shigchp5

7/28/2019 ShigChp5

1/20

7/27/20

Chapter 5: Failures

Resulting from Static

Loading

BAE 417-Design of Machine Systems

Failure of truck drive-shaft dueto corrosion fatigue

7/28/2019 ShigChp5

2/20

7/27/20

Impact failure of lawnmower

blade driver hub

Tensile failure of a bolt

7/28/2019 ShigChp5

3/20

7/27/20

Brittle failure due to stress

concentration

Valve spring failure caused by springsurge in an oversped engine

(Fracture exhibits classic 450 shear failure)

7/28/2019 ShigChp5

4/20

7/27/20

Static Strength

Ideally, the material being used in a design should be tested

for strength exactly as it will be used, i.e. alloy designation,

heat treatment, types of loading etc.

Sometimes such determinations are only done when large

numbers of machines will be manufactured, such as appliances

and automobiles where the cost of material testing can be

spread over many units.

Design Categories

1. Failure of a part would endanger human life, or

part is made in large quantities.

2. Part is made in large enough quantities that

moderate testing is justified.

3. Part is made in such small quantities or sorapidly that testing is not feasible.

4. Part has been found to be unsatisfactory and

more analysis is required to improve it.

7/28/2019 ShigChp5

5/20

7/27/20

Stress Concentration

When loads are static and the material is ductile, designers

set the geometric (theoretical stress concentration factor)

equal to unity.

Failures Theories

Ductile materials ( and Syt= Syc= Sy)

Maximum shear stress

Distortion energy

Ductile Coulomb-Mohr

Brittle materials

Maximum normal stress

Brittle Coulomb-Mohr

Modified Mohr

05.0f

7/28/2019 ShigChp5

6/20

7/27/20

Maximum-Shear-Stress Theory

for Ductile Materials

AP

This theory predicts thatyielding begins whenever the

maximum shear stress in any element equals or exceeds

the maximum shear stress in tensile-test specimen of the

same material at yielding.

2max

Recalling that for simple tensile stress

and the maximum shear stress occurs on a plane 450 from

the tensile surface with magnitude

So, maximum shear stress at yield is

2maxyS

Maximum-Shear-Stress Theoryfor Ductile Materials

321

2

31max

y

y

S

S

31

31

max22

For the general state of stress, the 3 principal stresses can be

determined and ordered so that

The maximum shear stress is then

Thus, for the general state of stress, the maximum-shear-stress

theory predicts yielding when

or

7/28/2019 ShigChp5

7/20

7/27/20

Maximum-Shear-Stress Theory


ysy SS 5.0

This implies that yield strength in shear is

For design, we incorporate a factor of safety (n) and

n

S

n

S

y

y

31

max2

Maximum-Shear-Stress Theoryfor Ductile Materials

BA

yA

A

BA

S

0,

0

31

yBA

BA

BA

S

31,

0

yB

B

BA

S

31,0

0

For plane stress problems where

Case 1:

Case 2:

Case 3:

7/28/2019 ShigChp5

8/20

7/27/20

Distortion-Energy Theory


3

321

avg

The distortion energy theorypredicts that yielding occurs

when the distortion strain energy per unit volume reaches

or exceeds the distortion strain energy per unit volume for

yield in simple tension or compression of the same material.

Ductile materials stressed hydrostaticallyexhibit yield strength

>> than indicated in simple tension or compression, i.e. yielding

related to angular distortion.



321

2

13

2

32

2

21

133221

2

3

2

2

2

1

23

1

2226

21

Euuu

Eu

vd

v

2

1u

3322112

1 u

133221

2

3

2

2

2

1 22

1

Eu

The element in (c) is

subjected to pure

distortion with no

volume change.

Strain energy per unit volume for simple tension is

Strain energy for element (a) is

2133

3122

3211

1

1

1

E

E

ESubstituting

for principal

strains (see

eq 3-19)

gives

7/28/2019 ShigChp5

9/20

7/27/20



3

321

avg

212

32

E

uavg

v

avg

321 ,,

Substituting for yields strain energyproducing only volume change, or

Substituting the square of

yields

133221232221 2226

21

E

uv

Distortion-Energy Theoryfor Ductile Materials

133221

2

3

2

2

2

1 2

2

1

E

u

133221

2

3

2

2

2

1 2226

21

E

uv

The distortion energy (ud) is the difference between uand uvor

23

12

13

2

32

2

21

Euuu vd

minus

gives

7/28/2019 ShigChp5

10/20

7/27/20



2

321

3

1

0,

yd

y

SE

u

S

yS

2/1

2

13

2

32

2

21

2

21

2

13

2

32

2

21

2'

'

yS

For simple tension,

and for the general stress state,

The single, equivalent or effective stressfor a general stress

state is called the von Mises stress, '



2/1222

2/1222222

3'

62

1'

zxyyxx

zxyzxyxzzyyx

2/122' BBAA For plane stress

Using x,y,zcomponents of 3-D stress

which is graphed here with

and for plane stress,

The distortion energy theoryis also called:

The von Mises or von Mises-Hencky theory

The shear-energy theory

The octahedral-shear-stress theory

yS'

7/28/2019 ShigChp5

11/20

7/27/20



2/12132322213

1 oct

An isolated element in which

the normal stresses on each

surface are equal to the

hydrostatic stress, is show here.avg

There are 8 surfaces symmetric to

the principal directions that contain

this stress. This forms the octahedron

and the shear stresses on these surfaces

are called octahedral shear stresses.

