Theories of FailureFailure of a member is defined as one of two conditions.

1. Fracture of the material of which the member is made. This type

of failure is the characteristic of brittle materials.

2. Initiation of inelastic (Plastic) behavior in the material. This

type of failure is the one generally exhibited by ductile materials.

When an engineer is faced with the problem of design using a specific

material, it becomes important to place an upper limit on the state of

stress that defines the material's failure. If the material is ductile, failure

is usually specified by the initiation of yielding, whereas if the material

is brittle it is specified by fracture. 1

These modes of failure are readily defined if the member is subjected to

a uniaxial state of stress, as in the case of simple tension however, if the

member is subjected to biaxial or triaxial stress, the criteria for failure

becomes more difficult to establish.

In this section we will discuss four theories that are often used in

engineering practice to predict the failure of a material subjected to a

multiaxial state of stress.

A failure theory is a criterion that is used in an effort to predict the

failure of a given material when subjected to a complex stress condition.

2

i. Maximum shear stress (Tresca) theory for ductile materials.

ii. Maximum principal stress (Rankine) theory.

iii. Maximum normal strain (Saint Venan’s) theory.

iv. Maximum shear strain (Distortion Energy) theory.

Several theories are available however, only four important theories are

discussed here.

3

Maximum shear stress theory for Ductile Materials

The French engineer Tresca proposed this theory. It states that a

member subjected to any state of stress fails (yields) when the

maximum shearing stress (τmax)in the member becomes equal to the

yield point stress (τy)in a simple tension or compression test (Uniaxial

test). Since the maximum shear stress in a material under uniaxial stress

condition is one half the value of normal stress and the maximum

normal stress (maximum principal stress) is max, then from Mohr’s

circle.

2max

max

4

5

If both of principal stresses are of the same sign tension

compression then

σ1 <σy (4)

σ2 <σy (5)

In case of Biaxial stress state

)3(

22

)2(22

21

21

max

21minmaxmax

y

y

y

6

A graph of these equation is given in the figure. Any given state of

stress will be represented in this figure by a point of co-ordinate 1 and

2 where 1 and 2 are two principal stresses. If these point falls within

the area shown, the member is safe and if outside then member fails as

a result of yielding of material. The Hexagon associated with the

initiation of yield in the member is known as “Tresca Hexagon”. (1814-

1885). In the first and third quadrant 1 and 2 have the same signs and

7

max is half of the numerically larger value of principal stress 1 or 2. In

the second and fourth quad, where 1 and 2 are of opposite sign, max is

half of arithmetical sum of the two 1 and 2.

In fourth quadrant, the equation of the boundary or limit

(yield/boundary stress) stress line is

1 - 2 = y

And in the second quadrant the relation is

1 - 2 = -y

8

y 21

y 1

y 1

y 2

y 21

y 2

9

Problem 01:-

The solid circular shaft in Fig. (a) is subject to belt pulls at each

end and is simply supported at the two bearings. The material has a

yield point of 36,000 Ib/in2• Determine the required diameter of the

shaft using the maximum shear stress theory together with a safety

factor of 3.

10

400 + 200 lb200 + 500 lb

11

12

0

42800

64

24200

.42006700

.36006600

64

2

3

4

4

y

x

x

B

x

d

d

d

inlbMc

inlbM

dI

dc

I

Mc

13

inlb

OR

inlb

T

d

d

d

J

Tr

xy

xy

.480024200

24)200400(

.480016300

16)200500(

480,24

32

24800

3

4

xyyx

yxxy

3

480,24

dxy 3

800,42

dx

x

14

2

2

21

max21

22

2

know weAs

xyyx

3

36000

2)(

2...

...

And

21

21

max

FOS

SOF

SOF

yield

yield

y

15

''76.1

480,24

2

428001036

480,24

2

42800

2

000,12

480,24

2

428002

3

000,36

480,24

2

428002

2

3

2

36

2

3

2

3

2

3

2

3

2

3

2

321

d

dd

dd

dd

dd

16

Problem 02

The state of plane stress shown occurs at a critical point of a steel

machine component. As a result of several tensile tests, it has been found

that the ten sile yield strength is y = 250 MPa for the grade of steel used.

