Theory of elasticity - Timoshenko.pdf

r

iJr

r

ao

r

r

a q, sin o 2

(37)

+ ao

2

2

~

(b)

ln the sarne manner we find

ª"'àr + .lr2a2q, a8 2

a2q, = a2 q, sin 2 0 + 2 a2 q, sino cos o + aq, cos2 8 ay 2 àr 2 ao àr r ar r _ aq, sin 8 cos O + à2 q, cos 2 O (e) 2 ao r2 à8 2 r2 Adding together (b) and (e), we obtain

a2q, àr 2

(38)

a2q, ax 2

(1 ª"')

a r ao. ar

a

where q, is the stress function as function of r and 8. This of couràe may be verified by direct substitution. A derivation of (38) is included in what follows. i To yield a possible stress distribution, this function must ensure that the condition of compatibility is satisfied. ln Cartesian coordinates (see page 26) this condition is

iJ4q, ax4

+

To get the second derivative with respect to x, it is only necessary to repeat the above operation; hence

=

These equations take the place of Eqs. (18) when we solve twodimensional problema by means of polar coordinates. When the body force Ris zero they are satisfied by putting

11

57

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

THEORY OF ELASTIC ITY

éJ4q, + 2 ax2 ày2

ª4"' + ay4 =O

+

a2 q, a2 q, ay 2 = àr 2

1 aq,

1 a2 q,

+ r àr + Ti 80

(d)

2

Using the identity

a4q, àx'

a4q,

a•q,

+ 2 ax2 ay2 + ày4 =

( 82 axz

a2 )

+ ayz

(ª2"' 2"') ax2 + ay2 0

and Eq. _(d), the compatibility equation (a) in polar coordinates becomes 2 a22) ( ara + r1 ara + Ti1 à0 2

(a)

(ªàr2+ r aq,àr + T2 aà0q,)2 2

"'

1

1

2

=

o

(39)

for the present purpose. we need this equation transfortned to polar roºrdinates.~ _Th~ relatiQn between pola:r·and Q.11.rtes_ian coordinate,s-is given by

From various solutions of this partial differential equation we obtain solutions of two-dimensional problema in polar coordinates for various boundary conditions. Several examples of such problems will be discussed in this chapter.

fróniwhich

The first. and second of the expressions (38) follow from Eqs. (b) and (e). If we choose any point in the plate, and let the x-axis pass through it, we have O = O, and
ar X = - =coso àx r · ' sin ao _ JL = iJx = r2 r -

iJr = '!!.. = sin O _ày _ r, ··-·0, aO X COS 0 ay = T2 = - r -

___

(J

=o,

59

THEORY OF ELASTICITY

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

This expression continues to represent
r = O. Hence, for a plate without a hole at the origin and with no body forces, only one case of stress distribution symmetrical with respect to the axis may exist, namely that when ur = u9 = constant and the plate is in a condition of uniform tension or uniform compression in all directions in its plane. If there is a hole at the origin, other solutions than uniform tension or compression can be derived from expressions (43). Taking B as zero, 1 for instance, Eqs. 43 become

58

and the third expression of (38) can be obtained likewise by finding the expressiori for -a•q,/éJx éJy analogous to (b) and (e).

26. Stress Distribution Symmetrical about an Axis. lf the stress distribution is symmetrical with respect to the axis through O perpendicular to the xy-plane (Fig. 40), the stress components do not depend on (J and are functions of r only. From symmetry it follows also that the shearing stress Tro must vanish. Then only the first of the two- equations of equilibrium (37) remains, and we have

au.+~+R =0 ar

(40)

r

If the body force R is zero, we may use the stress function cf>. When this function depends only on r, the equation of compatibility (39) becomes

2 1 dq,) d2 1 d) (d q, (dr 2 + rdr dr 2 + r dr

d4q,

2 d 3 q,

+ r dr

3 -

2 + 1 dcf> dr 2 dr -

1d

T2

ra

o

(41)

This is an ordinary differential equation, which can be reduced to a linear differential equation with constant coefficients by introducing a new variable t such that r = e1• ln this manner the general solution of Eq. (41) can easily be obtained. This solution has four const~n.ts of integration, which must be determined from the boundary cond1t10ns. By substitution it can be checked that cf> = A log r + Br 2 log r + Cr 2 + D (42) is the general solution. The solutions of all problems of symmetrical stress distribution and no body forces can be obtained from this. The corresponding stress components from Eqs. 38 are r

= -1acf> - = -A + B(l r ar r2

a2q, u8 = - 2 iJr

Tr8

= -

+ 20 (44)

A

-+ 2C r2

This solution may be adapted to represent the stress distribution in a hollow cylinder submitted to uniform pressure on the inner and outer surfaces 2 (Fig. 41). Let a and b denote theinner and outer radii of the cylinder, and p; and Po the uniform interna! and externa! pressures. Then the boundary conditions are:

(a)

= - 2A r

Substituting in the first of Eqs. (44), we obtain the following equations to determine A and C: A -a2 + 2C =-pi

A

b2

+ B(3 + 2 log r) + 2C

-pº

Substituting these in Eqs. (44) the following expressions for the stress components are obtained:,

(43)

= 0

+ 2C =

from which

+ 2 log r) + 2e

If there is no hole at the origin of coordinates, constants A and B vanish since otherwise the stress components (43) become infinite when:

'

U9

A r

= 2

(ur)r=a = -p;,

= dr4

u

Ur

a 2b2 (Po -

Ur

=

uo

= -

b2 - a2

Pi)

a 2b2 (Po -

b2 - a2

1 • r2

Pi)

+

Pob 2 b2 - a2

p;a 2

2

2

Pob . -1 + p;a .,____ - _;__ r2

(45)

b2 - a2

Proof that B must be zero requires consideration of displacements. See p. 68 The solution of this problem is due to Lamé, "Leçons sur la théorie . . . de l'élasticité," Paris, 1852. · 1

1

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

, It is interesting to note that the sum
27. Pure Bending of Curved Bars. Let us consider a curved bar with a constant narrow rectangular cross section 1 and a circular axis bent in the plane of curvature by couples M applied at the ends (Fig. 42). The bending moment in this case is constant along the length of the bar and it is natural to expect that the stress distribution is the sarne in all radial cross sections, and that the solution of the problem

a2p;

(

b2) 1 - r2


= b2 - a2


= b2a?;a2 ( l

+ ~9

(ue)max. is always numerically greater than the internai pressure and approaches this quantity as b increases, so that it can never be reduced below p;, however much material is added on the outside. Various applications of Eqs. (46) and (47) in machine design are usually discussed in elementary books on the strength of materials. 1 The corresponding problem for a cylinder with an eccentric bore is considered in Art. 66. It was solved by G. B. Jeffery. 2 If the radius of the bore is a and that of the externai surface b, and if the distance between their centers is e, the maximum stress, when the cylinder is under an internai pressure p;, is the tangential stress at the internai surface ai the thinnest part, if e < }-a, and is of the magnitude

,•·


=

2b 2(b 2 + a 2 - 2ae - e2) ] p; [ (a 2 + b2) (b 2 - a 2 - 2ae - e2) - 1

If e = O, this coincides with Eq. (47). See, for instance, S. Timoshenko, "Strength of Materiais," vol. 2, p. 236, 1941. Trans. Roy. Soe. (London), series A, vol. 221, p. 265, 1921. See also Brit. Assoe. Advancement Sei. Repts., 1921. 1 1

ll

o

(46)

These equations show that
!'

61

THEORY OF ELASTICITY

60

Fm. 42.

can therefore be obtained by using expression (42). Denoting by a and b the inner and the outer radii of the boundary and taking the width of the rectangular cross section as unity, the boundary conditions are (1)
(2)

f

(3)

ue dr = O, Tr8

=

Lb u9r dr =

-M

(a)

O at the boundary

Condition (1) means that the convex and concave boundaries of the bar are free from normal forces; condition (2) indicates that the normal stresses at the ends give rise to the couple M only, and condition (3) indicates that there are no tangential forces applied at the boundary. Using the first of Eqs. (43) with (1) of the boundary conditions (a) we obtain A 2 + B(l + 2 log a) + 2C = O a A

b2

1

+ B(l + 2 log b) + 2C =O

(b)

From the general discussion of the two-dimensional problem, Art. 15, it follows that the solution obtained below holds also for another extreme case when the dimens~on of the cross section perpendicular to the plane of curvature is very large, as, for mstance, in the case of a tunnel vault (see Fig. 10), if the load is the sarne along the length of the tunnel.

TWO-DIMENSlONAL PROBLEMS IN POLAR COORDINATES

THEORY OF ELASTICITY

62

Substituting the values (f) of the constants into the expressions (43) for the stress components, we find

From (2) of conditions (a) we find

fb a

u8 dr =

Jb a

a2q, - 2 dr = \ªct>\b = O ar ar a

ur

or substituting for e/> its expression (42), we find

[~ + B(b + 2b log b) + 2Cb] - [ ~ + B(a + 2a log a) + 2Ca J= O

<18 Tr8

(e)

Comparing this with (b), it is easy to see that (e) is satisfied, and the forces at the ends are reducible to a couple, provided conditions (b) are satisfied. To have the bending couple equal to M, the condition

ª

2

[ªb uer dr = [b c/>2 r dr = }, }a ar must be fulfilled. a

(d)

-M

W e have

fb

fb

a2q, -rdr = \ªct> - r\b ar 2 ar a

a

aq, \ªct> -dr= - r·1b - \e/> lb ar ar a a

and noting that on account of (b)

aq, r\b =O \ ar a we find from (d), or substituting expression (42) for e/>,

A log~

a

!: i i~

+ B(b 2 log b -

a 2 log a)+ C(b 2

a 2) = M

-

(e)

This equation, together with the two Eqs. (b), completely determines the constants A, B, C, and we find 2M 2 4M b B = - (b - a2 ) A = - N a 2b2 log a' N (f) M 2 2 2 2 C = N [b - a + 2(b log b - a log a)] where for simplicity we have put

N I'

1·

= (b 2 -

a

63

2 2 ) -

4a 2b2

(1og ~)

2

(g)

(a

r

ª)

2 2 b + b2 log b + a 2 log r = - 4M N 7 b log a

2 2 4M ( a b b r a = - N - T2 log + b2 log b + a 2 log

a

=

r + b2 -

) a2

(48)

Ü

This gives the stress distribution satisfying all the boundary conditions1 (a) for pure bending and represents the exact solution of the problem, provided the distribution of the normal forces at the ends is that given by the second of Eqs. (48). If the forces giving the bending couple M are distributed over the ends of the bar in some other manner the stress ' distribution at the ends will be different from that of the solution (48). But on the basis of Saint-Venant's principle it can be concluded that the deviations from solution (48) are very small and may be neglected at large distances from the ends, say at distances greater than the depth of the bar. It is of practical interest to compare solution (48) with the elementary solutions usually given in books on the strength of materials. If the depth of the bar, b - a, is small in comparison with the radius of the central axis, (b + a)/2, the sarne stress distribution as for straight bars is usually assumed. If this depth is not small it is usual in practice to assume that cross sections of the bar remain plane during the bending, from which it can be shown that the distribution of the normal stresses <18 over any cross sections follows a hyperbolic law. 2 ln all cases the maximum and minimum values of the stress u 8 can be presented in the forro M u9 = m(h) a2 1 This solution is due to H. Golovin, Trans. Inst. Tech., St. Petersburg, 1881. The paper, published in Russian, remained unknown in other countries, and the sarne problem was solved later by M. C. Ribiere (Compt. rend., vol. 108, 1889, and vol. 132, 1901) and by L. Prandtl. See A. Fõppl, "Vorlesungen über techniséhe Mechanik," vol. 5, p. 72, 1907; also A. Timpe, Z. Math. Physik, vol. 52, p. 348, 1905. 2 This approximate theory was developed by H. Résal, Ann. mines, p. 617, 1862, and by E. Winkler, Zivilingenieur, vol. 4, p. 232, 1858; see also his book "Die Lehre von der Elastizitãt und Festigkeit,'' Chap. 15, Prag, 1867. Further development of the theory was made by F. Grashof, "Elastizitãt und Festigkeit," p. 251, 1878, and by K. Pearson, "History of the Theory of Elasticity " vol. 2 pt. 1 p. 422 1893. ' ' ' '

THEORY OF ELASTICITY

64

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

The following table gives the values of the numerical factor m calcu~ lated by the two elementary methods, referred to above, and by the m

CoEFFICIENT

OF

EQ. (h)

b a

Linear stress distribution

Hyperbolic stress distribution

Exact solution

1.3 2 3

±66.67 ± 6.000 ± 1.500

+72.98, -61.27 + 7.725, - 4.863 + 2.285, - 1.095

-61.35 +73.05, + 7.755, - 4.917 + 2.292, - 1.130

exact formula (48). 1 It can be seen from this table that the elementary solution based on the hypothesis of plane cross sections gives very accurate results. .m 10 8

~\~ 6

l:ili

ba2a

i\.

~

2

tl

o

~

2

f

~

4

i'-...

0

r--...,._

ÔIJ

iJr

.......

61.0 1.1 t.2 1.3 1.4 1.5 . 1.6 1.1' 1.8 1.9 2.0 Valuesof ~

/J t.S

_.,,,

~

• 0.9 tl 0.6

C):'j

0.3

~---

/

.......

......

Frn. 43.

It will be shown later that, in the case of pure bending, the cross

sections actually do remain plane, and the discrepancy betwe~n the elementary and the exact solutions comes irom the fact that m. t~e elementary solution the stress component ur is neglected and it is The results are taken from the doctorate thesis, Univ. Michigan, 1931, of

V. Billevicz.

:.

·'

au

'

f

º1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 Values offr

1

y

nents of the displacement in the radial and tangential directions, respec:tively. If u is the radiaf displacemênt of the si de ad of the element abcd (Fig. 44), the radial displacement of the side bc is u + (au/ar) dr . Then the unit elongation of the element abcd in the radial direction is

gax.(/.070)

~ ...

/

b=Za'=>" f

X

r-... ,Neutralaxis(I=l44J)

4

.

assumed that longitudinal fibers of the bent bar are in simple tension or compression. From the first of Eqs. (48) it can be shown that the stress ur is always positive for the direction of bending shown in Fig. 42. The same can be concluded at once from the direction of stresses ue acting on the elements n - n in Fig. 42. The corresponding tangential forces give resultants in the radial direction tending to separate longitudinal fibers and producing tensile stress in the radial direction. This stress increases toward the neutral surface and becomes a maximum near this surface. This maximum is always much smaller than (ue)max.· For instance, for b/a = 1.3, (ur)max. = 0.060(ue)ma:.:.; for b/a = 2, (ur)max. = 0.138(ue)max.; for b/a = 3, (ur)max. = 0.193(ue)max.. ln Fig. 43 the distribution of ue andurfor b/a = 2 is given. From this figure we see that the point of maximum stress ur is somewhat displaced from the neutral axis in the direction of the center of curvature. 28. Strain Components in Polar Coordinates. ln considering the displacement in polar coordinates let us denote by u and v the compo-

..............

~1~1.2

65

Er

(49)

= ar

As for the strain in the tangential direction it shpuld be ebserved that it .depends not only on the displacement v b~t ~l~o- on the ~adÍ~l cfi;_ placement u. ·A.ssumjpg, ior instaf!ce, th!!t .:the pointsa ·and doUIJ,e element abcd. (Fig. 44) have only the radiál displacement u, the new le'ngth bf the a.rc adis (r + u) dO and the tangential strain is therefore (r

+ u) d O r dO

r dO

u

= -

r

The difference in the tangential displacement of the sides ab a:nd cd of

66

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

THEORY OF ELASTICITY

the element abcd is (av/ao) do, and the tangential strain dueto the displacement V is accordingly av/r ao. The total tangential strain is thus 1

u

EB

=

av

au

av ar

1

= E (u, -

Eg

=

'Yr9 =

1

E (<18

-

V
(52)

V
+ 2(1

::i::;

1[ E -

(1

+r v)A + 2(1

- v)B log r

+ (1

-3v) B

+ 2(1 -

v)C

]

- v)Br log r - B(l

+ v)r

= Fr,

f1(r)

f(o) do

+ f i(r)

(b)

f

f(O) dO -

!r f1(r)

=

O

(e)

f(O) = H sin O+ K cos O

(d)

where F, H, and K are constants to be determined from the conditions of constraint of the curved bar or ring. Substituting expressions (d) into Eqs. (a) and (b), we find the following expressions for the displacements. 1

u

=E1 [ -

v)r]

4B~

v = ']JJ

(1

+r v)A + 2(1 -

v)Br log r - B(l

+ v)r

+ f(O)

(a)

+ Fr + H cos O -

- v)r]

+ H sin O+ K cos O

(53)

. K sm O

in which the values of constants A, B, and C for each particular case should be substituted. Consider, for instance, pure bending. Taking the centroid of the cross section from which Ois measured (Fig. 42) and àlso an element of the radius at this point, as rigidly fixed, the conditions of constraint are u

The symbol Eg was used with a different meaning in Art. 10. It is assumed here that we have to do with plane stress, i.e., that there is no etress ª• perpendic-ul11r to the plane of the plate (see p. 11). ·· 2

!r af(O) + afi(r) + !r ao éJr

+ 2C(l

+ 2C(l 1

J

wheref1(r) is a function of r only. Substituting (a) and (b) in Eq. (51) and noting that 'Yre is zero since Tre is zero, we find

GTr(J

By integration we obtain u

-

1

we can obtain sufficient equations for determining u and v. 29. Displacements for Symmetrical Stress Distributions. Substituting in the first of Eqs. (52) the stress components from Eqs. 43, we find

au l[(l+v)A ar = E r2

4Bro v= 1 r

(51)

r

Substituting now the expressions for the strain components (49), (50), (51) into the equations of Hooke's law, 2 Er

from which, by integration,

from which

v

'YB=-+---

r ao

From the second of Eqs. (52), we

av = 4Br _ f(O) ao E

(50)

T + r ao

Considering now the shearing strain, let a'b' e'd' be the position of the element abcd after deformation (Fig. 44). The angle between the direction ad and a' d' is due to the radial displacement u and is equal to au/r ao. ln the sarne manner the angle between a'b' and ab is equal to av/iJr. It should be noted that only part of this angle (shaded in the figure) contributes to the shearing strain and the other part, equal to v/r, represents the angular displacement dueto rotation of the element abcd as a rigid body about the axis through O. Hence the total change in the angle dab, representing the shearing strain, is r

in whichf(O) is a function of O only. find, by using Eq. 50,

67

=o,

V=

0,

~ a+b ar= O for O =O and r = ro = - 2

Applying these to expressions (53), we obtain the following equations for calculating the constants of integration F, H, and K: 1 Equation (e) is satisfied only when f f(O) do is taken from (d) without an additive constant.

68

THEORY OF ELASTICITY

X({l + v)A + 2(1 E ro

v)Bro log To - B{l

TWO-DIMENSIONAL.PROBLEMS IN. POLAR COORDINATES

+ v)ro

+ 2C(l -

v)ro] + K Fro

= O

=O F =O

This means that the displacement of any cross section consists of a translatory displacement - K sin 8, the sarne for all points in the cross section , and of a rotation of the cross section by the angle 4B8 /E about . the center of curvature O (Fig. 42). We see that cross sect10ns remam plane in pure bending as is usually assumed in the elementary theory of the bending of curved bars. ln discussing the symmetrical stress distribution in a full ring (page 59) the constant B in the general solution (43) was taken as zero, and in this manner we arrived at a solution of Lamé's problem. Now, after obtaining expressions (53) for displacements, we see what is implied by taking B as zero. B contributes to the displacement v the term 4Br8/E. This term is not single valued, as it changes when we increase 8 by 211", i.e., if we arrive at a given point after making a complete circle round 45· Fw. the ring. Such a many-valued expression for a displacement is physically impossible in a full. ring, and so, for this case, we must take B = O in the general solution (43). A full ring is an example of a multiply-connected body, i.e., a body such that some sections can be cut clear across without dividing the body into two parts. In determining the stresses in such bodies we úsually arrive at the conclusion that the boundary conditions referring to the stresses are not sufficient to determine completely the stress distribution, and additional equations, representing the conditions that the displacements should be single valued, must be considered (see page 118). · The physical meaning of many-valued solutions can be explained by considering the initial stresses possible in a multiply-connected body. If a portion of the ring between two adjacent cross sections is cut out (Fig. 45). and the ends of the ring are joined again by welding or other

.

means, a ring with initial stresses is obtained, i.e., there are stresses in the ring when externai forces are absent. If a is the small angle measuring the portion of the ring which was cut out, the tangential displacement necessary to bring the ends of the ring together is

+H

From this it follows that F = H = O, and for the displacement v we obtain 4Br8 • (54) v="E-Ksm8

69

(e)

v =ar

The sarne displacement, obtained from Eq. (54) by putting 8 = 211", is V=

4Br 211"E

(f)

From (e) and (f) we find (g)

The constant B, entering into the many-valued term for the displacement (54) has now a definite value depending on the way in which the initial stresses were produced in the ring. Substituting (g) into Eqs. (f) of Art. 27 (see page 62), we find that the bending moment necessary to bring the ends of the ring together (Fig. 45) is

(b 2 M=-ªE 811"

-

a 2) 2

-

2(b 2

4a 2b2 (1og -

a 2)

~) a

2

(h)

From this the initial stresses in the ring can easily be calculated by using the solution (48) for pure bending. 30. Rotating Disks. The stress distribution in rotating circular disks is of great practical importance. 1 If the thickness of the disk is small in comparison with its radius, the variation of radial and tangential stresses over the thickness can be neglected 2 and the problem can be easily solved. 3 If the thickness of the disk is constant Eq. (40) can be applied, and it is only necessary to put the body force equal to the inertia force. 4 Then

(a) 1 A complete discussion of this problem and the bibliography of the subject can be found in the well-known book by A. Stodola, "Dampf- und Gas-Turbinen," 6th ed., pp. 312 and 889, 1924. 2 An exact solution of the problem for a disk having the shape of a flat ellipsoid of revolution was obtained by C. Chree, see Proc. Roy. Soe. (London), vol. 58, p. 39, 1895. It shows that the difference between the maximum and the minimum stress at the axis of revolution is only 5 per cent of the maximum stress in a uniform disk with thickness one-eighth of its diameter. 8 A more detailed discussion of the problem will be given !ater (see Art. 119). ' The weight of the disk is neglected.

70

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

THEORY OF ELASTICITY

where p is the mass per unit volume of the material of the disk and the angular velocity of the disk. Equation (40) can then be written in the forro

w

applied there, we have (
=C

- 3

t

V

pw2b2

71

=Q

from which

e= 3 +V pw2b2

(b)

8

and the stress components, from Eqs. (g), are This equation is satisfied if we derive the stress components from a stress function F in the following manner: T
= F,

dF dr

= -


+ pw2r

(e)

2

The strain components in the case of symmetry are, from Eqs. (49) and (50), u du E9 = T E, =dr' Eliminating u between these equations, we find E9 -

Er

+ r -de9 dr

(d)

= Ü

Substituting for the strain components their expressions in terms of the stress components, (52), and using Eqs. (e), we find that the stress function F should satisfy the following equation:

d 2F r 2 dr 2

+ r dF dr - F

+ (3 + v)pw r

2 3

= O

(e)

r

8

and from Eqs. (e) we find

(f9

3+v =8 - pw2(b2 - r2)


38 +V- pw2b2 - 1-+ 3v =8 - pw2r2

(55)

These stresses are greatest at the center of the disk, where
=


3+v 8

= - - pw 2b2

(56)

ln the case of a disk with a circular hole of radius a at the center the constants of integration in Eqs. (g) are obtained from the condi;ions at the inner and outer boundaries. If there are no forces acting on these boundaries, we have (u,),._ = O,

=O

(u,).....õ

(h)

from which we find that 3+v e = -8pw2(b2 + a2);

Substituting in Eqs. (g),

It can be verified by substitution that the general solution of this equation is 1 3 +V (f) F = Cr + C1 - - - - pw 2r3



3 +V - - - pw 2r2 8 1 1 + 3v 2 2 = C - C1-2 - - - - pw r r 8

=

C

+ C1 r-1

2

t

(b2 + a2 - a;~2 - r2) ue = 3 + v pw2 (b2 + a2 + a2b2 - 1 + 3v 2) 8 r2 3+ v r
= 3

v pw2

We find the maximum radial stress at r = (g)

The integration constants C and C 1 are determined from the boundary conditions. For a solid disk we must take C 1 = O since otherwise the stresses (g) become infinite at the center. The constant C is determined from the condition at the periphery (r = b) of the disk. If there are no forces

(O"r) max.

=

Vab,

(57)

where

3 +V - 8 - . pw 2 (b - a) 2

(58)

The maximum tangential stress is at the inner boundary, where 3+v pw 2 ( b2 + () 0"9 max. = - 4

1-v) +

3

V

a2

It will be seen that this stress is larger than (u,)max..

(59)

THEORY OF ELASTICITY

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

When the radius a of the hole approaches zero, the maximum tan.,. gential stress approaches a value twice as great as that for a solid di~k (56); i.e., by making a small circular hole at the center of a sohd rotating disk we double the maximum stress. This phenomenon of stress concentration ata hole will be discussed later (see page 78). Assuming that the stresses do not vary over the thickness of the disk, the method of analysis developed above for disks of constant thickness can be extended also to disks of variable thickness. lf h is the thickness of the disk, varying with radius r, the equation of equilibrium of such an element as shown in Fig. 40 is

A good approximation to the actual shapes of rotating disks can be obtained by dividing the disk into parts and fitting approximately to each parta curve of the type (m). 1 The case of a conical disk has been discussed by several authors. 2 Very often the calculations are made by dividing the disk into parts and considering each part as a disk of constant thickness. 3 1 ·31. Bending of a Curved Bar by a Force at the End. 4 W e begin .withJ4e shnple case shown in Fig. 46. A bar P _of a na,rrow rectangular cross section. and _with a circular axis is constrained at the lower end and bent by aforceP applied at theupper end in the radial direction. The bending moment at any cross section mn is propor· tional to sin O, and the normal stress uo, according to elementary theory of the bending of curved bars, is proportional to the bending moment. Assuming that this holds also for Y Fm. 46. the exact solution, an assumption which the results will justify, we find from the second of Eqs. (38) that the stress functjon q,, satisfying the equation

72

:r (hru,) - hus

+ hpw 2r2 = O

(k)

This equation is satisfied by putting

dF 2 2 hus = dr+ hpw r

hrur = F,

where F is again a stress function. Substituting these expressions for the stress components into the compatibility equation (d) we arrive at the following equation for the stress function F:

+ r dF d~ ~

r2 d2F

F

_hr ddh (r ddF - vF) =O (Z) r r ln this manner the problem of finding the stress distribution in a disk of variable thickness is reduced to the solution of Eq. (Z). ln the particular case where the thickness varies according to the equation h = Crn (m) in which C is a constant and n any number, Eq. (I) can easily be integrated. 1 The general solution has the form F = mrn+a +Arª+ Brfl in which (3 + v)pw 2C m = - (vn + 3n + 8) while

a

a2 1 a 1 a2 ) ( ar2+ r ar+ T2 ao 2

+ (3 + v)pw2hrª -

and (3 are the roots of the quadratic equation

x2

-

nx

+

vn - 1 = O

and A and B are integration constants which are determined from the boundary conditions. i This case was investigated by Stodola, Zoe. cit. Turbinenwesen, 1915.

