14.2):
w=o,
aw an
-=0
ow = aaw/on =
Thus, 0, 0 in Eqs. (14.52) and (14.53). For a simply supported edge, the deflection is zero and the slope of deflection normal to the edge is not zero (Fig. 14.2):
w=o, ow
oaw/on
aw
-#0
an
Thus, = 0 in Eq. (14.53) and is arbitrary in Eq. (14.52). Hence, the expression in braces, which will later be shown to be equal to the bending moment on the edge, must be zero. For a free edge. there is no restriction on the values of w, and and hence and are arbitrary. Hence, the expressions inside braces
ow/on
ow
oaw/on
14.3
o
Boundary Conditions
465
·-·--.x
a
w(a,y,l)
Fixed edge
ax
b
=0
(a,y,I)-
y
o~._._~x
I
l'
z
(a)
o
a
·_·_·~x
Simply supponed edge
a2w
iJ2w)j
rJX" <>
ay-, (a,y,l)
M=-D~+v~ ( x
or
b
aw a7
w (a, y,
a
2
I)
I
=0
-0 (a,y,I)-
=0
(b)
Figure 14.2 Boundary conditions: (a) fixed edge; (b) simply supported edge; (c) free edge; (d) edge supported on a linear elastic spring; (e) edge supported on a torsional elastic spring,
in Eqs. (14.53) arId (14.52), which will later be shown to be equal to the effective shear force and bending moment, respectively, on the edge, must be zero.
14.3 BOUNDARY CONDITIONS The equation of motion governing the trarIsverse vibration of a plate is a fourth-order partial differential equation. As such, the solution of the equation requires two boundary conditions on each edge for a rectarIgular plate. If the edges of the rectangular plate are parallel to the x arId y axes, the following boundary conditions are valid. 1. Clamped, fixed. or built-in edge. If the edge x = a is clamped, the deflection arId slope (normal to the edge) must be zero [Fig. 14.2(a)]: wlx=a = 0
awl
-
ax
-0 x=a -
(14.54) (14.55)
466
Transverse Vibration of Plates
o
a
·_·_·~x
iPW)\ (a,y,t)- - 0
()2w
a~ + v al
Mx = - D(
Free edge
Vx = Qx + b
=-
aMry
ay
D[~+
(2 - v)
a~~ 1I(a,y,t)=
0
I
YT ._._._._._._._._._._.~
o
a
x
a
·_·_·~x
zT (c)
o Edge supported on a linear elastic spring
M -
x- -
()2w
D( -+
ax2
Vx =_D[a3~w
b
a2w)\ -0 al (a,y,t) -
v -
3
+(2-v)
a w 1\ (a,y,t) axal
= - k w(a,y,t) 2
I
YT a
0r'-'-'-'-'-'-'-'-'-'-'-'
x
(d) Figure 14.2
(continued)
2. Simply supported edge. If the edge x = a is simply supported, the deflection and bending moment must be zero [Fig. 14.2(b)]:
(14.56)
wlx=a = 0
Mx=_D(02~+vo2~)1 oxoy
=0 x=a
(14.57)
B\Ilmdary .Comlitiolls_467
14.3
o
a
,-,-,~x
Edge supported on a torsional elastic spring
Mx
==-
b V =-D x
azW)1 (a.y.t)
iPW
D ( axl + v ~.2 k
awl ax
,2
(a. v,t)
a3w
a3w [-::;-T+(:!-v)--
ax-
axai
]1 (a.y,t)
=0
.. /aw
//~ ax -'-'x
I I
t
z
(e)
Figure 14.2
(continued)
y
z
.
-'
-'
\
\
Mxydy
\
f
\
I
d
aM
xy
\ J
Mxy+--
(Mxy+ a~XYdY) dy
.--." .
dy
d Y
(b)
(a)
aM ay
xy
--y
d
(c)
Figure 14.3
Replacing the twisting moment Mxydy by an equivalent vertical force.
468
Transverse Vibration of Plates
Since w = 0 along the edge x zero. Thus,
= a, all the derivatives
awl ay
2
aw
of w with respect to y are also
=0
(14.58)
=0
(14.59)
x=a
I
ay2 x=a In view of Eq. (14.59), Eq. (14.57) can be rewritten as
w\
2
-a -0 ax2 x=a -
(14.60)
3. Free edge. If the edge x = a is free, there will be no stresses on the edge [Fig. 14.2(c)]. Hence, it appears that all the force and moment resultants on the edge are zero; that is, Mxlx=a = 0
(14.61)
Qxlx=a = 0
(14.62)
Mxy Ix=a = 0
(14.63)
Equations (14.61)-(14.63) represent three boundary conditions, whereas the equation of motion requires only two. Although Poisson formulated Eqs. (14.61)-(14.63), Kirchhoff showed that the conditions on the shear force Qx and the twisting moment Mxy given by Eqs. (14.62) and (14.63) are not independent and can be combined into only one boundary condition. To combine the two conditions given by Eqs. (14.62) and (14.63) into one condition, consider two adjacent elements, each of length dy, along the edge x = a as shown in Fig. 14.3(a). Because of the shear stresses Lxy acting on the edge, a twisting moment Mxydy is developed on the element cghd and a twisting moment [Mxy + (aMxyjay)dy]dy is developed on the element ecdf. These moments can be replaced by vertical forces of magnitude Mxy and Mxy + (aMxyjay)dy, respectively, on the elements cghd and ecdf, each with a moment arm dy as indicated in Fig. 14.3(b). Noting that such forces can be visualized for all elements of length dy along the entire edge x = a, we find that an unbalanced force of magnitude (aMxyjay)dy exists at the boundary between two adjacent elements, such as line cd [Fig. 14.3(c)]. When this unbalanced force per unit length, aMxyjay, is added to the shear force resultant Qx that is present on the edge x = a, we obtain the effective shear force Vx per unit length as Vy = Qx
aMxy + --ay
(14.64)
In a similar manner, the effective shear force Vy per unit length along the free edge y = b can be expressed as
aMyx
Vy=Qy+--
ax
(14.65)
14.3 Boundary Conditions
~69
The effective shear force resultant, along with the bending moment Mx, is set equal to zero along the free edge x = a. Thus, the two boundary conditions, also known as Kirchhoff boundary conditions, valid for a free edge x = a are given by {Fig. 14.2(c)] Mx = -D
VX = Qx
2w (a--2 ax
aMx = + -_Y ay
I
a2w) +vz ay
-D
(14.66)
=0
x=a
3 aw axay
3 [a-3w + (2 ax
v)-. --2
JIx=a = 0
(14.67)
=
4. Edge resting on a linear elastic spring. If the edge x a, otherwise unloaded, is supported on a linear elastic spring that offers resistance to transverse displacement, the restoring force will be k2W, as shown in Fig. 14.2(d). The effective shear force Vx at the edge must be equal to the restoring force of the spring. Also. the bending moment on the edge must be zero. Thus. the boundary conditions can be stated as
(14.68)
or -D
3w a3w [a-3 ax '-.+ (2 - v)--2 ax ay
JI .~=a
= -k2Wlx=a
(14.69)
=
5 . Edge resting on an elastic torsional spring. If the edge x a, otherwise unloaded, is supported on a torsional spring that offers resistance to the rotation of the edge, the restoring moment will be ktz(aw/ax), as shown in Fig. 14.2(e). The bending moment, Mx. at the edge must be equal to the resisting moment of the spring. Also, the effective shear force on the edge, Vx, must be zero. Thus, the boundary conditions can be expressed as
Mxlx=a =
-D
(aa2~ + v aa2~) I x
Vxlx=a = (Qx +
y
a:;y )Ix=a
x=a
=
w
ktz aa X
Ix=a
(14.70) 3
=
-D [~:~
+ (2 -
v)
a: a: Jlx=a = 2
0
(14.71)
6. Boundary conditions on a skew edge. Let an edge of the plate be skewed with its outward drawn normal (n) making an angle e with the +x axis and s indicating the tangential direction as shown in Fig. 14.4. The positive directions of the shear force Qn. normal bending moment Mn• and twisting moment Mns acting on the edge are also indicated in Fig. 14.4(a). Noting that the normal and twisting moments on the skew edge are defined as h/2
Mn =
j
(1nZ
dz
(14.72)
dz
(14.73)
-h/2 h/2
Mns =
j
'l'nsZ -h/2
470
TransverseVibrationof Plates
b
/
(a)
(b)
/
r
y
Figure 14.4 Skew edge of a plate: (a) shear force and momentresultants; (b) stressesacting on the edges.
where an and
= O'x cos2e
+ O'y sin2 () +
(14.74) (14.75)
By substituting Eqs. (14.74) and (14.75) into Eqs. (14.72) and (14.73), respectively, and carrying out the indicated integrations over the thickness of the plate, we obtain Mn = Mx cos2
()
+ My
sin2
()
+ Mxy
sin 2()
Mns = Mxy cos 2() - ~ (Mx - My) sin 2()
(14.76) (14.77)
The vertical force equilibrium of the element of the plate shown in Fig. 14.4(a) yields
or Qn = Qx-
dy ds
dx
+ Q,,-. ds
= Qx cose
The effective shear force resultant on the skew edge
+ Qy
Vn
.
sme
(14.78)
can be found as (14.79)
14.4
Free VibratioJl.of.Rectangular.PJates,
471
The different boundary conditions on the skew edge can be stated as follows: (a) Clamped. fixed. or built-in edge. The deflection and slope (normal to the edge) must be zero:
w=o aw -=0
(14.80) (14.81)
an
(b) Hinged or simply supported edge. The deflection and the normal bending moment resultant on the edge must be zero:
w=o
(14.82)
Mn =0
(14.83)
(c) Free edge. The normal bending moment and effective shear force resultants must be zero on the edge: (14.84)
Mn =0
(14.85) Note that the boundary conditions of Eqs. (14.83)-(14.85) can be expressed in terms of deflection, w, using Eqs. (14.76)-(14.79) and the known expressions of MJC, My, Mxy, Qx, and Qy in terms of w from Eqs. (14.1)-(14.6).
14.4 FREE VIBRATION OF RECTANGULAR PLATES
=
Let the boundaries of the rectangular plate be defined by the lines x 0, a and y = 0, b. To find the solution of the free vibration equation, Eq. (14.8) with f 0, we assume the solution to be of the type W(x,y,t)
= W(x,
=
(14.86)
y)T(t)
and obtain the following equations from Eq. (14.8): 1 d2T(t) -----
T(t)
dt2
fJ2
=-w
2
y) = _w2
_-=I_V4W(x,
W (x , y)
(14.87) (14.88)
where w2 is a constant and fJ 2
1 -
.E... ph
(14.89)
472
Transverse Vibration of Plates Equations (14.87) and (14.88) can be rewritten d2T(t) --2-
+ w2T(t)
dt
V4W(x,
as
y) - ).4W(x,
y)
=0
(14.90)
=0
(14.91)
where (14.92) The general solution of Eq. (14.90) is T(t)
=
and Eq. (14.91) can be expressed
as
(V4 _ ).4) W(x,y)
+ B sinwt
A coswt
= (v2 + ).2)(V2
(14.93)
- ).2) W(x,y)
=0
(14.94)
By the theory of linear differential equations, the complete solution of Eq. (14.94) can be obtained by superposing the solutions of the following equations: (V2 +
).2)WI (x,y)
2
=
.,
(V -),-)W2(X,y)
.a2
=
a2
wI
a2WI
+ -2-
-2-
ax
ay
W2 ax2
+
+ ).2WI (x,y)
=0
(14.95)
0
(14.96)
a2 W2 2·· .. ay2 -'). W2(X,y)=
Equation (14.95) is similar to the equation obtained in the case of free vibration of a membrane [Eq. (c) of the footnote following Eq. (13.44)], whose solution is given by the product of Eqs. (13.46) and (13.47) as
=
WI (x,y)
+ A2 sin ax cos,By sin ,By + A4 cos ax cos ,By
Al sin ax sin,By
+ A3 cos ax
(14.97)
where). 2 = a2 + ,B2. The solution of Eq. (14.96) can be obtained as in the case of solution of Eq. (14.95) except that). is to be replaced by i).. Hence the solution of Eq. (14.96) will be composed of products of sinh and cosh terms. Thus, the general solution of Eq. (14.91) can be expressed as W(x,y)
=
Al sin ax sin,By
+ A2 sin ax
cos,By
+ A3 cos ax sin,By + A4 cos ax cos ,By + A5 sinh ex sinh ¢y + A6 sinh e x cosh ¢y + A7 cosh ex sinh ¢y + As cosh ex cosh¢y
(14.98)
where (14.99)
14.4
14.4.1
Free Vibration ofR.ectangular-Plates
'473
Solution for a Simply Supported Plate For a plate simply supported on all the sides, the boundary conditions to be satisfied are w(x.y.t) w(x,y,t)
= MX
These boundary conditions
= 0 = 0
for x = 0 and a} for y = 0 and b •
=0.
(d2~ +vd2~)1 dx dy 2 d W d2W) ( --2 + v-2dx dy
= 0,
W(a.y)
= 0,
W(x,O)
=0,
W(x,b)
(14.100)
in terms of W. using Eq. (14.86). as
can be expressed
W(O.y)
t :::0
(d2~ dy
+ v d2~)'1
(d2~ dy
+vd2~)1 dx
dx
=0. (0,)')
I
= 0, (14.101)
(a,y)
= o. (x,O)
=0. (x.b)
As W is a constant along the edges x = 0 and x = a, d2 W / dy2 will be zero along these edges. Similarly, d2Wjdx2 will be zero along the edges y = 0 and y = b. Thus, Eqs. (14.101) will be simplified as
d2 W
d2 W
W(O, y) = dT(O, y) = W(a, y) = dT(a, y) = 0 dlW ".. dlW W(x,O) = dy2 (x, 0) W(x, b) = dy2 (x, b) 0
=
(14.102)
=
When these boundary conditions are used, we find that all the constants Ai, except A I. in Eq. (14.98) are zero; in addition, we obtain two equations that ex and fJ must satisfy: sinaa = 0 sinfJb=O Equations
(14.103) represent
the frequency
(14.103)
equations
ama=mrr, fJnb = nrr,
whose solution is given by
m=I.2 •... n = 1, 2, ...
(14.104)
Thus, we obtain from Eqs. (14.104), (14.99), and (14.92) the natural frequencies of the plate as Wmn=A~n
(D)
1/2
2 [
=rr
ph
m
n
2
2] ( D ) 1/2
+(b)
(-;)
ph
m, n = 1,2, ... (14.105)
The characteristic Wmn(x,y)
function
Wmn(x, y) corresponding
= Almn
. mrrx -a-
SIO
and the natural mode as . mrr x wmn(x,y,t) = SIO --
a
. nrry SIO
. nrr y
SIO
--(Amn b
h'
to Wmn can be expressed as
m, n = 1, 2, ...
coswmnt
' + B mn SIOWmnt)
(14.106)
(14.107)
.
_______________
!!!!!!!!I!!!!__
I!!!!!!!!!!!!II
!!!!!!!!!!!!!'!'!_ •• !!!'!'! ••.~!!!'!'! .•.. !!!'!'!._.!!!'!'!.~._'!!"!"n~~
~J
•• !!!'!'! .•. _.!!!'!'! ••• !!!'!'!~.!!!'!'!.,,!!!!!!._
i
474
Transverse Vibration of Plates
The general solution of Eq. (14.8) with 00
w(x,y,t)
f
= 0 is given by the sumof
the natural modes:
00
. --(Amn nn y = "" ~ "" ~sm . mrr -- x sm m=1 n=1
. + Bmn smwmnt)
coswmnt
b
a
(14.108)
Let the initial conditions of the plate be given by w(x,y,O)
8w a;
=
wo(x,y)
=
wo(x,y)
(14.109) (x,y,O)
By substituting Eq. (14.108) into Eqs. (14.109), we obtain 00
00
""
""
Amn
b
a
m=1 n=1 00
(14.110)
00
""
""
B mnWmnsm . -mnx
a
m=1 n=1
. -nrry = sm b
Multiplying each of the equations in Eq. (14.110) by grating over the area of the plate leads to
41 l
b
D
Amn
.
Bmn
= -b a
= wo(x,y)
mnx sm . nrry sin ---
0
wo(x,y)
0
ll D
4
= -b-a Wmn
0
wo(x,y)
0
sin(mnx/a)
mnx nrry sin -sin -a b
b
mrrx
sin --
..
a
. wo(x,y)
sin(nny /b) and inte-
dx dy
(14.111) nny
sin --
b
dx dy
The mode shape, Wmn(x,y), given by Eq. (14.106) consists of m half sine waves in the x direction and n half sine waves in the y direction of the plate. The first few modes of vibration corresponding to the natural frequencies Wmn' given by Eq. (14.105), are shown in Fig. 14.5. 14.4.2
Solution for Plates with Other Boundary Conditions To solve Eq. (14.91) for a plate with arbitrary boundary conditions, we use the separation-of-variables technique as . W(x,y)
= X(x)y(y)
(14.112)
Substitution of Eq. (14.112) into Eq. (14.91) leads to x"I/y
+ 2X"Y" + Xyl/I/
- ).4XY
=0
(14.113)
where a prime indicates a derivative with respect to its argument. The functions X (x) and Y(y) can be separated in Eq. (14.113) provided that either Y"(y)
= -f32y(y),
y""(y)
=
_f32y"(y)
or X"" (x) = _a2 X" (x)
(14.114)
14,4 Free Vibration of Rectangular Plates y
475
y
/
~/ft/ mF
t + m.t •• , _x
/ +
~m.2".'
/
Nodal
Nodal
line
line
m~!~~---_---_-=-_---_-+-
~m~~:1---~---fm:nt/
I
Figure 14.5 Mode shapes of a rectangular simply supported plate. Dashed lines denote nodal lines other than the edges.
or both are satisfied. Equations (14.114) can be satisfied only by the trigonometric functions
SinamX} {cos amx
or
sin f3nY } {cos f3n Y
(14.115)
with mrr
nrr
f3n = b' n = 1,2, ... am = -, m = 1,2, ... , a We assume that the plate is simply supported along edges x 0 and x implies that
=
m = 1,2, ...
(14.116)
= a.
This
(14.117)
476
Transverse Vibration of Plates where A is a constant. Equation Xm(O)
(14.117) satisfies the conditions
= Xm(a) = X~(O) = X~(a) = 0
(14.118)
for any integer m, and hence the boundary conditions w(O,y,t)
= w(a,y,t) = V2w(0,y,t)
= V2w(a,y,t)
=0
(14.119)
Using the solution of Eq. (14.117), Eq. (14.113) becomes y""(y)
- (A4 - a~)Y(y)
- 2a~yll(y)
=0
(14.120)
It can be observed that there are six possible combinations of simple boundary conditions when the edges x = 0 and x = a are simply supported. The various boundary conditions can be stated, using the abbreviations SS, F, and C for simply supported, free, and clamped edges, respectively, as SS-SS-SS-SS, SS-C-SS-C, SS-F-SS-F, SS-C-SS-SS, SS-F-SS-SS, and SS-F-SS-C. Assuming that A4 > a~ in Eq. (14.120), its solution is taken in the form Y(y) = eSY
(14.121)
which yields the auxiliary equation: s4 _ 2s2a~ - (A 4 - a~)
=0
(14.122)
The roots of Eq. (14.122) are given by SI,2
=
±JA2
+ a~,
S3,4 = ±iJA2
- a~
(14.123)
Thus, the general solution of Eq. (14.120) can be expressed as Y(y)
= CI sinolY
+ C2 cos olY + C3 sinh02Y + C4 COSh02Y
(14.124)
= JA2 -
(14,125)
where
01
a~,
02
= JA2 + a~
Equation (14.124), when substituted-into the boundary condition relations on the edges Y = 0 and Y = b, we obtain four homogeneous algebraic equations for the coefficients CI, C2, C3, and C4. By setting the determinant of the coefficient matrix equal to zero, we can derive the frequency equation. The procedure is illustrated below for simply supported and clamped boundary conditions. 1. When edges Y = 0 and Y can be stated as W (x , 0) W (x, b)
=b
are simply supported.
The boundary
=0 =0
conditions (14.126) (14.127)
W
,My(x,0)=-D--2(o2
oy
W My(x, b) = -D (02 --2
oy
02W) +v-2
ax
o2W)1 + V-2
ax
I
=0
(14.128)
=0
(14.129)
(x.O)
(x.b)
14.4 Free Vibration of Rectangular Plates
=o
=
= aw lax = a2wlax2 = o will also be zero.
Since W along the edges Y Oandy b, Thus, the boundary conditions of Eqs. (14.126)-(14.129)
can be restated as follows:
Y (0)
=0
(14.130)
Y(b)
=0
(14.131)
d2 Y (0) --=0 dy2 2
d Y(b) dy2 Since Eqs. (14.132) Eq. (14.124),
and (14.133) involve
(14.132)
=0
(14.133)
the second derivative
d2 Y (y) 2. 2 ~. dy2 = -dt CI smdlY - dl C2 COSdlY + 0iC3 smh02Y Using Eqs. (14.124) and (14.134), the boundary conditions can be expressed as
(14.134)
of Eqs. (14.130)-(14.133)
+ C4 = 0
(14.135)
+ C2 cos8tb + C3 sinh82b + C4 cosh82b = 0 -8tC2 + 8~C4 = 0 c28t cos 8lb + C38~ sinh 82b + C48~ cosh 82b = 0
(14.136)
CI sinStb
sin 81b -
of Y, we find from
+ 822C4 cosh82Y
C2
-C18t
477
(14.137) (14.138)
Equations (14.135) and (14.137) yield C2
= C4 = 0
(14.139)
In view of Eq. (14.139), Eqs. (14.136) and (14.138) reduce to
+ C3 sinh82b
CI sindlb
=0
or (14.140) and -C18t
sin8lb
+ C38~ sinh82b
=0
(14.141)
=0
(14.142)
Using Eq. (14.140), Eq. (14.141) can be written as Cl (8t
+ 8~) sin81b
For a nontrivial solution, we should have sindlb
=0
or nrr
dl=-, b
n = 1,2, ....
(14.143)
~
',1
478
Transverse Vibration of Plates with the corresponding
mode shapes as (14.144)
Since 81 =
../)..Z -
a~, we have
~z
Z mn = am
R
RZ + I-'n'
__ nj(
I-'n -
(14.145)
b
This leads to the result
m,n
=
The mode shapes Wmn(x,y) Xm(x)Yn(y), Wmn of Eq. (14.146), are given by
=
1, 2, ...
corresponding
(14.146)
to the natural frequencies
m, n = 1, 2, ...
(14.147)
where Cmn is a constant. This solution can be seen to be the same as the one given in Section 14.4.1. 2. When edges y. 0 and y b are clamped. The boundary conditions can be stated as .
=
=
=0 (0) = 0
Y (0) dY dy
(14.148) (14.149)
=0
(14.150)
-(b) = 0
(14.151)
Y(b)
dY dy
Using Eq. (14.124) and dY(y)
~
..
= CI81 cos81y - Cz81 sm81y
the boundary conditions
+ C382 coshozy + C40z
of Eqs. (14.148)-(14.151)
CIOI
CIOI cos olb -
(14.152)
can be expressed as
+ C4 = 0
(14.153)
+ C30Z = 0
(14.154)
Cz
+ Cz cos 81b + C3 sinhozb + C4 coshozb CZOIsinolb + C30Z cosh 82b + C40Z sinhozb
CI sinolb
SlnhoZY
=0
(14.155)
=0
(14.156)
14.5 Forced Vibration of Rectangular Plates
479
Equations (14.153)-(14.156) can be written in matrix form as
COS~8Ib si382b COS~82b] [ S~lb 81cos81b -81 sin81b 82cosh82b 82sinh82b
I~:]I~] =
C4
(14.157)
0
By setting the determinant of the matrix in Eq. (14.157) equal to zero, we obtain the frequency equation, after simplification, as 28182(cos81b cosh82b - 1) -
a;' sinolb
sinh82b = 0
(14.158)
For any specific value of m, there will be successive values of A and hence w that satisfy the frequency equation (14.158). The natural frequencies can be denoted as WII, WI2,W13, ... , WzI, Wz2, W23, ... , whose values depend on the material properties E, v, and p and the geometry h and b/ a of the plate. The mode shape corresponding to the nth root of Eq. (14.158) can be expressed as Yn(Y) = Cn [(cosh 82b - cos8lb) (81sinh82Y - 82sin8ly)
- (81sinh82b - 82sin81b) (cosh 82Y - cos 81Y)]
(14.159)
where Cn is an arbitrary constant. Thus, the complete mode shape Wmn = Xm(x)Yn (y), corresponding to the natural frequency Wmn, becomes Wm~C~,y) = CmnYn(y)
sinamx
(14.160)
where Yn(y) is given by Eq. (14.159) and Cmn is a new (arbitrary) constant. The frequency equations and the mode shapes for other cases (with other edge conditions at Y = 0 and Y = b) can be derived in a similar manner. The results for the six combinations of boundary conditions are summarized in Table 14.1.
14.5
FORCED VIBRATION OF RECTANGULAR PLATES We consider in this section the response of simply supported rectangular plates subjected to external pressure f (x, y, t) using a modal analysis procedure. Accordingly, the transverse displacement of the plate, w(x,y,t), is represented as 00
=
w(x,y,t)
00
L L Wmn(x,Y)1]mn(t)
(14.161)
m=ln=1
where the normal modes are given by [Eq. (14.106)] Wmn(x,y)
. mrrx
= A1mn sm -a-
. nrry
sm b'
m, n = 1, 2, ...
The normal modes are normalized to satisfy the normalization condition
ll a
b
phW;n dx dy = 1
(14.162)
"'"c .g
:.ac o
U
C ce
"0 C
::I
o
~
o \I ..l::> N
"'C
.c
o \I
-
'"'
C
.v.; ..l::> "'C
.5
'" NE
~ I
-
'"'
I
..l::> N "'C
.c o '"
u ..l::>
O"'C
"
..l::>~~ "'C
C
N "'C
·00 ~
-
tntn tntn , , tntn tntn
uu:. thrn
tntn II)
'"
U
-
N
14.5
Forced Vibration ofRectangu!ar'Ptates.-481
which yields Almn = 2/.J phab. By using the normalized normal modes in Eq. (14.161) and substituting the result into the equation of motion, Eq. (14.8), we can derive the equation governing the generalized coordinates Tlmn(t) as ijmn(t)
+ w;'nTlmn(t)
where the generalized force Nmn(t)
=
1, 2, ...
(14.163)
is given by
ii a
Nmn(t) =
m, n
= Nmn(t),
b
Wmn(x,y)f(x,y,t)
(14.164)
dx dy
and the natural frequencies by [Eq. (14.105)] m, n = 1,2, ... The solution of Eq. (14.163) can be expressed as [see Eq. (2.109)] 17mn(t)
~mn(O) = 17mn(O) coswmnt + --Wmn
. smwmnt
+ --1
Wmn
it 0
.
Nmn(r) smwmn(t
(14.165)
- r)dr
(14.166)
Thus, the final solution can be written as
(14.167)
Example 14.1 Find the response of a simply supported uniform plate subjected to a concentrated force F(t) at the point x = Xo, Y = Yo. Assume the initial conditions to be zero. SOLUTION Since the initial conditions are zero, the response is given by the steadystate solution: 00
w(x,y,t)
=
00
L L Wmn (x,Y)Tlmn m=ln=l
(t)
(E14.1.l)
482
Transverse Vibration of Plates
where
t Nmn('r) sinwmn(t
TJmn(t) = _1_
10
Wmn
Nmn(-r) =
la lb
- -r)d-r
(EI4.1.2)
dx dy
(EI4.1.3)
Wmn(x,y)f(x,y,t)
The concentrated force F(t) can be expressed as (£14.1.4)
= F(t)8(x - xo, y - Yo)
f(x,y,t)
where 8(x - xo, y - Yo) is a two-dimensional spatial Dirac delta function defined by 8(x - Xo, Y - Yo) = 0
ll a
#- Xo
for x
and/or y
#- Yo (E14.1.5)
b
8(x-xo,y-yo)dxdy=1
The natural frequencies Wmn and the normal modes Wmn(x,y) are given by Eqs. (14.165) and (14.162), respectively. By substituting Eqs. (14.162) and (EI4.1.5), into Eq. (EI4.1.3), we obtain
2 Nmn(-r) = r;:r:;:r:F(-r) phab =
2F(-r) .jphab
11 a
0
b
0
n~y sin m~x - - sin -8(x a b
- Xo, Y - Yo) dx dy
. m~xo . n~yo sm- -sm--·· a, b
(EI4.1.6)
With the help of Eq. (E14.1.6), Eq. (EI4.1.2) can be written as 1 TJmn(t)= Wmn = If F(t)
l' 0
2F(-r) . m~xo . n~yo . r;:r:;:r:sm -sm -smwmn(t - -r) d-r vphab a b
2 . m~xo n~yo r;:r:;:r:sm -sin -Wmnv phab a b
= Fo = constant, Eq. (EI4.1.7) l)mn(t) =
l' .
(EI4.1.7)
F(-r) smwmn(t - -r) d-r
0
becomes
2Fo . m~xo n~yo r;:r:;:r:sm -sin --(1 wmnv phab a b 2
(EI4.1.8)
- coswmnt)
If F(t) = Fo sin nt, Eq. (EI4.1.7) becomes TJmn(t)=
2Fo 2 ~ (wmn - n )vphab 2
m~xo . n~yo sin -,sm --(wmn a b
. sm nt
- n sin wmnt) . (E14.1.9)
Once TJmn(r)is known, the response can be found from Eq. (EI4.1.l). Example 14.2 A rectangular plate simply supported along all the edges is subjected to a harmonically varying pressure distribution given by f(x,y,t)
= fo(x,y)
sin
nt
(E14.2.l)
14.5
where plate.
n is
Forced Vibration of Rectangular Plates
':483
the frequency of the applied force. Find the steady-state response of the
SOLUTION The equation of motion for the forced vibration of a rectangular plate can be expressed as (EI4.2.2)
where
= fo(x,y)
f(x,y,t)
We assume the response of the plate, w(x,y,t),
(El4.2.3)
also to be harmonic with
= W(x,y)
W(X,y,t)
sin nt
sin nt
(EI4.2.4)
where W (x, y) indicates the harmonically varying displacement distribution. Using Eq. (EI4.2.3), Eq. (EI4.2.2) can be written as n4W _ ,4W = fo(x,y) v D
E
( 14.2.5)
/I.
where (EI4.2.6) We express the pressure distribution fo(x,y) and the displacement distribution W(x,y) in terms of the mode shapes or eigenfunctions of the plate Wmn(x,y) as 00
L L Amn Wmn(x,y)
=
W(x,y)
00
(EI4.2.7)
m=l n=1 00
=
fo(x,y)
00
L L Bmn Wmn(x,y)
(EI4.2.8)
m=1n=1 where
ll ll a
Amn =
a
Bmn =
b
w(x,y)
Wmn(x,y)dxdy
fo(x,y)
Wmn(x,y) dx dy
b
(EI4.2.9) (EI4.2.10)
The eigenvalue problem corresponding to Eq. (EI4.2.2) can be expressed as m, n = 1, 2, ...
(EI4.2.1l)
where (E14.2.12)
484
Transverse Vibration of Plates
Let the eigenfunctions of the plate Wmn (x, y) be normalized as
Lal
b
dx dy = 1
W;n(X'Y)
Multiply both sides of Eq. (E14.1.4) by Wmn(x,y) plate to obtain
(EI4.2.13)
and integrate over the area of the
ro 10(b [v,4W(x,y) _ 4W(x,y)]Wmn(x,y) . 1 r (b =D fo(x,y) Wmn(x,y) dx dy .l-.
dx dy (EI4.2.14)
10 10
which upon integratioq..by parts yields B 4 4 A mn (.1-. mn _.1-. ) = ~D
or A
Bmn mn - D(.I-.~n _ .1-.4)
(El 2 1 ) 4.. 5
By substituting Eqs. (E14.2.15) and (E14.2.1O) into Eq. (E14.2.7), we obtain the displacement distribution of the plate as 00
W(x
00
) = '" '" Bmn Wmn(x,y) ,y L...- L...- D(.l-.4mn -.1-.4) m=l n=l =
.2.- .~ ~
w""
Wmn(x,y) If fo(x,y)
= =
= fo =
" 11 a
o
b 0
Wmn(X,y!
J; J; fo(x',
y') Wmn(X', y') dx' dy' (E14.2.16) w2 - n2 mn Note that Eq. (E14.2.16) is applicable to plates with arbitrary boundary conditions. In the case of a simply supported plate, the natural frequencies and normalized eigenfunctions are given by
L...- L...m=l n=l
ph
\ ....-
",/£[(:)' G)']. +
2
.
mrrx
m, n
= 1, 2, ...
(E14.217)
nrry
r::r:;;r: sm - - sin (EI4.2.18) '\! phab a b constant, the double integral in Eq. (E14.2.16) can be evaluated as
""
fo(x , y) Wmn(x, Y )dx
dy
=
.
2fo r::r:;;r: phab
'\!
=
11 a 0
b 0
,mrrx ,mn sm __ ' sm a b -,-.I
dx' dy'
2fo ab r;::;;:;;r:: -2- (l - cas mrr) (1 - cas nrr) '\! phab rr mn if m is even or if m is odd and
11
11
is even is odd (E14.2.l9)
14.6 Circular Plmes "'48S
Thus. the response of the simply supported plate can be expressed, using Eqs. (E 14.2.19), (EI4.2.16) and (EI4.2.4), as (
.
) _ 1610
W .\. V,r
.
-
-2-
rr
L: L: oo
oo
.
m=I.3 .... n=1.3 •...
sin(mrrxla) sin(nrrylb) sin Or 4 2 mn {rr D [(mla) + (nib)] 2 2 - phO 2} (EI4.2.20)
14.6 CIRCULAR PLATES 14.6.1
Equation of Motion Consider an infinitesimal element of the plate in polar coordinates as shown in Fig. 14.6. In this figure the radial moment Mr, tangential moment Me, twisting moments Mre and Mer, and the transverse shear forces Qr and Qe are shown on the positive and negative edges of the element. The equations of motion of the plate can be derived in polar coordinates by considering the dynamic equilibrium of the element shown in Fig. 14.6 as follows (see Problem 14.9): Moment equilibrium about the tangential «(}) direction: aMr
ar
+ ! aMre + Mr
r ao
- Me _ Qr
r
=0
t
y
(a)
(b)
Figure 14.6
Element of plate in polar coordinates.
(14.168)
486
Transverse Vibration of Plates
Moment equilibrium about the radial aMr(J
direction:
(R)
1 aM(J
2
--ar + -r ae + -Mr(J r Force equilibrium in the
. - Q(J =0
(14.169)
z direction:
aQr
1 aQe
Qr
a;- +;:-ae + -;- + f
-
a2w ph
at2
=0
(14.170)
Equations (14.168)-(14.170) can be combined to derive a single equation of motion in terms of the moment resultants Mr, M(J, and Mr(J. By substituting the moment resultants in terms of the transverse displacement w, the final equation of motion, shown in Eq. (14.183), can be obtained. The coordinate transformation technique can also be used to derive the equation of motion in polar coordinates from the corresponding equation in Cartesian coordinates, as indicated below. 14.6.2
Transformation
of Relations
The Cartesian and polar coordinates of a point P are related as (Fig. 14.7)
x = r cose,
y = r sin e
+l
r2 = x2
e = tan-1
(14.172) (14.173)
~
x
y
t I
·-·-·-x
Figure 14.7
(14.171)
Canesian and polar coordinates.
487
14.6 Circular Plates
From Eqs. (14.172) and (14.171), we obtain
ar x -ax = -r
ar
-
= cos 0,
ay
y . = - = sm 0
(14.174)
r
Similarly, Eqs. (14.173) and (14.171) give
ao
y
a;:- = - r2
ao
sin 0 = --r-
x
-----ay - r2 -
cosO
(14.175)
r
Since the deflection of the plate w is a function of rand 0, the chain rule of differentiation yields
aw ax aw ay
-
aw ar aw ar
= =
ar aw ao ax + -ao -ax ar + aw ao ay ao ay
-
aw cos 0 - -1 -aw smO . ar r ao aw sinO + aw cosO ar ao r
= -
(14.176)
=
(14.177)
For the expressions a2w/axZ, a2w/ax ay and azw/ayZ, the operations of Eqs. (14.176) and (14.177) are repeated to obtain
a/ax
and
a/ay
aZw a (aw) a (aw) 1 a (aw) sinO axz = ax ax = ar ax cosO - ~ ao ax azw Z aZwsin20 awsinzO awsinZO aZwsinzO =-cos 0------+-· ---+---+--z ar aoar r ar r ao r2 aoz rZ (14.178)
aZw = ~ (aw) = ~ (aw) sinO + ~ (aw) cosO ayZ ay ay ar ay ao ay r azw Z azw sin 20 aw cosz 0 aw sin 2B azw cosz 0 =-sin 0+-----+arz arao r ar ------+--r ao r2 aoz rZ (14.179)
a2w ax ay
a (aw) a (aw) 1 a (aw) . ax ay = ar ay cos 0 - ~ ao ay S10 0 azw sinZO a2w cos20 aw cosZO aw sinZO azw sinZO = --arz 2 + ---arao r - -ao --r2 - --ar Zr - -aoz --Zr2
=
(14.180) By adding Eqs. (14.178) and (14.179), we obtain (14.181)
____________________________________
••••, •••. _"'.. ,••• _.!<''''' •••... :....•••....,..•••••
----'1..·-~, ~","!'•'r?J~••••. •• _•.,
••••. _. __ -.-.-.
488
Transverse Vibration of Plates
By repeating the operation V2 twice, we can express 4
V w
.
2
= V (V w) = 2
a4w = or4
(02 or2
2 a3w + ; or3 -
4 02w 002
+ r4
1 a +;: or
1 02w r2 or2
+
+
1 (2) r2 002
(02W or2
1 oW
1 02W) 002
+;: a;: + r2
1 ow 2 04w 2 03w 3 2 2 2 r a;: + r or ao - r3 o020r
1 04w
+ r4
004
(14.182)
Using Eqs. (14.178), (14.179), and (14.180) in (14.8), the equation of motion for the forced transverse vibration of a circular plate can be expressed as DV4w
02w
+ ph ot2
=f
or
(14.183)
14.6.3 Moment and Force Resultants Using the transformation procedure, the moment resultant-transverse relations can be obtained as (see Problem 14.17)
(1
1
ow o2W)] - - -+-2 r or r ao2
Mr=-D
02W [ -+v or2
Me=-D
law 1 02w -+v( -r -+or r2 002
Mre
= Mer = -
02W) or2
0(1 ; ae ow)
(1 - v)D or
displacement
(14.184) (14.185) (14.186)
Similarly, the shear force resultants can be expressed as Qr
=;1
[ a
ar (rMr) - Me
a (02W = -D - 2 or or QIi
=;1
[a or (rMre)
ae
+ aMre]
(14.187a)
1 02W) + -r1 -aw + -ar r2 002
0 2 = -D -(V w) ar
ae
+ aMe] + Mre
1 a (02w = -D - .- 2 r 00 or
(14.187b) (14.188a)
W
1 02 ) + -r1 -ow + or r2 002
1 a 2 = -D - -(V w) r 00
(14.188b)
14.6 Circular Plates The effective transverse
I
shear forces can be written as
I-
aMrl} [a 2 Vr = Qr + -;:--ae- = -D ar (Y' W) + -rVI} = QB + aMrl} ar
489
v
+ (1 _
= -D [~~(Y'2W) r ae
(1
a a2w law) ] ae -;:ar J8 - r2 aii 2 v)~ (~ a w _ 2. aw) ] ar r ar ae r2 ae
(14.189)
(14.190) Note that the Laplacian operator appearing coordinates by Eq. (14.181).
14.6.4
in Eqs. (14.187)-(14.190)
is given in polar
Boundary Conditions 1. Clamped. fixed. or built-in edge. The deflection and slope (normal to the boundary) must be zero:
w=o aw ar
(14.191)
=0
(14.192)
where r denotes the radial (normal) direction to the boundary. 2. Simply supported edge. The deflection and bending moment resultant be zero: ~.
w=o Mr = -D
must
(14.193)
2 2 [a w2 + v (~aw + 2.2 a w) ]ar r ar r ae2 -
0
(14.194)
3. Free edge. The bending moment resultant and the effective shear force resultant on the edge must be zero:
Mr = -D
a2w [ -2 ar + v
(14.195)
r
V =
or -D
1-
(I
aMrB Qr + -rI -ae
1
a-(Y' 2 w)+---v a -----a2w aw) ] [ ar r ae r ar ae r2 af:)
=0
(14.196)
=0
(14.197)
4. Edge supported on elastic springs. If the edge is supported on linear and torsional springs all around as shown in Fig. 14.8. the boundary conditions can be stated as follows:
or (14.198)
490
Transverse Vibration of Plates
I
+w
Figure 14.8
Edge supported on elastic springs.
Vr = -kow or -D [a-(V 2 w)+---I-va ar r ae
14.7
(1
1
2 a w -----r or oe
aw)] r2 oe
=-kow
(14.199)
FREE VIBRATION OF CIRCULAR PLATES The equation of motion of a circular plate is given by Eq. (14.183): DV4w+ph-
a2w at2
=
f
(14.200)
where (14.201) For free vibrations of the plate, Eq. (14.200) gives, after separation of variables, d2T (t)
2
~+w
(14.202)
T(t) =0
V4W(r, e) - ).4W(r, e) = 0
(14.203)
where ).4
phw2 = __
(14.204)
D
Using Eq. (14.201), Eq. (14.203) can be written as two separate equations: a2w or2
loW
+
-;:a;:
1 02W
2
+ r2 ae2 +). W
=0
(14.205)
(14.206)
__________
·A91
14.7 Free Vibration of Circl.i1ar::,paates
By expressing W(r, 0) = R(r)e(O), Eqs. (14.205) and (14.206) can be rewritten [after dividing each equation by R(r)e(O)/r2] as 2
+ ~dR(r)
~ [d R(r) R(r) dr2
r
= __
± 'A.2]
2
I_d e = a2 e(O) d02
dr
where a2 is a constant. Thus,
e
d2
2 -+a e=0 2
(14.207)
d0
2
d Rid __ R + ( _+ dr2 r dr
±'A.2
__
2 a ) R=O r2
(14.208)
The solution of Eq. (14.207) is e(O) = A cosaO
+ B sinaO
(14.209)
Since W(r,O) has to be a continuous function, e(O) must be a periodic function with a period of 2rr so that W(r, 0) = W(r, 0 + 2rr). Thus, a must be an integer: a=m,
m
= 0,
(14.210)
1, 2, ...
Equation (14.208) can be rewritten as two separate equations: 2
d Ro'J dR -+~-+ 2 dr
r'dr
2
d Rid R -+--dr2 r dr
(2 (2
2
a)
'A.
--
'A.
a ) +r2
r2 2
R=O
(14.211)
R -_ 0
(14.212)
Equation (14.211) can be seen to be a Bessel differential equation [as in the case of a circular membrane, (Eq. (13.131»] of order m (= a) with argument 'A.r whose solution is given by (14.213) where Jm and Ym are Bessel functions of order m of the first and second kind, respectively. Equation (14.212) is a Bessel differential equation of order m (= a) with the imaginary argument i'A.r whose solution is given by (14.214) where 1m and Km are the hyperbolic or modified Bessel functions of order m of the first and second kind. respectively. The general solution of Eq. (14.203) can be expressed as W(r, 0) = [C~) Jm('A.r) + C~)Ym('A.r)
+ C~4)Km(Ar)](Am
cosmB
+ C~) 1m('A.r) + Bm sinmO),
m
= 0, 1,2, ...
(14.215)
where the constants c~l), ... , C~4), Am. Bm. and 'A.depend on the boundary conditions of the plate. The boundary conditions of the plate are given in Section 14.6.4.
i !!!!!!!!!!!!!!!
~~~!!!!!!!!!!!!!!ll
492
14.7.1
Transverse Vibration of Plates
Solution for a Clamped Plate For a clamped plate of radius a the boundary conditions, in terms of W(r, e), are (14.216)
W(a,e)=o
aw ar
(14.217)
-(a,e)=o
In addition, the solution W (r, e) at all points inside the plate must be finite. For this, the constants C;;) and C~4)must be zero as the Bessel functions of the second kind, Ym()"r) and KmO"r), become infinite at r = O. Thus, Eq. (14.215) reduces to W(r, e) = [C~) JmO...r)
+ C;;) ImO"r)](Am
cosme
+ Em sin me),
m=0,1,2, ... (14.218)
The boundary condition of Eq. (14.216) gives = _ JmCJ...a)C(I)
C(3)
m
ImO"a)
(14.219)
m
so that W(r, e) = [ JmCJ...r)-
JmCJ...a)] ImCJ...r) (Am cosme
1m () ..a)
. + Em slOme),
m = 0, 1, 2, ...
(14.220) where Am and Em are new constants. Finally, Eqs. (14.217) and (14.220) give the frequency equations: d JmCAa) d - ----ImO [ -JmCJ...r) dr ImCAa) dr
..r)
]
= 0,
m
= 0, 1,2, ...
(14.221)
r=a
From the known relations [13, 36]
d
m
-d Jm(Ar)
= AJm-1 (Ar)
- -Jm(Ar)
(14.222)
d -Im(Ar) dr
= )../m_I(Ar) - -Im(Ar)
(14.223)
r
r m r
the frequency equations can be expressed as m=0.1,2,
...
(14.224)
For a given value of m, we have to solve Eq. (14.224) and find the roots (eigenvalues) Amn from which the natural frequencies can be computed, using Eq. (14.204), as Wmn
= A~n
D ( ph
)1/2
(14.225)
As in the case of membranes, we find that for each frequency Wmn, there are two natural modes (except for m = 0 for which there is only one mode). Hence, all the natural
) ~~;i,;;~~iil.wo.:.:.~""~~'i>.~~~'~
•••~A.;.~~_~~~~~.o.i\~~~"'·"'J.W·,,"~i,d"~a.·'~~'~~ilikM"''j,«,H'''''''':·ii.'ict,~,"
14.7
Free Vibration of Circular Plates
.':t93
modes (except for m = 0) are degenerate. The two mode shapes are given by W~~(r, e) = [Jm()..r)ImU ..a) - Jm(Aa)/m (Ar)] cosme
(14.226)
W,~~(r, e) = [Jm(Ar)/m(Aa)
(14.227)
- Jm(Aa) 1m (Ar)] sinme
The two natural modes of vibration corresponding to w~~(r,
e, t)
= cosme[lm(Amna)Jm()"mnr)
Wmn
are given by
- Jm(>"mna)Im(Amnr)]
(14.228)
+ A~~ sinwmnt)
. (A~~ coswmnt
w~~(r, e, t) = sinme[lm(Amna)Jm(Amnr)
+ A~~ sinwmnt)
. (A~~ coswmnt
The general solution of Eq. (14.200) with 00
w(r, e, t) = L
- Jm(Amna)lm(Amnr)]
(14.229)
f = 0 can be expressed as
00
L[w~~(r,
e, t)
+ w;;~(r,
(14.230)
e, t)]
m=O n=O
and the constants A~~, ... ,A~~ can be determined from the initial conditions. Some of the first few roots of Eq. (14.224) are AOla = 3.196, A02a = 6.306, A03a = 9.439, Alia = 4.611, Al2a = 7.799, Al3a = 10.958, A2la = 5.906, A22a = 9.197, and A23a = 12.402. Note that the mode shape Wmn(r, e) will have m nodal diameters and n nodal circles, including the boundary of the circular plate. The first few mode shapes of the clamped circular plate are shown in Fig. 14.9. 14.7.2
Solution for a Plate with a Free Edge For a circular plate of radius a with a free edge, the boundary conditions are given by Eqs. (14.195) and (14.197): at
r =a (14.231)
at
r =a (14.232)
By substituting Eq. (14.215) into Eqs. (14.231) and (14.232), the frequency equation can be derived as [1] (Aa)2Jm(Aa)
+ (1- v)[AaJ~(Aa)
(Aa)2Im(Aa)
- (1 - v)[AaI~(Aa)
- m2Jm(Aa)] - m2Im(Aa)]
(Aa)2I:"(Aa)
+ (1 -
v)m2[AaJ~(Aa)
- Jm(Aa)]
(Aa)2I~(Aa)
- (1 - v)m2[Aal~(Aa)
- Im(Aa)]
= ..:-.:.....:-::.:...:..--:.....~_~~-.:.::...:.-:...-~~
(14.233)
494
Transverse Vibration of Plates ! z
I ~
~m.o".,
!z
I / ~
~m="
••
,
Nodal line /
_~m=2"=1
" Nodal line Nodal circle
/
m= l,n=2
/ Nodal line
Figure 14.9 the plate.
Mode shapes of a clamped circular plate. Dashed lines denote nodal lines within
where a prime denotes derivative with respect to the argument. When Aa > m, the frequency equation (14.233) can be approximated by the equation [16) Jm(Aa) J:"(Aa)
[(Aa)2 ::::::
+ 2(1-
v)m2)[Im(Aa)/I~(Aa») (Aa)2 -
2(1 -
- 2Aa(1 - v)
v)m2
The first few roots of Eq. (14.234) are given in Table 14.2.
(14.234)
14.8 Forced Vibratiun oLCircuJar Plates Table 14.2
v
Natural Frequencies
49;'
of Vibration of a Free Circular Plate with
= 0.33a Number
of nodal diameters,
m
Number of nodal circles,
n
o 3
12.23 52.91 111.3 192.1
5.253 35.25 83.9 154.0
20.52 59.86 119.0
9.084 38.55 87.80
1 2
3
2
0
Data from Refs. [1] and [16]. Values of (Aa)2 = wa2..fiiliTl5.
Source: a
14.8 FORCED VIBRATION OF CIRCULAR PLATES We shall consider the axisymmetric vibrations of a circular plate in this section. The equation of motion governing the axisymmetric vibrations of a circular plate is given by 02 p2 ( -;z r
I 0
+ --
r or
)2 w(r,
t)
I P
+ iiJ(r, t) = -h f(r,
t)
(14.235)
where D p2=_ ph
(14.236)
and a dot over w denotes a partial derivative with respect to time. The boundary conditions for a plate simply supported around the boundary r a are given by
=
w=O 02W
+ ~ow
or2
r or
=0
=a
at
r
at
r=a
(14.237) (14.238)
To simplify the solution, the boundary condition of Eq. (14.238) is taken approximately as [14] at
r =a
(14.239)
Equations (14.237) and (14.239) imply that the plate is supported at the boundary, a, such that the deflection and the curvature are zero. Thus, Eq. (14.239) will be satisfied to a greater extent for larger plates (with large values of a) than for smaller plates (with small values of a).
r
=
14.8.1 Harmonic Forcing Function The forcing function, f(r, t), is assumed to be harmonic, with frequency f(r, t) = F(r)emr
n, as (14.240)
496
Transverse Vibration of Plates
The solution of Eq. (14.235) is assumed to be of the form w(x, t) = W(r)emt
(14.241)
Using Eqs. (14.240) and (14.241), the equation of motion, Eq. (14.235), can be expressed as
)2
1d + --
-d2 ( dr2
r dr
1
W(r) - A4W(r)
(14.242)
= -F(r) D
where 4
Q2ph
Q2
(14.243)
A =2=-p D
Equation (14.242) can be solved conveniently by applying Hankel transforms. For this, we multiply both sides of Eq. (14.242) byrJo(Ar) and integrate with respect to r from o to a to obtain
[a
10
2 (d dr2
+ ~!£)2W(r)rJo(Ar)dr
_ A4W(A) = ~F(A) D
r dr
(14.244)
where W(A) and F(A) are called the finite Hankel transforms of W(r) respectively, and are defined as
i
and F(r),
a
W(A) =
rW(r)JO(Ar)
=i
(14.245)
dr
a
F(A)
To simplify Eq. (14.244), first consider the integral 1=
L
a
o
r
(d2W --2 dr
(14.246)
rF(r)Jo(Ar)dr
1 dW) + --d r
r
. (14.247)
Jo(Ar)dr
Using integration by parts, this integral can be evaluated as 1=
dW [ r-Jo(Ar) dr
- ArW J~(Ar)
]a - La rW JO(Ar) dr 0
The expression in brackets in Eq. (14.248) will always be zero at zero at r = a if A is chosen to satisfy the relation Jo(Aa)
=0
Jo(Aja)
= 0,
(14.248)
.1..2
0
r
= 0 and
will be (14.249)
or (14.250)
i = 1,2, ...
where Aja is the ith root of Eq. (14.249). Thus, Eq. (14.247) takes the form
L
a
o
r
(d2W
--2
dr
1 dW) + -r dr
Jo(Ar)dr
,,-
= -AiW(Aj)
(14.251)
14.8 Forced Vibration of Circular Plates ''497 When the result of Eq. (14.251) is applied twice, we obtain
fa
(d2
1d + ;. dr
o r dr2
)2
W(r)r loO ..r) dr = 1iW() ..j)
(14.252)
In view of Eq. (14.252), Eq. (14.244) yields W(1') __ 1 _F_(_1j_) • -D)"'~-14 I By taking the inverse Hankel transforms,
(14.253)
we obtain [13-15]
(14.254)
n
Note that as the forcing frequency approaches the ith natural frequency of vibration of the plate, )...7.j D / ph, the deflection of the plate W (r) -+ 00, thereby causing resonance.
14.8.2
General Forcing Function For a general forcing function f(r,·t),
the Hankel transforms of w(r, t) and f(r, t) are
defined as
l
a
W(1) = F(1) =
dr
(14.255)
t)loO ...r) dr
(14.256)
rw(r, t)lo(1r)
10t rf(r,
.
Using a procedure similar to the one used in the case of a harmonic forcing function, we obtain from Eq. (14.235), 2
d W(1') _ 1___ ' + 1~W() ..·) = -F(1·) 2 dt
The solution of Eq. (14.257), pressed as [14, 15]
•
I
D
(14.257) t
after taking the inverse Hankel transforms,
can be ex-
498
Transverse Vibration of Plates
where wo(r)
= w(r, t = 0)
wo(r)
=
ow at
-(r,
t
(14.259)
= 0)
(14.260)
denote the known initial conditions of the plate and = p)} =
Wi
I
{D).,} YPh
(14.261)
I
Note that the first term on the right-hand side of Eq. (14.258) denotes transient vibrations due to the initial conditions, and the second term represents the steady-state vibrations due to the forcing function specified. To illustrate the use of Eq. (14.258) for free (transient) vibration response, consider a plate subjected to the initial displacement wo(r)
=b
(1 - ::) ,
0::: r
::: a
(14.262)
where b indicates the displacement of the center of the plate (assumed to be small) and wo(r) = f(r, t) O. Using the initial condition of Eq. (14.262) and noting that
=
1
0
o
b
=2
rwo(r)JoU"ir)dr
a
1°
= 4bJ]
(a2 - r2)rJo().ir)dr
().ia) 3
(14.263)
a\
0
the free (transient) vibration response of the plate can be obtained from Eq. (14.258) as 8b
w ( r, t ) = - 3 a
'"
~ i=],2, ...
JOU"ir)
---
COSWit
---
J] ()I.]a)
(14.264)
A~
To illustrate the use of Eq. (14.258) for forced vibration response, consider the steadystate response of a plate subjected to a constant distributed force of magnitude do acting on a circle of radius Co suddenly applied at t = O. In this case, f(r,
where H
t)
= doH(co
denotes the Heaviside unit function defined by
(x)
H (x) = {O 1 Using the relation
1
0
o
(14.265)
- r)H(t)
r JO(Air) dr
l' 0
= do
l
1
r)
f(r
--'-
sin Wi (t
Wi
0
o
rJO(Air)H(co
- r)dr
for for
x < x :::
0 0
(14.266)
- t) dr
it
H(r) -0 Wi
sinwi(t
- t)dr
= do 1 -
COSWit 2
Wi
co
=
rJO(Air)dr
docoa JI (Ai co) 2
AiWi
(1 - coswit)
(14.267)
14.9 Effects of Rotary incnia and-5hear OefonnattQn .499
the forced response of the plate, given by Eq. (14.258), can be expressed as W ( r, t)
14.9
2doco = -aD
'"
L.,
i=I.2. ...
JOO·ir)J1 (AiCO) (1 ,'1 5 - cos P"jt a Ai [11 (Ai )]2
(14.268)
)
EFFECTS OF ROTARY INERTIA AND SHEAR DEFORMATION In the derivation of Eq. (14.8) we assumed that the thickness of the plate is small compared to its other dimensions, and the effects of rotary inertia and shear deformation are small. In the case of beams a method of accounting for the effects of rotary inertia and shear deformation, according to Timoshenko beam theory, was presented in Section 11.15. Mindlin extended the Timoshenko beam theory and presented a method of including the effects of rotary inertia and shear deformation in the dynamic analysis of plates [3,4]. We discuss the essential features of Mindlin plate theory in this section.
14.9.1 Equilibrium Approach Strain-Displacement Relations We assume that the middle plane of the plate lies in the xy plane before deformation and its deflection is given by w(x,y,t). A fiber AB oriented in the z direction takes the positions A' B' and A" B" due to bending and shear deformations, respectively, in the xz plane, as shown in Fig. 14.10. Thus, if 4>x
--.--.-.-. After deformation
/
w
z
t
A
I
L.
'-'-'-'-
B
Figure 14.10
-'-
x
Before deformation
Rotation of the normal AB.
•
_____________
~~·W"~."w,~,.c~.P.O
.•~..• c,.~,..
