AE321 - Solutions for Homework 9 Solution #1 a) Since we have only body forces, this is a traction prescribed BVP.
f x f y 0, f z g
1 POINT
We start with equilibrium equations. Of the three equations, only one remains:
z , z f z 0 z ,z g
2 PO POIN INTS TS
Since the effect of gravity is in z-direction, σij=0 for i ≠3 and j≠3. Integrate equilibrium equation:
z , z g z gdz gdz z gz
1 POINT
b) Use the constitutive equations to find strains:
1
gz
zz
zz ( xx yy ) zz E E E
xz
xz
yz
E yz E 1
0 3 POINTS
0
gz
z
xx
xx ( yy zz ) zz E E E
yy
1
xy
yy ( xx zz ) zz E E E xy E
0
All the strains are 1 st order functions, thus compatibility is satisfied.
1 POINT
Then solve for the displacements using the strain-displacement relations. Note: the easiest way to solve such problems is to find the displacements by solving the expressions for normal strains, and then substitute the displacements derived from normal strains into the equations that relate shear strains and displacements.
1
zz
gz E
xz 0
uz, z 1 2
u
x, z
gz E
uz
gz E
dz u z
u z , x 0 u x , z u z , x u x , z
gz 2 2E
f ( x, y )
gz 2 f (x, y ) x 2 E
f ( x, y ) f ( x , y ) ux z h( x, y ) x x f ( x, y ) g ( x, y ) yz 0 u y z y u x , z
xx
gz E
u x, x
gz E
2 POINTS
2 POINTS 2 POINTS
gz f ( x, y ) z h ( x , y ) x x E
2 f ( x, y ) g x 2 gz 2 f ( x, y ) h( x, y ) E z 2 x x E (all h( x, y ) x , y, z ) 0 x gx 2 C1 x C 2 f ( x, y ) 2 E 2 POINTS h( x, y ) C y C 3 4 must satisfy for
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Similarly
gy 2 C5 y C 6 gz f ( x, y ) y y 2 E E g ( x, y ) C x C 7 8
2 POINTS
Then, the unknown functions have the form:
f ( x, y)
g
x 2 E
2
y2 C1x C5 y C2 C 6 3 POINTS
g ( x, y) C7 x C 8 h( x, y) C3 y C4 After substituting the functions above into the expressions for the displacements:
gx C1 C3 y C4 E gy u y z C5 C7 x C8 E u x z
2
2 POINTS
Use the relationship for shear strain to determine a relationship between the constants:
xy 0
1 2
u
x, y
uy ,x 0 C3 C7
2 POINTS
Finally, the displacements are:
u x
g
u y
E g
xz yz
E g 2 u z z x2 y 2 2 E
where C9=C2+C6 and
C3 y C1z
C4
C5 z
C8
C3 x
C1x C5 y
1 POINT
C9
1 POINT
0 C3 C1 RBR w C3 0 C5 , C1 C5 0
c)
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C 4 RBT C 8 C 9
If we assume that point A(0,0, ℓ) is fixed (no rotations and not displacements):
xy (0,0, ) 0
1
u 2
x, y
uy ,x 0 C 3 0
xz (0,0, ) 0
1
x, z
uz , x 0 C 1 0
yz (0,0, ) 0
1
u 2
u 2
y,z
3 POINTS
uz , y 0 C 5 0
u x (0,0, ) 0 C 4 0 u x (0,0, ) 0 C 8 0 u x (0,0, ) 0 C 9
2
g 2
Substituting the above into the expressions for displacements:
3
3 POINTS
u x
g
u y
E g
xz 3 POINTS
yz
E g 2 u z z x 2 y2 2 2 E
(d-e) The deformed and undeformed shapes are shown below. The deformed shape can be constructed by
placing a number of points in the expressions for displacements. For instance, the displacements will be identically zero at point A, as it was found in part (c). At point 0, the will be a displacement only in zdirection, where elongation takes place: u z=-ρgℓ2/2E. At corner B, the body will contract the most by u x= νρgℓw/2E, the width of the bar is w, as the upper face will bear all the beam weight. The angles at the
beam corners will remain 90°, as there are no shear stresses at the corners. Finally, if the bar was colored with vertical stripes, the stripes would be tilted according to the equation for u y (evaluated at constant y and at different z), but they will remain straight, as shown below:
B
z A
3 POINTS
g
l
y
0 C x
Right angles remain right
4
1 POINT
(f) A fragile plate (panel) can carry axial loads but not bending loads. Hence, we should bond this flat panel to a surface that remains flat. In this case, we can bond it to any of the vertical planes that is either y-z or x-z.
1 POINT
(g) Maximum critical length is given by l cr
f g
, where σf , ρ and g are the failure stress, the density of
the material and the acceleration due to gravity, respectively. If the bar is 1.2 lcr long, failure occurs in the region where
2 POINTS
σzz is greater than the failure stress, σf . In
this case, failure occurs in the shaded region shown in the figure below.
