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Mechanics of MaterialsVon Mises
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1
ME111Instructor: Peter Pinsky
Class #21November 13, 2000
Today’s Topics
• The von Mises yield criterion.
Reading Assignment
Juvinall, 6.5 – 6.8
• Failure of ductile materials under static loading.
• The maximum shear stress criterion.
1. The stress (in kpsi) at a point is given byCalculate the factor of safety against failure if the material is:(a) Brittle with Sut = 50, Suc = 90 and using modified-Mohr
theory.(b) Ductile with Sy = 40 using (i) the max. shear stress criterion,
and (ii) von Mises criterion.
20,0,10 321 −=== σσσ
Problem Set #7. Due Wednesday, November 22.
2. Consider two designs of a lug wrench for an automobile: (a) single
ended, (b) double ended. The distance between points A and B is
12 in. and the handle diameter is 0.625 in. What is the maximumforce possible before yielding the handle if Sy = 45 kpsi?
3. A storage rack is to be designed to hold a roll of industrial paper. Theweight of the roll is 53.9 kN, and the length of the mandrel is 1.615 m.Determine suitable dimensions for a and b to provide a factor ofsafety of 1.5 if:(a) The beam is a ductile material with Sy = 300 Mpa(b) The beam is a brittle material with Sut = 150 Mpa, Suc = 570 Mpa.
5. Repeat Example 21.2 (a) and (b) using the maximum shear stresscriterion.
4. For the problem in Example 19.1 (torsion-bar spring), what diameterd will provide a factor of safety of N = 3 against yielding based onvon Mises with Sy = 150 MPa.
6. For the beam shown, determine the factor of safety for:
(a) Ductile material with Sy = 300 Mpa(b) Brittle material with Sut = 150 Mpa, Suc = 570 Mpa.
2
Problem 6
Problem 3
Problem 2
Recall Uniaxial Tension Test:
1σ
32,σσ
τ
σ
1dF
2dF
bF
• Ductile materials fail by yielding and are typically limited by theirshear strengths
• Brittle materials fail by cracking or crushing and are typically limited by their tensile strengths
Static Loads:
1dF2dFbF
21.1 Failure of Ductile Materials Under Static Loading
3
Recall Torsion Test:
1σ2σ
τ
σ
dF
bF
3σ
dF
bF
21.2 Three-Dimensional Stress States
• It’s useful to think about 3-d stress states using principal stresses:
xzττ
zyττ
yzττ
x
y
z
xσ
zσ
yσ
xyττ yxττ
zxττ
1
2
3
1σ
3σ
2σ
0322
13 =−+− III σσσσσσ
2223
2222
1
2 xyzzxyyzxzxyzxyzyx
zxyzxyxzzyyx
zyx
I
I
I
τστστστττσσσ
τττσσσσσσ
σσσ
−−−+=
−−−++=
++=
2
2
2
2332
1221
3113
σστ
σστ
σστ
−=
−=
−=
),,max( 322113max ττττ =
Principal stresses
Maximum shear stresses
4
1
2
3
1σ
3σ
2σ
1
2
3
mσ
mσ
mσ
1
2
3
mσσ −1
mσσ −3
mσσ −2
Mean stress producesvolumetric strain
Produces distortionalstrain responsible forplastic yielding
)(31
321 σσσσ ++=m
21.3 Split of Stress into Mean and Distortional Components
Given principal stressstate
21.4 Yielding of Ductile Materials
• A ductile material yields when a yield criterion is exceeded.
• Two yield criteria are important:
The von Mises Yield Theory
• The strain energy per unit volume of an elastic body has the form:
[ ]
[ ]13322123
22
21
13322123
22
21
31
)(26
21
σσσσσσσσσν
σσσσσσσσσν
−−−+++
=
+++++−
=
+=
EU
EU
UUU
d
m
dm
where is the strain energy per unit volume associated with purevolume change (dilation) due to the mean (hydrostatic) stress:
hU
)(31
321 σσσσ ++=m
and where is the strain energy per unit volume associated withpure distortion
dU
• Note that in a tensile test at yield:
0, 321 === σσσ yS
and, in this special case,
2
31
yd SE
Uν+
=
5
• The von Mises yield criterion predicts failure in a general 3-d stressstate when the distortion energy per unit volume is equal to thedistortion energy per unit volume in the tensile test specimen atfailure, i.e.
dU
2
31
yd SE
Uνν+
=
or,
13322123
22
21 σσσσσσσσσ −−−++=yS
[ ] 2133221
23
22
21 3
13
1yS
EEνν
σσσσσσσσσσσσσσσσσσνν +
=−−−+++
For plane stress,
2331
21 σσσσσσσσ +−=yS
02 =σσ Note, this is rather arbitrary but we’ll workwith this as the plane stress state.