Determine the factor of safety with respect to yield, using (a) the

maximum-shearing-stress criterion, and (b) the maximum-distortion-

energy criterion.

17

18

SOLUTION Mohr's Circle. We construct Mohr's circle for the given state of stress and find

MPaOC yxave 20)4080()( 21

21

MPaFXCFRm 65)25()60()()( 2222

Principal Stresses

MPaCAOCa 856520 MPaCAOCb 456520

a. Maximum-Shearing-Stress Criterion. Since for the grade of steel used

the tensile strength is ay = 250 MPa, the corresponding shearing stress at

yield is MPaMPaYY 125)250(2

121 19

MPaFor m 65 2.19.65

125. SF

MPa

MPaSF

m

Y

20

b. Maximum-Distortion-Energy Criterion. Introducing a factor of safety

into Eq. (7.26), we write

222

.

SFY

bbaa

For a = +85 MPa, b = -45 MPa, and y = 250 MPa, we have

2

22

.

25045458585

SF

19.2..

2503.114 SF

SF

21

Comment. For a ductile material with y = 250 MPa, we have drawn the

hexagon associated with the maximum-shearing-stress criterion and the

el lipse associated with the maximum-distortion-energy criterion. The

given state of plane stress is represented by point H of coordinates a =

85 MPa and b = -45 MPa. We note that the straight line drawn through

points O and H intersects the hexagon at point T and the ellipse at point

M. For each criterion, the value obtained for F.S. can be verified by

measuring the line segments in dicated and computing their ratios:

19.2.2.19. OH

OMSFb

OH

OTSFa 22

23

The solid shaft shown in Fig. 10-41a has a radius of 0.5 in. and is made

of steel having a yield stress of y = 36 ksi. Determine if the loadings

cause the shaft to fail according to the maximum-shear-stress theory

and the maximum-distortion-energy theory.

Example 10-12

Solution

The state of stress in the shaft is caused by both the axial force and the

torque. Since maximum shear stress caused by the torque occurs in the

material at the outer surface, we have

24

ksiin

ininkip

J

Tc

ksiin

kip

A

P

xy

x

55.162/.)5.0(

.)5.0.(.25.3

10.19.)5.0(

15

4

2

The stress components are shown acting on an element of material at

point A in Fig. 10-41b. Rather than using Mohr's circle, the principal

stresses can also be obtained using the stress-transformation equations,

Eq.9-5.

22

2,1

2

2

2,1

)55.16(2

010.19

2

010.19

22

xyyxyx

25

= -9.55 ± 19.11

CTI = 9.56 ksi

CT2 = -28.66 ksi

Maximum-Shear-Stress Theory. Since the principal stresses have

opposite signs, then from Sec. 9.5, the absolute maximum shear stress

will occur in the plane, and therefore, applying the second of Eq. 10-27,

we have

362.38

3666.2856.9?

21

Y

26

Thus, shear failure of the material will occur according to this theory.

Maximum-Distortion-Energy Theory. Applying Eq. 10-30, we have

Using this theory, failure will not occur.

12961187

)36(66.2866.2856.956.9

)(

2?

22

2221

21

y

27

Maximum Principal Stress theory or

(Rankine Theory)

According to this theory, it is assumed that when a member is

subjected to any state of stress, fails (fracture of brittle material or

yielding of ductile material) when the principal stress of largest

magnitude. (1) in the member reaches to a limiting value that is equal

to the ultimate normal stress, the material can sustain when subjected to

simple tension or compression.

The equations are shown graphically as

)1(1 ult )2(2 ult 28

ult

These equations are shown graphically if the point obtained by plotting

the values of 1 + 2 falls within the square area the member is safe.29

It can be seen that stress co-ordinate 1 and 2 at a point in the material

falls on the boundary for outside the shaded area, the material is said to

be fractured failed. Experimentally, it has been found to be in close

agreement with the behavior of brittle material that have stress-strain

diagram similar in both tension and compression. It cab be further

noticed that in first and third quad, the boundary is the save as for

maximum shear stress theory.

