Failure Theories

Chapter4- 1 Department of Mechanical & Aerospace Engineering

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Chapter 4

Criteria of Material Failure


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OUTLINE

4.1 Definition of Structural Failure.

4.2 Types of Failure.

4.3 Failure Criteria.

4.3.1 Maximum Principal Stress Criterion.

4.3.2 Maximum Shear Stress Criterion.

4.3.3 Maximum Distortion Energy Criterion.

(Von Mises Failure Criterion)

4.3.4 Coulomb-Mohr Failure Criterion.

4.4 Application of Failure Criterion.


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4.1 Definition of Structural Failure

Structural failure can be defined as any

change in size, shape or material

properties of a structure that makes it

incapable of satisfactorily performing its

intended function.


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4.2 Types of Failure

The common types of failure observed in

engineering practice are described as:

(1) Force/ temperature induced excess

(2) Excess plastic deformation (yielding).

elastic deformation.

(3) Ductile rupture (excess plastic deformation

(4) Brittle failure (excess elastic deformation

that causes the structure broken into pieces).

that causes the structure broken into pieces).


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4.2 Types of Failure (cont’d)

(5) Fatigue Failure

(6) Buckling failure

-- A certain size of crack initiation

resulting from the cyclic loading.

(7) Creep failure

-- Structure is buckled.

--The plastic deformation in structure under

the influence of stress over a period of time.

(8) Corrosion, wearing, impact etc.


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4.3 Failure Criteria

Stress analysis would not be able to predict

the failure of a given structure. To know about

how high stress can a structure be safely

sustained, a failure criterion is needed.

The criteria discussed in this section will be

focused to relate failures due to force

induced failure, yielding, brittle failure and

ductile rupture.


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4.3.1 Maximum Principal Stress Criterion

Failure occurs when the maximum principal

stress becomes equal or exceeds the maximum

principal stress (ultimate strength) in a simple

uniaxial tension/compression of the same

material.

Mathematically, we can write

01 t

u 03 c

uor if fails (4-1)


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Maximum Principal Stress Criterion (cont’d)

c

u

t

u ,

t

uc

u

),,( 321

Ultimate strength of the material

in tension or compression

1t

uc

u

c

u

t

u2

Any state of stress within the

envelope would not produce

failure.

* No effect of and interaction

For the plane stress condition, 03

SAFE

1 2


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4.3.2 Maximum Shear Stress Criterion

Failure occurs when the maximum shear stress

magnitude becomes equal to or exceeds the

maximum shear stress magnitude at yielding

failure in a simple uniaxial load.

Mathematically, we have

S2,1 S3,1 S3,2 (4-2)

2

212,1

2

32

3,2

2

31

3,1

2

0S 0

y at failure

01


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Maximum Shear Stress Criterion (cont’d)

22

21 y

22

01 y

22

0 2 y

03 for

y 21

y 1

y 2

1

2

y

y

y SAFE

(4-3)


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4.3.3 Maximum Distortion Energy Criterion (Von Mises Failure Criterion)

Failure occurs when the distortion energy

becomes equal to or exceeds the distortion

energy at yielding failure in a simple tension.

odoVo UUU

Strain energy per unit volume

due to distortion

Strain energy per unit volume

due to volume change

Total strain energy per unit volume


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Maximum Distortion Energy Criterion (cont’d)

3322112

1

2

1

2

1 oU

EEE

3211

(2-26)

133221

2

3

2

2

2

1 22

1

E

2

2

321 21

2

3

3

21

2

3moV

EEU

(2-30)

oVood UUU 2

13

2

32

2

213

1

2

1

E(2-31)

o

odod UU ( at failure )


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For uniaxial tension, y 1 032

2

3

1y

o

odE

U

or 22

13

2

32

2

213

1

3

1

2

1y

EE

22

13

2

32

2

21 2 y at failure

(4-4)

For plane stress, 03

22

2

2

1

2

21 2 y

2

21

2

2

2

1 y or (4-5)


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1

2

oo

o

o

Since,

yocto tensionuniaxialanin

3

2

3

2)( 1

QuickTime™ and a decompressor

are needed to see this picture.