,.!!!!'!II!!!!=." !!!!!!!!!!!!!!j
500
. Transverse Vibration of Plates
denotes the rotation in the xz plane of a line originally normal to the midplane before deformation, the displacement of a point C located at a distance Z from the midplane in the direction of the x axis is +z¢Jx. It can be seen that point C will not have any x component of displacement due to shear deformation. Similarly, point C will have a component of displacement parallel to the y axis, due to bending of the plate in the yz plane. Its value is +z¢Jy, where ¢Jy denotes the rotation in the yz plane of a line originally normal to the midplane before deformation. Thus, the complete displacement state of any point (like C) in the plate is given by = +z¢Jx(x,y,t) = +z¢Jy(x,y,t) = w(x,y,t)
u(x,y,t) v(x,y,t) w(x,y,t)
(14.269)
As in the case of a beam, the slope of the deflection surface in the xz and yz planes (ow/ox and ow/oy) will be increased by the shear angles Yx and yy, respectively, so that Yx = ¢Jx
+
ow
yy
ox'
= ¢Jy + -ow oy
(14.270)
=
=
Note that the classical or Kirchhoff plate theory can be obtained by setting Yx yy 0 or ¢Jx = -ow/ox and ¢Jy = -ow/oy. The linear strain-displacement relations can be expressed as 0 = -ou = -(+z¢Jx) ox ox ov 0 Byy = -oy = -(+z¢J oy
Bxx
exy
=
eyz
=
Bxz
=
ou oy + ov oz + OU oz +
=. +z- o¢Jx ox o¢Jy y ) = +z-. oy-
ov 0 ox = oy (+z¢Jx) ow 0 = oz (+z¢Jy) ow 0 ox oz (+z¢Jx)
ay
=
0
(o¢Jx oy
= +z ow ow + ay = +¢Jy + ay
+ ox
+
(+z¢Jy)
ow ox
+
o¢Jy) ox
(14.271)
ow
= +¢Jx + a;
Bzz = 0 Stress Resultants As in thin plates, the nonzero stress components are axx, ayy, axy, ayz, and axz. The force and moment resultants per unit length, Qx, Qy, Mx, My, and Mxy, are defined as in Eq. (3.27). Using the stress-strain relations
i t~~~w"'j,l:;;&~~~';;;;"'.4J,~~~~~~;,t.~~~~~"'~~~~~::C~'f$I,~,·.~"~."Jh'~'~',:l';"'iI'i'j,Azi>1t~~'·~'·.~)j::it<;,
E
axz
= -1-(exx + VByy) -v 2
ayy
=
1_
ax)'
=
Gexy
ayZ
= GeyZ
ax:
= Gexz
E v2
(eyy
+ VBxx) (14.272)
(-1..9 Effects of Rotary Inertia and Shear Deformation
and the strain-displacement can be written as
2
Qx
= lh/
-h/2
axzdz
2
Qy
=
2 h/ a dz yz l -h/2 2
= k Gh (+cPy
relations of Eq. (14.271), the force and moment resultants
= lh/2
-h/2
= k Gh (+CPX+
501
Gcxzdz
= G 1"/2 ( +CPx+ -aw)
dz
aw) = G 1"/2 ( +CPy+ -.-
dz
-"/2
ax
~:)
= lh/2
GcyZ dz
-"/2
-h/2
oy
+ ~~)
h/2 1h/2 E axxZ dz = -1--2(cxx + Vcyy)Zdz l -h/2 -h/2 - v h2 = _E_l / Z2 (aCPX + v acpy) dz = +D (aCPX + v acpy) 1 - v2 -h/2 ax oy ax ay
Mx =
(14.273)
h/2 1h/2 E ayyzdz = --2 (cyy + vcxx)zdz -h/2 1 - v l -h/2 h2 = _E_l / Z2 (Orfjy + v OCPx) dz = +D (acpy + v OCPx) 1 - v2 -h/2 ay '. ox . oy ax
My
=
Mxy
=
h/21h/2 axyzdz l -h/2
=
_ D(1 - v) (acpx 2 ay
-h/2
+
GSxyzdz = +G
1h/2 2 (acpx acpy) z -a- + -adz -h/2 Y x
acpy) _ ax - Myx
where E
G=---
2(1
+ v)
(14.274)
is the shear modulus of the plate. Note that the quantity k2 in the expressions of Qx and Q y is similar to the Timoshenko shear coefficient k and is introduced to account for the fact that the shear stresses axz and ayz are not constant over the thickness of the plate, -h/2 :::Z ::: h/2. The value of the constant k2 was taken as ~ by Reissner for static problems, while Mindlin chose the value of k2 as Jl'2/12 so as to make the dynamic theory consistent with the known exact frequency for the fundamental "thickness shear" mode of vibration. Equilibrium Equations The equilibrium equations for the Mindlin plate can be derived with reference to Fig. 14.11 and by considering the effect of rotatory inertia as follows:
502
Transverse Vibration of Plates
I
Mx+_dxl
ax
+ 0;;
dX)
.. = mertla
aMyx d y
Y d ----.;
Moment and shear force resultants on an element of a plate.
I. Vertical force equilibrium ( Qx
\
Myx+(Jy
I I
~
Figure 14.11
I \
aMx
in the
z
+ ( Qy + o~y
dy
direction: dY)
., . th ., lorce m ez dIrectIon
dx
+f
= ph dx
d
dx dy - Qx dy - Qy dx
Y
o2w
ot2
or oQx ox
+ oQy + f
= ph o2w
where p is the mass density of the plate and f(x,y,t) distributed force. 2. Moment equilibrium - ( Qy
+ oQv) oy' dy
- Mydx
= inertia
is the intensity of the external
about the x axis:
dx dy
- Mxydy
(14.275)
ot2
oy
y
+ (OM My )+ ay dy
+f
y
dx
OMx ) dx + ( Mxy + ---a;-
dy
dy dx dYT
= pIxdx
moment due to rotation CPy
o2cp dv--f . ot
ph3
,Pcp
= ---fdx 12 ot
dy
or oMv
-Qy
oMxy
+ ay + --a;-
ph302cpy
= 12
(14.276)
at2
where the terms involving products of small quantities are neglected, and (pIx dx d y) is 3 3 the mass moment of inertia of the element about the x axis, with Ix = (1)h = h / 1'2 denoting the area moment of inertia per unit width of the plate about the x axis.
i2
___________________
14.9 ~Effectsof Rotary Inertia and Shear Deformation 3. Moment equilibrium - ( Qx
Qx
a ) + ~dx
- Mxdy
=
about the y axis: yx
+ (oMr) Mx +
dy dx
- Myxdx
503
+ fdx
ax' dx
dy
aM ) + ( Myx + --aydY
dx
dx dy-
2
.. d' merna moment ue to rotatIOn 4Jx
= pI . d x v
d
2 a 4Jx Y -2at
J 2 ph a 4Jx -2- d x d y 12 at
=-
or aMx
-Qx
phJ a24Jx
oM.rv
+ a;- = 12at2
+~
(14.277)
by neglecting products of small quantities, and (ply dx dy) is the mass moment of inertia of the element about the y axis, with Iy (1)hJ hJ /12 representing the area moment of inertia of the plate per unit width about the y axis.
= /2
=
Substituting Eqs. (14.273) into Eqs. (14.275) to (14.277) yields the final equations of motion in terms of the displacement unknowns w, 4Jx, and 4Jy as
("2
k2Gh
D 2
[2(1 -
v)V 4Jx + (1 + v)-
a
(a4Jx ax ox
w+-+-a4Jx ox
Y)] + -o4J oy
O4JY) oy
+f
2 ( 4Jx + -aw)
k Gh
ox
2 =p- h a w ot2 (14.278) J
2
ph a 4Jx =-12 at2 (14.279)
D -
2
[2(1-v)V4Jy+(1+v)-
can be rewritten
(a4Jx o4JY)] -+oy ax oy
+ a2/oy2
where V2 = a2/ax2
2 ( 4Jy+- OW) -kGh
a
ay
is the Laplacian operator. Equations
(14.278) to (14.280)
as
v 2 4Jx + (1 + v)-0<1>] ax
D [ (1- v)V 2
2
4Jy
+ (1+ v)-a
+
2
(
2
(
k Gh
- k Gh
= ph-2aatw 2
k2Gh(V2w D [ (1 - v) 2
J 2 ph a 4Jy =--2 12 at (14.280)
<1»
4Jx +
4Jy
+
f
OW) -.ox
aw) + -;vy
J 2 ph a 4Jx 12 ot2
(14.282)
J 2 ph a 4Jy 12 at
(14.283)
= -=
(14.281)
--2
where (14.284)
II!!!!!!!!!!!IIII!!!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!======:=.d
,
1f""
504
Transverse Vibration of Plates
We can eliminate tPx and tPy from Eqs. (14.282) and (14.283) by first differentiating them with respect to x and y, respectively, and then adding them to obtain ph3 02<1> = -(14.285) 12 2 By eliminating <1>from Eqs. (14.281) and (14.285), we obtain a single equation in terms of w as DV2<1>_ k2Gh<1>- k2GhV2w
2-
(V
k~G
::2)
2
(DV
22
- ~~ 2
D
= ( 1 - k2Gh V2
ot
:t
+ ph
) w
O:t~
2
ph + 12k 2G
0 ) ot2
!
(14.286)
If shear deformation only is considered, the right-hand-side terms in Eqs. (14.279) and (14.280) will be zero, and Eq. (14.286) reduces to D
2
(
2 0 )
P
2
V - k2G ot2
V w
+ ph
02w ( D 2) ot2 = 1 - k2Gh V!
(14.287)
If rotary inertia only is to be considered, terms involving k2 should be neglected in Eqs. (14.278) to (14.280), in which case Eq. (14.286) becomes (
DV2---
ph3 02) 12 ot2
V2w+ph-=!
02w ot2
(14.288)
If both the effects of rotary inertia and shear deformation are neglected, terms involving k2 and ph3 /12 will be zero in Eqs. (14.278) to (14.280), and tPx and tPy will be replaced by -ow/ax and -ow/oy, respectively, and Eq. (14.286) reduces to the classical thin plate equation: 02W
DV4w + ph-2
=!
ot
(14.289)
For a plate with constant thickness, the equation of motion, Eq. (14.286), can be expressed as DV4w
pD
02
(V2w) __
k2G ot2 D
= ! - k2Gh
2
V
!+
ph3
12
V2_
02w ot2
2 3 04W 12k G ot4
p h + --2
+ ph- 02w2 8t
ph2 02! l2k2G ot2
(14.290)
For free vibration, Eq. (14.290) becomes ph 82 2 V w - k2Gh ot2 (V w) 4
ph3 202w l2D V ot2
ph
+ Ii
3
ph 04w 12k2Gh 8t4
2 8 w D 8t2
+ ph
=0 (14.291)
For harmonic motion at frequency w, w(x,y,t)
= W(x,y)eiW1
(14.292)
(14.293)
14.9 Effects of Rotary Inertia and Shear Deformation
14.9.2
50S
Variational Approach To derive the equations of motion and the associated boundary conditions using the variational approach, we first note that the displacement components are given by Eq. (14.269). The strain-displacement relations are given by Eq. (14.271). The force and moment resultants are given by Eq. (14.273). The strain energy due to the deformation of the Mindlin plate is given by (14.29'+)
where xx ayy a }
a =
c.u Cyy} 8 = Cxy
and
axy
(14.295)
{ CYZ
{ ayz azx
czx
The stress-strain relations of Eq. (14.272) can be expressed in matrix form as (14.296)
~ = [B]8 where (by using the factor k2 for ayz and azx) vE
E
l'-v2
vE
E
I-v22
o
o
[B] =
o o
0 0 (14.297)
GOO
o o o o
o o
0 0 k2G
0
0
k2G
Using Eq. (14.296), Eq. (14.294) can be written as 7r = ~
III
T
(14.298)
8 [B]8dV
v
Substitution of Eq. (14.297) into Eq. (14.298) leads to 7r = -1 2
Iff (
E --2cXX
v
2
I-v
vE + --2Cyy
2
I-v
2 + k 2GcyZ 2 + k 2Gczx 2) + Gcxy
dV
(14.299)
Using Eqs. (14.271), Eq. (14.299) can be rewritten as
I
h/2
7r - ~ - 2
[
dzlf
z=-h/2
+Gz2 (o¢x oy
~Z2
1- v
(o
2
+ ~Z2
1 - v2
(o
2
E o +2v_ _
A
.
+ O¢y)2 + k2G (¢y + ow)2 + k2G (¢x + OW)2] ox
oy
ox
dA (14.300)
506
Transverse Vibration of Plates
=
where dA dx dy denotes the area of a differential element of the plate. By carrying out integration with respect to z and using Eq. (14.274), Eq. (14.300) can be expressed as
7r=-1 2
If I
)2 -2(1-v)
y
D [(O¢x -+-ot/J
ox
oy
(o¢x ---- ot/Jy ox oy
1 4
)2)]
(ot/Jx ot/Jy -+oy ox
A
+k'Gh ((~,
+ ~:
Y + (~y + ~; y)}
dA
(14.301)
The kinetic energy T of the plate can be expressed as (14.302)
where p is the mass density of the plate and t is the time. Substituting Eq. (14.269) into Eq. (14.302), we obtain (14.303)
The work done by the distributed transverse load middle surfaces of the plate) can be expressed as
w=
ff
f(x,y,t)
(load per unit area of the
(14.304)
wf dA
A
The equations of motion can be derived from Hamilton's principle:
f
IZ
o
(T '- 7r
+ W) dt
=0
(14.305)
11
Substituting Eqs, (14,301), (14.303), and (14.304) into Eq, (14.305), we obtain
D(1 - v)
2
+ ( ¢y
(ot/Jx o¢y) [O(O¢x) 2G ~ +~ 0)' + O(0t/Jy)]_k ~ h [( ¢x + ow) ~ (0 ¢x + O(OW)) ~ y uX uX uX uX
aw) (0 O(OW))] hOW o(ow) ph3 [ot/Jx o(o¢x) + oy ¢y + ay +p + 12 7it-a-t-
at at
+ O¢y O(O¢Y)] + fowl ot
at
dAdt = 0
(14.306)
14.9 Effects of Rotary Inertia and Shear Deformation
507
By performing integration by parts. Eq. (14.306) yields
where C denotes the boundary of the plate.
Equations of Motion
otPy. and ow
By equating the coefficients of the various terms involving ocjJx. under the area integral in Eq. (14.307) to zero. we obtain
Equations (14.308)-(14.310) can be rewritten as
508
Transverse Vibration of Plates
D(1- v) \12t/J + D(1 + v) ~ (at/Jx 2 y 2 ay ax
+ at/Jy)
_ k2Gh
ay
2 a t/Jy 12 at2 (14.312)
(t/J + aw) = ph3 y
ay
+ at/Jx + at/Jy) + f = ph
k2Gh(V2W
ax
ay
a2~ at (14.313)
Equations (14.311)-(14.313) denote the equations of motion of the Mindlin plate and can be seen to be same as Eqs. (14.281)-(14.283), derived using the dynamic equilibrium approach. It can be seen that Eqs. (14.311)-(14.313) are coupled in the variables w(x,y,t), t/Jx(x,y,t), and t/Jy(x,y,t). The explicit functions t/Jx and t/Jy can be eliminated from Eqs. (14.311)-(14.313), and a single equation of motion in terms of w can be derived as indicated earlier [see Eqs. (14.281)-(14.286)]: 2 p a ) ( \12_ k2G at2
(
3
2 a w ( D 8t2 = 1- k2Gh
2
2 ph a ) DV -12 at2 w+ph
2
\12
ph
+ 12k2G
2 a ) at2
f
(14.314) Note that in the equations above, the terms containing k2G denote the effect of shear deformation, and the terms containing (ph3 /12)(a2 /at2) represent the effect of rotary inertia. General Boundary Conditions The boundary conditions can be identified by setting the line integral in Eq. (14.307) equal to zero:
f [( D
c
-
at/Jx y -+vax
at/J ) ot/Jxdy-D
(at/J at/Jx) -y +vot/Jvdx ay ax .
ay
D(1 - v) (at/Jx at/Jy) ~ -+-ot/Jxx 2 ay ax
+ D(1-2 v) (at/Jx ay -k2Gh
+ at/Jy) ax
d
oA. d +k2Gh 'l'y. Y
(t/Jy + ~;) ow dx ]
=0
(A. 'l'x
+ aw) ax
owd
y (14.315)
Using the expressions for the force and moment resultants given in Eq. (14.273), Eq. (14.315) can be expressed as
f c
(Mxot/Jx dy - Myot/Jy dx - Mxyot/Jx dx
+ Mxyot/J)'dy + Qxow dy - Qyow dx) = 0 (14.316)
Boundary Conditions on an Inclined Boundary Consider an arbitrary boundary of the plate whose normal and tangential directions are denoted n and s, respectively,
14.9
Effect~ of Rotary lnertia and Shear Deformation
509
s
Boundary, C n
x
n
(a)
(b)
n (c)
Figure 14.12 (a) coordinate systems on a boundary; (b) bending rotations at a boundary; (c) shear force and moment resultants on an inclined boundary.
as shown in Fig. 14.12(a). The (x,y) coordinates boundary are related as
and the (n,s) coordinates
on the
dx = -sint9ds
(14.317)
= cost9ds
(14.318)
dy
If
(14.319)
(14.320)
510
Transverse Vibration of Plates
The bending moment resultant normal to the boundary, Mn, and the twisting moment resultant at the boundary, Mns, can be expressed in terms of Mx, My, and Mxy as [see Fig. 14.12(c)] Mn = Mx cos2 0
+ My
Mns = (My - Mx)
+ 2Mxy cos 0 sin 0 cosO sinO + MXy(cos20 - sin20) sin2 0
Similarly, the shear force resultant acting on the boundary in the be expressed in terms of Qx and Qy as Qn = Qx cos 0
+ Qy
(14.321) (14.322)
z
direction, Qn, can
sin 0
(14.323)
Using Eqs. (14.317) and (14.318), Eq. (14.316) can be expressed as
f
(Mx cos 00
c
+ Qx
+ Qy
cosOow
sinOow)ds
(14.324)
=0
Substitution of Eqs. (14.319) and (14.320) into Eq. (14.324) results in
f
{(Mx cos2 0
+ My
sin20
+ 2Mxy
sinO cosO)o
c [(-Mx
cosO sinO + My cosO sinO)
+ (QxcosO + QysinO)ow}ds
+ MXy(cos2
0 - sin20)]o
=0
(14.325)
In'view of Eqs. (14.321), (14.322), and (14.323), Eq. (14.325) can be written as
f
(Mn0
+ Mnso
ds = 0
(14.326)
c Equation (14.326) will be satisfied when each of the terms under the integral is equal to zero. This implies that the following conditions are to be satisfied along the boundary of the plate: (14.327) Mn =0
or o
(
(14.328) (14.329)
Mns = 0 or
o
(
(14.330) (14.331)
Qn =0 or
ow =0
(w is specified)
(14.332)
Thus, three boundary conditions are to be specified along a boundary or edge of the plate in Mindlin theory. The boundary conditions corresponding to some common
14.9 support conditions
Effects of Rotary Inertia and Shear Deformation -$11
are as follows:
1. Fixed or clamped boundary:
= 0,
w =0
(14.333)
w=o
(14.334)
or hinged boundary: Mn =0,
= 0,
The support conditions of Eq. (14.334) are sometimes called hard-type simple support conditions. In some cases, the following conditions, known as sofHype simple support conditions, are used [7]:
w=o
Mns = 0,
Mn =0,
(14.335)
3. Free boundary:
14.9.3
(14.336)
Mns = 0,
Mn =0,
Free Vibration Solution For free vibration, f(x,y,t) is set equal to zero and the variables
=
=
w(x,y,t) so that Eqs. (14.311)-(14.313) D [(12
v)V2
D [(1 _ v)V2
2
(14.337)
= W(x,y)eiwt
become 3 2
+ (1 + v)~
- k2Gh (
+ (1 + V)~
_ k2Gh (
ox
+ 0'1W) + ph w
=0
(14.338)
=0
(14.339)
+
(14.340)
12
uX
3 2
oy
+ 0'1W) + ph w
k2Gh(V2W
uy
12
where o
ax
oy
(14.341)
To find the solution of Eqs. (14.338)-(14.340),
y -
+ -oH
oy
o1fr _ oH oy ox
(14.342) (14.343)
512
Transverse Vibration of Plates
where 1/1" (x, y) and H (x , y) can be shown to correspond to the dilatational and shear components of motion of the plate, respectively. By substituting Eqs. (14.342) and (14.343) into Eqs. (14.338)-(14.340), we obtain
-
o
-
o
ox
oy
(4 + (4
1) 1)
11-
W]
v 0 + ---('\7 2 oy
2
S
2 + 03)H
=0
(14.344)
W]
- ---('\7 v
0
2
2 + 03)H
=0
(14.345)
+ SktW
=0
(14.346)
[2'\7 1/1"+
Rkb - S
1/1"- -
[2'\7 1/1"
Rkb - S
1/1"- -
S
2
ox
'\72(1/1"+ W)
where
I
=h=
R
h2 12
(14.347)
represents the effect of rotary inertia and is equal to the square of the radius of gyration of the cross-sectional area of a unit width of plate, D S = k2Gh
(14.348)
denotes the effect of shear deformation, 4 _
kb
-
phw2
J)
(14.349)
indicates the frequency parameter in classical plate theory, and 02
_
3-
2(Rkt - 1/ S) I-v
(14.350)
To uncouple Eqs. (14.344)-(14.346), first differentiate Eqs. (14.344) and (14.345) with respect to x and y, respectively, and add the resulting equations to obtain 2 '\72 ['\7 1/1"+
( Rk;
- ~) 1/1"- ;]
=0
(14.351)
Next differentiate Eqs. (14.344) and (14.345) with respect to y and x, respectively, and subtract the resulting equations to obtain '\72('\72
+ o~)H
=0
(14.352)
Equation (14.346) gives '\721/1"= - '\72 W - SktW
By substituting Eq. (14.353) into Eq. (14.351), the resulting expressed as
(14.353)
equation can be (14.354)
where (14.355)
,...,-..;,""IZ",~-"",:".~,."~,_._!, .~ .••.
14.9
Effects of Rotary Inertia and Shear Deformation
'::513
Thus, the solution of Eq. (14.354) is given by W = WI
+ Wz
(14.356)
where Wj satisfies the equation (Vz
+ 8]) Wj = 0,
j
=
(14.357)
1, 2
It can be verified that Eqs. (14.346) and (14.351) are satisfied by
1/1 = (tL
(14.358)
- I)W
where tL is a constant. Substitution of Eq. (14.358) into Eqs. (14.346) and (14.351) yields two values for tL as 8~.1 = Rkt _ 1/ S
tLI.Z
(14.359)
Using Eqs. (14.356) and (14.358), the tenus in brackets on the left-hand sides of Eqs. (14.344) and (14.345) can be seen to be zero, so that the equation governing H becomes [from Eqs. (14.344), (14.345), and (14.352)] ('V'z
+ 8~)H
=0
(14.360)
Thus, the solution of Eqs. (14.338)- (14.340) can be written as q,x
= (tLl
-
q,y
= (tLl
-
W = WI
aWl
"
1)-
awz
+ (tLz -
ax
1)-
. aWl
ax
aH
+ -ay
awz
l)ay + (tLz -l)ay -
aH a;
+ Wz
(14.361) (14.362) (14.363)
where WI. Wz, and H are governed by Eqs. (14.357) and (14.360). 14.9.4
Plate Simply Supported on All Four Edges Let the origin of the coordinate system be located at the center of the plate as shown in Fig. 14.13 so that the boundary conditions can be stated as [20] a (14.364) W Mx q,y 0 at x ±W
=
=
=
= My
= q,x
=0
=
at
y
2
b
= ±2
(14.365)
The solutions of Eqs. (14.357) and (14.360) are assumed as WI (x,y) = CI sinalx sin.8IY
(14.366)
Wz(x,y) = Cz sinazx sin.8zy
(14.367)
= C3 cosa3x cos .83Y
H(x,y)
(14.368)
with the conditions a~ + .8~= 8~ I
I
I'
i
= I, 2, 3
(14.369)
.j.,.,
514
Transverse Vibration of Plates y A
I
/ ..•
T
a
~I
2"
T b
.
I .
oL._._._._
b
1 I""
J.._x
~I
a
Figure 14.13
-2
Rectangular plate simply supported on all edges.
By substituting Eqs. (14.366)-(14.368) into (14.361)-(14.363),
we obtain
(14.370) + C3a3 sina3x cosf33Y W = Cl sinalx sinf3lY
(14.371)
+ C2 sina2x
sinf32Y
(14.372)
Using Eqs. (14.370) and (14.371), the bending moments Mx and My [see Eqs. (14.273)] can be expressed as Mx = D[-Cl(J.tl
- l)(af
- C2(J.t2 - l)(al
+ vf3f)
+ vf3i)
sinalx sinf3lY
sin a2X sin f32y+C3a3f33(1
- v) sina3x sin f33Y] (14.373)
My = D[-Cl(J.tl
- l)(f3f
- C2(J.t2 - 1)(f3i
+ vaf)
+ vai)
sinalx sinf3lY
sina2x sin f32Y - C3a3f33(1 - v) sina3x sin f33Y]
In view of Eqs. (14.370)-(14.374), (14.365) yield
(14.374) the boundary conditions of Eqs. (14.364) and
j = 1,2,3 j
= 1, 2, 3
(14.375) (14.376)
where rj and Sj are even integers. Thus, Eqs. (14.366) and (14.367) denote modes or deflections odd in both x and )'. The function H does not represent any mode or deflection but produces rotations with the same type of symmetry obtained from WI and W2• Hence, all the modes or solutions given by Eqs. (14.366)-(14.368) will be
_________
14.9
Effects of Rotary Inertia and Shear Deformation''51S
odd in both x and y. If we want the modes even in x, we need to interchange sinajx and cos a j x in all the three equations (14.366)-(14.368) and r j are to be taken as odd integers. Similarly, for modes even in y, we need to interchange the terms sin P j y and cos P j y, with sj taken as odd integers. By substituting Eqs. (14.375) and (14.376) into Eqs. (14.369) and using the definitions of OJ, j = 1,2,3 [Eqs. (14.355) and (14.350)], we obtain
C~f
= ~ {I
+ I:
v
[1 + 2
V)] 1/IJ + (-l)jQj},
k (1 2
j
2 (~ ) = I + 1/If
= 1,2 (14.377)
where (14.378)
j = 1, 2, 3
Qj
I
= [ + 1
2 1 _ v (1
k 2 (1 - v) + 2 )
2
2
1/Ij] -
jl/2 8k 2 4 j I - v 1/I
j = 1,2,3
,
(14.379)
For any specific ratios a/ b and a/ h and the mode number j, the three frequencies W j, j = I, 2, 3, satisfy the inequaIities WI < W3 < C1J2, with Wz and W3 being much larger than WI except for very thick plates. The mode corresponding to WI is associated with WI and is most closely related to the one given by the classical plate theory. The mode associated with C1J2corresponds to W2 and denotes the thickness shear deformation mode. The mode shape associated with W3 contains no transverse deflection except for two components of rotation which are related to one another so as to denote twist about an axis normal to the plate. The types of modes or deflections generated by WI, W2, and H are called flexural, thickness-shear, and thickness-twist modes, respectively. The mode shapes corresponding to the three frequency ratios given by Eq. (14.377) are shown in Fig. 14.14.
14.9.5 Circular Plates Equations of Motion
Considering the dynamic equilibrium of an infinitesimal element shown in Fig. 14.6, the equations of motion of a circular plate, can be derived in terms of the force and moment resultants as
8Mr 8r
+ ~aMr(} + Mr r 80 aMr
1 8MIJ
3
ar
~!!!!!!!!!I!!!!!
Q _ ph3 a2
2M
a;- + -;:ae + -; 8Qr
2
- M(} _ Qr = ph a
(} -
12 at2
+ ~8Q(} + Qr = ph r 80
r
(14.380) (14.381)
2
a
w
at2
(14.382)
~~~!!!!!!!!!!!!!!!!!!!!!!!!!!!!!J
516
Transverse Vibration of Plates
r x
z
./ ./ ./
y
~:
./
(c)
Figure 14.14 Modes of a rectangular plate: (a) flexural mode; (b) thickness-shear mode; (c) thickness-twist mode. (From Ref. 20; reprinted with pennission from the publisher.)
where the displacement components have been assumed to be of the fonn
e, t) vCr, e, t) w(r, e, t) u(r,
e, t) = z¢o(r, e, t) = w(r, e, t) = Z¢r(r,
(14.383) (14.384) (14.385)
·517
14.9 Effects of Rotary Inertia;and'Shear 0e1'o'rmatioo
The moment and force resultants can be expressed in terms of
M
r
q,r, q,e and w as
= D [a
Me
=D
Mre
=
U (q,r + aa~)
(14.386)
+ v aa~ ]
D (1 _ v .[~ (aq,r _
(14.388)
ar
ae
r
Qr = k2Gh (r+ ~~) Qe
= k2Gh
(14.389)
The equations of motion, Eqs. (14.380)-(14.382),
D[ 2" D [
2"
(1 - v)V
(1- v)V
2
2
(14.390)
(e+ ~~;)
act>] q,r + (1 + v) a;-
act>] q,e + (1 + v)ae-
-
can be written as 3
2 ( aw) k Gh
3
2 ( 1 aw) k Gh q,e +;-ae
-
k2Gh(V2w
2
ph a q,r = 12 at2
2
a q,e = ph 12at2 a2w
+ <1»
= ph
(14.391)
at2
(14.392) (14.393)
where (14.394) By assuming harmonic solution as (14.395)
t)
=
e, t)
= W(r, e)eiwt
(14.397)
q,r(r,
e,
t)
q,e(r,
e,
w(r,
(14.396)
Eqs. (14.391)-(14.393) can be expressed as
D [ (1 2"
3 2 2 a] (aw) h w 2 v)V <1>r+ (1 + v) a;- - k Gh ct>r + a;:- + ~r
=0 (14.398)
-D
2
[2(1-
v)'V
ae
-
2 (
k Gh
r
3
2
ph w ae + --e 12
=0 (14.399)
k2Gh(V2W
+
=0 (14.400)
....
518
Transverse Vibration of Plates
where a
(14.401)
Free Vibration Solution Using a procedure similar to that of a rectangular plate, the solution of Eqs. (14.398)-(14.400) is expressed as W (r, e) = WI (r, e)
+ W2(r, e)
aWl
+ (J.t2
1 aWl
(14.402) aW2 - 1)ar
+ (J.t2
1 aH + --
r ae
1 aW2 aH - 1)-- - r ae ar
(14.403) (14.404)
where WI, W2 and H are solutions of the equations
+ 0;) Wj ('y2 + o~)H
('y2
= 0,
j = 1,2
(14.405)
=0
(14.406)
with (14.407)
0;,
j = 1, 2, 3, given by Eqs. (14.355) and (14.350), and J.tl and J.t2 by Eq. (14.359). Note that if the effects of rotary inertia and shear deflection are neglected, Eqs. (14.405) and (14.406) reduce to (\72
+ 82) WI
(\72 -
02)W2
=0 =0
(14.408) (14.409)
where 2
2
o =0
1
IR=S=O
2) 1/2
2 2 phw = -02IR=S=O = kb = ( -V
(14.410)
The solution of Eqs. (14.405) and (14.406) are expressed in product form as
= Rj(r)8j(e), e) = R3(r)83(e)
Wj(r, e) H(r,
j = 1,2
(14.411) (14.412)
to obtain the following pairs of ordinary differential equations: (14.413) (14.414)
14.9
E[fec{S~f-R~,ary1nertiaand Shear DeformatioR_,s19
where m2 is the separation constant. Noting that Eq. (14.413) is Bessel's equation. the solutions, Wj (r, ()) and H (r, ()), can be expressed as 00
Wj(r, ()) =
L Wjm(r,
differential
00
L [Ajm) Jm(ojr)
()) =
m=O
+ Bt)Ym(ojr)]cosm(),
j=1.2
m=O (14.415)
00
H(r, ()) =
00
L Hm(r, ()) = L [A~m)Jm(03r)
+ Bjm)Ym(03r)]
(14.416)
sinm()
m=O
m=O
where A jml and Bt), j = 1,2,3. are constants, Jm and Ym are Bessel functions of the first and second kind, respectively, of order m, and m corresponds to the number of nodal diameters. In terms of the solutions given by Eqs. (14.415) and (14.416), the moments Mr and Mr9 and the shear force Qr (for any particular value of m) can be expressed as
M)"'(r, e) = D
[~A}.' I
(U; - 1) [J~ (I,r)
+ ~2
Bj(m) { (O"j"'7,l) [ Y~(Ojr)
+ A~m) {(1-
+8j"'
1(1-
v) v)
[7 [7
+ ;J~(I;r)
v~'J. (l,r)]I CDS me 2
v . + ;Y~(Ojr)
vm - -,:r-Ym(Ojr)
] } cosm()
J~(03r) - ~ Jm(03r)]} cosm()
)]I me]
y~(13r) - ~ Y. (13r
M:;' (r. e) = D(I - v) {~A}.'
-
[
-7
(14.417)
CDS
J~(I,r)+ ~ J.(I,r)]
1)
(UI -
2
+
L Bj(m) [-7 Y~(Ojr)
+ ~ Ym(Ojr)]
(O"j-1)
i=1
+ A3(m)
[
(m) [.
+B3
1 (03r) /1 --Jm 2
m 1 ] + -J1m(03r)
- -2
1 " -2Ym(03r)
+ 2r1 Ym(03r)
- 2r2 Ym(03r)
Q$·'(r, e) = k'Gh { ~[A}.'
+A~m) m Jm(03r) r
2r
r
+ Bjm)m
r
Jm(03r)
2
m
I
U, J~ (I;r)
2
]}
.
smm()
(14.418)
+ 81"' U; Y~(I,r)]
)j·
Ym(03r
cosm()
(14.419)
520
Transverse Vibration of Plates
The transverse deflection of the plate can be expressed as
(14.420) The boundary conditions can be stated as follows: 1. Clamped or fixed edge: (14.421) 2. Simply supported edge (hard type): W
=
(14.422)
3. Simply supported edge (soft type): W
= Mr = Mre = 0
(14.423)
4. Free edge: (14.424)
14.9.6
Natural Frequencies of a Clamped Circular Plate For a solid circular plate, the constants Bt), j = 1, 2, 3, are set equal to zero in Eqs. (14.415) and (14.416) so as to avoid infinite displacements, slopes, and bending moments at r = O. The natural frequencies can be determined by substituting Eqs. (14.402)-(14.404), (14.415), and (14.416) into the appropriate equation in Eqs. (14.421)-(14.424). If the plate is clamped at the outer boundary, r = a, Eqs. (14.421) lead to the frequency equation in the form of a determinantal equation:
CII
C12 C13
C21 C22 C23
=0
(14.425)
C31 C32 C33
where C1j = (J..Lj- l)J~(oj a), j = 1,2 C2j = m(J..Lj - l)Jm(oj a), j = 1, 2 C3j Jm(oja), j = 1, 2 CI3 = m Jm(03a) C23 = J~(o3a)
=
C33
(14.426)
=0
Some of the natural frequencies given by Eq. (14.425) are shown in Table 14.3.
PI:a.e,~iift
14.10
Table 14.3
.£la:;{ic-Foundau{lu
521
Natural Frequencies of a Clamped Circular Platea h/a Number of nodal circles, n
Number of nodal diameters, m
I
0
2 I
2 0
2
I
0.05
0.10
0.25
10.145 38.855 21.002 58.827 34.258 80.933
9.941 36.479 20.232 53.890 32.406 72.368
8.807 27.253 16.521 37.550 24.670 47.650
Thin plate theory 10.216 39.771 21.260 60.829 34.877 84.583
So~ce:Re([71.~.~ _ aValues of a2
14.10
J phw2 / D.
v = 0.3, k2 =]'(2/12.
PLATE ON AN ELASTIC FOUNDATION The problem of vibration of a plate on an elastic foundation finds application in several practical situations, such as reinforced concrete pavements of highways and airport runways and foundation slabs of heavy machines and buildings. By assuming a Winkler foundation where the reaction force of the foundation is assumed to be proportional to the deflection, we can express the reaction force, R(x,y,t), as (Fig. 14.15): (14.427)
R = kw(x,y,t)
where k, a constant known as thefoundation modulus, can be interpreted as the reaction force of the foundation per unit surface area of the plate per unit deflection of the plate. Since the reaction force acts in a direction opposite to that of the external force f(x,y,t), the equation of motion governing the vibration of a thin plate resting on an elastic foundation can be expressed as 4 4 a4w 1 [ a2w(x,y,t)] -aa xw4 + 2 axa2awy 2 + -a 4 =f(x,y,t) - kw(x,y,t) - ph at2 (14.428) y D For free vibration,
f (x, y, t)
= 0 and the solution is assumed to be harmonic: (14.429) W(X,y,t) = W(x,y) eiwt
so that Eq. (14.428) becomes a4w
a4w
phw2 - k
o4w
-OX4 + 2-+ -i1y4 = --- D ax2oy2
W
=0
(14.430)
Defining -4
A
phw2
= ----
-
k
D
(14.431)
Eq. (14.430) can be express~d as V4W -
X4W = 0
(14.432)
522
Transverse Vibration of Plates ft.x,y,t) I I I
dy I I I I I
------------
dx
----------1-
I
I
(a)
R =kw(x,)',t) (b)
Figure 14.15
Plate on an elastic foundation.
Equation (14.432) can be seen to be similar to Eq. (14.91) and hence the solutions obtained earlier (Section 14.4) are applicable to plates on elastic foundation provided that X is used in place of A. For example, for a plate on elastic foundation, simply supported on all the edges, the natural frequencies of vibration and the natural modes can be determined using Eqs. (14.105) and (14.107) as
(14.433) or (14.434) and (14.435)
14.11 Transverse Vibration of Plates Subjected to In-Plane Loads
523
where Wmn(x,y)
. mrrx = sm-a
nrry sin-b
(14.436)
14.11 TRANSVERSE VIBRATION OF PLATES SUBJECTED TO IN-PLANE LOADS 14.11.1 Equation of Motion To derive the equation of motion governing the transverse vibration of a plate subjected to in-plane loads (see Fig. 14.16) using the equilibrium approach, consider an infinitesimal plate element of sides dx and dy. Let the element be subjected to the time-independent (static) in-plane or membrane loads Nx(x,y), Ny(x,y), and Nxy(x,y) = Nyx(x,y) per unit length as well as a time-dependent transverse load f(x,y,t) per unit area. The in-plane loads acting on the various sides of the element are shown in Fig. 14.17(a), and the transverse loads and moments acting on the sides of the element are shown in Fig. 14.17(b). The transverse deformation, along with the forces acting on the element, as seen in the xz and yz planes are shown in Fig. 14.17(c) and (d), respectively. The force equilibrium equation 'in the x direction gives [from Fig. 14.17(c)]
'" Fx = ( Nx + -aNx) dx .t-
ax
dy cosel ,( + Nyx
'.
aNyx) + -ay
dy
d x cos --el + e~
el +e'I = 0
- Nx dy cosel - Nyx dx cos __
2
where
Figure 14.16 Rectangularplate subjectedto in-planeloads.
2
(14.437)
524
Transverse Vibration of Plates
aMx dx M x+-
ax
(a)
(b)
Nx
aNx
Nx+-
ax
l--dx I
aNy dy Ny+iJy dx
rdy
I
-,
I
Qx
: I
aQx
Q, + -
ax
dx
-'-, NaN\, ,,+_U) . ay Q aQy d \'+Y . iJy
(c)
(d)
Figure 14.17 (a) In-plane loads acting on the sides of an element of a plate; (b) moment and shear force resultants acting on the sides on an element of the plate; (c) deformation in the xz plane; (d) deformation in the yz plane.
Since the deflections are assumed to be smail, 01 will be smail. so that COSO}~),
01 +0' cos -2-1
cosO; ~ 1,
~
1
(14.438)
In view of Eg. (14.438) and the fact that Nyx = Nxy, Eq. (14.437) can be simplified as aNx
aNy
-ax + -'ay
=0
(14.439)
14.11 Transverse Vibration of Plates Subjected to·In-Plane Loads
525
In a similar manner, the force equilibrium equation in the y direction yields [from Fig. 14.17(d)]: xy a:;-
--ay
"" Fy = ( Ny + aNy) dy
dx cos82/ + ( Nxy + aN ) dx
82 +8/
dy cos--2
82 + 8~ - Ny dx COS82- Nxy dx cos --=0 2
2 (14.440)
where 8~ = 82 + dBz = 82
a82
+ -ay
dy
Since the deflections are assumed to be small, 82 and 8~ will also be small, so that 82 + 8~ cos--~l 2
cos8~~1,
cosBz~l,
(14.441)
In view of Eq. (14.441), Eq. (~4.440) reduces to aNy
ay
+ aNxy
=0
(14.442)
ax
The force equilibrium equation in the z direction can be obtained by considering the projections of all the in-plane and transverse forces as follows [from Fig. 14.17(b), (c), and (d)]:
+ ( Nyx
+
aNyx) ay
+ ( Nxy
+
aNxy) a:;dx
+ (aQx + ~Qx
+
(Qy
+ aa~y
dy
dx
)
dY)
-/
dx sin81
-
-
Nyx dx sin81
-/
-
dy sin82 - Nxy dy sin82
dy - Qx dy
dx
+f
dx dy = 0
(14.443)
526
Transverse Vibration of Plates
By using the assumption of small displacements and slopes, we can express
aw ax aw
sin 61 ~ 61 ~ -, sin 62 ~ 62 ~
ay' (14.444)
The first two terms in Eg. (14.443) can be rewritten as ( Nx
aNx) + --a.;dx
.. dysm6f
- Nxdysm61
~ ( Nx
x + aN ax )dx
dy (aw ax
2
+ aaxw2 dx )
aw
- Nx dy ax a2w ~Nx-dxdy+ax2
aNx aw -dxdy ax ax (14.445)
by neglecting the higher-order term. Similarly, the next pair of terms in Eg. (14.443) can be rewritten as Ny (aN
+
. 2 ~ ( Ny a: dy ) dx sin6~ - Nydx Sm6
2
ay + aay2w dy )
+ aN a:
dy ) dx (aw
aw ay
- Ny dx -
~N
a2w -dxdy+y ay2
aNy aw -dxdy ay ay (14.446)
The components of the in-plane shear forces Nxy and (aNxyjax) dx, acting on the x edges (i.e ., edges whose normals lie parallel to the x axis) in the z direction can be expressed as
aNxy) ( Nxy + a:;-
dx
-I
dy sin62-Nx)'
_
(14.447)
dy sin62
where the slopes of the deflection surface in the y direction on the x edges, are given by Eg. (14.444): -
-I
-
62 ~ 62 +
a62
-ax
aw dx ~ ay
2
a w
+ -ax ay
dx
e2 and e;,
14.11
527
Transverse Vibration of Plates SubjeCted~·Ifl-P!ane ~s
Thus, Eq. (14.447) yields, by neglecting the higher-order term, a2w xy + N ax ay dx dy
aNxv aw ay dx dy
a?""
(14.448)
Similarly, the component of the in-plane shear forces Nyx and (aNyx/ay) dy, acting on the y edges (i.e ., edges whose normals lie parallel to the y axis) in the z direction can be written as
+ a~;x
(Nyx
dx sine~ - Nyx dx sinel
dY)
(14.449)
where the slopes of the deflection surface in the x direction on the y edges, eland are given in Eq. (14.444): _ aw 111 ~ 111 ~ -, ax
-I
-
111
111
~
ael
aw dy ~ ax
+ -ay
a2w
+ -ax ay
e~,
dy
Thus, the expression (14.449) can be simplified as a2w N x -dx dy y ax ay
aNyx aw
+ --ay -ax
(14.450)
dx dy
Thus, Eq. (14.443) can be expressed as a2w a2w xy v . -ax 2 +N . -ay 2 +2N
a Qx a Qy -a-+-a-+ f +Nx x y
+ (aNx -+-ax
aNyx) ay
3w --+ ax
(aNxy --+ax
aNy) ay
a2w -axya aw ay
-=
0
(14.451)
The expressions within the parentheses of Eq.(14.451) are zero in view of Eqs. (14.439) and (14.442) and hence Eq. (14.451) reduces to aQx
aQy
a.;- + ay + f
a2w + Ny ay2
a2w +Nx ax2
a2w xy +2N ax ay = 0
(14.452)
The in-plane forces do not contribute to any moment along the edges of the element. As such, the moment equilibrium equations about x and y axes lead to Eqs. (3.29) and (3.30). By substituting the shear force resultants in terms of the displacement w [Eqs. (14.5) and (14.6)], Eq. (14.452) yields a4w ax4
a4w +2 ax2ay2
4 a w 1 ( + ay4 = D
f
2 a w + Nx ax2
2 a w +Ny ay2
2 a w ) xy +2N ax ay
(14.453)
Finally, by adding the inertia force, the total external force is given by
f -
ph
a2w at
(14.454)
2
and the equation of motion for the vibration of a plate subjected to combined in-plane and transverse loads becomes a4w ax4
a4w +2 ax2ay2
a4w 1 ( + ay4 = D
f -
2 a w ph at2
2 a w x +N ax2
2 a w + Ny ay2
2 a w ) xy + 2N ax ay
(14.455)
528
Transverse Vibration of Plates
14.11.2
Free Vibration For free vibration, f is set equal to zero in Eq. (14.455) and the solution of the resulting equation is assumed to be harmonic with frequency w: = W(x,y)eiwt
w(x,y,t)
Substitution of Eq. (14.456) into Eq. (14.455) with 4
w
a ax4
4
a
4
w
+ 2 ax2ay2 +
w
a 1 ( ay4 = D ph{J}W
+ Nx
(14.456)
f = 0 leads to
w
a2 ax2
+ Ny
w
a2 ay2
+ 2Nxy
a2W) ax ay
(14.457) Introducing phw2
4
)..=--
(14.458)
D
Eq. (14.457) can be rewritten as V4W_)..4W=_ 14.11.3
2 1 (aNW X-2ax D
2 a +Ny- -+2N -2 ay xy
2 a W) ax ay
w
(14.459)
Solution for a Simply Supported Plate We consider the free vibration of a plate simply supported on all edges subjected to the in-plane forces Nx = N1, Ny = N2 and Nxy = 0, where N1 and N2 are constants. For this case, the equation of motion, Eq. (14.459), becomes
(a W D
V4W_)..4W=_
2
I
CPW)
N1--+N2-ax2
(14.460)
ay2
As in the case of free vibration of a rectangular plate with no in-plane forces, the following solution can be seen to satisfy the boundary conditions of the plate: 00
L
W(x, y) =
Amn
. m1rX sin-'n1rY
sm --
m,n=1
a
(14.461)
b
where Amn are constants. By substituting Eq. (14.461) into (14.460), we can obtain the frequency equation as [(m1r)2 -a
1 [ N1 + (n7r -b )2]2 + -Dab
D =-
[(mrr)2 I -- +(nrr -b )2]2 +a ph
(m1r)2
+ N2
(n1r
(mrr)2 a
+N2
-
-
)2]
-l
phw;'n =).. =-D
(14.462)
or w
2 mn
ph
[N
1
-
(nrr _ )2] . b
m, n
= 1,2,
'"
(14.463) If N1 and N2 are compressive, Eq. (14.463) can be written as m, n
= 1,2,
'"
(14.464)
14.12 Vibration of Plates with VariableThickness '·529 It can be seen that Wmn reduces to zero as the magnitude of Nt and/or N2 increases. For example. when N2 O. the value of Nt that makes Wmn O. called the critical or buckling load. can be determined from Eq. (14.464) as
=
(Nt)cri
=
a -D (mrr:)
=
2[
mrr: (~)
2
nrr:
+ (b)
2]2
Drr:2
=
-7
[
n m +n (;)
a
C;)
2]2 (14.465)
where the negative sign represents a compressive load. It is to be noted from Eq. (14"+64) that the fundamental natural frequency may not correspond to m 1 and n I but depends on the values of Nt. N2. and a/b. Using Eq. (14.465). the frequency given by Eq. (14.463) can be expressed for N2 = 0 as
=
=
(14.466)
It can be shown that the fundamental frequency in this case always occurs when n = 1 and not necessarily when m = 1 [17]. By introducing (14.467)
where Wref denotes the fundamental natural frequency of vibration of a square plate with no in-plane loads. and the square of the frequency ratio. (Wmn/Wref)2. can be written as
Using Eq. (14.465). Eq. (14.468) can be rewritten as
( - )2
4[
Wmn
mat --1+ -
Wref
-
4
(b)
2]2 [N---1
]
(14.469)
(NI)cri
The variation of the square of the frequency ratio (Wmn/Wref)2 with the load ratio NI!(Nt)cri is shown in Fig. 14.18 for different values of m and a/b (with n 1).
=
14.12 14.12.1
VmRATION OF PLATES WITH VARIABLE TIDCKNESS Rectangular Plates Let the thickness of the plate vary continuously with no abrupt changes so that it can be represented as h = hex. y). In this case the expressions given by Eqs. (14.1)-(14.3) for the moment resultants can be used with sufficient accuracy. However, the shear
530
Transverse Vibration of Plates 5 I I I
,
a
1i=4
,
I
I
---------~--------,
4 ---------~--_----I
m=21
,
I
I I
I I ,
3 ",11
---------~----------f--------m=l
I
~
'
13
I
E
I L
I
:
_
m=3
I
, 2
,
I
---------~----------+-----, , I I , I I
I I I I ,
'
I
I
--------~----------+----------~----
a
Ii=l
m= I
'
::
,,
o 0.25
0.50
0.75
1.00
Fraction of critical load, N1/(Nj)crj
=
Figure 14.18 Effect of the in-plane force Nt Nx on the fundamental frequency of a rectangular plate simply supported on all sides. (From Ref. [J 7].).
force resultants given by Eqs. (14.5) and (14.6) will be modified as 2 2 2 a a [ -(1-v)D-- a Qx=- a [ -D (a-+"v+ax ax2 ay2 ay ax ay 2 2 2 a (a a aD (a a2 = -D ax ax2 + v ay2 ax2 + v ay2
w""
w
w)]
w)
w]
a:; w
w)
a3w aD a3w - (1- v)-ax ay2 ay ax ay2 2 2 2 a a [ -(1-v)D-- a Qy=- a [ -D (a-+v+2 ax ay2 ax ax ax ay 2 2 2 a (a a aD (a a2 -D ay ay2 + v ax2 ay2 + v ax2 - (1- v)D--
w
=
w
w)]
w)
(14.470)
w]
ay
w
w)
a3w aD a2w ' - (1- v)--2 ax ay ax ax ay
- (1- v)D--
(14.471)
where aD = ~ [ Eh3(x,y) ] = Eh2(x,y) ah(x,y) ax ax 12(1 - v2) 4(1 - v2) ax aD = ~ [ Eh3(x,y) ] = Eh2(x,y) ah(x,y) ay ay 12(1 - v2) 4(1 - v2) ay
(14.472) (14.473)
14.12 Vibration of Plates':witb ·Variable-Thickness
531
Finally, by substituting Eqs. (14.470) and (14.471) into Eq. (14.7) yields the equation of motion of a plate with variable thickness as 2 2 2 2 04W a4w a4w) aD 0 (a w a w) oD 0 (a w a w) D ( -+2 ++2---++2--+ax4 ox2ay2 ay4 ax ax ox2 ay2 ay oy ax2 ay2 2
2 2 (a w a w) -+ax2 ay2 a2D a2w a2D a2w -- 2-- -( ax2 ay2 ax ay ax ay
D a2D) +(a-+ax2 ay2 - (1- v)
2
a D + -ay2
a2w) ay2
+ ph(x,y)-,
a2w at~
= f(x,y,t) (14.474)
Closed-form solutions of Eq. (14.474) are possible only for very simple forms of vari~ ation of thickness of the plate, h(x,y). In general, the solution of Eq. (14.474) can be found using either approximate analytical methods such as the Rayleigh-Ritz and Galerkin methods or numerical methods such as the finite element method.
14.12.2
Circular Plates As in the case of rectangular plates, the variable thickness of the plate in the case of a circular plate is denoted as h = h(r, e) and the bending rigidity as .
Eh3(r, e) ..D = 12(1 _ v2)
(14.475)
By assuming the expressions of moment resultants given by Eqs. (14.184)-(14.186) to be applicable with sufficient accuracy, the shear force resultants given by Eqs. (14.187a) and (l4.188a) will be modified by treating D as a function of r and e. The resulting expressions when substituted into Eq.(l4.170) lead to the equation of motion for the vibration of a circular plate of variable thickness. The resulting equation will be quite lengthy and the closed-form solutions will be almost impossible for a general variation of the thickness of the plate. However, if the plate is assumed to be axisymmetric, the thickness of the plate will vary with r only, so that h = h(r). Although the equation of motion can be derived as in the case of a rectangular plate with variable thickness, the equilibrium approach is used directly in this section. For this, consider the free-body diagram of an element of the circular plate of variable thickness, along with internal force and moment resultants and external force, as shown in Fig. 14.19(a). Note that the shear force resultant Qe and the twisting moment resultant M,e are not indicated in Fig. 14.19(a), as they will be zero, due to symmetry. The equilibrium of forces in the z direction gives (Q,
+a Q,
or
dr) (r
+dr)
de - Q,r de
+fr
2
dedr = ph
a w2 at
(14.476)
Noting that the component of Me in the tangential direction is given by [from Fig. 14.19(b)] . dB
. de
2
2
Mesm- +Mesm-;:::::
d Me e
(14.477)
532
Transverse Vibration of Plates
(a)
(b)
Figure 14.19
Circular plate element.
for small angles de, the moment equilibrium about the e direction leads to
aMr) dr ( Mr + a;:-
(r
+ dr)
de - M~~ de - Qrr de dr - Me dr de = 0
(14.478)
By dividing Eqs. (14.476) and (14.478) by dr de and neglecting small quantities of higher order, we obtain 1 a --;:ar(rQr)+ph
a2w(r,
t)
8t2
aM 1 -8r r + -(M r r
=f(r,t)
(14.479)
- Me) - Qr = 0
or
aMr
Qr = -')r
1
+ -(M r r
- M(j)
(14.480)
For the axisymmetric vibration of plates with variable thickness, the bending moment resultants of Eqs.(14.184) and (14.185) can be expressed as Mr = -D(r)
M(j
a2w \) aw) ( -8r2 +-r 8r 1
aw + \)-82W) ar2
= -D(r) ( -r ar
(14.481) (14.482)
where Eh3(r) D(r) - ---
- 12(1 -
'.'
.;;,:
\)2)
(14.483)
14.12
Vibration of Plates with Variable Thickness
533
Substituting Eqs. (14.481) and (14.482) into Eq. (14.480), we obtain Qr = _
[D aar 3
w
+ (aD +
ar
3
D)r aarw + (!:r aDar _ rD) aw] ar 2
2
2
(14.484)
When Eq. (14.484) is substituted for Qr in Eq. (14.479), we can obtain the equation of motion for the transverse axisymmetric vibration of a circular plate with axisymmetric variation of thickness. An alternative but identical equation of motion can be derived in a simple manner by first specifying Eq. (14.187b) for Qr for the axisymmetric case as
[1
2
Qr
=
laW) a a (aw)] -D(r)- ara (a-arw2 + -r ar == -D(r)- ar --r ar r-ar
(14.485)
and then substituting Eq. (14.485) into Eq. (14.479):
1
[1
2
-- a { rD(r)- a -- a ( r-aw)]} + ph-a w2 r ar ar r ar ar at 14.12.3
=
fer, t)
(14.486)
Free Vibration Solution According to the procedure used by Conway et al. [12], the external force, set equal to zero and Eq. (14.486) is rewritten as
[1
-ara { rD(r)-- ara .r-- ara ( r-aw)]} ar
fer, t), is
2
=
-phr- aatw 2
(14.487)
For a plate with thickness varying linearly with the radial distance, we have
her) = hor
(14.488)
where ho is a constant. Equation (14.488) leads to
D(r)
= Dor3
(14.489)
where Do = Using Eqs. (14.488)-(14.490), as
Eh3 0
12(1 - v2)
(14.490)
the equation of motion, Eq. (14.487), can be expressed
Assuming the value of v as ~, which is applicable to many materials, Eq. (14.491) can be reduced to (14.492)
534
Transverse Vibration of Plates
Assuming a harmonic solution with frequency w for free vibration as w(r, t) = W(r)eiwt
(14.493)
we can rewrite Eq. (14.492) as
d2
d W) 2
2
-
2 ( Z4 __
dz
= z2W
(14.494)
dz
where z = pr
(14.495)
with
= (32PW2) p
1/2
(14.496)
3Eh3
o
The solution of Eq. (14.494) can be expressed as W)(z =
C] h(2.jZ)
C2Y2(2.jZ)
---+ z
---z
C3h(2.jZ)
C4K2(2.jZ)
+---+ z
---z
(14.497)
where h and Y2 are Bessel functions of the second kind and h and K2 are modified Bessel functions of the second kind. For specificity, we consider the axisymmetric plate to be clamped at both the inner and outer radii as shown in Fig. 14.20, so that W(z)
dW = -(z) dz
= 0,
(14.498)
Using the relations [13, 18J :z [z-m/2 Jm (kZI/2)J = - ~kZ-
(14.499)
:z[Z-m/2Ym(kzl/2»)
= -~kZ-
(14.500)
:z [z-m/2Im(kzl/2»)
= _~kZ-
Im+1 (kZI/2)
(14.501)
d dz [z-m/2 Km(kzl/2»)
1 = -2kz-
Km+1 (kZI/2)
(14.502)
I I I -----_
I
/
_-----
------:=~lc=:-----/ Figure 14.20
I
rr=RI-1 Linearly tapered axisymmetric plate.