σzz > σf
0.2lcr
2 POINTS lcr
(h)
While designing a space cable, the stress in the cable reaches its maximum strength (for a given cross section) at l cr . As the length of cable increases beyond l cr , the cross sectional area should also increase so that the stress at any section does not exceed its failure stress σf as shown below:
A(z)
dz
1 POINTS
z
lcr
5
Force
The stress at any section that is at a distance z from origin is given by
Area
. The force applied on a
cross section with thickness dz is the weight of the cable hanging below it. Since cross section and gravity are functions of z, we calculate the resulting force by integration. The gravity of the earth g(z) varies as
9.8(1
2 z Re
) , where z is the distance from origin and Re is radius of the earth (6,400 km). Since the
problem asks us to minimize the material needed, we design the cable with minimum cross sectional area (Amin) till it reaches l cr and then we increase the cross sectional area to prevent failure. z
f g f g
F A
z
A( z) A( z) dz (1 0
A( z) Amin lcr Amin
mg
f 2 z Re
)
l cr 2 Re
A( z )
f g
A( z )dzg (1
0
2 z ) Re
2 POINTS
A( z ) l cr
A( z)
A
min
(1
0
z
A( z) dz (1 l cr
2 z ) Re
2z Re
z
) dz
2 z
A( z) dz (1 R ) e
l cr
2 POINTS
Differentiating both sides, we have:
f dA( z) g
dz
A( z )(1
A( z ) exp(
2z Re
)
(Re 2 z )2 4Re
f dA( z) g A( z)
) const where
A( z ) exp(
4 Re
4 Re
f
4 Re
)
Amin
) exp(
(Re 2 z )2
g
(Re 2l cr )2
exp(
A( z ) exp(
Re
) dz
Amin
At z=lcr , we have A(z) = Amin const
(Re 2 z )2
2z
(1
(Re 2l cr )2 4 Re
(Re 2l cr )2 4 Re
)
) Amin
1 POINT
(i) If the space cable if made of steel, σf =900 MPa, ρ=7800 Kg/m 3. We know that: lcr
f g
l cr 11.77 km. Substituting the values of l cr , Amin and Re in the expression
for A(z), we can calculate the cross sectional area at z=2,000 km to be 1.5131e+44 m 2. The radius of the cable = 6.9418e+21 meters, which of course is quite impractical.
1 POINT
6
Problem 2:
Before
After
y
Neutral axis
x Angle θ
Assuming that the curve segment with length dx is approximately linear:
y θ
θ
dv θ
dx
(a) tan
u y
, for small angles,
(b) u y y
xx
du dx
dv d v dx 2
x
(1)
1 POINT
(2)
dx 2
y
u y
2 POINTS
y R
1 POINT
(3)
1
d 2 v dv 1 Therefore, R=curvature= 2 , for dx dx 2 d 2 v dv If not, / 1 R dx 2 dx
1
Using constitutive equations, xx
NOT GRADED
3/ 2
(4)
Ev '' y
(5) 7
1 POINT
(c) Applied moments (M):
2 POINTS
h
M w y xx dy
y
w
h h
Ev '' w y 2 dy EIv ''
M
dy
M
x
h h
where, I
w y 2 dy
1 POINT
h
EIv '' (Euler – Bernoulli equation)
(6)
EIv '' M y I ( Ev '' y) xx
Then:
I
2 POINTS
y
(7)
(d) Assumptions:
Plane sections remain plane
3 POINTS
There is a neutral axis The beam maintains its thickness in the y direction
(e) The boundary conditions are: (a) Surface ±2:
n2 (0, 1, 0)
T i 0
y,v
1 POINT
n2
12 22 32 0
w M
x,u
(b) Surface ±3:
n3 (0, 0, 1)
T i 0 n3
l
1 POINT
z,w
13 23 33 0 (c) On surface ±1:
n1 ( 1, 0, 0) T2 n1 , T 3n1 0 n T 2 1 0 12 0
M
1 POINT
T 3n1 0 13 0 T 1n1 0 11 0
8
Therefore, we have traction boundary conditions. Next solve for stresses -> strains -> displacements. (f) We saw that using a pure geometrical description of beam bending: xx
I
y
Let’s use this as a starting point to try the following solution:
xx y , ij 0 (i, j 2,3)
1 POINT
(8)
Check for equilibrium conditions:
11,1 12,2 13,3 f 1 0 21,1 22,2 23,3 f 2 0 31,1 32,2 33,3 f 3 0
1 POINT
Assuming no body forces, we see that all the equilibrium equations are satisfied. Check compatibility: We can use the Beltrami – Michelle compatibility equations which are satisfied trivially as second order derivative of stress relations is zero.