Problem

A thin-walled cylindrical pressure vessel is subject to an internal

pressure of 600 lb/in2. the mean radius of the cylinder is 15in. If the

material ha a yield point of 39,000lb/in2 and a safety of 3 is employed, 30

determine the required wall thickness using (a) the maximum normal

stress theory, and (b) the Huber-von Mises-Hencky theory.

P = 600psi.

r = 15"

y = 39000psi

F.O.S = 3

t = ?

According to

c Circumferential stress / girth stress

where

Pr

tc

31

By comparing 1 = c.

Thus, According to the normal stress theory, maximum Principal stress

should be equal to yield stress/FOS i.e.

|1| ≤ ult |2| ≤ ult

f = maximum (|1|, |2|, |3| - ult = 0)

tt

t

stressallengitudinFor

tt

c

l

c

4500

2

"15600

2

Pr

9000"15600

32

As y = 39000psi

.69.0

900013000

9000

3

39000

Anst

t

t

Problem 02:-

The solid circular shaft in Fig. 18-12(a) is subject to belt pulls at

each end and is simply supported at the two bearings. The material has

a yield point of 36,000 Ib/in2• Determine the required diameter of the

shaft using the maximum normal stress theory together with a safety

factor of 3. 33

400 + 200 lb200 + 500 lb

34

35

0

42800

64

24200

.42006700

.36006600

64

2

3

4

4

y

x

x

B

x

d

d

d

inlbMc

inlbM

dI

dc

I

Mc

36

inlb

OR

inlb

T

d

d

d

J

Tr

xy

xy

.480024200

24)200400(

.480016300

16)200500(

480,24

32

24800

3

4

xyyx

yxxy

3

480,24

dxy 3

800,42

dx

x

37

2

2

2

2

2

1

22

22

xyyxyx

xyyxyx

2

3

2

331

20.24446

2

42800

2

42800

ddd

According to maximum normal stress theory .

2

3

2

33

1

20.24446

2

42800

2

42800

3

36000

ddd

ult

38

"48.1

51.10

10514.110144

1062.5971096.457

2

1092.91510144

20.24446

2

42800

2

4280012000

6

6

96

6

6

6

6

3

66

2

3

2

33

d

d

d

ddd

ddd

39

Mohr’s Criterion:-

In some material such as cast iron, have much greater strength

in compression than tension so Mohr proposed that is 1st and third

quadrant of a failure brokes, a maximum principal stress theory was

appropriate based on the ultimate strength of materials in tension or

compression. Therefore in 2nd and 4th quad, where the maximum shear

stress theory should apply.

Pure shear is one in which x and y are equal but of opposite sense.

40

cult

Failure or strength envelope

Pure shear

Smaller tension strength

Largest compressive strength

41

tension

compression

42

Problem 3:-

In a cast-iron component the maximum principal stress is to be limited

to one-third of the tensile strength . Determine the maximum value of

the minimum principal stresses ,using the Mohr theory. What would be

the values of the principal stresses "associated with a maximum shear

stress of 390 MN/m2? The tensile and compressive strengths of the cast

iron are 360 MN/m2 and 1.41 GN/m2 respectively

43

stress Shearingfourthand2

stress Normalquadrant thirdand1

criterionsMohr'perAs

?

nd

st

2

quadrant

2

)(1

/1203

360

3

thatindicatedAs

mMN

tult

26

2

/101414

/360

mMN

mMN

cult

tult

44

By using principal stress theory

2

2

2

1

/414

/360

)(

)(

mMN

mMNAs

b

a

cult

tult

ult

ult

21 /120 mMN

Maximum principal stress = 360/3 = 120 MN/m2 (tension). According

to Mohr's theory, in the second and fourth quadrant

From (a) and (b) 45

Therefore

22

2 /94011410360

120mMNand

The Mohr's stress circle construction for the second part of this problem

is shown in Fig. 13.7. If the maximum shear stress is 390 MN/m2, a

circle is drawn of radius 390 units to touch the two envelope lines. The

principal stresses can then be read off as +200 MN/m2 and -580

MN/m2•

1

11

21

21

culttult

culttult

and

46

cult tult

360tult1410cult

390max

47

Fig 13.7 48

Now from second state

21

21max

21

21

21cult

21tult

2max

2

2

1414

380

quadfourth in and

theorystress shearingBy

/390

R

Also

Thus

mMN

49

Example 10-11

The solid cast-iron shaft shown in Fig. 10-40a is subjected to a torque

of T = 400 Ib . ft. Determine its smallest radius so that it does not fail

according to the maximum-normal-stress theory. A specimen of cast

iron, tested in tension, has an ultimate stress of (σult)t = 20 ksi.