14.13
Rc~ent.c{Hltributions ~.53.5
Table 14.4 Natural Frequencies of Vibration of Axisymmetric Annular Plates Clamped at Both Edges Mode 1 2 3 4 5
Value of the frequency parameter 0 for": R2/Rt=3 R2IRt=4 R2/Rl=1O
R2/Rl=2 16.5 45.2 88.4 146 211
Source: Ref. [12]. = (wR2I ho) (2p/3E)tf2.
a~
8.04 21.9 42.6 70.3 104.8
V
=
5.84 15.8 30.6 50.4 75.0
3.32 8.71 16.7 27.3 40.5
1.
application of the boundary conditions of Eq. (14.498) in Eq. (14.497) leads to the frequency equation in the form of a determinantal equation: f2(f3) Y2(f3) h(f3) K2(f3) h({3) Y3 ({3) - h({3) K 3 ({3) J2(a) Y2(a) h(a) J3(a) Y3(a) -h(a)
K2(a) K3(a)
=0
(14.503)
where 2'
---,
{3 .'
= 4Z2= 4R2
(2)1/2 32pw
.
--3 3Eho
a =4z1 =4Rl
32pw2 --3 3Eho
2
(
(14.504) 1/2 )
(14.505)
The natural frequencies of vibration given by Eq. (14.503) for different values of the thickness ratio R2IRl are given in Table 14.4.
14.13 RECENT CONTRIBUTIONS Thin Plates A comprehensive review of the various aspects of plate vibration has been given by Liessa [1]. A free vibration analysis of rectangular plates was presented by Gorman [21]. Cote et al. [22] investigated the effects of shear deformation and rotary inertia on the free vibration of a rotating annular plate. Kim and Dickinson studied the flexural vibration of rectangular plates with point supports [23]. Vera et al. [24] dealt with a theoretical analysis of the dynamic behavior of a system made up of a plate with an elastically mounted two-degree-of-freedom system, a study based on an analytical model with Lagrange multipliers. Plates on Elastic Supports Gorman [25] conducted a comprehensive study of the free vibration of rectangular plates resting on symmetrically distributed uniform elastic edge supports. Xiang et al. [26] investigated the problem of free vibration of a moderately thick rectangular plate with edges elastically restrained against transverse and rotational displacements. The Ritz method, combined with a variational formulation
536
Transverse Vibration of Plates
and Mindlin plate theory was used. The applicability of the formulation was illustrated using plates with different combinations of elastically restrained edges and classical boundary conditions. Plates with Variable Geometry and Properties Gupta and Shanna [27] analyzed the forced motion of a plate of infinite length whose thickness, density, and elastic properties vary in steps along the finite breadth by the eigenfunction method. The numerical results for transverse deflection computed for a clamped-clamped plate subjected to constant or half-sine pulse load were plotted in graphs. Wang presented a power series solution for the vibration of plates with generalized variable thickness [28]. These solutions, represented by the recursive relations of the coefficients of the infinite power series, can be applied to various boundary conditions to obtain the resonant frequency spectra and mode shapes. Plates on an Elastic Foundation Horvath described different subgrade models of soil-structure interaction analysis [29]. In the Winkler model, the foundation was represented by continuous springs. In the Pasternak model, the shear deformation between the spring elements was also considered by connecting the ends of the springs to the plate with incompressible vertical elements that deform only by the transverse shear. The exact relationship between the ith natural frequency of a simply supported polygonal Mindlin plate resting on a Pasternak foundation and the corresponding natural frequency of a Kirchhoff plate has been derived by Liew et al. [7]. The effect of elastic foundation on the mode shapes in stability and vibration problems of simply supported rectangular plates was considered by Raju and Rao [30]. The results of this work are expected to be useful in the dynamic stability studies of plates resting on elastic foundation. Numerical solutions of the vonKarman partial differential equation governing the nonlinear dynamic response of circular plates on Winkler, Pasternak, and nonlinear Winkler elastic foundation and subjected to uniformly distributed step loading were presented by Smaill [31]. The effect of the foundation parameters on the central deflec. tion was presented for both clamped and simply supported immovable edge boundary conditions. The nonlinear effects of Pasternak and nonlinear Winkler foundations on the deflection of plates were also determined. Plates with In-Plane Forces Devarakonda and Bert [32] considered the flexural vibrations of rectangular plates subjected to sinusoidally distributed in-plane compressive loading on two opposite edges. The procedure involves first finding a plane elasticity solution for an in-plane problem satisfying all boundary conditions. Using this solution, flexural vibration analysis was then carried out. The free lateral vibrations of simply supported rectangular plates subject to both direct and shear in-plane forces have been considered by Dickinson [33]. Thick Plates The best shear coefficient and validity of higher spectra in the Mindlin plate theory has been discussed by Stephen [34]. The axisymmetric free vibrations of moderately thick circular plates described by Mindlin theory were analyzed by the differential quadrature method by Liew et al. [35]. The first 15 natural frequencies of vibration were calculated for uniform circular plates with free, simply supported, and clamped edges. A rigid point support at the plate center was also considered.
Referen~
. SJ7
Martincek [36] used a dynamic resonance method to estimate the elastic properties of materials in a nondestructive manner. The method was based on measured frequencies of natural vibration in test specimens in the shape of circular plates. Exact analytical solutions for the free vibration of sectorial plates with simply supported radial edges have been given by Huang et al. [37]. Wittrick [38] gave analytical, three-dimensional elasticity solutions to some plate problems along with some observations on Mindlin plate theory.
REFERENCES 1. A. W. Leissa, Vibration of Plates, NASA SP-160, National Aeronautics and Space Administration, Washington, DC, 1969. 2. K. F. Graff, Wave Motion in Elastic Solids, Ohio State University Press, Columbus, OH, 1975. 3. R. D. Mindlin, Influence of rotary inertia and shear on flexural motions of isotropic plates, Journal of Applied Mechanics, Vol. 18, pp. 31-38, 1951. 4. R. D. Mindlin and H. Deresiewicz, Thickness-shear and flexural vibrations of a circular disk, Journal of Applied Physics, Vol. 25, pp. 1320-1332, 1954. 5. E. H. Mansfield, The Bending and Stretching of Plates, Macmillan, New York, 1964. 6. R. Szilard, Theory and Analysis of Plates: Classical and Numerical Methods, Prentice-Hall, Englewood Cliffs, NJ, 1974.-- .. 7. K. M. Liew, C. M. Wang, Y. Xiang, and S. Kitipomchai, Vibration of Mindlin Plates: Programming the p- Version Ritz Method, Elsevier, Amsterdam, 1998. 8. E. Ventsel and T. Krauthammer, Thin Plates and Shells: Theory, Analysis. and Applications. Marcel Dekker, New York, 2001. 9. E. B. Magrab, Vibrations of Elastic Structural Members, Sijthoff & Noordhoff, Alphen aan den Rijn, The Netherlands, 1979. 10. E. Skudrzyk, Simple and Complex Vibratory Systems, Pennsylvania State University Press, University Park, PA, 1968. 11. S. S. Rao and A. S. Prasad, Vibrations of annular plates including the effects of rotatory inertia and transverse shear deformation, Journal of Sound and Vrbration, Vol. 42, No.3, pp. 305-324, 1975. 12. H. D. Conway, E. C. H. Becker, and J. F. Dubil, Vibration frequencies of tapered bars and circular plates, Journal of Applied Mechanics, Vol. 31, No.2, pp. 329-331, 1964. 13. K. G. Korenev, Bessel Functions and Their Applications, Taylor & Francis, London, 2002. 14. I. N. Sneddon, Fourier Transforms, McGraw-Hill, New York, 1951. 15. W. Nowacki, Dynamics of Elastic Systems, translated by H. Zorski, Wiley, New York, 1963. 16. R. C. Colwell and H. C. Hardy, The frequencies and nodal systems of circular plates, Philosophical Magazine, Ser. 7, Vol. 24, No. 165, pp. 1041-1055, 1937. 17. G. Herrmann, The influence of initial stress on the dynamic behavior of elastic and viscoelastic plates, Publications of the International Association for Bridge and Structural Engineering, Vol. 16, pp. 275-294, 1956. 18. A. Jeffrey, Advanced Engineering Mathematics, Harcourt/Academic Press, San Diego, CA, 2002. 19. T. Wah, Vibration of circular plates, Journal of the Acoustical Society of America, Vol. 34, No.3, pp. 275-281, 1962.
538
Transverse Vibration of Plates 20. R. D. Mindlin, A. Schacknow, and H. Deresiewicz, Flexural vibrations of rectangular plates, Journal of Applied Mechanics, Vol. 23, No.3, pp. 430-436, 1956. 21. D. 1. Gorman, Free Vibration Analysis of Rectangular Plates, ElsevierlNorth-Holland, New York, 1982. 22. A. Cote, N. Atalla, and J. Nicolas, Effects of shear deformation and rotary inertia on the free vibration of a rotating annular plate, Journal of Vibration and Acoustics, Vol. 119, No. 4, pp. 641-643, 1997. 23. C. S. Kim and S. M. Dickinson, The flexural vibration of rectangular plates with point supports, Journal of Sound and Vibration, Vol. 117, No.2, pp. 249-261, 1987. 24. S. A. Vera, M. Febbo, C. G. Mendez, and R. Paz, Vibrations of a plate with an attached two degree of freedom system, Journal of Sound and Vibration, Vol. 285, No. 1-2, pp. 457-466,2005. 25. D. 1. Gorman, A comprehensive study of the free vibration of rectangular plates resting on symmetrically distributed uniform elastic edge supports, Journal of Applied Mechanics, Vol. 56, No.4 pp. 893-899, 1989. 26. Y. Xiang, K. M. Liew, and S. Kitipornchai, Vibration analysis of rectangular Mindlin plates resting on elastic edge supports, Journal of Sound and Vibration, Vol. 204, No.1, pp. 1-16, 1997. 27. A. P. Gupta and N. Sharma, Forced motion of a stepped semi-infinite plate, Journal of Sound and Vibration, Vol. 203, No.4, pp. 697-705, 1997. 28. J. Wang, Generalized power series solutions of the vibration of classical circular plates with variable thickness, Journal of Sound and Vibration, Vol. 202, No.4, pp. 593-599, 1997. 29. J. S. Horvath, Subgrade models for soil-structure interaction analysis, in Foundation Engineering: Current Principles and Practices, American Society of Civil Engineers, Reston, VA, 1989 Vol. I, pp. 599-612 ... 30. K. K. Raju and G. V. Rao, Effect of elastic foundation on the mode shapes in stability and vibration problems of simply supported rectangular plates, Journal of Sound and Vibration, Vol. 139, No.1, pp. 170-173, 1990. 31. J. S. SmailI, Dynamic response of circular plates on elastic foundations: linear and non-linear deflection, Journal of Sound and Vibration, Vol. 139, No.3, pp. 487-502, 1990. 32. K. K. V. Devarakonda and C. W. Bert, Flexural vibration of rectangular plates subjected to sinusoidally distributed compressive loading on two opposite sides, Journal of Sound and Vibration, Vol. 283, No. 3-5, pp. 749-763, 2005. 33. S. M. Dickinson, Lateral vibration of rectangular plates subject to in-plane forces, Journal of Sound and Vibration, Vol. 16, No.4, pp. 465-472, 1971. 34. N. G. Stephen, Mindlin plate theory: best shear coefficient and higher spectra validity, Journal of Sound and Vibration, Vol. 202, No.4, pp. 539-553, 1997. 35. K. M. Liew, J. B. Han and Z. M. Xiao, Vibration analysis of circular Mindlin plates using the differential quadrature method, Journal of Sound and Vibration, Vol. 205, No.5, pp. 617-630,1997. 36. G. Martincek, The determination of Poisson's ratio and the dynamic modulus of elasticity from the frequencies of natural vibration in thick circular plates, Journal of Sound and Vibration, Vol. 2, No.2, pp. 116-127, 1965. 37. C. S. Huang, O. G. McGee and A. W. Leissa, Exact analytical solutions for free vibrations of thick sectorial plates with simply supported radial edges, International Journal of Solids and Structures, Vol. 31, No. 11, pp. 1609-1631, 1994.
··P4'Oblems
c5J9
38. W. H. Wittrick, Analytical, three-dimensional elasticity solutions to some plate problems and some observations on Mindlin plate theory, International Journal of Solids and Strucnlres, Vol. 23, No.4, pp. 441-464, 1987. 39. E. Kreyszig, Advanced Engineering Mathematics, 8th ed., Wiley, New York, 1999. 40. R. Solecki and R. 1. Conant, Advanced Mechanics of Materials, Oxford University Press, New York, 2003.
PROBLEMS 14.1 Determine the fundamental frequency of a square plate simply supported on all sides. Assume that the side and thickness of the plate are 500 and 5 mm, respectively, and the material is steel with E 207 GPa, v 0.291, and unit weight 76.5 kN/m3.
=
=
=
14.2 Determine the first two natural frequencies of a steel rectangular plate of sides 750 and 500 mm and thickness 10 mm. Assume the values of E and v as 207 GPa and 0.291, respectively, with a weight density 76.5 kN/m3 and the boundary conditions to be simply supported on all sides. 14.3 Derive the frequency equation for the free vibrations of a circular plate of radius a with a free edge. 14.4 Consider the static deflection of a simply supported rectangular plate subjected to a uniformly distributed load of magnitude fo. By assuming the solution as 00
w(x,y)
=
00
LL
1/mn Wmn(x,y)
m=ln=l
where W mn (x ,y) are the normalized modes (eigenfunctions), determine the values of the constants 1]mn·
14.5 Derive the frequency equation for a uniform annular plate of inner radius b and outer radius a when both edges are clamped. 14.6 Consider the equation of motion of a simply supported rectangular plate subjected to a harmonically varying pressure distribution given by Eq. (E14.2.1). Specialize the solution for the following cases: (a) static pressure distribution; (b) when the forcing frequency is very close to one of the natural frequencies of vibration of the plate, Q -+ Wmn; (c) static load, P, concentrated at (x
= xo. y = Yo)·
14.7 State the boundary conditions of a circular plate of radius a supported by linear and torsional springs all around the edge as shown in Fig. 14.21. 14.8 Derive the equation of motion of an infinite plate resting on an elastic foundation that is subjected to a constant force F that moves at a uniform speed c along a straight line passing through the origin using Cartesian coordinates. Assume the mass of the foundation to be negligible. 14.9 Derive the equation of motion of a plate in polar coordinates by considering the free body diagram of an element shown in Fig. 14.6 using the equilibrium approach.
Ie,
• r
a
Figure 14.21
540
Transverse Vibration of Plates
o \ \
I I \
I
\
I \
I \ \
I
yt----
'1
------.
I I I I I I
\
I I \ I \1
P
\0(
\ \
t y
\ Figure
14.10 Skew plates find use in the swept wings of airplanes and parallelogram slabs in buildings and bridges. Derive the equation of motion for the vibration of the skew plate shown in Fig. 14.22 using the transformation of coordinates. The oblique coordinates ~ and 1]are related to x and y as
14.22
14.13 Derive the frequency equation of a rectangular plate with SS-F-SS-F boundary conditions. 14.14 Derive the frequency equation of a circular plate of radius a with a free edge including the effects of rotary inertia and shear deformation. "
~=x-ytan8,
Y 1]=-cos8
Hint: Transform the Laplacian coordinate '\72
operator
to the oblique
system and derive
2 a2 +_a2 ) = cos -- 12 8 (a-a~2 _ 2sin8-- a~a1] a 1]2
14.11 Derive the frequency equation of an annular plate with both the inner and outer edges simply supported. Assume the inner and Outer radii of the plate to be band a, respectively.
14.15 Find the first four natural frequencies of vibration of a rectangular plate resting on an elastic foundation with all the edges simply supported. Compare the results with the frequencies of a simply supported plate with no elastic foundation. Size of the plate: a 10, b 20, h 0.2 in; unit weight: 0.283 Ib/in3; E = 30 X 106 psi; v 0.3, foundation modulus k 1000 Ib/in2-in.
=
= =
=
14.16 Find the steady-state response of a simply supported uniform rectangular plate subjected to the force
f(x,y,t) 14.12 Find the response of a circular plate subjected to a suddenly applied concentrated force Fo at r = 0 at t O.
=
=
= f08
(x - ~,
y - ~) cos Qr
14.17 Derive Eqs. (l4.184)-(I4.190) nate transformation relations.
using the coordi-
15 Vibration of Shells 15.1 INTRODUCTION AND SHELL COORDINATES A thin shell is a three-dimensional body that is bounded by two curved surfaces that are separated by a small distance compared to the radii of curvature of the curved surfaces. The middle surface of the shell is defined by the locus of points that lie midway between the two bounding curved surfaces. The thickness of the shell is denoted by the distance between the bounding surfaces measured along the normal to the middle surface. The thickness of the shell is assumed to be constant. Shells and shell structures find application in several areas of aerospace, architectural, civil, marine, and mechanical engineering. Examples of shells include aircraft fuselages, rockets, missiles, ships, submarines, pipes, water tanks, pressure vessels, boilers, fluid storage tanks, gas cylinders, civil engineering structures, nuclear power plants, concrete arch dams, and roofs of large span buildings. 15.1.1
Theory of Surfaces The deformation of a thin shell can be described completely by the deformation of its middle (neutral) surface. The undeformed middle surface of a thin shell can be described conveniently by the two independent coordinates a and {3shown in Fig. 15.1. In the global Cartesian coordinate system OXYZ, the position vector of a typical point in the middle surface, r, can be expressed in terms of a and {3as
r = r(a, {3)
(15.1)
which, in Cartesian coordinate system, can be written as
X=X(a,{3) (15.2)
Y = Y(a,{3)
Z= Z(a,{3) or
-r
= X(a, {3)7 + Y(a, {3)]
+ Z(a, {3)k
(15.3)
where 7, ], and k denote the unit vectors along the X, Y, and Z axes, respectively. The derivatives of the position vector with respect to a and {3,denoted as
a-r _ aa - .a a-r _
--r
a{3 =
r.{3
(15.4)
541
542
Vibration of Shells
x Figure 15.1
Curvilinear coordinates in the middle surface of a shell.
a and f3 coordinate .lines as
represent tangent vectors at any point of the surface to the shown in Fig. 15.1. '
15.1.2
Distance between Points in the Middle Surface before Deformation The distance vector between the infinitesimally separated points P and Q lying in the middle surface of the shell (before deformation) can be expressed as ~ dr=
ar ar ~ -da+-dR=r~da+rfJdR aa af3 P,~
~
(15.5)
,P
The magnitude of the distance vector (ds) is given by (ds)2
=
~ ~ dr· dr
ar ar = --(da)2
+ 2 --ar
aexaex
ar
aa af3
dexdf3 +
ar ar __
(df3)2
(15.6)
8f3 af3
which can be written as (ds)2
= A2(dex)2 + 2AB
cos y dexdf3 + B2(df3)2
(15.7)
where A and B denote the lengths of the vectors r,a and r,fJ' respectively: 2
= r,a . r,a = Ir,al2 2 B = r,fJ . r,fJ = Ir,fJ/2 A
and y indicates the angle between the coordinate r()f·rfJ
-'--'-
AB
(15,8) (15.9)
curves ex and f3 defined by
= cos Y
(15.10)
Introduction and Sh~lLGoor-dinate$ ·$43
15.1
Equation (15.7) is called the fundamental form or first quadratic form of the surface defined by ; = ;(a, 13) and the quantities A2, AB cos y, and B2 are called the coefficients of the fundamental form. If the coordinates a and 13 are orthogonal, y will be 90°, so that
;,a . 'l.P
=0
(15.11)
and Eq. (15.7) reduces to
(15.12) where A and B are called the Lame parameters. (dS)2 = dsr
Equation (15.12) can be rewritten as
+ dsi
(15.13)
where dSI = Ada
(15.14)
dS2 = B df3
and
denote the lengths of the arc segments corresponding to the increments da and df3 in the curvilinear coordinates a and {3.It can be seen that Lame parameters relate a change in the arc length in the middle surface to the changes in the curvilinear coordinates of the shell. Example 15.1 Determine the Lame parameters and the fundamental form of the surface for a cylindrical shell. SOLUTION The curved surface of a shell is defined by the two lines of principal curvature a and 13 and the thickness of the shell wall is defined along the z axis. Thus, at each point in the middle surface of the shell, there will be two radii of curvature, one a maximum value and the other a minimum value. In the case of a cylindrical shell, the shell surface is defined by the curvilinear (cylindrical) coordinates a x and 13 fJ, where a is parallel to the axis of revolution and 13 is parallel to the circumferential direction as shown in Fig. 15.2. Any point S in the middle surface of the shell is defined by the radius vector
=
; = x 7 + R cos (}. ]
+R
sin (}. k
=
(E15.1.1)
where R is the radius of the middle surface of the cylinder and 7, ], and k are unit vectors along the Cartesian coordinates x, y, and z, respectively, as shown in Fig. 15.2. The Lame parameters can be obtained from Eq. (E15.1.1) as
a;
----i
a;
aa - ax -
-
(E15.1.2)
544
Vibration of Shells x
• I
,,
/:
y
/ 7
/
Figure 15.2
A
I I ..
ar = Iii = 1 = aCt
ar
ar
-afJ = -ae = -R B
Cylindrical shell.
=
/::1 =
(E15.1.3)
....
sine}
+ Rcosek
[(-Rsine)2
(E15.1.4)
+ (Rcose)2]1!2 = R
(E15.1.5)
The fundamental form of the surface is given by (ds)2
=
2 A (dCt)2
+ B2(dfJ)2
=
(dx)2
+ R2(de)2
(E15.1.6)
The interpretation of Eq. (E15.1.6) is shown in Fig. 15.2, where the distance between two infinitesimally separated points P and Q in the middle surface of the shell can be found as the hypotenuse of the right triangle PQT, where the infinitesimal sides QT dx and PT R de are parallel to the surface coordinates of the shell.
=
=
15.2 Determine the Lame parameters and the fundamental form of the surface for a conical shell.
Example
SOLUTION Any point S in the middle surface of the shell is defined by the radius vector r taken from the top of the cone (origin 0 in Fig. 15.3) along its generator,
15.1
Introduction and Shell Coordinates
545
-+
.-.-.-.L.-.-.- .•.y
a
_---L--_ /
~~---
I'
-
----
as = x = I r I ,
+
as.
=xcosao
SlS2
= x sin ao cos 9 = x sin ao sin 9
SIS3
I
SIS =xsinao
t
x Figure 15.3
Conical shell.
and angle 0 formed by a meridional plane passing through the point S with some reference meridional plane (the meridional plane passing through the Z axis is taken as the reference meridional plane in Fig. 15.3). If the length of the vector r is x and the semi cone angle is ao, the radius vector can be expressed in the Cartesian coordinate system XYZ as
r =x
+ x sin ao sin 0 J + x sin ao cas 0 k
(EI5.2.1)
= x and fJ = 0, the Lame parameters can be obtained from or or ;- = - = cos aoi + sin ao sin 0 j + sin ao cas 0 k
(E15.2.2)
cas aoi
With a
ox A = orl ox = (cas a5 + sin2 ao sin2 0 + sin2 ao cos2 0) 1/2 = 1
va
or
-
l
-
or
-
- x
afJ - ao B
=
a'00ll
l
sinao
cosO
J-
x
sinao sin Ok
= [(x sinaocosO)2 + (-x
sinaosinO)2]1/2
(EI5.2.3) (EI5.2.4)
= x sinao
(E15.2.5)
546
Vibrationof Shells
The fundamental fonn of the surface is given by (E15.2.6) The interpretation of Eq. (E15.2.6) is shown in Fig. 15.3, where the distance between two infinitesimally separated points P and Q in the middle surface of the shell can be found as the hypotenuse of the right triangle PQT with the infinitesimal sides QT and PT given by dx and xsinaode, respectively. Example 15.3 Find the Lame parameters and the fundamental fonn of the surface for a spherical shell. SOLUTION A typical point S in the middle surface of the shell is defined by the radius vector;: as shown in Fig. 15.4. Using the spherical coordinates ¢ and () ·shown in the figure, the radius vector can be expressed as
;:= R cos ¢7 + R sin¢cose]
+ R sin¢ sinek
(E15.3.1)
where 7, ], and k denote the unit vectors along the Cartesian coordinates X, Y, and Z, respectively, and R is the radius of the spherical shell. Using a = ¢ and f3 = e, the
X .l
I
7t-
OS=R 04=Rcos¢ 02=Rsinct> 12
= R sin ct>
sin £I
32 = R sin ct> cos £I
·_·_~z ~ k
Rsin¢
Figure 15.4 Spherical shell.
d£l
Introduction and Shell Coordinates. '.547
15.1
Lame parameters can be determined as ar ar -aa = -a41 = -Rsin41i + Rcos41cosOj + Rcos41sinOk A
= I:; I = [(-R
ar ar -af) = -vo B =
sin 41)2+ (R cos>cos 0)2 -
= -Rsin41sinOj
I:; I = [(-R
(EI5.3.2)
+ (R cos 41sinO)2]1/2 = R
+ Rsin41cosOk-
sin41sin 0)2
+ (R
(E15.3.3) (E15.3.4)
= R sin 41
sin 41cos 0)2]1/2
(E15.3.5)
The fundamental form of the surface is given by (ds)2
= A2(da)2 + B2(df))2 = R2(d»2 + R2 sin2 41(dO)2
(E15.3.6)
Equation (E15.3.6) implies that the distance between two infinitesimally separated points P and Q in the middle surface of the shell is determined as the hypotenuse of the right triangle PQT with the infinitesimal sides QT and PT given by R d41 and R sin41dO, respectively (see Fig. 15.4). 15.1.3
Distance between Points Anywhere in the Thickness of a Shell before Deformation Consider two infinitesimally separated points P and Q in the middle surface of the shell and two more points P' and Q' that lie on the normals to the middle surface of the shell, passing through points P and Q, respectively, as shown in Fig. 15.5. Let ii denote the unit vector normal to the middle surface passing through the point P and z indicate the distance between P and P'. Then the point Q' will be at a distance of z + dz from the middle surface. The position vector of p'(ih can be denoted as R(a, {3, z) = r(a, f))
+ z ii(a,
(15.15)
f))
Since the position vector of Q was assumed to be r + dr, the position vector of Q' can be denoted as R + dR, where the differential vector d R can be expressed, using Eq. (15.15), as (15.16) with
_ aii dn(a, {3) = aa da
aii
+ af) df)
(15.17)
The magnitude ds' of d R is given by (ds')2
=
dR·
dR
= dr· dr
+ 2zdr·
= (dr + zdii + dz ii)·
(dr
+ z2 dii . dii + (dZ)2ii . ii dii + 2dzdr· ii + 2zdzdii·
+ zdii + dz ii) ii
(15.18)
548
Vibration of Shells
I I I I I I I I I I I
.•...•....•......
o
a~
a~
P'S" = ...! dCt, SS aCt
= zn:PQ QS = parallel PP'
"" ""
= ...! d{3 a{3 = dt P'O' = dR; P'S
I
"
'"
I
--------
v_- --
I
dr:
=
to PP'
Ct
art - art -, ST = z acxdCt,TQ = z ap d{3, QQ
~ = zn
Figure 15.5 Distance between arbitrary points P' and Q'.
=
Since ii is a unit vector, ii . ii 1, and because of the orthogonality of the coordinate system, dr· ii = 0 and dii . ii = O. Thus, Eg. (15.18) reduces to
(ds')2
= (ds)2 + Z2dii . dii + (dZ)2 + 2z dr.
dii
(15.19)
The second term on the right-hand side of Eg. (15.19) can be expressed as 2
~
~
z dn· dn = z
2
[oii oii - . -(da) oa oa
2
oii + -oii . -(d/3)
2
0/3 0/3
+2
J
oii oii . _ da d/3 Ocr 0/3
-
(15.20)
Because of the orthogonality of the coordinates,
oii
oii
-,-=0 ocr 0/3
in Eq. (15.20). The quantities z (oii/ocr) and the radii of curvature as
lor/ocrl
-- Rex
Iz(oii/ocr)1
= ---
z
(15.21) Z
(oii/o/3) can be expressed in terms of
10r/o/31
---Rp
Iz(oii/o/3)I
= ---
z
(15.22)
where Ra and Rp denote the radii of curvature of the cr and /3 curves, respectively. When the relations (15.23)
j
I~_"""'~~-""",,,,,,,,:,,,,,,~,-,,,,,,,",,,,,,,,,,,-,,,,,~,,,-,,-,,,,,,,-,,,,,,,,,,,,,-,-=,,,,,,,~~,=o."'",,,,,,",-~~~,,,,,,,,,,,,,,,,,,,,,~,~~'.,.
15.1 Introduction and Shell Coordinate~. 549
are used from Eqs. (15.8) and (15.9), Eqs. (15.22) yield
Iz an I = zB afJ Rfj
= zA.
z ani aa
I
Ra
(15.24)
Thus, Eq. (15.20) can be rewritten as 2
__
Z dn· dn =
Z2A 2 -2
2
(da)
Z2B2
2
+ -2-(dfJ)
Ra
(15.25)
R/3
The fourth term on the right-hand side of Eq. (15.19) can be expressed as 2z dF . dn = 2z (aF da aa
+ aF dfJ) afJ
aF an
+ an dfJ) afJ
aF + -.
an -dadfJ+-· a{3 aa
2
= 2z [ -·-(da) aa aa
. (an da aa
aF aa
an -dadfJ a{3
aF + -.
an -(d{3) a{3 a{3
2]
(15.26) Because of the orthogonality of the coordinates, Eq. (15.26) reduces to __ 2zdr· dn
aF = 2z -aa
an . -(da) aa
2
+ 2z
aF an - . -(d{3) a{3 a{3
2
(15.27)
Using Eqs. (15.23) and (15.24), we can express aii d a..·)2 -or . -( aa aa '.
ar II -aii I (d)2 = I-aa a = -A (d a) 2 aa Ra
an -or . -(dfJ) ofJ a{3
ar II -an I (dfJ) 2 = -(d{3) B2 = -a{3 a{3 R/3
21
2
(15.28)
2
Using Eqs. (15.12), (15.25), (15.27) and (15.28) in Eq. (15.19), we obtain (ds')2 = A2
15.1.4
(1 + :a) 2
(da)2
+ B2
(1 + :/3)2
(d{3)2 + (dZ)2
(15.29)
Distance between Points Anywhere in the Thickness of a Shell after Deformation Equation (15.29), which gives the distance between two infinitesimally separated points P' and Q' (before deformation), can be rewritten as (ds')2 = hll (a, fJ. z) (da)2
+ h22(a. fJ. z)
(dfJ)2
where
+ h33(a, {3. z)
(dZ)2
(15.30)
+
:aY
(15.31)
2 h22(a, fJ. z) = B ( 1 +
:/3) 2
(15.32)
hll(a.
2 fJ. z) = A (1
h33(a, fJ. z) = 1
(15.33)
Note that Eqs. (15.29) and (15.30) are applicable for a shell in the undeformed condition. When the shell deforms under extemalloads, a point P' with coordinates (a. {3. z)
550
Vibration of Shells in the undeformed condition will assume a new position p" defined by the coordinates (a + 7]1, f3 + 7]2, Z + '73) after deformation. Similarly, if a point Q' in the neighborhood of pi has coordinates (a + da, f3 + df3, z + dz) in the undeformed condition, it will assume a new position Q" with coordinates (a + da + '71 + d'71, f3 + d{3 + '72 + d'72, z + dz + '73 + d'73) after deformation. Let the physical deflections of the point pi along the a, f3, and Z directions resulting from the deformation of the shelI be given by ii, v and w, respectively. Since the coordinates of pi are given by (a, f3, z) and of p" by (a + '71, f3 + '72, z + '73), ui and '7i are related by Eq. (15.30) as
w = ..jh33 The distance (ds") between P" and Q", after deformation,
(a, {3, Z)TJ3
(15.34)
can be found as
+ '71, f3 + '72, z + '73)(da + d'71)2 + h22(a + '71, {3 + '72, z + '73)(df3 + d'72)2 + h33(a + '71, f3 + '72, z + 7]3)(dz + d'73)2
= hll (a
(dS")2
(15.35)
where, in general, hu, h22, and h33 are nonlinear functions of a + '71, (3 + '72, and For simplicity, we can linearize them using Taylor's series expansion about the point P'(a, f3, z) as
z + 7]3· hjj(a
+ '71, (3 + '72, z + '73)
;::: hii(a,
+
f3, z)
+ Ohii
'Ohii of3 (a, {3, Z)7]2
(a, f3, z) oa '71 Oh
ii + a;-(a,
{3, Z)173,
i
= 1,2,3 (15.36)
Similarly, we approximate Eq. (15.35) as
the terms (da
(da (df3 (dz
+ d17])2 + d172)2 + d173)2
+ d171)2,
(df3
+ d172)2,
and (dz
+ 2(da)(d17l) ;:::(d{3)2 + 2(df3)(d172) ;:::(dZ)2 + 2(dz)(d173)
+ d173)2
;:::(da)2
(15.37)
by neglecting the terms (d 171)2, (d 172f, and (d 173)2. Since TJivaries with the coordinates
a, f3, and z, Eq. (15.37) can be expressed as (da (df3
(dz
+ d17]) 2 ;:::(da) 2 + 2(da)
(0171 da
+ d172)2
(OTJ2 da oa
+
(0173 da oa
+ 0173 df3 + 0173 dZ)
+ dTJ3)2
;:::(df3)2
;::: (dZ)2
+ 2(d{3) + 2(dz)
in
oa
+ -0171 df3 + -(171)dz 0{3
OTJ2d{3 of3
of3
oz
+ OTJ2dZ) oz.
OZ
(15.38)
551
15.1 Introduction and Shell Coordinates
Using Eqs. (15.36) and (15.38), Eq. (15.35) can be expressed as
" 2 ( hll
(ds ) =
+
ll
[2(da)
ll
ohll oh Oh ) + a;;111 + aj3112 + ai:113
:~I
(a111
a;; da
1 d{3 + a : dZ)] o
a;
ah22 ah"2) + ( h22 + ah22 oa 111+ aj3T/2 + 113
+
+ 2(da)
:~2
[2
+ 2(d{3) (a112 a;; da
[2
+ 2(dz) (aT/3 a;; da
(d{3)
2 d{3 + °a: dZ)]
Oh33 ah33 ah33) + ( h33 + a;;T/1 + aj3T/2 + ai:113
+ :~
d{3 + °a~ dZ) ]
= ( hll +
Ohll oa T/I
+
ahll Ohll) a{3 T/2 + ai:113
aT/I ( aa da
+ 2h II(da)
+
2
(da)
az
OT/I (111) 0{3 d{3 + dz
Ohll --~\Ohll Ohll) ( aa T/I + 0{3 T/2 + ai:113
+ 2(da)
+ ( h22 +
ah22 oa T/I
+ 2h22(d{3) + 2(d{3)
Oh22
OT/2 ( oa da
ah22 ( -T/I oa
ah22)
+
OT'J2 0{3 d{3 +
a az
+ 2h33 (dz)
(o~ oa
da
+ -0{3
ah33 ( oa T/I
+
ah33 ah33) 0{3 T/2 + ai:113
az
aT'JI) dz
)
dz (0T/2 da oa
ah33)
aT/2) + -0112 d{3 + dz a{3 az
2
+ aj3T/2 + ai:T/3
o~
d{3 +
2
Oh22 ah22) + -T/2 + -T/3 0{3 az ah33
aT/I
+ aif
(d{3)
112
Oh33 oa 771 -
(aT/I oa da
+ aj3T/2 + ai:113
+ ( h33 +
+ 2(dz)
(dz)
(dz)
a~) az
d{3 + -
dz (0T/3 oa da
a 113
+ aif
d{3 +
az
aT/3) dz
(15.39) The terms involving products of partial derivatives of hjj and 77j on the right-hand side of Eq. (15.39) can be neglected in most practical cases. Hence, Eq. (15.39) can be expressed as (ds")2 ~
HII (da)2
+ 2H12(da)(d{3)
+ H22(d{3)2 + H33(dz)2 + 2H23 (d{3)(dz) + 2H13(da)(dz)
(15.40)
552
Vibration of Shells
where ohll oa
HII = hll
+
H22 = h22
+ ~711
+
ohll of3
+
Oh22 of3 712
+ --a;:713 + 2h22
+
Oh33 0/3 712
+ --a;:713 + 2h33
711
Oh22
Oh33 H33 = h33 + oa 711
0711
ohll
Oh22
Oh33
0711 oa
(15.41)
0712 0/3
(15.42)
0713 oz
(15.43)
0712
HI2
= H21 = hll-
H23
= H32 = h22- OZ + h33- 0/3
HI3
= H31 = h 11 - OZ + h33 -oa
0/3
+ --a;:713 + 2hll
712
+h22-
0712
(15.44)
oa 0713
0711
(15.45)
0713
(15.46)
Thus, Eq. (15.40) gives the distance between P" and Q" after deformation. Points P" and Q" indicate the deflected position of points P' and Q' whose coordinates (in the undeformed shell) are given by (a, /3, z) and (a + da, /3 + d/3, z + dz), respectively.
15.2 STRAIN-DISPLACEMENT
RELATIONS
The distance between two arbitrary points after deformation, given by Eq. (15.40), can also be used to define the components of strain in the shell. The normal strain (eaa == ell) along the coordinate direction a is defined as
ell
change in length of a fiber originally oriented along the a direction = original length of the fiber
(15.47)
Since the fiber is originally oriented along the a direction, the coordinates of the end points of the fiber, P' and Q', are given by (a, /3, z) and (a + da, /3, z), respectively, and Eqs. (15.40) and (15.30) give (d s")2
==
(d s")i
(ds')2
==
(ds')il
I
= HII (da)2
(15.48)
= hll(da)2
(15.49)
Using a similar procedure, for fibers oriented originally along the we can write II
,
(ds )22 = H22(df3 , ,
(ds )22 = h22(d/3) II
,
(ds )33
= H33(dz)
(ds')~3 = h33(dz)2
/3 and z directions,
)2
(15.50) 2
(15.51) 2
(15.52) (15.53)
15.2 Strain':::DisplacementRelation.:;· ,55.3
Since the normal strain along the coordinate direction ezz == e33). is defined as =
e22(e33)
fJ(z).
denoted by
epp
change in length of a fiber originally oriented along the fJ(z) .. onglOal length of the fiber
== e22
(or
direction (15.54)
the normal strains
eji.
'" = (dS")ii -
i = 1. 2. 3. can be expressed as
=
(ds')"
(ds')ji
=
1 + Hji - hii
(ds")" _ 1
=;
H" _ I
(ds')ji _
=
hii
+ Hii hii
hji
I
hii
- I
(15.55)
hii
The quantity (Hii - hii)/ hii is very small compared to 1 and hence the binomial expansion of Eq. (15.55) yields eii
1 Hii - hji ) = ( 1+ ----+ ... -1 2 hii
1 Hii - hii
~ ----. 2 hii
The shear strain in the afJ plane (denoted as e12
cap
== e12)
i = 1.2.3
(15.56)
is defined as
= change in the angle of two mutually perpendicular fibers originally oriented along the a and fJ directions
(15.57)
which from Fig. 15.6, can be expressed as, 1r
el2
where
812
= - -
2
(15.58)
(1t2
== 8aP' To be specific, consider two fibers, one originally oriented along the fJ. z) and P = (a + da, fJ. z) and the
a direction [with endpoint coordinates R = (a,
a
Figure 15.6
Angular change of fibers.
554
Vibration of Shells
other originally oriented along the f3 direction (with endpoint coordinates R and Q = (a, f3 + df3, z»). This implies that the distance PQ is given by
==
(ds')2
(ds')I2
= (a,
+ h22(df3)2
= hIJ (da)2
f3, z)
(15.59)
~y considering points P and Q to be originally located, equivalently, at (a, f3, z) and (a - da, f3 + df3, z) instead of at (a + da, f3, z) and (a, f3 + df3, z), respectively, we can obtain the distance P' Q' as (15.60) Using a similar procedure in the f3z plane (for £{3z == £23) and za plane (for Eza we can obtain
+ h33(dz)2
(dS')~3 = h22(df3)2 (ds")~3
(dS')~1 = h33(dz)2
+ hlJ
In Fig. 15.6, the sides RQ',
Q' P', and RP'
(15.62) (15.63)
- 2H31 (dz) (da)
(15.64)
are related by the cosine law:
+ (ds")b
= (ds")II
- 2H23(df3)(dz)
(da)2
= H33(dz)2 + HIJ (da)2
(ds")~1
(ds")I2
(15.61)
+ H33(dz)2
= H22(df3)2
== E31),
- 2(ds")1J
(ds"b
cos (}12
(15.65)
= 12, 23, 31
(15.66)
Equation (15.65) can be expressed in a general form as (ds");j
= (ds");;
+ (ds")]j
- 2(ds")ii
(ds")
jj
cos (}ij,
ij
Using Eqs. (15.60), (15.48), and (15.50), Eq. (15.65) can be expressed, as HIJ (da)2
+ H22(df3)2
= HIJ(da)2
+
- 2HI2(da)(df3)
H22(df3)2 - 2(..jHIJ
da)(..jH22
df3)
COS(}12
(15.67)
which can be simplified to obtain cos (}12 =
H12 -;:;::;=::=:::::::::
(15.68)
.JHIJH22
Equations (15.58) and (15.68) yield cos (}l2 = cos
(!: -
£12)
= sin £12
= -;::;::;::::HI2;;::;::=
2
.JHIJH22
(15.69)
In most practical cases, shear strains are small, so that the following approximations can be used:
.
sm £12
~
E12 =
M2
-;::;::;::::;;:== ~
.JHIJH22
M2 ---=-= .Jh
(15.70)
IIh22
A similar procedure can be used for the other shear strains, £23 (in the f3z plane) and (in the za plane). In general, the shear strains can be expressed as
£31
ij = 12,23,31
(15.71)
555
15.2 Slmin-DisplacementRelations
Strains in Terms of Displacement Components The normal strains can be expressed in terms of the displacement components, using Eqs. (15.56), (15.41)-(15.43) and (15.31)-(15.34), as
ell
-
-
e
I
1
-
-0
aa-2A2(1+z/Ra)2
+_a [(A2 a{3
a [
+ aa
oa
l+_Z Ra Ii
[A
)2]
A (1 + z/ Ra)
+ -;a)
Z
-
Ii ----A(1+z/Ra)
+ _a az
v__
B
(1 + z/ R(3)
[(A2
l+_Z Ra
)2]]
w
]
= -A-(-1_+I-z_/ R-a-){-a~ [ A (1 + -a~ [A (1
2 ( 1 + -Ra )2]
+ -;a ) ]
-A-(-1-:-z-/R-a-)
U ] -B-(I-+-z-/R-{3-)+
+ -A-(-I-+I-zR-/ a-)-:: - -a~ [A (1
-%. w ]
+ -;a)]
li (15.72) -A-2-(I-+- Z-/-R-a)-2
The following partial derivatives, mown as Codazzi conditions [9], are used to simplify Eq. (15.72):
!... [A a{3 !...[B aa Thus, the normal strain
(1 + ..:...)] (1 + ..:...) (1 + ..:...)] (1 +..:...) Ra
R{3
ell
=
=
Ra
(15.74)
aB aa
can be expressed as 1
811
(15.73)
aA R{3 a{3
= eaa = A (1 + z/ Ra)
(OU
Using a similar procedure, the normal strains 1
822
= epp = B (1 + z/R(3)
e33
= ezz = -oz
ow
oa
v aA A ) + Ii a{3 + w Ra
e22
and
833
(ou + A op
Ii aB aa
(15.75)
can be determined as B )
+ w R{3
(15.76) (15.77)
556
Vibration of Shells Similarly, using Eqs. (15.71), (15.44)-(15.46), and (15.31)-(15.34), can be expressed in terms of displacement components as
a [
A (1 + zl Ra)
El2
= Eap = B (1 + zl Rp) B
(1 + zl Rp)
+ A(1+zIRa) E -E 23 -
E31
=
-B
pz Eza
ofi A
[
-0
[
)
Ra
v
B (1 +zIRp)
V]
(1 + zl
az B
]
J
B(I+zIRp)
z ) -a ( 1+Rp az
= A ( 1 + -Z
+ zl Ra)
v
0 [
aa
u
(1
the shear strains
Ra)
(15.78)
J +B (1 +zIRp) 1
+
A
1 (1 + zl
_ ow
(15.79)
afi
ow _
(15.80)
Ra) oa
15.3 LOVE'S APPROXIMATIONS In the classical or small displacement theory of thin shells, the following assumptions, originally made by Love (1), are universally accepted to be valid for a first approximation shell theory: 1. The thickness of the shell is small compared to its other dimensions, the radii of curvature of the middle surface of the shell.
such as
2. The displacements and strains are very small, so that quantities involving second- and higher-order magnitude can be neglected in the strain-displacement relations. 3. The normal stress in the transverse (z) direction is negligibly to the other normal stress components.
small compared
4. The normals to the un deformed middle surface of the shell remain straight and normal to the middle surface even after deformation, and undergo no extension or contraction. The first assumption basically defines a thin shell. For thin shells used in engineering applications, the ratio hi Rmin, where h is the thickness and Rmin is the smallest radius of curvature of the middle surface of the shell, is less than and thus the ratio
to
llx,·
zl Hence, terms of the order neglected compared to unity in the strain-displacement Rmin will be less than
z
-«
Ra
1
'
z
-« Rp
hi Ri or zl Ri (i
=
Q',
fi)
can be
relations 1
(15.81)
The second assumption enables us to make all computations in the undeformed configuration of the shell and ensures that the governing differential equations will be linear. The third assumption assumes that the normal stress a~~ == a33 = 0 in the direction of z. If the outer shell surface is unloaded, a33 will be zero. Even if the outer shell surface is loaded, a33 can be assumed to be negligibly small in most cases. The fourth assumption, also known as Kirchhoff's hypothesis, leads to zero transverse shear strains
557
15.3. Love's Approximations
and zero transverse normal strain: 813
= 823 = 0
(15.82)
833
=0
(15.83)
It is to be noted that assumptions 3 and 4 introduce the following inconsistencies: (a) The transverse normal stress (0"33) cannot be zero theoretically, especially when the outer surface of the shell is subjected to load. (b) In thin shell theory, the resultant shear forces, QI3 == Qaz (acting on a face normal to the a curve) and Q23 == Qpz (acting on a face normal to the j3 curve), are assumed to be present. These resultants can be related to (caused by) the transverse shear stresses, <113 and 0'23, in the shell. If transverse shear stresses are present, the transverse shear strains cannot be zero. This violates the assumption in Eq. (15.82). Despite these inconsistencies, Love's approximations are most commonly used in thin shell theory. Equations (15.82) imply, from Hooke's law, that the transverse shear stresses are also zero: (15.84) To satisfy Kirchhoff's hypothesis (fourth assumption), the components of displacement at any point in the thickness of the shell are assumed to be of the following form:
+ z{h (a, j3) j3) + z82(a, j3)
ii(a, j3, z) = u(a, j3) v(a, j3, z)
= v(a,
w(a, j3, z) = w(a, j3)
(15.85) (15.86) (15.87)
where u, v, and w denote the components of displacement in the middle surface of the shell along a, j3, and z directions and 81 and (h indicate the rotations of the normal to the middle surface about the j3 and a axes, respectively, during deformation: 81 82
=
cfii(a, j3, z)
=
av(a, j3, z)
az az
(15.88) (15.89)
Note that Eq. (15.83) will be satisfied by Eq. (15.87) since the transverse displacement is completely defined by the middle surface component, w:
alE
833
aw
= -az = -az = 0
(15.90)
558
Vibration of Shells
Substituting Eqs. (15.85)-(15.87) into Eqs. (15.79) and (15.80), we obtain
£13
(1
:z
= A + :a) = A !... [u(a,
[A
fJ) + z8] ] A (1 + z/ Ra)
oz
+!..
+A
£23
u + Z 81 + Z (1)- A2(1 + z/ Ra)2
= B !...
1
Z )
]}
1 + Ra
ow
(15.91)
ooz [B
+!..
1 o( B (1 + z/ Rf3) oz v
+Z
1
ow
+B
ofJ
(82
_ v
v
= 82 - Rf3
B
B2 1
(1 :Z/Rp)
~;
ow(a, fJ) ofJ
8) 2
-
z)]}
V+Z82 0 [ ( B2 (1 + z/ Rf3)2 oz B 1 + Rf3
+.!. ow
+ Z 82 ~).
B
+B
(1 :Z/Rf3)]
[v(a, fJ) + Z 82] B (1 + z/ Rf3)
OZ
=B
0 [( oz A
+ -A -oa
+ :13)
= B (1
=B{
~:
ow oa u Ra
= 81 -
(1 :Z/Ra)
ow(a, fJ) oa
A
1 0 = A { A (1 + z/ Ra) oz (u 1
+A
(1 :Z/Ra)]
.. B ofJ
Rp
ow
+ -B -ofJ
(15.92)
To satisfy Eqs. (15.82), Eqs. (15.91) and (15.92) are set equal to zero. These give expressions for fh and 82:
ow oa ow
u 1 8] = - - - Ra A
v
1
82 = -
(15.93)
- - Rf3 B ofJ
(15.94)
Using Eqs. (15.81), (15.85)-(15.87), (15.93) and (15.94), the strains in the shell. Eqs. (15.75)-(15.80), can be expressed as 1
£11
=A
0
oa
(u
1 0 £22=--(V+Z82)+ B ofJ £33
=0
+ z8d +
v
u
+ z8, oA w - - +-
(15.95)
+ Z 01 -+0B w
(15.96)
AB
AB
ofJ
oa
Ra
Rp
(15.97)
15.3
a~
CI2=;
(V+BZO~)
559
(15.98)
0
(15.99)
=0
(15.100)
c23 = c31
a:
(u+AZOI)+~
Love's Approximations
The nonzero components of strain, Eqs. (15.95), (15.96), and (15.98) are commonly expressed by separating the membrane strains (which are independent of z) and bending strains (which are dependent on z) as
+ zkll C~2 + Zk22 C?2 + Zkl2
CIl = C?I
(15.101)
C22 =
(15.102)
CI2 =
(15.103)
where the membrane strains, denoted by the superscript 0, are given by
o cll
=
A
1 au act
+ AB
v aA afJ
B afJ
+ AB
olav c22 =
o cl2
u
A
a
= BafJ
(U)A
aB act B
w
+ Ra w
+ RfJ
a
+ A act
(V)B
(15.104) (15.105) (15.106)
and the parameters kll, k22,and kl2are given by 1 aOI kll=--+-A act 1
a (h
k22=--+-B a fJ
02 aA AB afJ 01
(15.107)
aB
(15.108)
AB act
kl2= A ~ (01) B afJ A
+B~
A act
(02) B
(15.109)
Notes 1. The parameters kll and k22denote the midsurface changes in curvature, and kl2 denotes the midsurface twist. 2. The strain-displacement relations given by Eqs. (15.101)-(15.109) define the thin shell theories of Love and Timoshenko. Based on the type of approximations used in the strain-displacement relations, other shell theories, those of such as (a) Byrne, Flugge, Goldenveizer, Lurye, arid Novozhilov; (b) Reissner, Naghdi, and Berry; (c) Vlasov; (d) Sanders; and (e) Mushtari and Donnell have also been used for the analysis of thin shells [5]. Example 15.4 Simplify the strain-displacement (15.109) for a circular cylindrical shell.
relations given by Eqs. (15.101)-
560
Vibrationof Shells
=
=
=
=
SOLUTION Noting that a x, fJ 0, A 1, B R, and the subscripts I and 2 refer to x and 0, respectively. Ra = Rx 00, RfJ = Re R, Eqs. (15.93) and (15.94) can be expressed for a circular cylindrical shell as
=
=
aw
Ox
= -- ax
Oe
=
(EI5.4.I) I
v R -
R
aw ae
(EI5.4.2)
The membrane strains are given by Eqs. (15.104)-(15.106):
au
o S x x
= -ax
(EI5.4.3)
olav see = R ao
o sxe The parameters kll,
k22,
k xx
k k
=
av ax
w
+R
(EI5.4.4)
I au ao
(EI5.4.5)
+R
and k12 can be determined from Eqs. (15.107)-(15.109) as _ aox __ a2w ax - ax2 (EI5.4.6)
1 aOe I au 1 a2w ee = R = R2 ao - R2 ao2
ea
aOe
I aox
1 av 2 a2w ax - R axao
a; + R ea = R
xe =
The total strains in the shell are given by Eqs. (15.101)-(15.103): au a2w Sxx s~x + zkxx Z -2 ax ax olav w z au z a2w See = eee + Zkee = R ae + R + R2 ae - R2 ae2
=
=- -
e - eO + k = au xe - xe z xe ax
+ 2. R
au ae
+~ R
2 av _ 2z a w ax R ax ae
(EI5.4.7) (E15.4.8)
(EI5.4.9) (EI5.4.1O) (EI5.4.H)
In Eqs. (EI5.4.1)-(EI5.4.II), u, u and w denote the components of displacement along the x, e, and z directions, respectively, in the midplane of the shell. Example 15.5 Simplify the strain-displacement (15.109) for a conical shell.
relations given by Eqs. (15.101)-
SOLUTION For a conical shell, a = x. fJ = 0, A = I, B = x sin ao, Ra = Rx = 00, and Rfl = Re = x tan ao· Using x and e for the subscripts I and 2, respectively, ex and ee can be obtained from Eqs. (15.93) and (15.94):
aw ax
ex
= --
ee
= --x tan ao
u
(EI5.5.1) 1 xsinaoae
aw
(EI5.5.2)
15.3
Love's 'Approx.im"tions
_.561
The strains in the shell are given by Eqs. (15.101)-(15.103): 8xx --
+ Z k xx + zk911 + Zk xII
0 8xx
= 8~9
899
8.t9 --
0 8 x(J
(E15.5.3) (E15.5.4) (EI5.5.5)
where the membrane and bending parts of the strain are given by Eqs. (15.104)(15.109):
o
8 xx
au
(EI5.5.6)
=ax
olav 89
9
o
u
= x----sin ao
a ()
au
1
8 x9
=----+x sin ao a ()
kx x
= =
=
ov ax
(EI5.5.8) (EI5.5.9)
o<;} ()()B
u
+ -x1.()x
cosao ov z x2 sin ao
ae -
~
ax
(()9) +
1
=--x
v
--x
ax
.1 x sm ao
kx9=X
(EI5.5.7)
= _ oZwZ
a ()x ax
k99
w
+ -x + x---tan ao
tan ao
x
x
xZ
1 sin ao
ov ax
1 oZw z sin ao a ()2
xZ
1 ow -
(EI5.5.10)
ax
:;
o()x
a ()
2v tan ao
-
1
oZw
. x sin ao
2
-+~--a ()Z x2 sin ao
ow
(EI5.5.11)
o()
Note that u,v, and w in Eqs. (EI5.5.1)-(EI5.5.11) denote the components of displacement along x, (), and Z directions, respectively, in the midplane of the shell. Example 15.6 Simplify the strain-displacement (15.109) for a spherical shell.
relations given by Eqs. (15.101)-
SOLUTION For a spherical shell, a =
()q, = R 09 =
..!. R
ow )
u - 0
(v __ ow) a 1_
sin
(EI5.6.1) (EI5.6.2)
562
Vibration of Shells
The membrane strains are given by Eqs. (15.104)-(15.106): o 1 au w Et/>t/> R a ¢ + R
=
o _
1 R sin ¢
Ee e ESe =
1
av
ae +
+
R
+ -1 -av _
au ae
-
Rsin¢
u cot ¢
(EI5.6.3)
w (EI5.6.4)
R
v cot ¢ R
R a¢
(EI5.6.5)
The total strains are given by Eqs. (15.101)-(15.103): E4>4>== E~4> + zk4>4>
(EI5.6.6)
+ Zke8 E4>8 = ES8 + Zk4>8 E88 = E28
(EI5.6.7) (EI5.6.8)
where k4>4>,k88, and k4>8 are given by Eqs. (15.107)-(15.109): 1 au 1 a2w k4>4>= R2 a ¢ - R2 a ¢2 k
88 -
1 av R2 sin ¢ a e 1
k4>8 = R2 sin ¢
. 1
1 R2 sin2 ¢
a2w a e2
cot¢
+ Ji2" u
1 a2w R2 sin ¢ a¢ a e
au a () -
a2wcos¢
R2 sin ¢ -8¢ ae
cos¢
-
Ji2"
(EI5.6.9) aw a¢
(EI5.6.1O)
1 av a¢
+ R2
aw
cot¢
+.R2 sin2 ¢ -a () - -v+ R2
cot¢
aw
R2 sin ¢ ae
Note that u, v, and w denote the components of displacement along ¢, directions, respectively, in the mid plane of the shell.
