AE 321 – Solution of Homework #10 Solution #1
Start by making the following assumptions:
Plane cross-sections sections remain plane
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The rotation angle θ is proportional to the distance from the fixed base: z The projection of each section on xy rotates rigidly (no θ dependence of the solution) y, v
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P’ r r θ
P
β
x, u
From the schematic calculate the displacements ux and uy using the definition X-x = u: u x r cos r cos u y r sin r sin
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Using the trigonometric identities for cos(a+b) and sin(a+b) and the transformations from cylindrical to Cartesian coordinates x=rcosβ x=rcosβ and y=rsinβ y=rsinβ: u x x cos 1 y sin u y x sin y cos 1 For θ<<1 the last equations become: u x y u y x Using the first assumption made above:
1
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u x yz u y xz
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u z 0
Once we have the expressions for displacements, the strains can be derived: xz
1
yz
x
u 2
x. z
1
y
2
2
u z, x y 0
2
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xx yy zz xy 0
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: xz
E
1 2 xz xz kk y 1 1 2
xz x
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xx yy zz xy 0 where
E
1
2
Check equilibrium (assuming no body forces): xx. x xy. y xz. z 0 xy. x yy. y yz. z 0 xz. x yz. y zz. z 0
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Finally, define the boundary conditions and prove that are all satisfied:
x y At the outer lateral face n cos , sin , 0 , , 0 and T 0 : r r
2
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x y T n n n 0 0 0 y 0 x xx x xy y xz z r r x y Ti ij n j Ty xy nx yy n y yz nz 0 0 0 x 0 r r x y T z xz nx yz n y zz nz 0 y r x r 0
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Thus, the boundary conditions on this face are trivially satisfied.
x y At the inner lateral face n cos , sin , 0 , , 0 and T 0 : r r
x y T n n n 0 0 xx x xy y xz z r 0 r y 0 x x y Ti ij n j Ty xy nx yy n y yz nz 0 0 0 x 0 r r x y T z xz nx yz n y zz nz 0 y x 0 r r
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Thus, the boundary conditions on this face are also satisfied.
At side face we have n 0,0,1 and dA dxdy .
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dF TdA
dM r dF ijk rj dFk ei dM ijk rj Tk dA
However, r has components only on the xy plane: dM zxk xTk dxdy zyk yTk dxdy
This part is FYI, it is not graded
The permutation symbol takes a non-zero value for y and z values respectively: dM zxy xTy dxdy zyx yTx dxdy xTy yTx dxdy
1
1
Apply Cauchy’s formula to determine the tractions:
T x xx nx xy ny xz nz Tx xz Ti ij n j T y xy nx yy n y yz nz Ty yz T n n n T xz x yz y zz z z zz z Thus the expression for the moment becomes:
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dM z x yz y xz dxdy dM z x2 y2 dxdy
M z x y dxdy 2
2
2 R2
r rdrd 2
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0 R1
M z
2
R
4 2
R14
In this case the torsional rigidity is given by:
J
R 2
4 2
R14
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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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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.
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(ii) St Venant’s 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.
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Figure 1
(b)
Figure 2
Referring to Figure 1 above, it can be seen, on the end faces nˆ 0
0 1
T
33 P
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Net Torque T
x 2 32 x1 dA
31
A
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Referring to Figure 2, on the outer surface nˆ cos sin 0
T
11 cos 12 sin 0 12 cos 22 sin 0
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Similarly, on the inner surface: nˆ cos
T sin 0
11 cos 12 sin 0 12 cos 22 sin 0 (c)
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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.
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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 c 0 0 0 0 0 P
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Therefore:
0 0 0 0 c t 0 0 0 0 0 0 P M z y J
0 Mz J
0 J M z x 0 J M 0 z y J
0
x
M z
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M z y J M z x J
0 0 M z J
x
P
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