Lecture 12
Fundamental Concepts in Structural Plasticity
Problem 12-1:
Stress yield condition
Consider the plane stress yield condition in the principal coordinate system
a)
1
,
2
between the Von-Mises and Tresca yield
Calculate the maximum difference
condition
b)
Show the locations on the plane stress graph where the maximum difference occurs
Problem 12-1 Solution:
Through observation, the maximum difference
occurs at either A or B, as indicated in
the below figure
Mises yield condition can be expressed as
2
1
2
1
2
2
2
y
1
The difference
at A
At Tresca yield surface
Tresca
2A
y
At Mises Yield surface
1
2
1
Substitute
1
1
2
y
into Mises yield condition, we have
1
2
2
1
2
y
2
y
Mises
2A
The difference
2
2
2
y
1.15
y
1.15
y
at A is
Mises
A
Tresca
2A
2A
0.15
A
The difference
y
y
0.15
y
y
at B
Tresca yield surface in the second quadrant is
2
2
1
2
1
2
2
y
The distance from Tresca yield surface B to the origin is
d BTresca
2
1
2
2
2
2
y
Mises yield surface in the second quadrant is
2
1
2
1
2
2
2
y
2
Also, at Mises yield surface
1
0
2
Combining the above two equations, we have the coordinate of Mises yield surface B
1
3
1
y
,
1
3
2
y
The distance from Tresca yield surface B to the origin is
d B Mises
The difference
2
1
d B Mises
d BTresca
B
A
y
at B is
B
Compare
2
3
2
2
and
B
2
2
0.11
2
3
y
0.15
0.11
y
y
, the maximum difference
max
y
occurs at A, where
y
3
Problem 12-2:
In the early twenties, passenger cars did not have electric starters.
crank to start the engine.
is shown below.
The crank is a solid rod of radius r and the geometry of the crank
3
Define the equivalent stress by
stress produced by bending and
a)
The driver had to use a
3
2
where
sigma is the
is the shear stress due to torsion.
Find the relationship between the maximum equivalent stress in the crank and the
magnitude of the crank load P. (Use the principle of superposition)
b)
Derive a formula for the elastic deflection under the load P in the direction of the
load P.
Problem 12-2 Solution:
a) Determine the magnitude of the load P causing first yield in the mostly stressed point on
the crank
We will use the Mises yield condition which simplified to
2
o
from the moment, and
We calculate
2
3
2
from the torsion
Mr
where I
I
Tr
where J
J
r4
4
r4
2
Because we are asked to find P causing first yield at the maximum stressed point, we
need to find
max
and
max
From physics we can see that the M max and Tmax will occur at the support
4
Pl
M max
Tmax
M max r
Plr
I
r4 4
Tmax r
PRr
J
r4 2
max
PR
4 Pl
r3
2 PR
r3
max
max
2
2
max
o
4 Pl
r3
2
max
2
2 PR
3
r3
2
4
P 2 4l 2 3R 2
r
2
o
P
3
2 6
o
2
r2
1
l
4
r
2
R
3
r
2
b) Calculate the elastic deflection in the direction of the load P
Beam AB
Beam CD
MA
M
T
P L R
PR
0
PL
0
Mo
M
M
M
Px1 at 0
Px1
x1
0
R
Mx
Mx
PL Px3
0
PL Px3 at 0
x3
L R
5
Beam CB
M
T
P R x2 at 0
PR at 0
x2
x2
R
R
In summary, here are the moment distributions
Torsion is constant along
Beams AB and BC: T PR
Beam CD: T 0
6
M2
dx
2EI
U bending
l
0
2
Px1
dx1
2 EI
P 2 x13
2 EI 3
2
P R x2
R
dx2
2 EI
0
R
2 Rx2 2
2
2
R x2
0
P L x3
L R
2EI
0
R
x23
3
2
L x3
0
2Lx32
2
2
dx3
x33
3
L R
0
after lengthly algebra
P2
R3
6EI
L3
P2
R3
6EI
U bending
L3
T2
dx
2GJ
U torsion
R
0
2
PR
dx2
2GJ
PR
2GJ
L R
0
2
PR
dx2
2GJ
2
R l R
2
U torsion
PR L
2GJ
Use Castigliano’s Theorem to calculate the deflection where the load is applied, in the
direction of the load
U
P
U torsion U bending
P
2
P
R3
P 6 EI
P
R3
L3
3EI
2
3
L
PR L
2GJ
R2 L
GJ
7
r4
and J
4
Recall that for solid circular cross-section I
Also, G
E
1
2
P
R3
L3
R2 L
GJ
3EI
R2 L 1
R 3 L3
P
3E r 4 4
Note, if R
r4
2
E r4 4
4 P R 3 L3
Er 2
3r 2
R2 L 1
r2
4 P R3 L3
Er 2
3r 2
R2 L 1
r2
0 , we have a cantilever beam whose deflection is
PL3
3EI
8
Problem 12-3:
Consider a thin-walled tube of radius r, thickness t and length L. The tube is fully clamped on
one end and free on the other. It is twisted at the free end by an axial torque T.
(a) Derive an expression between torsional moment and the relative end rotation.