See also H. Holzer, Z. ges.

should be proportional to sin O.

q,

(éPc/>

ar 2

1

aq,

1


+ r ar+ T2 ao2

=

o

73

(a)

Taking

=J(r) sin O ·

(b)

and substituting in Eq. (a), we find thatf(r) must satisfy the following ordinary di:fferential equation: See M. Grübler, V.D.!., vol. 50, p. 535, 1906. See A. Fischer, Z. oesterr. lng. Arch. Vereins, vol. 74, p. 46, 1922; H. M. Martin, Engineering, vol. 115, p. 1, 1923; B. Hodkinson, Engineering, vol. 116, p. 274, 1923; K. E. Bisshopp, J. Applied Mechanics (Trans. A.S.M.E.), vol. 11, p. A-1, 1944. 3 This method WJl.S developed.hy M.. Donath; seecJ:lls book, "Die Berechnung rotierender Scheiben und R.iííge," ~Berlin, 1912.---- It is described in English by H. Hearle in Engineering, vol. 1061 p. 131,_ 1918. A further development of the method was given by R. Grammel 1' Dinglers. Polytech,. J., _v-01. 338, p. 217, 1923. The case when materiar does not follÔw -Hooke's làw was investigated by M. Grübler, V.D.!., vol. 41, p. 860, 1897, and vol. 44, p. 1157, 1900. See also H. Schlechtweg, Z. angew. Math. Mech., vol. 11, p. 17, 1931, and lngenieur-Archio, vol. 2, p. 212, 1931. ' H. Golovin, Zoe. cit. 1

2

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

1'1

THEORY OF ELASTICITY

74 2

d ( dr 2

in which

..!) (ddrf + !.r dfdr _ L) =0 r 2

1 d _

+ r dr

Cc)

2

2

r2

N = a2

This equation can be transformed into a lin~ar differential ~qu~tion with constant coefficients (see page 58), and its general solut1on is

+ B ~ + Cr + Dr log r

f(r) = Ar 3

(d)


- !.r acJ> + _! a2cJ> ar r2 ao2

(2Ar - 2Brª + D) sin r

=

ª

2


TO= _ r

r

or, from Eqs. (60),

= Tro = O for r B 2 2Aa - -

{60)

2Ab -

D

3

O

W

O

The last condition is that the sum of the shearing forces distributed over the upper end of the bar should equal the force P. !ªk~ng the width of the cross section as unity-or P as the load per umt thicknes& of the plate-we obtain for O = O,

ib

Tro

dr =

-

ib :r (~::)dr l~ ::\: =

= \Ar2

or -A(b2 - a2) From Eqs. (e) and

p A= 2N'

+ ~ + C + D log

- a2) + B (b2 a2b?

b - D log ã

Cn we find Pa 2b2 B=---1 2N

ri:

D= -

t

=

P

= (a2

)

b log a

(J

p


+ b2)

(g)

2

3

(h)

+b

2) ]

= 7r /2,

=

Ü

p [ 3r = -N

1]

(k)

a2b2 - - r 3 - (a 2 ·+ b2) -r

The expressions (60) constitute an exact solution of the problem only when the forces at the ends of the curved bar are distributed in the 2.0

= a and r = b

+ -a = ~ 2B + !!. = b b

For the lower end,

ue

From the conditions that the outer and inner boundaries of the curved bar (Fig. 46) are free from external forces, we require that
2

[r + a2b2 - !. (a r r

N

Tr8

2

2

Ü

= - P

Tre

(J

~ + !!.) sin 9 r 2 ~ (!. acJ>\ = - (2Ar - ~ +Dr) cos 9 ar r ao) r

= ar2 = (6Ar +

+ (a + b

Substituting the values (g) of the constants of integration in Eqs. (60), we obtain the expressions for the stress components. For the upper end of the bar, (J = O, we find

· hi h A B e and D are constants of integration, which are deter1n w c , , , · (d)· mined from the boundary conditions. Substituting solut10n m expression (b) for the stress function, and using the general formulas (38), we find the following expressions for the stress components:
b2

-

l.S

~1t~'·º

~

1~"

~ ~

s .-..._

- -~ ~b~ v -:J..

//~~ V

11 / / [

1

i/ .,,ef.244(6'-a;J.

0.5

o ,.

""

::JJ/(6-a)

~ ,(}.4JS(b-a} ,:.soo(h-aJ

1 1

-

-1

().

1-

--

1

-~ ,__~ ....., ~

Ir-......... t--..... -.....

~~

-

......... ............ ~ ........ ~ ........ ....

1

.......

1

'

'~ ~

.......... ~

1

~~

.......... ~

1

~ "'11

....

fh'-a, FIG. 47.

manner given by Eqs. (h) and (k). For any other distribution of forces the stress distribution near the ends will be different from that given by solution (60), but at larger distances this solution will be valid by Saint-Venant's principie. Calculations show that the simple theory, based on the assumption that cross sections.remain plane during bending, again gives very satisfactory results. ln Fig. 47 the distribution of the shearing stress Tre over the cross section 9 = O (for the cases b = 3a, 2a, and l.3a) is shown. The abscissas are the radial distances from the inner boundary (r = a). The ordinates represent numerical factors with which we multiply the average shearing stress P/(b - a) to get the shearing stress at the

THEORY OF ELASTICITY

76

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

point in question. A value 1.5 for this factor gives the maximum shearing stress as calculated from the parabolic distribution for rectangular straight beams. From the figures it may be seen that the distribution of shearing stresses approaches the parabolic distribution when the depth of the cross section is small. For such proportions as are usual in arches and vaults the parabolic distribution of shearing stress, as in straight rectangular bars, can be assumed with sufficient accuracy. Let us consider now the displacements produced by the force P (Fig. 46). By using Eqs. (49) to (52), and substituting for the stress components the expressions (60), we find au sin () [ 2B - = -E 2Ar(l - 3v) - - 3 (1 ar r av a() = res -

'YrB

au = r ao

+

v)

+

+

av ar -

(m) and (n), are then

u

2D = - E() cos

v =

2

8+

D 8 sin 8 -

E

sin () [

JT D(l - v) log r

-D(l - v) log r ]

= -

cos () [Ar2(5

E

+

v)

r + ~ (1 +

v)

+ D(l

- v) log r]

+ f(O)

H = O,

(m)

(u)s=o

+ r~ (1 +

v) - D log r(l - v)

f

f(O) d()

+ F(r)

(n)

+ f'(O) + rF'(r)

- F(r)

(r)

L =

1;°

(s)

1; =

+ b2) + (a + b

P7r(a2 -

b2)

2

2

)

log

~]

(61)

The application of this formula will be given later. When b approaches ª'. and the depth of the curved bar, h = b - a, is small in comparison w1th a, we can use the expression log

~a

= log

(1 + ~)a = ~a - ! ah 2

2

2

+!

h 3

3 a3

_

Substituting in (61) and neglecting small terms of higher order, we obtain

4D cos O E

f} () cos () + K sin O+ L cos O

f (O) = - 2

=

E [ (a 2

(u)s=o = -

This equation is satisfied by putting F(r) = Hr,

+ Hr

The deflection of the upper end is, therefore, using (g),

in which F(r) is a function of r only. Substituting now (m) and (n) in the third of Eqs. (l) we arrive at the equation f(O) dO

cos () - L sin 8

(q)

The constant L is obtained from the condition at the built-in end (Fig. 46). For () = 7r/2 we have v = O; av/ar = O, hence, from the second of Eqs. (q),

V

+ D(l - v)] -

f

+ D(l E+ v) cos 8 + K

(u)o=o = L

(l)

where j(O) is a function of () only. Substituting (m) in the second of Eqs. (l) together with the expression for es and integrating, we find

v

- 3v)r2

. 8 + L cos 8 + B(l r2+ v)] + K sm cos () [A(5 + v)r2 + B(l + v) E r2

From the first of these equations we obtain by integration u = si;() [ Ar2(1 - 3v)

+ A(l

The radial deflection of the upper end of the bar is obtained by putting () = O in the expression for u, which gives

-Dr (1 - v) ]

U

17

(p)

in which H K 1 and L are arbitrary constants, to be determined from the conditi~ns of constraint. The components of displacements, frcm

3

~~ ; 3

which coincides with the elementary formula for this case.1 By taking the stress function in the form e/> = f(r) cos () 1

See S. Timoshenko, "Strength of Materiais," vol. 2, Art. 13, 1941.

THEORY OF ELASTICITY

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

and proceeding as above, we get a solution for the case when a :'ertical force and a couple are applied to the upper end of the bar (Fig. 46). Subtracting from this solution the stresses produced by the couple (see Art. 27), the stresses dueto a vertical force applied at the upper end of the bar remain. Having the solutions for a horizontal and for a vertical load, the solution for any inclined force can be obtained by superposition. In the above discussion it was always assumed that Eqs. (e) are satisfied and that the circular boundaries of the bar are free fro~ forces. By taking the expressions in (e) different fro~ zero, w~ obtam the case when normal and tangential forces proport10nal to sm ~ ~nd cos oare distributed over circular boundaries of the bar. Combmmg such solutions with the solutions previously obtained for pure bending and for bending by a force applied at the end we can approach the 1 condition of loading of a vault covered with sand or soil. 32. The Effect of Circular Holes on Stress Distributions in Plate~. Figure 48 represents a plate submitted to a uniform tension of magm-

These forces, acting around the outside of the ring having the inner and outer radii r = a and r = b, give a stress distribution within the ring which we may regard as consisting of two parts. The first is due to the constant component jS of the normal forces. The stresses it produces can be calculated by means of Eqs. (45) (page 59). The remaining part, consisting of the normal forces jS cos 20, together with the shearing forces -jS sin 20, produces stresses which may be derived from a stress function of the form

78

cf>

= f(r) cos 20

79

(b)

Substituting this into the compatibility equation

we find the following ordinary differential equation to determine f(r):

m1

The general solution is f(r) = Ar 2

+ Br + C r21 + D 4

The stress function is therefore y

~

s

Fm. 48.

tude S in the x-direction. If a small circular hole is made ín the middle of the plate, the stress distribution in the neighborh ood _of ~he hole will 1 be changed, but we can conclude from Saint-Venant s prmcipl~ that the change is negligible at distances which are large compared w1th a, the radius of the hole. Consider the portion of the plate within a concentric circle of radius b, large in comparison with a. The stresses at the radius b are e~ec­ tively the sarne as in the plate without the hole and are therefore g1ven by ( = S cos 2 O = jS(l + cos 20) (a) (r,e)r..b = - jS sin 20 1 Several examples of this kind were discussed by Golovin, loc. cit., and Ribiere, Zoe. cit., and Compt. rend., vol. 132, p. 315, 1901.

.il

cf> = ( Ar

2

+ Br + C~ + D) cos 20 4

(e)

and the corresponding stress components, from Eqs. (38), are 1 a cf> ( 6C 4D) r ar + -r -ao = - 2A + -r + -r cos 20 a cf> = ( 2A + 12Br + T4 6C) cos 20
1 acp = - ~

2

2

2

4

2

2

2

2

(1

Tre = - -a - -act>) =

ar r ao

( 2A + 6Br 2 -

2D) . r

-6C4 - r

(d)

2

.

sm 28

The constants of integration are now to be determined from conditions (a) for the outer boundary and from the condition that the edge of the hole is free from externa! forces. These conditions give

' ?

· THEORY OF ELASTICITY

80

6C 4D 1 +-+b4 b2 = --S 2 6 2A + C + 4D =O a4 a2 6C 2D 2A + 6Bb b4 b2 =

2A

2

2A

+ 6Ba

2

-

-

-

a4

1 - -2 S

-

-

2D a2 = O

Solving these equations and putting a/b = O, i.e., assuming an infinitely large plate, we obtain

s

a4

C = - -S,

B =O,

A=--, 4

4

Substituting these values of constants into Eqs. (d) and adding the stresses produced by the uniform tension ~S on the outer boundary calculated from Eqs. (45) we find 1

(i - ª2) + §_ (i + r4 = §. (i + ª §_ (i + r~ ) r 3 = - ~ (i - ~ + ~ )

=


r

§_

2

r2

3

2

<18

) 2

2

-

2

4

T,9

4~2) cos 28 r

3a4 -

2

2

4

cos 28

(62)

2

sin 28

If ris very large,
=

Tr8

= O,


= 8 - 28 cos 20

It can be seen that
<18

= -

s

so that thefe is a compression stress in the tangential dii-ection at these points. i This solution was obtained by Prof. G. Kirsch; see V.D.!., vol. 42, 1898. It has been well confirmed many times by strain measurements and by the photo~tic method (see Chap, 5 and the p90k;a, cited on P· 131).

81

For the cross section of the plate through the center of the hole.and perpendicular to the x-axis, 8 = 7í/2, and, from Eqs. (62), Tr8

-

6C -

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

=

Ü,

<19

= -8

2

(2

+ -a2r + 3 -ª4) r 2

4

It is evident that the effect of the hole is of a very localized character, and, as r increases, the stress
= 8 - 28 cos 28 - [8 - 28 cos (20 - 7í)]

For 8 = 7í/2 or 8 = 37í/2, i.e., at the points n and m, we find
See S. Timoshenko, Bull. Polytech. Inst., Kiew, 1907. We must take S equal to the load divided by the gross area of the plate. ~ See. paper by T. L. Wilson, The S.S. Leviathan, Damage, Repairs and Strength al_Ys1s, presented ata meeting of the American Society of Naval Architects and Manne Engineers, November, 1930.

83

THEORY OF ELASTICITY

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

It is often necessary to reduce the stress concentration at holes, such as access holes in airplane wings and fuselages. This can be done by adding a bead 1 or reinforcing ring. 2 The analytical problem has been solved by extending the method employed for the hole, and the results 2 have been compared with strain-gauge measurements. The case of a circular hole near the straight boundary of a semiinfinite plate under tension parallel to this boundary (Fig. 50) was analyzed by G. B. Jeffery. 3 A corrected result anda comparison with 4 photoelastic tests (see Chap. 5) were given later by R. D. Mindlin.

This stress should be compared with the tensile stress at the point m on the straight edge of the plate, given by the formula

82

s

<18

The stress at the hole, at the point n nearest the edge, becomes a very large multiple of the undisturbed tensile stress when mn is small compared with np. G. B. Jeffery also investigated the case of a uniform normal pressure Pi acting on the boundary of the hole. This is a pr?blem of practic~l interest. It gives the stresses near a rivet hole while the hot plast1c rivet is being forced home under pressure. If the hole is very far from the straight edge the stresses at the boundary of the hole, from Eqs. (46) (page 60), are
(u8)max.

= p;

+ r2

d2 d2 -

r2

1 See S. Timoshenko, J. Franklin I nst., vol. 197, p. 505, 1924; also S. Timoshenko, "Strength of Materials,'' 2d ed., vol. 2, p. 317. 2 S. Levy, A. E. McPherson, and F. C. Smith, J. Applied Meehanics
paper. 3 Trans. Roy. Soe. (London), series A, vol. 221, p. 265, 1921. • Proc. Soe. Expl. Stress Analysis, vol. 5, p. 56, 1948.

r2

(64)

= 0.758

at the point m. The method used in this article for FIG. 51. analyzing stresses round a small circular hole can be applied when the plate is subjected to pure bending.2 The cases of a row of circular holes in an infinite plate, 3,4,5 a row of holes in a strip, 5•6 and in a semi-infinite plate, 6 and a ring of holes in a plate 7 (under all-round tension) have also been investigated. A method devised by Hengst has been applied to the case of a hole in a square plate 8 under equal tension in both directions, and under shear 9 when the hole is plain or reinforced. Solutions have been obtained for the infinite plate with a circular hole when forces are applied to the boundary of the hole, 1 ºfor the corre1

R. C. J. Howland, Trans. Roy. Soe. (London), Series A, vol. 229, p. 49, 1930. Z. Tuzi, Phil. Mag., February, 1930, p. 210; also Sei. Papers Inst. Phys. Chem. Researeh (Tokyo), vol. 9, p. 65, 1928. The corresponding problem for an elliptieal hole was solved earlier by K. Wolf, Z. teeh. Physik, 1922; p. 160. The circular hole in a strip is discussed by R. C. J. Howland and A. C. Stevenson, Trans. Roy. Soe. (London), series A, vol. 232, p. 155, 1933. A proof of convergence of the series solutions is given by R. C. Knight, Quart. J. Math., o~ord series vol 5 .P 255 1934. "'J' ' • ' • ' 2

3

(63)

4p;r 2 d2 -

For d = r 0, the two expressions have the sarne magnitude. If d is greater than this the maximum stress is at the circular boundary, and if it is less, the maximum stress is at the point m. The case of a plate of finite width with a circular hole on the axis of symmetry (Fig. 51) was discussed by R. C. J. Howland. 1 He found, for instance, that when 2r = id, o- 8 = 4.38 at the point n and 0"8

FIG. 50.

=

M. Sadowsky, Z. angew. Math. Meeh., vol. 8, p. 107, 1928. : R. C. J. Howland, Proe. Roy. Soe. (London), series A, vol. 148, p. 471, 1935. K. J. Schulz, Proe. Nederl. Akad. van Wetensehappen, vol. 45, pp. 233, 341, 457, an~ 524, 19~2, vol. 48, pp. 282 and 292, 1945. R C. B. Ling, P. S. Wang, and K. S. Jing, Teeh. Rept. No. 9, Bur. Aeronaut. eseareh, Chengtu, China, Jan. 1944. : C. B. Ling and P. S. Wang, Teeh. Rept. No. 6, ibid., June, 1943. H. Hengst, Z. angew. Math. Meeh., vol. 18, p. 44, 1938. 9 · C. K. ~ang, J. Applied Meehanics (Trans. A.S.M.E.), vol. 13, p. A-77, 1946. 10 W. G. Bickley, Trans. Roy. Soe. (London), series A, vol. 227, p. 383, 1928.

84

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

sponding problem of the strip, 1 and for a row of holes parallel and near to the straight edge of a semi-infinite plate 2 (row of rivet holes). li an elliptical hole is made in an infinite plate under tension S, with one of the principal axes parallel to the tension, the stresses at the ends of the axis of the hole perpendicular to the direction of the tension are

the hole is small in diameter compared with the thickness between the ends, the disturbance will be confined to the neighborhood of the ends. But if the diameter and the thickness are of the sarne order of magnitude, the problem must be treated as essentially three-dimensional throughout. Investigations of this kind 1 have shown that tre remains the largest stress component and its value is very close to that given by the two-dimensional theory. 33. Concentrated Force ata Point of a Straight Boundary. Let us consider now a concentrated vertical force P acting on a horizontal p

(65)

where 2a is the axis of the ellipse perpendicular to the tension, and 2b is the other axis. This and other problems concerning ellipses, hyperbolas, and two circles are discussed in Chap. 7, where references will be found. A very slender hole (a/b large) perpendicular to the direction of the tension causes a very high stress concentration. 3 This explains why cracks transverse to applíed forces tend to spread. The spreading can be stopped by drilling holes at the ends of the crack to eliminate the sharp curvature responsible for the high stress concentration. When a hole is filled with material which is rigid or has elastic constants different from those of the plate (plane stress) or body (plane strain) itself, we have the problem of the rigid or elastic inclusion. .This has been solved for circular 4 and elliptic inclusions. 5 The results for the rigid circular inclusion have been confirmed by the photoelastic method6 (see Chap. 5). The stresses given by Eqs. (62) for the problem indicated by Fig. 48 are the sarne for plane strain as for plane stress. ln plane strain, however, the axial stress tr. = v(u, + uo) must act on the plane ends, which are parallel to the xy-plane, in order to make Ez zero. Removal of these stresses from the ends, to arrive at the condition of free ends, will produce further stress which will not be of a two-dimensional (plane stress or plane strain) character. li R. C. Knight, Phil. Mag., series 7, vol. 19, p. 517, 1935. 2 C. B. Ling and M. C. Hsu, Tech. Rept. No. 16, Bur. Aeronaut. Research,

1

Chengtu, China, February, 1945. a The problem of a narrow slot was discussed by M. Sadowsky, Z. angew. Math. Mech., vol. 10, p. 77, 1930. 4 K. Sezawa and G. Nishimura, Rept. Aeronaut. Research Inst., Tokyo Imp. Univ., vol. 6, no. 25, 1931; J. N. Goodier, Trans. A.S.M.E., vol. 55, p. 39 (1933). s L. H. Donnell, "Theodore von Kármán Anniversary Volume,'' p. 293, Pasadena, 1941. 6 W. E. Thibodeau and L. A. Wood, J, Research Natl. Bur, Standards, vol, 20, p. 393, 1938.

L11

85

THEORY OF ELASTICITY

y

A

O

B

.X

fa)

(6)

FIG. 52.

st~aig~t boundary AB of an infinitely large plate (Fig. 52a). The distnbut10n of the load along the thickness of the plate is uniform as indica~ed in Fig. 52b. The thickness of the plate is taken as unit; so that P is the load per unit thickness. The distribution of stress in this case is a very simple one2 and is called a simple ra
=

2P cos 6 r

(66)

S A. E. Green, Trans. Roy. Soe. (London), series A, vol. 193, p. 229, 1948; E. tern berg and M. Sadowsky, J. Applied Mechanics (Trans. A.S.M.E.), vol. 16, P. 27'1949. 2 1:he solution of this problem was obtained by way of the three-dimensional ~~~~ion ~f J. Boussines<_l (p. 362) by Flamant, Compt. rend., vol. 114, p. 1465, Bo '!'ar1s. The extens1on of the solution to the case of an inclined force is dueto M'ussmesq, Compt. rend., vol. 114, p. 1510, 1892. See also the paper by J. H. a~?hell, Proc. L~n ~ath. Soe., vol. 32, p. 35, 1900. The experimental investi~ ion of stre$s d1str1bution, which suggested the above theoretical work was done Y Carus Wilson, Phil. Mag., vol. 32, p. 481, 1891. '

THEORY OF ELASTICI_TY

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

The tii.ngential stress u 8 and the shearing stress T,e are zero. It is easy to see that these values of the stress components satisfy the equations of equilibrium (37) (page 56). The boundary conditions are also satisfi.ed because ue and 'Tro are zero along the straight edge of the plate, which is free from externai forces except at the point of application of the load (r = O). Here ur becomes infinite. The resultant of the forces acting on a cylindrical surface of radius r (Fig. 52b) must balance P. It is obtained by summing the vertical components urr d8 cos (} acting on each element r d(} of the surface. ln this manner we fi.nd

plane at any point M (Fig. 52a) are calculated from the simple compression in the radial direction,

2

.2

Ía

4:Ío

ur cos (} · r d8 = -

...

2

cos 2

(}

d8 = -P

To prove that solution (66) is the exact solution of the problem we must consider also the equation of compatibility (39). The above solution is derived from the stress function

p

u,, = u, cos 2 8 u 11

=

u, sin 2 8

rZ'll

=

Ur

q, = - -7r r8 sm (}

m

(66')

7rd

the point of application of the load. Taking a horizontal plane mn at a distance a from the straight edge of the plate, the normal and shearing components of the stress on this

i,

(67)

(}

2P . sm (} cos 3 (}

- -

r

7r

A

/

-- 7 /

o -;- -r-

cos 4 8

. 2P sin 8 cos 2 8 s1n 8 cos 8 = - -

y

2P cos

i.e., the stress is the sarne at all points on the circle, except the point O,

it

'Ira

~

which coincides with solution (66). Substituting the function (a) into Eq. (39), we can easily show that this equation is satisfied. Hence, (a) represents the true stress function and Eqs. (66') give the true stress distribution. ' Taking a circle of any diameter d with cénter on the x-axis and tangent to the y-axis at O (]'ig. 52a), we have, for any point C of the circle, d cos 9 = r. Hence, from Eq. (66), 2P = -

= - 2p sin 2 (} cos 2

(a)

(!

Ur

'Ira

'Ira

ln Fig. 53 the distribution of stresses u,, and 'Tzv along the horizontal plane mn is represented graphically. At the point of application of the load the stress is theoretically infinitely large because a finite force is acting on an infinitely small

We can verify .this by using Eqs. (38) as follows:

aq, 1 a2q, '1r = r ar + T2 ao2 = a2q, uo = -ar2 =o 'Tr8 = - ~ ar r aq,) ao = . o

2P

r

7r

IP

~

o

Pz .1111

01

02

V

.

i

2P cos 3 8

= - - -- = - -

87

-

M

V

~ f--8, 13'

'-7

"

l"I

lP 1fa

"" n

ss

JqB

~X

' ~ R' u.,

.......... h&

H

--

l,L1'.,,

""

,_

,.,

i..--

X FIG. 53.

area. ln practice, at the point 0f application there is always a certain yielding of material and as a result of this the load will become dis~ tributed over a finite area. Imagine that the portion of materiai which suffered a plastic flow is cut out from the plate by a circular cylindrical surface of small radius as shown in Fig. 52b. Then the equations of elasticity can be applied to the remaining portion of the plate. An analogous solution can be obtained for a horizontal force P applied to the straight boundary of the semi-infinite plate (Fig. 54). The stress components for this case are obtained from the sarne Eqs. (66'); it is only necessary to measure the angle (} from the direction of the force, as shown in the figure. By calculating the resultant of the forces acting on a cylindrical surface, shown in Fig. 54 by the dotted

88 lin~,

'

THEORY OF ELASTICITY

we find

we find

.._

- -2Plc" cos 2 8 d8 = -P 7r o

.,,1

. = -Pa (8 + sm

A

O

B

Fm. 54.

Fm. 55.

a horizontally (Fig. 55), the radial stress at any point C is, from Eqs.

(66'),

7

! [p

cos a cos 8

+ .P sin a cos (~ + o) J 2P

= - -

?rr

cos (a

+ 8)

(68)

Hence Eqs. (66') can be used for any direction of the force, provided in each case we measure the angle 8 from the direction of the force. The stress function (a) may be used also in the case when a couple is acting on the straight boundary of an mfinite plate (Fig. 56a). It is easy to see that the stress functi.on for the case when the tensile force P is at the point 01, at a distance a from the origin, is obtained from e/>, Eq. (a), regarded for the nioment as a function of x and y instead of r and 8, by writing y + a instead of y and also - P instead of P. This and the original stress function cf> can be combined, and ~e then obtain the stress function for the two equal and opposite forces applied at O ànd Oi, in tµe forro -cf>(x,y +a)

4>2 = 4>1 - (c/>1

1 a) + âc/> ây

=

2Ma

-1rT

l'ubstituting (a) in ~q. (b), and noting (see page 57) thaij

ª"' . ,;. ª"' am..

· itfJ "" fJr

8

+ ao oosr 8·

(69)

ây

cos 3 8

(70)

X

fa)

X

Fm. 56.

Having the distribution of stresses, the corresponding displacements can be obtained in the usual way by applying Eqs. (49) to (51). For a force normal to the straight boundary (Fig. 52) we have .

au

Er

'YrO

=

2P cos 8

=ar = - ?rE_r_ r

When a il!I very i;mall, this app:roaches the value ()y

+ ain 8 coa 8)

a3 4>1

-

Eo = '.!±

(b)

(8

If the directions of the couples are changed it is only necessary to change the sign of the function (70). A series of stress functions obtained by successive differentiation .has been employed to :solve the problem of stress concentration due to a semicircular notch in a semi-infinite plate in tension parallel to the edge. 1 The maximum tensile stress is slightly greater than three times the undisturbed tensile stress away from the notch. The strip with a semicircular notch in each edge has also been investigated. 2 The stress-concentration factor (ratio of maximum to mean stress at mini~um section) falls below three, approachmg unity as the notches are made larger.

+ cf>(x,y)

c/>i =,-a i>cf>

= -M11'.

in which M is the moment of the applied couple. . Reasoning in .the sarne manner, we find that by differentiation of 't'h .i. we obtain the stress f unct10n c/>2 for the case when two equal and opposite couples M are acting at two points O and 0 1 a very small distance apart (Fig. 56b). We thus P . find that

= -

ur = ...,...

8 coa 8)

1("

This resultant balances the externai force P, and, as the stress components Tro and u 9 at the straight edge are zero, solution (66') satisfies the boundary conditions. Having the solutions for vertical 11nd horizontal concentrated forces, solutions for inclined forces are obtained by superposition. Resolving the inclined force P into two components, P cos a .;ertically and P sin

$9

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

+

àu r ii{J

àv =

r ao

àv

+ àr -

V

2P cos 8

?rE

r

(e)

rv = 0

t F. G. Maunsell, Phil. Mag., vol 21 p 765 1936 2 C B · · . ' . ' . H p ·. · Ling, J. Ap?lied Mechanics (Trans. A.S.M.E.), vol. 14, p. A-275 1947; · oritsky, H. D. Smvely, and C. R. Wylie, ibid., vol. 6, p. A-63, 1939. '

straight boundary of the plate. The horizontal displacements are obtained by putting 8 = ±r/2 in the first of Eqs. (g). We find

Integrating the first of these equations, we find .. 2P u = - rE cos fJ log -r

where f(O) is a function of O only. (e) and integrating it, we obtain

v=

= - (l ~;)P 8 sin

8

+A

(d)

(u)

Substituting in the second of Eqs.

in which F(r) is a function of r only. third of Eqs. (e), we conclude that j(O)

+ f(fJ)

J

~: sin fJ + =~ log r sin fJ -

sin 8

f(fJ) d8

+ F(r)

(e)

Substituting (d) and (e) in the

+B

cos O,

F(r) = Cr

(f)

where A, B, and C are constants of integration which are to be determined from the conditions of constraint. The expressions for the displacements, from Eqs. (d) and (e), are u = - 2p cos

v=

rE 2vP - sin 8 rE

olog r

-

(l -Ev)P fJ sin 8 r

2P

.

