AE 321 – Solution Solution of Homework #11 Solution #1
(a)
∇4 0 0 It does satisfy equation we find that
(b)
but it gives a trivial solution for stresses. stresses. Using Cauchy’s .
∇4 0 2, 2, 0
It satisfies function.
. This means that ϕ satisfies the requirement for an Airy Stress y 2a
The tractions are
2 : ℎ2 :
x
-2b
2, 0 0, 2
2b -2a
Same with (-) sign on the negative sides x=-l/2 & y=-h/2.
(c)
> 0 ∇4 0 0, 0, <0
It satisfies
a
.
Use Cauchy’s formula:
∓ 2 : ∓ ℎ2 :
y
0, ± ±, 0
a
a
a
x
(d)
3
y
∇4 0 6, 0, 0 It satisfies
.
x
Use Cauchy’s formula:
± 2 : ∓ ℎ2 : (e)
(f)
±6, 0 0, 0
3 ∇4 0 ⟹ 0 ⟹ 4 4 >0 ∇4 0 12, 12, 0 It satisfies
.
2
12ax 12ay2
x -12ay2
Use Cauchy’s formula:
∓ 2 : ± ℎ2 :
±12, 0 0, ±12
y
-12ax2
Solution #2
For a cylinder with internal and external pressure, on e has:
− −+ ′ { − 1 − } ′ 1 ; ′ 1
(1)
This is a plane-stress solution which can be conve rted into plane-strain using
P
P
Inner cylinder ur I
Outer cylinder ur II
, , , ≈, 0, , 0
For the inner cylinder: For the outer cylinder:
(2)
Hence, we can compute the radial displacements for the two cylinders by putting the values of r 1, r 2, Pi and Po from (2) in equation (1). We would get displacements for the inner cylinder as ur I and for the outer cylinder as ur II as a function of radial distance r. The schematic below shows the inner and the outer cylinders when they are shrink- fitted. 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. ur I is the radial displacement for the inner c ylinder. It is negative because positive radial direction is always pointing outwards from the origin O. ur II is the radial displacement for the outer cylinder. The two cylinders get shrink-fitted along the red dash-dot line.
-urI urII O
e
Hence, from the geometry of the problem we can say that:
Putting the value of r=b in the expressions for ur I and ur II we would have the following after doing some algebra.
′ 1 1 2 1
Where
≈
.
Solution #3
From the solution for a hollow cylinder under pressure:
− Where 1
( 1) ( 1) 1 ( 1) ( 1) 1 0
For the torque problem of a hollow cylinder we have:
ℎ 2 4 4 The maximum stress for the combination can be found by solving the eigenvalue problem. In the simpler cases we have: (a)
≠ 0, 0 ⟹ 0 ⟹ |= ≤ ⟹ ≤ ⟹ ≤ |= For the material to survive,
comes from Mohr’s circle.
τ
45o
90o σ
crack
(b)
0, ≠ 0 ⟹ 0 |= 1 Again, for the material to survive,
≤ ⟹ 1≤ ⟹ ≤ (1)
crack σθ