(EI5.6.1l)
e,
and z
15.4 STRESS-STRAIN RELATIONS In a three-dimensional isotropic body, such as a thin shell, the stresses are related to the strains by Hooke's law as ElI
1 =E
[all
-
v (0'22
+ 0'33) ]
(15.110)
V (all
+ 0'33) ]
(15.111)
V (all
+ 0'22) ]
(15.112)
1
E22 = E [0'22 -
1
E33 = - [0'33 E EI2=~
0'12
G 0'23
E23 = G 0'13
E13=
-
G
(15.113) (15.114) (15.115)
l5.5
Force and Moment Resultants
where <111, <122, and <133 are normal stresses, <1t2 = <12" <1t3 = <13t. and shear stresses. Based on Kirchhoff s hypothesis, we have
<123
= <132 are (15.116)
=0
e33
563
(15.117) Equations (15.112) and (15.116) yield <133
=
V(<1tl
+ (122)
(15.118)
But according to Love's third assumption, <133 = 0, which is in contradiction to Eq. (15.118). As stated earlier, this is an unavoidable contradiction in thin shell theory. Another contradiction is that <1t3 and <123 cannot be equal to zero because their integrals must be able to balance the transverse shear forces needed for satisfying the equilibrium conditions. However, the magnitudes of <113 and <123 are usually very small compared to those of <111, <122 and <112. Thus, the problem reduces to one of plane stress, described by 1
ell
=
-(<111
-
V(122)
(15.119)
e22
=
-(<122
1 E
-
V<1l1)
(15.120)
. e12
E
<112
(15.121)
=-
. G
Inverting these equations, we are able to express stresses in terms of strains as
E
= -1--2(ell + Ve22) -v E <122 = -1--2(e22 + Veil) -v E 0'12 = 2(1 + v) e12
<111
(15.122) (15.123) (15.124)
Substitution of Eqs. (15.101)-(15.103) into Eqs. (15.122)-(15.124) leads to <111
= --2I-vE
0 [ell
+ ve220 + z(kll + Vk22)]
(15.125)
E
0 [e22
+ Vell0 + Z(k22 + Vkll)]
(15.126)
<122 = --2
I-v
<112 =
15.5
2(1
E
+ v)
0 . (e12
+ Zk12)
(15.127)
FORCE AND MOMENT RESULTANTS Consider a differential element of the shell isolated from the shell by using four sections normal to its middle surface and tangential to the lines a, a + da, fJ, and fJ + dfJ as shown in Fig. 15.7. Here the middle surface of the shell is defined by abco. The stresses
564
Vibration of Shells z
t I I
Figure 15.7
Differential element-of shell. ab = dsg = B dj3, be = ds2 = Ada.
acting on the positive faces of the element are also shown in the figure. For the stresses, the first subscript denotes the face on which it is acting (1 for the face normal to the a curve and 2 for the face normal to the f3 curve), and the second subscript denotes the direction along which it is acting (1 for the a direction and 2 for the /3 direction). The force and moment resultants induced due to the various stresses are shown in Figs. 15.8 and 15.9. These resultants can be seen to be similar to those induced for a plate. To find the force resultants, .first consider the face of the element shown in Fig. 15.7 that is perpendicular to the a-axis (called the a face). The arc length ab of the intercept of the middle surface with the face is given by dsg = Bd/3
(15.128)
and the arc length a' b' of intercept of a parallel surface located at a distance the middle surface is given by
Z
from
(15.129) The stress 0"1) multiplied by the elemental area dS{3d z gives the elemental force d N]). Integration of the elemental force over the thickness of the shell gives the total in-plane
______________________
15.5
Figure 15.8
Force and Moment Resultants
·,565
Force resultants in a shell (positive directions are indicated).
fJ
Figure 15.9
Moment resultants in a shell (positive directions are indicated).
normal force acting on the face along the a direction as total force
=
2
h/
= lh/2
(TlIdsfJdz
l -h/2
(TuB
(
-h/2
Z )
1+RfJ
d{3dz
(15.130)
The force resultant Nu (force per unit length of the middle surface) can be obtained by dividing the total force by the arc length ds~ = B d{3: NlI =
l
2 h/
(TlI
-h/2
!I!!!!!!!!!!!!!I!!!!!!I!!!!!I!!!!!!!!!!I!!!!!!
__
••
(
1+
-z )
dz
(15.131)
RfJ
!!!!!!!!'!!!!_!!!!.~!!!!.,
.•!!!!.!!!! .• !!!!u!!!!.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
566
Vibration of Shells
By neglecting
z/
RfJ as a small quantity compared to 1, Eq. (15.131) gives h/2
Nil
=
j
0"1l
dz
(15.132)
-h/2
Using Eq. (15.125) in Eq. (15.132), we obtain NlJ
=
jh/2
E --2
1-
V
= 1 -E v2
[E?I
-h/2
+ VE~2 + z(klJ + Vk22)]
[0(ElJ + 0
h/2
VE22)(zLh/2
dz
+ (klJ + Vk22)
(Z)
:2/2
;
]
= 1 Eh _ v2
0+ 0
(ElJ
v E22)
(15.133) By defining the middle surface or membrane stiffness of the shell (C) as Eh
C=--2
(15.134)
I-v
Eq. (15.133) can be written as (15.135) The stress 0"12 acting on the ex face multiplied by the elemental area dSfJ dz (or the stress 0"21 = 0"12 acting on the f3 face multiplied by the elemental area ds()[ dz) gives the elemental force N12. Integration of the elemental force over the thickness of the shell gives the total in-plane shear force, which when divided by the arc length for the ex face and ds2 for the f3 face, ~givesthe force resultant NI2 as
dsg
NI2
- V) = N21 = C(I-2-
The transverse shear force resultant, can be found as
acting on the
Q13,
0
(15.136)
EI2 ex
face due to the shear stress
0"13,
h/2
Q13
Similarly, by integrating
0"22
=
j
(15.137)
0"13 dz -h/2
on the f3 face, the force resultant
N22
can be obtained as (15.138)
By integrating the shear stress 0"23 on the f3 face, we can find the transverse shear force resultant, Q23, acting on the f3 face as h/2
Q23
=
j
0"23 dz
(15.139)
-h/2
To find the bending moment resultants, we again consider the ex face of the element in Fig. 15.7. The moment of the elemental force (O"lJ dSfJ dz) about the f3 line will be ZO"II dSfJ dz, and the bending moment resultant MIl (m~ment per unit length of the
______
15.5 Force and Moment Resultants
'567
middle surface) can be found by integrating the elemental moment over the thickness and dividing the result by the arc length 'ds~ = B df:3. Thus, we obtain total moment = fh/2
C1IlZB
(1 + ~)
C1UZ (1 +
Mil = fh/2
z/
~)
(15.141)
dz
Rp
-h/2
By neglecting the quantity
(15.140)
df:3dz
Rp
-h/2
Rp in comparison to unity, Eq. (15.141) gives h/2
=
MIl
f
(15.142)
C1\1zdz
-h/2
By substituting Eq. (15.125) in Eq. (15.142), we obtain E
h/2
Mu=
f
-h/2
-1--2 [S?1 - v
E [ = 1 _ v2 (e?1
+ ve~2 + z(kll + Vk22)]Z
+ ve~2)
(
~
2 )h/2
-h/2
dz
+ (kIl + Vk22)
(
~
3
)h/2
]
-h/2
Eh3
= 12(1 _ v2) (kl ~+Vk22)
= D(kIl
+ Vk22)
(15.143)
where D, the bending stiffness of the shell, is defined by Eh3
D = 12(1 _ v2)
(15.144)
Using a similar procedure, the moment resultants due to the stresses C122and can be found as M22 = D(k22
+ vkll)
C112
= C121
(15.145)
v)
1 MI2 = M21 = D ( -2-
kl2
(15.146)
Notes
1. Although the shear strains e13 and S23 are assumed to be zero (according to Kirchhoff's hypothesis), we still assume the presence of the transverse shear stresses C113 and C123 and the corresponding shear force resultants, Q13 and Q23, as given by Eqs. (15.137) and (15.139). 2. The expressions of the force and moment resultants given in Eqs. (15.132)(15.146) have been derived by neglecting the quantities z/ Ra and z/ Rp in comparison to unity. This assumption is made in the thin shell theories of Love, Timoshenko, Reissner, Naghdi, Berry, Sanders, Mushtari, and Donnell. 3. If the quantities z/ Ra and z/ Rp are not neglected, we will find that NI2 =/: N21 and MI2 =1= M21 (unless Ra = Rp), although the shear stresses causing them are the same (i.e., C112 = (121)'
, I
i
I
I
I
~~,~"'~_~
I
568
Vibration of Shells
Example 15.7 Express the force and moment resultants in terms of displacements for a circular cylindrical shell. SOLUTION The force resultants are given in terms of membrane strains by Eqs. (15.135), (15.136), and (15.138). Using the strain-displacement relations given in Example 15.4, the force resultants can be expressed as) au ax
Nxx
= C(£~x + v£211) = C (
Nxll
- V) = Nllx = C(1-2-
o
0 + v£xx)
NIIII = C(£IIII
0
£xll (
=C
v av ae
+R
- V) = C(1-2-
1 av R ae
V)
+RW (av ax
(EI5.7.1) 1 au) + Rae
au) + wR + v ax
(EI5.7.2) (EI5.7.3)
where C is given by Eq. (15.134). The moment resultants, given by Eqs. (15.143), (15.145) and (15.146), can be expressed using the strain-displacement relations of Example 15.4 as Mxx = D(kxx
+ vkllll)
2 D (a w = - ax2
Mxe
(EI5.7.4)
v av ae
(EI5.7.5)
+ R2
1 aee = D ( -+ v-aex ) R ae ax 1 av 1 a2w a2 ( D R2 ae - R2 ae2 - v ax2
Mell = D(kell
=
a:; + Rvaell) &e 2 v a w) - R2 ae2 aex
=D (
+ vkxx)
w)
= Mex = D (1-- -2 v) = D (~)
(EI5.7.6)
2
e kxll = D (1-- - v) (ae-2 ax
(~av _ ~~) R ax
R axae
(EI5.7.7)
+ --1 aex) R ae
(EI5.7.8) (EI5.7.9)
where D is given by Eq. (15.144).
Example 15.8 Express the force and moment resultants in terms of displacements for a circular conical shell. SOLUTION Equations (15.135), (15.136), and (15.138) give the force resultants in terms of membrane strains. By using the strain-displacement relations of Example 15.5, )For a cylindrical shell. u, II and w denote the components of displacement along the axial (x), circumferential (9), and radial (z) directions, respectively, as shown in Fig. 15.10.
15.5 Force and Moment Resullants,S69
Figure 15.10
Coordinate system of a thin circular cylindrical shell.
the force resultants can be expressed as2
+ ve~lI)
Nxx = C(e~x
=
C[_au + (_1__ avao + ~ + _w_)] v
ax
x sinao
C (_1_;_V)
x
Nx(}
= N(}x
NIIII
= C (_l_;_V) (xs~ao :~ + :~ -;) = C(8~(} + ve~x)
=
= C ( __1_ x sinao
(E15.8.1)
x tanao
8~(}
(E15.8.2)
av +'~ + __w_ + v au)
ao
x
x tanao
(E15.8.3)
ax
where C is given by Eq. (15.134). The moment resultants are given by Eqs. (15.143), (15.145), and (15.146). Using the strain-displacement relations of Example 15.5, the moment resultants Can be expressed as
(E15.8.4)
cosao av 1 a2w 1 aw a2w) = D ( x2 sin2 ao ao - x2 sin2 ao a02 - -; ax -v ax2 Mxll
= Mllx = D (_1 _;_v) =D
(_l_;_V)
(E 15.8.5)
kxll
(-x-ta-~-a-o-:-~- -x""'2-~-v-a-o--x si~ao
~:~
+ x2 s~nao ~;) (E15.8.6)
where D is given by Eq. (15.144). 2For a circular conical shell. u, v. and w denote the components of displacement along the generator (x). circumferential (6), and radial (z) directions, respectively. as shown in Fig. 15.11.
I-
570
Vibration of Shells -A-r
I \\II- 1\ I_ I \\ I \\ \-
I
\\
I
\\ \ I \
~
Figure 15.11 Components of displacement in a circular conical shell. u, v, and w are along the x, e, and z directions.
Example 15.9
Express the force and moment resultants in terms of displacements for a spherical shell. --'
SOLUTION Using the force resultant-membrane strain relations given by Eqs. (15.135), (15.136) and (15.138), and the strain-displacement relations given in Example 15.6, the force resultants can be expressed as N¢¢ = C(£~¢
+ v£29)
1 au
= C [ R a¢
w
+ R +v
(I-V)
N¢B = N9¢ = C -2= C (12 N&B
V)
(1 R sin ¢
av
u cot¢
ae + -R-
W)]
+R
(EI5.9.1)
£¢9
(_I_au R sin ¢
ae
+.!.. av _ R a¢
vcot¢) R
(EI5.9.2)
= C(£29 + v£~¢) = C [_1_ R sin ¢
au +.!.. av _ ae
R a¢
vcot¢ R
+ v (!!.. _ R
2. aw)] R a¢
(EI5.9.3)
where C is given by Eq. (15.134). Using the moment resultant-bending strain relations given by Eqs. (15.143), (15.145), and (15.146) and the strain-displacement relations
_________
Strain Energy. Kinetic Ene~gy. and Work Done by External Forces
15.6 given in Example
M¢¢ = D(k¢¢
15.6, the moment resultants can be expressed
1 a2w R2 a4J2
(
+v
1 av R2 sin 4J ae -
2
1 aw R2 sin2 4J ae2
aw)]
cot4J u _ cos4J R2 R2 a4J
Moo = D(koo
(EI5.9.4)
+ vk¢¢) 2 I a_ _w+ _co_t_4J u R2 sin2 4J ae2 R2
= D [ __ I __ av R2 sin4J ae _ cottj) aw R2 a4J
M¢o
as3
+ vkoo)
1 au = D [ R2 a4J -
+
571
v
+
(_I
I
2 au __ a w)] R2 a4J R2 a4J2
(EI5.9.5)
= Mo¢ = D (_I _;_V) k¢o I
-
V) (
= D ( -2-
+ _I__ a_v _ R2 a4J
2 1 au 1 a w R2 sin4J ae - R2 sin4J a4J ae
1 .a2 w R2 sin 4J a4J ae
+ _c_os_4J__ R2
sin2
a_w __co_t_4J v 4J ae R2
+ _c_o_t4J __
a_w) R2 sin 4J ae (EI5.9.6)
where D is given by Eq. (15.144).
15.6 STRAIN ENERGY, KINETIC ENERGY, AND WORK DONE BY EXTERNAL FORCES To derive the equations of motion of the shell using Hamilton's principle, the expressions for strain energy, kinetic energy. and work done by the external forces are required. These expressions are derived in this section.
15.6.1 Strain Energy The strain energy density or strain energy per unit volume of an elastic body is given by (15.147) The strain energy (rr) of the shell can be found by integrating the strain energy density over the volume of the shell: rr = ~
f ff
(0'11£11
+ 0'22£22 + 0'33£33 + 0'\2£12 + 0'23£23 + 0'31£31)
dV
(15.148)
v 3For a spherical shell, U, II, and w denote the components of displacement along the 4J, 6, and respectively, as shown in Fig. 15.12.
!!!!!!!!!!!!!!!~_~~~~"~~_,,
z
directions,
.••••~ ..
~J
I
572
Vibration of Shells
Figure 15.12 Components of displacement in a spherical shell. u, v, and w are along the cp, e, and z directions.
where the volume of an infinitesimal shell element is given by (15.149) By neglecting the quantities expressed as
z/
Ra
z/
and
R{3
in comparison to 1, Eq. (15.149) can be
dV = ABdctdf3dz
(15.150)
Note that the transverse normal stress 0"33 is assumed to be zero according to Love's third assumption in Eq. (15.148). At the same time, the transverse shear strains 823 and 831 are retained in Eq. (15.148) although they were assumed to be zero earlier [Eqs. (15.99) and (15.100»). By substituting Eqs. (15.91), (15.92), (15.95), (15.96) and (15.98), respectively, for the strains 813,823,811,822, and 812, and Eq. (15.150) for dV into Eq. (15.148), we obtain the strain energy of the shell as
7r=-111 2
a
[ {0"11[1- (ou -+z{3
A
l
1 +0"22 [ B
(OV-
of3
Iou
+ 0"l2 [ Ji of3 +~
0V A oct
_
+ 0"31( OJ -
+z-
U oA AB of3
0B AB Oct
~
U
Ra
(02) of3
1 oA +--(v+z~)+AB of3
(01) Oct
oct
1 oB + --(U
AB oct
z oOJ
+ EM
+!:...002 _ A Oct
+ AlOW)} Oct
-
+ZOI)
w ] Ra
+-.w] Rp
zOJ oA AB of3
02 0 B] AB Oct Z
ABdctdf3dz
+ 0"23(02
w)
_ ...!:... + ~ 0 R{3 B of3
(15.151)
Performing the integration with respect to z and using the definitions of the force and moment resultants given in Eqs. (15.135)-(15.139), (15.143), (15.145), and (15.146),
________
-I
15.6 Strain Energy. Kinetic Energy. and Work Done by External Forces
573
the strain energy [Eq. (15.151)] can be expressed as
111
rr = -
2
a fJ
(au NIIB-
aa
aA + NttAB+ MIIB- a(}t + NII-vaA + MII-{}.z oa
OV o~ + NzzA-{3 + MzzAo a{3 ou oA + NI2A- Ntz-u a{3 a{3 + NI2B-
o{3
oB
aB
+ Nzz-uoa + Mzz-(}t oa o(}t
o(}z
+ MtzB- oa
w
+ NZ2AB-RfJ
MI2-(}t
aB - MI2-(}z aa
aw
V
+ QZ3AB(}z - QZ3AB-
RfJ
+ Q13AB(}l - QI3AB-
w Ra
oA a{3
+ MI2A- a{3 -
ov oB - Ntz-V aa aa
a{3
u
Ra
+ QZ3A-
a{3
+ QI3B- aw)
oa
(15.152)
da d{3
15.6.2 Kinetic Energy The kinetic energy of an infinitesimal element of the shell can be expressed as dT
-
1
·2
== 2P(U
·2
+v
. 2
(15.153)
+W )dV
where P indicates the density of the shell and a dot over a symbol denotes the derivative with respect to time. By substituting Eqs. (15.85)-(15.87) for ii, v, and w into Eq. (15.153), and integrating the resulting expression over the volume of the shell, we obtain the total kinetic energy of the shell as T =
~p
2
[ [1[U2 lalfJ z
+ ii + w2 + a~(Ot + Oi)
+ 2a3(UOl + vOz)]AB
(1 + :a) (1 + :fJ)
da d{3dz
(15.154)
By neglecting the terms z/ Ra and z/ RfJ in comparison to 1 in Eq. (13.154) and performing the integration with respect to z between the limits -h/2 to h/2 yields Z T= ~P li[h(U2+v2+w )+
(15.155)
~~(Ot+Oi)]ABdad{3
15.6.3 Work Done by External Forces The work done by external forces (W) indicates the work done by the components of the distributed loads fa, ffJ' and fz along the a, {3, and z directions, respectively, (shown in Fig. 15.13), and the force and moment resultants acting on the boundaries of the shell defined by constant values of exand {3(shown in Fig. 15.14). All the external
..,
I
~
__
~~!!'!'!!!!!!!!!!J
574
Vibration of Shells
j
Figure 15.13 Distributed loads fa, fp, and fz acting along the ex, f3, and middle surface of a shell.
Figure 15.14
z directions in the
Prescribed force and moment resultants at the boundary of a shell.
loads are assumed to act in the middle surface of the shell. The work done (W d) by the components of the distributed load is given by Wd
== =
Lh Lh
(fau
+ !fJv + !zw)
(fau
+ !fJv + !zw)AB
ds2 dsg (15.156)
da df3
The work done by the boundary force and moment resultants (Wb) can be expressed as Wb =
L
(N22V
+ {
Jp
(Nuu
+ N21U + Q23W + M2202 + M2]OdA + N]2V + Ql3w+
MlIO]
+ M1202)B .
da df3
(15.157)
15.7 Equations of Motion from Hamilton's Principle
575
where a bar over a symbol denotes a prescribed or specified quantity. Thus, the total work done (W) by the external forces is given by (15.158)
15.7 EQUATIONS OF MOTION FROM HAMILTON'S The generalized
Hamilton's 8
PRINCIPLE
principle can be stated as
1'2
L dt = 0
'I
1'2
(T -
+ W)
1f
dt
=0
(15.159)
'1
where L denotes the Lagrangian, and the kinetic energy T, strain energy 1f, and \\iork done by external forces, W, are given by Eqs. (15.155), (15.152), and (15.158), respectively. Equation (15.159) is rewritten as
1
'2
(oT - 01f
=0
+ 8W)dt
(15.160)
'1
T,
For convenience, integrals of the variations of as described next. 15.7.1
1f,
and
W are evaluated individually
Variation of Kinetic Energy
/,:2 8T dt
From Eq. (15.155),
['2
1"
can be written as
8T dt = p~
+
l',)'21alpr [u 8u + ti 8ti + W 8w 2 h (01801
12
+ 02 8(2)]
(15.161)
AB da d{3dt
The integration of individual terms on the right-hand side ofEq. (15.161) can be carried by parts. For example, the first term can be evaluated as
1'211 ,)
u8udad{3dt a pap
= =
11 1'2 11 ( 1 dad{3
p
a
au a(8u) a-a -dt
t
,)
dad{3
t
'2 a2u
-
-a 2 8udt tat
,)
au
+ -ou
1(2)
(15.162)
,)
Since the displacement component, u, is specified at the initial and final times (tl and t2), its variation is zero at tl and t2, and hence Eq. (15.162) becomes
1'21 r ,)
uou da d{3dt = -
alp
['21 r
1,)
U ou da d{3dt
(15.163)
alp
Thus, Eq. (15.161) can be expressed as
['2
1"
oT dt = -ph
+
____________________
!!!!!I!!!
['21 r
1,)
[u 8u
+ ii8v + wow
alp
2 h (018611 + 82 0(12)] AB da d{3dt 12
!!!!!!!!!_!!!!!!!!!
(15.164)
'._~'"!'!!!-.!!!!!!!!,~,!,!!!,,!,!,!!!!!,!!!!!,!!!!,!!!!!!!!!!!!,!!!!,!!!!!!!~~'!!!!!!
576
Vibration of Shells
Note that the terms involving 81 and 8z denote the effect of rotatory inertia. By neglecting the effect of rotatory inertia (as in the case of Euler-Bernoulli or thin beam theory), Eq. (15.164) can be expressed as
1
1 1r (u '2
'28T
dt = -ph
~
15.7.2
~
+ w 8w)AB
8u + v 8v
da df3dt
(15.165)
ex~
Variation of Strain Energy From Eq. (15.152),1,:2 8n dt can be written as
1
'2
8ndt =
1'211
'1
,)
ex
[8(8U) NllB--
+ MllB-- 8(881)
oa
(3
8A
8a
8A
+ Nll-8v
8f3
+ Nll-8w
8f3
8(8v)
+ NzzA--
AB
+ Mll-Mh
8f3
Rex
8 (88z)
+ MzzA--
8B
AB
8a
Rp
8f3
8B
+ Nzz-8u 8a
+ Mzz-881 + Nzz-8w 8(ou)
+ NlzA--
8A - N1z-8u 8f3 8f3
8A
- Mlz-cS81 .. 8f3
8(cSv)
+ NIZB-- 8a
8(88z)
+ M1zB-- 8a
8(Mh)
+ M1zA--
8B - Mlz-;-88z va
8f3
8B - Nlz-cSv 8a
+ QZ3ABo8z
AB 8(cSw) - QZ3R{3ov + QZ3A + QJ3AB081
M
AB
-
QI3-
Rex
8(OW)] da df3dt 8a
oU + Q13B--
(15.166)
The terms involving partial derivatives of variations, 8u, ov, 8w, o8b and 88z, in Eq. (15.166) can be evaluated using integration by parts. For example, the first term can be evaluated as
1'211 1 [-1 r oa
NllB--8(8u) dadf3dr
,)
ex
8a
{3
'2
=
~(NllB)Ou
,)
ex J{3
da df3 +
r
J{3
NIIB8u df3] dt
(15.167)
where the second term on the right-hand side of Eq. (15.167) denotes the contribution to the boundary condition. Thus, Eq. (15.166) can be expressed as
1" 1'211[ - ondt =
I)
'1
ex
(3
-- 0 (NllB)cSu - -(M 0 llB)881 8a
8a
8A
8A
of3
of3 -
+ NII-cSv + Mil-of),
AB
+ Nll-8w
Rex
15.7 Equations of Motion from Hamilton·s Principle ',577 o - -(N22A) o~
0 8v - -(M22A) o~
oB
AB
+ M22-Mh
Rp
oA - N12-8u o~
0 - -(MI2A) o~
o - -(NI2B)8v oa oB - M12-802 oa
[
101
+
AB - QI3-8u
Ra
[(NIlB
8u
+ MuB
802 0 - -(Q23A) o~
0 - -(Q13B)8w
]
oa
+ M22A 802 + NI2A
(N22 A 8v
12
8w
dad~dt
8u
+ MI2A
801 + Q23A 8w) da dt
801 + NI2B 8v
+ MI2B
802 + QI3B 8w) d~ dt
p
II
15.7.3
oA 801 - M12-80I o~
0 - -(MI2B) oa AB Q23AB 802 - Q23-8v Rp
12
II
oa
oB - N12-8v oa
+Q13AB801
1 +1 l
oB
+ N22-8u
0 - -(NI2A)8u o~
+ N22-8w
oa
+
8fh
(15.168)
Variation of Work Done by External Forces
1,:2 8W dt
From Eq. (15.158),
1
128W
dt
=
II
1
12 (8Wd
II
1 +h
+
can be expressed
+ 8Wb)
as
1 lot 1 12
dt =
[[
I(
[(fa
8u
+ fp
8v
+ fz
8w) AB da d~
p
(N228v
+ N218u + Q23 8w + M22Mh + M2180J)Ada
(N1l8u
+ NI28v + Q13 8w + MI1801 + MI28fh)Bd~]
dt (15.169)
15.7.4
Equations of Motion Finally, by substituting Eqs. (15.165), (15.168) and (15.169) into Eq. (15.160), Hamilton's principle can be expressed as follows:
___________
I!!!!!!
!!!!!II!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!~~===~~====~:::::I.J
578
Vibration of Shells
+ [O(QI3B) + O(Q23A) _ oa
AB _ N22AB Rp
Nil
of3
Ra
+ (fz
0 + [ -(MllB) oa
0 + -(M21A) of3
oB - M22~ va
+ M12-oA of3
+
+ ~(M22A) of3
- Mll :~
+ M21 oB oa
+
[~(MI2B) oa /2 1[(N21
l
'1
N22) av
]
Mh
Mh} da
- Q23AB]
+ (Q23 -
- M22) 002 + (M21 - M21) aOdAda
1J /2
{[(N11
'1
+ (N22 -
Q13AB
ow
df3 dt
Q23) oW
a
+ (M22 +
- N21) ou
VfJ
_ PhW)AB]
N11)OU + (N12 - N12)av
-
dt
+ (Q13 -
Q13)aw
p
+ (M 11 -
M11) 001 + (M
M12) 002]B df3 dt = 0
12 -
(15.170)
To satisfy Eq. (15.170), the terms involving the triple and double integrals must be set equal to zero individually. By setting the term involving the triple integral equal to zero, we can obtain the equations of motion of the shell. When the terms involving the double integrals are set equal to zero individually, we can derive the boundary conditions of the shell. First we set the triple integral term equal to zero.
h{[
[/21
+ :f3 (N21A) + N12 ~;
oOa(N11B)
-N22-
oB
oa
+ Q13-AB + (fa
O(NI2B) oa
+[
+
O(N22A) of3
oB
oa
_ N11 AB _ N22AB Ra Rp
of3
a + -(M2IA) af3
0 + [ -(M11B) aa 0 (MI2B) aa
+ [-
Q23AB]
aB - M22~ va
a
+ -(M22A)
Mh}
oA
u]
AB
+ N21 oa - Nll of3 + Q23 Rp + (fp - phv)AB
+ [O(QI3B) + O(Q23A)
-
.] - phii)AB au
Ra
-
of3
da df3 dt
aA af3
MII-
+ (f
+ M12-oA of3
_ z
ov
h U)AB] ow P w
Q13AB
]
801
aB aa
+ M21-
=0
(15.171)
In Eq. (15.171), the variations of displacements, OU,OV,ow, 001, and 082, are arbitrary and hence their coefficients must be equal to zero individually. This yields the following equations, also known as Love's equations: a(N11B) oa
-
a(N21A) of3·
oA of3
- N12-
oB
+ N22-oa
- AB-
Q13 Ra
+ ABphii
= ABia (15.172)
15.7 Equations of Motion from Hamilton's Principle ~579
a(NI2B) ____ aa
- a(N"2A)aB - N21afJ aa a(Q13B) aCt
+ NIl- aA
- AB-~Q"3 afJ RfJ
+ AB
_ a(Q23A) afJ
a(MIlB) a(M2IA) ---+ ---+ aCt afJ
MI2-
a(M12B) a(M22A) ---+---+
M
aa
afJ
(Nil Ra
+ ABphiJ ' = ABf{3
+ N22) + ABw
= ABf.
R{3
aA afJ
(15.174)
-
aA aB Q - M22nAB afJ aa
aB 12--MIl-aa
(15.173)
Q
23A
= B
=
0
(15.175)
0
(15.176)
Here Eqs. (15.172)-(15.174) denote the equations of motion of the shell for motions in the a, fJ, and z directions, respectively, and Eqs. (15.175) and (15.176) indicate how the transverse shear force resultants Q 13 and Q23 are related to the various moment resultants. 15.7.5
Boundary Conditions Next, each of the tenns involving the double integral is set equal to zero in Eq. (15.170) to obtain
(2 [ [(N21
1'1 1a
+ (M21 (2 [ [(N
1'1 1{3
_ N21)OU
- M21)o(h+
11 -
+ (M II
+ (N22
-
Nil) ou
+ (Q23
(M22 - M22)082]Ada
+ (N
Mil) 081
- N22) ov
12 -
+ (M 12 -
N12) ov
+ (Q13
- Q23) OW dt
=0
(15.177)
- Q13) OW
M12) 082]B dfJ dt = 0
(15.178)
It appears that Eqs. (15.177) and (15.178) will be satisfied only if either the variation of the displacement component (ou, OV, ow, 0810 or 082) or its coefficient is zero in each of these equations: N 21 - N21 = 0 N22 - N22 = 0 Q23 - Q23 0 M21-M21 =0 M22-M22=0
or or or or
N 11 - Nil = 0 N 12 - NI2 0 Q 13 - Q 13 = 0 Mil - Mil 0 M 12 - M12 = 0
or or or or
=
=
=
OU= 0 OV 0 ow 0 081 =0 or 082=0
= =
at fJ
= fi = constant
(15.179)
at
= a = constant
(15.180)
= =
OU 0 OV 0 0W = 0 OBI= 0 or 082 = 0
Ct
Equations (15.179) and (15.180) indicate that there are 10 boundary conditions for the problem. However, from the force resultant/moment resultant-stress relations, stress-strain relations, and strain-displacement relations, we can find that the three'
___________________________________
.... -----------""'!!,.d
580
Vibration of Shells
equations of motion, Eqs. (15.172)-(15.174), together involve partial derivatives of order 8 involving spatial variables. This indicates that we can have only four boundary conditions on any edge. Before developing the actual boundary conditions, we assume that the boundaries coincide with the coordinate curves, that is, the a and f3 coordinates. Let a boundary of the shell be defined by the line f3 constant. The five force resultants (including the moment resultants) acting on this boundary are given by N21, N22, Q23, M21, and M22. The deformation on this boundary is characterized by the five displacements (including the slopes), u,v,w'(h, and 02, which correspond to N21, N22, Q23, M21, and M22, respectively. Equations (15.179) indicate that either one of the five force resultants or its corresponding displacement must be prescribed on the boundary, f3 = = constant. However, this is not true because the five quantities indicated (either force resultants or the displacements) are not independent. For example, the slope 02 is related to the displacements wand v in order to preserve the normal to the middle surface after deformation to satisfy Kirchhoff's hypothesis (Love's fourth assumption). Thus, the number of independent displacements (and hence the corresponding generalized forces) will only be four. Hence, only four boundary conditions need to be prescribed on each edge of the shell. To identify the four boundary conditions for the edge, f3 constant, we rewrite Eg. (15.177) by expressing 01 in terms of u and w using Eg. (15.93) as
= -p =
-p
= -p =
1/ 1 2
{(N21 - N21)
/1
ou + (N22
- N22)
ov + (Q23
- Q23)OW
a
+ (M21
- M2l)
[ -ou
Ra
1 0 - --(ow) A oa
The term involving o(ow)joa
]
+ (M22
- M22) 002 } Adadt
=0
(15.181)
can be integrated by parts as
(15.182)
where the first term on the right-hand side is equal to zero since M21 = M21 along the edge on which a varies. Using the resulting equation (15.182) in Eq. (15.181) and collecting the coefficients of the variations OU, ov, ow, and 002, we obtain
[' f.{[ + [(-
(N21
Q23
+
~:1)~
+ A1 ~aM21)
(N21
- ( Q23
+(M22 - M22) 09,} Ad. dt
+
~:1)]ou +
aM2l)] + A1 ---a;-
=0
(N22 - N22) 0"
ow
(15.183)
15.7 Equations of Motion from Hamilton's Principle
581
fh in tenns of
v and
A similar procedure can be used with Eq. (15.178) by expressing w using Eq. (15.94) to obtain
f £ (Nil + [(-
Q13
+ (M
II -
- NII)8u
~:2)- (N12 + ~:2)]
+ [(N12 +
8M 12) + B1 ----a{3 -
(
8,
8MI2)] + B1 ----a{3 . 8w
QI3
Mil) 89,} B dP dt ~ 0
(15.184)
Defining the effective inplane shear force resultants FI2 and F21 as F12 = N12
+ -M12
(15.185)
F21 = N21
+ -M21
(15.186)
1 8M12 + ---
(15.187)
Rp
Ra and the effective transverse shear force resultants VB and V23 as VB = Q13 ". V23
B
8fJ
1 8M21
+ ---
= Q23
(15.188)
'" A 8a Eqs. (15.183) and (15.184) can be expressed as
1
2
- F21) 8u
1[(F21
t
+ (N22
- N22) 8v
a
t1
+ (V23
1 lp
- V23) 8w
+ (M22
2
t
{[(N
u - Nu) 8u
+ (F
- M22) 8lh]Ada dt = 0 12 -
F12) 8v
+ (V 13 -
(15.189)
VB) 8w
t1
+ (M u -
(15.190)
Mu) 8ellB dfJ dt = 0
Equations (15.189) and (15.190) will be satisfied only when each of the displacement variations or its coefficient will be zero. Noting that the variation in a displacement will be zero only when the displacement is prescribed. the boundary conditions can be stated as follows:
= = =
= = =
F21 F 21 or u ii } N22 = N22 or v = V V23 V23 or w w M22 M 22 or ~ 92
at
fJ =
-p = constant
(15.191)
Nu = Nu F12 = F12 V13 = VI3 Mu = Mu
at
a=
a = constant
(15.192)
or u =u or v =v or w=w or el = 91
}
582
Vibrationof Shells
el,
where ii, V, w, and e2 denote the prescribed values of u, v, w, OJ, and 02, respectively. Equations (15.191) and (15.192) represent the four independent boundary conditions to be satisfied in solving the equations of motion of shells. The boundary conditions for some of the commonly encountered edges can be stated as follows:
= -p =
1. For the edge defined by f3
constant:
(a) Clamped or fixed edge: u
=0
= 0,
v
w = 0,
02
=0
(15.193)
(b) Hinged or simply supported edge with the support free to move in the normal direction: u =0,
(15.194)
v =0,
(c) Hinged or simply supported edge with no motion permitted in the normal direction: w=O,
v =0,
u =0,
=0
M22
(15.195)
(d) Free edge:
= 0,
N22
= 0,
F2l
2. For the edge defined by ex
=a=
V23
= 0,
M22
=0
(15.196)
constant:
(a) Clamped or fixed edge: u =0,
v
= 0,
w=O,
01
=0
(15.197)
(b) Hinged or simply supported edge with the support free to move in the normal direction: u =0,
v =0,
Mll
= 0,
Vl3
=0
(15.198)
(c) Hinged or simply supported edge with no motion permitted in the normal direction: u =0,
v
= 0,
w=O,
=0
Mil
(15.199)
(d) Free edge: Nll
15.8
= 0,
F12
= 0,
Vl3
= 0,
Mil
=0
(15.200)
CIRCULAR CYLINDRICAL SHELLS For a cylindrical shell, x, 0, and z are used as the independent coordinates, as shown in Fig. 15.10. The components of displacement parallel to the x, 0 and z directions are denoted as u, v and w, respectively. The radius of the shell is assumed to.be R. The parameters of the shell are given by (see Example 15.1): CI.
=x,
A=l,
f3
= 0,
B=R
(15.201)
The radius of curvatures of the x and 0 lines are given by Ra
= Rx = 00,
RfJ = R(} = R
(15.202)
15.~
15.8.1
Circular Cylindrical
Shells
583
Equations of Motion The equations of motion, Eqs. (15.172)-(15.174)
reduce to
x .!. aNo f ae + x -
aNx.x ax
+R
---ao + R + = 1 f + ---ao - R + • = 1 aNoo
aNxo
~
Qo.
+R
aQx:
~
"
JO
aQoz
R
Noo
hit
(15.203)
h" P v
(15.204)
h" P w
(15.205)
P
The relations between the transverse shear force resultants and the moment resultants, Eqs. (15.175) and (15.176), reduce to aM
Q
(15.207)
ae
+
1 - v a2ex 2R2
1 + v 02eo ) + 2:R ax ae 0200 1 + v 020x ) ae2 + 2:R ax ae
(15.208)
ae2
1
v 02eo
1-Qoz = D (
R
Qxz and Qo. [Eqs. (15.206) and (15.207)] can be
a2ex D ( ox2
=
(15.206)
+ .!. aMoo
_ oMxo o. ax
Using Eqs. (EI5.7.5)-(EI5.7.7), expressed as Qxz
1 aMox
xx = a;+ Rae
Qxz
7':\ ox2 + R2
(15.209)
Substituting Eqs. (EI5.4.!) and (EI5.4.2) into Eqs. (15.208) and (15.209) leads to
+
1 + v a2v 2R2 oxoe
1 - v a2v
1 02v
a3w Qxz = D ( - ax3
= D ( 2:R
Qo.
ax2
+ R3
1 03W) - R2 axae2
1 a3w
(15.210)
1 a3w)
oe2 - R3 ae3 -
R ax2 ao
(15.211)
Finally, using Eqs. (EI5.7.1)-(EI5.7.3) and (EI5.7.5)-(EI5.7.7) and Eqs. (15.210) and (15.211), the equations of motion, Eqs. (15.203)-(15.205), can be expressed in terms of the displacement components u,v, and w as a2u C ( ax2
+
v a2u
12R2
1 - v a2v c ( -2- ax2
oe2
1 - va2v
+D (
a4w D ( - ax4 -
+ R ax + 2:R
1 a2v
+ R2
2R2 ax2
1 + v a2v )
v aw
ae2
+ R4 ae2
+~
ax oe
1 a3w
1
oe
1 a2v
-
h ..
(15.212)
=P u
a3w) R4 ae3 - R2 ax2 ae
a3v 2 04W 1 a4w R2 ax2 ae - R2 ax2 ao2 - R4 ae4
1
+
+ Ix
1 + v a2u )
1 aw
+ R2
ax ae
c ( R21 av w v au) ae + R2 + R ax +
"
Jz
= p h w..
+ 10 = phv
.. (15.213)
v)
1 a3
+ R4 ae3
(15.214)
584
Vibration of Shells
15.8.2
Donnell-Mushtari-
Vlasov Theory
The equations of motion of a cylindrical shell, Eqs. (15.212)-(15.214), can be simplified using the Donnell-Mushtari-Vlasov (DMV) theory. The following assumptions are made in the DMV theory in the context of vibration of cylindrical shells: 1. The contribution of in-plane displacements u and v to the bending strain parameters kll, k22, and k12 [in Eqs. (E15.4.6)-(EI5.4.8)] is negligible. 2. The effect of the shear term (1/ R)Q(Jz in the equation of motion corresponding to v [Eq. (15.204)] is negligible. This is equivalent to neglecting the term involving D in Eq. (15.213). The equations of motion corresponding to the DMV theory can be expressed as follows: 1 :...va2u v aw 1 + v a2v (1 - v2)p ,j2u + 2R2 ae2 + R + 2R ax ae E at2 1 - v a2v 1 a2v 1 aw 1 + v a2u (1 - v2)p a2v -2ax2 + R2 ae2 + R2 7iB+ 2R ax ae E at2 a2u ax2
a;
=
=
au
V
-
(
R ax
_ h2 (a4w 12 ax4
15.8.3
1
+ R2 +
av
(15.215) (15.216)
w )
+ R2
ae
a4w R2 ax2 ae2
2-
2- a4w)
+ R4
ae4
_-
2
2
(1- v )p a w E at2
(15.217)
Natural Frequencies of Vibration AccordiIig to DMV Theory Let the circular cylindrical shell, oflength 1, be simply supported on its edges as shown in Fig. 15.15(a). The boundary conditions of the shell, simply supported at x = 0 and x 1, can be stated as follows:
=
e, t) w(O, e, t) v(O,
Nxx(O,e,t)=c
=0
(15.218)
=0
(15.219)
au
(
v av Rv)
ax+Rae+
a2w Mxx(O, e, t) = D ( - ax2
+
(O,e,t)=o
(15.220)
v av v a2w) R2 ae - R2 ae2 (0, e, t) = 0
(15.221)
W
v(l,e,t)=0
(15.222)
w(l,e,t)=o
(15.223)
Nxx(l,e,t)
v av + -wv) = C (au-ax + -R ae R
a2w Mxx(l, e, t) = D ( - ax2
+
(l,e,t)
=0
v av v a2w) R2 ae - R2 ae2 (l, e, t) = 0
(15.224) (15.225)
15.8 Circular Cylindrical Shells
585
(a)
,..... --/
,'-~
/
/
1-.-'-'-'-'-'-'-'/'--
',
/
,
'-~
/
/
~--,
.-
/
'--'"
m=4 (b)
Figure 15.15 (a) Cylindrical shell (simply supported); (b) Displacement pattern during vibration. The solution of the equations of motion corresponding to DMV theory, Eqs. (15.215)(15.217), is assumed to be in the following form: u(x. 0) =
~~
L.-, L.-, Amn
mrrx cos -/cos nO cos wt
(15.226)
m '. n
v(x, 0) =
~
m
w(x, 0) =
~
L.-, L.-, Bmn ~
. mrr x.
(15.227)
mrrx -/cosnO coswt
(15.228)
sm
n17
n
~
L.-, L.-, Cmn sin m
II
cos wt
sm -/-
n
where m is the number of half-waves of displacement in the length of shell, n is the number of half-waves of displacement in the circumference of the shell [see Fig. 15.15(b)]. and Amn. Bmn, and Cmn are constants. Note that the assumed solution [Eqs. (15.226)-(15.228)] satisfies the boundary conditions [Eqs. (15.218)-(15.225)] as well as the periodicity condition in the circumferential (0) direction. Substituting Eqs. (15.226)-(15.228) into the equations of motion, Eqs. (15.215)-(15.217), yields the following equations:
+ Q)Amn + (a2A.n)Bmn + (VA.)Cmn = 0 . 2 2 (a2A.n)Amn + (-alA. - n + n}Bmn + (-n)Cmn = 0 + (-n)Bmn + (-1 - A.41J.- 2A.2n21J. - n41J. + Q)Cmn = 0 (_A.2
(vA.)Amn
-
aln2
(15.229) (15.230) (15.231)
where mrrR
A.=-/ h2 1J. = 12R2
(15.232) (15.233)
586
Vibration of Shells
n=
(1 - v2)R2p
W2
(15.234)
E I-v
al=--
(15.235)
a2
(15.236)
2 l+v = -2
For a nontrivial solution of the constants Amn, Bmn, and Cmn: the determinant of their coefficient matrix in Eqs. (15.229)-(15.231) must be zero. This leads to -A2-aln2+n a2An
VA
An
a2
-a1A2 - n2
-n
VA
+n
-n -}
- A4f.l-
2A2n2f.l-
n4f.l
+ n)
=0 (15.237)
The expansion of Eq. (15.237) leads to the frequency equation n3+bln2+b2n+b3
(15.238)
=0
where
+ 2n2 f.l) - n2(1 + al) - A4f.l - n4 f.l - 1 b2 = A6 f.l(1 + al) + A4(al + 3n2f.l + 3aln2 f.l) + A2(1 + n2+ ain2 - a~n2:- v2 + al + 3aln4f.l + 3n4f.l) + n6f.l(1 + al) + n4al + n2al' b3 = -A8ali.t - A6n2f.l(1 + 2al - a~ + ai) - A4(al + 2aln4 f.l + 2ain4 f.l + 2n4 J.L - 2a~n4 f.l - a2v2) - A2[n6 f.l(1 + ai + 2al - a~) + n2( -a~ + ai + 2a2v - v2)] bl
= -A 2(1 + al
(15.239)
(15.240)
- n8alf.l
(15.241)
It can be seen that Eq. (15.238) is a cubic equation in n and that the coefficients bl, b2, and b3 depend on the material properties (E, V and p), geometry (R, I, and h), and the vibration mode (m and n). It can be shown that all the roots of Eq. (15.238) are always real and the positive square roots of n can be used to find the natural frequencies of the shell. Thus, there will be three values of the natural frequency (w) for any specific combination of values of m and n. For any specific natural frequency of vibration (w), the ratio between the amplitudes, (e.g., Bmn/Amn and Cmn/Amn) can be computed from any two of the equations among Eqs. (15.229)-(15.231). These ratios provide the ratios between the amplitudes of the longitudinal, tangential, and normal displacements of the shell for any specific natural frequency w.
15.8.4
Natural Frequencies of Transverse Vibration According to DMV Theory It has been observed that, of the three natural frequencies given by Eq. (15.238), the frequency corresponding to the transverse mode of vibration will have the smallest
15.8
Circular CylindricaIShells.;587
Table 15.1 Natural Frequencies of Transverse Vibration of a Cylindrical Shell (h O.lin.)
=
Natural frequency (radls)
Mode shape m
1 1 1 2 2 2 3 3 3
n
From Eq. (15.238)
1
5, 187.4575 2,298.6262 1,295.3844 11,446.168 6,632.3569 3.984.2207 15. 160.400 10,466.346 7.072.3271
2 3 1 2 3 1 2 3
From Eq. (15.243) 6,567.4463 2,494.8096 1,360.8662 13.839.495 7,386.4487 4.267.3979 16,958.043 11,527.878 7,612.5889
value. Hence the square and cubic terms of Q will be comparatively small and can be neglected in Eq. (15.238). This gives the approximate value of Q as
b3 b2
Q ~ - -
(15.242)
or (15.243) Example 15.10 Find the natur~ frequencies of vibration of a circular cylindrical shell simply supported at x 0 and x I, using DMV theory for the following data: E 30 x 106 psi, v = 0.3, p = 7.324 x 1O-41b-sec2/in4, R = 10 in, 1 = 40 in, and
=
=
=
h=O.lin SOLUTION The natural frequencies computed using DMV theory [Eqs. (15.238) and (15.243)] are given in Table 15.1. Only the smallest natural frequency given by Eq. (15.238), corresponding to the transverse mode, is given. It can be seen that the natural frequencies given by Eq. (15.243) are slightly larger than the corresponding values given by the DMV theory. 15.8.5
Natural Frequencies of Vibration According to Love's Theory Consider a cylindrical shell of radius· R, length 1, and simply supported at both ends, x = 0 and x = 1. The boundary conditions are given by Eqs. (13.218)-(13.225). In the absence of external forces (f;c ff) f1. = 0), the equations of motion, Eqs. (13.212)-(13.214), reduce to
=
=
2 c(J2U 1 - v a2u ~ aw 1 + v J v ) _ hu .ax2 + 2R2 ae2 + R ax + 2R ax ae - P 2 1 - v a2v 1 a2v 1 aw 1+ v a u ) -2ax2 + R2 ae2 + R2 ao + '2R ax ao D (1 - v a2v 1 a2v 1 a3w a3w) + R2 -2- ax2 + R2 ao2 - R2 ao3 - ax2 ao = phv
(15.244)
c(
(15.245)
588
Vibration of Shells
(15.246) The solution is assumed to be harmonic during free vibration as
u(x, e, t) vex, e, t) w(x, e, t)
U(x, e)eiwt = vex, e)eiwt = W(x, e)eiwt =
(15.247) (15.248) (15.249)
where w is the frequency of vibration.In view of Eqs. (15.247)-(15.249), the boundary conditions, Eqs. (15.218)-(15.225), can be stated as
a2w v av ( - ax2 + R2 ae
v (0, e) = 0
(15.250)
W (0, e) = 0
(15.251)
v a2W) R2 ae2
-
~w)
av + (auax + ~ R ae R
a2w v av ( - ax2 + R2 ae
(0,
e)
=0
(15.252)
(0
e) = 0
(15.253)
'
V(l,e)=o W(l,e)=o v a2W) R2 ae2 (l, e) = 0
-
+.:'.w) (l e) (auax + ~R av ae R '
(15.254) (15.255) (15.256)
=0
(15.257)
Substitution of Eqs. (15.247)-(15.249) into Eqs. (15.244)-(15.246) leads to
(a2U 1 - v a2u v aw 1 + v a2v) , ax2 + 2R2 ae2 + Ii ax + 2R ax ae + phw-U c (1 - v a2v 1 a2v 1 aw I + v a2U) -2- ax2 + R2 ae2 + R2 ae +"2i? ax ae
C
D (1 - v
+ R2 C (_~
-2-
a2 v
1
a2 v
1
a3 W
a3 W
(15.258)
=0
)
ax2 + R2 ae2 - R2 ae3 - ax2 ae + phw2V
=0
(15.259)
av _ W _ .:'. au) +!2 .. (~R2a4W ~ _ 2 a4w R2 ae R2 R ax R2 ax4 + ax2 ae ax2 ae2
a w 1 a v) - R2 ae4 + R2 ae3 1
4
3
+ phw2W = 0
(15.260)
15.8 Circular Cylindrical Shells ':589
The following solutions are assumed to satisfy the boundary conditions of Eqs. (15.250) -(15.257): mrrx (15.261) U(x, 0) = Cl cos -cos n(O - >0) I
V(x,
.
= C2 SIn --
0)
.
W(x, 0) = C3 SIn
C
mrrx
sin
I
mrrx
--
C3
(15.262)
n(O - >0)
(15.263)
COS n (0 - >0)
I
where l, C2, and are constants and >0 is the phase angle. Using Eqs. (15.261)(15.263), Eqs. (15.258)-(15.260) can be expressed as Cl [-C
2 - C 12~2v (n )
(mt)2
+ C2 (c I +
2R
Cl
(c
+ C3
Cl (C
v mrrI n) + C3 (c ~R mrr) I
+ C3
_ E... R2
(mrr)2 -t-
1- v -2-
R2
R2
-
R2
mrr) + C2 [-C I
[_.£ _
D
R2
4
n R2
(15.264)
R
n2
1- v 2
D R2 - R2
C
[-c !:. _ E... n3.~ E... (mrr)2 n] =
.!:. R
=0
!:)
1+v ~ 2 t
+ C2 [ -C
+ Phw2]
. R2
!:. _ E... R2
R2
I
(~)2I
+ PhW2]
2
--
D n -+phw R2 R2
0
n _
(mrr)4 _ E... (2) (mrr)2 I R2 I
(mrr)2 I
2] (15.265)
E... ~] R2 R2
n2 (15.266)
=0
Equations (15.264)-(15.266) can be written in matrix form as
[
~~w2 - du Phw~l: d22 ~~~ ] 2 d31 d32 phw - d33
where
du = C (m1rrf
= d21 = C -21-
d22 = C -2-
v .(mrr)2 -t-
o~}
(15.267)
(~f
+C 1; v
1 + v -1mrr v mrr d13 = d3l = C R -td12
{C~3~} = {
(15.268)
n
(15.269)
R
+C
(15.270)
(nR )2 + R2D
1-
-2-
v (mrr)2 t
E... (~)2
+ R2
R
(15.271)
590
Vibration of Shells
d23
= d32 = -
d33
=
C
R2
Cn Dn (mrr)2 R2 - R2 -Z-
Dn (n)2R
(15.272)
R2
-
+ D (mrr)4 -z- + 2D (mrr)2 -z-
(n)2 R
(15.273)
+D
For a nontrivial solution of CI, C2, and C3, the determinant of their coefficient matrix in Eq. (15.267) must be equal to zero. This leads to the frequency equation:
=0
(15.274)
or cv 6 + PI
CV
4
+ P2CV 2 + P3 = 0
(15.275)
where 1
PI
=-
P2
1 = p2h2
P3
= p3h3
ph
+ d22 + d33)
(dl1
122
(15.276) 2
2
2
(dlld22
+ d22d33 + dud33
(dlld23
+ d22dl3 + d33d122 + 2d12d23dl3-dud22d33)
- d12-d23
-
dl3)
(15.277) (15.278)
It can be seen that the frequency equation, Eq. (15.275) is a cubic equation in cv2 (as in the case of DMV theory). Example 15.11 Find the natural frequencies of transverse vibration of a circular cylindrical shell simply supported at x 0 and x Z using Love's theory for the following data: E 30 x 106 psi, v = 0.3, p = 7.324 x 1O-4lb-sec2 /in4, R 10 in., Z 40 in., and h 0.1 in.
=
=
= =
SOLUTION ·The natural frequencies shell are given by the smallest roots of Eq. (15.275) for different combinations results can be compared with the values Table 15.2
Natural Frequencies
Mode shape m
n
1 1
1 2
1
3
2
1
2
2
of Transverse
Natural frequency from Eq. (15.275) (rad/s) 4,958.2515 2,375.8223 1,321.9526 9,878.3359 6,595.9062
Vibration
=
corresponding to transverse vibration of the Eq. (15.275). The values of cv obtained from of m and n are given in Table 15.2. These given by the DMV theory in Example 15.10. of a Cylindrical
Mode shape
m
=
n
2 3 3
3 1 2
3
3
Shell (h = 0.1 in.)
Natural frequency from Eq. (15.275) (rad/s) 4,105.0059 12,989.213 10,086.031 7,181.8760
J5.9 Equations of Motion of Conical and Spherical Shells
59J
15.9 EQUATIONS OF MOTION OF CONICAL AND SPHERICAL SHELLS In this section, the general equations of motion of a shell are specialized for circular conical and spherical shells. The equations of motion are obtained by substituting the proper values of a, {3. A. B. Ra. and Rp into Love's equations of motion, Eqs. (15.172)-(15.176). 15.9.1
Circular Conical Shells Using a x. {3 e. A 1, B x sinao. Ra R~ 00, (see Example 15.5) and Nx x, Nx (}, No x. No o. Mx x' Mx Qo z for Nll, N12, N21. N22. Mll, M12, M21. M22, Q13, Eqs. (15.172)-(15.176). we obtain the equations governing shell as follows: aNxx 1 aNox 1 -a-+ . -ae +-(Nxx-Noo)+lx=ph-a2 x x sm ao x
=
=
aNxo
=
2 N
-a;- + -;
aQxz -+ -x1 ax
+x
ox Q
xz
+.x
=
=
1 aNoo sin ao ae
1 tan ao Qoz
+x
1 aQoz sm ao ae
=
1 ---Noo x tan
ao
aMxx Mxx' 1 aMox Moo --+-+'--------ax x x sin ao ae x aMxo -+ -x2 ax
M
ox
=
a2u t 2 a v
+ 10
(15.279)
= pha;r 2 + Iz = ph-aa tw2 Q
(15.280) (15.281)
0
(15.282)
Qo = 0 . <:
(15.283)
xz
a Moo + x sm .1 ao -ae
=
and Rp Ro x tan ao o. Mo x. Mo o. Qx z. and and Q23, respectively, in the vibration of a conical
=
where u, v. and w denote the components of displacement along the x. e, and z directions, respectively. By using the expressions of Nxx• Nxo = Nox, Noo, Mxx. Mxo MIJx. and MIJo in terms of the displacements u, v. and w given by Eqs. (EI5.8.1)-(E15.8.8) the equations of motion. Eqs. (15.279)-(15.281) can be expressed in termsofu, v, and w (see Problem 15.1).
=
15.9.2
Spherical Shells Using a = tjJ. {3 e. A = R, B Rsin tjJ. Ra = RI/J R, and Rp Ro R (see Example 15.6) and NI/JI/J,NI/JIJ.NIJI/J.Noo, MI/JI/J.MI/Jo, MOI/J.MIJIJ, QI/Jz, and Qoz for Nll, N12, N21, N22, Mll, M12, M2t. M22, Ql3, and Q23, respectively, in Eqs. (15.172)(15.176), we obtain the equations governing the vibration of a spherical shell as follows:
=
a (
=
.
-NI/JI/Jsm tjJ) + atjJ
a NOI/J-ae
=
lYIJIJcOSOf'+ A.
AT
Q
=
=
. I/Jzsm tjJ
a2u
+ RII/J sin tjJ = R sin tjJ ph at2
(15.284)
. tjJ) + ao aNoo aatjJ (NI/Jo sm + lYOI/JcostjJ + Q'IJzsm tjJ AT
+ RIIJ sin tjJ = R sintjJ ph
a2v at2
(15.285)
592
Vibration of Shells
B . - (Q,pz smq,) Bq,
+ Rfz
Be
(N,p,p
BM(),p Be BM99
+ N(9)
(15.286)
. M99 cosq, - Q,pz R smq,
=0
(15.287)
. - Q9z R smq,
=0
(15.288)
+ --
-
+ --
+ M9,p cosq,
ae
. smq,
B2w Bt2
sin q, = R sin q,ph
B . - (M,p,p smq,) Bq, B . - (M,p9 smq,)
aq,
BQh + --
where u, v, and w denote the components of displacement along the q" e, and z directions, respectively. By using the expressions of Nt/>,p,N,p() N9,p, N9(), M,p,p, M,p9 M9,p, and M99 in terms of the displacements u, v, and w given by Eqs. (E15.9.1)-(E15.9.8) the equations of motion, Eqs. (15.284)-(15.286), can be expressed in terms of u, v, and w (see Problem 15.2).
=
15.10
=
EFFECT OF ROTARY INERTIA AND SHEAR DEFORMATION In the vibration of thick shells, the effects of shear deformation and rotary inertia play an important role. In this section, the equations of motion of a shell are derived by including the effects of shear deformation and rotary inertia using an approach similar to that of Timoshenko beams and Mindlin plates [5,14,15]. The effects of shear deformation and rotary inertia become increasingly important as the thickness of the shell (or the value of hi Ra or hi R(3) increases. These effects can be significant even for thin shells in higher modes. Thus, the effects of shear deformation and rotary inertia can be significant when dealing with short wavelengths, especially those that have the same order as the thickness of the shell or less. As in the case of beams and plates, the effect of shear deformation is incorporated through generalization of the strain-displacement relations, and the effect of rotary inertia is incorporated through the basic equations of motion (in the dynamic equilibrium approach).
