1
AE 321 Solution of Homework #10
Solution #1
Start by making the following assumptions:
- Plane cross-sections sections remain plane
- The rotation angle is proportional to the distance from the fixed base: z u o =
- The projection of each section on xy rotates rigidly (no dependence of the solution)
From the schematic calculate the displacements u
x
and u
y
using the definition X-x = u:
( )
( )
cos cos
sin sin
x
y
u r r
u r r
| u |
| u |
= +
= +
Using the trigonometric identities for cos(a+b) and sin(a+b) and the transformations from
cylindrical to Cartesian coordinates x=rcos and y=rsin:
( )
( )
cos 1 sin
sin cos 1
x
y
u x y
u x y
u u
u u
=
= +
For <<1 the last equations become:
x
y
u y
u x
u
u
=
=
Using the first assumption made above:
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y, v
x, u
P
P
r
r
2
0
x
y
z
u yz
u xz
u
o
o
=
=
=
Once we have the expressions for displacements, the strains can be derived:
( ) ( )
. ,
1 1
0
2 2 2
2
0
xz x z z x
yz
xx yy zz xy
y
u u y
x
o
c o
o
c
c c c c
= + = + =
=
= = = =
The compatibility equations are satisfied automatically as none of the components of the strain
tensor is a function higher than first order.
Then use constitutive equations to find the stresses:
( )( )
( ) 1 2
1 1 2
0
xz xz xz kk
xz
xx yy zz xy
E
y
x
t v c vo c o
v v
t o
t t t t
= + = (
+
=
= = = =
where
( )
2
1
E
v
=
+
Check equilibrium (assuming no body forces):
. . .
. . .
. . .
0
0
0
xx x xy y xz z
xy x yy y yz z
xz x yz y zz z
t t t
t t t
t t t
+ + =
+ + =
+ + =
Finally, define the boundary conditions and prove that are all satisfied:
At the outer lateral face { } cos , sin , 0 , , 0
x y
n
r r
u u
= =
`
)
and 0 T =
:
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( )
( )
( ) ( )
0 0 0 0
0 0 0 0
0 0
x xx x xy y xz z
i ij j y xy x yy y yz z
z xz x yz y zz z
x y
T n n n y
r r
x y
T n T n n n x
r r
x y
T n n n y x
r r
t t t o
o t t t o
t t t o o
= + + = + +
= = + + = + +
= + + = + +
Thus, the boundary conditions on this face are trivially satisfied.
At the inner lateral face { } cos , sin , 0 , , 0
x y
n
r r
u u
= =
`
)
and 0 T =
:
( )
( )
( ) ( )
0 0 0 0
0 0 0 0
0 0
x xx x xy y xz z
i ij j y xy x yy y yz z
z xz x yz y zz z
x y
T n n n y
r r
x y
T n T n n n x
r r
x y
T n n n y x
r r
t t t o
o t t t o
t t t o o
| | | |
= + + = + +
| |
\ . \ .
| | | |
= = + + = + +
| |
\ . \ .
| | | |
= + + = + +
| |
\ . \ .
Thus, the boundary conditions on this face are also satisfied.
At side face we have { } 0, 0,1 n = and dA dxdy = .
dF TdA
ijk j k i ijk j k
dM r dF r dF e dM r T dA c c
=
= = =
However, r has components only on the xy plane:
zxk k zyk k
dM xT dxdy yT dxdy c c = +
The permutation symbol takes a non-zero value for y and z values respectively:
( )
1 1
zxy y zyx x y x
dM xT dxdy yT dxdy xT yT dxdy c c
= =
= + =
Apply Cauchys formula to determine the tractions:
x xx x xy y xz z x xz
i ij j y xy x yy y yz z y yz
z xz x yz y zz z z zz
T n n n T
T n T n n n T
T n n n T
t t t t
o t t t t
t t t t
= + + =
= = + + =
= + + =
Thus the expression for the moment becomes:
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This part is FYI, it is
not graded
4
( ) ( ) ( )
( ) ( )
( )
2
1
2 2
2
2 2 2
0
4 4
2 1
2
z yz xz z
R
z
R
z
dM x y dxdy dM x y dxdy
M x y dxdy r rdrd
M R R
t
t t o o
o o u
ot
= =
= + =
=
} } } }
In this case the torsional rigidity is given by:
( )
4 4
2 1
2
J R R
t
=
The solution for the hollow cylinder that was just derived differs from the solution of the solid
cylinder (derived in problem 1) only in terms of the torsional rigidity.
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5
Solution #2
(a) (i) Superposition: Since all the equations for small strain elasticity are linear, we can add
two or more solutions derived for each of the boundary conditions of the given problem
and we can get the solution to our more complicated problem.
(ii) St Venants Principle: If the distribution of surface forces on a portion of the surface of
the body is replaced by a different distribution, having the same resultant force and
moment, then the effect of both surface forces at the points of the body sufficiently
removed from the region of applications of force is the same. This means that, if we are
far enough from the point of application of load, i.e. 4-5 times the dimension where the
boundary condition is applied, only the net resultant of the loading matters and not its
exact distribution.
Figure 1 Figure 2
(b) Referring to Figure 1 above, it can be seen, on the end faces | |
T
n 1 0 0 =
33
P o =
( )dA x x T
A
}
o o = =
1 32 2 31
Torque Net
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Referring to Figure 2, on the outer surface | |
T
n 0 sin cos u u =
0 sin cos
0 sin cos
22 12
12 11
= u o + u o
= u o + u o
Similarly, on the inner surface:
| |
T
n 0 sin cos u u =
0 sin cos
0 sin cos
22 12
12 11
= u o + u o
= u o + u o
(c) We solve this problem by using the principle of superposition. We split the problem into
two: One of hollow cylinder at the ends of which we apply uniform compressive stress,
and a second problem which is problem 2 of this homework. The sum of the two stress,
strain and displacement solutions are the complete solution to the given problem. Using
superposition we add the stress tensors for the two different types of loading.
The solution for the torsional case is given in problem (2), while for the compression problem
can be easily shown to have the following solution:
0 0 0
0 0 0
0 0
c
P
o
(
(
=
(
(
Therefore:
0 0 0 0
0 0 0
0 0 0 0 0 0 0
0 0
0
z z
z z
c t
z z z z
M M
y y
J J
M M
x x
J J
P
M M M M
y x y x P
J J J J
o o o
( (
( (
(
( (
(
( (
= + = + =
(
( (
( ( (
( (
( (
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