+ r-E log r sm 8 + (l

+A

(1 - v)P r

~Ev)P sin

8

E

+A

sin 8

+B

cos 8

O cos 8

= -

2P rE log r

(g)

cos O - B sin 8 + Cr

+B

(h)

To find the constant B let us assume that a point of the x-axis ata distance d from the origin does not move vertically. Then from Eq. (h) we find

B

,.. = _ (1 - v)P, 2E

2P

= rE log d

Having the values of all the constants of integration, the displacements of any point of the semi-infinite plate can be calculated from Eqs. (g). Let us consider, for instance, the displacements of points on the

(u)

9 =2

,, = _ (1 - v)P 2E

9 = -2

(7 l)

The straight boundary on each side of the origin thus has a constant displacement (71), at all points, directed toward the origin. We may regard such a displacement as a physical possibility, if we rémember that around the point of application of the load P we removed the portion of material bounded by a cylindrical surface of a small radius (Fig. 52b) within which portion the equations of elasticity do not hold. Actually of course this material is plastically deformed and permits di8placement (71) along the straight boundary. The vertical displacements on the straight boundary are obtained from the second of Eqs. (g). Remembering that vis positive if the displacement is in the direction of increasing 8, and that the deformation is symmetrical with respect to the x-axis, we find for the vertical displacements in the downward direction ata distance r from the origin (v)

Assume that the constraint of the semi-infinite plate (Fig. 52) is such that the points on the x-axis have no lateral displacement. Then v = O, for 8 = O, and we find from the second of Eqs. (g) that A = O, C = O. With these values of the constants of integration the vertical displacements of points on the x-axis are (u)e=o

91

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

THEORY OF ELASTICITY

90

9=

,.. = - (v)

-2

.. = 2p log ~ - (l

9 =2

rE

r

+

v)P rE

~72)

"

At the origin this equation gives an infinitely large displacement. To remove this difficulty we must assume as before that a portion of material around the point of application of the load is cut out by a cylindrical surface of small radius. For other points of the boundary, Eq. (72) gives finite displacements. 34. Any Vertical Loading of a Straight Boundary. The curves for (j"' and -rZll of the preceding article (Fig. 53) can be used as influence lines. We assume that these curves represent the stresses for P equal to a unit force, say 1 lb. Then for any other value of the force P the stress (j"' at any point H of the plane mn is obtained by multiplying the ordinate HK by P. If several vertical forces P, P 1, P 2, • • • , act on the horizontal straight boundary AB of the semi-infinite plate, the stresses ori the horizontal plane mn are obtained by superposing the stresses produced by each of these forces. For each of them, the (j"' and -rZ!I curves are obtained by shifting the (]',, and -rZll curves, constructed for P, to the new origins 01, 02, . • . . From this it follows that the stress(]',, produced, for instance, by the force P 1 on the plane mn at the point D is obtained by multiplying the ordinate H 1K 1 by P 1. ln the sarne manner the (j"' stress at D produced by P 2 is H 2 K 2 • P 2, and so on. The total normal

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

THEORY OF ELASTICITY

92

stress at Don the plane mn produced by P, P1, P2, ""' = DD1 ·

P

+ H1K1

· P1

+ H2K2

is

· P2

---

another manner by means of a stress function in the form (a)

+ ··

Hence the
in which A is a constant.

The corresponding stress components are

a2q, "' = r ar + T2 ao2 = 2A8 a2q,
}-Atr

X

93

1

(b)

(!

Applying this to the semi-infinite plate we arrive at the load distribution shown in Fig. 57a. On the straight edge of the plate there acts a uniformly distributed shearing force of intensity -A and a uniformly distributed normal load of the intensity A7r, abruptly changing sign at the origin O. The directions of the forces follow from the positive directions of the stress components acting on an elemente. By shifting the origin to 01 and changing the sign of stress function cp, we arrive at the load distribution shown in Fig. 57b. Superposing the two cases of load distribution (Figs. 57 a and 57b), we obtain the case of uniform loading of a portion of the straight boundary of the semi-infinite plate shown in Fig. 57 e. To obtain the given intensity q of uniform load, we take 2A1r

= q,

1 A= - q 211"

The stress at any point of the plate is then given by the stress function 1

q, =

FIG. 57.

Having these curves, the stress components at D for any kind of vertical loading of the edge AB of the plate can easily be obtained. If, instead of concentrated forces, we have a uniform load of_ intensity q, distributed over a portion 8s of the straight boundary (Fig. 53), the normal stress
Jl !ii11

A(r 28 - r 12 8 1) = ~ (r 28 - r 1 2 81)

2

(e)

From Eqs. (b) we see that the first term of the stress function (e) gives, at any point M of the plate (Fig. 58a), a uniform tension in all directions in the plane of the plate equal to 2A8 anda pure shear -A. ln the sarne manner the second term of the stress function gives a uniform compression -2A8 1 anda pure shear A. The uniform tension and compression can be simply added together and we find a uniform compressive stress p = 2A8 - 2A8 1 = 2A(8 - 81) = -2Aa

(d)

in which a is the angle between the radii r and r 1• ln superposing the two kinds of pure shear, one corresponding to the direction r and the other to the direction r 1 , we shall use Mohr's circle (Fig. 58b), which in this case has a radius equal to the numerical value of the pure shears A. By taking 1 This solution of the problem is dueto J. H. Michell, Proc. London Math. Soe., Vol. 34, p. 134, 1902.

94

THEORY OF ELASTICITY

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

the two diameters, DD1 parallel to r andFF1 perpendicular to r, as the r-and
When r1 - r = O, the angle f3 determines the direction of one of the pri:i:icipal stresses at M. From the figure we see that r and Tt are numerically equal if 2(3 = 2(a - (3), from which f3 = a/2. The direction of the principal stress therefore bisects the angle between the radii r and r 1• The magnitudes of the principal stresses are therefore

±2
= ±2A sin 2{3 = ±2A sin a

95

(e)

Combining this with the uniform compression {d) we find for the total values of the principal stresses at any point M, ' -2A(a +sina),

-2A(a - sina)

(j)

Along any circle through O and 01 the angle a remains constant, and so the pri:i:icipal stresses (f) are also constant. At the boundary, between the points O and Oi; x·--~---l-----...i

(a)

Frn. 59.

(Fig. 58a), the angle a is equal to .,,., and we find, from (f), that both principal stresses are equal t? -:-2'll"A = -q. For the remaining portions of the boundary a = O, and both pnnc1pal stresses are zero. Hence if an arbitrary load distribution (Fig. 59) is regarded as composed of a large n~mber of loads of varying intensities on short elements of the boundary, the. horizontal stress
H (b)

FIG. 58.

The sarne circle can be used also to get the stress components due to pure shear Considering again the plane mini, and noting that the normal to this plane makes an angle a - f3 with the direction r1 (Fig. 58a), it appears that the stress components are given by the coordinates of the point H of the circle. To take care of the sign of the pure shear corresponding to the r 1-direction, we must change the signs of the stress components, and we obtain in this manner the point H 1 on the circle. The total stress acting on the plane mini is given by the vector CK, the components of which give the normal stress - (
in the direction r 1 (see page 16).

+ .,.

1

ll,11111

all along the straight boundary. . Se_veral other cases of distributed load on a straight boundary of the semiinfinite plate were discussed by S. D. Carothers, 1 and by M. Sadowsky.1 Another manner of solving this problem will be discussed later (see page 125).

The deflections of the straight boundary of the plate can be found for any load distribution by using Eq. (72) obtained for the case of a co~centrated force. If q is the intensity of vertical load distribution (Fig. 59), the deflection produced at any point O ata distance r from the shaded element q dr of the load, from Eq. (72), is 2q log
(l

+ v)q d

~E

r

~ Proc. Roy. Soe. (London), series A, vol. 97, p. llO, 1920. Z. angew. Math. Mech., vol. 8, p. 107, 1928.

;~;,;~1~ -;~~~E;M\t .

t

ti

t-\

\'.)OARA

8\BUOTEC~ CENTIULl . ...

-·

.

96

THEORY OF ELASTICITY

and the total deflection at O is 2 7rE

Vo = -

1l+x q log -d "'

r

1

+

V

dr - - -

7rE

1l+x q "'

dr

(h)

ln the case of a uniformly distributed load, q is constant and we find Vo

= -2q [ (l 7rE

d + x) log-l +X

d] +1--- q V xlogl X 7rE

The constant k will now be adjusted so as to satisfy the condition of equilibrium at the point O. Making the resultant of the pressures on the cylindrical surface (shown by the dotted line) equal to -P we find . -2 from which

h

2

8

ª kP c;s k

(i)

rde = - kP (a + ~ sin 2a) = - p

=

1

a+ i- sin 2a

Then, from Eqs. (a),1

ln the sarne manner, for a point under the load (Fig. 00), we find

2 vo = q [
97

TWO-DIMENSIONAL l'ROBLEMS IN POLAR COORDINATES

+ xlog~] + X

"' = - r(a 1

- v ql 7rE

(j)

P cose sin 2a)

+t

(73)

By_ making a = 7r /2 we ar.rive at solution (66) for a semi-infinite plate, wh1ch has already been d1scussed. lt may be seen that the distribu-

Equation (h) can be used also for finding the intensity q of load distribution, which produces a given deflection at the straight boundary,

p

1 .___,-t-----'•I X

o

y Fm. 60.

Fm. 61.

FIG. 62.

Assuming, for instance, that the deflection is constant along the loaded portion of the straight boundary (Fig. 61), it can be shown that the distribution of pressure along this portion is given by the equation 1

p q = ----;:===

7rVa2 - x2

35. Force Acting on the End of a W edge. The simple radial stress distribution discussed in Art. 33 can be used also in investigating the stresses in a wedge due to a concentrated force at its apex. Let us consider a symmetrical case, as shown in Fig. 62. The thickness of the wedge in the direction perpendicular to the xy-plane is taken as unity. The conditions along the faces, 8 = ±a, of the wedge are satisfied by taking for the stress components the values "' = 1

Sadowsky, loc. cit.

kP cose r

Fm. 63.

tion of normal stresses over any cross section mn is not uniform and the ratio of the normal stress at the points m or n to the maxi~um stress at the center of the cross section is found to be equal to cos4 a. If the force is perpendicular to the axis of the wedge (Fig. 63), the sarne solution (a) can be used if 8 is measured from the direction of the force. The constant factor k is found from the equation of equilibrium ~+a 2 [

2

from which

k and the radial stress is <1r

"º = o,

T,9 = Ü

(a)

= -P

u, cos 8 · rd8

-a

=

=

1 1

a -

2 Slll

.

2a

P cose t sin 2a)

r(a 1 This solution is di.ie to Michell, loc. cit.

chaussées, 1901,

. .. ·

'·

~-:.'

•.,•. '

-...

(74)

See also A. Mesnager, Ann. por$~

98

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

THEORY OF ELASTICITY

clockwise.

The normal and shearing stresses over any cross section mn are q

-

-

!/ -

T

:ti/ -

Pyx sin 4 O y 3 (a - t sin 2a) Px 2 sin 4 O y 3 (a - t sin 2a)

Ur

-,-~~---.,.-

-

Tre

ln the case of a small angle a, we can put . (2a) 3 2a - sm 2a = - 6 Then writing I for the moment of inertia of the cross section mn we find from (b) that 4 tan U11 = ( -a- Sin 0 T:l;IJ

=

ª)ª. tan ª)ª . ( --

a

Sin

The stresses are

=-

(b)

-,--~---=--,-

(e)

4

0

For small values of a, the factor (tan a/a) 3 sin 4 Ocan be taken as nearly unity. Then the expression for u11 coincides with that given by the elementary beam formula. The maximum shearing stress occurs at the points m and n and is twice as great as that given by the elementary theory for the centroid of a rectangular cross section of a beam. Since we have solutions for the two cases represented in Figs. 62 and 63, we can deal with any direction of the force P in the xy-plane by resolving the force into two components and using the method of superposition.1 It should be noted that solutions (73) and (74) represent an exact solution only in the case when, at the supported end, the wedge is held by radially directed forces distributed in the manner given by the solutions. Otherwise the solutions are accurate only at points at large distances from the supported end. The problem of the wedge loaded by a bending couple M, in the plane of the wedge, and concentrated at the tip, is solved by the stress function. 2 _ M sin 20 - 20 cos 2a (d)

99

=

M · -4.sm 20' 2(sin 2a - 2a cos 2a) r 2

M

2

uo

=O (e)

· 2a - 2 a cos 2 a ) · 2r (cos 20 - cos 2a) 2 (s1n

36. Concentrated Force Acting on a Beam. The problem of stress distribution in a beam subjected to the action of a concentrated force is of great practical interest. It was shown before (Art. 22) that in continuously loaded beams of narrow rectangular cross section the stress distribution is obtained with satisfactory accuracy by the usual elementary theory of bending. e N ear the point of application of a concentrated force, however, a serious local perturbation in stress dis:Y tribution should be expected and a further investigation of the problem Fm, 64. is necessary. The first study of these local stresses was made experimentally by Carus Wilson. 1 Experimenting with a rectangular beam of glass on two supports (Fig. 64) loaded at the middle, and using polarized light (see page 132), he showed that at the point A, where the load is applied, the stress distribution approaches that produced in a semi-infinite plate by a normal concentrated force. Along the cross section AD the normal stress
Loc. cit. Wilson, loc. cit.; also G. G. Stokes, "Mathematical and Physical Papers," vol. 5, p. 238. 2

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

THEORY OF ELASTICITY

100 tion p /2 and subtracting the moment of all the radially directe~ tens~le forces applied to one-half of the beam. This .latter mo~ent IS eas1ly calculated if we observe that the radially distr1buted tensile forces are statically equivalent to the pressure distribution over the .quadrant ab of the cylindrical surface abc at the point A (Fig. 65c) or,.usmg Eq. (66), are equivalent to a horizontal force P /7r and a vertical force P /2, p

A better approximation is obtained if we observe that a continuously distributed load is applied to the bottom of the beam (Fig. 65b) and use Eqs. (36'). The intensity of this load at the pointD, from Eq. (66), is P /7rc. Substituting this in (36') and combining with the value of u,. above, we obtain, as a second approximation, u,. = 3P

2c

z--;

mr-t

p

Uy

a

A

e

~ ~~~~::,.,-~~---iq-r

e

3

= 27rC

(.!:.2 _ ~) y + 27rC _f_ + P 'Ir

p

+

1rC

Uz;

=

Then the bending moment, i.e., the moment

applied at A (Fig. 65d). about the point O, is

~l 2

-

~e 'Ir

and the corresponding bending stresses are

u,.' =

1

j (~ - ;) y = ;~ (~ -

;) y

To these bending stresses the uniformly distributed tensile stress p /27rc produced by the tensile force P /ir sh?uld ?e a~ded. The normal stresses over the cross section AD, as obtamed m th1s elementary way, are therefore q,.

3P

= 2c3

(zZ - :;;:e) y + 27rCP

1

As before we take p as the force per unit thickness of the plate.

(a)

4c - 4c 3

Ü,

Uy

=

2P --(c_+_y_)

(b)

-'Ir

21rC

P

lP

+ 27rC + 51rC

P

= - 0 ·254 C

while the more accurate solution gives only -0.133(P /e). The first attempt to get a more accurate solution of the problem was made by J. Boussinesq. 1 He used Flamant's solution (see Art. 33) for the semi-infinite plate. To annul the stresses over the boundary np (Fig. 65a), he superimposes an equal and opposite system of stresses and uses again the Flamant solution, i.e., considers the beam as a semi~nfinite plate extending above the line np. This corrective system mtroduces extra stresses over the top of the beam, which again can be removed by repeated application of Flamant's solution, and so on. This process is too slowly convergent and did not lead to a satisfactory result. A solution of the problem by means of trigonometric series was obtained by L. N. G. Filon. 2 He applied this solution to the case of concentrated loads and made calculations for severa! particular cases (see Art. 23), which are in good agreement with more recent investigations. 1

This coincides with the formula given by Stokes.

y)

3 10 C

_

as for a semi-infinite plate, in order to obtain the total stresses along the section AD. A comparison with a more accurate solution, given below (see table, page 106), shows that Eqs. (a) and (b) give the stresses with very good accuracy at all points except the point D at the bottom of the beam at ' which the correction to the simple beam formula is given as 3P

Frn. 65.

(3y yª)

'lrC

3

( y 2c 3

These stresses should be superposed on the stresses

-

y

101

2

Compt. rend., vol. 114, p. 1510, 1892. L. N, G, Filon, Trans. Roy. Soe. (London), series A, vol. 201, p. 63 1 1903,

THEORY OF ELASTICITY 102 Further progress in the solution of the problem was made by H. Lamb.1 Considering an infinite beam loaded at equal intervals by equal concentrated forces acting in the upward and downward directions alternately, he simplified the solution of the two-dimensional problem and obtained for several cases expressions for the deflection curves. It was shown in this manner that the elementary BernoulliEuler theory of bending is very accurate if the depth of the beam is small in comparison with its length. It was shown also that the correction for shearing force as given by Rankine's and Grashof's elementary theory (see page 43) is somewhat exaggerated and should be diminished to about 0.75 of its value. 2 A more detailed study of the stress distribution and of the curvature near the 3 point of application of a concentrated load was made by T. v. Kármán and F. Seewald. a Kármán considers an infinitely long beam and makes use of the solution for a semi-infinite plate with two equal and opposite couples acting on two neighboring points of its straight boundary (Fig. 56b). The stresses along the bottom of the beam which are introduced by this procedure can be removed by using a solution in the forro of a trigonometric series (Art. 23) which, for an infinitely long beam, will be represented by a Fourier integral. In this manner Kármán arrives at the stress function Ma (ac cosh ac + sinh ac) cosh ay - sinh aC sinh ay. ay da = 7 }o sinh 2ac + 2ac cos ax Ma f 00 (ac sinh ac + cosh ac) sinh ay - cosh ac cosh ay · ay da (e) - 7 }o sinh 2ac - 2ac cos ax a This function gives the stress distribution in the beam when the bending moment diagram . consists of a very narrow rectangle, as shown in Fig. 66. For the most general loading of M the beam by vertical forces applied at the top ;:._ _ _ _ _-J--1--c- . of the beam 4 the corresponding bending-moL----+=;...-----•--'~c X ment diagram can be divided into elementary rectangles such as the one shown in Fig. 66, Y and the corresponding stress function will be FIG. 66. obtained by integrating expression (e) along the length of the beam. This method of solution was applied by Seewald to the case of a beam loaded by a concentrated force P (Fig. 64). He shows that the stress O',, can be split into two

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

y=-C

3.0

2.5 2.0 1.5

H. Lamb, Atti IV congr. intern. matemat., vol. 3, p. 12, Rome, 1909. s Filon came to the sarne conclusion in his paper (loc. cit). a Abhandl. aerodynam. lnst., Tech. Hochschule, Aachen, vol. 7, 1927. 'The case of a concentrated load applied half way between the top and bottom of the beam was discussed by R. C. J. Howland, Proc. Roy. Soe. (London), vol. 124, p. 89, 1929 (see p. 115 below), and pairs of forces within the beam by K. Girkmann, lngenieur-Archiv, vol. 13, p. 273, 1943. Concentrated longitudinal forces in the web of an 1-beam are considered by Girkmann in Oe$terr. lngenieur-Archiv, vol.1, p. 420,

LO

0.5

0.5

1.0 1.5 2.0 2.5 3.0 (a)

y=-f ~

<:§

~

~

e::::;

e::::;

1

1

\(>

~

~

<::>

, I

.3.0 2.5 2.0

1.5

1.5 2.0 2.5 3.0 (b)

y=O

" "' ....

~~~~~ i-'~c::::;~~

,.-f.;:: 'r 'r

;2;

~

e::::;

e::::;

1

1

~

~

g;

~

<::) 1

!º 3.0 2.5 2.0 1.5

1.0 0.5

0.5 1.0

.o

3.0 2.5 2.0

.X

,~,.,'1-Ç::>

.....

')>'ji<:jl~~

<::)

~. ~~~

1.5 1.0 0.5

0.5

1.0

X

e

1.5 2.0 2.5 3.0 (e)

'1-

~ ;.

.....

~

c::::; ;.

~

e

~ <::) ;.

1.5 2.0 2.5 3.0

(d}

y=+C

1

1946.

.2;

e

r"'

1

o:., =,8· C 1!. X

~

~ ~

~

1'

;.

c::::;

3.0 2.5 2.0

1.5

1.0 0.5

~

c::::;

1.5 2.0 2.5 (eJ FIG.

67.

~

~

c::::; ;.

3.0

e

103

THEORY OF ELASTICITY

104

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

parts: one, which can be calculated by the usual elementary beam formula; and another, which represents the local effect near the point of application of the load. This latter part, called
1.5

2.0

1.0

0.5

o

1.0

0.5

1.5

2.0 2.5

....

yc-f 3.0

p 3.0 2.5

p .......

2.0

1.5

1.0

0.5

~ ~~ ~ ·~ <:><:> ... ~ 1/ ~

~

~

<:)

1

~

1

~

-f----.,,

1.0

1.0 0.5

l.S

~

K

<:)

...

~l/

2.0 2.s 3.0

l§l:g~

'r'i'~'r~

~ ;;:::

<::! ;.

<:)

<::i

;.

~

e

<:)

...

3.0

2.0

1.5

1.0

0.5

1.5 2.0 2.5 3.0

~

(e)

1

~

1 That is, stresses which must be superposed on those obtained from the ordinary beam formula.

1"

1 \.

3.0

l

V

~ i::;

<::!

<::i

<::i <::!<::!

0.5

1

(õ}

~

l..I

. 3.0

1.0 1.5 2.0

~

-'-

~

the point of application of the load and at a distance equal to the depth of the beam are usually negligible. Using the values of the factor J3 for x = O, the local stresses at five points of the cross section AD under the load (Fig. 64) are tabulated below. For comparison the local stresses,1 as obtained from Eqs. (a) and (b} (page 101), are also given. It is seen that these equations give the local stresses with sufficient accuracy. Knowing the stresses, the curvature and the defiection of the beam can be calculated without any difficulty. These calculations show that the curvature of the defiection curve can also be split into two parts--one as given by the

~

2.0

1/ ...,~ ...,~~"'> ...,~ o

fl~~

FIG. 68.

:i .~\

l.O l.S

I 1.0

~~~~~

i

I

;.

o;::>

y=O

22

fJ

o.s o o.s .....

~

i

0.5

<:$

<:)

~U

~

(b)

1.0

.. ..

~

::;:;

~

/J ._..._

~

§i

1.5

<::! ;.

o

y=O

= z

<::i <::i <::i <::i ;. ;.

1

~

- '-

(a}

2.5 2.0

~

(a)

<::!

3.0

~~

X

1

1.5

-

e

\~ ~1

2.0

!'-

~~ ~ 'I:. ~~

3.0

li

r\

3.0 2.5

\

~

y=+f

~J

<::i

2,0

1.5 ·-

L-0 . 0.5 -

-- -

-·-

JI

IJ º-

li ·-'-

::s

~

a~ .,..! e::>~ I/~ <:::s <::! <::! <::i !:<>

~

l/v

-·-··

0.5 - l.O -1.s 2.0

(C)

&1

<::i

3.0

--

X

e

X

e

105

THEORY OF ELASTICITY

106

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

TABLE OF F ACTORS fl FOR THE MIDDLE

e

-e

y=

The corresponding deftection at the middle is

e

o

-2

e

2

ô1

=

Uu

=

0.121 -0.456

0.428 -1.23

00

-0.136 -0.145

-0.133

Cfy

=

0.426 -1.22

00

1

0.159 -0.477

-o .108

-0.254 1

ô = ôo

elementary beam theory and the other representing the lo~al effect of the concm.1.trated load P. This additional curvature of the center line can be represented by the formula 1 p - ==a(d)

r

Ecz

Taking

'

L...

-~ 3.0

2.5 2.0

l.S

1.0

,/

f:=a~z ~ .... """ ~ !'. ~~

~ ~ ~i::) V~~ """f\ ~ <::i<::j~i'.:)

o.s

....,

"<

i- ~

~C::S<::i<::S ..............

o.s

1.0

1

~

~ e::; 1

~~ 1

1

p

E

-

Pl

3

ô2 = 48EI

( 3 + Pl 4c 4G

3 311) P - lOE - 4E - 0·21 E

(75)

= 0.3, this gives 3 [

1

+ 2.85 (2c) T -

0.84 (2c) T

2

ªJ

(75')

The elementary Rankine-Grashof theory (see page -43) gives for this case 3

PlEJ ô = 48

[

1

+ 3.90 (2c)2] T

(g)

It appears that Eq. (g) gives an exaggerated value for the correction dueto shear. ln these formulas the defiection due to p local deformation at the supports is not taken into account.

e::; 1

11

+ ôi

Pl ô = 48EI

in which a is a numerical factor varying along_the length of the beam .. Several values of thi'l factor are given in Fig. 70. It is seen that at cross sect1ons at a

a

(f)

Denoting now by ôo the defiection as calculated by using the elementary theory, the total deflection under the load is

o

-0.155 1

1

(2 - -2_ - ª") lOE 4E

4c 4G

ô2 = 0.21

Approximate solution

0.573

4

o

1

qz' =

= !! = Pl

From this defl.ection a small further correction ô2, removing the sharp che.nge of slope at A, should be subtracted. This quantity was also calculated by Seewald and is equal to

Exact solution tl:c'

107

+:&.

e

1.5 2.0 2.5 3.0

FIG. 70.

distance greater than half of the depth of the beam the additional curvature is negligible.

37. Stresses in a Circular Disk. Let us begin with the simple case d of two equal and opposite forces -=--t--=-~-\-~__,.~~~~--rD=-if-x P acting along a diameter AB (Fig. 72). Assuming that each of the forces produces a simple radial stress distribution [Eq. (66)], we can find what forces should be applied at the circumference of the disk in y order to maintain such a stress disFIG. 72. tribution. At any point M of the circumference we have compressions in the directions of r and r1 equal to 2P cos 8 2P cos 81 . 1 ~ -r- and -r -r, respectively. Since r and r 1 are perpendic1 ular to each other and fl

On account of this localized effect on the curvature, the two branches of the deflection curve AB and AC (Fig. 71) may be considered to meet at an angle eque.I to (e)

cos 8 - r-

cos 81

1

= --;:;--- = d

1

It is e.ssumed that P is the force per unit thickness of the disk.

(a)

109

THEORY OF ELASTICITY

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

where d is the diameter of the disk, we conclude that the two principal stresses at M are two equal compressive stresses of magnitude 2P /7íd. Hence the sarne compressive stress is acting on any plane through M perpendicular to the plane of the disk, and normal compressive forces of the constant intensity 2P /7íd should be applied to the circumference of the disk in order to maintain the assumed pair of simple radial stress distributions. If the boundary of the disk is free from externai forces, the stress at any point is therefore obtained by superposing a unif orm tension in the plane of the disk of the magnitude 2P /7íd on the above two simple radial stress distributions. Let us consider the stress on the horizontal diametral section of the disk at N. From symmetry it can be concluded that there will be no shearing stress on this plane. The normal stress produced by the two equal radial compressions is

The normal MN to thetangent at M is the diameter of the disk; hencé MAN and MBN are right-a:iigled triangles and the angles which the normal MO makes with r and r1 are?r/2 - 81 and?r/2 - 8, respectively. The normal and shearing stresses on an element of the boundary at M are then

108

- 2 . 2P cos 8 . cos2 8 7í r

3

= _ 4P cos 8

11

7í

+ 2P ?rd

r

or, using the fact that cos 8 =

d

-===== 2 2

yd + 4x

we find 2P [

4d4

"11 = ?rd 1 - (d2 + 4x2)2

FIG. 73.

]

(b)

The maximum compressive stress along the diameter CD is at the center of the disk, where 6P
r

r

r1

.,,.