15.10.1
Displacement Components The displacement components of an arbitrary point in the shell are assumed to be given by (15.289) u(a, f3, z) = u(a, f3) + z1/1! (a, f3) V(a, f3, z)
=
v(a, f3)
w(a, f3, z)
=
w(a, f3)
+
Z1/12 (a, f3)
(15.290) (15.291)
where 1/11 and 1/12 denote the total angular rotations, including the angular rotations due to shear, of the normal to the middle surface about the f3 and a axes, respectively. Note that 1/11 and 1/12 are different from e1 and e2 used in Eqs. (15.85) and (15.86). e1 and e2 were used to denote the rotations, with no shear deformation, of the normal to the middle surface, about the f3 and a axes, during deformation. Equations (15.289)-(15.291) assume that straight lines normal to the middle surface remain straight lines after deformation, even if they no longer are normal. This assumption is consistent with the Timoshenko beam and Mindlin plate theories.
~ .-"'!"'
.•••.•• " ••••
15.10
r '
15.10.2
Effect of Rotary Inertia and Shear Defonnation
-59.3
Strain-Displacement Relations When shear deformation is considered, the fourth assumption in Love's theory (Kirchhoff's hypothesis), which states that normals remain normal, will not be considered. This implies that Eq. (15.82) is to be dropped and the shear strains eaz and e{3z will no longer be zero. Thus, 1/11 and 1/12 will no longer be related to u, v, and w [as in Eqs. (15.93) and (15.94)]; they need to be treated as independent variables in addition to u, v, and w. At the same time, Love's third assumption, which states that the transverse normal stress o"z z is negligible, is included. Thus, the normal stress 0":: is ignored and the transverse shear strains e13 and e23 are retained in the analysis. The strain-displacement relations in curvilinear coordinates are given by Eqs. (15.75)-(15.80). By substituting Eqs. (15.289)-(15.291) into Eqs. (15.75)-(15.80), and imposing the assumption of Eq. (15.81), the strain-displacement relations can be expressed as 0
ell
-
ell
e22
=
e22
el2
-
0 e12
e23
-
1/12
el3
-
1/11
e33
=
0
+ + +
0
zkll
(15.292)
Zk22
(15.293)
Zkl2
(15.294)
v R{3
u Ra
+
1 aw -B af3
(15.295)
+
1 aw -A aa
(15.296) (15.297)
where e?l' e~2' and e?2 denote membrane strains (which are independent of and k12 indicate curvatures given by
z)
and
kll,
k22,
0 ell 0
e22 0
= =
v
aA
AB
af3
+
u aB -AB aa
+
1 au A aa
+
1 au -+ B af3 B
a
(V)
a
A B af3
w
(15.298)
Ra
w
(15.299)
R{3
(U)
(15.300)
e12
= A aa
kll
=
1/12 aA 1 a 1/11 --+ AB af3 A aa
(15.301)
k22
=
1/11 aB 1 a 1/12 ---+ AB aa B af3
(15.302)
k12
= B A aa
!.-
B
(1/12)
B
+
+
A
A!.(1/11) B af3 A
(15.303)
Notes 1. The shear strains the shell.
e23
and
el3
denote the shear strains in the middle surface of
594
Vibration of Shells
2. Equations (15.295) and (15.296) are set equal to zero in Love's shell theory, and 1/11 and 1/12 (61 and 62 in the notation of Love's shell theory) are expressed in terms of u, v, and w.
15.10.3
Stress-Strain Relations The stress-strain relations are given by Eqs. (15.122)-(15.124) and (15.113)-(15.115). The transverse shear stresses given by Eqs. (15.114) and (15.115) are valid only at the middle surface of the shell. These stresses must diminish to zero at the free surface of the shell (at z ±h/2). The average values of the transverse shear stresses, denoted
=
(713
and
(723,
are defined as
(15.304) (15.305)
= k(f23 = kGs23
a23
where k is a constant less than unity, called the shear coefficient.
15.10.4
Force and Moment Resultants Force and moment resultants due to
(fll,
and
(f22,
(f12
are given by Eqs. (15.135),
(15.136), (15.138), (15.143), (15.145), and (15.146): Nll = C(S?l + VS~2)
+
(15.306)
(15.307)
VS?I)
N22
=
C(S~2
N12
=
N21 D(kll
+
Vk22)
M22
= =
D(k22
+
Vkll)
M12
=
M21
= ~
Mll
1 -
=
v
0
~
CS12
1 -
(15.308) (15.309) (15.310)
V
Dk12
(15.311)
where S?l' s~2' S?2' kll, k22, and k12, are given by Eqs. (15.298)-(15.303). The transverse shear force resultants Q13 and Q23 due to 0'13 and (f23, respectively, are defined by [see Eqs. (15.137) and (15.139)) h/2
=
Q13
f f
0'13
dz
(15.312)
-h/2 h/2
Q23
=
0'23 dz
(15.313)
-h/2
In view of Eqs. (15.304) and (15.305), Eqs. (15.312) and (15.313) can be written as Q13
= a13h = kGhs13
(15.314)
(15.315)
15.10 Effect of Rotary lnertia and Sheaf-Deformation
,595
Noting that the shear strains el3 and e23 can be expressed in terms of u,v,w,1/11and 1/12, using Eqs. (15.295) and (15.296), Eqs. (15.314) and (15.315) can be expressed as
Ra
1 aw) + -A aa
(15.316)
Q23 = kGh ( 1/12 - . Rf3
aw) +-B afJ
(15.317)
= kGh
QI3
U (
1/11 -
-
vI
15.10.5
Equations
of Motion
The generalized Hamilton's principle is used to derive the equations of motion: 8
12
1
112
Ldt=8
(15.318)
(T-rr+W)dt=O
11
11
The variation of the kinetic energy term in Eq. (15.318), including the effect of rotary inertia, is given by Eq. (15.164) by replacing Oi by 1/1i(i= 1,2):
1~
8T dt
12
1~ 1[ ah 12
= -ph +
2 h (lfr1 81/11 12
[it 8u
+ ii 8v + W 8w
+ lfr2 81/f2)] AB
da dfJdt
(15.319)
The strain energy term in Eq. (15.318) can be obtained from Eq. (15.168), by replacing OJby 1/1i(i 1,2) and expressing Q13 and Q23 in terms of e13 and e23 using Eqs. (15.314) and (15.315), as
=
112 11211 [ 8rrdt =
tl
a
11
f3
a
a
--(NllB)8u aa
aA
- -(MllB) act
aA
81/11
AB
+ Nll-8va{3 + Mll-81/12 + Nll-8w a{3 Ra a - -(N22A) a{3
a 8v - -(M22A) a{3
aB
AB
act
Rf3
+ M22-81/11 + N22-8w aA - Nl2-8u a{3
aB
8u + N22-aa
a
- -(Nl2A) a{3
8u
a aA - -' (Ml2A) 81/11 - Ml2-81/fl a{3 ,a{3
a
- -(Nl2B)8v aa
aB
- MI2-81/f2
aa
a
81/12
aB
- N12-8v aa
+ kGhe23AB
a
- -(M12B) aa
AB
81/f2- kGhe23-8v
+ kGhe13AB
- -(kGhe23A) a{3
8w
-kGhel3-8uAB Ra
a - -(kGheI3B)
aa
81/12
Rf3
81/11 8w ] da d{3dt
I
I
596
Vibration of Shells
{'21
+
}'1
(N22A ov
+ M22Ao1/12+ N21A ou + M21A 01/11
a
+ kGhE:23A OW)dadt
{'21
+
}'1
(NuB OU
+ MuB
01/11+ N12B OV+ M12 B01/12
p
+ kGhE:13B OW)d~ dt
(15.320) The term related to the work done by the external forces in Eq. (15.318) can be obtained fromEq. (15.169) by replacing 0; by 1/1; (i = 1,2). By substituting Eqs. (15.319), (15.320) and the modified Eq. (15.169) into Eq. (15.318), the equations of motion corresponding to the variables u, v, W, 1{;1, and 1/12 can be identified as [14] o 0 oA oB - -(NuB) - -(N21A) - N12+ N22oa o~ o~ oa -
AB
o - -(N12B) oa
0 - -(N22A) o~
oB - N21oa
_ ~; E:23kGh + ABphv
o
+ AB
(Nu Ra
o aa (MuB)
oA
+ Nll- o~
= ABfp
+: N22) Rp
+ ABphw
(15.322)
0
= ABfz
oA
+ o~ (M21 A) + M12 a~ ph3 ..
- kGhE:13AB - ABU1/11 o . oa (Ml2B)
(15.321)
8 - kGh-(E:23A) o~
- kGh-(E:13B) oa .
0
aB M22 oa (15.324) 8A
oB
ph3 ..
- Mll
=
a~
=0
Boundary Conditions The boundary conditions of the shell on the edges a constant can be identified as follows: or v = v N22 = N22 or 1/12 = 1/1 2 M22 M22 at ~ or u = ii N21 = N21 Q23 = kGhE:23 = Q23 or w=w or 1/11 1/1 1 M21 = M21
=
(15.323)
=0
+ o~ (M22A) + M21 oa
- kGhE:23AB - ABU1/12
15.10.6
.
Ra E:13kGh + ABphu = ABfa
(15.325)
= a = constant = -p = constant
and ~
= -p =
(15.326)
15.10 Effect of Rotary Inertiaatld Shear Det~ion
Nll=Nll NI2 NI2 QI3 = kGhsl3 Mll Mil
=
MI2
15.10.7
= = MI2
=
or u=u or u=v or w=w or 1/1"1 = or 1/1"2
Qn
a
at
= 'ii = constant
597
(15.327)
1/1"I
= 1/1"2
Vibration of Cylindrical Shells For a cylindrical shell, x, e, and z are used as the independent coordinates. From the fundamental form of the surface, we find that a = x, f3 = e, A = I, B = R, Ra = Rx = 00, and Rf3 = Re = R(see Example 15.4). In this case, the equations of motion, Eqs. (15.321)-(15.325), become (using the notation 1/1"x for 1/1"1 and 1/1"e for 1/1"2)
+ -.-= aNex
Nxx Ra--
ax
a Nee + -+ Qez
aNxe R-ax aQxz R-ax
ae aQez + -ae
-
(15.328)
.. = Rphu - Rie
(15.329)
..
(15.330)
= RPh3 .i~
(15.331)
aMee _ RQ = RPh3 .i~ ae ez 12 'l'e
(15:332)
ax + ..ae +
R puh··
= Rphw - Riz
Nee
RaMxx ... aMex _ RQ RaMxe ax
Rf x
ae
12 'I' x
xz
The force and moment resultants can be expressed in terms of the displacement components, from Eqs. (15.306)-(15.311), (15.316) and (15.317), as: ( --+1 au w)] N =C [au -+v x ax R ae R 1 au
Ne = C ( - R ae Nxe
(15.333)
+ -wR + v-au) ax
1- v C = Nex = -2-
( 1 au Rae
(15.334) au)
+ ax
a1/1"x va1/1"e) Mx = D ( ax + R ae Me=D
1 a1/1"e ( --+vR ae
Mxe=Mex=-2-D
a1/1"x) ax
(15.335) (15.336) (15.337)
1- (1
(15.338)
+ 1/1"x)
(15.339)
v
a1/1"x a1/1"e) RM+ ax
Qx
= kGh
(~:
Qe
= kGh
[~ ~; - (~ - 1/1"e )]
(15.340)
598
Vibration of Shells
where C and D are given by Eqs. (15.134) and (15.144). By substituting Eqs. (15.333)- . (15.340) into Eqs. (15.328)-(15.332), we obtain the equations of motion in tenns of the displacement components as a2u 1 - v 1 a2u 1 + v 1 a2v v aw p(1 - v2) a2u ax2
+ -2- R2
1 a2v R2 ae2
1 - v a2v
+ -2- ax2
k
v
2
a2v
E
at2
+
w =
a21/1x ( ax2
+ -2- R ax ae + Ii~ = 1 + v 1 a2u 1 aw k + -2- R ax ae + R2 ae + Ii
p(1 - v2)
= _(
af)2
at2
(
v R
1/16 -
(15.341) 1 aw)
+ Rae
(15.342)
a1/1x 1 a1/16 ax + R
ae - R21 aeav)
p(1 - v2)
a2w
E
at2
+
E
1 - va21/1x 2R2 ae2
1 (1 av - R R ae
. w R
+
au)
+ v ax
(15.343)
+
1 + v a21/16) _ 12k (aw 2R ax ae h2 ax
p(1 - v2) a21/1x E at2 a21/16 1 - v a21/16
,/,)
+ 'l'X (15.344)
=
2-
( R2
ae2 =
+
2
ax2
+
1 + v a21/1x) _ 12k 2R ax ae h2
(.!.
aw R ae
p(1 - v2) a21/16
E
,Ir
+ '1'6
_~)
R (15.345)
at2
where k
I-v = --k 2
2
vw = 15.10.8
a2w ax2
(15.346) 1
+ R2
a2w
(15.347)
ae2
Natural Frequencies of Vibration of Cylindrical Shells Consider a cylindrical shell of radius R, length 1, and simply supported at both ends, x = 0 and x = 1. The boundary conditions of the shell can be expressed as v(O,e,t)=o
(15.348)
w(O,e,t)=o
(15.349)
Mxx(O,e,t)=D
a1/1x ( ax
Nxx(O,e,t)=c
au [ ax+v
1/16(0, e, t)
=0
+
va1/16) Rae (laV Rae+
(O,e,t)=o w)] R
(O,e,t)=o
(15.350) (15.351) (15.352)
15.10
Effect of Rotary Inertia and Shear Defoonation .. 599
v(l,e,t)=o
(15.353)
w(l,e,t)=o
(15.354) Olfrx ( ox
Mxx(l,e,t)=D
v OlfrlJ) + RM
+v(i :~ +
Nxx(l,e,t)=c[::
(15.355)
(l,e,t)=o
(15.356)
~)](l,O,t)=O
(15.357)
lfrlJ(l, e, t) = 0 For free vibration,
the solution is assumed to be harmonic as
e)iwr
(15.358)
v(x, 0, t) = V(x, e)e
iwr
(15.359)
w(x, e, t) = W(x, e)eiwr
(15.360)
lfrx(x, e, t) = \{lAx, O)eiwr
(15.361)
lfrlJ(x, e, t) = \{I1J(x, e)eiwr
(15.362)
U(x,
e, t) ==
U(x,
where (J) is the frequency of vibration. In view of Eqs. (15.358)-(15.362), conditions of Eqs. (15.348)-(15.357) can be restated as
o \{Ix ( ox
[ ~~ +
v (~
+ ~~
V(O, e) =0
(15.363)
W(O, 0) =0
(15.364)
O\{lIJ) (0 e) oe '
~
R
+ ;)]
the boundary
=
0
(15.365)
(15.366)
\lJ1J(O,e)
=0 =0
V(l,e)
=0
(15.368)
(0. e)
W(l,e)=0 O\{lx ( ox
[~~ +
v (~
+ ~
O\{lIJ) (l e) = 0 oe '
R ~~
+ ;)]
(l, e)
= 0
\lJ1J(l,e) = 0
(15.367)
(15.369) (15.370)
(15.371) (15.372)
The following solutions are assumed to satisfy the boundary conditions ofEqs. -(15.372): mrrx U(x, e) C1 COS -cos n(9 - (/J)
(15.363)
= C2sm.
(15.374)
=
V(x,9)
1
mrr x . (Ll ) -sm n u-4>
1
(15.373)
600
Vibration of Shells W(x,O)
= C3 sin ~
\lIx(x, 0)
= C4COS
z
(15.375)
m1l' x --
z
= C5 sin ~
\lie (x, 0)
cos n(O - 4» cos
n(B -
4»
(15.376)
sin
n(B -
4»
(15.377)
z
Substituting Eqs. (15.358)-(15.362) and (15.373)-(15.377) into the equations of motion, Eqs. (15.341)-(15.345), we obtain dlJ d21 d31
0 0
n2
d22 d32
n2
d23
n2
. d33 -
0
d35
d34
n2
d45
n2
d55 -
d54
d53
d52
d25
d44 -
d43
n~}~{~}
0
0 0
d13
d12
(15.378)
where dlJ =
(mz1l')2
+
+
1
v
1;
d12
= d21 = - -- 2
d13
= d31
d22
= 1 ;
d23
1 n = d32 = - RR
d25
= d52 = --R
v
(15.379)
(j)2
v n m7C
-R - Z
(15.380)
m7C
(15.381)
== ...- -R - Z
f
v (mz
7C
n
+
(j f + ;2
k
(15.383)
+ -R -R
k
d33=k
(15.382)
(15.384)
{(m;f +(~f}+
d34
- nt7C = d43 = k -
d35
= d53 = -k
d44
=
d45
= d54 = - -2
d55
=
~2
(15.385) (15.386)
/
-
12k h2
+
n -
(15.387)
R
1 -
-2-
v (n .R
1- v n
12k
h2 +
R
1 -
2
)2(nt7C)2 + -/-
m1l'
-
Z
v (m 7C)2 z
(15.388) (15.389)
+ (_Rn)2
(15.390)
Effect of Rotary Inertia and Shear Deformation ';·601
15.10
where (15.391)
k is given by Eq. (15.346). For a nontrivial solution of Cl, Cz, ... , Cs, the determinant of the coefficient matrix in Eq. (15.378) must be zero. This yields the frequency equation as a fifth-degree polynomial equation in nZ• The roots of this polynomial equation give the natural frequencies of the cylindrical shell wZ• For every combination of m and n, there will be five distinct natural frequencies. The mode shapes of the shell can be determined by first substituting each natural frequency into the matrix equations (15.378), and then solving for any four constants among Cl, Cz •... , Cs in terms of the remaining constant. For example, by selecting Cs as the independent constant, we can find the values of CJ/Cs, Cz/Cs, C3/C5, and C4/CS. and
15.10.9
Axisymmetric Modes In the particular case of axisymmetric modes (n 0), the five equations of motion will be uncoupled into two sets: one consisting of three equations involving u, w, and 8x and the other consisting of two equations involving v and 88. Thus, the first set of equations leads to a cubic frequency equation and describes flexural or radial modes and the second set leads to a quadratic frequency equation and describes circumferential modes. For n = 0, Eqs. (15.379),-(15.390) reduce to
=
dll
=
.·(m7r)Z
(15.392)
-/-
= dZl = 0 v dl3 = d3l = - R
(15.393)
dlZ
1 -
dZ2
=
dZ3
= d32 =
m7r
m7r)Z
V
(
2
/
k
+ RZ
(15.395) (15.396)
0
k
(15.397)
d2S = d52 = --
R
_ ( m7r ) Z
d33 = k
(15.394)
-/-
1
+ R2
-m7r
(15.398)
d34 = d43 = k -l-
(15.399)
d35 = dS3 = 0
(15.400)
d44 =
12k h2
+
(m7r)Z -l-
(15.402)
d45 = dS4 = 0
12k dS5 =
h2
+
(15.401)
- v 2
(15.403)
602
Vibrationof Shells In this case, Eq. (15.378) can be written as 0
dll - Q2
o
d22
-
0 0
d31
o o
dS2
{i~ }
0 0
d)3
Q2
0
d33 - Q2 d34 d43 d44 - Q2
0
= { ~}
0
(l5.MM)
This equation can be rewritten as a system of two uncoupled matrix equations as
0] {Cl} { 0 }
dJl - Q2 d)3 d31 d33 - Q2 d34 [ d43 d44 - Q2
C3
o
=
C4
d22 - Q2 d2S ] [ dS2 dss - Q2
{CCs2}
(15.405)
0 0
= {0} 0
(15.406)
Equations (15.405) and (15.406) lead to the following cubic and quadratic frequency equations, respectively: Q6 _ (dll
+ d33 + d44)Q4 +
+ (dJld33d44 Q4 _ (d22 +
- dlldj4 dSS)Q2
(dlld33
+ dlld44 + d33d44
- df3 - dj4)Q2
- d44df3 ) = 0
+ (d22dSS
(15.407)
- dls) = 0
(15.408)
Notes 1. The equations of motion of a circular cylindrical shell, considering the effects of rotary inertia and shear deformation, were derived by several investigators [15-17, 23]. Naghdi and Cooper derived a set of equations that are more general than those presented in Sections 15.10.8 and 15.10.9 (see Problem 15.14). 2. The equations of motion for the axisymmetric vibration of cylindrical shells and resulting equations for free vibration, Eq. (15.406), were derived as a special case of the more general equations by Naghdi and Cooper [15] and from general shell theory equations by Soedel [14]. Example 15.12 Find the natural frequencies of axisymmetric vibration of a circular cylindrical shell simply supported at x 0 and x I, considering the effects of rotary inertia and shear deformation for the following data: E 30 x 106 psi, G 12 X 2 4 4 6 10 psi, v 0.3, p = 7.32 X 10- lb-sec jin. , R 10 in., I 40 in., h 0.1 in., and
=
=
k-~-
=
=
=
=
=
=
6"
SOLUTION The natural frequencies of axisymmetric vibration of the cylindrical shell can be determined from the roots of Eqs. (15.407) and (15.408). The smallest values of the natural frequencies found for different modes (m = 1, 2, ... , 10) are given in Table 15.3. If the smallest natural frequencies of the cylindrical shell are to be determined by neglecting the effect of shear deformation, we need to set the value of k equal to 0 in Eqs. (15.407) and (15.408)
15.11
Recent Contributions
~3
Table 15.3 Smallest Values of Natural Frequencies Given by Eqs. (15.407) and (15.408) Smallest value of natural frequency, m 1 2 3 4 5 6 7 8 9 10
WI
(rad/see)
With shear defonnation and rotary inertia 24,057.074 29,503.338 36,832.332 45,135.996 53,966.332 63,102.639 72,429.156 81,880.961 91,419.211 101,019.41
15.11 RECENT CONTRIBUTIONS A complete solution for the dynamic response due to time-dependent mechanical and/or thermal loading of spherical and cylindrical shells with arbitrary time-dependent boundary conditions was presented by Pilkey [19]. The solution was obtained in the form of a series expansion of the products of modes of free vibration and a generalized coordinate. The generalized FourIer transform was used to determine the generalized coordinate which contains all physically admissible boundary conditions. The free flexural vibrations of cylindrical shells stiffened by equidistant ring frames was investigated by Wah [20] using finite difference calculus. The theory accounts for both in-plane flexural and torsional vibration of the ring stiffeners. Tables of natural frequencies and graphs of representative mode shapes of harmonic elastic waves propagating in an infinitely long isotropic hollow cylinder have been presented by Armenakas et al. [21]. The free vibration problem of a homogeneous isotropic thick cylindrical shell or panel subjected to a certain type of simply supported edge conditions was investigated by Soldatos and Hadjigeorgiou [22]. The governing equations of three-dimensional linear elasticity were employed and solved using an iterative approach, which in practice leads to the prediction of the exact frequencies of vibration. In the case of a flat or a complete shell, excellent agreement was found between the results given by this approach and those given by other exact analysis methods. The role of median surface curvature in large-amplitude flexural vibrations of thin shells was studied by Prathap and Pandalai [23]. It was shown that whereas the nonlinear behavior of flat plates and straight bars is generally of a hardening type, the behavior of thin structural elements involving finite curvature of the undeformed median surface in one or both principal axis directions may be of the hardening or softening type, depending on the structural parameters as well as on whether the shell . is open or closed. The free vibration of circular cylindrical shells with axially varying thickness was considered by Sivadas and Ganesan [24]. The free vibration of noncircular cylindrical shells with circumferentially varying thickness was discussed by Suzuki and
.,
j
604
Vibration of Shells
Leissa [25]. Soedel summarized the vibration analysis of shells in Ref. [26]. The free vibration of thin cylindrical shells, consisting of two sections of different thicknesses but with a cornmon mean radius, was investigated by Warburton and AI-Najafi [27]. They determined the natural frequencies and mode shapes using two types of ring finite elements as well as by solving the shell equations using the boundary conditions and the continuity condition at the intersection of the two thicknesses. Sharma and Johns [28] determined the circumferential mode vibration characteristics of clamped-free circular cylindrical shells experimentally. A two-dimensional higher-order shell theory was applied to the free vibration problem of simply supported cylindrical shell subjected to axial stresses by Matsunaga [29]. Using a power series expansion of displacement components, the dynamical equations were derived, including the effects of rotary inertia and shear deformation, from Hamilton's principle. Thin-walled regular polygonal prismatic shells were used in several practical applications, such as honeycomb cores of sandwich plates, guide supports of welding frameworks, and high piers of highway bridges. The free vibration of regular polygonal prismatic shells has been presented by Liang et al. [30] using a novel plate model and beam model on the basis of geometric symmetry. Analytical solutions were obtained by combining the vibration theories of Euler beams and thin-walled plates.
REFERENCES 1. A. E. H. Love, A Treatise on the Mathematical Theory of Elasticity, 4th ed., Dover, New York, 1944. 2. W. Fltigge, Stresses in Shells, Springer-Verlag, Berlin, 1962. 3. A. L. Goldenveizer, Theory of Thin_Shells, translated by G. Herrmann, Pergamon Press, New York, 1961. 4. H. Kraus, Thin Elastic Shells, Wiley, New York, 1967. 5. A. W. Leissa, Vibration of Shells, NASA SP-288, National Aeronautics and Space Administration, Washington, D C, 1973. 6. W. Soedel, Vibration of Shells and Plates, 2nd ed., Marcel Dekker, New York, 1993. 7. A. C. Ugural, Stresses in Plates and Shells, McGraw-Hill, New York, 1981. 8. S. P. Timoshenko and S. Woinowsky-Krieger, Theory of Plates and Shells, McGraw-Hill, New York, 1959. 9. E. Ventsel and T. Krauthammer, Thin Plates and Shells: Theory, Analysis. and Applications, Marcel Dekker, New York, 2001. 10. V. V. Novozhilov, Thin Shell Theory, translated by P. G. Lowe, Noordhoff, Groningen, The Netherlands, 1964. II. L. H. Donnell, Beams, Plates, and Shells, McGraw-Hill, New York, 1976. 12. V. Z. Vlasov, General Theory of Shells and Its Applications in Engineering, NASA-TT-F99, National Aeronautics and Space Administration, Washington, DC, 1964. 13. N. A. Kilchevskiy, Fundamentals of the Analytical Mechanics of Shells, NASA-TT-F292, National Aeronautics and Space Administration, Washington, DC, 1965. 14. W. Soedel, On the vibration of shells with Timoshenko-Mindlin type shear deflections and rotatory inertia, Jou17Ullof Sound and Vibration, Vol. 83, No.1, pp. 67-79, 1982.
15. P. M. Naghdi and R. M. Cooper. Propagation of elastic waves in cylindrical shells, including the effects of transverse shear and rotatory inertia, Journal of the Acoustical Society of America, Vol. 28, No. I, pp. 56-63, 1956. 16. T. C. Lin and G. W. Morgan. A study of axisymmetric vibrations of cylindrical shells as affected by rotary inertia and transverse shear, Journal of Applied Mechanics, Vol. 23, No. 2, pp. 255-261, 1956. 17. I. Mirsky and G. Herrmann, Nonaxially symmetric motions of cylindrical shells, Journal of the Acoustical Society of America, Vol. 29, No. 10, pp. 1116-1123, 1957. 18. 1. Mirsky, Vibrations of orthotropic, thick, cylindrical shells, Journal of the Acoustical Society of America, Vol. 36, No. I, pp. 41-51, 1964. 19. W. D. Pilkey, Mechanically andlor thermally generated dynamic response of thick spherical and cylindrical shells with variable material properties, Journal of Sound and Vibration, Vol. 6, No. I, pp. 105-109, 1967. 20. T. Wah, Aexural vibrations of ring-stiffened cylindrical shells, Journal of Sound and Vibration, Vol. 3, No.3 pp, 242-251, 1966. 21. A. E. Armenakas, D. C. Gazis, and G. Herrmann, Free Vibrations of Circular Cylindrical Shells, Pergamon Press, Oxford, 1969. 22. K. P. Soldatos and V. P. Hadjigeorgiou, Three-dimensional solution of the free vibration problem of homogeneous isotropic cylindrical shells and panels, Journal of Sound and Vibration, Vol. 137, No.3, pp. 369-384, 1990. 23. G. Prathap and K. A. V. Pandalai, The role of median surface curvature in large amplitude flexural vibrations of thin shells, Journal of Sound and Vibration, Vol. 60, No. I, pp. 119-131,1978. 24. K. R. Sivadas and N. Ganesan, Free vibration of circular cylindrical shells with axially varying thickness, Journal of Sound and Vibration, Vol. 147, No.1, pp. 73-85,1991. 25. K. Suzuki and A. W. Leissa, Free vibrations of non-circular cylindrical shells having circumferentially varying thickness, Journal of Applied Mechanics, Vol. 52, No.1, pp. 149-154, 1985. 26. W. Soedel, Shells, in Encyclopedia of Vibration, S. Braun, D. Ewins, and S. S. Rao, Eds., Academic Press, San Diego, CA, 2002, Vol. 3, pp. 1155-1167. 27. G. B. Warburton and A. M. J. AI-Najafi, Free vibration of thin cylindrical shells with a discontinuity in the thickness, Journal of Sound and Vibration, Vol. 9, No.3, pp. 373-382, 1969. 28. C. B. Sharma and D. J. Johns, Natural frequencies of clamped-free circular cylindrical shells, Journal of Sound and Vibration, Vol. 21, No.3, pp. 317-318, 1972. 29. H. Matsunaga, Free vibration of thick circular cylindrical shells subjected to axial stresses, Journal of Sound and Vibration, Vol. 211, No.1, pp. 1-17, 1998. 30. S. Liang, H. L. Chen, and T. X. Liang, An analytical investigation of free vibration for a thin-walled regular polygonal prismatic shell with simply supported oddleven number of sides, Journal of Sound and Vibration, Vol. 284, No. 1-2, pp. 520-530,2005.
PROBLEMS 15.1 Derive the equations of motion of a conical shell in terms of the components of displacement u, v, and w from Eqs. (15.279)-(15.281).
15.2 Derive the equations of motion of a spherical shell in terms of the components of displacement u, v, and w from Eqs. (15.284)-(15.286).
606
Vibration of She))s
15.3 Specialize the equations of motion of a cylindrical she)) given in Section 15.8.1 to the case of a rectangular plate. [Hint: Use R dO y and 1/R 0.]
=
=
15.4 Specialize the equations of motion of a cylindrical she)) given in Section 15.8.1 to the case of a circular ring. 15.5 Derive the frequency equation (15.238) using the solution given by Eqs. (15.226)-(15.228) in the equations of motion, Eqs. (15.215)- (15.217). 15.6 Using Donne))-Mushtari- Vlasov theory, find the natural frequencies of vibration corresponding to m, n 1,2,3 for an aluminum circular cylindrical she)) with simple supports at x 0 and x 1. E 71 GPa, v 0.334, unit weight 26.6 kN/m3, R 0.2 m, 1 2 m, and h 2 rom.
=
=
=
=
=
=
=
=
=
15.7 Derive the frequency equation (15.275) using the solution given by Eqs. (15.261)-(15.263) in the equations of motion, Eqs. (15.258)-(15.260). 15.8 Using Love's theory, find the natural frequencies of vibration of the cylindrical shell described in Problem 15.6. 15.9 Solve Problem 15.6 considering the material of the shell as steel with E = 207 GPa, v = 0.292, unit weight 76.5 kN/m3, R 0.2 m, 1 2 m, and h=2mm ..
=
=
=
thickness 0.025 in., Poisson's ratio 0.3, Young's modulus 30 x 106 psi, and unit weight 0.283 Ib/in3• Find the number of half-waves in the circumferencecorresponding to the minimum natural frequency of vibration according to DMV theory. 15.13 Show that the solution given by Eqs. (15.373)(15.377) satisfies the boundary conditions of Eqs. (15.363) -(15.372). 15.14 According to Naghdi and Cooper [15], the force and moment resultants in a cylindrical shell, by considering the effects of rotary inertia and shear deformation, are given by Nx=C tv, o N
=
_ xO -
W)
OU ( 1 OV --+[ -+v ox R 00 R
C[IR
Mx
= D (01/1x
15.11 (a) Derive the equation of motion for the axisymmetric vibrations of a thin cylindrical shell from the general equations.
Qx
= kGh
R 80
00
LC
mrx
n
sin -/-
sin wt
R ox
1
1- v Mxo = -2-D = 1- v D
M
(~:
ox
R2
(OV -+W)] 80
(1Rae + --a.; + R1 (2. + -2-. o1/fx
81/fx R 80
81/10
ov) 8x
81/10_ 8U) ox R2 00
+ 1/fx)
(b) Assuming the transverse deflection w(x, t) to be
=
ae
+
R 00
[1
2
w(x, t)
2 h ( 1 OU 1 81/1x)] 12R R2 00 - R
81/10 81/1x 0=D --+v---
OX
=
2 h 01/10] - 12R2
+ 2:.. 01/10 + 2. ou)
ox
·....·.1
= =
W
+ R + v 8x
tv, _1-v C [ 1 OU 8v Ox -2 R 00 + ox
=
=
8u
ae 1 - v C ( 1 OU OV h2 81/1x) -2R 80 + 8x + 12R ae
15.10 Using Love's theory, find the natural frequencies of vibration corresponding to m, n 1,2,3 for a steel circular cylindrical shell with simple supports at x 0 and x 1. E 207 GPa, v 0.292, unit weight = 76.5 kN/m3, R 0.2 m, 1 = 2 m, and h = 2mm.
=
OV 80
2
h O1/1x] +-12R ox
Qo = kGh [~ ~; -
(*-
1/10)]
m=l
where x is the axial direction, 1 is the length, and w is the frequency of axisymmetric vibration of a cylindrical she)) simply supported at x = 0 and x = 1, find an expression for the natural frequency w. . 1.5~12Consider a cylindrical shell simply supported at x = 0 and x = I, with radius 6 in., length 15 in., wall
Express the equations of motion, Eqs. (15.328)-(15.332), in terms of the displacement components u, v, w, 1/1.• , and 1/10. 15.15 A cylindrical shell is loaded by a concentrated harmonic force 1 (x, 0, t) = 10 sin nt in the radial direction at the point x Xo and 0 = 00. Determine the amplitude of the resulting forced vibration.
=
16 Elastic Wave Propagation 16.1 INTRODUCTION Any localized disturbance in a medium will be transmitted to other parts of the medium through the phenomenon of wave propagation. The spreading of ripples in a water pond, the transmission of sound in air, and the propagation of seismic tremors in Earth are examples of waves in different media. We consider wave propagation only in solid bodies in this chapter. Although the propagation of a disturbance in a solid takes place at a microscopic level, through the interaction of atoms of the solid, we consider only the physics of wave propagation by treating properties such as density and elastic constants of the solid body to be continuous functions that represent the averages of microscopic quantities. In solid bodies, compression and shear waves can occur. In compression waves, the compressive and tensile stresses are transmitted through the motion of particles in the direction of the wave motion. In shear waves, shear stress is transmitted through the motion of particles in a direction transverse to the direction of wave propagation. Three types of waves can occur in a solid body: elastic waves, viscoelastic waves, and plastic waves. In elastic waves, the stresses in the material obey Hooke's law. In viscoelastic waves, viscous as well as elastic stresses act, and in plastic waves, the stresses exceed the yield stress of the material. We consider only elastic waves in this chapter. Elastic waves in deformable bodies play an important role in many practical applications. For example, oil and gas deposits are detected and Earth's geological structure is studied with the help of waves transmitted through the soil. The waves generated by Earth's tremors are used to detect and study earthquakes. Properties of materials are determined by measuring the behavior of waves transmitted through them. Some recent medical diagnosis and therapy procedures are based on a study of elastic waves transmitted through the human body.
16.2
ONE-DIMENSIONAL WAVE EQUATION The one-dimensional wave equation is given by 282cp 8cp c-=2 8x 8t2
(16.1)
where cp = cp(x, t) is the dependent variable and x and t are the independent variables. Equation (16.1) represents the equation of motion for the free lateral vibration of strings, 607
608
Elastic Wave Propagation Table 16.1
Physical Significance of ¢ and c in Eq. (16.1) Significance of c
Significance of ¢
Type of problem Lateral vibration of strings
Longitudinal vibration of bars
Torsional vibration of rods
Lateral or transverse displacement of a string
c=.JP/p
Longitudinal or axial displacement of the cross section of a bar Angular rotation of the cross section of a rod
c
= mass per unit length
= .JE/p = Young's modulus = mass density
c=.JG/p G p
a C
= tension
P p E p
a
= shear modulus = mass density
has dimensions of linear velocity.
longitudinal vibration of bars, and torsional vibration of rods. The study of propagation of waves in a taut string is useful in the manufacture of thread and in understanding the characteristics of many musical instruments and the dynamics of electrical transmission lines. The longitudinal waves in a bar have application in seismic studies. Waves transmitted through the Earth are used to detect and study earthquakes. The physical significances of 4> and c in different problems are given in Table 16.1.
16.3 TRAVELING-WAVE SOLUTION 16.3.1
D'Alembert's Solution D'Alembert derived the solution of Eq. (16.1) in 1747 in a form that provides considerable insight into the phenomenon of wave propagation. According to his approach, the general solution of Eq. (16.1) is obtained by introducing two new independent variables, ~ and YJ as ~= x - ct
(16.2)
+ ct
(16.3)
YJ
=x
By expressing the dependent variable wI in terms of ~ and T/ (instead of x and t), we can obtain the following relationships using Eqs. (16.2) and (16.3):
a~ = 1
a~
aw
a~ _
'
aw
-c
at -,
a~
aw aT/
1
aYJ ax -
aw
aw
-=--+--=-+ax a~ ax aYJ ax a~ aYJ a2w = __ a2w _ ax2
a~2
(16.4)
,
(16.5)
a~ + a2w aYJ + a2w a~ +__ a2w aYJ = -+2--+a2w a2w a2w ax a~ oYJ ax a~ aYJax aYJ2 ax a~2 a~ aYJ aT/2
(16.6)
IThe dependent variable, t/J, in Eq. (16.1) is denoted as w in this section.
16.3 Traveling-Wave Solution
aw at a2w
aw a~ a~ at (a2w
aw aT} aT} at
aw a~
-=--+--=-e-+e-
at2 = -e
a~2
a~
at +
a2w aT}) a~ aT}
at
aw aT}
..be9
(16.7)
2 ( aZw a~ a waT}) + e a~ aT} + aT}2
at
at
zazw 2 2 a2w za2w = e a~z - e a~ aT} + e aT}2
(16.8)
Substituting Eqs. (16.6) and (16.8) into Eq. (16.1), we obtain the one-dimensional wave equation in the form (16.9) This equation can be integrated twice to obtain its general solution. Integration of Eq. (16.9) with respect to T} gives
aw 8f
= h(~)
(16.10)
where h(~) is an arbitrary function of ~. Integration of Eq. (16.10) with respect to ~ yields _ow =
f
h(~) d~
+ geT})
(16.11)
where geT}) is an arbitrary function of T}. By defining f(~)
=
f
h(~) d~
(16.12)
the solution of the one-dimensional wave equation can be expressed as w(~, T}) = f(~) = f(x
+ geT}) - et) + g(x + et)
(16.13)
The solution given by Eq. (16.13) is D'Alembert's solution [see Eq. (8.35)]. Note that f and g in Eq. (16.13) are arbitrary functions of integration which can be determined from the known initial conditions of the problem. To interpret the solution given by Eq. (16.13), assume that the term g(x + et) is zero and the term f(x - et) is nonzero. Assume that at t 0, the function f(x - et) f(x) denotes a triangular profile as shown in Fig. 16.1 with the peak of f(x), equal to f(O), occurring at x = O. At a later time, the peak f(O) occurs when the value of the argument of the function f(x - et) is zero. Thus, the peak occurs when X2 - etz = 0 or t2 = xz/e, as shown in Fig. 16.1. Using a similar argument, every point in the triangular profile can be shown to propagate in the positive x direction with a constant velocity e. It follows that the function f (x - et) represents a wave that propagates undistorted with velocity e in the positive direction of the x axis. In a similar manner, the function g(x + et) can be shown to represent a wave. that propagates undistorted with velocity e in the negative direction of the x axis. Note that the shape of the disturbances, f(x - et) and g(x + et), which is decided by the initial conditions specified, remains the same during wave propagation.
=
=
I
I
610
Elastic Wave Propagation
fix -
t
C1)
Wave velocity,
C
C
)
I
-10 0 0 jAtt)
jAltz
=0
= ctz
~X2
Figure 16.1
16.3.2
)
i·
) x
iAtt3
=
C(13 - (2)-1
(X3-X2)
Propagation of the wave f(x
- ct) with no distortion.
Two-Dimensional Problems For a two-dimensional problem (membrane), the equation of motion is given by [see Eq. (13.2)] 2
C2(82W + 8 W) = 8x2
8y2
2 8 w 8t2
(16.14)
where w(x, y, t) denotes the transverse displacement of the membrane. The solution of Eq. (16.14) is given by [3]
+ my
w(x, y, t)=f(lx
~ ct)
+ g(lx + my + ct)
(16.15)
where f and g denote plane waves propagating in a direction whose direction cosines are given by 1 and m. The transverse displacement w can be shown to be constant along the line Ix + my = constant at each instant of time t.
16.3.3
Harmonic Waves By redefining the variables ~ and
TJ
as
1 x ~= - - (x - ct) = t - -, c c
TJ
= -c1 (x + C1)
=t
x
+ -c
(16.16)
the solution of the one-dimensional wave equation can be expressed as w(x, 1) =
(t - ~)
f
+g
(1
+ ~)
(16.17)
where f and g denote, respectively, the forward- and backward-propagating waves. A wave whose profile (or shape or displacement configuration) is sinusoidal is called a harmonic wave. In general, harmonic waves moving in the positive and negative x directions can be represented, respectively, as
_ {A sin w (t - ~) w(x,t)-
. Asmw
(
X)
t+-
c
or or
(1 - ~) A cos w (1 + ~)
A cos w
(16.18) (16.19)
16.4 Wave Motion in Strings :;:.611
where A denotes the amplitude of the wave. By defining the wavelength A as 2;rrc A= w
(16.20)
we note that sinw (t
± x:
cos w (t
±x
A)
= sin [w (t ± ~) + 2;rr] = sinw (t ± ~)
: A) = cos [w (t
± ~) + 2;rr]
= cos w (t
(16.21) (16.22)
± ~)
Thus, the wave profile given by Eq. (16.18) or (16.19) repeats itself at regular intervals of the wavelength (x A). The reciprocal of the wavelength. known as the wave number (n).
=
n=-
1 A
(16.23)
denotes the number of cycles of the wave per unit length. The period of the wave (r) denotes the time required for a complete cycle to pass through a fixed point x so that ).. 2Jr (16.24) r=-=c w The frequency of the wave (j) is defined as the reciprocal of the period: A
1
c
(16.25)
f=;="i.
It can be seen that the frequency (j) and the wavelength ()..)are related as
fA = c
(16.26)
16.4 WAVEMOTION IN STRINGS 16.4.1 Free Vibration and Harmonic Waves The boundaries of a string of finite length invariably introduce complications in wave propagation due to the phenomenon of reflections. Hence, we first consider a long. infinite or semi-infinite string, where the problem of boundary reflections need not be considered. Using the separation-of-variables approach. the solution of the wave equation (free vibration solution of the string) can be obtained as [see Eq. (8.87)] w(x. t) = (A cos ~x
+ B sin ~x)
(C coswt
+ D sinwt)
w w .. -x cos wt + A2 cos -x smwt c C w + A4 sin -x sin wt
= AI
COS
w
+ A3 sm -x
coswt
C
(16.27)
C
,~
I
.,;
612
Elastic Wave Propagation w(X,t)
x
Figure 16.2
Vibration pattern of a string given by Eq. (16.28) (standing wave).
where Al = AC, A2 = AD, A3 a typical term in Eg. (16.27): WI
= BC,
and A4
(x, t) = Al
= BD
are the new constants. Consider
W
COS
-x coswt c
(16.28)
The deflections of the string at successive instants of time, given by Eg. (16.28), are shown in Fig. 16.2. It can be seen that certain points (called nodes) on the string undergo zero vibration amplitude, whereas other points (called antinodes) will attain maximum amplitude. The nodes and antinodes occur at regular spacings along the string and remain fixed in that position for all time as indicated in Fig. 16.2. This type of vibration is called a stationary or standing wave. The solution given by Eg. (16.27) can also be expressed, using trigonometric identities, as
W(X,t)=Blsin(~X
+wt)+B2sin(:x
+ B3 cos (~X where BI, B2, B3, and B4 are constants.
WI(X,t) = B4cos(:x
+ wt) + B4 COS
-wt) (:x
- wt)
(16.29)
The term
-wt)
= B4COSW(~ -t)
(16.30)
can be seen to denote a wave propagating in the positive x direction. We can see that as time (t) progresses, larger values of x are needed to maintain a constant (e.g., zero) value of the argument w[(xjc) - t). In this case, the deflection pattern of the string at successive instants of time, given by Eg. (16.30), appears as shown in Fig. 16.3. It can be observed that the constant (wj2rrc) is the wave number, the argument w[(xjc) - t] is the phase
16.4 Wave Motion in StrIngs
Phase velocity,
o
.~ ,;<.....
I( Figure 16.3
C
x
"
,,'
,613
/
-;.... ~._-><_./. Wavelength,
-\
A
Vibration pattern of a string given by Eq. (16.30) (propagating wave).
We can see that Eq. (16.31) represents a wave propagating in the negative direction. The solutions given by the other terms of Eq. (16.29) are similar in nature. Thus, the solution given by Eq. (16.29) denotes a propagating wave solution, whereas the solution given by Eq. (16.27) represents a standing-wave solution. In fact, the standing-wave solution of Eq. (16.27) can be obtained from constructive and destructive interference of waves propagating to the right and left. The validity of this aspect can be seen by considering the sum of two waves of equal amplitude propagating in different directions as w(x, t)
= A cosw (~ + t) + A cosw (~ - t) ....
wx
= 2A cos -
cos wt c where the term on the right-hand side represents a standing wave. 16.4.2
(16.32)
Solution in Terms of Initial Conditions If the initial displacement and velocity of the string are specified as U(x) and V(x), respectively, we have w(x,O) = U(x)
aw -(x, at
Using the general solution of Eq. (16.13) at f(x)
+ g(x)
(16.33)
0) = V(x)
= 0, we obtain
t
= U(x)
aw 0) af a(x - ct) -(x, = at a(x - ct) at
ag
+ a(x + ct)
a(x
+ ct)
at
(16.34) = V(x)
or -cf'(x)
+ cg'(x)
= V(x)
(16.35)
where a prime denotes differentiation with respect to the argument. Integration of Eq.(16.35) yields
11
X
f(x) - g(x) = -c
XQ
V(y) dy
(16.36)
614
Elastic Wave Propagation
where Xo denotes an arbitrary lower limit introduced to eliminate the constant of integration. The solution of Eqs. (16.34) and (16.36) yields
l +- l
x
1 = -U(x) - -1
f(x)
2
2cxo
V(y) dy
(16.37)
V(y)dy
(16.38)
x
1 g(x) = -2U(x)
1
2c
Xo
To express the general solution for t =1= 0, we replace x by x - ct in Eq. (16.37) and x + ct in Eq. (16.38) and add the results to obtain
+ g(x + ct)
w(x, t) = f(x - ct)
1 = -U(x:2
1
ct) - 2c
lx-ct
1 + -U(x + ct) + -1
V(y)dy
~
2
lx+ct
2c ~
V(y) dy
(16.39) Noting that -
ct xl
V(y)dy
+
~
lx+ct
V(y)dy
=
~
lx+ct
V(y)dy
~a
(16.40)
and defining x+ct
l
V(y)dy
X-ct
= R(x +ct)
- R(x - ct)
(16.41)
the displacement solution can be expressed as w(x, t)
1
= -2 [U(x - ct)
1
+ U(x + ct)] + -[R(x 2c
- ct)
+ R(x + ct)]
(16.42)
It can be seen that the motion, given by Eq. (16.42), consists of identical disturbances propagating to the left and the right with separate contributions from the initial displacement and initial velocity. 16.4.3
Graphical Interpretation of the Solution To interpret the solution given by Eq. (16.42) graphically, consider a simple case with zero initial velocity [with V (x) = 0 so that R (x - ct) = R(x + ct) = 0]: w(x, t) = HU(x - ct)
+ U(x + ct)]
(16.43)
This solution denotes the sum of waves propagating to the right and left which have the same shape as the initial displacement U (x) but have one-half its magnitude. Let the initial displacement, at t 0, be assumed as
= w(x, t) = HU(x) + U(x)] = U(x)
==
p(x)
(16.44)
with a peak value of 2po as shown in Fig. 16.4(a). The displacement distribution at any particular time can be obtained by superposing the waves propagating to the right
16.4 Wave Motiondn,~ngs
-615
W(X,t)
2po
1.75Po 1.5po 1.25po Po O.75po O.5Po O.25po x
a
0
-a
(a) W(X,t)
2Po 1.75po W(X,t)
l.Spo
= t p(x -
1)
+
t
p(x
+ 1)
1.25Po Po O.75po O.5Po O.25po
- ia 3
-a
-~3
0 (b)
Figure 16.4 Propagation of displacement disturbance with time: (a) at ct = 0; (b) at ct = a/3; (c) at ct ~a; (d) at ct a; (e) at ct 1a.
=
=
=
and left as shown in Fig. 16.4(b)(d). Notice that the separation of the two waves at any value of ct would be twice the distance traveled by one wave in the time ct. As such, for all values of ct > a, the two waves will not overlap and travel with the same shape as the initial disturbance but with one-half its magnitude (see Fig. 16.4e). Next, consider the case with initial velocity, V(x), and zero initial displacement [U(x) = 0], so that the solution becomes w(x, t)
1 = -[-R(x 2c
- ct)
+ R(x + ct)] = -1
2c
l
x ct
+
x-ct
V(y)dy
(16.45)
Po
O.75po O.5Po O.25Po
-1a -20 3
-a
-la 3
0
1a 3
x a
20
(e)
Figure 16.4
(continued)
The graphical interpretation of Eq. (16.45) is based on the integral of the initial velocity distribution. At any specific position x and time t, the solution given by Eq. (16.45) depends on the initial velocity distribution from x - ct to x + et. For example, if the initial velocity is assumed to be a constant as
ow 7it(x,O)
= Va
== qo
(16.46)
the displacement given by Eq. (16.45) at different instants of time will appear as shown in Fig. 16.5. It can be seen that when t < ale, the maximum displacement is given by Vat in the interval Ix I ::: a - et. In the ranges a - et :::x ::: a + et and - (a _
16.5 Reflection of Waves in One-Dimensional Problems
l~ I~
-a
•
x
a
0
617
(a)
/.
_.±.a 3
-a_~a
,~
.1. a
0
3
3
a
.±.a 3
•
x
(b)
2qoa
3C qoa
3C _2..a 3
-a '. _-La 3
0
-La
a
3
x
(c)
Figure 16.5 Displacement given by Eq. (16.45) at (a) t = 0, (b) ct = ~, (c) ct = ~a, (d) ct
= a and (e) ct = ~a.
et) ~ x ~ -(a + et), the displacement decreases linearly to zero. When t > ale, the maximum displacement is given by Voa/e in the interval-(et - a) ~ x ~ (et - a). As et -+ 00, the string gets displaced at a uniform distance Voa/c from its original position.
16.5 REFLECTION OF WAVESIN ONE-DIMENSIONAL PROBLEMS The investigation of propagation of waves in one-dimensional problems such as strings and bars with infinite domain does not require consideration of interaction of waves at boundaries. We consider the reflection of waves at fixed and free boundaries in this section. For simplicity, an intuitive approach is presented instead of a mathematical approach.
16.5.1
Reflection at a Fixed or Rigid Boundary Consider a semi-infinite
bar fixed at x = 0. The boundary condition at the fixed end is
given by u(O, t)
=
°
(16.47)
618
Elastic Wave Propagation
x
(e)
Figure 16.5
(continued)
where u is the axial displacement. An image displacement pulse system is introduced to satisfy the boundary condition ofEq. (16.47). Consider a wave approaching the rigid boundary at x = 0 from the right as shown in Fig. 16.6(a). Imagine the rigid boundary at x = 0 as removed and extend the bar to infinity. Now assume an "image" wave to the original propagating wave as shown in Fig. 16.6(b). The image wave is placed symmetrically with respect to x = 0, is opposite in sense to the original propagating wave, and propagates to the right. As the original and image waves approach x = 0, they interact as shown in Fig. 16.6(c). As they pass, their displacements will mutually cancel at x = 0, yielding u(O, t) = 0 always. Thus, the rigid boundary condition of the semiinfinite bar is always satisfied by the image wave system in the infinite bar. Mter some time, the interaction stage will be completed and the image wave propagates into x > 0, while the original "real" wave propagates into x < 0 as shown in Fig. 16.6(d). Thus, the reflected wave propagates along the positive x axis with the sign of the wave reversed. 16.5.2
Reflection at a Free Boundary Consider a semi-infinite bar free at x = O. Since the longitudinal zero, the boundary condition is given by O"xx(O, t)
AU = £-(0, ax
t)
stress at a free end is
=0
or
AU
-(O,t)=O
ax
(16.48)
16.6 Reflection and Transmission of Waves aUheinletface
Fixed boundary
of Two Etlstic Materiais
::619
c-.-1/:/ o-·_-·---·_-·--~·_-·_-· ..... ~
f"'-......
_-·_-·+~ ....;.
(a)
_'~"7'::'='::'='7"-=--:=:--=--:::,=,+_/:_,-,-,----,-C • ~---'.. L
-.::
------------
',/
, , '" •
~_+...,;: ••••
0
c (b)
c
/
.
_. ~ "7'::'=.::'=' 7"-=--: =:--=- -:::.= . L
---
,
.••.••• (
~x
0, I
c (c)
c -+----;- ..
..
"" ... ..•....;--~ ~------'-----_ '
I
.•••..•.• ",
-. 4_:...-=-..:.==·==..:.~·==.:.._=_..:
o
.+.~
' c
•
(d)
Figure 16.6 Sequence of events during reflection of a wave at a rigid boundary. (From Refs. [1] and [4]..)
Consider a wave approaching a free boundary from the right as shown in Fig. 16.7(a). Imagine the free boundary at x = 0 as removed and extend the bar to negative infinity. Now assume a "mirror image" wave to the original propagating wave as shown in Fig. 16.7(b). The image wave is symmetrically placed with respect to x = 0 and propagates to the right. As the original and image. waves approach x 0, they interact as shown in Fig. 16.7(c). As they pass, their slopes will mutually cancel at x 0 yielding always. au(O, t)/ax = o. Thus, the free boundary condition of the semi-infinite bar is always satisfied by the image wave system in the infinite bar. As time passes, the interaction stage will be completed and the image wave propagates into x > 0 while the original real wave propagates into x < 0 as shown in Fig. 16.7(d). Note that the sign of the original wave remains unchanged after completion of the reflection process.
=
16.6
=
REFLECTION AND TRANSMISSION OF WAVES AT THE INTERFACE OF TWO ELASTIC MATERIALS Consider two elastic half-spaces made of two different materials bonded together at their bounding surfaces as shown in Fig. 16.8(a), where the plane x = 0 denotes the
620
Elastic Wave Propagation
-) -0
c
•
(a)
+
-----~c ,
,.
c •
".-.:- - -'... - - - - - - --. .c-L_~--=-"':=-:"-==-;"~'==~-=--:_ . n
o
(b)
- c
c
,,,,,""
,."
.-x
_.E-::::::.::.:.:::.: ::-.-
o
(c)
c
c
~----I I
I .•.•
....
_ ..c7'=':: L -:.:!. ~..::::
... ....
.--
=-= ~~ -: ==.- -
x
_
o (d)
Figure 16.7 and [4].)
Sequence of events during reflection of a wave at a free boundary. (From Refs. [1]
interface between the two materials. The two materials are designated 1 and 2. Let a specified incident wave propagate in the positive x direction in the half-space 1. When the wave reaches the interface between the two materials, it gives rise to a reflected wave propagating in material 1 in the negative x direction and a transmitted wave propagating in material 2 in the positive x direction as shown in Fig. 16.8(b). Denoting the incident wave as (16.49)
and the reflected wave as r(1JI)
==
r
(1 + :J
(16.50)
the displacement in material 1 can be expressed as ul
= P(~I) + r(1JI) = P (1 -
~)
+r
(1 + ~)
(16.51)
16.6
621
Reflection and Transmission of Waves at the Interface of Two Elastic Materials J
•.•...-.-.-.
\--~
x
__
Plane, x
.~.
_.~.\-~x Material 2
Malerial I
Malerial 2
Malerial I
-
C2
Ct
=0
Plane, x = 0
(inlerface of two malerials) (b)
(a)
Free boundary
Rigid boundary
I
c,
.-.-.-.
- -'\AL
~
x
-
Malerial 2 (ri id)
Malerial I
Plane. x
_.K
f-.-.-.Material 2
Malerial I
-
=0
I .\-
Plane, x
=0
(d)
(c)
Figure 16.8 Reflection and transmission of waves at the interface of materials: (a) Incident wave; (b) reflected and transmitted waves; (c) reflection at a rigid boundary; (d) reflection at a free boundary.
By considering only the transmitted wave (s), the displacement in material 2 can be written as U2
= S(~2)
== S
(t - :J
(16.52)
Since the incident wave is specified, the function r is known. However, the functions s and p corresponding to the reflected and transmitted waves are to be determined using the following conditions:
1. The displacements in the two half-spaces must be equal at the interface: (16.53) 2. The normal stress in the two materials must be equal at the interface: O"XXt(O, t)
= O"XX2(0,
(16.54)
t)
Using Eq. (A.16), Eq. (16.54) can be rewritten as aUt
(A.+ 2tth -(0,
ax
aU2
t) = (A.+ 2tth-(0,
ax
t)
or, equivalently, (16.55)
'Z:·I."