Solution

The maximum or critical stress occurs at a point located on the surface

of the shaft. Assuming the shaft to have a radius r, the shear stress is

34max

..8.055

)2/(

)/.12)(.400(

r

inlb

r

rftinftlb

J

Tc

50

51

52

Mohr's circle for this state of stress (pure shear) is shown in Fig. 10-

40b. Since R = max, then

The maximum-normal-stress theory, Eq. 10-31, requires

|1| ≤ ult

3max21

..8.3055

r

inlb

23

/000,20..8.3055

inlbr

inlb

Thus, the smallest radius of the shaft is determined from

..535.0

/000,20..8.3055 2

3

Ansinr

inlbr

inlb

53

The solid shaft shown in Fig. 10-41a has a radius of 0.5 in. and is made

of steel having a yield stress of y = 36 ksi. Determine if the loadings

cause the shaft to fail according to the maximum-shear-stress theory

and the maximum-distortion-energy theory.

Example 10-12

Solution

The state of stress in the shaft is caused by both the axial force and the

torque. Since maximum shear stress caused by the torque occurs in the

material at the outer surface, we have

54

ksiin

ininkip

J

Tc

ksiin

kip

A

P

xy

x

55.162/.)5.0(

.)5.0.(.25.3

10.19.)5.0(

15

4

2

The stress components are shown acting on an element of material at

point A in Fig. 10-41b. Rather than using Mohr's circle, the principal

stresses can also be obtained using the stress-transformation equations,

Eq.9-5.

22

2,1

2

2

2,1

)55.16(2

010.19

2

010.19

22

xyyxyx

55

= -9.55 ± 19.11

CTI = 9.56 ksi

CT2 = -28.66 ksi

Maximum-Shear-Stress Theory. Since the principal stresses have

opposite signs, then from Sec. 9.5, the absolute maximum shear stress

will occur in the plane, and therefore, applying the second of Eq. 10-27,

we have

362.38

3666.2856.9?

21

Y

56

Thus, shear failure of the material will occur according to this theory.

Maximum-Distortion-Energy Theory. Applying Eq. 10-30, we have

Using this theory, failure will not occur.

12961187

)36(66.2866.2856.956.9

)(

2?

22

2221

21

y

57

Maximum Normal Strain or Saint Venant’s

Criterion

In this theory, it is assumed that a member subjected to any state of

stress fails (yields) when the maximum normal strain at any point

equals, the yield point strain obtained from a simple tension or

compression test (y = σy/).

Principal strain of largest magnitude |max| could be one of two

principal strain 1 and 2 depending upon the stress conditions acting in

the member . Thus the maximum Principal strain theory may be

represented by the following equation.58

)1(2max

1max

y

y

As stress in one direction produces the lateral deformation in the other

two perpendicular directions and using law of superposition, we find

three principal strains of the element.

x=

y=

z=

σx

σx / E

-μσx / E

-μσx / E

σy

μσy / E

σy / E

-μσy / E

σz

–μσz / E

-μσz / E

σz / E

x=

y=

z=

x=

y=

z= 59

x = σx / E -μσy / E -μσz / E

= σx / E -μ / E (σy + σz)

y = σy / E -μσx / E -μσz / E

= σy / E -μ / E (σx + σz)

z = σz / E -μσy / E -μσx / E

= σz / E -μ / E (σx + σy)

(2)

Thus

)3(

)(

)(

)(

123

3

312

2

321

1

EE

EE

EE

60

Also

)6(

)5(

Then

and

4and1Equating

)4()2(

12y

21y

122

21

1

211

EE

EEE

dForEE

y

y

61

The yield surface ABCD is the straight under biaxial tension or biaxial

compression Individual principal stresses greater than σy can occur

without causing yielding.