1 POINT Therefore, (8) is a solution to this problem. Next, find the strains:
11
1 E
11 ( 22 33 )
22 33 ij (i j )
y E ij 2
E
(9)
3 POINTS
0
(10) (11)
Next, solve equations (9-11) for displacements (HANDOUT GIVEN IN CLASS):
11
du dx
du
12
dy
y
E
(12)
dv
0
(13)
dx
du dw 0 dz dx dv y 22 13
dy
(14) (15)
E
9
33
dw dz
dv
23
dz
y
E
(16)
dw
0
(17)
dy
Solving for u,v,w from above equations:
xy
(12) u (13)
dv
(14)
dw
(15)
dv
(16)
dx
x du dz
x
dw
(18)
df y, z
v
dy
df y, z dz
d 2 f y, z dy
2
d 2 f y , z
x
dz
E
dx dy
f y, z
E
dz 2
2E
w x
dg ( y, z)
dh ( y , z )
dy dz
df y, z
x2 x
df y, z dz
y
E
g ( y, z)
dy
h( y, z )
(19)
(20)
(21)
(22)
y E
(10) and (11) hold for all x:
d 2 f y, z dy 2
d 2 f y, z dz 2
0 f ( y, z ) a1 z b1 y c1 yz d 1
(23)
Also, (21) + (23):
dg ( y, z ) dy
y
y
E
y 2
g ( y, z )
V0 ( z )
(24)
yz W0 ( y )
(25)
2E
And (22) + (23):
dh( y, z ) dz
E
h( y , z )
E
Summary:
v ( x, y , z )
2 E
x x( b1 c1 z) 2
w( x, y, z ) x( a1 c1 y)
yz E
y 2 2 E
V0 ( z)
(26)
W0 ( y)
From (17) and (26),
dv
23
dz
dw
z
0 xc1 V0 ' ( z ) xc1 W ' 0 ( y) 0 dy E
The last is true for any x, y, z: 10
c1 0 V0' ( z )
z E
R1
W ' 0 ( y ) R1 By integrating, we obtain:
V0 ( z )
z 2
R1 z d 2 2 E W0 ( y ) R1 y d3
(27)
where R 1 is a constant. If we substitute
4 POINTS
I
, then we have:
M RBR RBT xy EI b1 y a1 z d1 u 0 M M M v (28) x2 y2 z 2 b1 x 0 R1 z d2 2 EI 2 EI 2EI w a1 x R1 y 0 d3 M yz EI
Notes:
NOT GRADED
(a) If I fix the body at (0,0,0), (d 1,d2,d3)=(0,0,0)
b1 a1 0 (b) The matrix: b1 0 R1 is skew-symmetric a1 R1 0
(c) From v in (28), we can calculate v ' and v '' 2 d 2 v dv -1 In general, curvature κ = R = 2 / 1 dx dx
d 2 v For v ' <<1 , κ = R = EI dx 2 -1
M
11
3/ 2
EIv '' (Euler Bernoulli equation)
(g) To remove the effect of RBR and RBT, consider no rotations and no rigid body translations at (x,y,z)=(0,0,0):
uvw0 dv dx
dw dx
dw dy
2 POINTS
0
By substituting the BCs, we see that constants in RBR and RBT become zero.
(h) A saddle forms at the top of the beam where we have compressive deformation and the curvature is called “anti-elastic curvature”.
y,v
3 POINTS
z,w
Right angles remain right
1 POINT
Next, consider any cross section x= constant. After deformation, the points on this cross section will be
x ' x u c u c (
M EI
cy)
2 POINTS
This means under pure bending the cross-section remains plane, i.e. the location of the deformed points is only a function of y. Next, consider the beam sides z b y
At x=C z
2h
2b
12
After bending, we have:
z ' b w b(1
M EI
y)
2 POINTS
Thus, the beam sides are a linear function of y after the beam deforms and, therefore, become inclined, as shown in the figure in the previous page. The top and bottom sides of the beam at y h are:
y ' h v h
M 2 EI
2 POINTS
[ c2 ( h2 z 2 )]
This is the equation of a parabola. If we take a cut such that y = f = constant and v = g = constant, then we find:
g
M 2 EI
[ x 2 z 2 ]
M 2 EI
f 2 x 2 z 2 = constant, which describes hyperbolas with asymptotes:
x z 0 2
2
NOT GRADED
(i) The figure below shows how vertical stripes could change upon beam bending. On the x-y plane vertical stripes would become tilted according to the equation for u x (evaluated at constant x and for different y), but they will remain straight. Similarly, on y-z plane, the stripes would become tilted according to the equation for u z (evaluated at constant z and for different y), but they will remain straight.
y
2 POINTS x
y,v
2 POINTS
z,w
Right angles remain right 13