622
Elastic Wave Propagation
where PI and P2 are the densities and CI and C2 are the velocities of dilatational (compressional) wave propagation in materials 1 and 2, respectively. Equations (16.51) and (16.52) yield OUI
-
oX
0P(SI) OSI OSI
OU2 _ OS(~2) OS2 __
ox -
OS2
01]1
or(1]I)
+ ---oX 01]1 oX
= ----
oX -
1 dp(SI) = ----
1
dSI
CI
~ dS(~2)
(16.56) (16.57)
d~2
C2
dr(1]I)
+ --CI d1]1
The boundary conditions, Eqs. (16.53) and (16.54), can be expressed, using Eqs. (16.51), (16.52), (16.56), and (16.57), as pet)
+ r(t)
= set)
(16.58)
d pet)
dr(t)
ds(t)
dt
dt
dt
---+--=-a--
(16.59)
where a
= P2 2 C
(16.60)
PICI
depends on the properties of the two materials. Equation (16.59) can be integrated to obtain
+ r(t)
-pet)
=
+A
-as(t)
(16.61)
where A is an integration constant. By neglecting A, Eqs. (16.58) and (16.61) can be solved to obtain r(t) and set) in terms of the known function pet) as . r
(x)+ -
= --p 1+a
(t _ ~)
= _2
t
Cl
s
C2
I-a ( + -x) 1+ a
t
(16.62)
CI
p
(t _ ~)
(16.63)
C2
Note that if the two materials are the same, a = 1 and there will be no reflected wave, but there will be a transmitted wave that is identical to the incident wave. 16.6.1
Reflection at a Rigid Boundary Consider an elastic material 1 (first half-space) bonded to a rigid material 2 (second half-space) as shown in Fig. 16.8(c). Let a specified incident wave propagate in the positive x direction in the half~space 1. The wave reflected from the rigid boundary (interface) can be found from Eq. (16.62) using the relation P2C2 --+ 00 for material 2 so that a ~ 00: (16.64) Equation (16.63) shows that there will be no transmitted wave in material 2, and Eq. (16.64) indicates that the reflected wave is identical in form to the incident wave but propagates in the negative x direction. The reflection process at a rigid boundary is shown graphically in Fig. 16.8(c).
16.7 Compressional and Shear Waves
16.6.2
623
Reflection at a Free Boundary Consider an elastic half-space 1 (material 1) with a free boundary at x = 0 as shown in Fig. 16.8(d). Let a specified incident wave propagate in the positive x direction in material 1. The wave reflected from the free boundary can be found from Eq. (16.62) using the relation P2C2 -+ 0 for material 2 so that a -+ 0: (16.65) Equation (16.65) indicates that the reflected wave is identical in form to the incident wave and propagates in the positive direction. The reflection process at a free boundary is shown graphically in Fig. 16.8(d). Note that the processes of reflection at fixed and free boundaries are similar to those shown for a bar in Figs. 16.6 and 16.7.
16.7 16.7.1
COMPRESSIONAL
AND SHEAR WAVES
Compressional or P Waves Consider a half-space with the x axis pointing into the material and the yz plane forming the boundary of the half-space as shown in Fig. 16.9(a). The half-space can be disturbed and compressional waves can be generated by either a displacement or a normal stress boundary condition. as indicated in Fig. 16.9(b) and (c). In Fig. 16.9(b). the half-space is assumed to be initially undisturbed with u (x, t) 0 for t ~ 0, and the boundary is then given a uniform displacement in the x direction so that
=
(16.66)
.. u(O, t) = r(t)
where r(t) is a known function of time that is zero for t ~ O. The motion resulting from this displacement boundary condition (with no motion of the material in the y and z directions) can be described by the components of displacement: u = u(x, t)
(16.67)
v=O W=O
For this one-dimensional motion. the equation of motion, Eq. (A.31), reduces to 82u
8t2 (x, t) = ex
where
282u(x,t)
8x2
(16.68)
'•.•" __(A +
2J.L)1/2 (16.69) P0 Po is the density of the material in the reference state, and A and J.L are Lame constants, related to Young's modulus E and Poisson's ratio v of an isotropic linear elastic materi ,as vE (16.70) A= (l
+ v)(1 E
J.L
- 2v)
= 2(1 + v) == G
(16.71)
624
Elastic Wave Propagation y
y
o
o
x
(U
x
(O,t)=r(t)
(a)
(b)
y
o
x
( un (O,t) = S (t) (c)
, Figure 16.9 (a) Undisturbed half-space; (b) displacement boundary condition; (c) normal stress boundary condition.
and G is the shear modulus. Equation (16.68) is known as a one-dimensional wave equation and a, in Eq. (16.69), denotes the wave velocity. Thus, the one-dimensional motion of the material, given by Eq. (16.67), is governed by Eq. (16.68). The solution or waves given by Eq. (16.68) are called compressional or P waves and ex is called the compressional or P wave speed [1, 4]. The displacements of a set of material points due to a compressional wave at a specific time t are shown in Fig. 16.10. Note that the material points move only in the x direction and their motions depend on x and t only. In Fig. 16.9(c), the boundary of the initially undisturbed half-space is given a uniform normal stress in the x direction so that axX<0, t) = s(t)
(16.72)
where $(t) is a known function of time that is zero for t ::::0. Equation (16.72), called the normal stress boundary condition, can be expressed in terms of the displacement u. using the stress-strain and strain-displacement relations. as axx(O, t) = (A
+ 2tL)sxX<0,
011(0, t)
t)
= (A + 2tL) ax
_
= s(t)
(16.73)
or
ou ox
-(0, t)
= s(t)
(16.74)
16.7 Compressional and Shear Waves
;625
y
o
x
(a) y
-
Direction of wave propagation I I
I I
I I
I I
I I
I I
~
~ ,I
I
,
x
-I
A.
Displacement boundary condition (b)
Figure 16.10 (a) Set of material points in the undisturbed half-space; (b) displacements of the set of material points due to a compressional wave;
where set) =
1
)..+ 2IJ-
_(
s t)
(16.75)
In this case also, compressional waves are generated. 16.7.2 Shear or S Waves The half-space shown in Fig. 16.11(a) can be disturbed and shear waves can be generated either by a displacement or a shear stress boundary condition, as indicated in
626
Elastic Wave Propagation y
y
st 0
~t ~t ';:'t t
x
0
x
(0)
(b) y
t St I., t 11
~t 0 8...f<'t t
x
(c)
Figure 16.11 (a) Undisturbed half-space; (b) displacement boundary condition; (c) shear stress boundary condition.
Fig. 16.11(b) and (c). In Fig. 16.11(b), the initially undisturbed uniform displacement in the y direction, so that
= r(t)
v(O, t)
half-space is given a
(16.76)
where r(t) is a known function of time that is zero for t ::: O. The motion resulting from this displacement boundary condition (with no motions along the x and z directions) can be described by
u=O v=v(x,t)
(16.77)
w=O For this one-dimensional
motion, the equation of motion, Eq. (A.33), reduces to 8v(x.t) 8t1
=
f3
282v(x,t) 8x2
(16.78)
where f.J.
f3= ( -
Po
)1/2
(16.79)
16.7 Compressional and Shear Waves
627
Equation (16.78) is also called a one-dimensional wave equation. and /3. given by Eq. (16.79). denotes the wave speed. The solution or waves given by Eq. (16.78) are called shear or S waves and /3 is called the shear or S-wave speed [1. 4]. The displacements of a set of material points due toa shear wave at a specific time t are shown in Fig. 16.12. Note that the material points move only in the y direction and their motions depend on x and t only. In Fig. 16.11(c), the boundary of the initially undisturbed half-space is given a uniform shear stress in the y direction. so that (16.80)
't"xy(O, t) = s(t)
y
o
x
...
'.
I (a)
y
t t t
~ Dm:ction of particle monon ,
t
,
t t t
....
p.....,
1'-_ '-
--
--
b-~
b-~
- ...•b-~
~" ""
~~
- - - ...•
""
- - -....
~~
-'"
- - ....
,
•... ,,
,
,, ,
"" -I
A
Direction of wave .propagation
x
•... , , I
Displacement boundary condition (b)
Figure 16.12 (a) Set of material points in the undisturbed half-space; (b) displacements of the set of material points due to a shear wave.
628
Elastic Wave Propagation
where s(t) is a known function of time that is zero for t ::: O. Equation (16.80), called the shear stress boundary condition, can be expressed in terms of the displacement v, using the stress-strain and strain-displacement relations, as 't"xy(O, t)
[auay
= j1.Exy(O, t) = j1. -(0, t)
av
av] + -(0, ax
t)
_
= j1.-(0, t) = s(t)
(16.81)
av ax
(16.82)
ax
or -(0, t) = s(t)
where
L
s(t) = -s(t)
(16.83)
j1.
In this case also, shear waves are generated.
16.8 FLEXURAL WAVES IN BEAMS The equation of motion for the transverse motion of a thin uniform beam, according to Euler-Bernoulli theory, is given by
a4w(x,
t)
ax
4
1
+c2
a2w(x,
t)
=0
at
2
(16.84)
where (16.85) It can be observed that Eq. (16.84) differs from the one-dimensional wave equation, Eq. (16.1), studied earlier in terms of the following: 1. Equation (16.84) contains a fourth derivative with respect to x instead of the
second derivative. 2. The constant c does not have the dimensions of velocity; its dimensions are in2/sec and not the in./sec required for velocity. Thus, the general solution of the wave equation, w(x, t)
= f(x
- ct)
+ g(x + ct)
(16.86)
will not be a solution of Eq. (16.84). As such, we will not be able to state that the motion given by Eq. (16.84) consists of waves traveling at constant velocity and without alteration of shape. Consider the solution of Eq. (16.84) for an infinitely long beam in the form of a harmonic wave traveling with velocity v in the positive x direction: 2JT
w(x, t) = A cos -(x
A
- vt)
==
A cos(kx - wt)
(16.87)
l6.S
Hcxural Waves in Beams
629
where A is a constant, A is the wavelength, v is the phase velocity, k is the wave number, and W is the circular frequency of the wave, with the following interrelationships: W
= 2rrf = kv
(16.88)
k
= 2rr
(16.89)
A
Substitution of Eq. (16.87) into Eq. (16.84) yields the velocity, also called the wave velocity or phase velocity, as
v = 2" c = 2"
A
A
J
£1
(16.90)
pA
Thus, unlike in the case of transverse vibration of a string, the velocity of propagation of a harmonic flexural wave is not a constant but varies inversely as the wavelength. The material or medium in which the wave velocity v depends on the wavelength is called a dispersive medium. Physically, it implies that a nonharmonic flexural pulse (of arbitrary shape) can be considered as the superposition of a number of harmonic waves of different wavelengths. Since each of the component harmonic waves has different phase velocity, a flexural pulse of arbitrary shape cannot propagate along the beam without dispersion, which results in a change in the shape of the pulse. A pulse composed of several or a group of harmonic waves is called a wave packet, and the velocity with which the group of waves travel is called the group velocity [4,5]. The group velocity, denoted by g .• is the velocity with which the energy is propagated, and its physical interpretation can be seen by considering a wave packet composed of two simple harmonic waves of equal amplitude but slightly different frequencies W + !:i.w and w - Aw. 'The waves can be described as
v
- Wit)
(16.91)
W2(X, t) = Acos(k2X - Wzt)
(16.92)
Wl(X, t) = Acos(kiX
where A denotes the amplitude and
+ !:i.k
(16.93)
k2 = k - !:i.k
(16.94)
Wi
=w+Aw
(16.95)
Wz
=w-
(16.96)
k1 = k
Aw = !(Wi - Wz)
(16.97)
Ak = !(k1 - k2)
(16.98)
The wave packet can be represented
WP
=
Wi
(X, t)
+
W2(X, t)
Aw
by adding the two waves:
=
A [COS(kiX -
Wi
t)
+
COS(k2x
Wz t)] (16.99)
630
Elastic Wave Propagation
) Group velocity, vg Wave packet (group) composed of waves 1 and 2
Figure 16.13
Wave pa~ket and group velocity.
Equation (16.99) can be rewritten as wp(x, t) = 2A cos(~kx - ~wt) cos(kx - wt)
(16.100)
The wave packet, given by Eg. (16.100), is shown graphically in Fig. 16.13. It can be seen that it contains a high-frequency cosine term (at frequency w) and a low-frequency cosine term (at frequency ~w). The high-frequency oscillatory motion is called the carrier wave and moves at a velocity v, known as the phase velocity, given by w v = k
(16.101)
while the low-frequency oscillatory motion propagates at a velocity vg, known as the group velocity, given by Vg
=
(16.102)
The wave motion given by Eg. (16.100) is called an amplitude-modulated carrier and is shown in Fig. 16.13. It can be seen that the low-frequency term acts as a modulator on the carrier (denoted by the high-frequency term). Accordingly, the factor cos (kx - wt) represents the carrier wave and the factor cos (D.kx - D.wt) indicates the envelope
Wave Propagation in an lntinite Elastic Medium
16.9
631
moving at the group velocity. The behavior is similar to that of beats observed in coupled oscillators (see Section 1.9.2). The group velocity can be represented in an alternative form using the basic relation w = kv as v
g
=
.
t:.w
dw
d(kv)
t:.k
dk
dk
hm-=-=--
Ak ••.•O
= v
+
dv k dk
(16.103)
where v is considered a function of k. Noting that k = 21CjA, Eq. (16.103) can also be expressed in terms of the wavelength A as dv A dA
vg = v -
(16.104)
In the case of propagation of flexural waves in beams, the group velocity is giyen by dv = v - A (21CC - -2 ) = v dA
A
vg = v - A -
+ -21CC A
= 2v
HI/
= -41C A --pA
(16.105)
Equation (16.105) shows that for flexural waves in beams, the group velocity is twice the wave velocity.
16.9
WAVE PROPAGATION IN AN INFINITE ELASTIC MEDIUM In this section the elastic wave propagation in solid bodies is considered. Since the body is infinite, boundary interactions of waves need not be considered. The equations governing waves in infinite media are derived from the basic equations of elasticity. Two basic types of waves, dilatational and distortional waves, can propagate in an infinite medium. These two types of waves can exist independent of one another.
16.9.1 Dilatational Waves To see the simplest types of waves generated in the solid body, differentiate the equations of motion (A.31), (A.33), and (A.34) with respect to x, y, and z, respectively, and add the resulting equations to obtain (16.106) or 2 13
at
where
CI,
t:.
2
= cfV'2 t:.
called the velocity of wave propagation, CI
=
(A:2/LY/2
(16.107)
is given by (16.108)
Equation (16.106) is called a wave equation that governs propagation of the dilatation t:. through the medium with a velocity c.
632
16.9.2
Elastic Wave Propagation
Distortional Waves The dilatation (t.) can be eliminated from the equations of motion (A.31), (A.33), and (A.34) to obtain the equation governing distortional waves. For this, differentiate Eqs. (A.31) and (A.34) with respect to z and y, respectively, and subtract the resulting equations one from the other to obtain 2 (
/-L"V .
a2 (ou--+OW) at2 az ay
au OW) --+OZ
=p-
oy
Noting that the quantity in parentheses denotes twice the rotation be rewritten as
(16.109) wx,
Eq. (16.109) can
(16.110) By using a similar procedure, the other two equations of motion can be written in terms of rotations wyand Wz as /-L"V2w
y
2 /-L "V
Wz
=
a2w
p--y
at2
a
2 w = P -2-z
at
(16.111) (16.112)
Each of Eqs. (16.110)-(16.112) denotes a wave equation that governs the propagation of rotational or equivoluminal or distortional or shear waves through the elastic medium with a velocity of wave propagation given by C2
=
(~Y/2
(16.113)
It can be seen from Eqs. (16.108) and (16.113) that the velocity of rotational waves is smaller than the velocity of dilatational waves. The dilatational and rotational waves denote the two basic types of wave motion possible in an infinite elastic medium. 16.9.3
Independence of Dilatational and Distortional Waves In general, a wave consists of dilatation and rotation simultaneously [2, 4]. The dilatation propagates with velocity c] and the rotation propagates with velocity C2. By setting dilatation t. equal to zero in Eqs. (A.31), (A.33), and (A.34) one obtains the equations of motion: (16.114) (16.115) (16.116)
16.9 Wave Propagation in an·lafinite·ElastKMedium.
-~3
Equations (16.114)-(16.116) describe distortional waves that propagate with a velocity C2 given by Eq. (16.113). In fact, the motion implied by Eqs. (16.114)-(16.116) is that of rotation. Thus, the wave equations (A.31), (A.33), and (A.34), in the absence of dilatation, reduce to those obtained for rotation, Eqs. (16.110)-(16.112). Next, we eliminate rotations wx, wyand Wz from the equations of motion. For this, we set Wx wy Wz 0:
=
=
=
aw
av
au
aw
av
au
0
ay - az = az - ax = ax - ay and define a potential function
-
U--
(16.117)
=
such that
t/J
at/J
at/J
at/J ax
.
v--
(16.118)
w=-
az
ay'
Using Eq. (16.118), the rotation wx, for example, can be seen to be zero: w _ ~ x - 2
(aw _ av) _ ~ (a2t/J _ a2t/J) ay az - 2 ayaz azay
=0
(16.119)
Using Eq. (16.118), the dilatation can be written as
au + av + aw . ax ay az
~=
=
V2t/J
(16.120)
Differentiation of Eq. (16.120) with respect to x gives
a~ = ~(V2t/J) = V2 (at/J) = v2u ax ax ax
(16.121)
and using a similar procedure leads to
all
2
(16.122)
v2w
(16.123)
-=Vv
ay
a~
az
=
Substitution of Eqs. (16.121), (16.122), and (16.123) into the equations of motion, (A.31), (A.33), and (A.34) respectively, yields (A.
+ 2JL)V2u
(A.
+ 2JL)V2v
(A.
+ 2JL)V2w
a2u at2 a2v = P-"2 at a2w = p at2
=p
(16.124) (16.125) (16.126)
Equations (16.124)-(16.126) describe dilatational waves that propagate with a velocity Cl given by Eq. (16.108).
634
Elastic Wave Propagation
Notes 1. If the displacement components u, v, and W depend only on one coordinate, such as x, instead of x, y, and z, the resulting waves are called plane waves that propagate in the direction of x. Assuming that u = u(x, t),
V=V(x,t)
W=W(x,t)
(16.127)
the dilatation is given by
ou ox
A =-
(16.128)
and the equations of motion (A.31), (A.33) and (A.34) reduce to 02u
(A
+ 2/.L) ox2
02u
= p ot2
02v
ox2
/.L
(16.129)
02v
= Pai2
02w
(16.130)
02w
/.Lox2 = P ot2
(16.131)
Equations (16.129) describes a longitudinal wave in which the direction of motion is parallel to the direction of propagation of the wave. This wave is a dilatational wave and propagates with a velocity CI. Each of Eqs. (16.130) and (16.131) describes a transverse wave in which the direction of motion is perpendicular to the direction ·of propagation of the wave. These are distortional waves and propagate with a velocity C2. 2. Dilatational waves are also called irrotational, longitudinal, or primary (P) waves, and distortional waves are also known as equivoluminal, rotational, shear or secondary
(S) waves.
3. All the wave equations derived in this section have the general form c2'i;12TJ = P-
02TJ
ot2
(16.132)
Consider a specific case in which the deformation takes place along the x coordinate so that Eq. (16.132) can be written as 202TJ C
02TJ
ox2 = ot2
(16.133)
The general solution (or D' Alembert's solution) of Eq. (16.133) can be expressed as
= f(x
TJ(x,t)
- ct)
+ g(x + ct)
(16.134)
where f denotes the wave traveling in the +x direction and g indicates the wave traveling in the -x direction. If the disturbance occurs only at one point in the elastic medium, the deformation depends only on the radial distance (r) from the point. The radial distance r can be expressed in terms of the Cartesian coordinates x, y, and z as r
= J x2 + y2 + Z2
(16.135)
· 16.10 Rayleigh or Surface Waves and hence the derivatives
of
TJ
with respect to
x, y
and
z
can be expressed as
a27] x2 a2TJ r2 - x2 aTJ ax2 = r2 ar2 + r3 ar a27]
y2 a2TJ r2 - y2 aTJ + --:-=----'. r2 ar2 r3 ar
-
= --
a27] az2
=
ay2
Addition of Eqs. (16.136)-(16.138)
635
Z2 a2TJ r2 - Z2 aTJ r2 ar2 + -r-3 - ar
(16.136)
(16.137)
(16.138)
yields 2
'il2TJ = a TJ+ ~a7] ar2 r ar
(16.139)
and hence Eq. (16.132) can be written as c2
2 (a 7]
ar2
+ ~aTJ) r ar
= 'il27]
(16.140)
or 2
c Equation (16.141) is known can be expressed as
a2(rTJ)
a;:z
as the spherical
r7] = f(r
- ct)
=
a2(rTJ)
---at2
wave equation,
+ g(r + ct)
(16.141) whose general solution
(16.142)
In Eq. (16.142), f denotes a diverging spherical wave and g indicates a converging spherical wave from the point of disturbance. The amplitude of the wave (TJ) is, in general, inversely proportional to the radial distance r from the point of disturbance.
16.10
RAYLEIGH OR SURFACE WAVES As seen earlier, two types of waves, dilatational and distortional waves, can exist in an isotropic infinite elastic medium. When there is a boundary, as in the case of an elastic half-space, a third type of waves, whose effects are confined close to the bounding surface, may exist. These waves were first investigated by Rayleigh [7] and hence are called Rayleigh or surface waves. The effect of surface waves decreases rapidly along the depth of the material, and their velocity of propagation is smaller than those of P and S waves. The discovery of surface waves was closely related to seismological studies, where it is observed that earthquake tremors usually consist of two minor disturbances, corresponding to the arrival of P and S waves followed closely by a third tremor that causes significant damage. The third wave (surface wave) is found to be associated with significant energy that is dissipated less rapidly than the P and S waves and is essentially confined to the ground surface. To study these surface waves, consider a semi-infinite elastic medium bounded by the xy plane, with the z axis pointing into the material, as shown in Fig. 16.14. Assume a wave that is propagating on the bounding surface along the x direction with its crests parallel to the y axis. Hence, all the components of
636
Elastic Wave Propagation
x
Bounding plane
I --.-.-.-.--+-
Z
Semi-infinite elastic medium
Figure 16.14
Semi-infinite medium.
displacement u, v, and w are independent of y. For this case, the equations of motion (A.31), (A.33) and (A.34) Can be written as
aA
a2u
at2
(16.143)
J1,\l2v
= p-aat2v
(16.144)
aA + J1,)az + J-L\l2w
= P-2aatw
(16.145)
(A
+ J-L)a; + J-L\l2u
=p
2
(A
2
where
au aw ax + -az a2 a2 = - 2 +- 2 ax az
A = -
\72
(16.146) (16.147)
Note that Eq. (16.144) describes distortional waves that are not confined just to the bounding surface. Also, Eqs. (16.143) and (16.145) do not contain v. Hence, we assume that v = 0 and consider only Eqs. (16.143) and (16.145) in the analysis. Define two potential functions ¢J and 1/1 such that
a¢J
a1/f
ax + -az
u =-
a¢J
a1/f
az
ax
(16.148)
w =- - Since
v
and y are not considered, dilatation
A and rotation
au aw . ax az = ~2 (auaz _ aw) ax
/),.=-+-=\l¢J
iV y
(16.149) iVy
are given by
2
(16.150)
= ~\l21/1 2
(16.151)
16.10
= ~ (OW _
W
6."
=0
(16.152)
= ~ (:: - :~) = 0
(16.153)
oy
2
x
Wz
ov) oz
Rayl~igh or Surface Waves
It can be seen that the definitions of
(16.155) Equations (16.154) and (16.155) will be satisfied if equations (can be verified by direct substitution):
()..+ 2fJ-)V2
02
and
are the solutions of the
1/f
= P ot2
or
2V2
(16.156)
021/f ot2
or
2V21/f _ c2 --
(16.157)
ot
_0
2
fJ- V
1/f = P
cP1/f ot2
where Cl and C2 denote the velocities of dilatational and distortional waves given by Eqs. (16.108) and (16.113), respectively. Next, the wave described by Eqs. (16.156) and (16.157) is assumed to be harmonic or sinusoidal propagating in the positive x direction with velocity C at a frequency w. The velocity can be expressed as
w c=n
== w)..
(16.158)
where n is the wave number and ).. is the wavelength (n = 1/),,). The solutions of Eqs. (16.156) and (16.157) are assumed to be of the following complex harmonic form:
= 2[ (z)ei(wt-nx)
(16.159)
1/f
= 22 (z)ei(wt-nx)
(16.160)
where the functions 2[ (z) and Z2(Z) indicate the dependence of the amplitudes of and 1/f on the coordinate z. When Eq. (16.159) is used in Eq. (16.156), we obtain
c2 1
(~
ox2
+ ~)
OZ2
Z (z)ei(wt-nx) 1
= ~[2 ot2
[
(z)ei(wt-nx)]
or (16.161)
638
Elastic Wave Propagation
The solution of this second-order ordinary differential equation can be expressed as (16.162) where
a = (n2 _ ;;Y/2
(16.163)
For real values of the amplitude of ¢, a must be positive. For positive values of a, the second term of Egs. (16.162), and hence the amplitude of the wave, increases with increasing values of z. This is not physically possible from the point of view of conservation of energy. Hence, C2 must be zero in Eq. (16.162) for a surface wave. By substituting Eq. (16.160) into Eq. (16.157) and proceeding as in the case of ¢, we obtain the solution of 22(z) as
22(Z) = C3e-fiz + c4efiz where
C4
(16.164)
= 0 and ~=
(n2 _ ;;Y/2
(16.165)
Using Eqs. (16.162) and (16.164) in Eqs. (16.159) and (16.160), the solutions of ¢ and t/I of Eqs. (16.156) and (16.157) representing a harmonic wave propagating in the positive z direction with frequency CI) and wave number n and decaying with increasing values of z can be expressed as ¢ =--Cl e-az+i(wt-nx) t/I
(16.166)
= C3e-fiHi(wt-nx)
(16.167)
where a and ~ are given by Eqs. (16.163) and (16.165), respectively, and Cl and C3 are constants to be determined from the boundary conditions. The free surface conditions applicable to the boundary surface (z 0) of the elastic half-space are
=
(16.168) Using the stress-strain relations of Eg. (A.16) and the strain-displacement of Eg. (A.2), the normal stress (Tzz can be expressed as (T---
= AD.
au
+ 2,,-ax
relations
(16.169)
I""
where D. is given by Eg. (16.150). Using Eqs. (16.148) and (16.149) for u and w, Eg. (16.169) can be written as (Tzz
or
2 2 2 = A (a ¢2 + a t/1 + a ¢ _ a2t/1) + 2J1- (a2¢ ax axaz az2 axaz az2
_
a2t/1) axaz (16.170)
16.10 Substitution
Rayleigh or Surface Waves
639
of Eqs. (16.166) and (16.167) into Eq. (16.170) gives (16.171)
The condition O'zz(z = 0) = 0 can be written as (16.172) The shear stress O'xz can be expressed using Eqs. (A.16) and (A.8) as O'xz Using Eqs. (16.148)
= J.tsxz = J.t (auaz + aw) ax
and (16.149) for u and w, Eq. (16.173) can be written as a2
Substitution
(16.173)
+
a21fr a21fr) az2 - ax2
(16.174)
of Eqs. (16.166) and (16.167) into Eq. (16.174) yields (16.175)
The condition O'xz(z
= 0) = 0 can
be expressed
+ (/32 + n2)c3
(2ina)cI
Finally the shear stress O'yz can be expressed,
O'yz
as
= J.tSyz = J.t
(16.176)
=0
using Eq. (A.16), as
aw av) ( ay + az
(16.177)
Nothing that u and w are independent of y and v is assumed to be zero, Eq. (16.177) gives O'yz = 0 throughout the elastic half-space. Hence, the boundary condition O'yz(z 0) 0 is satisfied automatically. Equations (16.172) and (16.176) represent two simultaneous homogeneous algebraic equations with CI and C3 as unknowns. For a nontrivial solution of Cl and C3, the determinant of their coefficient matrix is set equal to zero to obtain
=
=
or (16.178) Squaring both sides of Eq. (16.178) and using Eqs. (16.163) and (16.165) for a and /3, respectively, we obtain
[ -(A +
2
cf2 + 2J.tn ]2 ( 2n
(J)
2J.L)
2 c~2)2 (J)
-
242 (2) cf ( 2 c~2) w
(J)
=
16J.L n
n
-
n
-
(16.179)
640
.Elastic Wave Propagation
Dividing both sides of Eq. (16.179) by
f..L2n8
leads to
(16.180) Note the following relationship from Eqs. (16.108) and (16.113) and Eqs. (A.22) and (A.23): c~
E
f..L
cr = )..+
2f..L
{
= 2(1 + v).
1
+ v)(1
[v£/(1
-
2v)J
}
+ 2£/2(1 + v) =
1 _ 2v
2-
2v
(16.181) Introducing the notation c~ 1 - 2v ----y c21 - 2 - 2v -
2
(16.182)
w2
2
P =-2
(16.183)
c2 n2
we can express
(16.184) 'Substituting Eqs. "(16.181)-(16.184) into Eq~ (16.180) gives (2 -
p2)2(2 - p2)2
= 16(1 _
y2 p2)(1 _ p2)
or (16.185) Since p2
i= 0, Eq. (16.185) leads to p6 -
8p4
+ 8(3 -
2y2)p2
+ 16(y2
_
1) = 0
(16.186)
To investigate the roots of Eq. (16.186), we first note that it is a cubic equation in p2, and second, that the roots depend on Poisson's ratio [see Eq. (16.182)]. Since p is
defined as the ratio of the velocity of the surface wave to that of the distortion wave in Eq. (16.183), the velocity of propagation of the surface wave will be a constant for any given material. The roots of Eq. (16.186) can be real, imaginary, or complex, depending on the range of Poisson's ratio [1, 4J: 1. For
11
less than about 0.263; all three roots
(p2)
are real.
2. For v about 0.263; one root is real and two roots are complex conjugates. 3. Complex roots will not be valid in the present case since this will result in the attenuations of ¢> and 1/f with respect to time, as if damping were present, which is not the case in the present problem [1, 4].
16.10
Rayleigh or Surface Waves
641
Example 16.1
Investigate the variation of the velocity of propagation of surface waves for different values of the Poisson's ratio (v). SOLUTION First we consider the values of v = 0.25. It can be verified that this value of v corresponds to a surface wave with A.= /.L and is often used for rock. For v = 0.25, Eq. (16.182) gives 2 1 - 2v 1 y ----- 2 - 2v - 3
(EI6.1.l)
56 3
(EI6.1.2)
and Eq. (16.186) becomes 6
4
2
P -8p --p
32 3
--=0
The roots of Eq. (EI6.1.2) are given by 2
P 2-4 - , and the corresponding values of found as follows:
(X
c1n2
n2
c2
'--,
2
(EI6.1.3)
v'3
and {3,given by Eqs. (16.163) and (16.165), can be
(X2 . (Ji w2 -2=1--2-=I--2--2=I-yp
n
2--
2 + v'3'
C~
2
2
c1
=1--
p2
3
{32 w2' 2 n2 = 1 - c2n2 = 1 - p
(EI6.1.5)
2
For p2 = 4: (X2
-
n2
4 1 - 1 - - - -3 3'
(X
- = imaginary n
-
{32
"2 = 1-4=
-3,
n
{3
..
- = unagmary n
For p2 = 2 + 2/.J3: (X2
-
n2
2 2 = 1- - - ~ -0.051567, 3 3v'3
{32
"2 = 1 - 2 n
2 ;;:; ~ -2.15470,
",3
(X
••
- = lmagmary n {3
..
- = lmagmary n
For p2 = 2 - 2/.J3: (X2 2 - = 1- 2 n 3
{32
+-
2
3v'3 ~
(X
~ 0.71823,
2
"2 = 1 - 2 + M. n ",3
(EI6.1.4)
0.15470,
- ~ 0.84748 n
!!.. ~ 0.39332 n
642
Elastic Wave Propagation
It can be seen that the larger two values of p2 in Eq. (EI6.1.3) give imaginary values of aln and fJln which violate the requirement of aln > 0 and I3/n > O. Hence, the only valid value of p2 yields 2 ) ( 2-- ~
~=p= C2n
1/2
=0.91940
or w
- = 0.91940C2 n
(EI6.1.6)
The variations of WlC2n (and c2/cI = l/y) for different values of Poisson's ratio (v) are shown graphically in Fig. 16.15. Since the velocity of propagation, C = win, is independent of the frequency for any given material, the surface wave propagates with no dispersion at a velocity slightly smaller than the velocity of propagation of distortion or S waves.
Example 16.2 material (z).
Derive expressions for the variations of u and w with the depth of
SOLUTION The displacement components u and w can be expressed, using Eqs. (16.148), (16.149), (16.166), and (16.167) as -. :"'az+i(wt-nx)fJ u = -Cline - C3 e -/3z+i(wt-nx)
w = _clae-az+i(wt-nx)
_ c3ine-/3Z+i(wt-nx)
(EI6.2.1) (EI6.2.2)
Equation (16.176) gives
(EI6.2.3) (EI6.2.4) (EI6.2.5) The physical displacements u and w are given by the real parts of Eqs. (EI6.2.4) and (EI6.2.5) as
U
W
- n (-a. = CI -e
= cIa (_e-az
- -
+
2afJ e _RZ)' I' sm (wt - nx ) fJ2 + n2
(E16.2.6)
2n2 fJ
2
+n
2e-/3z) cos(wt - nx)
(EI6.2.7)
16.11
Recent Contributions
643
1.0 I I I I I
O~
------~------~------~-----: I I
.f 0.6 )
c2 ci
=.!.Y :I
:
I I I I
t I I
----I------T-----I I I I L
______
c; 0.4
I I I I
_
1.:
L
_
I
I t I I I I .L.
_
I I I I I I
()'
Q2
------~------~I I I I I
o
0.2
I I I I I
I I I I I
0.4 _v
___
I I I I I I
0.8
0.6
(a)
0.96 I I I I I . I
------r------ ------r------
0.94
I I I I I I
I I I I I I
I
I __ L
t I I I I L
I I I I I L
_
I I I I I I L
I I I I I I L
I I I I I I .L.
_
I I I I I I
I I I I I I
I I I I I I
0.4
0.6
------1- ----I------T------
0.92
I
0.90
_____
0.86
I
I I I I I I
o
0.2
_v
0.8
(b)
Figure 16.15
16.11
Variations of
czlct. w/nq, and W/nc2
with v. (From Ref. [8].)
RECENT CONTRIBUTIONS Comprehensive Studies A brief history of the study of wave propagation in solids and a presentation of wave motion in elastic strings, rods, beams, membranes, plates, shells, semi-infinite media, and infinite media have been given by Graff [4]. Kolsky summarized the basic developments in the field of stress waves in Ref. [9].
644
Elastic Wave Propagation
Longitudinal Waves A theory of propagation of longitudinal stress waves in a cylindrical rod with several step changes in the cross-sectional area was developed by Beddone [10]. The analysis obtained a transient solution of the one-dimensional wave equation by means of Laplace transform methods based on the concepts of traveling waves and reflection and transmission coefficients. Wave Propagation in Periodic Structures The problem of free coupled longitudinal and flexural waves of a periodically supported beam was studied by Lee and Yeen [11]. It was shown that the characteristic or dispersion equation can be factorized into product form, which simplifies the analysis and classification of the dynamic nature of the system. Sen Gupta [12] studied the propagation of flexural waves in doubly periodic structures consisting of the repetition of a basic unit that is a periodic structure in itself. The analysis is simplified by introducing a direct and a cross-chain receptance for multispan structures and by utilizing the concept of the equivalent internal restraint. Wave Propagation Under Moving Loads Ju used the three-dimensional finite element method to simulate the soil vibrations due to high-speed trains moving across bridges in Ref. [13]. He first analyzed a bridge system passed by trains. Then the pier forces and moments calculated were applied to a pile cap to simulate wave propagation in the soil. Waves Through PiLlte or Beam Junctions In a study of elastic wave transmission through plate-beam junctions by Langley and Heron [14], a generic plate-beam junction was considered to be composed of an arbitrary number of plates either coupled through a beam or coupled directly along a line. The effects of shear deformation, rotary inertia, and warping were included in the analysis of the beam, and due allowance was made for offsets between the plate attachment lines and the shear axis of the beam. Vrbration Analysis Using a Wave Equation Langley showed that the vibrations of beams and plates may be analyzed in the frequency domain by using a wave equation instead of the conventional differential equations of motion provided that certain assumptions are made regarding the response of the system in the vicinity of a structural discontinuity [15]. Measurement of Wave Intensity Halkyard and Mace [16] presented a Fourier series approach to the measurement of flexural wave intensity in plates. The approach is based on the fact that in regions sufficiently remote from excitation and discontinuities, the flexural motion of a plate can be expressed as the sum of plane propagating waves. lilmb Waves With the aim of clarifying the manner in which the dispersion curves for real- and imaginary-valued Lamb modes in a free, isotropic, elastic plate vary with the Poisson ratio, Freedman presented a set of Lamb mode spectra at fixed values of the Poisson ratio covering its full range [17].
REFERENCES 1. A. Bedford and D. S. Drumheller, Introduction to Elastic Wave Propagation. Wiley, Chichester, West Jeessey, England, 1994. 2. S. K.Clark, Dynamics of Continuous Elements, Prentice-Hall, Englewood Cliffs, NJ, 1972. 3. E.Volterra and E. C. Zachmanoglou, Dynamics of Vibrations, Charles E. Merrill, Columbus, OH, 1965.
Problems
645
4. K. F. Graff, Wave Motion in Elastic Solids, Ohio State University Press, Columbus, OH 1975. 5. L. Meirovitch, Analytical Methods in Vibrations, Macmillan, New York, 1967. 6. H. I. Pain, The Physics of Vibrations and Waves, 4th ed, Wiley, Chichester, West Ieessey, England, 1993.. 7. 1. W. S. Rayleigh, On waves propagated along the plane surface of an elastic s61id, Proceedings of the London Mathematical Society, Vol. 17, pp. 4-11, 1887. 8. A. Knopoff, On Rayleigh wave velocities, Bulletin of the Seismological Society of America, Vol. 42. pp. 307-308. 1952. 9. H. Kolsky, Stress Waves in Solids, Dover, New York, 1963. 10. Beddone, Propagation of elastic stress waves in a necked rod. Journal of Sound and Vibration, Vol. 2, No.2, pp. 150-164, 1965. 11. S. Y. Lee and W. F. Yeen. Free coupled longitudinal and flexural waves of a periodically supported beam, Journal of Sound and Vibration, Vol. 142, No.2, pp. 203-211, 1990. 12. G. Sen Gupta, Propagation of flexural waves in doubly-periodic structures, Journal of Sound and Vibration, Vol. 20, No.1, pp. 39-49, 1972. 13. S. H. Iu, Finite element analysis of wave propagations due to high-speed train across bridges, International Journal for Numerical Methods in Engineering, Vol. 54, No.9, pp. 1391-1408, 2002. 14. R. S. Langley and K. H. Heron, Elastic wave transmission through platelbeam junctions, Journal of Sound and Vibration, Vol. 143, No.2, pp. 241-253, 1990. IS. R. S. Langley, Analysis of b.eam and plate vibrations by using the wave equation, Journal of Sound and Vibration, Vol. ISO, No. I, pp. 47-65, 1991. 16. C. R. Halkyard and B. R. Mace, A Fourier series approach to the measurement of flexural wave intensity in plates, Journal of Sound and Vibration, Vol. 203, No. I, pp. 101-126,1997. 17. A. Freedman, The variation, with the Poisson ratio, of Lamb modes in a free plate, I: General spectra, Journal of Sound and Vibration, Vol. 137, No.2, pp. 209-230, 1990.
PROBLEMS 16.1 Determine the velocity of propagation of longitudinal, torsional, and bending waves in a steel bar with E = 207 GPa, G = 79.3 GPa, and a unit weight of 76.5 kN/m3• 16.2 Consider a bar with density 7500 kglm3 and Lame constants A. = 1.15 X 1011Pa and J.L = 0.75 X 1011 Pa. Determine the compressional and shear wave velocities. 16.3 Consider the following initial conditions for a long, transversely vibrating string: 2, -I < x < 1 w(x,O) = { 0, Ixl> 1 tiJ(x,
0)
=0
Show graphically the propagation of the wave along the string at different instants of time.
16.4 (a) Derive the equation of motion of a taut string resting on a foundation of elastic modulus k. (b) Derive the condition to be satisfied between y and (J) for the propagation of a wave of the form w(x, t) = AeYx-W' in the string. 16.5 Derive the equation of motion for the longitudinal vibration of an inhomogeneous bar for which Young's modulus E and density p are given by E = Eo(l + E,x3)and
p
= Po(l
+ E2X).
16.6 The initial conditions of an unbounded elastic material are given by u(x,O)
u(x, 0)
=
0,
1
=0
Asin7l'x,
0,
x
646
Elastic Wave Propagation
where A is a constant. Plot the displacement of the material at t 0, t 1/(2e), t lie, t 21e, and t 41e.
=
=
=
=
=
16.7 (a) Show that the ratio of velocities of P and S waves depends only on Poisson's ratio of the material. (b) Determine the ratio of velocities of P and S waves for the following materials:
=
Aluminum, v 0.334 Brass, v 0.324 Carbon steel, v = 0.292 Cast iron, v = 0.211
where A is a constant and y is the wave number, the substitution of Eq. (P16.2) into Eq. (P16.1) leads to EI
-y pA
4
I (
- A
1+
-kGE)
4 2
2 2
y e - y e
pI + --y kAG
4 4
e
=0
(P16.3) with
=
Copper, v 0.326 w= yc (P16.4) Glass, v 0.245 Stainless steel, v = 0.305 By considering the cases y _ 00 and y _ 0 in Eq. (PI6.3) separately, derive expressions for deter16.8 Consider a thin steel beam of circular cross mining the wave velocity c. section with Young's modulus 207 GPa, radius 1 em, 16.10 The potential and kinetic energy densities of a and unit weight 76.5 kN/m3. Determine the flexural string are given by wave velocity and the group velocity in the beam.
=
=
16.9 Consider a thick beam for which the equation of motion, according to Timoshenko theory, is given by 4
El8 w pA 8x4
I (
-
E)
A 1+ kG
82w
4
8 w 8x28t2
+
at2
pI
w
84
+ kAG
at4 =0
(P16.1) Assuming the propagation of harmonic waves in an infinite beam with solution w(x, t)
=
Aei(yx-wl)
(P16.2)
v(x,t)=-
P (8W)2 2
k(x, t) = -p
ax
2
(aw)2 at
The wave propagating in the positive x direction can be expressed as w(x, t)
= f(x
- ct)
= ",;
where e Pip is the phase velocity. Show that the potential and kinetic energy densities are equal for a propagating wave.
,, I
I'''''''''';'''~'ne:u-.~""""""*''''''''''''''''~'''''''".,eti'''''''''''''''''_i'''''''''''~_~
~_~,_"""",""
•.•.••••• """,,,-_,, •._ ••.•.• ~""'""'_.~.~,~~
Approximate Analytical Methods 17.1
INTRODUCTION The exact solutions of problems associated with the free and forced vibration of continuous systems have been considered in earlier chapters. Exact solutions are usually represented by an infinite series expressed in terms of the normal or principal modes of vibration. In many practical applications, the solution of the vibration problem is dominated by the first few low-frequency modes, and the effect of high-frequency modes is negligible. In such cases the solution may be expressed in terms of a finite number of normal modes or in terms of assumed polynomials that describe the deformation shape of the continuous system. Exact solutions are possible only in relatively few simple cases of continuous systems. The exact solutions are particularly difficult to find for two- and three-dimensional problems. Exact solutions are often desirable because they provide valuable insight into the behavior of the system through ready access to the natural frequencies and mode shapes. Most of the continuous systems considered in earlier chapters have uniform mass and stiffness distributions and simple boundary conditions. However, some vibration problems may pose insurmountable difficulties either because the governing differential equation is difficult to solve or the boundary conditions may be extremely difficult or impossible to satisfy. In such cases we may be satisfied with an approximate solution of the vibration problem. Several methods are available for finding the approximate solutions of vibration problems. The approximate methods can be classified into two categories. The first category is based on the expansion of the solution in the form of a finite series consisting of known functions multiplied by unknown coefficients. Depending on the particular method used, the known functions can be comparison functions, admissible functions, or functions that satisfy the differential equation but not the boundary conditions. If a series is assumed to consist of n functions, the corresponding eigenvalue problem will yield n eigenvalues and the corresponding eigenfunctions. The second category of methods is based on a simple lumping of system properties. For example, the mass of a system can be assumed to be concentrated at certain points, known as stations, and the segments between consecutive stations, called fields, are assumed to be massless with uniform stiffness distribution. This model, with n stations, can be used to derive an algebraic eigenvalue problem of size n whose solution yields n eigenvalues and the corresponding eigenvectors. The first class of methods is more analytical in nature and the second is more intuitive in nature. All the approximate methods basically convert a problem described by partial differential equations into a problem described by a set of ordinary differential equations. This essentially converts a differential eigenvalue problem into
647
648
Approximate Analytical Methods
an algebraic eigenvalue problem. There are two classes of methods that are based on series expansions: Rayleigh-Ritz methods and weighted residual methods. In this chapter we consider the Rayleigh and Rayleigh-Ritz method and the closely related assumed modes method as well as several weighted residual methods, inclUding the Galerkin, collocation, and least squares methods.
17.2
RAYLEIGH'S QUOTIENT The expression for Rayleigh's quotient and the stationary property of Rayleigh's quotient can be discussed conveniently by considering a specific system. Consider the torsional vibration of a shaft. In the absence of external torque, the expressions for the potential and kinetic energies of the shaft are given by 7rp(t)
=
T(t) =
1 (I 210 Glp(x)
1 (I 210 pIp
[ao(x, t)]2 ax dx
(x)
(17.1)
[ao(x t)]2 at' dx
(17.2)
The angular displacement is given by O(x, t) = X(x)f(t)
(17.3)
where X (x) is a trial function used to denote the maximum angular displacement at point x and f(t) indicates the harmonic time dependence, f(t)
= eiwt
(17.4)
where w is the frequency of vibration. Substituting Eqs. (17.3) and (17.4) into Eqs. (17.1) and (17.2), the potential and kinetic energies can be expressed as
1
1
iwt
7rp(t)
.
= -e 2
Glp(x)
0
eiM 2 T(t) = T(-w )
r
[dX(X)]2 -
dx
dx
(17.5)
(I
10
plp(x)[X(x)f
dx
(17.6)
Equating the maximum values of potential and kinetic energies leads to 7rPm"" = 1 1
Glp(x)
[d~;X)
dx = w211 plp(x)[X(x)]2
dx
Tmax
or
(17.7)
Rayleigh's quotient, R, is defined as . R(X(x»
= A = w 2 = f~Glp(x)[dX(x)/dxf 1 fo
plp(x)[X(x)]2
dx dx
(17.8)
It can be seen that the value of Rayleigh's quotient depends on the trial function X (x) used. To investigate the variation of R with different trial functions X (x), the trial
17.2 Raykigh's Quotient
649
function X (x) is expressed as a combination of the normal modes of the shaft, 8; (x), using the expansion theorem as 00
X(x) =
L
(17.9)
Ci8;(X)
;=1
where
C;
are unknown coefficients. Substituting Eq. (17.9) into Eq. (17.8) gives 2
=
R(CI,C2,"')=A=W
=
J~Glp(x) L~I
c;[d8;(x)f
dx]
/
Jo plp(x) L~I c;8;(x)
L~l Cj[d8j(x)f L~l cj8j(x) dx
L~IL~I C;Cj J~Glp(x)[d8j(x)f dx][d8j(x)f L~I L~I C;Cj J~8;(x)8j(x)dx
dx]dx
dx]dx
(17.10)
The orthogonality conditions of the normal modes of the shaft are given by [similar to Eqs. (EIO.3.8) and (E1O.3.9)]
1/
/
1 o
= C;j'
plp(x)ej(x)8j(x)dx
Glp(x)
d8j(x) dx
d8j(x)
dx = AjC;j,
.. dx
= 1,2, ...
(17.11)
i, j = 1,2, ...
(17.12)
i, j
wf
where Aj = is the ith eigenvalue of the shaft. Using Eqs. (17.11) and (17.12), Eq. (17.10) can be expressed as
If the trial function X(x) closely resembles any of the eigenfunctions 8k(X), it implies that all the coefficients Cj other than Ck are small compared to Ck, so that we can write i
= 1,2,
.. , , k - 1, k
+ 1, ...
(17.14)
where Cj are small numbers. By substituting Eq. (17.14) into Eq. (17.13) and dividing the numerator and denominator by c~ and neglecting terms in C; of order greater than 2, we obtain
650
Approximate Analytical Methods Note that if the trial function X (x) differs from the kth nonnal mode 8k (x) by a small quantity of order 1 in e [i.e., X(x) = 8k(X) + O(e)], Rayleigh's quotient differs from the kth eigenvalue by a small quantity of order 2 in e [i.e., R(CI, C2, •.. ) = Ak + O(e2»). This implies that Rayleigh's quotient has a stationary value at an eigenfunction Gk(X), and the stationary value is the corresponding eigenvalue Ak. The stationary value is actually a minimum value at the fundamental or first eigenvector, 8) (x). To see this, let k = 1 in Eq. (17.15) and write 00
R(X(x»
= A) + L(A;
- A)e;
(17.16)
;=2 Since the eigenvalues
satisfy the relation A) .:: A2 :: A3 :: •..
(17.17)
Eq. (17.16) leads to R(X(x» which shows that Rayleigh's quotient Equation (17.18) can also be interpreted
A)
2:: Al
(17.18)
is never smaller as follows:
= wi = min R(X(x»
than the first eigenvalue.
= R(8)(x»
(17.19)
which implies that the lowest eigenvalue is the minimum value that Rayleigh's quotient can assume and that the minimunivalue of R occurs at the fundamental eigenfunction, X(x) G)(x). Equation (17.19) denotes Rayleigh's principle, which Can be stated as follows: The frequency of vibration of a conservative system vibrating about an equilibrium position has a stationary value in the neighborhood of a natural mode. This stationary value, in fact, is a minimum value in the neighborhood of the fundamental natural mode.
=
17.3
RAYLEIGH'S METHOD In most structural and mechanical systems, the fundamental or lowest natural frequency is the most important. For a quick estimate of the dynamic response of the system, especially during the preliminary design studies, the fundamental natural frequency is used. In such cases, Rayleigh's method can be used most conveniently to find the approximate fundamental natural frequency of vibration of a system without having to solve the governing differential equation of motion. The method is based on Rayleigh's principle. It can be used for a discrete or continuous conservative system. In Rayleigh's method, we choose a trial function, X(x), that resembles closely the first natural mode, 81 (x), substitute it into the Rayleigh's quotient of the system, similar to Eq. (17.10), carry out the integrations involved, and find the value of 2 R A w . Because of the fact that Rayleigh's quotient has a minimum at the fundamental or lowest eigenfunction, the fundamental natural frequency of the system can be taken as w = v'R. Because of the stationarity of Rayleigh's quotient, the method gives remarkably good estimates of the fundamental frequency even if the trial function
= =
I 1~~"''''''''''''''","","--"~~4_'''''''';'''"''-'''''''''''''''''"'''~''~~''''''''~'''~''
__
••"'"' ..••.•.• _-~~~~~l:;
•••••.•
~l7.3
llilyleigh'$ Method
651
does not resemble the true fundamental eigenfunction too closely. In fact, the natural frequency thus computed will be one order of magnitude closer to the true fundamental natural frequency WI than the trial function X (x) is to the true fundamental natural mode 91 (x). Of course, the closer the trial function resembles the first eigenfunction, the better the estimate of the fundamental frequency . Usually, selecting a suitable trial function for use in Rayleigh's quotient is not difficult. For example, the static deflection curve under self-weight can be used for bar, beam, plate, or shell structures. Even if the system has nonuniform mass and stiffness distributions, the static deflection curve of the system found by assuming a uniform mass and stiffness distributions can be used in Rayleigh's quotient. Similarly, even if the system has complex boundary conditions, the static deflection curve, found with simple boundary conditions, can be used in Rayleigh's quotient. For example, for simplicity, a free end condition can be used instead of a spring-supported end condition of a beam. The strain and kinetic energies required in defining Rayleigh's quotient are given in Table 17.1 for some uniform structural elements. Example 17.1 Determine the fundamental frequency of transverse vibration of a uniform beam fixed at both ends (Fig. 17.1) using Rayleigh's method. Use the following trial functions for approximating the fundamental mode shape: (a)
=C
.. X(x)
2it' X
(
1 - cos -Z-
)
(EI7.1.1)
where C is a constant. This function is selected to satisfy the boundary conditions of the beam: X(O) 0, dX(O)/ dx 0, X(l) = 0, and dX(l)/ dx O.
=
=
=
(b) X(x)
= C(x2)(Z
(E17.1.2)
- x)2
with C = wo/24E I. This function is the static deflection curve of a fixed-fixed beam under a self-weight of Wo per unit length. SOLUTION The expressions for the strain and kinetic energies of a uniform beam are given by it'
= ~EI 2
1 T=-pA 2
t [o2W(X, 10 ox
t)]2
dx
(EI7.1.3)
2
l' [OW(X,ot t)]2 ---
d
----I Figure 17.1
(EI7.1.4)
x
0
Fixed-fixed beam.
I
c
.8
U o o
l:l:: "0
a
:; 00
§
....--..
N
~I~ ~
co'"
~IN " ~ '~
E
o
tiS
e.2 to)
oS tr.l o
c
E
o
.8
tr.l
.•..o
U o
l:l::
o
~ '§l
"0
..'" o
o c LLl
c
0
.~ ~ .c c c ~ £ ~ c
~
" " "
Q...
..
Cl)
•..
E o E
o c
•..o
.;;;
C
.5
E
o
tiS '--
~
.c
o
652. ..
.~
'-' ~~
o
c c
~
0
";;; VJ
a ES.
'Xl
N
8
•.. •..
~
....---
M
JI~ <'C)<'C)
_1
M
•••
+
~l'"
<'C)<'C) -I ••.
+
JIM ... <'C)<'C)
'-'
0,:
~
M
'"
~ i:::lIN
....--M
~IMq:,
<'C) <'C)
_I"'•..
+ ;:1'"
<'C)<'C) -I ••.
'-'
;:1'"•.. '"<'C)<'C) '""' :>
-
'-' I
N
....--~I'"
M
<'C)<'C) -I ••.
+
JIM... <'C) <'C)
'-' L..--...l 0,:
~
•.. •.. ,..---, ~
~l'"
<'C)<'C) -I ••.
;:1'"•..
'"<'C)
<'C)
'""' :>
I
N i:::l
I
l::
...
¢)
.0
5 5 ¢)
653
c::
., c::
.8
'2
-a
0
>. ~ c::u .c Q)
.S
0"
~
Q)
> §'"
Q)
Q)
•....
~
N'"l::t
E .8 '0 E
'K >. ~
~
c::
.::!
"§ 0
•.. •..
9/ ...
--=: Q.IN
Q.
c: ::s
,--.
•..
~I-
0..
E -c "
-
N
N
'" '"~
:---..
~ ~ ~
I::
Q)
~
'"l::t
::s
~
"
N
Q.
'2
~
E ~
N~I:a
-•..
~
"E.
~
~ ~
'"l::t
N
.~
~
~
'"l::t
0
ll::
"
N
•.... c::
N
0..
Q.
Q)
::s
-c ~
~ .t: .•... ~ "0
I::
N
::::::
~ c:: ~
•..
•..
Q)
:E
>
c: '"
'"l::t
"0
~
N~I:a
Q)
Q)
•...1:10
'"l::t
::s c:
Q)
cc
~
::s
e!l Q)
u ~ ~ "0 ~ '"•..
~
"0
::s.:=
~
.0
U ~ cc
u
Q)
0..
~
'" '"~
ll::
E
~
"
Q.IN
Q.
•.... . '"l::t
~
'"l::t
r---, N I::
>. e!l Q)
c:
~
c:
.~ r/.l
,--. 91 •....
~
N +
,--. 91~
~
--=:
~
~ .... r...:
•.. 0
:;
E~
~ •.. ~
~
::c ~
•...
I::
N
N
,--.
c: '"
c: '"
~I'"
c:
0 '" 'v: :.a::s c: ~ E •..
ll::
9" "
" "
l:l... I:l:::
Cl)
.c
....
654
•..
8
'-'
•..
..
'"l::t
Q)
c:: U Q) ~ •.. cc .c Q) "0 E Q) Q) E •.. '" c:: Q) .8 > 0
~ ~ ~ :---.. E .t: ~ ~ -c " ~ l:l...IN l:l... ~ l:l.../N
L---.J
.!
c:
.8
E ~ E
E
Q)
u:i
•.. ~
-
-5 .~ u
'-'
bIlQ)
c: c::
U.c Q)
~
E
Q)
r...: E
a
Q)
'C:: ~
:;§E,-.,. ~ 15 E c:: E ~.g
o
Q)
.-
oOE~E
0
17.3 Raykigh's Method ·~.(j.5S
where the transverse deflection function, w(x, t), can be assumed to be harmonic: W(x, t) = X(x) coswt
(EI7.1.5)
where w is the frequency of vibration. Rayleigh's quotient for a beam bending is defined by maximum strain energy = maximum kinetic energy Substitution of Eq. (EI7.L5) into Eqs. (EI7.1.3) and (EI7.1.4) leads to Jrmax
I = -EI 2
1 0
~PAw2
Tmax=
2
Equating Jrmax and
Tmax,
2 [d X(X)]2 2
1
(EI7.1.6)
dx
dx
t [X(x)]2dx
(EI7.L7)
10
Rayleigh's quotient can be derived as !EI f6 (d2X/ dx2)2 dx
2
= w = =------''-'-1---'---
R(X (x»
.