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Substituting the above equation into (4-4),

we have

20

22

2

3223

oct

yoct

o

octoct (4-7)

(4-6)

Hence, the distortion energy criterion can

be stated as

“Failure occurs when the octahedral shear stress

becomes equal to or exceeds the octahedral

shear at yield failure in a uniaxial loading.”

Rewriting equation (4-6), yyoct 47.0

3

2


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4.3.4 Coulomb – Mohr Failure Criterion

In general, c

u

t

u

The Coulomb – Mohr failure criterion is given as

1

2

t

u

c

u

allall

21 ,

c

u

t

u

121 c

u

t

u


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Coulomb – Mohr Failure Criterion (cont’d)

Consider a state of pure shear. The pure

shear stress state is equivalent to

=

2

1

)0( 112

Therefore,

111

1

c

u

t

u

all

c

u

t

u

c

u

t

uall

1


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4.4 Application of Failure Criterion

Example 1

A steel part is subjected to a state of stress,

(principal stress) = 35 ksi, (principal stress)

= -25 ksi. The part has ultimate strength in

tension, = 100 ksi, ultimate strength in

compression, =100 ksi and ultimate strength

in torsion, S = 60 ksi. Estimate the safety factor

using the following failure criteria:

1

2

c

u

t

u

(a) Maximum normal stress criterion.

(b) Maximum shear stress criterion.

(c) Maximum distortion energy criterion.


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Example 1 (cont’d)

Solution:

(1) Determine the maximum failure strength

(allowable) from the failure envelope.

(2) Compute factor of safety StressActual

StrengthFailureFS ..

(a) Maximum normal stress criterion

tS

tS

cS

cS

35

25

86.235

100.. FS

allow

allow

or2

1

2

1

25

35

100.7

5

7

512 allowallow

86.225

7/500..

FS


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(b) Maximum shear stress criterion

100

100

100

100

1100100

21 allowallow

7

5

25

35

2

1

127

5

10021 allowallow 1007

511 allow

ksiallow 33.581 67.135

33.58.. FS

or ksiallow 67.412 67.125

67.41..

FS


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2

021

2

2

2

1 . allowallowallowallow

127

5

2

0

2

1

2

1

2

1 .7

5

49

25 allowallowallow

ksiallow 05.67100.109

71

92.135

05.67.. FS


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Example 2

A force P0 kips applied by a lever arm to the shaft as shown

in the figure produces stresses at the critical point “A”

having the values shown on the element in the figure. The

shaft is made of steel with the yield strength, Y= 36ksi, and

the ultimate strength, Tult=75 ksi, C

ult =75ksi.

Determine the load that would cause

the shaft failure

• according to the maximum shear

failure criterion

• according to the maximum distort

energy criterion


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Solution:

(1) Find the stress at “A” related to radius, r

Ksir

bP

I

brP

I

Mr ooA

x 104

3

Let x along the shaft, and z perpendicular to the surface

“A”.

Ksir

aP

r

T

J

Tr oA

xy 14.1422

33

0A

y

a= distance from P to the

connection point

b= distance the connection pint to

point “A”


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The principal stresses are

014.14

14.1410

02

24

3

33

r

aPr

aP

r

bP

o

oo

A

ksiA 201

The state of stress at “A” gives

ksiA 102


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(a) Maximum shear stress criterion

3636

36

13636

21 allowallow

210

20

2

1

12 5.0

3621 allowallow 365.1 1 allow

ksiallow 241 2.120

24.. FS

36

allall

12 5.0

Hence, the failure load will be at 1.2Po .


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2

21

2

2

2

1 . y

allowallowallowallow

12 5.0

ksiallow 21.273675.1

11

36.120

21.27.. FS

allowallow

12 5.0

The failure load is 1.36P0.

Failure Theories

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