(b) Assuming L/R=10 and R/t=10, give the expression for the critical torque that will
cause the tube stress to reach yield in shear.
Problem 12-3 Solution:
(a)
Express T as a function of
T
x
rdA
d
dA
dx
Gr 2
d
r 2 dA
dx
d
GJ
dx
G
x 0
0
max
x L
T x
GJ L
(b) The distribution of torsional shear stress can be expressed as
x
Tr
J
Where
9
J
ro 4
2
ri 4
ro 4
2
ro t
4
Given
r
t
J
10
ro
2
4
t
r
10
ro
ro
10
0.071 ro 4
J
Tr
0.071 ro 4
x
4
5.8T
ro 3
Plane stress yield condition states:
2
xx
2
xx
yy
yy
3
xx
yy
0
2
xy
y
In pure shear
y
xy
3
5.8T
ro3
x
y
3.2
y
3
T
ro3
10
Problem 12-4: Consider the following key ring problem
a)
b)
c)
d)
Derive the out of plane displacement where the force is applied.
Determine the magnitude and distribution of the bending stress and the shear stress
along the ring.
Find the location of the maximum equivalent plastic strain.
Determine the critical opening force for which first yield would occur. Consider the
plane stress
Problem 12-4 Solution:
a) Derive the out of plane displacement where the force is applied.
We can use Castigliano’s Theorem to calculate the displacement where the force is applied.
In addition, to calculating the strain energy contribution from the moment, we must also
account for the contribution from torsion
U
1.
1 M2
dx
2 EI
1 T2
dx
2 GJ
Calculate M, T
From the geometry above, we can see that
M
Pa
T
Pb
where
a
b
R sin
R R cos
11
So
M
PR sin
T
2.
PR 1 cos
Calculate strain energy
U
1 M2
dx
2 EI
0
1 T2
dx
2 GJ
2
1 PR sin
2
EI
Rd
2
PR R
2
EI
0
PR R
GJ
0
PR R
GJ
0
3
GJ
Apply Castigliano Theorem, the out of plane displacement where the force is applied is
U
P
PR 3
1
EI
3
GJ
Determine the magnitude + distribution of bending stress + shearing
Mz
I
Tr
J
c)
2
sin 2
4
2
P 2 R3 1
2
EI
U
b)
2sin
2
PR R
2
EI
3.
Rd
2
sin 2
4
2
2
1 PR 1 cos
2
GJ
PR sin r
r4 4
PR 1 cos
r4 2
4
PR sin
r3
r
2
PR 1 cos
r3
Find the location of maximum plastic strain
Recall that relation of plastic strain and plastic stress is
n
A
The maximum plastic strain occurs at location of the maximum Mises equivalent stress,
1. Calculate the Mises equivalent stress
The stress on the outer surface is planar, so we will use the plane stress condition
2
xx
Where
o
2
xx
yy
yy
3
2
xy
2
o
is Mises equivalent stress
12
In our case,
0
yy
So
2
2
3
o
2
2
4
PR sin
r3
2.
3
Calculate the Maximum Mises equivalent stress
omax
will occur when we have
setting it equal to 0, find
d
d
2
8
d
Set
d
PR
r3
omax
. We can find the maximum by taking the derivative,
2
where
4
PR
r3
o
2
omax
2
2
PR 1 cos
r3
occurs and then go back to get
omax
2
2sin cos
3
2
PR
r3
omax
2
2 1 cos
sin
2
sin
cos
3
sin
cos
2
o
0 , we have
3
0
0,
0 , where
Minimum Mises equivalent stress occurs at
Maximum Mises equivalent stress occurs at
d)
, where
0
o
o
The critical opening force for which first yield would occur when
y
o
4 3
PR
r3
o
y
, that is
4 3
Pcr R
r3
So we have
Pcr
r3
4 3R
y
13
Problem 12-5: Plasticity
Consider the four –point bending of a beam of length L. The beam is loaded by two rollers
parted by a distance of L/3. The material of the beam is rigid, perfectly plastic. Determine the
load capacity of the beam under two different end conditions.
a) Write an expression for fully plastic bending moment of a beam of rectangular
cross-section b h .
b) Ends of the beam are simply supported
c) Ends of the beam are clamped
Problem 12-5 Solution:
y
h 2b
a)
Mo
b)
Moment distribution
M
M
M
4
Px
2
Pl
6
P
l x
2
0
x
l
x
3
2l
x l
3
We can assume that within
M
Mo
l
3
2l
3
y
l
3
x
2l
3
h 2b
4
14
The rate of change of internal energy is
2l 3
U
l 3
2l 3
MKdx
Where geometrically
2
o
l 3
wo
l 3
Mo
6
d
dx
dx
Mo
x 2l 3
x l 3
Mo
o
wo
l
Rate of work balance
Mo
o
Mo 6
wo
l
Pc
c)
2
Pc
wo
2
Pc wo
6M o
l
If the ends are clamped, we have two plastic hinges at the supports
1
wo
l 3
2
3
wo
l
Rate of work balance
i
M oi
Mo
o
Mo
M o 12
Pc
o
2
1
Mo
wo
l
Pc
wo
2
2
Pc wo
Pc wo
12M o
l
15
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2.080J / 1.573J Structural Mechanics
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