2

r

2

_ 2P (cos 8 sin Ih .,,.

r

r

+ cos 81 sin r1

• 81 cos 8 = - 2P -.,,. (cos -r-8 sm 1

-

cos

2

º)

01 -~

8)

(e)

• sm 8 cos 8)

These equations can be simplified if we observe that, from the triangles MAN andMBN, r = d sin 81, r 1 = d sin 8 Substituting in Eqs. (e), we find

in which r is the distance AN, and 8 the angle between AN and the . vertical diameter. Superposing on this the uniform tension 2P /7íd, the total normal stress on the horizontal plane at N is
f)i) _2P.,,. cosr1 81 cos 2 (~2 _

" = _ 2P cos 8 cos 2 (~ :


2P

= - ?rd sin (8

+ 81),

r = O

(d)

From Fig. 73 it may be seen that sin (O + 81) remains constant around the boundary. Hence uniformly distributed compressive forces of the intensity 2P j.,,.d sin (O + 01) should be applied to the boundary in order to maintain the assumed radial stress distributions. To obtain the solution for a disk with its boundary free from uniform compression it is only necessary to superpose on the above two simple radial distributions a uniform tension of the intensity 2P /.,,.d sin (O + 01). The problem of the stress distribution in a disk can be solved for the more general case when any system of forces in equilibrium is acting on the boundary of the disk. 1 Let us take one of these forces, acting at A in the FIG. 74. ?irection of the chord AB (Fig. 74). Assummg again a simple radial stress distribution we have at point M a simple radial compression of the magnitude (2P /7í) cos fh/r 1 acting in the direction of AM. 1

The problems discussed in this article were solved by H. Hertz, Z. Math. Physik, Vol. 28, 1883, or "Gesammelte Werke," voI. 1, p. 283; and J. H. Michell, Proc. London Math. Soe., vol. 32, p. 44, 1900, and vol. 34, p. 134, 1901. The problem corresponding to Fig. 72 when the disk is replaced by a rectangle is considered by J: N_. Goodier, Trans. A.S.M.E., vol. 54, p. 173, 1932, including the effects of distnbution of the load over small segments of the boundary .

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

THEORY OF ELASTICITY

110

'I'he moment of all the externa! forces with respect to O, from Fig. 74, is

Let us take as origin of polar coordinates the center O of the disk, and measure 9 as shown in the figure. Then the normal and the shearing components of the stress acting on an element tangential to the boundary at M can easily be calculated if we observe that the angle between the normal MO to the element and the direction ri of the compression is equal to 7r/2 - 92. Then
2P cos 9i . = - - - - s1n 2 92 7r ri 2P cos 9i •

Tr8 =

-

-

7r

l

(e)

- - Slil 92 COS 92 Ti

p sin (Ili = - 1rd

+ 92)

cos (9i

+ 92)

p

'TrfJ

= - 1rd

-

pd sin

1r

(92 -

9i)

p

- 1rd cos (92 -

9i)

obtained by superposing the following three stresses on the element. (1) A normal stress uniformly distributed along the boundary:

sin (9i

+ 92)

(g)

(2) A shearing stress uniformly distributed along the boundary: p - 1rd cos (9i

+ 92)

(h)

(3) A stress of which the normal and shearing components are (k)

and

Observing that the angle between the force P and the tangent at M is Oi - 92, it can be concluded that the stress (k) is of magnitude P /7rd and acts in the direction opposite to the direction of the force P. Assume now that there are several forces acting on the disk and each of them produces a simple radial stress distribution. Then the forces to be applied at the boundary in order to maintain such a stress distribution are: (1) A normal force uniformly distributed along the boundary, of intensity

-I:i~~+~

ro

on the simple radial distributions. By using this general method, various other cases of stress distribution in diskR can easily be solved. i We may select, for instance, the case of a couple acting on the disk (Fig. 75), balanced by a couple applied at the center of the disk. Assuming two equal radial stress distributions at A and B, we see that, "in this case, (Z) and the summation of (k) are zero and only shearing FIG. 75. forces (m) need be applied at the boundary in order to maintain the simple radial stress distributions. The intensity of these forces, from (m), is 2P 2Mt (n) - -d cos (Oi + 92) = - 1r 7rd• wher~ Mt

is the moment of the couple. To free the boundary of the disk from forces and transfer the couple balancing the pair of forces p from the crrc_umfe~en?e o~ the disk to its center, it is necessary to superpose on the simple radial distnbut1ons the stresses of the case shown in Fig. 75b. These latter stresses, produced by pure circumferential shear, can easily be calculated if we observe that for each concentric circle of radius r the shearing stresses must give a couple Me. Hence, Mt Tr921rT 2 = M1, Tr9 = (p) 27rr 2 s~earmg

These stresses may aleo be derived from the general equations (38) by taking as the stress function

"'= ~;9

(2) Shearing forces of intensity

~

-I:i

"

1 i

cos (9i

+ 92)

(m)

(3) A force, the intensity and direction of which are obtained by vectoria~ summation of expressions (k). The summation must extend over all forces acting on

i

l •

:'

~,.1

füt> boundary .

+ 92)d

(f)

This stress acting on the element tangential to the boundary at point M can be

-:i

p cos (9~

and, as this moment must be zero for a system in equilibrium we conclude that the shearing forces (m) are zero. The force obtained by summ~tion of the stresses (k), proportional to the vectorial sum of the externa! forces, is also zero for a system in equilibrium. Hence i~ is onl:y necessary to apply p at the boundary of the disk a uniform compression (Z) in order to maintain the simple radial distributions. If the boundary is free from uniform compression, the stress at any point of the disk is obtained by superposing a uniform tension of magnitude

Since, from the triangle AMN, ri = d sin 92, Eqs. (e) can be written in the forro q,

111

from which Ur 1

== 0'9 = O,

M1 Tr9 = 27rT2

Several interestini examples are discussed by J. H. Michell, loc . .U.

(q)

::

·"

f

~i1

112

1'WO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

1'HEORY OF ELAS1'ICI1'Y

38. Force at a Point of an lnfinite Plate. If a force P acts in the middle plane of an infinite plate (Fig. 76a), the stress distribution can easily be obtained by superposition of systems which we have already discussed. We cannot, however, construct a solution by simple superposition of two solutions for a semi-infinite plate as shown in Figs. 76b and 76c. Although the vertical displacements are the same in both these cases, the horizontal displacements along the straight boundaries are different. While in the case 7Gb this displacement is away from

113

the case of a se~i-infinite plate. The displacement along the straight boundary of t~Is plate in the direction of the shearing force acting on the boundary IS, from Eq. (61), D'll' (b)

E

1:he constant of int~gration D must now be adjusted so as to make the displacement resultmg from (a) and (b) vanish. Then (e)

With the result of superposing cases 76b , 76e, 76d , an d · this· adjustment fi · e IS an m mte p 1 ate loaded at a point, Fig. 7Ga. 76 The stress distribution in the plate is now easily obtaI'ned b y super· h . posmg t e stresses m a semi-infinite plate produced by a normal load P /2 · ·at theh boundary (see Art. 33) on the stresses in the curved b ar conammg · . t e constant of integration D · ObservI'ng the d'ff I erence m t measun~g the angle fJ in Figs. 46 and 76 and using Eqs. (60), the stresses m the curved bar are, for fJ as in Fig. 76,

(a) .X

D cos fJ

r

ur= uo

= D cos r

fJ

- D sin 6

Tro -

---

r

1 - v P cos fJ

-~--r-

1 - v p cos fJ

=~--r-

1 - v P sin () =- --471'

r

Combining this with stresses (66) calculated for the load P/2 bt · the f n · t d' . , we o ain o owmg s ress Istnbution in the infinite plate: O'r

(C)

471' r 1 - vP cos fJ

FIG. 76.

o-o=~

the point O, in the case 76c it is toward the point O. The magnitudes of these displacements in both cases, from Eq. (71), is

Tro

1 -

)1

-4E ·P

= 1 - PP COS () _ P COS (}

(a)

~

(3

+

v) P COS ()

471'

(76)

r

1 - vP sin (} 471' r

= -----

~:u:~:~n~ out fr~m t~e plate at the point O (Fig. 76a) a small element ct yh a cy~md~Ical surface of radius r, and projecting the forces g on t e cylmdncal boundary of the element on the x - and y-axes, we find

actin This difference in the horizontaldisplacements may be eliminated by combining the cases 76b and 76c'with the cases 76d and 76e in which shearing forces act along the straight boundaries. The displacements for these latter cases can be obtained from the problem of bending of a curved bar, shown in Fig. 46. Making the inner radius of this bar ap\)roach zero and the outer radius increase indefinitely, we arrive at

r

lo" (
X = 2

Tro

sin O)r d() = p

Y

Tro

cos O)r dfJ = O

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

THEORY OF ELASTICITY

114

we can obtain thef solution of the problem shown in Fig. 78. stress components or this case are

i.e., the forces acting on the boundary of the cylindrical element represent the load P applied at the point O. By using Eqs. (13) the stress components, in Cartesian coordinates, are found from Eqs. (76): p cos 8

o-,,= 4ir-r- [-(3 p cos 8 o- = 4ir-r- [1 11

'TzY

• + v) + 2(1 + v) sm . v - 2(1 + v) sm 8]

47r

81

r

+ 2(1 + v) cos

2

(77)

TzY

81

,=

O"z -

o-,,= dP [-(3 41íT 2

11

TzY

+ v) cos 8 + (1 2

= dP [(1 - v) cos 2 8 + (1 41íT 2

=

dP [ -(6 41íT 2

- v) sin 2

8 + 8(1 + v)

sin

2

8 cos

+ 3v) sin2 8 - 8(1

+ 2v) + 8(1 + v) sin

2

2

O"r

dP

= -2(1 - v)-1 47rr 2

8]

v) sin 2 8 cos 2 81

8] sin 8 cos 8

It can beseen that the stress components decrease rapidly, asr increases, and are negligible when r is large in comparison with d. Such a result is to be expected in accordance with Saint-Venant's principle if we have two forces in equilibrium applied very near to each other. By superposing two stress distributions such as given by Eqs. (78) 1

0"9

X

FIG. 78.

= 2(1 - v) 4irr2' ~ d

Tro

= O

(79)

This solution can be made to agre_e with solution (46) for a thick cylin~er subm1tted to the action of internai pressure if t_he o_uter diameter of _the cylinder is taken as mfimtely great. Y ln the sarne manner we can get a- solution for the case shown in Fig. 79a. The stress components are 1 O"r

=

0"9

=O,

(80)

They represen_t the stresses produced by a couple M apphed at the origin (Fig. 79b).

X

FIG. 79.

(78)

+

Y-t~------!õ

-2(1 - v) dP2 sin 28 47rr

( O"x +ao-,, d) = - d -ao-,, = - d (ao-,, sin 8) -ar COS 8 - -ao-,, ax ax a8 -r-

Thus the stress components for the case of Fig. 77 are obtained from Eqs. (77) by differentiation. ln this manner we find

u

=

The

-2(1 - v) dP2 (1 - 2 sin 2 8) 47rr

The sarne stress distribution expressed m polar coordinates is

From solution (77), for one concentrated force, solutions for other kinds of loading can be obtained by superposition. Take, for instance, the case shown in Fig. 77, in which two equal and opposite forces acting on an infinite plate are applied at two points O and 0 1 a very small distance d apart. The stress at any point M is obtained by superposing on the stress produced by the force at O the stress produced by the other force at 0 1 . Considering, for instance, an element at M perpendicular to the x-axis and FIG. 77. denoting by o- x the normal stress produced on the element by the force at O, the normal stress o-,/ produced by the two forces shown in the figure is O"z

=

u11 = 2(1 - v) 4irr dP2 (1 - 2 sin2 8)

2

P sin - 8 [1 - v

= - -

2

O"z

115

FIG. 80.

If instead an infinite p1~t e ~e h ave to d eal with · an infmitely long strip subjected to the of action (77) as if the la of a l~ng1t~d1~al for~e p ~Fig. 80), we may begin with solution the strip resulEngt~ wer:~-nfimte m all direct1ons. The stresses along the edges of t lS procedu~e can be annulled by superposing an equal and opposite system mined by using ~h s r~sses pro uced by this corrective system can be deterR. C. J. Howland• :;oenera method described in Art. 23. Calculations made by p diminish rapidly aswt~ha~~~e loca~ stresses produced by the concentrated force increases and at d" e lS ance rom the point of application of the load stresses ~ver the c:;tance~.gre~ter tha_n the wi~th of the strip the distribution of i A ss sec lOil lS pract1cally umform. ln the table below several

;i:

• ~cE. :S:t . LoSeve, "Theory of Elasticity,'' p. 214, Cambridge 1927 ealsoapape r b Y E · MI ' ·vol. 5, p. 314 1925, . c i . e an, Z. angew. Math. Mech.,

117

THEORY OF ELASTICITY

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

. t d on the assumption that the values of the stresses u. and u • are g1v~n, c~1cu1~ e. 1 . . fixed at the end x = + oo and P01sson s rat10 is •· s t np1s

The first three terms in the first line of this expression represent the solution for the stress distribution symmetrical with respect to the origin of coordinates (see Art. 26). The fourth term gives the stress distribution for the case shown in Fig. 57. The fifth term gives the solution for pure shear (Fig. 75b). The first term in the second line is the simple radial distribution for a load in the direction 8 = o. The remaining terms of the second line represent the solution for a portion of a circular ring bent by a radial force (Fig. 46). By a combination of ali the terms of the second line the solution for a force acting on an infinite plate was obtained (Art. 38). Analogous solutions are obtained also from the third line of expression (81), the only difference being that the direction of the force is changed by .,,./2. The further terms of (81) represent solutions for shearing and normal forces, proportional to sin n8 and cos n8, acting on the inner and outer boundaries of a circular ring. We had an example of this kind in discussing the stress distribution around a small circular hole (Art. 32). In the case of a portion of a circular ring the constants of integration in expression (81) can be calculated without any difficulty from the boundary conditions. If we have a complete ring, certain additional investigations of thc displacements are sometimes necessary in determining these constants. We shall consider the general case of a complete ring and assume that the intensities of the normal and shearing forces at the boundaries r = a and r = b are given by the following trigonometrical series:

116

u,2c

p

y=O

.. . . .

.....

+0.159

+0.511

0.532

0.521

0.500

=

0.110

0.364

.....

.....

.....

=

p

7r

30

18

=

.....

= =

e

y=O

u,2c

p

u,2c

p


p

7r

7r

~=

y=c

o

-0.992

u,2c

y=O

- 30

-0.118

p

y=O

7r

- 18

=

u,2c

y=c

-9

-3

e

7r

7r

7r

X

- =

7r

00

7r

9

3

2

.....

1.992

1.118

1.002

0.479

0.468

0.489

0.841

0.973

.....

.....

-0.364

-0.110

-0.049

ao log r + bor2 + cor2 log r + dor28 + ao'8 + ~ r8 sin 8 + (b,rª + a1'r-1 + b,'r log r) cos 8 2 - ~ r8 cos 8 + (dirª + ei'r-1 + di'r log r) sin 8 2

+

l.,

(anr"

+ b,.rn+2 + an'r-n + bn'r-n+2) cos n8

n=2

+

l.,

(c,.r"

+ d,.rn+2 + c,.'r-n + d,.'r-n+2) sin n8

(81)

n=2

1 Z. angew. Math. Me~h., vol. 12, p. 2 This solution was given ~y J.

34~, 1!~2Proc.

London Math. Soe., vol. 31, Physik vol. 52, p. 348, 1905. An ~llip:c~l ring w~s given by A. Timpe, M ath.

Hi ~c~h

p. 1100, 18991. t1~oi: f:~s~h~.c~~mJe~n ana ogous so u z., vol. 17, p. 189 1 1923.

(ur)r..a = Ao

l +l ., + l + l

+

.,

A,. cos nlJ

l

+·

n=l

e distance Stresses produced in a semi-infimte plate by a force apphed at som d' d b E Melan 1 from the edge have been iscuss~ ~ . nsionitl Problem in Polar Coordinates. 39. General Solution of th~ lwo- me f the two dimensional problem in polar . d' d various part1cu ar cases o . f th Havmg iscusse . .t. t write down the general solut1on o e . t re now in a pos1 ion o . . h d coor ma es we a . f the stress function , sat1sfymg t e comproblem. The general express1on or 2 patibility equation (39) is =

.,

(ur)r-1> = Ao'

(rre)r-b = Co'

+

B,.' sin n6

n=l

C,. cos n8

n=l

'

l ., + l + l

A,.' cos nlJ

n=l

helr-a = Co

B,. sin nlJ

n=l

(a)

D,. sin n6

n=l

C,.' cos nlJ

n=l

D,.' sin n8

n=l

in which the constants Ao, A,., B,., . . . , are to be calculated in the usual manner from the given distribution of forces at the boundaries (see page 49). Calculating the stress components from expression (81) by using Eqs. (38), and comparing the values of these :.iomponents for r = a and r = b with those given by Eqs. (a), we obtain a sufficieL:t number of equations to determine the constants of integration in all cases with n ;:;: 2. For n = O, i.e., for the terms in the first line of expression (81), and for n = 1, i.e., for the terms in the second and third lines, further investigatiom are necessary . Taking the first line of expression (81) as a stress function, the constant ao' is determined by the magnitude of the shearing forces uniformly distributed along the boundaries (see page 111). The stress distribution given by the term with the factor do is many valued (see page 93) and, in a complete ring, we must assume do = O. For the determination of the remaining three constants ao, b0 , and cr we have only two equations, and

(ur)r-b

= Ao'

119

THEORY OF ELASTICITY

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

The additional equation for determining these constants is obtained from the consideration of displacements. The displacements in a complete ring should be single-valued functions of 8. Our previous investigation shows (see Art. 26) that this condition is fulfilled if we put co = O. Then the remaining two constants ao and bo are determined from the two boundary conditions stated above. Let us consider now, in more detail, the terms for which n = 1. For determining the eight constants a1, b1, . . . , dl' entering into the second and the third tines of expression (81), we calculate the stress components u, and Tr9 using this portion of . Then using conditions (a) and equating corresponding coefficients of sin n8 and cos n8 at the inner and outer boundaries, we obtain the following

E'.qs. (q), we find the following many-valued terms in the d1splacements u and v, respectively: expressions for the

118

eight equations:

(a1 + bl')a- 1 + 2b1a - 2ai'a- 3 = A 1 (a 1 + b1')b- 1 + 2b 1b - 2a1'b- 3 = A 1' (c1 + dl')a- 1 2d1a - 2c1'a- 3 = B1 (c 1 d 1')b-1 + 2d1b - 2c1'b- 3 = Bi' 2d1a - 2ci'a- 3 + d1'a-1 = -C 1 2d1b - 2c1'b- 3 + d1'b- 1 = -C,' 2b1a - 2a1' a- 3 + b1' a- 1 = D 1 2b 1b - 2a 1'b-a + b1'b- 1 = D 1'

8sin 9

2 - r (J cos (J + --i

(J

a1 l -

11

a1 l -

2b'

•

2b'

11

cos

(J

These terms must vanish in the case of a com Plet e nng, · h ence a1 l -

2b'

2 - r +--i =o 11

or (b)

+

+

2 - r 8sm 8 +--i

b' 1

=

_a1(l-11)

(j)

4

Considering the third line of expression (81) in th e sarne manner, we find te)

d ' = _ c1(l - 11)

1

Comparing Eqs. (b) with (e) it can be seen that they are compatible only if a1a- 1 = Al - Dl aib- 1 = Ai' - D,' c1a- 1 = B1 + C1 c1b-1 = Bi' + C1'

(d)

(g)

4

Equations (j) and (g) together with E (b) d qs.. an (e), are now sufficient for determining all the constant~ in the stre .d . f . ss funct1on represented by the second and th e thir 1mes o express1on (81). We conclude that in the case of a complete rin th b d .. not sufficient for the determination of the stress arydc?td1tions (a) are consider the displacements The d" 1 t . ' an 1 is necessary to valued and to satisfy this ~ondition I~e a~:~te~a:~n a complete ring must be single

Jistr~bu~~o:

from which it follows that (e)

a(A1 - Dl) = b(Ai' - Dl'),

Co

It can be shown that Eqs. (e) are always fulfilled if the forces acting on the ring are in equilibrium. Taking, for instance, the sum of the components of all the forces in the direction of the x-axis as zero, we find

fo

1! i

2

" { [b(ur)r-b

- a(ur)r-al

COS

8 -

[b(Tr9)r-; - a(Tr9)r-al sin 8} d8 =

Q

Substituting for ur and r,8 from (a), we arrive at the first of Eqs. (e). ln the sarne manner, by resolvingall the forces along the y-axis, we obtain the second of Eqs. (e). When a1 and c1 are determined from Eqs. (d) the two systems of Eqs. (b) and (e) become identical, and we have only four equations for determining the remaining six constants. The necessary two additional equations are obtained by considering the displacements. The terms in the second line in expression (81) represent the stress function for a combination of a simple radial distribution and the 1 bending stresses in a curved bar (Fig. 46). By superposing the general expressions for the displacements in these two cases, namely Eqs. (g) (page 90) and Eqs. (q) (page 77), and, substituting ai/2 for -P /'Ir ..'n Eqs. (g) and bl' for D in 1 It should be noted that 8 + (7r/2) must be substituted for 8 if the angle is measured from the vertical axis, as in Fig. 52; instead of from the horizontal axis, as in Fig. 46.

bl' = -

= 0,

ª' (1 4-

li)

d

1

'

'

= _ c1(l - 11) 4

(82)

!~: s:;et~ed:~:~r~!~on ~~ ~ecomple~e ~ing will usually depend on the.ela!~i:o~~~~~~ We see that the constants bl' and dl' depend on Poisson's ratio

ª'

and e, vanish so that if [see Eqs. (d)] '

.

fro:~es 1~2 ep~~d~nt ~f q. (

Al = Dl

the elas~ic con~tants only when ), 1 - d 1 = O. Th1s particular case occurs and

B 1 = -C 1

et

Wethe have such ring va:i~~:sdit"i~;h t~e r:sultant of the forces applied to each boundary of tion of forces applied to th eh or ~ns ance, the res~ltant component in the x-direce oun ary r =a. Th1s component, from (a), is

Ío2r

(ur

cos

(J -

Tr9

sin 8)a d9 = a,,-(A1 - D1)

If it vanishes we find A - D I the y-direction we obta; -B :_ ~ the sarne manner, by resolving the forces in we may concl~de h m 1 - - 1 when the y-component is zero. From this of the elastic const!n~: the stress d~st~ibution in a complete ring is independent boundary is zero Th of the material if the resultant of the forces applied to each ' e moment of these forces need not be zero,

I'

o;==~-

1

THEORY OF ELASTICITY

120

These conclusions for the case of a circular ring hold also in the most general case of the two-dimensional problem for a multiply-conneeted body. From general investigations made by J. H. Michell, 1 it follows that, for multiply-connected bodies (Fig. 81), equations analogous to Eqs. (82) and expressing the condition that the displacements are single valued should be derived for each independent eircuit such as the circuits A and B in the figure. The stress distributions in such bodies generally depend on the elastic constants of the material. They are independent of these constants only if the resultant force on each boundary vanishes. 2 Quantitatively the effect of the moduli on the maximum stress is usually very small, and in practice it can be neglected. 3 This conclusion is of practical importance. W e shall see later that in the case of trans:FIG. 81. parent materials, such as glass or bakelite, it is possible to determine the stresses by an optical method, using polarized light (see page 131) and this conclusion means that the experimental results obtained with a transparent material can be applied immediately to any other material such as

121

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

effect · · an mit1al · · . t . by making a cut along a vertºma1 radºms and imposmg . · is obtained isplacemen ot one end of the ring with respect to th th · · · 0 d (Fi 82b) Th · · · e er m the vertical directº 10n g. . .F·igs. 82 a and 82b d .to th e m1tial stresses produced in the cases shown m correspon . e many-valued terms of the general solution when Eqs (jº) d (g) are not sat1sfied. · an Thet complete of these problems can be obt ame . d b Y app1ymg . the results 31 Th solution dº of Ar . · . d t e fisplacements dº · · given by Eqs. (q) of Art · 31 will b e foun d to have · (see p rob. 4, page 126). th e reqmre ype o IBcontmmty when applied to a rmg

r.

,,

\

'~

p

1\, ~,

!'),,

10~

~'

f/'.:~

V,-+>'. tí? !lo.

~~ ~ ~ b 0.9b

steel if the external forces are the sarne.

~o. ~

lrO.lb/ ~-6~

'/ª b

N~ ~~ f0 ~ ~ !/.% ,,~

p

,,

\

'

B»

1~' i;'l, f0 0, ~ \

b

0.9b

o.

I(}

l(Gl ]b

\

''-.:~ ~ ~ ~ r~ 0,

(a)

~ I~

o. b

I\' ~F/,:

l'S 10 8=90º

:/,tl.6b

·~ 10 V',,; ~ '~ ~1% ~ (\~ 1/,1.~

8=0º

'~

~. ~

'~ ~

1\ \

(h}

(a) FIG. 82.

It was mentioned before (see page 68) that the physical meaning of manyvalued solutions can be demonstrated by considering initial stresses in a multiplyconnected body. Suppose, for instance, that Eq. (f) above is not satisfied. The corresponding displacement is shown in Fig. 82a. Such a displacement can be produced by cutting the ring and applying forces P. If now the ends of the ring are joined again by welding or other means, a ring with initial stresses is obtained. 4 The magnitudes of these stresses depend on the initial displacement d. A similar

Loc. cit. It must be remembered that the body forces were taken as zero. a An investigation of this subject is given by L. N. G. Filon, Brit. Assoe. Advancement Sei. Rept., 1921. · See E. G. Coker and L. N. G. Filon, "Photo-elasticity," 1

2

Arts. 6.07 and 6.16. A discussion of such stresses is given by A. Timpe, Z. Math. Physik, vol. 52, 4 p. 348, 1905. A general theory is given by V. Volterra, Ann. éeole norm., Paris, series 3, vol. 24, pp. 401-517, 1907. See also A. E. H.Love, "Mathematical Theory of Elasticity,'' 4th ed., p. 221, 1927; J. N. Goodier, Proe. Fifth InterP··

Congr. Applied Meehanics, 1938, p. 129.

(cJ

(ó)

FIG. 83.

40. Applications of the General Solution in p 1 . application of the general solution of the t o-d· º.ar ~oordmat~s. As a first nates let us consider a circular rº w imens1ona problem m polar coordiacting along a diameterl (Fig ;:~ co~r~se~ by_ two equal and opposite forces t . a . . e egm w1th the solution for a solid disk (Art. 37). By cutti with normal and sh:a~i~~ f:r~~:~i~!;~~ ~o~ of radius a in this disk, we are left forces can be annulled b . I u e round the edge of the hole. These This latter system can b; r:u~erpo:1~g ~n equal. and opposite system of forces. . . p esen e w1th sufficient accuracy by using the first few terms of a F . obtained by usin ouner the series e . The ~ th e correspondmg stresses in the ring are together with the ~tresse~ ;:ira\ stol~t10~ of the_ pr~vious article. These stresses cu a e as or a sohd d1sk constitute the total stresses 1 See S. Timoshenko B ll p l h p. 1014, 1922. See als~ Ku Wieo ~~e . In~t. Kiew, 1910, and Phil. Mag., vol. 44, II, p. 1119, 1915. . g rdt, Sitzber. Akad. Wiss., Wien, vol. 124, Abt.