(EI7.L8)
~pAfo[X(x)]2dx
(a) In this case, X(x)
d2X(x) dx2
t (d2X)2
10
dx
= _ -
c( C
1- cos
(EI7.L9)
(2Jr)2 2Jrx 1 cos 1
= C2 (2Jr)4
dx2
10t [X(x)f
2~X) {I cos2 21l'x dx
10
1
= c2
(EI7.1.10)
10t (1 -
= C2 (2Jr)4
1
I
2 2JrX)2 3C 1 cos -1dx = -2-
~ 2
=
2
8C Jr4 13 (E17.1.1I)
(E17.1.12)
Thus, Eq. (EI7.L8) gives R
= w2 =
16Jr4 E I
!E I (8C2Jr4 /13) 2
= ---3
~pA(3C21/2)
pA14
or
(El
w = 22.792879 V
PAf4
(EI7.1.13)
(b) In this case, X(x) = Cx2(1d2X(x)
--- 2 dx
..
(EI7.1.14)
x)2
= 2C(6x2 - 61x
+ 12)
(E17.1.15)
656
Approximate Analytical Methods
1/(~:~) 1/
2
dx
= 4C21/
(6x2 - 61x
= C21/
2
(X (x)) dx
4
(X
-
21x3
+ 12)2dx =
+ x212)2 dx
SC215
(E17.1.l6)
= 6~OC219
(E17.1.l7)
Thus, Eq. (E17.1.8) gives R
= (Ii =
lEI(1C2~) 2 5 219) IpA (...LC 2 630
or
w
= 22.449944
= 504- EI
pAl4
1 f1I. ~
pAL
(E17.1.l8)
The exact fundamental natural frequency of the beam is given by (see Fig. 11.3)
wi = (fhl)2
f1I.1=
(4.730041)2
~ pAL
f1I.1= --4
pAL
22.373288
f1I.1 --4
pAL
(E17.1.l9) and the exact fundamental natural mode is given by [see Eq. (11.70)J ...
W)(x) = C [ SInh,8IX':" sin,8Ix
,
+
sinh,8Il- sin,8)l .] R (cosh,8lx - COS,8IX) cos fJIl - cosh ,8ll (E17.1.20)
where C is a constant. It can be seen that the fundamental natural frequencies given by Rayleigh's method are very close to the exact value and larger than the exact value by only 1.875410% in the first case [with Eq. (E17.1.9)J and 0.342623% in the second case [with (Eq. (E17.1.l4))J.
Example 17.2 Determine the fundamental frequency of transverse vibration of the uniform beam shown in Fig. 17.2. The beam is fixed at x = 0 and carries a concentrated mass m and rests on a linear spring of stiffness k at x l. Use Rayleigh's method with
=
I~ Figure 17.2
Beam fixed at one end and carrying a spring-supported mass at the other.
17.3 Rayleigh' s Method
657
the trial function X(x) = Cx2(31- x), where C = Ff6EI, which corresponds to the deflection shape of a cantilever beam fixed at x 0 and subjected to a concentrated force F and x I.
=
=
SOLUTION Considering the strain energy due to the deflection of the spring and the kinetic energy due to the motion of the end mass, the strain and kinetic energies of a uniform beam can be expressed as 2
1
Jr
= "2E1
10[I [a
w(x, t)]2 ax2
dx
1
1
1 T=-pA 2
0
1
+ "2k[w(l,
d x+-m 1
[aw(x, t)]2 --at
2
(E17.2.1)
t)]2
[aW(I, ~~- t)]2 at
(E17.2.2)
where the transverse deflection function, w(x, t), is assumed to be harmonic: (E17.2.3)
w(x, t) = X(x) coswt
where w is the frequency of vibration. Substitution of Eq. (E17.2.3) into Eqs. (E17.2.1) and (E17.2.2) yields
10t [d
2
1
"2 E..I
Jrmax
=
Tmax
= -pAw22
By equating Jrmax and
Tmax,
I'll
20
X(X, t)]2 dx2 dx
[X(x)]2 dx
1
+ "2kX2(l)
1 2X2(l) + -mw
(E17.2.4) (E17.2.5)
Rayleigh's quotient can be derived as (E17.2.6)
Using X(x) = Cx2(31 - x) d2X(x) -dx-2-
= 6C(1- x)
(E17.2.7) (E17.2.8)
we can find (E17.2.9) (E17.2.10) (E17.2.11) (E17.2.12)
658
Approximate Analytical Methods
Thus, Eq. (E17.2.6) gives R =It}
2 3 2 6 2l.EI(l2C 1 )+2kC 1 2
=
!pA (33C 17/35)
+
=
2mC216
6EI+2k13 ~pA14
_
+ 2 ml3
(E17.2.13)
Thus, the fundamental frequency of vibration is given by W
= ( 420E 1 + 140kl3 ) 33pAl4 + 140 ml3
1/2
(E17.2.14)
Note: If k and m are set equal to zero, the beam becomes a cantilever beam and the fundamental frequency given by Rayleigh's method, Eq. (E17.2.14), reduces to
W
= 3.567530
/£. ~
1
(E17.2.15)
pAL
This value can be compared with the exact value
WI
= (1.875104)2
/£.1 ffI.1 -14 pA
= 3.516150
--4
pAL
Thus, the frequency given by Rayleigh's method is larger than the exact value by 1.461257%.
Example 17.3 Find the fundamental natural frequency of longitudinal vibration of the tapered bar fixed at x 0 and connected to a linear spring of stiffness k at x shown in Fig. 17.3 using Rayleigh's method. Assume the variation of the cross-sectional area of the bar to be A (x) = Ao (l - x /21) and use the trail function X (x) = C sin(n x /2/) for the mode shape.
=
=/
SOLUTION The expressions for the strain and kinetic energies of a uniform bar, including the strain energy due to the deformation of the spring at x can be
= /,
o
·_·_._.~x
Figure 17.3
','~'
17.3
Rayleigh's Method
659
expressed as rr T
1/
A(x)
[au(x, ---
t)]2
=!p t A(x)
[au(x,
t)]2
1 = -E 2
2 where the longitudinal
ax
0
Jo
deflection
at
Ao
0
2
2
(E17.3.1)
(I, t)
dx
(E17.3.2)
= X (x) cos wt
where w is the frequency of vibration. Substitution and (E17.3.2) leads to
1/ (
1 + -ku
function, u(x, t), is assumed to be harmonic: u(x, t)
1 rrmax = - E , 2
dx
of Eq. (E17.3.3) into Eqs. (E17.3.1)
X) [dX(X)]2, --
1- 2/
(E17.3.3)
dx
dx
1 + -kX 2
2
(E17.3.4)
(I)
t (
* 1 X) 2 Tmax = 2P Jo Ao 1 - 21 X (x)dx
(E17.3.5)
Using "
rrx
'~(x)
= C sin
2/
(E17.3.6)
dX Crr rrx -(x) = -cosdx 21 21
(E17.3.7)
we can obtain
1/ ( o
Rayleigh's
Ao
X)
2
1 2 2 rrx 1- C sin - dx = AoC 21 21 2
(3 1) - - -2
4
rr
(E17.3.9)
quotient is given by
G + 1/rr2) +kC /2 G - 1/rr2) 2
2 rrmax (EAoC2rr2/16/) R - w - -- ------'-'-------(pAoC21/4)
-
- T;;ax -
= Thus, the natural frequency 1
I --2
pAol
(3.238212EAo
+ 3.063189kl)
(E17.3.1O)
of vibration is given by
w = [ --2 (3.238212EAo pAol
+ 3.063189kl)
]
1/2
(E17.3.1l)
660
Approximate Analytical Methods Example 17.4 Estimate the fundamental frequency of axisymmetric transverse vibration of a circular plate of radius a fixed along the edge r a using Rayleigh's method with the trial function
=
= c (1 _ ::)
W (r)
2
(EI7.4.1)
where c is a constant. SOLUTION The trial solution of Eq. (EI7.4.1) satisfies the boundary conditions of the plate at the edge r = a. The strain and kinetic energies of a circular plate of thickness h in axisymmetric vibration are given by (from Table 17.1) V=7rD
T = 7rph
1 1°
[(02W 10W)2 -+-2 or r or
o2w lOW] -2(l-v)-_rdr or2 r or
0
o
(~;)2rdr
(EI7.4.2)
(EI7.4.3)
If the plate is fixed at r = a, the integral of the second term on the right-hand side of Eq. (E17.4.2) will be zero and hence the equation reduces to
1 -+-(o2w or2
10W)2 rdr r or
0
V=7rD
o
(EI7.4.4)
Assuming harmonic variation of w(r, t) as
= W(r)Coswt
w(r,t)
(EI7.4.5)
where w is the frequency of vibration, we can obtain
T~ax
o
= 7rph
Rayleigh's quotient, R(W(r»,
1 -+-1° (d2 W dr2
0
Vrnax=7rD
[W(r)fr
1 dW)2
r dr
dr
rdr
(EI7.4.6) (EI7.4.7)
is given by R(W(r»
Vrnax = w 2 =_ Tr:;ax
(E17.4.8)
From Eq. (EI7.4.1), we have dW dr
= c (_
2
= c(-~2 + 12r2)
dW dr2
4r a2
a
+ 4r3) a4
a4
(EI7.4.9) (EI7.4.1O)
17.4
Raykigh-Ritz Method
661
Equations (E 17.4.6) and (E 17.4.7) can be evaluated as Vrnax =
=
rr:D
fa
10
[c (_~
2
a2
+ 12r4 ) +:: (_ a
r
4:a- + a 3)]2 4r
4
rdr
32rr:Dc2
(EI7.4.1l)
----:c-
3a2
T~a = "ph
f HI-::)']'
(EI7.4.12)
r de ~ "p~~'c2
Thus, the estimate of the fundamental natural frequency of the plate can be found from Eqs. (EI7.4.8), (E17.4.1l), and (E17.4.12) as
or w '" 10.3279
17.4
j
D pha
(E17.4.13)
4
RAYLEIGH-RITZ METHOD As stated earlier, exact solution or'eigenvalue problems of many continuous systems is difficult, sometimes impossible, either because of nonuniform stiffness and mass distributions or because of complex boundary conditions. At the same time, information about the natural frequencies of the system may be required for the dynamic analysis and design of the system. For most systems, only the first few natural frequencies and associated natural modes greatly influence the dynamic response, and the contribution of higher natural frequencies and the corresponding mode shapes is negligible. If only the fundamental natural frequency of the system is required, Rayleigh's method can be used conveniently. If a small number of lowest natural frequencies of the system is required, the Rayleigh-Ritz method can be used. The Rayleigh-Ritz method can be considered as an extension of Rayleigh's method. The method is based on the fact that Rayleigh's quotient gives an upper bound for the first eigenvalue, Al =
wr:
?: Al
R(X(x))
(17.20)
where the equality sign holds if and only if the trial function X (x) coincides with the first eigenfunction of the system. In the Rayleigh-Ritz method, the shape of deformation of the continuous system, X (x), is approximated using a trial family of admissible functions that satisfy the geometric boundary conditions of the problem: n
X(x)
=
L:>i4>i(X)
(17.21)
i=1
where CI. C2, .•. , Cn are unknown (constant) coefficients, also called Ritz coefficients, and 4>1 (x), 4>2 (x), ... , 4>n (x) are the known trial family of admissible functions. The
662
Approximate Analytical Methods
functions ¢i (x) can be a set of assumed mode shapes, polynomials, or even eigenfunctions. When Eq. (17.21) is substituted into the expression for Rayleigh's quotient, R, Rayleigh's quotient becomes a function of the unknown coefficients C), C2, ••• , Cn: (17.22) The coefficients C), C2, necessary conditions:
•••
, Cn
are selected to minimize Rayleigh's quotient using the
8R 8cj -
--0
i = 1,2, ... , n
,
(17.23)
Equation (17.23) denotes a system of n algebraic homogeneous linear equations in the unknowns C), C2, ••• , Cn· For the coefficients C), C2, ••• , Cn to have a nontrivial solution, the determinant of the coefficient matrix is set equal to zero. This yields the frequency equation in the form of a polynomial in (J} of order n. The roots of the frequency equation provide the approximate natural frequencies of the system W), W2, •.. , W . n Using the approximate natural frequency Wi in Eq. (17.23), the corresponding approx. . d (&lor l. = 1, 2 , ... , n.) It can be Imate mo de shape c)(i) ,c2(i) , ••• , Cn(i) can be detefffilne seen that a continuous system which has an infinite number of degrees of freedom is represented by a discrete model, through Eq. (17.21), having only n degrees of freedom. The accuracy of the method depends on the value of n and the choice of the trial functions, ¢i(X), used in the approximation, Eq. (17.21). By using a larger value of n, the approximation can be made more accurate. Similarly, by using the trial funco. tions ¢i (x), which are closer to the true eigenfunctions of the continuous system,. the approximation can be improved: The fundamental natural frequency given by the Rayleigh-Ritz method will be higher than the true natural frequency. The reason is that the approximation of a continuous system with infinitely many degrees of freedom by an n-degree-of-freedom system amounts to imposing the constraints Cn+)
= Cn+2 = ... = 0
(17.24)
on the system [in Eq. (17.21)]. The addition of constraints is equivalent to increasing the stiffness of the system, and hence the estimated frequency will be higher than the true fundamental frequency, When a larger number of trial functions are used, the number of constraints will be less and hence the fundamental natural frequency given by the Rayleigh-Ritz method, although higher than the true value, will be closer to the true value. If Rayleigh's quotient is expressed as R-w -
2
1l'max
N
----_ - T~ax - D
(17.25)
where N = 1l'max and D = T~ax denote, respectively, the maximum strain energy and reference kinetic energy of the system. The reference kinetic energy (T~ax) is related to the maximum kinetic energy (Tmax) as T.max --
T*max
W2
(17.26)
•.
,Rayleigb-RitzMethod
-I.1A
663
The maximum strain energy and the reference kinetic energy can be expressed as (17.27) n
* =D T malt
=
-21
n
L L c·c'm" I
}
T = -c 2L - [m]c
I}
(17.28)
i=l j=l
where [k] = [kij] is the stiffness matrix, [m] = [mij] is the mass matrix,
(17.29)
and the stiffness and mass coefficients kij and mij can be evaluated in terms of the stiffness and mass distributions of the system. For example, in the case of longitudinal vibration of bars, kij and mij are given by k..
..I}
'.
mij
/ 1 ~'1 =
0
.
d¢i d¢j
EA-d d dx x x
(1730)
pA¢i¢j
(17.31)
/
dx
and in the case of transverse vibration of beams, kij and mij are given by ki' = }
mij =
/
1 0
1/
d2¢i d2¢ j
EI----dx 2 2
(1732)
pA¢i¢j
(1733)
dx
dx
dx
Using Eqs. (17.27) and (17.28), Eq, (17.25) can be expressed as (1734) since the numerator N and denominator D depend on the coefficients cr, The necessary conditions for the minimum of Rayleigh's quotient are oR = D(oN lOCi) - N(oDloci) OCi D2
= 0,
i
= 1,2, ...
C2,··.,
cn•
,n
or
.!.. (aN D
OCi
_ NOD) D OCi
=
.!.. (aN D
OCi
_
>..(n)
aD) OCi
= 0,
i = 1, 2, .
H
,n (1735)
664
Approximate Analytical Methods
where). (n) is the eigenvalue and the superscript n indicates that the eigenvalue problem corresponds to a series of n terms in Eq. (17.21). Equation (17.35) can be written in matrix form as
(17.36) where
aNjac
and
aDjac
can be expressed,
aN
using Eqs. (17.27) and (17.28), as -T
ac
= c [k]
aD
-T
ac = c
(17.37)
[m]
(17.38)
and hence Eq. (17.36) leads to
or [[k] -). (n)[m]]
C=0
(17.39)
Equation (17.39) denotes an algebraic eigenvalue problem of order n. For a nontrivial solution of the vector the determinant of the coefficient matrix must be zero:
c,
I[k] - ). (n)[m]1
=0
(17.40)
Equation (17.40) denotes the frequency equation, which upon expansion results in a polynomial in ).(n) of order n. The roots of the polynomial, ).~n), i 1,2, ... , n, give the natural frequencies as
=
i = 1,2, ... , n For each natural frequency Wi, the corresponding vector of Ritz coefficients determined to within an arbitrary constant by solving the linear simultaneous neous equations:
(17.41) can be homoge-
c(i)
(17.42) which can be written in scalar form as ~(n)
k 12
-Ai
k 22
-
Ai
k n2
-
Ai
dn)
~ (n)
m12 m22
mn2
'"
i = 1, 2, ... , n
(17.43)
17.4 Rayleigh-RitzMethod
665
e~i)} (i)
Once
c(i)
=
e~
is determined, the eigenvector or normal mode associated with the
{ (i)
en frequency i can be determined using Eq. (17.21) as n
X(i)(x)
=
L cji)
= e~i)
+ e~)
i = 1,2, ... , n
j=1
(17.44) Note Once the natural frequencies and the corresponding mode shapes are determined using Eqs. 07.41) and 07.44), the dynamic or vibration response of the continuous system (with an infinite number of degrees of freedom) can be represented by an equivalent n-degree-of-freedom system. The vibration response of the continuous system can be expressed, using the separation-of-variables technique, as n
u(x, t) =
L
X(i) (x)r/i
(17.45)
(t)
i=l
where T/i(t), i = 1,2, ... , n, are the time-dependent coefficients or generalized coordinates . .-.-
Example 17.5 of the tapered Rayleigh-Ritz A(x) = AoO -
Determine the first three natural frequencies of longitudinal vibration bar fixed at x 0 and free at x 1 shown in Fig. 17.4 using the method. Assume the variation of the cross-sectional area of the bar as x/2l). Use the following functions as trial functions:
=
=
SOLUTION
x3
x2
x
= Z'
= (i'
= [3
(E17.5.1)
For the fixed-free bar, the geometric boundary condition is given by u(O)
=0
=Ao (1-
ft)
(EI7.5.2)
Ao A(x)
o
.-.-.-.-.-
x
I~
.1 Figure 17.4
666
Approximate Analytical Methods
and the natural boundary condition by
8u 8x (I) = 0
(E17.5.3)
Note that the trial functions in Eq. (E17.5.1) are admissible functions since they satisfy only the geometric boundary condition, Eq. (E17.5.2), but not the natural boundary condition, Eq. (E17.5.3). Assuming the longitudinal deflection function, u(x, t), to be harmonic as u(x, t) = X(x)coswt
(E17.5.4)
where w is the frequency of vibration, the Rayleigh quotient of the bar can be expressed as
R -
-
JTmax
r;;ax
= -
!fJ EA(x)[dX(x)/dx]2dx
I
I
210 pA(x)[X(x)]
Z
(E17.5.5)
dx
Using 3
X(x)
=~
x
Ci¢i(X)
1=1
d X (x)
CI
xZ
x3
= cII + cZ"j2 + c373
2czx
(E17.5.6)
3C3Xz
~=T+f2+[3
(E17.5.7)
7rmax and r;;ax Can be evaluated as foiIows:
(E17.5.8)
(E17.5.9) Rayleigh's quotient is given by R
= wZ =
JTmax(CI,
Cz, C3)
(E17.5.1O)
r;;ax(c], Cz, C3) The necessary conditions for the minimization of R are given by 8R = r;;ax(87rmax/8c;).;... 8Ci
r;;ix
7rmax(8r;;ax/8ci)
=0
~I
17.4 Rayleigh-Ritz Method
667
or i
= 1,2,3
(E17.5.1l)
Using
(3 + 43 + 45) (5 + 4 + 9) = -(21 + 5 + 9) 5 + 3 + 7) -27 + 3 + 4) = -9 + 4 7)
a Jrmax EAo ~ =U a Jrmax
--
aC2
EAo 21
3
aJrmax EAo -= -aC3
-CI
-C2
4
aT~ax pAol ( ~ = 12Ci aT~ax
--
aC2
.)
-CI
10
pAol ( -C2 2 30
(E17.5.14)
-c, 5 -
(E17.5.15)
1Oc2
30C3
-CI
-C3
(E17.5.16)
-C2+
-CI
(EI7.5.17)
10
aT~ax pAol ( -= --C3 aC3 2 56
(E17.5.B)
-:;CJ
3
-C3
21
(E17.5.12)
C3
C2
lCI
21
21
30
Eq. (EI7.5.11) can be expressed as E Ao 21
[! i 3
3
4:
:s
5
~ ] {CI}
9
21
C2
..2..
C3
(E17.5.18)
56
or [k]c
= A [m]c
(EI7.5.19)
where
[k]
=
U I !] ~:~~] :s
(EI7.5.20)
10
12
[mJ =
[
~ 30
4 21
(EI7.5.21)
9 56
pl2w2 A=--
E
(E17.5.22)
The solution of Eq. (E17.5.19) is given by
~=
3.2186 } 24.8137 { 100.8460
(E17.5.23)
668
Approximate Analytical Methods
The mode shapes can be found as c(l)
=
-0.871OJ -0.2198
1
,
c(2)
0.4394
Example 17.6
0.3732 -0.8370 0.4001
=
I
J,
c(3)
=
-0.2034J 0.7603
1
(E17.5.24)
-0.6170
Solve Example 17.5 using the following as trial functions: .
irX
= sm 2l'
3ir x = sin -2'
ifJ2(x)
/
= sin-5ir
2/
X
(EI7.6.1)
SOLUTION It can be seen that the trial functions in Eq. (EI7.6.1) are comparison functions since they satisfy both the geometric boundary condition, Eq. (EI7.5.2), and the natural boundary condition, Eq. (EI7.5.3). As in Example 17.5, the longitudinal deflection function, u(x, t), is assumed to be harmonic as u(x, t)
= X(x)coswt
(EI7.6.2) where w denotes the natural frequency of vibration. Rayleigh's quotient of the bar can be expressed as irmax
~
R=--= T~ax
J~EA(x)[dX(x)/ I ~ Jo pA(x)[X(x)]2
dxf
dx
(EI7.6.3)
dx
Using 3
X(x)
.
. irX + C2sm. 3irx . 5irx = L Ci
i=1
d X (x)
~
CI7r
=
7rX
c237r
2/
37rX
c357r
2/
(EI7.6.4)
5ir X
21 cos 2l + 21cos 21 + 21cos 21
(EI7.6.5)
[7.4
Rayleigh-Ritz Method
669
The Rayleigh's quotient is given by R = W2 =
;rmax(C), c2, C3)
(EI7.6.8)
T;;ax(CI, C2, CJ)
The necessary conditions for the minimization of R are given by (EI7.6.9) or O:Tmax
_
w2 oT;;ax = 0,
OCi
(EI7.6.1O)
i = I, 2, 3
OCi
Using ~
O:Tmax
EAo [(3;r2 = -2- c) 16l
~
O:Tmax
EAo [(27;r2 = -2- C2 16l
O:Tmax
EAo [(75;r2 = -2- C3 16l
~
oT;;ax = pAol aCI 2
[CI (~
_
4;r2
1)
+ 4l
1) 1)
(3) (15)] (15) ( 5 )] i) (_1 )] 1) ( 1) ( 1)] 1 ) _ (_1 ) + +
4l
+CI
4l
+C3
4l
+
4l
+C2
4l
+CI
36l
+ C2 (~)
[(3
oT;;ax = pAol OC3 2
[C3 (~ __ 4 25;r2
2
C2
---
4
_ C3
;r2
aT;;ax pAol --=-aC2
( 4l3) +C3 ( 36l5 )]
+C2
9;r2
+CI
-
1l'2
CI
91l'2
(EI7.6.11) (EI7.6.12) (EI7.6.13) (EI7.6.14)
1l'2
(EI7.6.15)
C2 (~)]
(EI7.6.16)
+C3
9;r2
1l'2
Eqs. (EI7.6.9) can be expressed as 3;r2
1
16+4 EAo 2l
3 4
3 4 271l'2
1
16+4
5 36 3
4w2 p Aol = 2
5 36
15 4 1 1l'2
15 4 75;r2
1
16+4 1 ;r2
3 4
1 --9;r
(EI7.6.17)
2
1
3 1 ---4 251l'2
670
Approximate Analytical Methods
or [k]2 = A[m]2
(E17.6.l8)
where 1
31l'2
3 4
"16+4 [k] =
3 4
"16+4
5 36
15 4
3
4[m]
5
36
271l'2
1
15 4 751l'2
1
1l'2
1l'2
- 91l'2
3 1 --4
1l'2
1
1
91l'2
1
1
- 91l'2
1l'2
1
"16+4
1
=
(E17.6.19)
3
4-
(E17.6.20) 1
251l'2
w2 p[2
A=_
E
(E17.6.21)
The solution of Eq. (EI7.6.17) is given by .,
i-~
{ 3.2189 } 23.0627 62.7291
(E17.6.22)
The mode shapes can be computed as
2(1)
= { -0'9996} 0.0288
,
0.0017
2(2)
0.1237 } = { -0.9912., 0.0461
2(3)
=
{
-0.0356} 0.0948 -0.9949
(E17.6.23)
17.5 ASSUMED MODES METHOD The assumed modes method is closely related to the Rayleigh-Ritz method. In fact, the discrete model obtained with the assumed modes method is identical to the one obtained with the Rayleigh-Ritz method. The main difference between the two methods is that the Rayleigh-Ritz method is commonly used to solve the eigenvalue problem, whereas the assumed modes method is generally used to solve the forced vibration problem. In the assumed modes method, the solution of the vibration problem of the continuous system is assumed in the form of a series composed of a linear combination of admissible functions
17.5 Assumed Modes Method
()71
time-dependent generalized coordinates, 17i (t). Thus, for a one-dimensional continuous system, the displacement solution is assumed to be n
(17.46)
w(x, t) = I>Pi(X)17i(t) i=l
where ¢i (x) are known trial functions that satisfy the geometric boundary conditions (admissible functions) and 17i(t) are unknown functions of time, also called generalized coordinates. For a forced vibration problem, the expressions of strain energy (Jr), kinetic energy (T), and virtual work of nonconservative forces, 8Wnc, are expressed in terms of the assumed modes solution, Eq. (17.46), and then Lagrange's equations are used to derive the equations of motion of the equivalent n-degree-of-freedom discrete system of the continuous system. For specificity, consider a tapered bar under longitudinal vibration subjected to the distributed load, f(x, t) per unit length. The strain energy, kinetic energy, and virtual work of nonconservative forces of the system are given by 1 r 210
l
Jr(t) =
T(t)
8Wnc
EA(x)
[au(x,t)]2 ax
1 t [aua;(x, = 210 pA(x)
=
i
t)
]2
dx
(17.47)
dx
(17.48)
l
f(x,
(17.49)
t)8u(x, t) dx
By substituting Eq. (17.46) into Eqs. (17.47)-(17.49), we obtain
where
kij
denote the symmetric stiffness coefficient, given by
-1 1 L L [t10 2 ~ f; 1
k IJ.. -- k··JI
-
EA(
o
1
T(t) =
1
)d¢i(X) dx
i(X);'i(t)
i=1
n
n
d4>j(x) d x, dx
i, j = I, 2, ... , n
(17.51)
n
n
pA(x)
o
=
X
;'i (t);'j (t)
¢j(X);'j(t)
dx
j=1
pA(X)¢i(X)¢j(X)
dx
]
=
1 n
2~
n
f;miji7i(t)i7j(t) (17.52)
672
Approximate Analytical Methods where ~i(t) mij
=
dt and mij indicate
d"li(t)f
1/
= mji =
io
n
f(x,
The Lagrange equations
Substituting system
nonconservative
force corresponding
t)¢i(x)dx,
can be expressed
~ (:;) aT fa"li
f(x,
Qinc(t)O"li(t)
i=l
where Qinc (t) denotes the generalized eralized coordinate "Ii (t), given by
1/
,n
given by (17.53)
n
i=l
=
mass coefficients,
= 1,2, ...
¢i (X)O"li (t) dx = L
t) L
o
Qinc(t)
i, j
pA(X)¢i(X)¢j(x)dx,
t
=
oWnc(t)
the symmetric
- :: + :~
i = 1, 2, ... , n
(17.54) to the gen-
(17.55)
as
= Qinc'
i =
1,2, ...
,n
(17.56)
Eqs. (17.50), (17.52), and (17.55) into Eq. (17.56), and noting that ... , n, we Can derive the equations of motion of the discretized
= 0, i = 1,2, as n
n
+
Equations
Lmij1jj(t)
Lkij"lj(t)
j=l
j=l
=
Qinc(t),
i = 1,2, ... , n
(17.57)
(17.57) can be expresse~ ~n matrix form as [m)ry(t)
+ [k)ry(t)
=
O(t)
(17.58)
where d2"11 (t) dt2 d2"12 (t) ry(t)
=
dt2
Notes 1. If O(t) is set equal to 6, Eq. (17.58) denotes the equivalent n-degree-of-freedom free vibration equation of the continuous system: [m)ry(t)
If all
"Ii
+ [k)ry(t)
= ()
(17.59)
(t) are assumed to be harmonic in Eq. (17.59) as "Ii(t)
=
Xi
coswr
(17.60)
17.7 Galcrkin's Method
673
where Xi denotes the amplitude of TJi(t) and w indicates the frequency of vibration, the resulting equations define the eigenvalue problem of the discretized system: w2[m]X
= [k]X
(17,61)
2. If the same trial functions
17.6 WEIGHTED RESIDUAL METHODS The Rayleigh and Rayleigh-Ritz methods of solving the eigenvalue problem are based on the stationarity of Rayleigh's quotient. These methods can be classified as variational methods because Rayleigh's quotient is related to the variational methods. There is another class of methods, known as weighted residual methods, for solving vibration problems. The Galerkin, collocation, subdomain collocation, and least squares methods fall into the category of weighted residual methods. The weighted residual methods work directly with the governing differential equation and boundary conditions of a problem. Let the eigenvalue problem of the continuous system be stated by the differential equation ....AW
= ABW
inD
(17.62)
with the boundary conditions Ej W = 0,
j = 1,2, ... , ponS
(17.63)
where A, B, and Ej are linear differential operators, W is the eigenfunction or normal mode (or displacement pattern), A is the eigenvalue, p is the number of boundary conditions, D is the domain, and S is the boundary of the system. In all the weighted residual methods, a trial solution, ($, is assumed for the problem. In general, the trial solution does not satisfy the governing equation, Eq. (17.62), and hence a measure of error is defined: for example, for a one~dimensional problem involving x as
R«$, x)
=
A
(17.64)
where R(
17.7 GALERKIN'S METHOD The Galerkin method is the most widely used weighted residual method. In this method, solution of the eigenvalue problem is assumed in the form of a series of n comparison functions which satisfy all the boundary conditions of the problem: (17.65)
where the Ci are coefficients to be determined and the 4>; (x) are known comparison functions. When Eq, (17.65) is substituted into the differential equation (17.62), the
674
Approximate Analytical Methods
resulting error or residual is defined as R = A¢j(n) _ A (n)B¢j(n)
(17.66) where A(n) is the estimate of the eigenvalue obtained with an n-term trial solution, Eq. (17.65). Note that the residual will be zero from the boundary conditions, Eqs. (17.63), since the trial solution is composed of comparison functions which satisfy all the boundary conditions. In the Galerkin method, the selection of the coefficients of the trial solution is based on the criterion of making the residual small. Specifically, we multiply the residual by the comparison functions
t R(
10
-(n)
i = 1,2, ... ,n
)
(17.67)
It can be seen that in Eq. (17.67), the integral of the weighted residual is set equal to zero, with the functions
C(i)} I C)
c(i)
=
I
c{... '
i
= I, 2, ...
,n
(17.68)
(i)
Cn
The Rayleigh and Rayleigh-Ritz methods are applicable to only conservative systems. However, the Galerkin method is more general and is applicable to both conservative and nonconservative systems. Example 17.7 Find the natural frequencies of vibration of a fixed-fixed beam of length L, bending stiffness EI, and mass per unit length m (Fig. 17.1) using the Galerkin method with the following trial (comparison) functions: 27l'
L -1
¢Z(x) = cos
L -I
47l'
SOLUTION
x
(E17.7.l)
x
(EI7.7.2)
The equation governing the free vibration of a beam is given by d4W -4 dx
-
p4W
=0
(E17.7.3)
where (EI7.7.4)
17.7 Galerkin' s Method
675
with w denoting the natural frequency of vibration of the beam. Using the trial functions ofEqs. (E17.7.1) and (E17.7.2), the approximate solution of the beam vibration problem is assumed as _
W(x)
21TX X ) = CI
(E17.7.5)
Substitution of Eq. (E17.7.5) into Eq. (E17.7.3) gives the residual as R(Cl,
C2) =
[ (T )4 - ] 21T
Cl
4
fJ
21TX cos L
4
+ ClfJ + C2
[()4T 41T
- fJ
4
]
41TX cos L
+ C2{3
4
(E17.7.6) The Galerkin method gives
[L
0,
R
lx=o
i
=
1,2
(E17.7.7)
which can be expressed, using Eqs. (E17.7.6), (E17.7.1), and (E17.7.2), as
[L ( cos L 21TX lx=o -
Cl
41T [ (T)-- 4]:-
+ c2
4
fJ
[L ( cos L 41TX lx=o -
+
) ( 1
) 1
or
CI
[(T)4]41T
C2
I
4
fJ
- {34] cos L21TX + ClfJ 4
[(21T)4
T
cos
L
41TX
4
] dx = 0
(E17.7.8)
- fJ 4] cos L21TX + clfJ 4
[.(21T)4
T
cos
+ C2fJ
41TX L + c2fJ
4
]
dx = 0
Hen'-
c, ( p'] - p'] - c,p' = -c,p' + c, (~[(~)' - fi'] - p'] =
(E17.7.9)
0
(E17.7.10)
0
(E17.7.11)
For a nontrivial solution of Eqs. (EI7.7.1O) and (EI7.7.11), the determinant of the coefficient matrix of CI and C2 must be zero. This gives
Her)'- p'] -{3
4
-p'
p' 1
41T
2 [( T ) - fJ 4
=0 4]
- fJ
4
(E17.7.12)
676
Approximate Analytical Methods
Simplification of Eq. (E17.7.12) yields the frequency equation:
+ 7, 771, 000 = 0
({3L)8 - 15, 900({3L)4
(E17.7.13)
The solution of Eq. (E17.7.13) is {3L = 4.741
or
11.140
(E17.7.14)
Thus, the first two natural frequencies of the beam are given by = 22.48/ EI L2 m
WI
Wz = 124.1/ EI L2 m
(E17.7.15)
The eigenvectors corresponding to WI and Wz can be obtained by solving Eq. (EI7.7.1O) or (E17.7.11) with the appropriate value of {3. The results are as follows. For WI: = {23.0} 1.0
CI}(}) { C2
(E17.7.16)
For cvz: CI } (2) {
=
{-0.69}
(E17.7.17)
1.00
C2
Example 17.8 Derive the equations of motion for the free vibration of a viscously damped tapered beam using the Galerkin method. The governing equation is given by w A(x) 02 (x, t) P ot2
+ d(x)
ow(x, t) at
+ £ [EI ox2
(x) 02w(x, t)] ox2
=0 '
O
(E17.8.l)
°
=
with two boundary conditions at x = as well as at x I. In Eq. (E17.8.l), the term t)/otJ denotes the viscous damping force per unit length of the beam.
d(x)[ow(x,
SOLUTION
The transverse deflection function w(x, t) is assumed to be of the form w(x, t) = W(x)eAt
(EI7.8.2)
where ). is complex and W (x) is the deflection shape (or mode shape). When Eq. (EI7.8.2) is substituted into Eq. (E17.8.1), we obtain 2
pA(x»).2W(x)
d + d(x»). W(x) + -2 dx
2 [ E1 (x) d W(X)] 2 dx
= 0, (E17.8.3)
17.7 GaJerkin' s Method
677
with two boundary conditions at each end. We assume the solution, W(x), in the form of a series of n comparison functions
= I>i
W(x)
(EI7.8.4)
i=l
where each function
Multiplying Eq. (EI7.8.5) by
j
= 1,2, ...
,n (E17.8.6)
Defining the symmetric stiffness, damping, and mass coefficients kij, dij. and mij, respectively, as d Jot dx2 2
kij =
1
[
E/(x)
d2
(E17.8.7)
1
dij =
(E17.8.8)
1
d(X)
mij =
(E17.8.9)
pA(X)
Eq. (EI7.8.6) can be written as n
)..2LmijCi i=1
n
n
+).. I>ijCi + i=l
LkijCi i=l
= 0,
j = 1,2, ... , n
(EI7.8.10)
or in matrix form as (E17.8.Il) Solution of Eq. (EI7.8.Il)
Three approaches can be used for solving Eq. (EI1.8.Il).
678
Approximate Analytical Methods
For a nonzero solution of 2 in Eq. (E17.8.l1), we
Approach J: Direct solution:
must have
2 IA [m]+A[d]+[k]1
=0
(E17.8.12)
Equation (E17.8.12) leads to a polynomial equation in A of order 2n, whose solution yields the roots Ai, i = 1,2, ... , 2n. For each Ai, Eq. (E11.8.H) can be solved to find the corresponding vector 2(i). This procedure is, in general, tedious and inconvenient to handle; hence the following procedures are commonly used to solve Eq. (E17.8.11). Approach 2: Proportional damping: In this approach, the damping matrix is assumed to be given by a linear combination of the mass and stiffness matrices: [d]
= arm]
+ .8[k]
(E17.8.13)
where a and .8 are known constants. This type of damping is known as proportional . damping because [c] is proportional to [m] and [k]. As in the case of a multidegree-offreedom system, the modal matrix of the corresponding undamped discretized system, [X], is defined as (E17.8.14) where 2(i), i = 1,2, ... , n, denote the modal vectors that satisfy the undamped eigenvalue problem: [k]c
= A[m]c
(E17.8.15)
2
where A = w and w is the natural frequency of the discretized undamped system. Assuming that the modal vectors are normalized with respect to the mass matrix, we have
[xf
[m] [X] = [I]
(E17.8.16)
= [A] = [:'
[X]'[k][X]
A2 ... A~]
(E17.8.17)
= wf,
where Ai i = 1,2, ... , n. Substituting Eq. (E17.8.13) into Eq. (E17.8.11) and using the transformation
c = [X]p
(EI7.8.18)
we obtain A 2[m][X]p
+ A(a[m] + .8[k])[X]p
+ [k][X]p
=6
(E17.8.19)
Premultiplication of Eq. (EI7.8.19) by [X]T and use of Eqs. (E17.8.16) and (E17.8.17) results in A
2
P + A(a[1] + .8[A]) P + [A]p = 6
(E17.8.20)
17.7 Galerkin's Method
679
Defining
S1W a[1]
+ fJ[A]
= [y]
== [2
l
2S2W2
...
o
0]
(E17.8.21)
2Snwn
where Si is called the damping ratio in mode i, Eq. (E17.8.20) can be written as (E17.8.22) or, in scalar form, i = 1, 2, ... , n
(E17.8.23)
Each equation in (E17.8.23) denotes a quadratic equation in the eigenvalue Awhich can be solved to find the ith eigenvalue, Ai, of the proportionally damped continuous system. Substituting this eigenvalue Ai along with Eq. (E17.8.13), into Eq. (E17.8.1l) and solving the resulting linear equations gives the eigenvectors c(i) of the proportionally damped system. In general, the eigenvalues and the eigenvectors of the proportionally damped system occur in complex-conjugate pairs. Approach 3: General viscous damping: In the case of general viscous damping, damping will not be proportional and hence the undamped modal matrix [X] will not diagonalize the damping matrix [d]. In such a case, we transform the eigenvalue problem, Eq. (E17.8.11), as indicated by the following steps:
(a) Define the identity AC
= AC
(EI7.8.24)
(b) Rewrite Eq. (E17.8.1l) as A2C = -A[mr1[d]c (c) Define a vector
y
- [mr1[k]c
(E17.8.25)
of dimension 2n as: (E17.8.26)
(d) Combine Eqs. (E17.8.24) and (E17.8.25) as
C}
A { AC
=
[[0]
-[mr1[k]
[1]]
-[m]-l[d]
{ACC }
(E17.8.27)
or
AY = [B]y
(E17.8.28)
680
Approximate Analytical Methods
where [BJ denotes a 2n x 2n matrix: [OJ
[BJ
=[
-[mrl[kJ
[l]
]
-[mJ-1 [dJ
(E17.8.29)
The transformed algebraic eigenvalue problem of order 2n defined by Eq. (E17.8.28), is solved to find the eigenvalues Ai and the corresponding mode shapes j;(i), i = 1,2, ... , 2n.
17.8 COLLOCATION METHOD In the Galerkin method, the integral of the weighted residual over the domain of the problem is set equal to zero where the weighting function is the same as one of the comparison functions used in the series solution. The collocation method is similar to the Galerkin method except that the weighting functions are spatial Dirac delta functions. Thus, for a one-dimensional eigenvalue problem, an approximate solution is assumed in the form of a linear sum of trial functions ¢i (x) as n -en)
¢
(x)
= LtCi¢i(X) '"
(17.69)
i=1
where the Ci are unknown coefficients and the ¢i (x) are the trial functions. Depending on the nature of the trial functions used, the collocation method may be classified in one of the following three types: 1. Boundary method: used when the functions ¢i (x) satisfy the governing differential equation over the domain but not all the boundary conditions of the problem. 2. Interior method: used when the functions ¢i (x) satisfy all the boundary conditions but not the governing differential equation of the problem. 3. Mixed method: used when the functions ¢i (x) do not satisfy either the governing differential equation or the boundary conditions of the problem. When the integral of the weighted residual is set equal to zero, the collocation method yields
,
1
o 8(x -
-en)
Xi
)R(¢
(x» dx
= 0,
i
= 1,2, ...
,n
(17.70)
where 8 is the Dirac delta function and Xi, i = 1, 2, ... , n, are the known collocation points where the residual is specified to be equal to zero. Due to the sampling property of the Dirac delta function, Eqs. (17.70) require no integration and hence can be expressed as -en)
R(¢
(Xi))
= 0,
i = 1, 2, ... , n
(17.71)
This amounts to setting the residue at XI, X2, .•• , Xn equal to zero. Equations (17.71) denote a system of n homogeneous algebraic equations in the unknown coefficients
.681
--L7.S CollocationMethod
Cl, C2, ••• , Cn and the parameter A. In fact, they represent an algebraic eigenvalue problem of order n. It can be seen that the selection of the collocation points XI, X2, ... , Xn is important in obtaining a well-conditioned system of equations and a convergent solution. The locations of the collocation points should be selected as evenly as possible in the domain and/or boundary of the system to avoid ill-conditioning of the resulting equations. To see the nature of the eigenvalue problem, consider the problem of longitudinal vibration of a tapered bar. The governing differential equation is given by [see Eq. (9.14) with harmonic variation of u(x, t)]:
-d [ EA(x)-- dU(X)] dx dx
- ApA(x)U(x)
=0
(17.72)
By assuming the trial functions CPi(x) in Eq. (17.69) as comparison functions, substituting the assumed solution into Eq. (17.72), and setting the residual equal to zero at Xi, we obtain d ~n Cj {[ dx
dcpj(x) EA(x)~
= 0,
] - ).,pA(x)cpj(X) }
J-
i = 1,2, ... , n
~~
(17.73) or
n
I:>ijCj
n
i = 1,2, ... ,n
= kLmijCj,
j=1
(17.74)
j=l
where)., is the eigenvalue of the problem and kij and mij denote the stiffness and mass coefficients, respectively, defined by k·. = ~ [EA(X.)dCPj(Xi)] IJ
dx
1
mij = pA(Xi)CPj(Xi)
dx
(17.75) (17.76)
Equations (17.74) denote an algebraic eigenvalue problem which can be expressed in matrix form as [k]c = A[m]c
(17.77)
where
is an n-dimensional vector of the coefficients, and [k] = [kij] and em] = [mij] are the stiffness and mass matrices of order n x n. It can be seen that the main advantage of the collocation method is its simplicity. Evaluation of the stiffness and mass coefficients involves no integrations. The main disadvantage of the method is that the stiffness and mass matrices, defined by Eqs. (17.75) and (17.76), are not symmetric although the system is conservative. Hence,
682
Approximate Analytical Methods
the solution of the nonsymmetric eigenvalue problem, Eq. (17.77), is not simple. In general, we need to find both the right and left eigenvectors of the system in order to find the system response. Example 17.9 Find the natural frequencies of transverse vibration of a unifonn fixed-fixed beam shown in Fig. 17.1 using the collocation method with the approximate solution (E17.9.1)
where rP}(x) = 1 - cos-
2JTX (E17.9.2)
I ¢>2(x) = 1 - cos-
4JTX (EI7.9.3)
I
SOLUTION It can be seen that the trial functions satisfy all the boundary (geometric) conditions of the beam: rPi(X drPi -dx-(x
= 0) = rPi(X = I) = 0,
i
= 1,2
drPi
= 0) = -(x dx
= 1) = 0,
i
(EI7.9.4)
= 1,2
(E17.9.5)
Since the assumed solution has two unknown coefficients, we need to use two collocation points. The collocation points are chosen as x} = 1/4 and Xz = 1/2. The eigenvalue problem is defined by the differential equation d4W(x) EI
dx 4
pAa}W(x)
-
=0
(E17.9.6)
When Eq. (EI7.9.1) is substituted in Eq. (EI7.9.6), the residual is given by =
R(X(x»
d4 X (x) dx4
-
(EI7.9.7)
AX(X)
where pAwz A=--
Using Eqs. (EI7.9.1)-(EI7.9.3),
4 d R = dx4
[(c}
- A [ c} =-
c}
(
(2JT)4 ,
- A [ c}
(
EI the residual can be expressed as
2JTX) 1 - cos -,-
+ cz (
4JTX)] 1 - cos _,_
2JTX) 1 - cos -1-
+ cz (
4JTX)] I - cos _,_
2JTX
cos --
I
2JTX) 1 - cos -1-
-
cz
(4JT)4 I
+ Cz (
(EI7.9.8)
4JTX
cos __
,
4JTX)] I - cos _,_
(EI7.9.9)
17.ll Collocation Method
By setting the residual equal to zero at C, [- (~)'
Co'i
- «I-CO'
= 1/4 and
XI
~)]
+C2 [-
683
= 1/2, Eq. (EI7.9.9) gives
X2
(4n'
co,,,
=0
-
(EI7.9.1O)
c, [- (~)'
co,,, -
(4;)'
CO,2" -
«I - CO,2,,>] ~ 0 (EI7.9.11)
Equations (EI7.9.1O) and (EI7.9.11) can be simplified as
[en' -2<] =0
C,(-<)+C2
c, [(~)'
-
2<]
+02 [_
(4;)'] =0
(EI7.9.12)
(EI7.9.13)
By setting the determinant of the coefficient matrix in Eqs. (EI7.9.12) and (EI7.9.13) equal to zero, we obtain the frequency equation as --)...
2;
(
)4
=0
(EI7.9.14)
-2)",
which can be simplified as (EI7.9.15) The roots of Eq. (EI7.9.15) are given by )...1.2
= 5.258246
(Tt,
194.741754
(Tr
(EI7.9.16)
Using Eqs (EI7.9.8) and (EI7.9.16), the natural frequencies can be obtained as
W2
(El
y PAi4
= 137.730878
(EI7.9.17)
The exact values of the first two natural frequencies of a fixed-fixed beam are given by
UJI
(El y PAi4 and
= 22.3729
W2
(El y PAi4
= 61.6696
(EI7.9.18)
It can be seen·that the first and the second natural frequencies given by the collocation method are larger by 1.1581% and 123.3367%, respectively.
('
'1
684
Approximate Analytical Methods
=
=
=
=
Note: If we use the points XI l/4 and X2 3l/4 or XI l/3 and X2 2l/3 as collocation points, it would not be possible to compute the natural frequencies. Because of symmetry, use of the points xl l/4 and X2 l/2 in the half-beam would be sufficient.
=
The mode shapes can be determined,
C2
={
using (E 17.9.12), as
(¥ / -
l
)
2A
) CI
(EI7.9.19)
to Al = 5.258246(il"4/ l4), is given by
Thus, the first mode shape, corresponding
ci
=
= 0.0214199c?)
or CI}(J) { C2
{
2
)
}
(1)
cI
(EI7.9.20)
to A = 194.741754(il"4/ l4), is given by
The second mode shape, corresponding
ci
I
= 0.0214199
= -1.458920c~2)
or CI
. { C2}
17.9
(2)
=
{
1 -1.458920
}
(2)
cI
(EI7.9.21)
SUBDOMAIN METHOD In this method, the domain of the problem, D, is subdivided into n smaller subdomains DiU 1,2, ... , n), so that
=
(17.78) and the integral of the residual
1;
over each subdomain
R(¢(x»
dx = 0,
is set equal to zero:
i = 1,2, ... , n
(17.79)
where ¢(x) denotes the assumed solution in the form of a linear sum of trial functions (such as comparison functions), Eq. (17.69). Equations (17.79) indicate that the average value of the residual in each subdomain is zero. Obviously, in this method, negative errors can cancel positive errors to give least net error, although the sum of the absolute value of the errors is very large. The subdomain method can be interpreted as a weighted residual method where the weighting functions, 1/Ji(x), are defined as 1/Ji(X)
= { 6:
if X is in Di otherwise
(17.80)
The stiffness and mass matrices given by the subdomain method are also non symmetric, and hence the resulting algebraic eigenvalue problem is difficult to solve.
17.I)
Subdomain Method
685
17.10 Find the natural frequencies of transverse vibration of a uniform fixed-fixed beam using the subdomain method with the approximate solution
Example
(EI7.1O.1) where 21l'X
¢\ (x) = 1 - cos -
(EI7.10.2)
I
41l'x = 1 - cos-·
¢2(X)
(EI7.1O.3)
t
SOLUTION As seen in Example 17.9, the trial functions satisfy all the (geometric) boundary conditions of the beam. Since the assumed solution has two unknown coefficients, we need to use two subdomains. Because of the symmetry, we choose the subdomains in the first half of the beam as D\ (0, I/4) and D2 (//4, I/2). For the approximate solution given by Eq. (EI7.1O.1), the residual is given by [Eq. (EI7.9.9)]:
=
=
(E17.lO.4) By setting the integral of the residual over the two subdomains equal to zero, we obtain [1/4 o R dx =
10t/ .[ -Cl 4
-
ACl
(21l')4
+ ACl
-t-
cos 21l'X I
COS --
-
21l'X
-t- -
(41l')4 C2
41l'X
-t- cos -t-
+ AC2 cos -41l'X] I
AC2
dx
=0
or (EI7.10.5) and 1/2 [
Rdx =
['/2 [
1/4
-C\
-
AC\
(21l')4 -
21l'X cos--
I
1/4
+ Act
21l'x cos -I
I AC2
-C2
(41l')4 -
41l'X cos--
I
+ AC2 COS -41l'X] t
I
dx = 0
or (EI7.1O.6)
¥.' j
686
Approximate Analytical Methods
For a nontrivial solution of c) and C2 in Eqs. (El 7. 10.5) and (E17.1O.6), the determinant of the coefficient matrix must be zero: AI _ A..L
(21T)3 _ / 3
- en
4
-Ai
2tr
4
Ai + A~ -Ai
-
=0
(E17.1O.7)
Equation (E17.10.7) can be simplified as
(E17.1O.8) The roots of Eq. (E17.1O.8) are
16,,4
A) = 0,
A2 =
14"
(E17.1O.9)
The mode shapes corresponding to A) and A2 can be determined using Eq. (E17.1O.5) as
+ 1/2,,)
(2"/1)3 - A (1/4 =
C2
A (1/4)
c)
(E17.1O.1O)
Thus, for A) = 0, the mode shape is given by
For
)..2
= 16,,4/14,
1i.l0
_
-
{ Cl } (I) _
C2
O} I c
{
-
(I)
2
(E17.10. ll)
the mode shape is given by
-(2)
C
-(I)
C
(2)
Cl
=
{
C2
1.0
{
}
=
}
183.4405//3
(2) c)
(E17.1O.12)
LEAST SQUARES METHOD The least squares method can be considered as a variational method as well as a weighted residual method. Because the method is also applicable to problems for which a classical variational principle does not exist, it is considered more as a weighted residual method. Basically, the least squares method minimizes the integral of the square of the residual over the domain:
L
2
R d D = minimum
(17.81)
where R is the residual of the governing differential equation and D is the domain of the problem. Assuming the approximate solution in the form ofEq. (17.69), Eq. (17.81) can be expressed as
1/
2
R «(f)(x)) dx =
1/
R2(c),
C2, •..
,
cII) dx = minimum
(17.82)
17.10
Least Squares Method
687
The minimization is carried with respect to the unknown coefficients C(, C2,... , Cn· The necessary conditions for the minimum of the integral in Eq. (17.82) are given by
(1/ ) 1/ /
-8 8Ci
or
1 o
R2dx
=2
0
0
8R R-dx
=0
i = 1, 2, ... , n
= 0,
8Ci
8R R-dx 8Ci
(17.83)
.
Equation (17.83) indicates that the least squares method is a weighted residual method where the weighting functions, 1/ri(X), are given by 1/ri(X)
= -,8R
i = 1,2, ... , n
(17.84)
8Ci
To see the nature of the algebraic eigenvalue problem given by the method of least squares, consider the problem of the longitudinal vibration of a tapered bar: - d [ EA(x)-- dU(X)] dx dx When the approximate can be expressed as
- ApA(x)U(x)
=0
(17.85)
solution given by Eq. (17.69) is used, the residual ofEq.
L Cj
R =
n
d
{[.-
j=l
EA(x)
dx,
dl/J·(x) J
]
ApA(x)4Jj(x)
-
}
(17.85)
(17.86)
dx
and hence
-8R = -d [ EA(x)-- dl/Ji(X)] - ApA(x)4Ji(X) 8Ci
dx
(17.87)
dx
Using Eqs. (17.86) and (17.87), Eq. (17.83) can be expressed as
to f; {d[ n
Cj
dx
EA(x)
4Ji(X)] d [dEA(x)~ . { dx
d4J~x'(X)]
- ApA(x)4Jj(x)
- ApA(x)4Ji(X) } dx
}
=0
or ~ ~ j=l
c.{ J
t
10
.!!... dx
4J
_ At.!!...
10
4Ji
[EA(X)d4Jj(X)].!!... [EA(X)d dx dx [EA(X)d
dx
/X)] dx
(X)] dx
dx
pA(X)4Ji(X)dx
_ A t PA(X)l/Jj(X).!!... [EA(X) d4Ji(X)] dx
10
+ )..21/
dx
dx
p2A 2(x )4Ji(x)4Jj (x) d x} = 0,
i = 1, 2, ... , n
(17.88)
688
Approximate Analytical Methods Defining the n x n matrices ki} =
t ~[EA(X)
10 dx
mil
=
hi}
=
1/
p2 A2(x)¢;(X)¢j(X)
o
Lkijcj
-}.. Lhijcj j=l
dx
(17.89)
dx
(17.90)
p-A(x)¢;(x)- d [ EA(x)-j d¢ ] dx dx dx
(17.91)
as
n
j=l
~ [EA(X) d¢j(X)] dx dx
1/
Eq. (17.88) can be expressed n
d¢;(X)] dx
n
n
-}.. Lhj;cj
+}..2 LmijCj
j=l
= 0,
i
= 1,2,
... , n
j=l
(17.92) or, equivalently,
in matrix form as [[k) - J..([h) + [hf)
+ }..2[m))c = 6
(17.93)
Equation (17.93) denotes a quadratic eigenvalue problem because it involves both }..and This equation can be seen to be similar to the one corresponding to the eigenvalue problem of a damped system, Eq. (E17.8.11). 2 }.. .
To reduce Eq. (17.93) to the form of a standard following vectors and matrices: "
b2nXl ~ -{ ~
=[
[Bhnx2n
=[
problem,
}
-[k) [m) (0)
define the
(17.94)
+ [h)T
[h) [Ahnx2n
eigenvalue
-[k) ] (0)
(0)]
-[k)
(17.95)
(17.96)
Using Eqs. (17.94), (17.95), and (17.96), Eq. (17.93) can be rewritten as
-
[A)b = }"[B)b
-
(17.97)
which can be seen to be a standard matrix eigenvalue problem of order 2n. Thus, the least squares method requires the solution of an eigenvalue problem of twice the order of that required by other methods, such as the Rayleigh-Ritz and Galerkin methods.
Example 17.11
Find the first two natural frequencies and mode shapes of a fixedfixed uniform beam using the least squares method. The free vibration equation of a uniform beam is given by
(E17.11.l)
l 7. 10
Least Squares Melhod
689
where
mw2 A = (34 = -
(El7.Il.2)
EI
solution for W (x) as
Assume an approximate
W(X)=CI
SOLUTION
( l-cos- - 21rX) +C2 ( l-cos- - 41rX) t
(EI7.11.3)
t
The least squares method requires that
(EI7.1l.4)
where R(W(x» given by
is the residual. The conditions
'1
for the minimum in Eq. (EI7.11.4)
are
1
o
With the assumed
oRR(W(x»-(W(x»dx
_
(21r)4 -
- (34
[(CI
The necessary conditions, For i 1:
=
oR(W(x» OCl
2Jl'X) + C2 ( -t-
1 - cos
t [(21r)4-tCl
+Cl{34
(
1 - cos
2Jl'
[(
-t-)4
cos
(4Jl')4
t
t
cos-
41rx
t (EI7.11.6)
can be expressed as follows.
cos
21rx
(4Jl')4
-t- + C2 -t-
21rX) + c2{3 4 ( -t-
21rX
-
41rX)] -t-
1 - cos
given by Eq. (E17.11.5),
dx = 10
21rx
COS--C2
t
dx
o
(EI7.11.5)
solution, the residual becomes
_ d4W R(W(X»=---AW=-Cl 4
t R(W(x» -
i = 1,2
= 0,
OCi
1 - cos
4
-t- + {3
(
cos
1 - cos
21rx
-t- )]
41rx
-t-
41rX)] -tdx = 0 (EI7.1l.7)
Upon integration
and simplification,
Eq. (EI7.11.7),
yields
(EI7.1l.8)
",
690
Approximate Analytical Methods
For i = 2:
1/ o
R(W(x»
oR(W(x»
1/
dx=
0
OC2
+ cd] .
