Scribd
Upload
91 views6 pages

AE 321 - Solution of Homework #11

This document contains solutions to homework problems related to stresses and displacements in cylindrical structures. Solution 1 considers stresses and displacements for a cylinder under internal and external pressures. Equations are derived relating radial displacement to applied pressures. Solution 2 considers shrink fitting of an inner and outer cylinder. Radial displacement equations are applied to determine the geometry after shrink fitting. Solution 3 considers stresses in a hollow cylinder under combined internal pressure and torque loading. Equations for maximum stresses are presented. Failure criteria are applied for different loading conditions.

Uploaded by

Arthur Ding
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
91 views6 pages

AE 321 - Solution of Homework #11

This document contains solutions to homework problems related to stresses and displacements in cylindrical structures. Solution 1 considers stresses and displacements for a cylinder under internal and external pressures. Equations are derived relating radial displacement to applied pressures. Solution 2 considers shrink fitting of an inner and outer cylinder. Radial displacement equations are applied to determine the geometry after shrink fitting. Solution 3 considers stresses in a hollow cylinder under combined internal pressure and torque loading. Equations for maximum stresses are presented. Failure criteria are applied for different loading conditions.

Uploaded by

Arthur Ding
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 6

AE 321 Solution of Homework #11

Solution #1

(a)

It does satisfy but it gives a trivial solution for stresses. Using Cauchys equation
we find that .



(b)

It satisfies . This means that satisfies the requirement for an Airy Stress function.





The tractions are





Same with (-) sign on the negative sides x=-l/2 & y=-h/2.



(c)

It satisfies .





Use Cauchys formula:




x
y
2b
2a
-2b
-2a
x
y
a
a
a
a
1+1+1=3 POINTS
1+1+1=3 POINTS
1 POINT
2 POINTS
2 POINTS
2 POINTS
1+1+1=3 POINTS
1 POINT
2 POINTS
2 POINTS
2 POINTS
(d)

It satisfies .






Use Cauchys formula:







(e)





(f)

It satisfies .








Use Cauchys formula:









x
y
x
y
12ay
2

-12ay
2

-12ax
2

12ax
2

1+1+1=3 POINTS
1 POINT
2 POINTS
2 POINTS
2 POINTS
1+1+1=3 POINTS
1 POINT
2 POINTS
2 POINTS
2 POINTS
Solution #2

For a cylinder with internal and external pressure we have:



This is a plane-stress solution which can be converted into plane-strain by substituting:
















For the inner cylinder:


For the outer cylinder:


Hence, we can compute the radial displacements for the two cylinders by substituting the values
of r
1
, r
2
, P
i
and P
o
from (2) in equation (1). The radial displacement for the inner cylinder is u
r
I
and for the outer cylinder is u
r
II
.

The schematic below shows the inner and the outer cylinders after they are shrunk-fit. The inner
cylinder is marked in blue and the boundaries are denoted by dashed blue lines. The outer
cylinder is marked in grey and boundaries are denoted by bold black lines.

u
r
I
is the radial displacement for the inner cylinder. It is negative because positive radial direction
is always pointing outwards from the origin O. u
r
II
is the radial displacement for the outer
cylinder.

The two cylinders get shrink-fitted along the red dash-dot line.

P
P
Inner cylinder u
r
I
Outer cylinder u
r
II

(1)
(2)
1 POINT
1+1 = 2 POINTS
1+1 = 2 POINTS
2 POINTS
2 POINTS

Hence, from the geometry of the problem we can say that:



Substituting r=b in the expressions for u
r
I
and u
r
II
we would have the following after
doing some algebra.





Where .

















O
-u
r
I

u
r
II

e
2 POINTS
2 POINTS
2 POINTS
Solution #3

From the solution for a hollow cylinder under pressure:









Where

For the torque problem of a hollow cylinder we have:



The maximum stress for the combination can be found by solving the eigenvalue problem. In the
simpler cases we have:

(a)
For the material to survive,


comes from Mohrs circle.














90
o

45
o

crack
3 POINTS
1+1 = 2 POINTS
4 POINTS
2 POINTS
2 POINTS
(b)



Again, for the material to survive,



crack


4 POINTS
2 POINTS

You might also like