122

THEORY OF ELASTICITY

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES 123

in the ring. The ratios a 8 : 2P /irb, calculated in this man~er f?r various points o~ the cross sections mn and m1n1 for the case b = 2a, are g1ven m the table below.

r =

b

0.9b

O.Sb

0.7b

0.6b

tion that there are only normal pressures acting on the inner and outer boundaries having the magnitudes: 1

_ 2P cos o for _ "!: ~ e ~ '.!: ,.. a 2"' "'2 (
0.5b

Exact theory

mn m,n,

1.477 -2.185

2.610 -3.788

-0.1131 -0.594

-2.012 1.240

-4.610 4.002

-8.942 10 .147

-4.806 5.108

-8.653 11.18

-4.86 5.20

-7.04 8.67

i.e., the pressures are distributed along the lower half of the inner edge an:l the upper half of the outer edge of the eye-shaped end of the bar. After expanding

Hyperbolic stress distribution

mn m1n1

2.885 -7.036

1.6021 -5.010

0.001 1 -2.060 -2.482 0.772

Linear stress distribution

mn m1n1

1.71 -5.20

3.90 -8.67 1

-0.48 -1.73

-2.67 1.73

1

For comparison we give the values of t~e sarne stre~ses calculated from the .two elementary theories based on the followmg assumpt1ons: (1) that c~oss sections remain plane; in which case the normal stresses over the cross sect10~ fo!low a hyperbolic law; (2) that the stresses are d1stnbuted according to a linear law. The table shows that for the cross section mn, which is at a comparatively large distance from the points of application of the loads P, the hyperbolic stress distribution gives results which are very nearly exact. The error in the maxim~m stress is only about 3 per cent. For the cross section m 1n 1 the errors of the approximate solution are much larger. It is interesting to note that the resultant of the normal stresses over the cross section m1n1 is P /.-. This is to be expected if we remember the wedge action of the concentrated force illustrated by Fig. 65d. The distribution of normal stresses over the cross section mn and m 1n 1 calculated by the three above methods is Frn. 84. Th h d li d shown in Figs. 83b and 83c. e met o app e above to the case of two equal and opposite forces can be used for the general case of loading of a circular ring by concentrated forces. 2 As a second example we consider the end of an eyebar 3 (Fig. 84). The distribution of pressures along the edge of the hole depends on the ~mount of clearancf, between the bolt and the hole. The following results are _obtamed on the assumpThe thickness of the plate is taken as unity. . . L. N. G. Filon, The Stresses in a Circular Ring, Se~e~ted E"!gineering Papers, No. 12, London, 1924, published by the Institution of Civil Engmeers.. a H. Reissner, Jahrb. wiss. Gesellsch. Luftfahrt, p. 126, 1928; H. Reissner, and F. Strauch, lngenieur-Archiv, vol. 4, p. 481, 1933. i

2

+0.068

+J.85

+0.169

Frn. 85.

these distributions into trigonometric series, the stresses can be calculated by using the general solution (81) of the previous article. Figure 85 shows the values of the ratio ae :P /2a, calculated for the cross sections mn and m1n 1 for b/a = 4 and 2 b/a = 2. It should be noted that in this case the resultant of the forces acting on each boundary does not vanish, hence the stress distribution depends on elastic constants of the material. The above calculations are for Poisson's ratio 11 = 0.3. 41. A Wedge Loaded along the Faces. The general solution (81) can be used also for polynomial distributions of load on the faces of a wedge. a By calculating 1

P is the force per unit thickness of the plate. For experimental determinations of the stress distribution in eyebars by the Photoelastic method see E. G. Coker and L. N. G. Filon, "Photo-elasticity," Art. 6.18, and K. Takemura and Y. Hosokawa, Rept. 12, 1926, Aeronaut. Research lnst., Tôkyô Imp. Univ. The stress distribution in steel eyebars was investigated by J. 3Mathar, Forschungsarbeiten, No. 306, 1928. See S. Timoshenko, "Theory of Elasticity," Russian edition, p. 119, St. Petersb urg, 1914. 2

124

'J'HEORY OF ELAS'l'ICI'l'Y

'J'WO-DIMENSIONAL PROBLEMS IN POLAR COORDINA'l'ES

the stress components from Eq. (81) in the usual way, and taking only the terms containing rn with n :;:: O, we find the following expressions for the stress components in ascending powers of r: ue = 2b 0 + 2do0 + 2a2 cos 20 + 2c2 sin 20 +6r(b 1 cos O + d 1 sin O + aa cos 30 + ca sin 30) + 12r2(b 2 cos 20 + d, sin 20 (n

from which (writing k = tan f3 - f3) we find do = -

+ a, cos 40 + C• sin 40)

+ 2)(n + l)rn[bn cos no + dn sin no + ª"+' cos -do

+ r(2b

1

(n

+ 2)0 + e"+' sin (n + 2)01

+ 2a2 sin 20 - 2c2 cos 20 sin O - 2d 1 cos O + 6a, sin 30 - 6cs cos 30) + r 2(6b 2 sin 20 - 6d 2 cos 20 + 12a. sin 40 - 12c, cos 40)

+ rn[n(n

+ l)bn sin nO

+

- n(n + l)dn CQS nO + (n + l)(n 2) an+• sin (n + 2)0 - (n + l)(n 2)cn+• cos (n

+

+ 2)0]

Thus each power of r is associated with four arbitrary parameters so that, if the applied stresses on the boundaries, O == a and O = {3, are given as polynomials in r, the stresses in the wedge included between these boundaries are determined. If, for instance, the boundary conditions are 1 (ue)e-a (ue)e-fJ he)e-a (-rro)e-iS

=No+ N,r + N.r• + · =No' + N,'r + N.'r 2 + = So + S,r + S2r 2 + · = So' + S,'r + S2'r 2 +

ue = 'l_ ( - k k q (1 Tr9 = k 2 -

+ -21 tan f3

q( -k u, = k

+ 21 tan {3

+

+

+

(a)

(b)

(n

+ 2)a1

=

N,.

with three other groups of equations for ue at O = {3 and Tr8 at O = a and o = {3. These equations are sufficient for calculating the constants entering into the solution (83). Let us consider, as an example, the case shown in Fig. 86. A uniform normal pressure q is acting on the face O = O of the wedge and the other face O = {3 is free from forces. Using only the first tines in the expressions (83) for ue and -r,8 the equations for determining constants bo, do, a2, and c2 are FIG. 86, 2bo

+

+ 2as

= -q 2dof3 2a2 cos 2{l 2c2 sin 2{l = O -do - 2c2 =O -d 0 + 2a 2 sin 2{3 - 2c2 cos 2{J = O

+

+

2bo

1 The terms No, N 0', S 0 , S 0' are not independent. oorner r = O and only three can be assigned.

1.

O-

-

1 ) 2 cos 20 1. 2 sm 20 + 21 tan {3

(e)

cos 20 )

-

;

( 7í -

O+

Fm. 87.

~ sin 20)

TrO

=

- !1- (1 - cos 20)

O'r

=

- ; ('lí -

21í

O-

(d)

~ sin 20)

1. Verify Eq. (d) of Art. 25 in the case =

+ 2)a + Cn+• sin

-

+ -21 sm 20)

Problems

= No Cs sin 3a) = N,

(n

- o - -1 tan f3' cos 20 2

. 20 21 tan {3 sm

ue =

·· ··

+ C2 sin 2a)

+ 2)(n + l)[b,. cos na+ dn sin na+ an+• cos

2k

'lí·

and generally (n

-q+qtantl

2bo

The stress éomponents for any other term in the polynomial load distribution (a) may be obtained in a similar manner. The method developed above for éalculating stresses in a wedge is applicable to a semi-infinite plate by making the angle f3 of the wedge equal to The stresses for the case shown in Fig. 87, for instance, are obtained from Eqs. (e) by substituting f3 = 7í in these equations. Then

we have, by equating coefficients of powers of r, 2(bo d 0a a2 cos 2a 6(b 1 cosa +d, sina + a 3 cos 3a

!1-,

2k

Substituting in Eqs. (83), we obtain'

(83) Tr9 =

125

They represent stress at the

x• - y• = (x 2

+ y 2)(x'

- y 2 ) = r 4 cos 20

2. Examine the significance of the stress function Co where C is a constant. Apply it to a ring a < r < b; and to an mfinite plate. A ring is fixed at r = a and subjected to a uniform circumferential shear at r = b forming a couple M. Using Eqs. (49), (50), (51), find an expression for the circumferential displacement v at r = b. 3. Show that in the problem of Fig. 45, if the inner radius a is small compared with the outer radius b, the value of u9 at the inside is given by

aE (1 - 2 log ~) 47r a and sois large, and negative when a is positive (the gap is being closed). 1

This solution was obtained by another method by M. Levy, Compt. rend., vol· 126, p. 1235, 1898. See also P. Fillunger; Z. Math. Physik, vol. 60, 1912. An ~pplic~tion of stress functior_is of this type to tapered box beams is given by · ~eIBsner, J. Aeronaut. Sei., vol. 7, P• 353, 1940. Other loads on wedges are cons1dered by C. J. Tranter, Quart. J. Mechs. and Appl. Math., voL 1, p. 125, 1948.

126

THEORY OF ELASTICITY

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

What is the largest gap (value of a) which can be closed without exceeding the elastic limit, if b/a = 10, E = 3 X 10 7 p.s.i., elas ti e limit = 4 X 10 4 p.s.i.? 4. Use the results of Art. 31 to obtain formulas for the stresses due to closing the parallel gap a in Fig. 88a, and due to the sliding of amount a in Fig. 88b, in terms of a. 5. Find by superposition from Eqs. (62) the stresses in the infinite plate with a hole when the undisturbed stress at infinity is uniform tension S in both the x- and y-directions. The results should correspond with Eqs. (45) for the special case b/a-+ oo, P• =O, (a) (hJ Po = -S. Use this as a check. FIG. 88. 6. Find expressions for the displacements ..iorresponding to the stresses (62), and verify that they are single-valued. 7. Convert the stress function (a) of Art. 33 to Cartesian coordinates and hence derive the values of ""'' "•• 'Tzu which are equivalent to the stress distribution of Eqs. (66'). Show that these values approach zero as the distance from the force increases in any direction. 8. Verify that in the special case of a = "Ir /2 the stress function (d), page 98, agrees with Eq. (69), and investigate whether the stress distribution (e), page 99, tends to agree with elementary bending theory for small a. 9. Show by evaluating the force resultants that the stress distribution (e), page 99, does in fact correspond to loading by a pure couple M at the tip of the wedge. 10. A force P per unit thickness is applied by a knife-edge to the bottom of a 90-deg. notch in a large plate as indicated in Fig. 89. Evaluate the stresses, and the horizontal force transmitted across an are AB.

y FIG. 91.

13. Determine the value of the constant C in the stress function = Cr 2 (cos 28 - cos 2a)



required to satisfy the conditions

o,

"8 =

= s on 8 =a = -s on 8 =

Tr8

o,

"8 =

Tr8

-a

corresponding to uniform shear loading on each edge of a wed e d" t d f th t y en·ry that no concentrated force or couple acts g , rrec e away rom ~ ver ex. on the t 14. Fmd the stress function of the type ver ex. a,r 3 cos 38

+ brr

3

cos 8

which satisfies the conditions

o,

"8 =

"8

sr on 8 = -sr on 8

Tr8 =

=o,

r,8

=

a

=

-a

s being ~ constant. Sketch the loading for positive s. 15. Fmd the stress function of the type

which satisfies the conditions

o,

r,8 =

=O,

Tr8 =

"8 = "8

B

11. Find an expression for the stress ""' on the section mn indicated in Fig. 90. The wedge theory of the present chapter and the cantilever theory of Chap. 3 give different stress distributions for the junction rs. Comment on this. 12. Determine the value of the constant C in the stress function

= C[r2(a - 8)

sr 2 on 8 = a -sr 2 on 8 =

-a

s being a ~onstant. Sketch the loading. 16. Derive the stress distribution

FIG. 89.

+ r2 sin 8 cos 8 -

r2

cos 2

8 tan a]

required to satisfy the conditions on the upper and lower edges of the triangular plate shown in Fig. 91. Evaluate the stress components
127

section mn. Draw curves for the case a = 20 de and · the curves given by elementary beam theory. g. draw also for companson

"., = - '!!_ (arctan "Ir

"•= -

J!. X

+ x2___!?Jf__) + y• ' '!!_ (arctan 1r

J!. X

Tzy

=

- '!!_

~) + y2

x•

from the stress function [see Eq. (a), Art. 341 =

-

ir ((x + y arctan ~ - xy) 2

2 )

7r

y2

x•

+ yt

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

THEORY OF ELASTICITY

128

and show that it solves the problem of the semi-infinite plate indicated in Fig. 92, with axes as shown. The load extends indefinitely to the left.

Pl 111 ! 11 ~ _,,

21. Show that if the pressure loading of p b 20 · 1 replacing p, the appropriate stress function i:o . lS rep aced by shear loading, s. =

()=ore ftm~

O

2:a [ xy• log (x•

+ y•) + (x2y

;y

~ [!2 y2 log

~

(x 2

+ y2) + xy arctan 'X!!

p {

left. Show that "• grows without limit as O is approached from any direction. (This is due to the discontinuity of load at O. A finite value is obtained when this is smoothed out, depending on the loading curve in the neighborhocd FIG. 93. of O.) 18. By superposition, using the results of Prob. 16, obtain "•• "ui"•v,forpressurc p on a segment -a < x
arctan '!!. X

+ !3 y

3

log (x 2

+y

2

)

-

! x•y] 3

solves the problem of the semi-infinite plate indicated in Fig. 94, the linearly increasing pressure load extending indefinitely to the left.

- ;

3

2

xy r2 - 3a• 1ºg r1 2 -

[a•4 + 21 (x• + y•) ( 1 -

y

x• 6a•

+ 2ay• )] ª + 23 ax{J + 21 ay ( 1 2

Fm. 96.

for pressure, and ; { ::. (3a• - 3x•

+ y•) log ~:: + ~ ay{J + ::. (x•

- 3y• - 3a•)a

+

4::'}

for shear, where

+

"' =

r1 2 = (x - a)• y•, 2ay 01 - O, = arctan x• y• - a•'

+

r2 2

+ a)• + y• + 02 = arctan

= (x

fJ = 01

x2

-

Zxy y• - a•

24. Show that in the problem f F' 72 h . along the vertical d' t o lg. t ere IS a tensile stress "• = 2P /.,,.d the semicircular pa~~m;;~ ~xcept a~A. and B. Account for the equilibrium of d d y cons1 ermg small semicircles about A and B in the manner of Fi s. 65 g e an . 26 • Verify that the stress function

"' == -

FIG. 94.

J

23. Show that the parabolic loading indicated in Fig 96 is . b th function · given Y e stress

s..;-=c:-=--P-==-=J=:"'l-----=x=

(!3 x' + xy 2)

- 3xy•

FIG. 95.

- y•]

solves the problem of the semi-infinite plate indicated in Fig. 93, the uniform shear loading s extending from O indefinitely to the

q, = - J!_ [

~

~

p

Examine the value of .,• ., (a) approaching O along the boundary Ox, (b) approaching O along the y-axis (the discrepancy is dueto the discontinuity of loading at 0). 17. Show that the stress function =

- y 3) arctan

22. Show how the distributions of loadindicat d· F' · superposition from loading of the type indicated ~n 1;'ig.1~4~5 may be obtamed by

:JC

Fra. 92.

q,

129

E{,,,,. cos º _i c1 -

11)r log r cos O -

!2 rO sin o

+ 4a log r

-

a• 32

(3 - 11)

r1 coso }

THEORY OF ELASTICITY

130

tisfies the boundary conditions for a force p acting in a hole in an infinite plate

ªª. · finiºty , and that the circumferential stress round the bole 18 with zero st ress a t ll1 p [2 + (3 - v) cos B} 'll"d

CHAPTER 5

except at A (Fig. 97). . Show that it also corresponds to single-valued d1splacements.

Fm. 97.

Deduce from Prob. 25 by integration the circumferential stress round the 26 hole .due to uniform pressure p in the hole, and check the result by means of Eqs.

(4~7

F" d the general form of f(r) in the stress function Bf(r), and ~d the expresll1 Co ld uch a stress funct10n apply to a u s sions for the stress components .,.,, "º' "•O· closed ring? •

THE PHOTOELASTIC METHOD 42. Photoelastic Stress Measurement. The boundaries of the plates so far considered have been of simple geometrical form. For more complex shapes the difficulties of obtaining analytical solutions become formidable, but these difficulties can be avoided by resorting to numerical methods (which are discussed in the Appendix) or to experimental methods, such as the measurement of surface strains by extensometers and strain gauges, or the photoelastic method. This method is based on the discovery of David Brewster1 that when a piece of glass is stressed and viewed by polarized light transmitted through it, a brilliant color pattern dueto the stress is seen. He suggested that these color patterns might serve for the measurement of stresses in engineering structures such as masonry bridges, a glass model being examined in polarized light under various loading conditions. This suggestion went unheeded by engineers at the time. Comparisons oí photoelastic color patterns with analytical solutions were made by the physicist Maxwell. 2 The suggestion was adopted much later by C. Wilson in a study of the stresses in a beam with a concentrated load, 3 and by A. Mesnager in an investigation of arch bridges. 4 The method was developed and extensively applied by E. G. Coker 5 who introduced celluloid as the model material. Later investigators have used bakelite, and more recently, fosterite. 6 ln the following we consider only the simplest form of photoelastio apparatus. 7 Ordinary light is regarded as consisting of vibrations iD 1

D. Brewster, Trans. Roy. Soe. (London), 1816, p. 156.

2

J. Clerk Maxwell, Sei. Papers, vol. 1, p. 30.

3

C. Wilson, Phil. Mag., vol. 32, p. 481, 1891. A. Mesnager, Ann. ponts et ehaussées, 4• Trimestre, p. 129, 1901, and 9° Series, vol. 16, p. 135, 1913. 6 The numerous publications of Prof. Coker are compiled in his papers: Gen. Elee. Rev., vol. 23, p. 870, 1920, and J. Franklin Inst., vol. 199, p. 289, 1925. See also the book by E. G. Coker and L. N. G. Filon, "Photo-elasticity," Cambridge University Press, 1931. 6 M. M. Leven, Proe. Soe. Expl. Stress Analysis, vol. 6, no. 1, p. 19, 1948. 7 More complete treatments may be found in the following books: "Handbook of Experimental Stress Analysis," 1950; M. M. Frocht, "Photoelasticity," 2 vols., 1941 and 1948; and the book cited in footnote 5. 131 4

THE PHOTOELASTIC METHOD THEORY OF ELASTICITY

132

. · f th ray By reflection ali directions transv.erse to the d irect10n o e . . t b from a iece of plate glass covered on one si~e with black ~am ' or y ?. h a olarizer-a Nicol prism, or Polaroid plate-we transIDiss10n throug p . f ·1· ht . which transverse vibra-· bt · ore or less polarized beam o lg m . .

r~~~;~s:~:yfi~!~~;~~::~º:r ~:~:~~~tio~e ~~:~sc~::~l~l::~ft~d;:::elcdy. . . f t We sha cons1 er on in the photoelastic mvest1gat1on o s ress. monochromatic light.

t: ~ : ~ :

L

M

p



A

S

L

PQp

M

º.4A

The displacement (a) in the plane OA is resolved into components with amplitudes OB = a cos a and OC = a sin a in the planes Ox, Oy respectively. The corresponding displacement components are x = a cos a cos pt,

(b)

h

t1 = -,

(e)

V,,

(b)

Frn. 98.

i.

Since the light waves are transmitted without change of forro, the x-displacement, x1, of the light leaving the plate at time t corresponds to the x-displacement of the light entering the plate at a time t 1 earlier. Thus (d) x 1 = a cosa cos p(t - t1), Y1 = a sin a cos p(t - t 2) On leaving the plate, therefore, these components have a phase dijference .1 = p(t 2 - t 1). It was established experimentally that for a given material ata given temperature, and for light of a given wave length, this phase difference is proportional to the difference in the principal stresses. It is also proportional to the thickness of the plate. The relationship is usually expressed in the forro .i

(C)

Frn. 99.

I F' 99a abcd represents a small element of the left-ha~d face of n ldg.l M 'the directions of the principal stresses 11,,, 11y be~ng ~rawn t h e mo e ' ·· f r ht 1 nzed m the vertical and horizontal for convenience. !1- ra! o lg p~aª in Fi 99 1 ne OA (Fi . 99) arrives from P, the d1~ec~1on of the Y_ g. pb ~ throug: the paper. The vibration lS s1mple harmoruc and may eing " d' 1 t" be represented by the transverse lSp acemen (a) s = a cos pt h d' t' OA where p is 211' times the frequency, depending on , . . . . · 1n t e irec 10n the color of the light, and t is the time. .

y = a sin a cos pt

The effect of the principal stresses 11,, and 11u, acting at the point O of the plate, is to change the velocities with which these components are propagated through the plate. Let v,, and Vy denote the velocities in the planes Ox and Oy. If h is the thickness of the plate, the times required for the two components to traverse the thickness are

s

. t• 11 a plane polariscope. A F'gure 98a represents d iagramma lCa y . h 1 . . .. L th ugh a polanzer p t en beam of light ongmat~tng adt 1 .Jª~~:h ~~difies the light acc~rding through the transparen mo. e . · · . . A t a to the stress, then. through an analyzer-a11;other P?lanzer - o . .. S n which a pattern of i11terference frmges (F1gs. 100 to 104) lS screen , o - .. · · · formed.

(aJ

133

27íh

= T · C(11,, - 11y)

(e)

where À is the wave length (in vacuo), and C the experimentally determined stress-optical coefficient. C depends on the wave length and temperature as well as the material. The analyzer A transmits on}y vibrations or components in its own plane of polarization. If this is at right angles to the plane of polarization of the polarizer,1 and if the model is removed, no light is transmitted by A and the screen is dark. W e now consider what occurs when the model is present. The components (d) on arrival at the analyzer may be represented as X2

= a cos

a cos if;,

Y2 = a sin a cos (if; - .i)

(j)

since they retain the phase difference .1 in traveling from M to A. Rere if; denotes pt constant.

+

1

The polarizer and analyzer are then said to be "crossed."