We can show that,



0, 321 yS yoct S3

2

yS

2/1

2

13

2

32

2

21

2

By the octahedral-shear-stress theory, failure occurs whenever

octahedral shear stress for any stress state equals or exceeds

octahedral shear stress for the simple tension-test specimen

at failure.

For a tensile test, and

when, for the general stress case, yoct S

3

2

3

1 2/1213

2

32

2

21

which reduces to

7/28/2019 ShigChp5

12/20

7/27/20



1

yxy S2/12

3

The MSS theory ignores the contribution of normal stresses on

the surfaces 450 from the direction in a tensile test specimen.

However, these stresses are P/2Aand notthe hydrostatic

stresses which are P/3A. This is the difference between the MSS

and DE theories.

syy

y

xy SSS

577.03

n

Syzxyzxyxzzyyx

2/1222222

62

1'

Thus, the DE design equation is

For shear failure, we have

or, finally

Example 5-1Hot-rolled steel with Sy= 100 kpsi and true strain at fractureof 0.55. Estimate factor of safety for the following states of

principal stress (kpsi):

a) 70, 70, 0; b) 30, 70, 0; c) 0, 70, -30; d) 0, -30, -70; e) 30, 30, 30

7/28/2019 ShigChp5

13/20

7/27/20

Coulomb-Mohr Theory for Ductile Materials

For materials in which yield strength is different in tension and

compression, different failure theories apply.Mohrs theory of failure was to construct 3 circles corresponding

to failure conditions in tension, compression and torsional shear

as shown:

LineA-B-C-D-Edefines the failure envelope for such a material

and the line need not be straight.


321

1 3

The Coulomb-Mohr theory holds that the envelope B-C-Dis

straight.

Conventional ordering of

principal stresses such that

and given that the largest

Coulomb-Mohr circle connects

13

1133

12

1122

OCOC

CBCB

OCOC

CBCB

and

We can show that

7/28/2019 ShigChp5

14/20

7/27/20


1

22

22

22

22

31

31

31

ct

tc

tc

t

t

SS

SS

SS

S

S

which is

and reduces to

where either yield strength or ultimate strength can be used.


BA

0 BA

BA 0 BA 0

1c

B

t

A

SS

tA S

cB S

For plane stress when

the situation is similar

to MSS theory, i.e.

Case 1:where

and

Case 2:

where

and

0,31 A

BA 31 , B 31 ,0

Case 3:

where

and

7/28/2019 ShigChp5

15/20

7/27/20


nSS ct

131

syS 31

sySmax

31

ycyt

ycyt

sySS

SSS

Finally, for design equations, divide all strengths by n

For pure shear,

Torsional yield strength is

Substituting into

gives

131 ct SS


This graph shows

measurements at

failure of ductile

materials compared

to the DE and MSS

theories.

Either theory can beused.

When St is not equal

to Sc, Coulomb-Mohr

theory is best.

7/28/2019 ShigChp5

16/20

7/27/20

Example 5-3

Maximum-Normal-Stress Theory

for Brittle Materials

321

ucB S

BA

The maximum-normal-stress (MNS) theory states that failure

occurs when one of the 3 principal stresses equals or exceeds

the strength.

In the general case where

failure is predicted whenever

orutS1

utA S

ucS3

For plane stress where

failure occurs whenever

or

7/28/2019 ShigChp5

17/20

7/27/20



Plotting failure in plane

stress using the MNS

theory looks like (fig. 5-18)



n

SutA 0 BA

BA 0n

SucB

BA 0

BA 0

ut

uc

A

B

S

S

ut

uc

A

B

S

S

BA

Failure criteria can be converted to

design equations where

Two sets of load lines are:

Load line 1

Load line 2

Load line 3

Load line 4

7/28/2019 ShigChp5

18/20

7/27/20

Modification of Mohr Theory


n

SutA 0 BA

nSS uc

B

ut

A 1

BA 0

n

SucB BA 0

Brittle-Coulomb-Mohr (BCM)

The BCM theory expands

the 4th quadrant

Modification of Mohr Theoryfor Brittle Materials

n

SutA 0 BA

BA 0 1A

B

nSSSSS

uc

B

utuc

Autuc 1

BA 0 1A

B

n

SucB BA 0

Modified Mohr

and

and

7/28/2019 ShigChp5

19/20

7/27/20

Failure of Brittle Materials

Example 5-5

ASTM grade 30 cast iron

Find force F leading to

failure by:

a) Coulomb-Mohr theory

b) Modified Mohr theory

7/28/2019 ShigChp5

20/20

7/27/20

Selection of Failure Criteria

shigchp5

Documents