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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.

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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

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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

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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

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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

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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