·.

~~~ :-:-~~~

'

THE PHOTOELASTIC METHOD THEORY OF ELASTICITY

134

. t d by mn in Fig. 99a, for The plane of polarization ofd4 isl reptre~e; e The components of the . It is set perpen icu ar o · convemence. "tt d b A are the components a1ong vibrations (f) which are transm1 e y Om which are, using Eqs. (f)' ' • _ cosa = --!a sin 2a cos (lf; - Ã) 1 x 2 sin a = "2"a sm 2a cos 1/1, Y2 The resultant vibration along mn is therefore 1 . ., _ a sm 2a [cos .,,1

_

(·'· _

cos .,,

Ã)] = -a sin 2a sin

)

~2 sin (lf; - ~

. ( Â)

2 ºh t the simple harmonic variation w1t The factor sm 1/1 - 2 represen s time.

The amplitude is

•

a sin 2a sm

.

·n

Â

2

135

The fringe value may be determined by loading a strip in simple tension. Since the stress is uniform there are no fringes, the whole piece appearing uniformly bright or dark on the screen. At zero load it will be dark. As the stress is increased, it will brighten, then darken as the stress difference (here simply the tensile stress) approaches the fringe value. On further increase of load it brightens once more, then darkens again when the stress is twice the fringe value, and so on. Similar cycles of brightness and darkness will clearly occur at any point of a nonuniform stress field as the load· is increased-provided the stress difference at the point reaches a multiple of the fringe value. These cycles at individual points correspond, in the view of the whole field, to gradual movement of the fringes, including the entrance of new

(g)

eh the screen unless either sin 2a = O

It follows that some hght w1 reah dº tº ns of the principal stresses . If · 2a = O t e irec 10 or sm Ã/2 = O. sm . directions of polarization of p and are parallel to the (perpendicular) h h oints of M will be extin-

Thus rays which pass throug. sue ph S will be dark. dº omts on t e screen ished and the correspon mg P · d" ted by a dark gu 11 r n one or more curves, m ica These points usua y ie o . l d " . cl1"nic " Very short lines S h rve is cal e an iso . . band on S. uc a cu d A be drawn at numerous points on it parallel to the axes of p and. ~ay f the principal stresses at these th (parallel) irect1ons o . t" to recor d e A . dºff t (perpendicular) orienta 10ns, . B tt" p and m i eren · · d pomts. y se mg . Th hort lines then cover the fiel different isoclinics are o~tamed. e s t and it is possible to draw f · fihngs over a magne , . h . t t the principal axes of stress. like a pattern o iron curves which are tangential at eac pom . o . l stresses The latter lines are trajectories of the prmc1paO 1 2 . When . / th  = 2n?r where n = , ' •· · · · 0 If sm  2 = ' en Points where this occurs are  = O, the principal stresses a_re efqual. b dark Points at which . . . . . t and w1ll o course e called isotrop1c pom s, . e of the first order, points for wh1ch n = 1 forro a dark band, or frmg ' d These fringes are called n = 2 a fringe of the second ordehr.'tanl. shot º1.ns. used they correspond to · (b e when w I e ig ' I isochromatics ecau.s ' 1 gth d therefore to a color band). t anf . - 2 has twice the value extinction of acertam wave en th t _ u on a rmge n . . Y d To evaluate the prmc1pal follows from E q. (e) a u,, · n 1 an so on. of u,, -
A

'li

·

FIG. 100.

fringes, as the load is increased. The orders of the fringes may therefore be determined by observing this movement and counting the fringes. For instance a strip in pure bending gives a fringe pattern as shown in Fig. 100. The parallel fringes accord with the fact that in the portion of the strip away from the points of application of the loads, the stress distribution is the sarne in all vertical cross sections. By watching the screen as the load is gradually increased we should observe that new fringes appear at the top and bottom of the strip and move toward the middle, the fringes as a whole becoming more and more closely packed. There will be one fringe at the neutral axis which remains dark throughout. This will clearly be the fringe of zero order (n =O). . 43. Circular Polariscope. W e have seen that the plane polariscope ~ust discussed provides, for a chosen value of a, the corresponding isoclinic as well as the isochromatics or fringes. Figure 100 should

136

THEORY OF ELAST!CITY

THE PHOTOELASTIC METHOD

137 Before examining the effects of QA and A u on th . . . p . . e hght it Is convenient to represent the motion (k) . Th. . b as a superposit10n of two circular is may e do f ll . t/t' _ (â/ ) b . ,11 mo t 10ns. - ;o ne as o ows. Denotmg 2 y 'f'1 ' an d a / v 2 by b, Eqs. (k) give

therefore show darkness wherever the orientations of the principal stresses coincide with the orientations of the polarizer and analyzer. Figure 100 was actually obtained in a circular polariscope, which is a modification of the plane polariscope designed to eliminate the isoclinics. 1 I t is indicated diagrammatically in Fig. 98b, which corresponds to Fig. 98a with the addition of two quarter-wave plates QP, QA. A quarter-wave plate is a crystal plate having two polarizing axes, which affect the light like a uniformly stressed model, introducing a phase difference à as in Eq. (f) - but the thickness of the quarter-wave plate is chosen so that  = 7r/2. Using Eq. (f) with this value of à for the light leaving QP, we observe that a simple result is obtained by choosing a, now denoting the angle between the plane of polarization of P and one of the axes of QP, as 45 deg. Then we may write I

a:2

a ./, = y'2 cos '1'1

y 2,

=

Va 2 cos (·'·'f'

-

z7r')

=

. ·'· Va 2 sm 't'

.

a

V

=

2

COS

{3

COS

1/1,

Ya

= _;

2

sin {3 cos (1/1 -

Â)

will merely result in a cha~ge ocf tahnge ho axes flor a circular motion t e p ase ang e ift" b Thus the clockwise circular motion can be represented b y a cons ant. of the forro Y components

a

= -

.

--=::: Sln

y2

. {3 Sln

i/t

where 1/1' = i/t

=

+ {3.

Va 2 COS .1,1 't',

X5

if;,

= c cos t/t,

Ys = c sin

(t/t -

~)

= -c cos 1"

(o)

1/1 ha ving again changed by a constant.

'

Ya = _; cos {3 sin (1/1 - â)

2

(j)

Ya = _; sin (1/1' - Ã)

2

(k)

lf the polarizer and analyzer rotate, their axes remaining perpendicular, the fringes remain stationary and the isoclinics move. If the rotation is rapid the isoclinics are no longer visible. The circular polariscope achieves the sarne effect by purely optical means. 1

C COS

(i)

Adding the components in Eqs. (i) and (j) we find for the light leaving the model Xa

=

Y4 = c sin i/t (n) aiong the axes of QA, where i/t is again of the f r Identifying X4 with the fast axis of Q we h II oh m pt + constant. A, s a ave on emergence from QA X4

and for the light dueto Y2' only Xa

(m)

pri!!~:~:;~:se~!s :a~;J~ E~.(~ and (7) are along the directions of

+

Xa

(!)

whwh represent the superposition of . 1 . . (â/2), clockwise in Fig. 99b (where th: circu ar modt10n of rad1us b cos th ) d . ray passes ownward through e paper ' an a circular motion of radius b sin (â/2) t 1 k . ha · 'coun erc oc w1se We m h th 1 ª"! _now s ow t t if the polarizing axis of A is set at 45 deg t~ e po anzmg axes of QA one of th . 1 . t th S ' ese c1rcu ar motions is transmitted o e screen ' the other extinguished d th d . e esired result-isochromatics without isoclinics-is obtain~d~n

(h)

Here x 2' corresponds to the "fast" axis of the quarter-wave plate. A point moving with these displacement components (1/1 always having the forro pt constant for a given point along the ray) moves in a circle. Such light is therefore described as circular polarized. The components (h) are along the axes of polarization of Qp. Using {3 for the angle between x2' and the direction of u., in the model (Fig. 99b), and  once more for the phase difference caused by the stressed element, we have, for the light leaving the model, dueto x2' only

x, ~ bcoo (.v' + i) ~ b(•o• i co, f" - ''" i ''" r) y, ~ bffin (r -i) ~ b(co'i•inf" - •in~co•r)

If we now set the anaiyzer (A) axis a t 45 de to Ox . 99c), the components of the dispiacements (o) !iong it ~~:d Oy4 (Fig.

c cos 45º cos i/t - c cos 45º cos i/t or ~ero._d T_hus ~he clockwise circular motion is extinguished ons1 ermg m the sarne th . motion 0 f E (l) . way e counterciockwise part of the qs. and (m), i.e., w

XI 4

• • /.

= -c sm 't',

yi = -c cos 1"

(n')

e find that the tram•mitted dispiacement along the analyzer axis is

-c cos 45º sin i/t - c cos 45º sin i/t

-- -.. -º-,.-.. .

f lt~STITUTUL

--,.·-~--

. ,._.

POUTEHNIC ·.

llMl':·OARA ~ENTRALl

amuoTECf..

138

THEORY OF ELASTICITY

THE PHOTOELASTIC METHOD

139 ln such cases the maximum stress is at the b . obtained directly by the optical method beca~undary, and It ~a~ be stresses vanishes at the free boundary. e one of the prmc1pal . Figure 101 shows the fringe pattern of a curved b i M. The outer radius is three times th . ar bent by couples the fringes marked on the right-ha d edmhner. The. order numbers of n en s ow a max1mum of 9 at both

and the amplitude is thus

v2c

or

V2 b sin ~

or

. t;.

asm

2

(p)

remembering that b denotes a/V2 and that a is the amplitude leaving the polarizer. No account has been taken of course of loss of light in the apparatus. Comparing this result with the result (g) for the plane polariscope, we observe that the factor sin 2a is now absent, and therefore the isochromatics appear on the screen, but no isoclinics.

Fm. 101.

If t;. is zero, the amplitude (p) is also zero. Thus if there is no model, or if the model is unloaded, the screen is dark. W e have a dark field setting. If the analyzer axis is turned through 90 deg. with respect to QA we should have a light field and light fringes taking the place of the former dark fringes. The sarne effect is brought about in the plane polariscope by baving the polarizer and analyzer axes parallel instead of at right angles. 44. Examples of Photoelastic Stress Determination. The photoelastic method has yielded especially important results in the study of stress concentration at the boundaries of holes and reentrant corners.

top and bottom. The re ui . tribution of bendin t g . ar hspacm~ corresponds to the linear disg s ress m t e stra1ght b k Th . marked along the top ed h h s .ªn.. e frmge orders ge s ow t e stress d1stribution in the curved 1

E. E Weibel T •

'

rans.

A

.

s .M.E., vol. 56, p.

637, 1934.

140

THEORY OF ELASTICITY

part (the complete model continues above this top edge, which is its axis of symmetry), indicating a compressive stress at the inside represented by 13.5, anda tensile stress at the outside represented by 6.7. These values are in very close agreement, proportionally, with the theoretical "exact-solution" values in the last line of the table on page 64. Figures 102, 103 represent 1 the case of bending of a beam by a force applied at the middle. The density of distribution of dark fringes indicates high stresses near the point of application of the load. The number of fringes crossing a cross section diminishes as the distance of

3

.J

\

STRESS CONCENTRATJON FACTORS TENSJON

b

o-D/d=J o-D/d=l.S

\

d\ ~

1

o

0.1

"'... 0.2

·... ~

1 2

M. M. Frocht, Trans. A.S.M.E., vol. 53, 1931. See paper by Weibel, loc. cit.

t

1

0.3

R~tioR/d lcrn. 105.

Frn. 106.

V

D 1

~

~

Frn. 104.

the cross section from the middle of the beam increases. This is due to decrease in bending moment. Figure 104 represents the stress distribution in a plate of two different widths submitted to centrally applied tension. It is seen that the maximum stress occurs at the ends of the fillets. The ratio of this maximum stress to the average stress in the narrower portion of the plate is called the stress-concentration factor. It depends on the ratio of the radius R of the fillet to the width d of the plate. Severa! values of the stress-concentration factor obtained experimentally 2 are given in Fig. 105. It is seen that the maximum stress is rapidly increasing as the ratio R/d is decreasing, and when R/d = 0.1 the maximum stress is more than twice the average tensile stress. Figure 106 repre-

141

THE PHOTOELASTIC METHOD

(

d -.

-

0.4

1--

,.,

0.5

0.6

143

THEORY OF ELASTICITY

THE PHOTOELASTIC METHOD

sents the sarne plate submitted to pure bending by a couple applied at the end and acting in the middle plane of the plate. Figure 107 gives the stress-concentration factors for this case. 46. Determination of the Principal Stresses. The ordinary polariscope, as we have seen, determines only the difference of the principal stresses and their directions. When it is required to determine the ' stresses throughout the model, or at a boundary where there principal is unknown loading, further measurement, or calculation, is required.

forro an air film with thickness variations determined by the thickness variations in the plate, yields the required information in a single photograph. The differential equation satisfied by the surn of the principal stresses, Eq. (b) on page 26, is also satisfied by the deflection of a membrane of constant tension, such as a soap film, and if the boundary values are made to correspond, the deflection represents
142

STRESS CONCENTRATION FAC.TORS PURE BENOING

2.6 1

-u ~

'º

e o +: 2

b

IS

....

1:Q) 1..)

6

c[[):::D

"~ ..........

1..)

"'

V)

......... .a.

Q)

L.

j)

1

o

+

o-D/d=J D-D/d=lS

\_

1...

O.J

0.2

-

~

~

0.3 0.4 Ratio R/d

-

o.s

0.6

FIG. 107.

Many methods have been used, or proposed. Only a brief description of some of these will be given here. 1 The sum of the principal stresses can be found by measuring the changes in the thickness of the pla.te. 2 The decrease in thickness due to the stress is

t.h = hv E (
+

)


+

(a)

whence O",, O"y may be calculated if t.h is measured at each point where the stresses are to be evaluated. Several special forros of extensometer have been designed for this purpose. 3 The pattern of interference fringes forrned when a model is placed against an optical fl.at, so as to For further information see the references cited in footnote 7 on p. J31. This method was suggested by Mesnager, Zoe. cit. a See M. M. Frocht, "Photoelasticity," vol. 2.

1

2

+

1 1

J. P. Den Hartog, Z. angew. Math. Mech., vol. 11, p. 156, 1931.

See R. D. Mindlin, J. Applied Phys., vol. 10, p. 282, 1939. See the article by D. C. Drucker in "Handbook of Experimental Stress Analysist which gives a comprehensive account of three-dimensional photoelasticity. J. Clerk Maxwell, Sei. Papers, vol. 1, p. 30. 3

•

144

THEORY OF ELASTICITY

THE PHOTOELASTIC METHOD

sarne property. The explanation is that these materiais have the structure of a strong elastic skeleton, or molecular network, which is unaffectecÍ by heat, the spaces being filled by a mass of loosely bonded molecules which softens on heating. When the hot specimen is loaded the elastic skeleton bears the load and is elastically deformed without hindrance. On cooling, the softened mass in which this skeleton is embedded becomes "frozen, " and holds the skeleton almost to the 1

145

NUT WITH TAPERED LIP

FIG. 108.

sarne deformation even when the load is removed. The optical effect is likewise substantially retained, and is not disturbed by cutting the specimen into pieces. A three-dimensional specimen may therefore be cut into thin slices, and each slice examined in the polariscope. The state of stress which produced the optical effect in the slice is not plane stress, but the other components Txz, Tyz,
CONVCNTIONAL NUT

FIG. 109.

of the sarne type, obtained from a (bakelite) model of a bolt and nut fastening. 1 The lower nut was a conventional type. The upper one has a tapered lip and shows a lower stress concentration than the conventional nut. 1

M. Hetényi, J. Applied Mechanics (Trans. A.S.M.E.), vol. 10, p. A-93, 1943. Results for severa! other forms of nut are given in this paper.

STRAIN-ENERGY METHODS

sponding to a larger amount of work. Thus a net amount of work would have been gained from the element in a complete cycle. The calculation of the work done is simplest if the forces, or stresses, ali increase simultaneously in the sarne ratio. Then the relation between each force and the corresponding displacement is still linear, as in Fig. llOb, and the work done by all the forces is

CHAPTER 6

STRAIN-ENERGY METHODS 47. Strain Energy. When a uniform bar is loaded in simple tension the forces on the ends do a certain amount of work as the bar stretches. Thus if the element shown in Fig. ... .... 110 IS -1 dA . sub"Jec t t o norma1 st resses ~ u., only, we have a force u., dy dz ~x dx which does work on an extension o B €.x E., dx. The relation between these (aJ (b) two quantities during loading is Fm. 110. represented by a straight line such as OA in Fig. llOb, and the work done during deformation is given by the area i(u., dy dz) (E., dx) of the triangle OAB. Writing dV for this work we have dV = iu.,E., dx dy dz (a)

dV = Vodxdydz

1 If this were not so there would be a substantial difference between adiabatic and isothermal moduli of elasticity. The actual differences are very slight. See G. F. C. Searle, "Experimental Elasticity," Chap. 1.

1

~i

i

l11

l

'

'

'

1,'

1il1:1

146

(b)

where

rd.

It is evident that the sarne amount of work is done on all such elements, if their volumes are the sarne. We now inquire what becomes of this work-what kind or kinds of energy is it converted into? In the case of a gas, adiabatic compression causes a rise of temperature. When an ordinary steel bar is adiabatically compressed there is an analogous, but quite small, rise of temperature. The corresponding amount of heat is, however, only a very small fraction of the work done by the compressive forces. 1 For our purposes it is sufficiently accura~e to disregard this small fraction. Then none of the work done IS accounted for by heat, and we may say that it is all stored within the element as strain energy. It is assumed that the element remains elastic and that no kinetic energy is developed. The sarne considerations apply when the element has all six components of stress u.,,
147

Vo = i(u.,E.,

+
(e)

Thus Vo is the amount of work per unit volume, or strain energy per unit volume. ln the preceding discussion the stresses were regarded as the sarne on opposite faces of the element, and there was no body force. Let us now reconsider the work done on the element when the stress varies through the body and body force is included. Considering first the force u, dy dz on the face 1 of the element in Fig. IlOa, it does work on the displacement u of this face, of amount i(u,u)i dy dz, where the subscript 1 indicates that the functions u,, u must be evaluated at the point 1. The force u, dy dz on the face 2 does work -i(u,u)2 dy dz. The total for the two faces is the sarne, in the limit, as 1 â

2 ox (u,u)

dx dy dz

(d)

Computing the work done by the shear stresses Txy, -r., on the faces 1 and 2, and adding to (d), we have the work done on the two faces by all three components of stress as 1 a 2 oX (u,U + TxyV + TxzW) dx dy dz

where v and w are the components of displacement in the y- and z-directions. The work done on the other two pairs of faces can be similarly expressed. We find, for the total work done by the stresses on the faces, I

2

[a

OX (uxU

+ TxyV + TxzW) + oya (uyV + TyzW + TxyU) + :z (u,w + -r.,u + TyzV)] dx dy dz

(e)

As the body is loaded the body forces X dx dy dz etc. do work

iCXu

+ Yv + Zw) dx dy dz

'J'.he total work done on the element is the sum of (e) and (!). differentiations in (e) we find that the total work becomes

(f)

On carrying out the

THEORY OF ELASTICITY

148 au

l [

2 tTx é)z

av (ªv au) (ªw av) + tTy éJy + tTz aw é)z + Txy ÕX + éJy + Tyz é)y + az + Txz (ªU + ÕW) + U (iJtTx + ÕTxy + iJTxz + x) ()z iJX iJX iJy (}z + v (ªtTy + iJTy• + iJTxy + y) + W (iJtTz + iJTxz + ÕTy• + z)] dz dy dz ay az ax az ax ay

But on account of the equations of equilibrium (127) derived in Art. 76 the brackets multiplying u, v, w are zero. The quantities multiplying the stress components are, from Eqs. 2, Ex, • • • , 'Yxy, • . . • Thus the total work done on the element reduces to the value given by (b) and (e). These formulas therefore continue to give the work done on the element, or strain energy stored in it, when the stress is not uniform and body forces are included.

By means of Hooke's law, Eqs. (3) and (6), we can express Vo, given by Eq. (e), as a function of the stress components only. Then V

0

= ~

2

(
+ q,y2 +
-

i

(
+
+ 2~ (r:r;y2 + Tya2 + T:r:a2)

(84)

Alternatively we may use Eqs. (11) and express Voas a function of the strain components only. Then

V 0 = .p,e2 + G(e:r: 2 + Ey 2 + Ez 2)

+

j-G("'(:r:y

in which

e = E:r:

+ Ey + e.,

}..-

- (1

2

+

"'(yz

2

+

2

'Y:r:z )

,,

Vo = Xe

OE:r;

f

+ 2Ge:r: = u

!i' 1

(g)

:r:

= O,

we have (86)

'l

~

The total strain energy of a deformed elastic body is obtained from the strain energy per unit volume Vo by integration:

i

!

V=

~ '1

'

1

'

'

ill:1

Jf fVo dx dy dz


+ p,


+ p,

u,

=

u.'

+p

where (h)

u.'

+ u.' + u.'

= O

the stress condition u.', u.', u.' produces only distortion, and the change in volume depends entirely 3 on the magnitude of the uniform tension p. The part of the total energy dueto this change in volume is, from Eq. (8),

- 2v)

For the case of plane stress, in which
1

The quantity of strain energy stored per unit volume of the material is sometimes used as a basis for determining the limiting stress at which failure occurs. 1 ln order to bring this theory into agreement with the fact that isotropic materials can sustain very large hydrostatic pressures without yielding, it has been proposed to split the strain energy into two parts, one due to the change in volume and the other due to the distortion, and consider only the second part in determining the strength.2 We know that the volume change is proportional to the sum of the three normal stress components [Eq. (8)], so if this sumis zero the deformation consists of distortion only. We may resolve each stress component into two parts,

(85)

Ev

+ v) (1

It represents the total work done against interna! forces during loading. If we think of the body as consisting of a very large number of particles interconnected by springs, it would represent the work done in stretching or contracting the springs. By using Eq. (84) or (85) it can be represented either as a function of stress components or as a function of strain components. The application of both these forros will be illustrated in the following discussion.

Since, from this,

This forro shows at once that Vo is always positive. It is easy to show that the derivative of Vo, as given by (85), with respect to any strain component gives the corresponding stress _component. Thus taking the derivative with respect to E:r: and usmg Eq. (11), we find 0

149

STRAIN-ENERGY METHODS

(87)

ep = 3(1 - 2v) 2 = 1 - 2v ( 2 2E p 6E trx

+ trv + u,)

(i)

2

Subtracting this from (84), and using the identity tTxtTy

+ tTytTz + tTz
= -j-[(trx - u.)2

+ (try

- u.) 2

+ (u,

-
+ (trx + tTy2 + u,•) 2

we can present the part of the total energy due to distortion in the forro

V

o-

1 - 2v

~

(ux

+ tTy + u,)

2

1

+v

= ~ [(trx - try) 2

+ (trz

- tTx) 2 ]

+ (try

- u,) 2

+ 21G (rx.2 + Txz 2 + Ty.2)

(88)

1 The various strength theories are discussed in S. Timoshenko, "Strength of Materials," vol. 2. 2 M. T. Huber, Czasopismo technizne, Lw6v, 1904. See also R. v. Mises, Gõttingen Nachrichten, Math.-phys. Klasse, 1913, p. 582, and F. Schleicher, Z. angew. Math. Mech., vol. 5, p. 199, 1925. 3 • The shearing components Txy, r ••, Txz produce shearing strains which do not ·nvolve any change of volume.

THEORY OF ELASTICITY

STRAIN-ENERGY METHODS

ln the case of simple tension in the x-direction,
The above argument can be repeated without change for a load with non-zero resultant, so long as there is a fixed surface element within or near to the loaded part. Thus if a deformable material is bonded to a rigid one, pressure applied to a small part of the former near to the attachment will produce only local stress.t

150

from which (j)

Experiments with steel show 1 that the ratio between the yield point in tension and the yield point in shear is in very good agreement with that given by (j). Saint-Venant's principle (see p. 33) can be shown by consideration of strain energy to be a consequence of the conservation of energy. 2 According to the principle, two different distributions of force having the sarne resultant, acting on a small part of an elastic body, will produce the sarne stress except in the immediate neighborhood of the loaded part. If one of these distributions is reversed, and combined with the other, there will be zero stress except in this neighborhood. The combined load::s are self-equilibrating, and the principle is in fact equivalent to the statement that a self-equilibrating distribution of force on a small part of an elastic solid produces only local stress. Such a distribution of force does work during its application only because there is deformation of the loaded region. Let one surface element of this region be fixed in position and orientation. If p denotes the order of magnitude (e.g., average) of the force per unit area, anda a representative linear dimension (e.g., diameter) of the loaded part, the strain components are of order (p/E) and the displacements within the loaded part are of order pa/E. The work done is of order pa 2 • pa/E or p 2a 3 /E. On the other hand, stress components of order p imply strain energy of order p2/E per unit volume. The work done is therefore sufficient only for a volume of •)rder a 3, in accordance with the statement of the principle. It has been supposed here that the body obeys Hooke's law and is of solid form. 'l"he former restriction may be dispensed with, E in the above argument then denoting merely the order of magnitude of the slopes of the stress-strain curves of the material. lf the body is nota solid form, as for instance a beam with a very thin web, or a thin cylindrical shell, a self-equilibrating distribution of force on one end may make itself felt at distances many times the depth or diameter. 3 i' 1 See the papers by W. Lode, Z. Physik, vol. 36, p. 913, 1926, and Forschungsarbeiten, No. 303, Berlin, 1928. 2 J. N. Goodier, Phil. Mag., series 7, vol. 24, p. 325, 1937; J. Applied Phys., vol. 13, p. 167, 1942. ªV. Z. Vlasov, "Thin Walled Elastic Bars," Moscow, 1940; J. N. Goodier and M. V. Barton, J. Applied Mechanics (Trans. A.S.M.E.), vol. 11, p. A-35, 1944; N. J. Hoff, J. Aeronaut. Sei., vol. 12, p. 455, 1945.

1

!I

!

~

1

.

.

tilLi·

151

48. Principie of Virtual Work. ln the solution of problems of elasticity it is sometimes advantageous to use the principle of virtual, work. ln the case of a particle, this principie states that if a partich• is in equilibrium the total work of all forces acting on the particle in anv virtual displacement vanishes. As a virtual displacement of a particl~, free to move in any direction, any small displacement can be taken. If ou, ov, ow are components of a virtual displacement in the x-, y-, and z-directions and ~X, ~Y, ~z are the sums of projections on the sarne directions of forces, acting on the particle, the principie of virtual work gives ou ~X= O, ov ~y =o, ow ~z =o These equations are satisfied for any virtual displacement if ~X=

O,

~y

=O,

~z

=O

Thus we arrive at the known equations of equilibrium of a particle. ln applying the principie of virtual work the acting forces are considered as constant during a virtual displacement. If some of the forces acting on a point are elastic reactions, as reactions of bars in the case of a hinge of a truss, we assume that virtual displacements are so small that the change in magnitudes or directions of reactions can be neglected. An elastic body at rest, with its surface and body forces, constitutes a system of particles on each of which acts a set of forces in equilibrium. ln any virtual displacement the total work done by the forces on any particle vanishes, and therefore the total work done by all the forces of the system vanishes. A virtual displacement in the case of an elastic body is any small displacement compatible with the condition of continuity of the material and with the conditions for the displacements at the surface of the hody, if such conditions are prescribed. If it is given for instance that a ~e~tain portion of the surface of the body, say a b~ilt-in end of ~ heam, is immovable or has a given displacement, the virtual displaceIi'.lent for this portion must be zero. . Let us consider, as an example, the case of a plane stress distributior; ma plate. Denote by u and v the components of the actual displace~ents due to the loads and by ôu and óv the components of a virtual displacement from the loaded position of equilibrium. These latter 1

Goodier, J. Applied Phys., Zoe. cit.

152

THEORY OF ELASTICITY

STRA/Joí-ENERGY METHODS

components are arbitrary small quantities satisfying the conditions of continuity of an elastic deformation, i.e., they are continuous functions of x and y. For any system of displaccmcnts thc work done against the mutual actions bet,veen the particlcs is equal to thc strain energy stored, i.e., the strain energy corresponding to the dü;placements. If we change u and v by óu and fív, therefore, the work done against the mutual actions bet'\veen the particles is the difference betv;een the strain energy corresponding to u llu, v + óv, and the strain energy corresponding to u and v. 1'he virtual displacements llu, 6v produce the rhange in strain cornponents

plate. The condition that the total work done during the virtual dis-placement vanishes, no\v takes the form, from (b), (e), and (d), f(X !u

ó

&UfVo dx dy - ff(Xu

iJx

ày

The corresponding change in strain energy per unit volume, from exprcssion (85), is

avo

l!Vo = ,,.----- l!E,,, vE,,,

avo l!Ey + ~ avo
u,,,
+ u,,
(a)

The change of the total strain energy of the body is then f f (J Vo dx dy, in ,vhich the integration is taken over the whole area of the plate of unit thickness. As already stated, this change in strain encrgy measures thc work clone against the mutual actions between the particles. ln order to get the work clone by the mutual actions the sign must be reversed. Hence the 'vork clone by thesc forces during the virtual displacement is

-ff !Vodxdy

(e)

in which the integration is taken along the houndary s of the plate,_ Similarly the work dane by the body forces is

+Y

!v) dx dy

(d)

in which X and Y are the components of the body force per unit volume Qf the plate, a.nd the integration is taken over the whole area of the

+ Y~) dx dy

- f(X-u

+ Yv) ds]

=

o

(89')

An equation analogous to (89) can easily be written down for a threedimensional stress distribution. The principle of virtual \Vork is especially useful for finding the deformation of an elastic body produced by given forces. To ~ t illustrate the application of the s A = 4 : - d x ln S method let us consider here a -"f-l:::l=t+:-ltJIJl[[!jt:LJ-i:::J:~lf"---'!.. fe\y simple examples, the solu-i 1-.-~s tions of which are already well Fra. 111. kno\vn. The first is the deflection curve of a pcrfectly flexible clastic string AB stretched by forces S bctween fixed points A and B (Fig. 111) and loaded by a distributed vertical load of intensity q. We assume that

(b)

ln calculating the v.rork clone by externa! forces during a virtual displacement, consideration must be given the forces applied at the boundary of the plate and the body forces. Denoting by X and Y the components of the boundary forces per unit area, the work clone by thcsc forces on the virtual displacements having components l!u and liv may be \Vritten do\vn at once as

ff(X !u

(89)

'rhe first 1:erm in the hracket is the potential energy of deformation. The se
=àóv+àllu

'Y"'11

ff 8V0 dxdy ~O

Since, in applying the principie oi virtual work, the given forces and the actual stress components corresponding to the position of equilibrium are considered as constar.t during a virtual displat'ement, the variation sign & can be put before the integral signs in Eq. (89), and, changing signs throughout, we have

+

a !u ôe~=ax'

+ i' ôv) d•+ ff(X ôu + Y !v) dxdy -

153

1

•

We neglected them when we a.ssumed that stress components and forces reu1ain con.stant during any virtual displaccment.

THEORY OI/ ELASTICITY

STRAIN-ENERGY METHODS

the initial tension of the string is so la.rge that the increase in tensile force due toadditional stretchingduring the deflection can be neglected. Then the increase in strain energy due to the deflection is obtained by multiplying the initial tensilc forces S by the stretch of the string due to the deflection. Taking coordinatcs as sho,vn in Fig. 111 1 we find

1'his equation will be satisfied for any virtual displacement Qy, only if

154

ds-dx=dx

[1+ (dy)']' -dx=z1(ªy)' dx

dx

=

J

2

f' (ªd~)'

sd'v+q-o

(g)

dx'

Thus v.re obtain the kno-wn differential equation of a verlically loaded <1tring. T'he principle of virtual work can be used not only for establishing a differential equation for the deftection cur,·c, as in the example given above, ~ but also for the direct determination of deflections. 1 Take, for examplc, a ~ .x pri:,;matical bar supported at the ends /-and loaded by a force P (Fig. 112). Y I n the most general case the deflection 1''ro. 112. cu~ve of such a bar can bc represcnted in the form of a trigonometric series, , rx + . 2-irx a 3 sin -3rx y = a1 s1n T a2 s1n - (h) 1 1

F__

dx

The stretching of the string is (' (ds - dx) }o

155

c_=L---''---------

dx

and the corresponding increase in strain energy of the string is

('(ª")' dx dx

!'2 } 0

(<)

To get the total strain energy of the string the constant strain energy dueto initial stretching would have to be added to expression (e). The principle of virtual work in this case gives the following equation, analogous to Eq. (89):

+

+···

Substituting this in the well-known formula for the strain energy of bending of a prismatical bar, 2 we find

(fJ

V=

2E! lo('

(ª'y)' dx2

dx

~ 413 4

Efr'

=

n4a,.2

(k)

n~l

Calculating the variation of thc second term, we find

('ô

lo

(dy)2 dx dx

=

2 (ldy ady dx dx dx

=

lo

2 [ldyd l'iy dx dx dx

lo

Integrating by parts and taking into account that at the ends of thP string f'iy = Ü, "\Ve find 2

f' (dy) dôy dx

lo

dx

dx

or

1

111:1

ôy = ôa,. SIIl • -nrx -

1

ôV Efr4 ôa,. ila,. = 2l3 n 4a,. ôan

2

~

P ua,. 1

"

d' )' l'iydx =0 sdx;+q

(m)

and the work dane by the externai force P during the virtual displacemcnt (l) is

lo('d'd~ l'iy dx + lo(' q l'iy dx =O lof'(

(l)

'fhe corresponding change of strain energy, from Eq. (k), is

- 2 (ldydx ôyl'o - lo(' d'y ôy dx) dx

Substituting into Eq. (f), v.•e obtain S

Lct us consider a virtual displacement from t.he actual deflection curve obtained by giving to any coefficient a,. in t.he series (h) a variation ilan. Then

. nrc

Sln -

1

(n)

See S. Timoshenko, Bull. Polytech. lmt., Kiew, 1909. Sec also S. Timoshenko ' Strength of Materiais," vol. 2, p. 44, 1941. • See S. Timoahenko, "Strength of Materiais," vol. 1, p. 297, 1941.

THEORY OF ELASTJCITY

156

STRAIN-E.llfERGY METHODS

By using (m) and (n) the equation of virtual work becomes EJ,.

P

4

2l3 n 4a,. lla,.

-

. n11"c

lia,. s1n -

1-

ceed as in the case of the beam and take virtual di.spla.cements in the form

~o

ôu = ôA,,.,. sin mrx sin nT1J a b

from '\rhich 2Plº .

a,.

nrc

ôv = ôB,,.n sin mrx sin nT1J a b

"" -l-

=

~~~~~

El1í4n 4

The equation (89) of virtual displacements then gives

Substituting in the series (h), '\\·e find the deflection curve .,

_ 2Pl 3 y - Ef11'4

"

f..t

·-·

•

nTC

•

n1'X

sin -1- "" -ln4

óAm,. (o}

óB.,,"

This series rapidly converges, and a fe,v terms give a satisfactory approximation. Taking, for instance, the load P at the middle of the span (e = Z/2), the deflection undcr the load is

(y)

~ 2PI' ,,_21 Ef1r4

157

!! ff

n~ dx dy = Y sin m;x sin n-i;; dx dy =

X sin m;x sin

a:m,. ! ! V

ôBmn

Õ~mn

ff

0

dx dy

(q)

y dx dy 0

For the calculation of strain energy in the case of plane stress we use the formula

v~Jjv.axay~JJ[ 2(1 E-

(i + _!.34 + _!.5' + ...)