[(21r)4 CI -
4(
/
21rX COS-+C2 /
(41r)4 /
/
21rX) + c2fJ4 ( 1 - cos _/_ 4JrX)] 1 - cos -/-
4Jr 4Jrx 4 4Jrx -/- )4 cos -/- + fJ ( 1 - cos -/-)
[(
cos-4Jrx
]
dx
=0
(EI7.11.9) Upon integration and simplification, Eq. (EI7.11.9), yields CI(/fJ8)
+ C2
(
215Jr8 44Jr4 fJ4 /7 /3
--
3
+ -/fJ8 2
)
== 0
(EI7.11.10)
For a nontrivial solution of CI and C2in Eqs. (EI7.11.8) and (E17.11.10), the determinant of their coefficient matrix must be equal to zero. This gives
which upon simplification gives the frequency equation 16 5/ fJI6- 1632Jr4/12fJI2+ 213, 760Jr8/8fJ8 - 2, 228, 2241rI2/4fJ4 + 224JrI6 = 0 (EI7.11.12) Defining
(EI7.11.13) Eq. (EI7.11.12) can be rewritten as 5X4 - 102):3 + 835P - 544): + 256 = 0
(EI7.11.14)
The roots of Eq. (EI7.11.14) are given by (using MATLAB) AJ.2= 9.8671 ± 7.4945i,
A3.4= 0.3329
± 0.4719i
(EI7.11.15)
The natural frequencies can be computed using Eqs. (EI7.11.2) and (EI7.11.13) as - A"
m/4 w2I EJ. 24Jr4
/4 fJ~, __
, - 24Jr4 or Wi
~(EJ)I/2 m/
= 39.478602y A.i
4
'
i = 1,2,3,4
(EI7.11.16)
I7.10
Least Squares Method
69 J
Noting that
/il.2 = 3.3360 ±
Ji3,4
(E17.11.17)
1.1233i
= 0.6747 ± 0.3497i
(E17.1Ll8)
Eq. (EI7.11.16) gives natural frequencies as
+ 44.346314i)
(:l~) (:l~)
WI
= (131.700617
Wz
= (131.700617 - 44.346314i)
W3
= (26.636213
W4
EI = (26.636213 - 13.805667i) ( ml4
+ 13.805667i)
EI ) ( ml4
1/2
(EI7.11.19)
1/2
(E17.11.20)
1/2
(EI7.11.21)
)1/2
(E17.11.22)
To find the mode shapes, Eq. (EI7.11.11), is rewritten in terms of). as
[
t - ).+ fP P
).2
i28 -:-16X
+ fX2
] {CI} {oJ C2
=
°
(E17.11.23)
The first equation of (EI7.11.23) can be written in scalar form as (EI7.11.24) or C2
= ( -~
+
I-2~2)
(EI7.11.25)
CI
By substituting the value of ).i given by Eq. (EI7.11.15) into Eq. (EI7.11.25), the corresponding eigenvector c(i) can be expressed as
. = 1 c(i)] = 11.0 + -f + b - ~ O.Oi
1
-(I) C
(i)
C2
A
-(1)
=
i = 1,2,3,4
XI
= 9.8671
+ 7.4945i,
1 ci I { c~l) l)
(EI7.11.26)
2A
Thus, the eigenvectors are as follows: For
C
] (')1 c1 '
1.0 + O.Oi } (I) = -1.4366 - 0.0457i c1
(E17.11.27)
For ).2 = 9.8671 - 7.4945i, -(2)
C
=
1 ci I { c~2) 2)
=
1.0 + O.Oi } (2) -1.4366 + 0.0457i c1
(E17.11.28)
692
Approximate Analytical Methods
For
X3
= 0.3329
+ 0.4719i,
-(3)
c
For
X4
=
Ic?) ] = { I I{ c~3)
1.0 + O.Oi } (3) 0.0010 _ 0.0026i c)
(E17.l1.29)
+ O.Oi } (4) + 0.0026i c)
(E17.11.30)
= 0.3329 - 0.47l9i, -(4)
c
=
1.0 = 0.0010
c~4) c~4)
Thus, the natural frequencies of the beam are given by Eqs. (E17.l1.19)-(E17.11.22) and the corresponding mode shapes by Eqs. (E17.l1.27)-(E17.l1.30) Notes 1. The eigenvalues (and the natural frequencies) and the corresponding eigenvectors in the least squares method will be complex conjugates. Although it is difficult to interpret the complex eigenvalues and eigenvectors, usually the imaginary parts are small and can be neglected. 2. The least squares method leads to a quadratic eigenvalue problem. That is, the size of the eigenvalue problem will be twice that of the problem in the Rayleigh-Ritz method. 3. The matrices involved in the least squares method are more difficult to compute. 4. In view of the foregoing features, the least squares method is not as popular as the other methods, such as the Rayleigh-Ritz and Galerkin methods for solving eigenvalue problems. However, the least squares method works well for equilibrium problems, as indicated in the following example. Example 17.12 Find the deflection of a fixed-fixed uniform beam subject to a uniformly distributed load fo per unit length using the least squares method. The goveming differential equation for the deflection of a beam is given by d4w dx4
-
fo = 0,
0
=:: x =::
I
(E17.12.l)
where Po
fo=-
(E17.l2.2)
EI
Assume an approximate solution for w (x), using comparison functions, as w(x)
SOLUTION
= c}X2(x
-1)2
+ C2x3(X
_1)2
(EI7.12.3)
The least squares method requires the minimization of the function I
=
1/
R2(w(x))
dx
(EI7.l2.4)
The necessary conditions for the minimum of I are given by
r
.fo
R(w(x))
oR(w(x))
oc;
dx
= 0,
i = 1,2
(EI7.l2.5)
17.11 Recent Contributions
693
With the assumed solution ofEq. (E17.12.3), the residual and its partial derivatives are given by
+ (l20x
R(iiJ(x» = 24ct
- 48l)C2 - 10
aR
(EI7.12.6) (EI7.12.7)
-=24 aCt
aR
-
(EI7.12.8)
= 120x - 48l
aC2
Thus, Eq. (EI7.12.5) can be expressed as
1/[ 1/[
24Ct+ (120x - 48l)C2 - 10](24) dx = 0 24cl + (120x - 48l)C2 - 10](120x - 48l) dx
(E17.12.9)
=0
(E17.12.1O)
Equations (E17.12.9) and (E17.12.1O) can be evaluated to obtain
+ 12lc2 = 10
(E17.12.11)
24cl - 80lc2 = 10
(E17.12.12)
24cl
The solution of Eqs. (E17.12.11) and (E17.12.12) gives fo ct
= 24'
C2
=0
(E17.12.13)
Thus, the deflection of the beam, in view of Eqs. (E17.12.3) and (E17.12.13), becomes (E17.12.14) It can be seen that this solution coincides with the exact solution.
17.11
RECENT CONTRIBUTIONS Dunkerley's Method The basic idea behind Dunkerley's method of finding the smallest natural frequency of a multidegree-of-freedom elastic system was extended by Levy [6]. to determine all the frequencies of the system simultaneously. The method is found to converge fast when the frequencies are not close to each other. The method is demonstrated with the help of several lumped-parameter systems. Badrakhan presented the application of Rayleigh's method to an unconstrained system [7]. Frequency in Terms of Static Deflection Radhakrishnan et al. [8] developed a method to estimate the fundamental frequency of a plate through the finite element solution of its static deflection under a uniformly distributed load without the associated eigenvalue problem. The results computed in the case of a clamped rectangular
694
Approximate Analytical Methods
plate with a central circular hole were found to be in reasonable agreement with experimental results. The method is useful for determining the fundamental frequency of elastic plates of arbitrary geometry and boundary conditions. Bert [9J proposed the simple relation
(17.98)
where C is a dimensionless constant and g is the acceleration due to gravity for estimating the natural frequency (w) in terms of the static deflection (8) of a linear system. Nagaraj [IOJ showed that Eq. (17.98) also holds for a rotating Timoshenko beam if C is selected properly. A variational formulation of the Rayleigh-Ritz method was presented by Bhat [llJ. The stationarity of the natural frequencies was investigated with respect to coefficients in the linear combination of the assumed deflection shape as well as natural modes. Beams The frequencies of beams carrying multiple masses using Rayleigh's method were considered by Low [12J. The solution methods for frequencies of three massloaded beams are presented with both the transcendental characteristic equation and Rayleigh estimation. Gladwell [13J presented a method of finding the natural frequencies and principal modes of undamped free vibration of a plane frame consisting of a rectangular grid of uniform beams. A general method of finding a set of assumed modes for use in the Rayleigh-Ritz method was given. The resulting equations, expressed in matrix form, were solved for the case of a simple frame, to illustrate the method. Membranes The dynamic stability of a flat sag cable subjected to an axial periodic load was investigated by Takahashi using the Galerkin method [14J. The results include unstable regions for various sag-to-span ratios and ratios of wave speeds. The transient response of hanging curtains clamped at three edges was considered by Yamada and his associates [15]. A hanging curtain was replaced by an equivalent membrane for deriving the equation of motion. The free vibration of the membrane was analyzed theoretically, and its transient response when subjected to a rectangularly varying point force was studied using Galerkin's method. Plates The use of two-dimensional orthogonal plate functions as admissible deflection functions in the study of flexural vibration of skew plates by the Rayleigh-Ritz method was presented by Liew and Lam [16J. Free vibration analysis of triangular and trapezoidal plates was considered by the superposition technique [17, 18J. The superposition method was extended by Gorman for the free vibration solution of rectangular plates resting on uniform elastic edge supports [19]. The elastic edge supports were assumed to be uniform elastic rotational and translational supports of any stiffness magnitudes in terms of eight stiffness coefficients. The vibrations of circular plates with thickness varying in a discontinuous fashion were studied by Avalos et al. [20]. The free vibration of a solid circular plate free at its edge and attached to a Winkler foundation was considered by Salari et al. [21]. The free vibrations of a solid circular plate of linearly varying thickness attached to a Winkler foundation were considered by Laura et al. [22] using linear analysis and the Rayleigh-Schmidt method.
References
695
Avalos et al. [20] presented general approximate solution for vibrating circular plates with stepped thickness over a concentrated circular region. Approximate values of the fundamental frequencies of vibration of circular plates with discontinuous variations of thickness in a non concentric fashion were determined by Laura et al. [23]. The Ritz method and Rayleigh's optimization procedure were used in finding the solution of plates whose edges are elastically restrained against rotation.
REFERENCES 1. S. H. Crandall, Engineering Analysis: A Survey of Numerical Procedures, McGraw-Hill, New York, 1956. 2. S. S. Rao, Applied Numerical Methods for Engineers and Scientists, Prentice Hall, Upper Saddle River, NJ, 2002. 3. S. S. Rao, The Finite Element Method in Engineering, 3rd ed., Butterworth-Heinemann, Boston. 1999. 4. S. S. Rao, Mechanical Vibrations, 4th ed., Prentice Hall, Upper Saddle River, NJ, 2004. 5. 1. N. Reddy, Energy and Variational Methods in Applied Mechanics, Wiley, New York, 1984. 6. C. Levy, An iterative technique based on the Dunkerley method for determining the natural frequencies of vibrating systems, Journal of Sound and Vibration, Vol. 150, No.1, pp. 111-118,1991. 7. F. Badrakhan, On the application of Rayleigh's method to an unconstrained system, Journal of Sound and Vibration, Vol. 162, No.1, pp. 190-194, 1993. 8. G. Radhakrishnan, M. K. Sundaresan, and B. Nageswara Rao, Fundamental frequency of thin elastic plates, Journal of Sound and Vibration, Vol. 209, No.2, pp. 373-376, 1998. 9. C. W. Bert, Application of a version of the Rayleigh technique to problems of bars, beams, columns, membranes and plates, Journal of Sound and Vibration, Vol. 119, No.2, pp. 317-326, 1987. 10. V. T. Nagaraj, Relationship between fundamental natural frequency and maximum static deflection for rotating Timoshenko beams, Journal of Sound and Vibration, Vol. 201, No. 3,pp. 404-406,1997. 11. R. B. Bhat, Nature of stationarity of the natural frequencies at the natural modes in the Rayleigh-Ritz method, Journal of Sound and Vibration, Vol. 203, No.2, pp. 251-263, 1997. 12. K. H. Low, Frequencies of beams carrying multiple masses: Rayleigh estimation versus eigenanalysis solutions, Journal of Sound and Vibration, Vol. 268, No.4, pp. 843-853, 2003. 13. G. M. L. Gladwell, The vibration of frames, Journal of Sound and Vibration, Vol. I, No.4, pp. 402-425, 1964. 14. K. Takahashi, Dynamic stability of cables subjected to an axial periodic load, Journal of Sound and Vibration, Vol. 144, No.2, pp. 323-330, 1991. 15. G. Yamada, Y. Kobayashi, and H. Hamaya, Transient response of a hanging curtain, Journal of Sound and Vibration, Vol. 130, No.2, pp. 223-235, 1989. 16. K. M. Liew and K. Y. Lam, Application of two-dimensional orthogonal plate function to flexural vibration of skew plates, Journal of Sound and Vibration, Vol. 139, No.2, pp. 241-252, 1990. 17. H. T. Saliba, Transverse free vibration of fully clamped symmetrical trapezoidal plates, Journal of Sound and Vibration, Vol. 126, No.2, pp. 237-247, 1988.
696
Approximate Analytical Methods 18. H. T. Saliba, Transverse free vibration of simply supported right triangular thin plates: a highly accurate simplified solution, Journal of Sound and Vibration, Vol. 139, No.2, pp. 289-297, 1990. 19. D. J. Gorman, A general solution for the free vibration of rectangular plates resting on uniform elastic edge supports, Journal of Sound and Vibration, Vol. 139, No.2, pp. 325-335, 1990. 20. D. Avalos, P. A. A. Laura, and H. A. Larrondo, Vibrating circular plates with stepped thickness over a concentric circular region: a general approximate solution, Journal of the Acoustical Society of America, Vol. 84, No.4, pp. 1181-1 185, 1988. 21. M. Salari, C. W. Bert, and A. G. Striz, Free vibrations of a solid circular plate free at its edge and attached to a Winkler foundation, Journal of Sound and Vibration, Vol. 118, No. 1, pp. 188-191, 1987. 22. P. A. A. Laura, R. H. Gutierrez, R. Carnicer, and H. C. Sanzi, Free vibrations of a solid circular plate of linearly varying thickness and attached to a Winkler foundation, Journal of Sound and Vibration, Vol. 144, No.1, pp. 149-161, 1991. 23. P. A. A. Laura, R. H. Gutierrez, A. Bergmann, R. Carnicer, and H. C. Sanzi, Vibrations of circular plates with discontinuous variation of the thickness in a non-concentric fashion, Journal of Sound and Vibration, Vol. 144, No.1, pp. 1-8, 1991.
PROBLEMS 17.1 The eigenvalue problem for finding the natural frequencies of vibration of a taut string, shown in Fig. 17.5, is given by d2·W dy2
+ AW = 0,
W(y) = cly(l
0
with the boundary conditions W(y)
=0
at
y
17.2 Solve Problem 17.1 and find the natural frequencies of the string using the trial solution - y)
+ c2y(l
_ y)2
17.3 Solve Problem 17.1 and find the natural frequencies of the string using the trial solution
= 0 and y = 1
where pL2 A=--
ui
p
p is the mass per unit length, L is the length, P
is the tension, (J) is the natural frequency, y is the . nondimensional length = x / L, and W is the transverse deflection shape of the string. Find the natural frequency of vibration of the string using the Galerkin method with the following trial solution: W(y) = cJy(l-
17.4 Find the natural frequency of vibration of the string described in Problem 17.1 using the subdomain collocation method with the trial solution W(y) =Cly(l
Assume that the subdomain is defined by y
y)
Wer)
.-.-._.-.J._._._._._._._._._._.
j.-y~
- y)
L
Figure 17.5
= 0 to ~.
Problems 17.5 Find the natural frequencies of vibration of the string described in Problem 17.1 using the subdomain collocation method with the trail solution
= Cty(l
W(y)
- y)
+ c2y(l
_ y)2
Assume that the subdomains are defined by Yt and Y2
= (i, n·
= (0, k)
17.6 Find the natural frequencies of vibration of the string described in Problem 17.1 using the subdomain collocation method with the trail solution
Assume that the subdomains are defined as Yt = (0, Y2 = (~, and Y3 = 1).
n,
k),
G,
17.7 Consider the eigenvalue problem of the taut string described in Problem 17.1. Find the natural frequency of vibration of the string using the least squares method with the trial solution
= cIy(l
W(y)
= Cty(1
- y)
+ c2y(1
_ y)2
= Cty(1
+ c2y(1
- y)
- y)2
+ c3y(1
where
p is the mass per unit length, I is the length, P is the tension, (J) is the natural frequency, y is the nondimensional length = x / I, and W is the transverse deflection shape of the string. Find the natural frequencies of vibration of the string using the Rayleigh-Ritz method with the following trial solution:
= cIy(1
= k.
17.11 Find the natural frequencies of vibration of the string described in Problem 17.1 using the collocation method with the trail solution W(y)
= cIy(1
- y)
+ c2y(1
Assume the collocation points to be Yt
_ y)2
= k and Y2 = t·
17.12 Find the natural frequencies of vibration of the string described in Problem 17.1 using the collocation method with the trail solution W(y)
= Ciy(1
-
y)
+ c2y(1
- y)2
+ c3y(1
- y)
+ c2y(1
_ y)2
= Cty(1
- y)
+ c2y(1
- y)2
+ c3y(1
_ y)3
17.15 Find the natural frequency of the transverse vibration of the string described in Problem 17.13 using Rayleigh's method with the trial solution W(y)
= cIy(l
- y)
17.16 The natural frequencies of vibration of a tapered bar in axial vibration are governed by the equation
- y)
Assume the collocation point to be YI
= cIy(1
17.14 Solve Problem 17.13 with the trial solution
17.10 Find the natural frequency of vibration of the string described in Problem 17.1 using the collocation method with the trail solution W(y)
k, Y2 = t, and
17.13 Rayleigh's quotient corresponding to the transverse vibration of a string, shown in Fig. 17.5, is given by
W(y)
_ y)3
be Yl =
= S'
W(y)
17.9 Solve Problem 17.7 and find the natural frequencies of the string with the trial solution W(y)
Y3
(0
5
- y)
17.8 Solve Problem 17.7 and find the natural frequencies of the string with the trial solution W(y)
Assume the collocation points
697
_ y)3
d -
dx
[
dU(X)] EA(x)--
dx
+m(x)U(x)w2 =
°
where E is Young's modulus, A(x) is the crosssectional, area, U (x) is the axial displacement shape, m(x) is the mass per unit length, and (J) is the natural frequency. Find the natural frequencies of axial vibration of the bar shown in Fig. 17.6 using the Galerkin method with the trial solution
U(x)
3rrx = CI sin rrx - + C2 sin-21 21
698
Approximate Ana]ytica] Methods Assume the collocation points to be /12.
y]
=
114
and
yz
=
17.21 The natura] frequencies of transverse vibration of a tapered beam are governed by the equation 2
d
~I Figure 17.6
17.17 Ray]eigh's quotient corresponding to the axial vibration of a nonuniform bar is given by R(U(x))
I~EA(x)[dU(x)1 = w Z = --/--
dxf
10 m(x)[U(x)]2
dx2
[
Z
d
E/(x)
W(X)]
dxz
-
=0
pA(x)W(x)wz
where W(x) is the deflection shape, E is the Young's modulus, / (x) is the area moment of inertia of the cross section, p is the density, A(x) is the cross-sectional area, and w is the natural frequency. Find the natura] frequencies of vibration of the tapered beam shown in Fig. 17.7 using the Ga]erkin method with the trial solution
_dx
dx
Find the natura] frequency of vibration of the tapered bar shown in Fig. 17.6 using Ray]eigh's method with the tria] solution •
U(x) =c)sm-
W(x)
'J'{X
2/
17.18 Ray]eigh's quotient corresponding to the axial vibration of a nonuniform bar is given by R(U(x))
=w
Z
I~EA(x)[dU(x)1 ------
=
~-l
dxf
10 m(x)[U(x)]2
dx I ------~
dx d(x)
=
U (x)
. 3'J'{x = c] sm. -'J'{X + c, sm -2/ 2/
17.19 Find the natura] frequencies ofaxia] vibration of the tapered bar described in Prob]em 17.16 and shown in Figure ]7.6 using the subdomain collocation method with the tria] solution U (x)
. 3'J'{x = c] sm. -'J'{x + c, sm -21 21
Assume the subdomains for collocation as and X2 = (114./12).
y]
U(x)
'J'{x
.
Figure 17.7 17.22 Find the natural frequencies of transverse vibration of the tapered beam described in Problem 17.2] and shown in Figure 17.7 using the collocation method with the tria] solution
-TX)4 + cZIX
(
1-
Assume the collocation points to be
XI
=
W(x) = c] ( 1
IX)4 114
and
Xz
= (0, 114) 17.23 Find the natural frequencies of transverse vibration of the tapered beam described in Problem 17.21 and shown in Figure 17.7 using the subdomain collocation method with the trial solution W(x)
= (1- 7)4 +cz7 (1- 7)4 c)
3'J'{x
= c) sm-2/ + c,- sm--21
=
112.
17.20 Find the natura] frequencies ofaxia] vibration of the tapered bar described in Problem 17.16 and shown in Figure 17.6 using the collocation method with the tria] solution .
dox
I
Find the natura] frequencies of vibration of the tapered bar shown in Fig. ]7.6 using the Ray]eigh-Ritz method with the trial solution
Assume the subdomains to be
(0.114)
and
(114, 112).
Problems 17.24 Rayleigh's quotient corresponding to the transverse vibration of a nonuniform beam is given by
Find the natural frequency of the tapered beam shown in Fig. 17.7 using Rayleigh' s method with the trial solution W(x)
=C\
(1- Tf
= (1- Tf + CI
T (1- Tf
c2
where
C
= pAj24E
I.
17.28 Find the natural frequencies of axisymmetric transverse vibration of a circular plate of thickness Iz and radius R clamped at the edge r R using the Rayleigh-Ritz method with the trial solution W(r)
where
17.26 Consider a fixed-free beam in the form of a wedge with width b, maximum depth d, and length L, as shown in Fig. 17.8. Estimate the fundamental natural frequency of vibration of the beam using Rayleigh's method with the following function for transverse deflection: W(x)
17.27 Estimate the fundamental natural frequency of vibration of a uniform fixed-fixed beam using Rayleigh's method. Assume the deflection function to be the same as the static deflection shape under self-weight:
=
17.25 Find the natural frequencies of transverse vibration of the tapered beam shown in Fig. 17.7 using the Rayleigh-Ritz method with the trial solution W(x)
699
=c(I-COS~) . 2L
CI
and
= C2
CI
(1_ ~:)
2
+ C2
(1- ~: y
are constants.
17.29 Estimate the fundamental natural frequency of transverse vibration of a rectangular plate of thickness h and dimensions a and b clamped on all four edges using Rayleigh's method with the trial solution W(x, y)
. rrx
'= csm -
a
. rry
sm-
b
, where c is a constant. 17.30 Estimate the transverse vibration dimensions a and b at the edges using solution
fundamental natural frequency of of a rectangular membrane of under uniform tension P clamped Rayleigh's method with the trial
W(x,y)
. rr x . rry = csm-sma b
where c is a constant.
Free
Fixed
end
end Figure 17.8
.; ~
t •.;.
A Basic Equations of Elasticity A.I
STRESS The state of stress at any point in a loaded body is defined completely in terms of the nine components of stress: CTxx, CTyy, CTZZ' CTxy, CTyx, CTYZ' CTZy, CTzx, and CTxz, where the first three are the normal components and the latter six are the components of shear stress. The equations of internal equilibrium in terms of the nine components of stress can be derived by considering the equilibrium of moments and forces acting on the elemental volume shown in Fig. AI. The equilibrium of moments about the x, y, and z axes, assuming that there are no body moments, leads to the relations (AI)
Equations (A. I) show that the state of stress at any point can be defined completely by the six components CTxx, CTyy, CTzz, CTxy, CTyZ' and CTzx•
A.2
STRAIN-DISPLACEMENT
-
RELATIONS
The deformed shape of an elastic body under any given system of loads can be described completely by the three components of displacement u, v, and w parallel to the directions x, y, and z, respectively. In general, each of these components u, v, and w is a function of the coordinates x, y, and z. The strains and rotations induced in the body can be expressed in terms of the displacements u, v, and w. We shall assume the deformations to be small in this work. To derive the expressions for the normal strain components Exx and Eyy and the shear strain component Exy, consider a small rectangular element OACB whose sides (of lengths dx and dy) lie parallel to the coordinate axes before deformation. When the body undergoes deformation under the action of external load and temperature distribution, the element OA CB also deforms to the shape 0' A'C' B', as shown in Fig. A2. We can observe that the element OACB has two basic types of deformation, one of change in length and the other of angular distortion. Since the normal strain is defined as change in length divided by original length, the strain components Exx and Eyy can be found as Exx
= change in length of the fiber 0 A which lies in the x direction before deformation original length of the fiber
{dx
= 700.:
+ [u + (aujax)dx] dx
- u} - dx
au
= ax
(A2)
A.2
Strain-Displacement
Relations
z
t I I dz I I I I
u
au az
+I--E., dz zt
: I
uyy+
.
I
/dx
t
/
/" x
Figure A.I
Stresses on an element of size dxdydz.
l-
u+ ~
-
dy -:
•.-------
I I I I
IE'
v+ .2!..dy
ay
dy
Figure A.2
Deformation of an element.
C'
au"" ~dy uy
701
702
Basic Equations of Elasticity
= change in length of the fiber
E
0 B which lies in the y direction before deformation
original length of the fiber 0 B
yy
= {dy
+ [v + (ovjoy)dy]
- v} - dy _ ov
dy
(A3)
oy
The shear strain is defined as the decrease in the right angle between fibers OA and OB, which were at right angles to each other before defonnation. Thus, the expression for the shear strain Exy can be obtained as
~=~+~~ dx
[v
+ (ovjox)dx]
+
[u
+
- v
(oujox)dx]
If the displacements are assumed to be small, ou Exy
[u
+
- u
= oy
dy
Exy
+ (oujoy)dy]
+ [v + (ovjoy)dy]-
ow = -oz
Eyz
=
Ezx
=-
ow oy ou
A.3
oz
(A.4)
v
can be expressed as
ov
+ ox
The expressions for the remaining nonnal strain component ponents EyZ and Ezx can be derived in a similar manner as Ezz
- u
(A.5) Ezz
and shear strain com-
(A6) ov
+ oz
(A.7)
ow
+ -ox
(A8)
ROTATIONS Consider the rotation of a rectangular element of sides dx and dy as a rigid body by a small angle, as shown in Fig. A3. Noting that A'D and C' E denote the displacements of A and C along the y and -x axes, the rotation angle a can be expressed as ov
ou
a -- -ox -- - -oy
(A9)
Of course, the strain in the element will be zero during rigid-body movement. If both rigid-body displacements and defonnation or strain occur, the quantity Wz
= ~ (:: - :;)
(A. 10)
can be seen to represent the average of angular displacement of dx and the angular displacement of dy, and is called rotation about the z axis. Thus, the rotations of an elemental body about the x, y, and z axes can be expressed as Wx
= ~ (~; - :~)
(All)
W
= ~ (ou _ ow). 2 oz ox
(AI2)
= ~ (OV _ OU)
(AI3)
y
w. -
2
ox
oy
AA
Stress-Strain Relations
703
B'
ct _u_u_:
uu---uu--------
c'
A;u"B
E
T
I I
:
~
,
+-.J._._.~,
a u
o
D
I-
dx
Figure A.3
A
-I
Rotation of an element.
A.4 STRESS-STRAIN RELATIONS The stress-strain relations, also known as the constitutive relations, of an anisotropic elastic material are given by the generalized Hooke's law, based on the experimental observation that strains are linearly related to the applied load within the elastic limit. The six components of stress at any point are related to the six components of strain linearly as axx
Cll
C12 Cn
C16
Exx
ayy
C21
C22
C23
C26
Eyy
azz
C31
C32
C33
C36
ayZ
=
azx axy
Ezz
(A. 14)
EyZ Ezx C61
C62
C63
C66
Exy
where the Cij denote one form of elastic constants of the particular material. Equation (A.14) has 36 elastic constants. However, for real materials, the condition for the elastic energy to be a single-valued function of the strain requires the constants Cij to be symmetric; that is, Cij = Cji. Thus, there are only 21 different elastic constants in Eq. (A.14) for an anisotropic material. For an isotropic material, the elastic constants are invariant, that is, independent of the orientation of the x, y, and z axes. This reduces to two the number of independent elastic constants in Eq. (A.14). The two independent elastic constants, called Lame's elastic constants, are commonly denoted as A and J1.. The Lame constants are related
704
Basic Equations of Elasticity
to Cij as follows: = C22 = C33 = A + 2f.J, C)2 = C21 = C31 = Cl3 = C32 = C23 = A C44 = C 55 = C66 = f.J, C))
all other
(A. 15)
0
Cij -
Equation (A.I4) can be rewritten for an elastic isotropic material as = ),,6. + 2/.LExx = A~ + 2J1.Eyy crzz = AD. +2J.LEzz
(Jxx (fyy
(A.I6)
uY1. = f.J,£y1. u1.x = f.J,£1.x uxy = f.J,Exy
where A = Exx
+ Eyy + E1.1.
(A.I7)
denotes the dilatation of the body and denotes the change in the volume per unit volume of the material. Lame's constants A and f.J, are related to Young's modulus E, shear modulus G, bulk modulus K, and Poisson's ratio 11 as follows: E = f.J,(3A+ 2f.J,)
(A.I8)
A+f.J, G=f.J,
(A.19)
K = A + ~f.J,
=_
(A. 20)
A
11
(A.2I)
2(A + f.J,)
or A=
liE
(A.22)
(1 + 11)(1 - 211)
E
f.J,-G - 2(1 + 11),-
A.5
(A.23)
EQUATIONS OF MOTION IN TERMS OF STRESSES Due to the applied loads (which may be dynamic), stresses will develop inside an elastic body. If we consider an element of material inside the body, it must be in dynamic equilibrium due to the internal stresses developed. This leads to the equations of motion of a typical element of the body. The sum of all forces acting on the element shown in Fig. A.I in the x direction is given by ~ LFx=
( uxx+--a;-dx auxx)
+
(
Uzx
+ au. a~x)dz
auu
= -'ax
dxdydz
( uxy+
dydz-uxxdydz+
dxdy auxy
+ --dxdydz ay
xv) au ay'dy
dxdz-uxydydz
- u1.xdxdy au
zx + -dxdydz az
(A. 24)
A.6
Equations of Motion in Temls of Displacements _-1:1.)5
According to Newton's second law of motion, the net force acting in the x direction must be equal to mass times acceleration in the x direction:
L Fx = pdxdydz
a2u
(A.25)
at2
where p is the density, u is the displacement, and a2u/at2 is the acceleration parallel to the x axis. Equations (A.24) and (A.25) lead to the equation of motion in the x direction. A similar procedure can be used for the y and z directions. The final equations of motion can be expressed as
aaxx oaxy -+--+-=pax ay aaxy aayy -+-+-=pax ay aazx aayZ -+-+-=pax ay
a2u at2 a2v at2 a2w at2
aazx az aayZ az aazz az
(A.26)
(A.2?)
(A.28)
where u, v, and w denote the components of displacement parallel to the x, y, and z axes, respectively. Note that the equations of motion are independent of the stress-strain relations or the type of material ..
A.6
EQUATIONS OF MOTION IN TERMS OF DISPLACEMENTS Using Eqs. (A.16), the equation of motion, Eq. (A.26), can be expressed as
a
a
a
a2u
ax ()..D. +2J.LExx)+ay (J.LExy)+ az (J.LExz)= P at2 Using the strain-displacement Eq. (A.29) can be written as
(A.29)
relations given by Eqs. (A.2), (A.4), and (A.8),
2 au)] +-a [ J.L(aw au)] =p_.a u -a ( AD.+2J.L-au) +-a [ J.L(av -+-+ax ax ay ax ay az ax az at2 (A.30) which can be rewritten as (A.31 ) where !:J. is the dilatation and V2 is the Laplacian operator:
a2
a2
a2
ax
ay2
az2
V2=-+-+2
(A.32)
706
Basic Equations of Elasticity
Using a similar procedure, the other two equations of motion. Eqs. (A.27) and (A.28), can be expressed as 0.6.
().+ J-L) - oy + J-LV2v ().+ J-L)
az + 8.6.
2
J-LV w
02v = P -0 2 t
(A.33)
02w
= P ot2
(A.34)
The equations of motion, Eqs. (A.31), (A.33), and (A.34), govern the propagation of waves as well as the vibratory motion in elastic bodies.
B Laplace and Fourier Transforms Table B.l
Laplace Transfonns
=
F(s)
Serial number
f(t)
L[f(t)]
1'' '
=
st
e- f(t)dt
s 2
S2
n!
3
t", n = 1, 2, ...
4
ta, a > -1
r(a
5
eat
~,s
6
t"eat, n = 1, 2, ...
(s -a)"+
7
sin at
8
cos at
9
t sin at
10
t cos at
11
eat sin bt
12
eat cos bt
13
sinh at
14
cosh at
s"+1
+
sa+
1)
,s
1
> a
1 > a
n! l's
> a
a s2
+
s2
+
a2
s a2
2as (S2
+ a2)2
s2 _
(s2
+
a2
a2)2
b (s - a)2
+
b2
s - a (s - a)2
+
b2
707
."
J
708
Laplace and Fourier Transforms
Table B.t
(continued)
=
F(s) Serial number
15
t sinh at
16
t cosh at
17
H(t
18
/j (t -
19
.' - afl (t)
L[f(t)]
:::: 1
00
f(t)
e-st f(t)dt
2as (S2
""'t'~
a2)2
_
+
s2
(s2 -
a2 a2)2
e-as
- a)
2: a
--,S
s
a)
+
bf2(t)
20
f(at)
21
eat f(t)
F(s
22
f'(t)
sF(s)
-
a) -
f(O)
sn F(s) - sn-I f (0) - sn-2 f' (0)
24
l 1
- ... - f(n-I)(o) f(u)
F(s)
du
s
t
25
f(u)g(t
- u) du
F(s)G(s)
fit)
26
'Vvvv . o
27
T
2T
3T
:~. o
-
1
TS2
T
2T
3T
TS
tanh 2
Laplace and Fourier TransfonllS
Table B.l
(continued)
=
F(s)
L[f(t)]
1
00
f(t)
Serial number
=
j{t) 1
28
o
3r
4r
Sf
1 TS t - tanh -
6r
2
S
-1
29
It It IV<' j{t)
30
e-as S
0
a
j{t)
31
-as
_e_(1_ s ,
0
j{t)
32
0
=
{sin
!!!..,
a
e-U)
t
a+'t
O::;t::;a
a
rra(1
a2s2 ) t a
+ e-as) + rr2
e-sr f(t)dt
709
710
Laplace and Fourier Transforms Table B.2
Fourier Transform Pairs
Serial number
I{x)
F(w)
_1_100 ./2ii
=
l(x)e-iWXdx
-00
I, { 0,
4 5
{ e -ax , 0, ax { e , 0,
f!sin aw
(a > 0)
x>O x
1
e-iaw _ e-ibw
./2ii
iw 1
1 ----./2ii a + iw
(a > 0)
b
e(a+iw)c_ e(a+iw)b
1
(a > 0)
./2ii
e-a1xl, a > 0
a - iw Ha 7r
6
7
H H
xe-a1x" a > 0 iax { e , 0,
w
7r
a
{ 1, 0,
2 3
Ixl a
8
e-a2.x2, a> 0
9
Jo(ax), a > 0
10
8 (x - a), a real
2iaw ;; (a2 + (2)2
-
Ixl b
a2 + w2
sin b(w - a) w-a
7r
--e1 ../ia
-(w2/4a2)
H
H(a -Iwl) ;; (a2 _ (2)1/2 --e1
-iaw
./2ii
12
sin ax x iax e
13
H(x)
i4
H(x - a)
11
15
If
H (a -
./2ii 8(w -
If If'2
iwa [e-i7rw H-~ 7r
-i
x 17
xn
18 19
1 (x - a)n
-
2
-i
x-a -i
If
'If If''2 If .
-I
a)
L:w + 8{W)]
H(x) - H(-x)
16
Iwl)
+ 8(w) ] w sgn w
[(-iW)n-1 sgn w] (n - I)! e-,wa sgnw
- e-'OW 2
(_iw)n-J sgn w (n - I)!
·..·1
Lapiace,and-F.ourier
Table B.3
Transforms
711
Fourier Cosine Transforms
1
00
Serial number
f(x)
1
e-ax, a >0
2
xe-ax
3
e-a2x2
4
H(a - x)
~a ~
+ w2
-; a2
2 2 a _ w 2)2 -; (a2
+(
1
_",2/402
../2 a ~
sin aw 11:
W
~ - r(a) - 1 cos -a11: 11: wa 2 2
2 (wcos 4a
6
cos ax2
-- 1
7
sin ax2
2 .;a
Table B.4
cos(wx) f(x)dx
--e
xa-I,O
5
F(w) = ~
2.;a
+ sin -(4a)
2
1
2 ( )
(w
cos 4a - sin 4a
Fourier Sine Transforms
1
00
Serial number
f(x)
1
e-ax, a >0
2
3
xe-ax,
xa-I,
6 7
a>
~
0
r(a)
- -- a 11: w 1
.Ji
.JW' a > 0
xe-a2x2 x
+ x2
w 2 + w
2aw -; (a2
~
sin(wx) f(x)dx
; a2
1
x-Ie-ax,
a2
=~ ~
0
4
5
F(w)
+ (2)2 .
11:a
sm -
2
w > 0
W ~ - tan- I -, w > 0 11: a 2-3/2
!!!.. e-",2/4a1 a3
~
Ie
-a",
oj
Index A Admissible function. 163 Amplification factor, 36 Amplitude, 1, 22 Amplitude-modulated carrier, 630 Amplitude ratio. 36 Antinodes, 612 Approximate analytical methods. 647 Assumed modes method, 139,670 Axial force in beams, 352 Axisymmetric modes of shells, 601 B Bar in axial vibration. 69, 234 Basic equations of elasticity. 700 Beam in transverse vibration. 71. 317 under axial force, 352 bending-torsional vibration, 380 boundary conditions, 323 continuous, 359 on elastic foundation, 364 Euler-Bernoulli theory, 317 flexural waves, 628 free vibration. 185, 197 infinite length, 385 under moving loads, 350 Rayleigh's theory, 369 Timoshenko theory, 371 transform method, 385 Beat, 23 Beat frequency, 23 Bessel functions, 445, 491 Bessel's differential equation, 445, 491 Bishop's theory, 260 Boundary conditions bar in axial vibration, 71, 239 beam in transverse vibration, 73, 323 circular plate, 489 extremization of functional, 100 forced, 10 1 free, 10 1 geometric, 10 I kinematic, 101 Kirchhoff, 469
membrane, 421 natural. 10 I plate. 79, 465 shaft. 281 shell, 579, 596 skew edge of a plate. 469,508 string, 209,211 thick plate, 508 C Calculus of a single variable, 85 Calculus of variations, 86 Cauchy-residue theorem, 193 Characteristic equation, 46, 219 Characteristic vector, 46 Circular frequency, 22 Circular membrane, 444 forced vibration, 448 free vibration, 444 mode shapes, 447 Circular plate, 485 equation of motion, 485 forced vibration, 495 free vibration, 490, 533 mode shapes, 494,516 variable thickness, 531 Circular rings, 393 classification of vibration, 397 equations of motion, 393 extensional vibration, 407 in-plane flexural vibrations, 398 torsional vibration, 406 twist-bending vibration, 402 Classical plate theory, 457 Collocation method, 144,680 boundary method, 680 interior method, 680 mixed method, 680 Comparison function, 163 Complementary energy. 105 Complex frequency response, 41 Complex number representation, 20 Compressional waves, 623
713
'J
714
Index Concept of vibration, 1 Conical shell, 544, 560, 591 Constitutive relations, 703 Constraints, 96 Continuous beams, 359 Continuous systems general, 132 history, 8 literature, 29 notable contributions, 12 Convolution integral, 42 Convolution theorem, 190 Coupled bending-torsional vibration, 380 Critical damping constant, 35 Curved beams, 393 thick, 414 thin, 408 Cylindrical shell, 543, 559, 582 axisymmetric modes, 601 Donnell-Mushtari- Vlasov theory, 584 equations of motion, 583 natural frequencies, 584, 598 rotary inertia and shear deformation, 592 D D' Alembert force, 69 D'Alembert's principle, 69 D' Alembert's solution, 210, 608 Damped harmonic response, 40 Damped vibration, 15 Damping matrix, 43 Damping ratio, 35, 171 Definitions, 21 Degree of freedom, 11 Derivation of equations, 85, 123, 125 Dilatational waves, 631 Dirac delta function, 152, 680 Discrete systems, 11, 13, 33 Dispersive medium, 629 Distortional waves, 632 Distributed systems, 14 Donnell-Mushtari- Vlasov theory of shells, 584 natural frequencies, 584 Duhamel's integral, 42 E
Eigenfunctions orthogonality, 133, 161, 165,246 properties, 160 Eigenvalue problem, 45, 163
formulation, 130 positive definite, 164 self-adjoint, 163 semidefinite, 164 Eigenvalues, 46 properties, 160 Eigenvector, 46, 55 Elastic element, 1 Elastic foundation, 364, 521 Elasticity, 700 basic equations, 700 constitutive relations, 703 equations of motion, 704 Lame's elastic constants, 703 rotations, 702 strain-displacement relations, 700 stress, 700 stress-strain relations, 703 Elastic wave propagation, 607 in infinite elastic medium, 631 traveling-wave solution, 608 Equation(s) of motion, 704 beam in transverse vibration, 71, 371 circular cylindrical shell, 583 circular plate, 485, 515 circular ring, 393, 398, 402 conical shell, 591 coupled bending-torsional vibration of beams, 380 curved beam, 408 integral, 127 longitudinal vibration of bars, 69, 235, 236, 258,260 membrane, 420, 444 plate with in-plane loads, 523 plate in transverse vibration, 73, 457 shell, 575 spherical shell, 591 thick beam, 371 thick shell, 595 thin beam, 317 torsional vibration of shafts, 271 transverse vibration of string, 205 Equilibrium approach, 68 bar in axial vibration, 69, 234 beam in transverse vibration, 71 D'Alembert's principle, 69 membrane, 420 Newton's second law, 68 plate with in-plane loads, 523 plate in transverse vibration, 73 457
shaft vibration, 271 string vibration, 205 thick plate. 499 thin ring. 393 Equivoluminal waves, 634 Euler-Lagrange equation, 89, 92. 95, 96, 109 Exceptional part, 123 Excitations. 17 Expansion theorem. 48, 151, 161, 166 Extensional vibration of circular rings, 407 Extremization of functional, 86, 96 F Finite-dimensional systems, 13 First approximation shell theory, 556 Flexural rigidity of plate, 78, 458 Flexural waves, 628 Forced response, 169 Forced vibration beam, 198, 344 beam on elastic foundation, 366 circular plate, 495 definition, 16 finite string, 183 longitudinal vibration of bar, 254, 264 membrane, 438, 448 under moving load, 367 multidegree-of-freedom system, 52, 53, 54 plate, 479 shaft in torsional vibration, 292 single-degree-of-freedom system, 36 string, 227 Foundation modulus, 364,521 Fourier integral, 26 Fourier series, 24, 175 Fourier transform pair, 27 sine and cosine, 178 Fourier transforms, 175, 707 membrane vibration, 441 string vibration, 213 Free vibration beam, 185, 325 beam on elastic foundation, 364 circular plate, 490 definition, 16 finite string, 181, 194,217 infinite beam, 385 longitudinal vibration of bar, 236, 249 membrane, 426, 444 Mindlin plate, 511 multidegree-of-freedom system, 47, 52
plate with in-plane loads. 528 plate with variable thickness. 533 rectangular plates. 471 shaft in torsional vibration. 276, 288 single-degree-of-freedom system, 33 string of infinite length, 210 transverse vibration of beams. 341 Frequency, I Frequency equation. 46 plate, 480 string, 219 Frequency ratio. 39 Frequency spectrUm,25 Functional, 86, 91. 93. 95 Fundamental frequency, 24, 46 G Galerkin equations, 674 Galerkin method, 143,673 Gauss points, 146 General force, 41 Generalized coordinates, 48, 109 General viscous damping, 54, 679 Green's function, 126. 131, 132 Group velocity, 629 H Hamilton's principle, 107 applications, 115 conservative systems, 109 continuous systems, III discrete systems, 108 generalized, III longitudinal vibration of bars, 235 nonconservative systems, 110 shell, 575 system of masses, 110 Timoshenko beam, 371 torsional vibration of a shaft, 115 transverse vibration of a thin beam, 116, 317 use of generalized coordinates, 109 Harmonic analysis, 24 Harmonic force, 36 Harmonic functions, 18 Harmonic motion definition, 1 representation, 18 Harmonics, 25 Harmonic waves, 610 Heaviside, 174
716
Index Homogeneous equations, 153 solution, 153 I
Importance of vibration, 4 Impulse response function, 126 Inertial element, 1 Infinite beam, 385 Infinite-dimensional systems, 14 Infinite elastic medium, 631 Initial conditions, 48 beam, 341 membrane, 421 shaft, 289 string, 209, 613 Initial excitation longitudinal vibration of bars, 249 In-plane flexural vibrations of rings, 398 In-plane loads on a plate, 523 Integral equation, 123 approach, 123 assumed modes method, 139 classification, 124 collocation method, 144 first kind, 124 Fredholm type, 124 Galerkin method, 143 homogeneous, 124 iterative method, 134 linear, 124 nonlinear, 123 normal, 125 numerical integration method, 146 Rayleigh-Ritz method, 139 second kind, 124 singular, 125 solution, 133 third kind, 124 Volterra type, 124 Integral transform, 174 Integral transform methods, 174 Integrodifferential equation, 123 Inverse transforms, 42, 193 Irrotational waves, 634 Isoperimetric problem, 96 Iterative method, 134 K Kernel, 123, 174 Kinetic energies of structural elements, 652
Kirchhoff boundary condition, 469 Kirchhoff's hypothesis, 556 L Lagrange equations, 109, 140,672 Lagrangian, 107, 109 Lame parameters, 543, 703 Lame's elastic constants, 543, 703 Laplace equation, 303 Laplace transforms, 41, 188,707 convolution theorem, 190 partial fraction method, 191 properties, 189 shifting property, 189 string vibration, 215 Least squares method, 686 Left eigenvector, 55 Linear frequency, 22 Linear vibration, 16 Literature on vibration, 29 Longitudinal vibration of bars, 234 Bishop's theory, 260 boundary conditions, 236, 238, 239 equation of motion, 235, 236 forced vibration, 254 free vibration, 236, 237 initial excitation, 249 mode shapes, 247, 259, 262 natural frequencies, 236, 259, 262 orthogonality of eigenfunctions, 246 Rayleigh theory, 258 support motion, 257 wave solution, 237 Longitudinal waves, 634 Love's approximations, 556 Lumped-parameter systems, 13 M Magnification factor, 36 Mass element, 1 Mass matrix, 43 Membrane analogy, 308 Membranes, 420 circular, 444 forced vibration, 438, 448 Fourier transform approach, 441 free transverse vibration, 130 irregular shape, 452 mode shapes, 430 partial circular, 453 rectangular, 426 Method of undetermined coefficients, 134
Mindlin theory of plates. 499 circular plate. 515 free vibration. 511 Modal analysis approach. 151. 167 forced vibration. 292 forced vibration of bars. 264 free vibration. 289 membranes, 438. 448 multidegree-of-freedom system. 52. 54 in state space. 54 torsional vibration of shafts, 289, 292 Modal coordinates, 48 Modal vector. 46 Mode shapes beam. 326 circular plate. 494 cylindrical shell, 601 longitudinal vibration of bar. 259. 262 membrane, 430, 447 rectangular plate. 480 string, 220 Moving load on a beam, 350, 367 Multidegree-of-freedom system, 43 eigenvalue problem, 45 equations of motion, 43 expansion theorem. 48 forced vibration analysis. 52, 53, 54 free vibration analysis, 47 generalized coordinates, 48 modal analysis, 47, 54 modal coordinates, 48 modal matrix. 47 orthogonality of modal vectors. 46 N Natural frequency beams. 326 circular plate, 520 circular rings. 398 cylindrical shell. 584. 598 definition, 23 Donnell-Mushtari- Vlasov theory of shells. 584 longitudinal vibration of bars. 259, 262 Love's theory of shells, 587 rotating beam. 359 shaft in torsional vibration. 277 single-degree-of-freedom system, 33 thick beams, 377 thick rings. 40 1
n-degree-of-freedom system. 44 Newton's second law of motion, 68 longitudinal vibration of bars, 234 Nodal lines. 430 Nodes. 612 Noncircular shafts, 295 torsional rigidity. 303 Nonhomogeneous equation. 167 Nonlinear vibration, 16 Nonperiodic function, 26 Nonperiodic motion. I Nonnalization, 46 Nonnal modes, 46 orthogonality, 339 Nucleus. 123 Numerical integration method. 146
o Orthogonality of eigenfunctions. 162. 165 in integral fonnulation. 133 longitudinal vibration of bars, 246 torsional vibration of shafts, 286 transverse vibration of beams. 339 Orthogonality of modal vectors, 46
P Partial fraction method. 191 Period,22 Periodic functions, 24 Periodic motion, I Phase angle. 22. 41 Phase difference, 23 Phase velocity. 612 Plate in transverse vibration. 73, 457 additional contributions. 80 boundary conditions, 79 circular plate, 485 equations of motion. 78 flexural rigidity, 78, 458 forced vibration, 495 free vibration, 471. 490 initial conditions, 79 with in-plane loads. 523 Mindlin theory, 499 mode shapes, 480. 494 moment-displacement relations, 78 on elastic foundation, 521 rotary inertia and shear defonnation, 499 state of stress, 75 strain-displacement relations. 76 variable thickness. 529
718
Index Potential energy, 104 Prandtl's membrane analogy, 308 Prandtl stress function, 303 Primary (P) waves, 634 Principle of minimum complementary energy, 105
Principle of minimum potential energy, 104 Principle of stationary Reissner energy, 106 Propenies eigenfunctions, 160 eigenvalues, 160 Proponional damping, 53, 678 P waves, 623 R Random vibration, 1 Rayleigh-Ritz method, 139,661 Rayleigh's method, 650 Rayleigh's principle, 650 Rayleigh's quotient, 648 Rayleigh theory, 258, 369 Rayleigh waves, 635 Recent contributions approximate analytical methods, 693 circular rings and curved beams, 416 elastic wave propagation, 643 integral equation approach, 147 integral transform methods, 201 longitudinal vibration of bars, 267 membrane vibration, 453 modal analysis approach, 171 multidegree-of-freedom systems, 60 vibration of plates, 535 torsional vibration of shafts, 313 transverse vibration of beams, 387 variational approach, 119 vibration of shells, 603 vibration of strings, 231 Rectangular plate, 471 boundary conditions, 465 on elastic foundation, 521 equation of motion, 457 forced vibration, 479 free vibration, 471 frequency equations, 480 with in-plane loads, 523 mode shapes, 480 rotary inenia and shear deformation, 499 subjected to in-plane loads, 523 variable thickness, 529 Reference kinetic energy, 662
Reflection of waves, 617, 619, 622 Regular part, 123 Reissner energy, 106 Residual, 144,673 Resonance, 38 Right eigenvector, 55 Ritz coefficients, 661 Rotary inertia, beams, 369 circular rings, 399,403 plates, 499 shells, 592 Rotational waves, 634 Rotations, 702 S Saint-Venant's theory, 295 Scotch yoke mechanism, 19 Self-adjoint eigenvalue problem, 163 Separation of variables, 153 bar vibration, 237 beam vibration, 325 membrane vibration, 426, 444 rectangular plate, 471 string vibration, 217 Shafts torsional propenies, 310 in torsional vibration, 271 Shear correction factor, 372 Shear deformation beams, 371 circular rings, 399, 403 curved beams, 414 plates, 499 shells, 592 Shear waves, 623, 625 Shell coordinates, 541 Shells, 541 boundary conditions, 579 conical shell, 544, 560, 568, 591 cylindrical shell. 543, 559, 568, 582 Donnell-Mushtari- Vlasov theory, 584 first quadratic form of surface, 543 force and moment resultants, 563 kinetic energy, 573 Kirchhoff's hypothesis, 556 Lame parameters, 543 Love's approximations, 556 rotary inenia and shear deformation, 592 spherical shell, 546, 561, 570, 591 strain-displacement relations, 552
-- ··1
strain energy, 571 stress-strain relations, 562 theory of surfaces. 541 Single-degree-of-freedom system, 33 critically damped. 36 damped harmonic response, 40 forced vibration, 36, 41 free vibration, 33 under general force, 41 under harmonic force. 36 overdamped, 36 underdamped, 35 Skew plate, 540 Solid mechanics. 104 Spectral diagram, 25 Spherical shell, 546, 561, 591 Spring element, 1 Standing wave, 612 State space, 54 State vector, 54 Static deflection, 39 Stiffness matrix, 43 Strain-displacement relations, 700 Strain energies of structural elements, 652 Strain energy, 104 Stress, 700 Stress-strain relations, 703 String boundary conditions, 211 finite length, 194 forced vibration, 183 free vibration, 181 harmonic waves, 611 infinite, 210 transverse vibration, 205 traveling wave solution, 210 wave motion, 611 Sturm-Liouville problem, 154 classification, 155 periodic, 155 regular, 155 singular, 155 Subdomain method, 684 Support motion longitudinal vibration of bars, 257 Surface waves, 635 S waves, 625 Synchronous motion, 23 T Terminology, 21 Theory of surfaces. 541
Three-dimensional vibration of circular ring, 393 Timoshenko-Gere theory. 300 Torsional properties of shafts. 310 Torsional rigidity, 303 Torsional vibration of circular rings, 406 Torsional vibration of shafts. 115, 271 elementary theory, 271 forced vibration, 292 free vibration, 276, 289 noncircular shafts, 295, 299 Timoshenko-Gere theory, 300 Transformation of relations. 486 Transform method in beams, 385 Transform pair, 175 Transient motion, I, 4 Transients, 42 Transmission of waves, 619 Transverse vibration of plates, 457 boundary conditions, 465, 489 circular plates, 485 on elastic foundation, 521 equation of motion, 457 forced vibration, 479 free vibration, 471, 511 frequency equations. 480 Mindlin theory, 499 mode shapes, 475, 480 rotary inertia and shear deformation, 499 with variable thickness, 529 Transverse vibration of strings, 205 Transverse vibration of thin beams, 71,116, 185,317 under axial force, 352 coupled bending-torsional vibration, 380 on elastic foundation, 364 equation of motion, 317 Euler-Bernoulli theory, 317 flexural waves, 628 forced vibration, 344 frequencies and mode shapes, 326 infinite length, 385 with in-plane loads, 523 on many supports,359 under moving load, 350, 367 orthogonality of normal modes, 339 Rayleigh's theory, 369 response due to initial conditions, 341 rotating, 357 Timoshenko theory, 371 transformation methods, 385
720
Index Traveling wave solution, 210, 608
U Uncoupled equations, 48 Undamped system, 47, 52 Undamped vibration, 15 Underdamped system, 35
V
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Variational approach, 85 membrane, 423 plate, 458 shaft, 272 string, 235 thick plate, 505 Variational methods, 85 in solid mechanics, 104 Variation operator, 89 Vibration analysis, 16 beams, 317, 369,371 beams on elastic foundation, 364 circular cylindrical shell, 582 circular rings, 393,406 concept, 1 continuous beam, 359 curved beams, 393, 408 developments, 5 forced, 52, 53 free, 47 history, 8 importance, 4 membranes, 420 multidegree-of-freedom system, 43 origins, 5 plates, 457, 485, 499 problems, 15 rotating beam, 357 shafts, 271 shells, 54]
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Wave equation D'A]embert's solution, 608 membrane, 421 one-dimensional, 607 string, 207 traveling-wave solution, 608 two-dimensional,610 Wavelength, 612 Wave number, 612 Wave packet, 629 Wave propagation, 607 in infinite elastic medium, 631 traveling wave, 608 Waves compressional, 623 dilatational, 631 distortional, 631 equivo]uminal, 634 flexural, 628 harmonic, 611 irrotationa], 634 ]ongitudinal, 634 primary, 634 P, 623, 634 Rayleigh, 635 rotational, 634 shear, 623, 625 standing, 612 surface, 635 S,625 traveling, 210, 608 Wave solution compression a] waves, 623 dilatational waves, 631 distortional waves, 632 flexural waves, 628 graphical interpretation, 614 group velocity, 629 interface of two materials, 619 membrane, 425 P waves, 623 Rayleigh waves, 635 reflection of waves, 617, 622 shear waves, 623, 625 string, 210, 611 surface waves, 635 S waves, 625 transmission of waves, 619 wave packet, 629 Weighted residual methods, 673