2)(6)()()( 222222
zxyzxyxzzyyxyS
ττττττσσσσσσσσσσσσ +++−+−+−=
In terms of zyx and, stresses
In terms of zyx and, stresses
222 3 xyyxyxyS ττσσσσσσσσ +−+=
1σσ
3σσ
Special cases:
Pure tension
032 == σσσσ
1σσ=yS
Pure shear (torsion)
0, 2max31 ==−= σσττσσσσ
max1 33 ττσσ ==ySσ
ττ
ττ
σ
Pure tension
Pure shear (torsion)
Von Mises ellipse (plane stress)
No yielding
Yield surface
6
Von Mises Effective Stress
13322123
22
21 σσσσσσσσσσσσσσσσσσσσ −−−++=′
(The von Mises effective stress is defined as the uniaxial tensilestress that would create the same distortion energy as is createdby the the actual combination of applied stress)
σσ ′
dU
Convenient to introduce the von mises effective stress
von Mises (distortion energy) Yield Criterion
The von Mises yield criterion predicts failure when:
σ ′=yS
Factor of Safety Against Yielding
σ ′=N
S y
σσ ′
Example 21.1
For the bracket shown, determine the von Mises stress and factorof safety against yielding at points (a) and (b) if Sy=10,000.
40
2030
5
4
3
A
B
C
x
yy MR ,zz MR ,y
z
xx MR ,
5
From previous analysis we found:
At (a):
At (b):
7590,271 31 −== σσσσ
343,5096 31 −== σσσσ
729,72331
21 =+−=′ σσσσσσσσσσ
3.1729,7000,10
==′
=σσ
ySN
274,52331
21 =+−=′ σσσσσσσσσσ
9.1274,5000,10
==′
=σσ
ySN
(b)
(a)
A
7
Example 21.2
A thin-walled cylindrical pressure vessel is subject to an axial forceand torque loads as shown:
t
r
p P T
(a) Given:
Determine the range of values of the axial load P which will provide afactor of safety against yielding of at least 1.4 based on the von Misescriterion.
MPaSmmkNT
mmtmmrMPap
y 290,450,3,35,15
=⋅=
===
(b) Given:
Determine the range of values of the radius r which will provide afactor of safety against yielding of at least 1.4 based on the von Misescriterion.
MPaSkNPmmkNT
mmtMPap
y 290,100,800,5,20
==⋅=
==
21.5 Maximum Shear Stress Criterion (Tresca Condition)
• Another important criteria is based on the theory that shear stresscontrols yielding (in contrast to von Mises theory based on distortionenergy).
• This theory was developed before the von Mises criterion and inpractice is slightly more conservative.
• The theory is easy to use in an “analytical” setting but is not wellsuited to use in finite element codes because of the corners inthe yield surface (see below).
• The maximum shear stress theory states that yielding will occurwhen the maximum shear stress reaches the shear stress in auniaxial test specimen at yield, i.e.
2)
2,
2,
2max( 133221 yS
=−−− σσσσσσ
oryS=−−− ),,max( 133221 σσσσσσ
For plane stress with
yS=− ),,max( 1331 σσσσ
02 =σ
Factor of Safety Against Yielding
),,max( 1331 σσσσ −=N
S y
Remark: von Mises is the preferred theory
8
1σσ
3σσ
Pure tension
Pure shear (torsion)
Von Mises and Tresca for Plane Stress
Von Mises yield surface
Tresca yield surface
31 σσ −=
von Mises: yS3
11 =σ
Tresca: yS21
1 =σ
03 =σ
von Mises: yS=1σ
Tresca: yS=1σ
Stress states inside the yieldsurface have not yielded
Example 21.3
For the bracket shown, determine the factor of safety against yieldingusing the maximum shear stress theory at points (a) and (b) if Sy=10,000.
40
2030
5
4
3
A
B
C
x
yy MR ,zz MR ,y
z
xx MR ,
5
From previous analysis we found:
At (a):
At (b):
7590,271 31 −== σσσσ
343,5096 31 −== σσσσ
(b)
(a)
A
27.1861,7000,10
),,max( 1331
==−
=σσσσ
ySN
84.1439,5000,10
),,max( 1331
==−
=σσσσ
ySN
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