ôAmn

(Ez '+

v2)

E11

'+ 2 l'E~~ )

+ 4(1 ~

By taking only the first term of this series, we obtain

J

v) 'Y""2 dx dy

(r)

Substituting in itl

We have a factor 48.7 while the exact value is 48, so that the error made in using only the first term of thc series is about 1! per cent. x In the preceding discussion we had to consíder 0 displaccment in only one direction and ,,,.e represented it by a sine series (h). A similar method can be applied in more complicated cases. Let us consider a rectangular plate with fixed Y edges, Fig. 113, and acted upon by body forces Fio. 113 · parallel to its plane. General expressions for the displacements u and v can be taken in the forra of series,

µ

u

=

V=

\.' \.'

. m.x . n.-y

\.' \.' B

. m.x . n.-y

LtLt.4 mns1nas1nb LtLt

1111.

nr . mrx nry Lt "f..t TB,,.,. sina cos -b-

_ r'Jv _ " ay -

tu -

- r'Ju

1'>;11 -

õy

(p)

mnBlllªSlilT

+ âx r'Jv " " n1r • m"lrX n1ry = Lt Lt T Âmn Slll a cos -b+

ll:

m'B ~ mn a

COS

m1fx. n1fy -- "" - a b

and performing the integrations we find

V_ •'Eab 4

Each term of these series vanishes at the boundary, so the boundary conditions are sa.tisfied. To calculate the coefficients A,,.n, Bmn we pro-

1.

mr A,,.,. cosmrx n1fy Lt "f..t a -sinb 0

r'Ju = " ~~ = ax

['-'\.'A , (. 1 Lt Lt ''"' 2(1 -

m' v2) a 2

1 n')

+ 4(1 + v) b2

+ Lt '.' Lt '.' B mn , ( 2(1

1 n' - v2) b2

+ 4(1 1+ v) m')] a2

Su~t. is not ~~ays legitimate to differentiate a Fouriex series tcrm by term. Clent condit1ona may be found in the book "Modera Anal · "b E T "~' ta.ker d G N W ySJll y · · J1Itan · · ataon, p. 169. ThCBC conditiona are fulfilled in the present problem,

THEORY OF ELASTICITY

STRAJN-ENERGY METHODS

Substituting this expression for the strain energy into Eqs. (q) we

sions (s). The calculation of the variation of the strain energy ca.n b simplified if '"e observe that e

158

obtain Eabr' (

A •• - 4

m' a 2 (l-

112 )

Eafrir2 ( m2 B •• - 4 - 2a2(1+11)

+ 2 b (ln'+·;)")

=

2

n2

+ b2(1 -

)

v2)

fºf' · m.x · ••Y fººf as1nb 0

ª Xs1nas1nbdxdy 1>

=

•

ª Ys1n

m1t'X .

n'lry

" ~ .;.(x,y)

+l

=

Õam l/>m(x,y),

b• .p.(x,y)

ób"'

we can write as many equations of equilibrium, similar to equations (q) in the preceding examplc, as the number of the coefficients in the series (s). These equations 'vill be linear with rcspect to a1, . . . , a,,., b1, • • • , b,,,, and solving them we will find the values of the coefficients in the series (s), reprcsenting the approximate solution of the problem. 1 ln using the principle of virtual displacements (89) it is assumed that t.he strain energy per unit volume V0 is represented as afunction of the strain components [Eq. (r)] and these are calculated by using expres1 This method of solving problema in ele.sticity we.s proposed by W. Ritz and was successfully used by him in an investigation of bending of rectangular ple.tes. See J. reine u. angew. Math., vol. 135, pp. 1-61. See e.ISO "Gellll.mmelte Werke," p. 192, Paria, 1911.

11

ÓEy

+ r"ll Õy"ll) dx dy

f f [crx:x õu + Uu a~ õv +rzv(:y õu +

f

0~õv)]dxdy

õu and õv vanish at the bound-

f[(~:;+ª;;)ou+(~~+ª;;)õv]dxdy

'faking for the virtual displacements expressions (t) vle then obtain the necessary equations for ealculating the coefficients a 1, • • • ' a''" bi, . . . , bm in thf' following form:

m

õu,,.

Õ~r + u

óV= -

Óam

(t)

(ur

Integrating by parts and ohserving that ary we obtain

(')

which satisfy the prescribed boundary conditions. For example, v:e can select !f>o and ,P0 so that they 'vill give at the boundary the required displacements, and the rest of the functions q, and ,P can then vanish at the boundary. For the calculation of the coefficicnts a1, . . . , am, b1 , • • • , b,.. v:e use the principle of virtual displacements (89). Taking virtual displacements in the form

ff

=

+ l am.P... (x,y) m

=

dxdy

W e see that for any kind of volume forces the coefficients in exprcssions (p) can be readily calculated and the complete solution of the problem can be obtaincd. The method of virtual displacements can be used for finding approximate solutions of tv;•o-dimensional problems 1•,..hen displaccments at the boundary are given. Assume that the displacements u and v can be represented with sufficient accuracy by series u = ef>.(x,y)

õV

159

º'

f f X!j>,.,(x,y) dxdy f f (~~ + ~;) !f>m(x,y) dxdy f f Yt.f;,,.(x,y) dx dy f f (~~+a::) f,,.(x,y) dx dy JJ(~~ + ª;;; + x) lf>,,.(x,y) dxdy f f (~~ + ª;; + ~O =

-õa,,.

=

-Õbm

=O

(90)'

Y).;.(x,y) dxdy

~s an example of the application of these equations let us again con~1der a rectangular platc (Fig. 113) and assume that three sides of it. are fixed and along the fourth side (y = b) the displacements are given bv

t.he equations

· u=O,

v=Cbsin11"X a

The boundary conditions will be sat.isfied by taking U=

v =

. m.x . n"11 4\ ' \ ' A mnSin7s1n-b 4 Cy sin 11"X + \' \' B • ffl"lrX . n7r'!J a 44 ,,.,, s1nas1nT

(u)

E<;iuations of virtual displacements in this form are sometimes ealled Galerkin'~ equattons. liffwever both forms of equations representcd by Eqe ( ) d E (90) ·a· d .. .qan qs. are tn teate by W. llitz 1n the above-mentioned paper. f\ee "C'n·~ammõlt~ Werke," p. 228. · '" " 1

STRAIN-ENERGY .METHODS

THEORY OF ELASTICITY

160

The corresponding stress components will be

rr,.

=

~ (ªu 1_

"2

+ " ay av)

ax

~ (\' \'

=

cos m'll"X sin n'fr'Y L.t L.t A,,.,. m1f a a b Bm" nb1f sin m;x cos n;u + vC sin ":)

l -

11 2

L:L ~ "2(:~ + v!:) ~ "2(2:2: +"

ª"

=

n; sin m;x cos n; mr nry) +Csina+j) LtLtA,,... acosas1n-b-

1

1

=

TX

'T%11 -

2(1

Bmn

\' \'

ffl11"X

•

av) ~ E ( \ ' \ ' A,,.,. nr sin m1fx cos nny + v) (ªu ay + ax 2(1 + v) LtLt b a b "\' m• mrX . n"1J + e ' ") + "\' LtLtB,,.,.acosas1nb Ya;cosa

E

Substituting into Eqs. (90) and assuming ilu = 1iA,,.

•

a

ffl?l"X

Slll

n7r11

•

Bill

T'

&;

= óB,,.

•

ffl'l
•

Bill - . - SIIl

n1í'U

-b

've obtain, after integration, _

ET

2

ab (

4

m' a 2 (l

J12)

+ 2b2(1n'+ v) )

A .....

. ffl1fX s1nb"" . n'lry dxdy + of.b X s1n-J.

º

0

2

-

E1t ab ( -4-

b2 (1

n' v 2)

+ 2a (1m'+ v) ) 2

~o

a

B,,.,.

. 'll"X . ffl'll"X ,·n n"1J -c11"2J.ºf,b ysm-s1n--• b dx dy a2 0 0 a a

+

. JJ

mrx . n7rJJdxd

Ys1nasm-b-

Y ~

O

If the body forces vanish we find that all coefficients A,,.,, vanish also. The coefficients B,,.,. are different from zero only when m = 1. Thcn

E•'•b( n' - 4 - b2(1

111 )

+ 2a 2(1 1+ v) )s ~

1 "

e,' f,b y Slll . -b nry dy

- 2a

•

=

General considerations of the total energy of a system were applied by_A. A. Griffith in developing his theory of rupture of brittle materials.1 It is known that matcrials always show a strength much smaller than might be expected from the molecular forces. For a certain glass Griffith found a theoretical strength in tension of the order of 1.6 X 10° p.s.i., 1vhile tensile tests "'ith glass rods gave only 26 X ios p.s.i. He sho\ved that this discrepancy bet,veen theory and experimcnts can be explained if \Ve assume that in such materials as glass there exist microscopic cracks or fiaws producing high stress concentrations and consequent spreading of the cracks. For purposes of calculation Griffith takes a crack in the form of a very narrow elliptical hole, the major axis of tvhich is perpendicular to the
av dl = ;.ts.; ai ~ ~--dl r 2

C?rb 2cos nr

2{1-n-

Determining B 1 ~ from this equation and substituting into the formulas (u) 1,ve obtain the displacements produced by the assumed displacements at the boundary.

161

dl

1

2E

A. A. Griflith, Trans. Roy. Soe. (London), series A, vol. 221, p. 163, 1921. See also his paper in Proc. Inlern. Congr. A.ppl. ~fech., Delft, 1924. A bibliography of the subject can 1>e found in "Handbuch der physikalischen und technischcn Mechanik," vol. 4, part 2, 1931, 11.rticle by Adolf SmekaJ. 1 See p. 201.

162

THEORY OF ELASTICITY

STRAIN-ENERGY METHODS

from which

163

in houndary forces; then, from boundary conditions (20), we find

(w) Experiments in which cracks of kno,vn length were formed with a glasscutter's diamond sho?rcd a ve:ry satisfactory agreement '\Vith Eq. (w). It '\'as also shown experimentally that, Ü precautions are taken to eliminate microscopic cracks, a much higher strength than usual can be obtained. Some glass rods tested by Griffith shovred an ultimate strength of the order of 900,000 p.s.i., which is more than half of the theoretical strcngth mentioned above. 49. Castigliano's Theorem. In the previous article the equilibriu1n configuration of an elastic body submitted to given bo in Eqs. (29) vre may find many diffcrent stress dist.ributions satisfying the equations of equilibrium and the boundary conditions, and the question arises: What distinguishes the true stress distribution from all the other statically possible stress distributions? Let u,,, uu, T%'11 he the true stress components corresponding to the position of equilibrium and liu,., liuu, lirey small variations of these components such that the new stress components u,, liu%, Uu liuy, Tey lir"ll satisfy the sarne equations of equilibrium (18). Then, by subtracting the equations for one set from those of the other, we find that the changes in the stress components satisfy the following equations of equilibrium:

+

+

+

(•) Corresponding to this variation of stress components there will be some variation in the surface forces, Let !iX and õY be these small changes

+ lirey m = + liT,,u l =

liu,, l liuu m

liX liY

(b)

Consider no": thc change in strain encrgy of thc body due to the above changcs 1n stress components. Taking the strain energy per u~t volume. as a function of thc stress components (86), the change of th1s energy 1s

(') 1

é!Vo

FJ

iJu,, =

(u,, -

lltru) = Ez

Õuu =

FJ (uu -

Ptr,,) = Eu

av º

1 ()--;Tey =

é!Vo

··~

giving us

•

liVo =

1

~

t,.

liu,,

+ E.y liuu + 'Y""J liT~ 1

and the total changc in the strain cnergy due to changes of stress components is liV

')'ry

= JJ liVo dxdy = ff(E,, li11x +Eu liuu + 'Yx liTrv) dx dy 11

,-~~~~~~~x

B

(d)

A ds

N

Lct us calculate this change in Y energy, taking into consideration the F1G. 11s. boundary conditions (b). The first tcrm in (d) gives, integrating by parts,

ff·~*•-f·J~~*-f•~~ -

ff

U

----a;; º'"·d

X

dy

(e)

in which ~he expression lu &r,,I represents thc diffcrence of the valucE" of th~ funct1on u liu"' at two opposite points of the boundary1 such as the po1nts A and B in Fig. 115. We then have dylu

&r,,I = dy(u liu,,).1

- dy(u liu,Js

= ds(u liu:. cos Nx).1

+ ds(u &r~ cos Nx)

8

(f)

165

THEORY_ OF ELASTICITY

STRAIN-ENERGY METHODS

\ovhere cos Nx = lis the_cosine of the angle between the extemal normal N and the x-axis, and ds is an element of the boundary. Summing up such expressions as given in (f} we find

The right side of this equation represents the '\Vork produced by the ehanges of externa! forces on the actual displacements. The true stresses are those ""'·hich satisfy this equation. An analogous equation can be obtaincd for the three-dimensional case, '"ith a third term W óZ added in the bracket in Eq. (91), and the integral extending over the boundary surface instead of the boundary curve. If ""'"e have concentrated loads instead of a continuous distribution of surface forces, the integration in Eq. (91) should be replaced by a summation. Letting P1, P2, . . . , be indcpendent concentrated loads and d1, d2, . . . , the actual displacements of the points of applications of the loads in the directions of these loads, Eq. (91) becomes

164

and Eq. (e) becomes

JJ Brr~dxdy J =

e:

u Bu:lds -

Jf

5 u ªa ::dxdy

(g)

in which the first integral is extended along the boundary and the second over the area of the plate. In the sarne manner the second and the third terms on the right side of Eq. (d) may be transfonned and we find

JJ'•

J J:: ~ J !u,I - JJ J JJvªa~ dxdy JJ J J:~ + J J:; ~ J ''~1- JJ ::~dxdy + J "'~I Jf :;rir f +f JJ ª"'~ JJ .... ~ dx vªa~"dxdy =

!u,dxdy

i'ru lh-Zll dx dy =

dy

dyl"

!u,dy

ua

dx dy -

=

dxl"

11

v&r11 mds -

lrrru dx

,a

dx

ll-r"!I dy

• (h)

dxlu

v 8rZll l ds

u

lircy

v iJx"'lldxdy-

m ds

u-a:y-dxdy

Substituting (g) and (h) in Eq. (d) we obtain óV =

J

[u(OO: l

(92)

+ 8-rq m) + v(llu"' m + lh-xy l)] ds

_JJ[u (ªaª::+ a~ru) + v(ªa~u +a ;:~)]dxdy in '\\•hich the first integral is extended along the boundary and the second over the area of the plate. Making use of Eqs. (a) and (b) we finally obtain the follo-.,ving expression for the change of strain energy dueto variation of stress components: (91) óV = f(u óX + v óY) ds

ln this discussion we have taken the most general variations of the stress components fulfilling the equations of equilibrium (a). Let us consider no'\v a special case when the variations of the stress components are such that they can be actually produced in an elastic body by proper changes in the externai forces. We assume that the stress components are expressed as functions of thc externai loads P 1 , P 2 , • • • , and we take those changes of stress components which are dueto the changes óP1, llP2, ••• , of these forces. Considering only cases when the stress components are linear functions of the externa! loads 1 P 1, P 2 , • • • , and substituting these functions in Eq. (84), we obtain the expression for the strain energy as a homogeneous quadratic function of the externai forces. It should be noted that the reactions at the supports, which can be determined from the equations of equilibrium of a rigid body, '"ill be expressed as functions of the given loads P1,P2, ••• and ,~·ill not enter into thc expression for the strain energy. If there are redundant constraints, the corresponding reactions should be considercd togethcr \Vith the loads P 1, P 2, • • • as statically independent forces. Having an expression for the strain energy in tcrms of the externai forces, the change of this energy due to changes in the forces is

av

llV = llP1 aP1

av lJP2 + · • · + -aP2

1 We exclude for instance such C!IBCS as the bcnding of thin hars by lateral forces with simultaneous axial tension or comprel:lllion. In these cases the stresses produced by the axial force dcpend on thc def\ections due to the lateral forccl;'I aud are I!Ot linear functions of the externai loads.

166

THEORY OF ELASTICITY

STRAIN-ENERGY METHODS

Substituting this in Eq. (92}, we find

(av - d1) r,p1 + (ªv - d11) iJP1

iJP2

l!P2 • · •

167

forces or couples acting in redundant elements or at redundant constraints of an elastic system, the magnitudcs of thcse statically indeterminate quantities can be calculated from the condition that the strain energy of the system, represented as a function of X, Y, z, must be a minimum, i.e., we have the equations

=o

The forces P 1 , P 2, • • • are, as explained above, statically independent and their changes 8P 1, lJP 2, • , • are completely arbitrary. We can take ali but one of them equal to .zero; hence Eq. (i) requires

av

av

ax =o,

av

az =o,

aY =O,

(95)

(93)

We see that if the strain energy V of an elastic system is represented as a function of statically independent externai forces P1, P2, . . . , the partial dcrivatives of this function with respect to any of these forces give the actual displacement of the point of application of the force in the direction of the force. 1'his is the well-known Castigliano' s theorem. 50. Principie of Least Work. ln deriving Eq. (91) 've assume any changes in stress components sati:;íying the equations of equilibrium. If we assume now that the cbanges are such that the surface forces remain unchanged, then, instead of Eqs. (b) of the previous article, we obtain lluJ frr,;vm = O lluum + llr,,yl =O and Eq. (91) becomes (94)

+

This means that if we have a body with given forces acting on the houndary, and if wc consider such changes of stress components as do not affect the equations of equilibrium and the boundary conditions, the true stress components are those making the variation of strain energy vanish. It can be sho,vn that these correct values of the stress components make the strain energy a minimum. Then Eq. (94) expresses tl1e so-called principle of least work. This equation holds also if a portion of the boundary is held rigidly fixed by the constraints and the changes of stress components are such that there are variations of surfacc forces along this constrained portion of the boundary. Since the displacement along the constrained boundary is zero, the right sidc of Eq. (91) vanishes, and we arrive again at Eq. (94). The principle of least \Vork is used very oftcn in elementary treatments of statically indeterminate systems. 1 If X, Y, Z, ... are 'See, for instance, S. Timoshenko, "Strength of Materials," vol. 1, 1940, or Timoshenko and Young, "Theory of Structures."

I:

•

Several applications of the principle of least work in the solution of two-dimensional problems will be discussed in the follo"ring articles. õl. Applications of the Principie of Least Work-Rectangular Plates. As an example let us consider a rectangular plate. Previously (page 46) it has been sho"-'Il that by using trigonometric series the conditions on two sides of a rectangular plate can be satisfied. Solutions obtained in this "\\'ay may be of practical interest \\Chen applied to a plate whose width is small in comparison with its length. If Loth dimensions of a plate are of thc sarne order, the conditions on all four sides must be considered. ln the solution of problems of this kind the principie of minimum cnergy can sometimes Lc successY fully applied. Fio. 116. Let us consider the case of a rectangular platc in tension, \vhen the tensile forces at the ends are distributed according to a parabolic lav;d (Fig. 116). The boundary conditions in this case are: Forx =±a,

For y

1'"'11

=o,

Tzt1

= Ü,

±b, rly

=

o

(a)

The strain energy for a plate of unit thickness is, írom Eq. (86),

lt should be noted that for a simply connected boundary, such as \V(~ have in the present case, the stress distribution does not depend on the 1

See S. Timoshenko, Phil. Mag., vol. 47, p. 1095, 1924.

1~8

THEORY OF ELAS7'ICITY

STRAIN-ENERGl" Jf1'.'1'HOD,<;

elastic constants of the material (sce page 25) and further calculations can thcrefore be simplificd hy taking Poisson's ratio v as zero. Then, introducing the stress function q:,, and substituting in (b)

functions; the second derivative of this expression ,,.,·ith respect to x vaniF>hes at the sides y = ± b, and the second deriva tive ,,.,·ith respect to y vanishes at the sides x = ±a; the second derivative a2 /ax ay vanishes on all four sides of the plate. The stress function can then bc taken as

p

-..ve find 1 V = 2E

f f [(ª'")' + (ª'•)' + (ª'" )'] ay2

iJx2

2

axay

=o

=à

rp

dx dy

Sy 2 ( 1 -

~ ~:) + (x2 -

a2)2(y2 _ b2)2(a 1

(')

169

+ 0:2X2 + asy2 + .. ·)

Only evcn po,vers of x and y are taken in the series because thc stress distribution is symmetrical "'rith respect to the x- and y-axes. Limiting uurselves to the first term a 1 in 02 04 06 08 series (f), we have d xs /.

1-'he correct expre8sion for the stress function is that satisfying conditions (a) and making the strain energy (e) a minimum. If \VC apply variational calculus to determine the minimum of (e), we shall arrivc at Eq. (30) for the stress function cfi. Instead of this wc shall use the follo\~·ing procedure for an approximate solution of the problem. Vire take the stress function in the form of a series,

'º

1 1

(d)

such ihat the boundary conditions (a) are satisfied, o: 1, o: 2, o: 3, being constants to be determined later. Substituting this series in expression (e) 1vc find Tr as a function of the second degree in o: 1, ai, as, . . . • The magnitude of the constants can then be calculated from the minimum conditions

av Üas

0

=

1'he first of Eqs. (e) then becomes /

'

" 1

1

Illll

y =

b) we find 0:1

Fio. 117.

r1z

=

(f~ =

6b

sinC'e this gives

, must be chosen so that the 1'hc rcmaining functions t/>1, t/>2, , stresses corresponding to thcm vanish at the boundary. To ensure this "·e take the expression (x 2 - a 2) 2 (y 2 - b2) 2 as a factor in all these

=

0.04253 ; a

and the stress components are

1 2 ( 1--:iL 1 2') -Sy 2

""

(e)

,,.,·hich \Vill be linear equations in 0:1, 0:2, aa . • . • By a suitable choice of the functions $1, $2, . . . , we ca.n usually get a satisfactory approximatc solution by using only a fe\v terms in the series (d). ln our case the boundary conditions (a) are satisfied by taking

t/>o

•

For a square platc (a =

(f)

7zir

=

S (1 -

~:)

- 0.1702,q (1 -

?;

2

-0.17028 (1 -

)

(1 -

~ ) (1 - a'x')' 2

3

~)

2

xy ( x') (1 - y') -a2

-0.680581 - a2 a2

The distri.bution of rr~ on the cross section x = O is represent.ed b~ curve 11 1 (Fig. 117). To obtain a closer approximation, we now take three terms in the series (f). Then Eqs. (e), for calculating the constants o: 1, o: 2, o: 3, are 1

Curve I representa the paraholic stress distribution at the ends of the plate.

ª

1

64 ( 7

+ 256 b' + 647 b') + a,a ' (64 + 64 b') 49 a a 77 49 a 2

+ ª

1

64 ( 11

+

64 b4\ 7 a~}

+

(192 a:ia 143 2

2

alf(l

64 (7

64 b 4)

+ TI a4 + ª ª 2

2

(64 77

+

ª

s

64 b5) 77 a 6 = a4iJ!

+

4 )

2

77 a

7 a4

2

2

64 ( 77

b' + 7764 ai;b')

a2

s

=

a4b2

(g)

=

4

+

64 b ) 77 a 4 2

3

64 b2 ( 49 a2

+ 256 b + !92 b + uaa

41

To deal with other symmetrical distributions of forces overthe edges ±a we have only to change the form of thc function 4>o in expression (f). Only the right-hand expressions in Eqs. (g) have to be changed. As an example of stress distribution nonsymmetrical with respect to the x-axis, let us consider thc case of bending shown in Fig. 118 1 in 'vhich the forces applied at the ends are (11,,),,_"°,, = Ay 3 (curve b in Fig. 118b). Clcarly, the stress system "'rill be odd "\vith respect to thc

x

4

4

'(192 b ª 7 a2

4

+

256 b 77 ã4

+ 192 b

6

143 a6

)

y

:For a square plate these givc

ª'

=

'fhe distribution of

{u,,)....,0 = S

s 0.04040 6'

( ay') 1 -

a2

a

U"x

2

= aa =

0.16168

(

Fia. 118.

a

x-axis and even vrith respect to the y-axis. These conditions are satis·· fied by taking a stress function in the form

y')

~

y' + y')

- 12 a2

15 -a"

ln Fig. 117 this stress distribution is shov.•n by the curve III. 1 As the length of the plate incrcases, the stress distribution over Lhe cross section x = O becomes more and more uniform. If v.•e take for instance a = 2b, we find, from Eqi!!. (g),

s

a 1 = 0.07983 a 4 b2 '

s

0.12.50 a 6b2'

a2 =

The corresponding values of below: 0.2 "" = 0.6908 0.6848

11,,

4> =

Jrr Ays + (x2

1 - 3 a2

+ 0.0235 (1

a3

=

fb)

(a)

s 0.01174 8

on thc cross section z =O is given by -

171

STRAIN-ENERGY METHODS

1'HA'ORY OF ELASTICITY

170

s

0.01826 a 6b2

over the cross section x = O are given

0,4

0.6

0.8

1.0

0.6698

0.6538

0.6498

0.6758

This distribution is represented in Fig. 117 by the dotted line. We see that in this case the deviation from the average stress, jS, is very small. 1 Similar resnlts were obt11.ined by C. E. Inglis, Proc. Roy. Soe. (Ltmdon), series A, vol. 103, 1923, 11.nd by G. Piekett, J. Applied Mechania (Trans. A.S.M.E.) vol. 11, p. 176, 1944,

_ a2)2(y2 _ b2)2 (a1y a2yx 2

+

+ a3y3 + a(X2y 3 + · · ·)

(h)

The first term, as before, satisfies the Loundary conditions for ip. lJsing F:q. (h) '\\o"ith four coefficients 0:1, , a4 in Eqs. (e), '"e find for a square plate (a = b)

u,.

=

ay ª'•

2

= 2Aa 3

l'

2

T/ 3

-

(1 - t 2) 2[0.08392(5T/ 3

'

-

+ 0.004108(21T/º - 2011 + 311)] ~ 2 ) 2 [0.07308(511 3 311) + 0.04179(2111

311)

3

-

~ 2 (1

-

-

5

-

2011 3

+ 311)]}

(k)

where t = x/a and 71 = y/b. The distribution at the middle cross "lec tion x =O is not far from being linear. It is shown in Fig. 118b by curve a. 62. Effective Width of Wide Beam Flanges. As 11.nothcr ex11.mple of thc application of the minimum--cnergy principle to two-
172

STRAIN~ENERGY

THEORY OF ELASTICITY

proportional to the distance fron1 the neutral axis, i.e., that the stresses do not change along the ·width of thc flange. Bnt if this 'vidth is vcry large it is known that parts of the :lianges at a distance from the web do not take thei:r full sharc in resisting bending mon1ent, and the beam is weaker than the elementary theory of hending indicatcs. It is thc usual practice in cal!'ulating stresses in such beams to replace the actual l\ridth of the flanges by a ccrtain reduccd width, such that the elementary theory of bending applied to such a transformed beam cross section gives the correct value of maximum bending stress. This reduced width of flange is called the ejfectú•e widlh. ln the following discussion a theoretical basis for dctermining the effcctivc width is given.t

173

METHODS

can be taken for our symmetrical case in the form of the series

l

11~.,



f,.(y) cos n;x

(b)

·-' in whiehf,.(y) ar11 íunetions of y only. ing expression for fn(y):

Substituting in Eq. (a), we find the follow-

(o)

'l'o satisfy the eondition that stresses must VH11ish for an infinite value of y, \Ye take C,. = D,. = O. The expression for the stress function is then

(d)

h The coefficients A,. and B,. will now he detennined from the eondition that the true stress distribution is that making the strain energy of the :flange together ,,,·ith that of the '\\'eh a miuin1um. Substituting

y~,---~.--,,-,~~~~~ 1

1

2l


1

.

-

éJ2
fb)

in the expression for strain energy

1

!

..

,

= Oy''

1

~ X

and using Eq. (d) for the stress function, the strain encrgy of the flange isl

Fui. 119.

To make the proble1n as sirnple as possible it is assume d that we have an infinitely long continuous bea1n on equidistant supports. Ali spans are equally loaded by loads ~yn1mctrical >vith respect to the middle of the spans. One oí the supporta of the ~pan shov•n in Fig. 119 is taken as the origin of coordinates, with the :e-axis in the direction of the axis of the beam. Due to symmetry, only one span and oue half of the flange, say that corresponding to positive y, need be considered. The width of the flange is assumerl infinitely large and its thickness h very small in eomparWon >vith the depth of the beam. Bending of thc flangc as a thin plate can then be neglected, and it can be assumed that during hl:nding of the beam the forces are transmitted to the flange in its middle plane so that the stress distribution in the flange presents a two-dimensional problem. Thc eorresponding stress function , satisfying the differcntial cquation O'
éJ44'

O'

ax•+ 2 ax•ay•+ay•

=O

(o)

1 The suhicct wasinvestigated by T. v. Kármán;see "FestschriftAugustFõppls," p. 114, 1923. Also G. Schnadel, Werft u.nd Reederei, vol. 9, p. 92, 1928; E. Reiasner, Der Stahlbau., 1934, p. 206; E. Chwalla, Der Stahlbau, 1936; L. Beschkine, Publ3. Intern.. Assoe. Bridge and Slruciu.ral Engineering, vol. 5, p. 65, 1988.

V1

=

. (B . + f...i1

2h \'

n3.,.-3 2

E

~

A .. R .. 2G

+ A ..") 2G

(•)

n~l

ln considering the strain energy of the web atone, let A be its cross-sectional area, l its moment of inertia about the horiionta.J axis through the centroid C, ande the distance from the centroid of the ·web to the middle plane of the flangc (Fig. 119). The total bending n1oment transmitted at any eross section by the web together with the flangc ean be represented for our symmetrical case by the series

M = llfo

+ J.!

1

cos

1!f + M, cos 2 ~X + ...

lfl

ln this series 11-f o L~ a statically indctenninate quantity depending on the magnitude of the bending moment at the supports, and the other coefficients J.f 1, M ,, . . . , are to be calculated from the conditions of loading. Letting N denote the compressive force in the flangc (Fig. 119c), the bcnding mornent 11-f can be divided into 1 The integrais entering into the expression for strain energy are calculated in the paper by Kármán, l-Oc. cit.

• 174

175

THEORY OF ELASTICITY

STRAIN-ENERGY METHODS

two parts: a part M' taken by the web and a part J.f", equa.l to Ne, dueto the longitudinal forces N in the web and fln.nge. From staties the normal stresses

Adding this to the strain energy (e) of the fiange, and introducing in this la.tt.er the notations n. X n• 2h-yA,.=

over any cross section of the complete bcam give a couple M, hence

fo a~dy =0 2he lo . a~ dy M 00

N +2h li!' where -2he

(g)

,~·e find

the following expression for the total strain energy:

=

lo"' ax dy ~ M" is thepartofthe bendingmomenttaken bythell11.nge.

V=

ÍhE

í

n[Y,.•

+ (1

1E f X,.•

+ v)X,.Y .. + (1+v)Xn'l+ 2

n~l

n-l

The strain energy of the \\'-eb is

...

+ 11.fo'l +-'\' EI 2El L

(h)

r· ,,.~ dy

=

l"I'

1

.. + eX..)!

(l)

Thc quantities Mo, X,., Y .. are to be determined from the minimum condition of the ~train energy (1). It can be seen that Mo appears only in the term M o'I/ b'l, and from the minimum requirement for (1) it follows that Mo = O. From the condition

From the first of Eqs. (g) we find

N = -2h )o

(M

-2h }o .,.. ay• dy = 2h ay ..

From expression (d) for the strosa function it may be seen that

•

(~~)~-~

=o,

.

Hl!noo

\ ' nr

Jf' = M

2Y,.

zA,.cos-1

+ 2he }o {

00

"~ dy ~ J.f

+ Ne

M

=O

+V x .. 2

Substituting this and M 0 = O in Eq. (l) we get the following expression for tbe atrain energy:

•

=

+ (1 + v)X,.

Y .. - - 1

n:irx

..f..i,

N = 2h

it follows that

\' n,..nrx L --yA~cos-1

.. ,

~

... x + 2he \'n,.. Li TA~ cos n-,-

,,. 3+2v-v 1 V = 2hE. 4

11=1

.

l

n=l

n~l

or, using the notation

.

Ln. X,.• + 2AE \'x• 1.. ..

\'

+ 211

-..ve may write

l..• ,

(M,.

+ eX,.)•

(m)

From the oondition that X,. should make Va minimum it follows that

M' = M

...,lx.

+e

•

•

oos

n;x

=Mo+ "L (M,.

ax• = o

.!f_

(k)

+ eX,.) cos nrz -,-

from which we find

•·•

x.

Substituting in (h) and noting that

f 21 eos 1 ~ dx

}o

l

(2! n,..-x mor:e Jo cos -,-
= l,

.

we obtain

\'x•+Mo'l+-'-"
l

"

(n)

Let us oonaider a particular case when the bending-moment diagram ia a simple line, say M = M, cos (:rX/l). Then, from Eq. (n),

~oaine

x,

=

176

THB'ORY OF ELASTICITY

STRAIN-ENERGY METHODS

and, from Eq. (k), the rnoment dueto the force N of the flauge iK

"

Taking, for instance, v ~ 0.3, we find 2X = 0.181(21)

(p)

lff" = - e1\' = -eX1COB/

177

for the a:ssumed bending-momcnt diagram the effective vádth of the fiange is · i.e., The distribution of the stress
A, =_IX,,

2,h

This distribution of
u,=•Tc+-y

(q)

and, from the equations of statics,

2Ah
+ u,A

= O 2/Jin,t, = 11,f"

The expressions for the t\vO portions of the bending moment, from Eqs. (q) and (r), are llf'

=

~

(u, - u,) =

~ ( 1 + 2~h) u,

Jl,[ 11 = 2Xheu,

The ratio of M" to the t
2Xheu,

M-+M" - - - '( '") 2MM, +e 1 +A "•

1

1

(•)

To makc this ratio E!q~~J.to th~-ratio M'';Aj Obtained from thé exact solution (:o), we must take 1 1rl3+2v-v• 2Xhe• = lie 2i 4. From this we obtll.in the following exprcssion for the effcclfre width 2X:

approximatcly 18 per eent of the span. . 1 the case of a: continuous beam ·with equal concentrated forces at the nuddlr of t~c spans, thc ben
4l

2X = 0.85 . ;r(3

+ 2v

- v')

i.e., somewhat less than it is for the

1'1c. 120. case of a moment diagram in the form of a eosine line. 63. Shear Lag. A problem of the sarne general nature ~s that discussed in Art. 52 occurs in aircraft structurcs. Considera box be,am, ~g.121, fornu•;d from . eJi~nnels ABJt'E and DCGII to "rhich are attachcd tbm sheets ABCD a:nd t ~Ou D "bili" EFGH, by riveting or ·welding along the edges. If the ·whole heam :3 u 1n at the left-hand end, and loaded as a cantilcver by two forces P applied to t~e channels at the other end, the elementary bending theory v1:ill give a tens1le hendiug stress in thc sheet ABCD uniform across any section parallel t.o R_C. Actually, hovocver, the shcct acqurres 1ts tcnsile stress from shear streSBCs on its edges communicated to it bythe cha:nnels, ~~;iJ~Z:--:;?.:::::-.,Jc as indicated in Fig. 121, and the distribu:_ tion of tensile stress across the ·width will p not be uniform, but, asin Fig. 121, higher a:t the edges than at the middle. This departure from the uniformity assumed by the ele1ni:ntary theory is known as Fia. 121. "shear lag,'' since it involves a sbcar deformation in the sheets. Thc problem has been a:nalyzed by strain-energy and other methods, voith the l1l'lp of simplifying assumptions.'

Problema

1. Find an exprcssion in terrns of u,, ""' T,~ for the strain energy V per 11nit thiekncss of a cylinder or pri5111 in plane strain (•, = O). 2. \Vrite dov.'11 the integral for thc strain energy V in tertns of polar coordinates and polar stress components for the case of plane stre>;S [ef. Eq. (11), Art. 51]. The stress
178

THEORY OF ELASTICITY

balancing couple to the outside. Evaluate the strain ene:rgy in the ring, and by equating this to the work done during loading deduce the rotation of the outside circlc when the ring is fu:ed 11.t the inside (cf. Prob. 2, page 125). M 3. Evaluate the strain energy per unit length of a cylinder a < r < b subjected to internai pressure p; [scc Eqs. (46)]. Deduce the radial displaccmcnt of the inner surface. Obtain the same result by use of Eq. (50) (taking v = O) and the stress-strain relations of plane stress. b 4. Intcrpret thc cquation

CHAPTER 7

Q

ffVo dx dy FIG. 122.

=

Íff(Xu

+ Yv) dx dy + j-J(X-u + Y-v) ds

and give thc justification of thc factors

j- on the right.

li. Sho\V from Eq. (84) that if \•;e have a case of plane stress anda corres}Xlnding case of plane strain (•. =O) in which the stresses ux, ou, Txu are the sa.me, thc strain encrgy is greater (per unit thickness) for the plane stress. 6. In Fig. 123, (a) represents a strip under compression, in which thc stress thercfore extends throughout. ln (b) the deformable strip is bonded to rigid plates

(a)

f6)

fc)

FIG. 123.

on its top and bottom edgcs. 1Vill there be stress throughout the strip or only locally at thc cnds? In (e) the upper edge is free, as in (a), hut the lower edge ili fixed, as in (b). \\'ili thc stress be local or not? 'l. From the principie that a system in stable equilibrium has less pot.entia.l energy than that corresponding to any neighboring confi.guration, show without calculation that thc strain energy of the plate in Fig. 114 must either deereBBe or remain the sarne whcn a fine cut AB is made. 8. Sta.te the Castigliano theorem expressed by Eq. (91) in a form suitable for use in polar coordinatcs, thc boundary forces 2 and Y being repla.ced by radial and tangcntial components R and T, and the diaplacement C(lmponents by the polar componcnts u and i; of Chap. 4. 9. "Equation (91) is vatid ·whcn av, a:X, aYresult from any small changes in the stress components which satisfy thc conditions of cquilibrium (a) Art. 49, v.'hether these changes violate the conditions of compatibility (Art. 15) or not. ln the latter case the changes in the stress are those which actually occur when the boillldary forces are changed by 62, i'lP"." Is tbis statement correct? Assuming that it is, show that thc radial displaccmcnt of Prob. 3 can be calculated from the formula

av

1 (u)r-a = --- · 2,..-a Op;

TWO-DIMENSIONAL PROBLEMS IN CURVILINEAR COORDINATES 64. Functions of a Complex Variable. For the problems solved so far, rectangular and polar coordinates have prove
Fio. 124.

+ i2xy

sincc i•

=

-1

C'.-0nverting to polar coordinates, as in Fig. 124,

z = x +iy = r(cos () +isin fJ)

(a)

Since

e'8 = 1

+ ifJ + 2~. (ifJ)' + -3\ . (ifJ)ª + 4.~ (ie)• + i2

=

-1, i3

=

-i, i' = 1, etc.,

We have

1 1 -21°· +41°

1

+ i(fJ

-

-

~ (JI

+

=cosfJ+iBinfJ

Ftom Eq. (a) therefore z=x+iy=rei~ 1

(b)

The definitions repreaent operations on pairs of real numbers the use ol i being

~erely a convenience. See for instance E. T. Whittaker and G. N, \Vatson Modem Analysis," 3d ed., pp. 6-8.

' 179

180

PROBLEMS IN CURVILINEAR COORDI1'.'A'l'b'8

Algebraic, trigonomctric, exponential, logarithmic, and other functions can bc formcd from z as well as from a real variable, provided an analytical rather than a geometrical defiuition IB adoptcd. Thus sin z, cos z, =d e• may be dcfincd by thcir power series. Any such function can be separated into "real" and "imaginary" parta, that is, put in the forro a(x,y) + ip(x,y) \vhere a(x,y), the real part, and fJ(x,y), the imaginary part,' are ordinary real functions of x and y (thcy do not contain i). For instance if the function of z,f(z), is 1/z, we have

derivatives which depend on z only, being the sarne for ali directions (of dz) at the point z. Such functions are eallcd analytic. The quantity x - iy may be regarded as a function of z, in the scnse that if z is given, x and y are given, and so x - iy is determined. However, x - iy cannot be formed from z as for instance Z", e• are formed. Its derivative with respect to z is the limit of (Ax - i Ay)/(fJ.:I; + i D.y) as D.x, D.y---> O. This is not indcpcndcnt of the direction of the shift 8:1:, D.yi li we take this shift in thc x-direction, so that Ay = O, we obtain 1 as the va\ue of the limit. li we take the shift in the y-dircction, fJ.:I; = O and the limit is -1. Thus x - iy is not an analytic function of x + iy. Analytic functions togethcr with x - iy will be used !ater in the construction of stress functions. Any function involving i will bc rcferrcd to as a "complex function." An annlytic function f(z) will have an indefinite integral, dcfincd n~ thc function having /(z) as its derivative 'vith re~pect to z, an
1

f(z) = x

X -

+ iy

=

iy

(XTii/}(X --~i/j = x'

Similarly, observing that coah iy e±•v = coa y ± i sin y, vre lind

=

X

+ y' + i

• (

x'

!(e'• +e-••), sinh iy =

-y)

+ yt

~(é•

- e-••), and

sinh (:i: + iy) = ~inh x cosh iy + co~h x sinh iy sinh x cos y + icosh xsin y cosh z = cosh (x + iy) = cosh x cosh iy + sinh x Binh iy = cosh x cos y + isinh xsin y sinh z

= =

J~dz=logz+C

As an illustration of the general method for converting a complex denominator to a real one, consider the function coth z. "\\Te havc coth z = ~~h~ = cosh (x + iy) . sinh (x - iy) sinh z sinh (x + iy) sinh (x iy) (cosh x coa y + i sinh x sin y)(sinh x cos y - i cosh x sin y) (sinh x co~ y + i cosh x slii y)(sirlh-Xcos y-~-{ cosh x sin y)

the additive const.ant C being now a complex number A + iB, containing t~'º real arbitrary coristants A and B. 55. Analytic Functions and Laplace's Equation. An analytic function /(z) can be regarded as a function of x and y, having partia! derivatives. Thus

..i_ f(z)

The denominator is the snmc as the real quantity (sinh x cos y)' + (cosh x ain y)I. Vl'hen the numcrator is multiplied out, and i' replaced by -1, the separation into real and imngiuary parts is completed. The result can be simplified t-0

=

iJx

since i!z/&x = 1.

.'dz!:. f(z) iJx
=

sinh 2x - i sin 2y cosh 2x - cos 2y

(o)

An alternntive procedure is indicated by Y.q. (p) of Art. 62. The dcrivativc of f(z) \1•ith rcspcct to z is by defuiition df(z) = lim f(z dz O

+ D.z)

- /(z)

D.z

whcrc 8:11 = ó.X + i D.y and fu"l---> O means, of course, both /J.x---> O and D.y---> O. We ean always think of x, y aa thc Cartesian coordinaWs of a point in a. plane. Then D.x, D.y represent a shift to a neighboriug point. It rnight be expected at first that (d) could bc diílerent for different directions of the shift. Nevertheless, the liinit in (á) is calculable directly in terms of z and 8z justas if these were real numbcrs, and the corresponding resulta, such ru:;

dz

(z')

=

2z,

d •

dz Slll Z

ilx

ily

sincc az/ay = i. But if /(z) is put in the forrn a(x,y) =

2

"'


(o)

(b)

+ i{3(x,y), or for brcvity"' + i(J, we have a i!yf(z) =ª"'+i'J(j iJy iJy

+ iiJfJ iJx

(e)

Comparing Eqs. (e) v.ith :Eq. (a) and Eq. (b) yields i (~~

àx

+ i àx iJfJ)

=

iJ"'

iJy

+ i ày a{3

(d)

Rcmembering that "'' fJ are real, i• = -1, and that the equality irnplies that real and imaginary parts are separately equal, \Ve find àa

-

d

f'(z) !!_.z_ = f'(z)

!_ f(z) = /'(z) iJz = i/'(z)

!_/(z) iJx (d)

=

Similarly f)y

coth z

iJx

iJfJ

=-, ày

= COS Z

must appear, ind('pendent of the choice of 8.z, and of Ax and ày. We may say, therefore, that all the functions we 1nay form from z in the usual way will he.ve 1 It ahould be observe
l,i

181

THEORY OF ELASTICITY

These nre called the Cauchy-Riemann eqnations. F.liminating fJ hy
182

THEORY OF ELASTICITY

PROBLEMS IN CURVILINEAR COORDINATEB

An equation oí this form is called Laplace's equation and any aolution ia called a h.ar7Wmic function. ln the same way elimination of a from Eqs. (e) yields

S. If r is a complex variable, and z = e cosh

d Slll . h

dz

(g)

Thus if two functions a and {3 of x and y are derived as the real and imaginary parta of a.n analytic function f(z), ea.ch will be a solution of Laplace's equation. Laplace's equation is encountcrcd in many physical prohlems, including those oí elasticity [see for instance Eq. (b), Art. 16]. The functions a and {3 are called conju.gale harmonic functions. It is evident that if we are giv('n any harmonic function a, Eqs. (e) will, but for a constant, determine another function {3, \vhich ·will bc the conjugate to a. As examples of the derivation of harmonic :functions from analytic functions of z, considerei"', zn, log z, n being a real constant. '\\'e have

showing that mi. cos nx, e---no sin nx are harmonic functions. Chang:ing n to -n we find that en~ cos n:t., e"" sin nx are a\so harmonic, and it follows that sinh ny sin nx,

cosh ny sin nx,

sinh ny cos nx,

cosh ny cos nx

(h)

are harmonic since they can he formed by addition and subtraction of the foregoing functions with factors j-. From

+ +

r, find

nr

\Yriting r = E i11 find the real and imaginary parts of this derivative when e and n are real. 4. If z = x iy, r = E i".,, and z = i"a coth ír where a is re11.l, show that

+

a sin 11 x = cosh ~ cos .,,'

a sinh

y=cosh~

e

cos,,

56. Stress Functions in Terms of Harmonic and Complex Functions. If 1/; is any function of x and y, '\\'e have by differentiation

ª' + ayà')2 (xi/;) = x (ª"' ôif; (ax2 ax2+ •'•) ay 2 + 2 ,ax If ip is harmonic, the parenthesis on the right is zero.

.

. . (à'ax 2+ ayª')2 (ª") ax

harmon1c funct1on, s1nce

=

r"

cos n9,

(ª'" •'•)

axà ax2+ ay2

=

O.

Thus another application of the Laplacian operation to (a) yields

(::2 + ::2) (:: + ::2) (xY,) =O

(b)

which is the sarne as

rn sin n9,

~

cos n9,

From log z = log rei9 = log r

r-.. sin n9

(•)

+ ie

we find the harmonie functions log r,

'

lt is eaaily verificd that the functions (i) and (j) satisfy Laplace's equation in polar coordinatcs [see Eq. (d), page 57], i.e., éll,f &r•

(a)

Also ay, / ax is a

2

we find the harmonic functions

183

+ .! élif + l. {J~if r àr

r• ao•

=

o

(k)

Problems 1. Determine the real functions oí x and y which are the real and imaginary parts of thc complcx íunctions z2, ~. tanh z. [x• - y•, 2xy; x• - 3xy 1, 3x'y - y•;

sinh 2x(cosh 2x + oos 2y)-I, sin 2y(cosh 2x +coo 2y)-1] 2. Determine the real functions of r and () which are the real and imaginary parts of the complex functions .r•, z log z. [r-• coa 2fJ, r-• sin fJ; r log r cos 9 - rfJ sin 9, r log r sin fJ

+ r9 coa 9]

(:;, + 2 ax~;y2 + ::4) (xi/;)

=O

Cornparison with Eq. (a), page 29, sho\vs that xY, rnay be used as a stress function, 1/; being harmonic. The sarne is true of yY,, and also, of course, of the function Y, itself. It can easily be shown by differentiation that (x 2 y 2)Y,, that is r 2 Y,, also satisfies the sarne differential equation and rnay therefore be taken as a stress function, f being harmonic. For instance, taking the two harmonic functions

+

sinh ny sin nx,

cosh ny sin nx

from the functions (h), page 182, and multiplying them by y, we arrive by superposition at the stress function (d), page 47. Taking the harmonic functions (i) and (J), page 182, as they stand or multiplied by x, Y, or r 2 , we can reconstruct ali the terms of the stress function in polar coordinates given by Eq. (81), page 116. The question of whether any stress function at ali can be arrived at in this fashion remains open, and will be answered irnmediately, in the

i

~' .

THEORY 01'' ELASTICITY

PROBI,EMS li'./ CURVIL!J.'EAR COOill)lh'ATES

process of expressing the general stress function in terms of t"\\-·o arbitrary analytic functions. Denoting the Laplacian operator

Equation (96) "\\'ili prove useful later, but it may be observed that the use of both the functions p and q is not neccssary. Instead of Eq. (g) "'e can write v2 (q, - 2xp) = V 2 c/i - 4 ap = o

184

a2

a2

ax

- +ay2 -

iJx2

uy V2, Eq. (a) on page 29 can be 'vritten V2(V 2.P) = O or V4q, =O. Writing P for v 2q,, ,vhich represents u,, + rfy, \Ve observe that P is a harmonic function, and so \vill have a conjugate harmonic function Q. Consequently P + iQ is an analytic function of z, and -..ve may \Vrite (e)

Thc integral of this function \vith respect to z is another analytic function, 4 if(z) say. Then, \Yriting p and qfor thc real and imaginary parts of if(z), \Ve have if(z) = p iq = tff(z) dz (d)

+

so that if/(z) = ±.f(z).

We have also

ª" + iaq ~ ~f(') ~ f'(') ª' iJx

ax

iJx

~ ~f(,) ~ ~(P + iQ) 4 4

àx

showing that cJi - 2xp is harmonic, say cqual to p2, so that any stress function must be expressible in the form (h)

v.·hcre p and p 2 are suitably choscn harmonic functions. Sirnilarly, considering q, - 2yq, Vi'e may sho"\\' that any stress function must also be expressible in the form

c/i

x(z)

(x - iy)(p

(n

Recalling that P = v 2q,, Eqs. (e) and (f) enable us to show that q, - xp - yq is a harmonic function. For

v2(q, _ xp - yq)

=

ap

aq

v 2q, - 2 ax - 2 ay =O

(g)

,p

~

+ iq1

+ iq) + P1 + iq,

Rc ["/(')

zz 1/l(z)

q,-xp-yq=p1 Consequently

(96) c/i = xp + yq +Pi which shows that any stress function can be formed from suitably chosen conjugate harmonic functions p, q anda harmonic function P1·

Thus any stress func-

+ x(,)]

(97)

-v,·here Re means "real part of," Z denotes x - iy, and f(z) and x(z) are suitably chosen analytic functions. Conversely (97) yiclds a stress function, that is a solution of Eq. (a), p. 29, for any choice of 1/l(z) and x(z). It is applied later to the solution of several problems of practical interest. '\Vriting the "complex stress function" in Lrackets in (97) as

Thus for any stress function c/i we have where p 1 is some harmonic function.

= P1

is identical \vith thc right-hand side of Eq. (96). tion is expressible in the form 1

Since p and q are conjugatc functions, they satisfy Eqs. (e) of Art. 55, and so

4

+ Pa

Thcn it is easily vcrified that the real part of

(e)

aq~!.p

-2yq

=

\Vhere q and p 3 are suitably chosen harmonic functions. Rcturning to the form (96), let us introduee the function q, \Vhich is the conjugate harmonic to p 1 , and 'vrite

E4uating real parts of the first and last mcmbers we find

ay

185

'

+ x(z)

and observing that Zz = r 2 , and 1/l(z)/z is still a function of z, v,"e find that any stress funetion ean also bc cxpressed as r~p1

where p4, Pr. are harmonic. 1

+ Pr.

E. Goursat, Bull. soe. math. France, vol. 26, p. 206, 1898. Math. Ann., vol. 107, pp. 282-312, 1932.

X. L '°'Iuschelillvili,

~i .

THEORY OF ELASTICITY

186

PROBLEJ.fS IN CllllVIL!NEAR COORDil\fA'l'ES

57. Displacement Corresponding to a Given Stress Function. It \\·as shown in Art. 39 that the detcrmination of the stress in a multiplyconnected region requires thc cvaluation of displacement to ensure that it is not discontinuous, that is, to cnsure that the stress is not partly due to dislocations. For this reason, as wcll as for cases vrhere the displaccments are of interest in themselves, \VC require a mcthod of finding the displacement functions u and v >•rhen a stress function is g1ven. The stress-strain relations for plane stress, Eqs. (22), (23), may be written

au

E iJx

=

u,. -

ª'

E ay

vu 11 ,

= u11

a(ª' ax +ªu) ay ~'

(b)

ey

Inserting thc stress function into the first, and rccalling that P \Ve have E

au

a2q, _ "a 2.p

=

ay 2

ax

=

ax 2

(P _a2q,) _

j)

= 'V 2 .p,

(e)

and similarly ay

=

-

(1

+ JJ) a2q, + P

(d)

ay2

But from Eqs. (f) and (g) of Art. 56, v.•e can replace P in Eq. (e) above by 4 iJp/iJx, and in Eq. (d) by 4 iJq/iJy. Then, after division by 1 + JJ,

2G au = _ iJx

a2q, + _4_ ap, iJx

1

2

2G

+ JJ iJx

av

ay

=

_ a2q, + _4_ aq iJy 2

1

+

li

iJy

(')

and these imply, by integration, 2fJu ~ -

•• 4 - +· - p iJx

l+JJ

+ f(y) '

20,

~

- :: + 1

'vhere f(y) and f1(x) are arbitrury functions. in the left of Eq. (b), "\•re obtain

_ o'
1

+

j)

iJx

••

2Gu = - iJx

2dy

!, + q

f,(x)

(f)

4

2dx

~

, w

(g)

But the first term on the left is equal to rw, and the parenthesis vanishes because p and gare conjugate harmonic functions satisfying

(h)

on thc understanding that a rigid-body displaccment can be added. These equations enable us to find u and v 'vhcn q, is kno"\vn. W e ha ve first to find P as V 2cP, determine thc conjugate function Q by means of the Cauchy-Riemann cquations

aP iJy =

--·ox aQ

forro the functionj(z) = P + iQ, and obtain p and q by integration of f(z) as in Eq. (d), Art. 56. The terms of Eqs. (h) can then be evaluated. The usefulness of Eqs. (h) will appear in la ter applications, for -..vhich the mcthod of determining displacements used in Chaps. 3 and 4 is not suitable. 68. Stress and Displacement in Tenns of Complex Potentials. So far the stress and displacemcnt components have been expresscd in terms of the stress function q,. But since Eq. (97) expresses q, in terms of two functions Y,(z), x(z), it is possiblc to express the stress and displacement in terms of thcse two "complex potentials." Any complex functionj(z) can be put into the forma+ i{3 "\\'here a and {3 are real. To this there corresponds the conjugale, 2 a - i{3, the value taken by j(z) when i is replaced, whcrcvcr it occurs in j(z), by -i. This change is indicated by the notation J(•) ~ a - i~

Thus if f(z) df,

-A

+ 1 + vp'

If these are substituted

(•v + aq) + !e df + !e ay

df,

'"here A is a constant. It follows that the termsf(y) andf1 (x) in Eq. (f) represent a rigid body displacement. Discarding these terms \Vfl may write Eqs. (f) as 1

axi

+ P) ª'• ax2 + p

dx

dx =

aP aQ -=-, iJx ày - (1

E av

dy

\Vhich implies

a2q,

ax 2

Hence

df+ªh~o

(a)

vu,.

-

the Cauchy-Riemann equations (Art. 56).

187

1

=

(a)

ein• '\\'e have

(b) A. E. H. Love, "11athematical Thcory of Elasticity," 4th cd., Arts. 144, 146,

1926. 1

The v.·ord is used here with a significance quite distinct from that in the term "conjuga.te harmonic functioos.''

189

PROBJ,f<.'}.18 IN CURVILINEAR COORDINATES

THPJORY OF Ef.ASTIC!TY

188

Differentiating it "'·ith respect to y and multiplying by i we have

This may bc contrasted with

0~ :y - ~~ = 2

!(•) ~ ""'

i

to illustrate the significance of the bar over the f in Eq. (a). Evidently

-

1f/(z)

+ X"(Z)

(e)

(~hanging i

+

to - i on both sides of Eq. (101) yields the altcrnative form 11v -

+ x(') + 'f<') + iW)

2