62

Maximum Shear Strain Energy (Distortion

Energy Criterion (Von MISES Criterion)

According to this theory when a member is subjected to any state of

stress fails (yields) when the distortion energy per unit volume at a

point becomes equal to the strain energy of distortion per unit volume

at failure (yielding) in uniaxial tension (or compression).

The distortion strain energy is that energy associated with a change in

the shape of the body.

The total strain energy per unit volume also called strain energy

density is the energy in a body stored internally throughout its volume

due to deformation produced by external loading. If the axial stress63

arising in a tension test is “” and the corresponding axial strain is ε

then the work done or a unit volume of the test specimen is the product

of mean value of the force per unit area (/2) times the displacement in

the direction of the force, or ε. The work is thus

)1(2

U

This work is stored as internal strain energy.

When an elastic element subjected to triaxial loading as show in Fig the

stresses can be resolves in to three principal stresses σ1,σ2 and σ3. where

1,2 and 3 are the principal axes. These three principal stresses will be

accompanied by these principal strain related to the stresses by

equations 3,4and 564

If it is assumed that loads are applied gradually and simultaneous then

stresses and strain will increase in the same manner. The total strain

energy per unit volume in the sum of energies produced by each of the

stresses (as energy is a scalar quantity )

E

u

E

)( 3211

E

u

E

)( 3122

E

u

E

)( 2133

(2)

(3)

(4)

65

Where ε1, ε2, and ε3, are the normal strain in the direction of principal

stresses respectively of strain are expressed in term of stresses than

equation (5) taken the following form

)5(,2

1,,

2

1,,

2

13322 U

EEEEEEEEEU 213

3312

2321

1 2

1

2

1)(

2

1

Which can be reduced to

)6()(22

1323121

33

22

21

EU

66

As the deformation of a material can be separated in to two parts

(a) Change in volume, (b) change in shape or distortion.

Similarly the total strain energy can be broken in to two parts. One part

representing the energy needed to cause volume change of the element

with no change in shape (Uv) and the other part representing the energy

needed to distort the element (Ud).

)7(dv UUU

The principal stresses σ1,σ2 and σ3 of Fig. (01) can be resolved in to two

states of stresses in Fig. (02) b & c. the state of stress shown in Fig. b

represents a hydrostatic stress condition in which all these principal

stress are equal to the quantity σ. 67

1

3

1

(a)

(b)

1-

3-

2-

(c)

68

Some materials were subjected to hydrostatic pressure resulting in

appreciable changes in volume but no change in shape and no failure by

yielding. Therefore the remaining portion of the stress (1 - ), (2 - )

and (3 - ) will result in distortion only (No volume change) and the

algebraic sum of the three principal strains produced by the three

principal stresses (1 - ave), (2 - ave) and (3 - ave) must be equal to

zero. i.e (ε1+ε2+ε3)d=0 (9)

Expressing strains in term of stresses.

69

0213

132

321321

dE

2

2

2

213

132

321

2

22

213

132321

63333222111

63222 332211

70

a) One part that causes Volumetric (Uv)

b) Change and one causes distortion an (Ud)

)4(dv UUU

a) As according to theory of distortion only energy due to

distortion is responsible for failure. Some experimental

evidence supports this assumption some materials were

subjected to hydrostatic pressure result in appreciable changes

in volume but no change in shape and no failure by yielding.

The hydrostatic pressure is the average of three principal

stresses 1 2 and 3 known as average stress. )5(

3321

max

71

This principal strains ε in the material.

Due to there Principal stresses (1, 2, 3) three principal strains ε1, ε2,

and ε3, are produced. The state of stress in Fig below will result in

distortion only (No volume change) if sum of three normal strains is

zero. That is

)6(0

2

221)(

213

3123221321

u

EE

Which eh reduces to

)5(3/

03)21(

321

321

72

The normal strains corresponding to these stresses are found from three

dimension from of Hook’s law given by the following equation.

)7()21(E

Strain energy resulting from these stresses or strains can be obtained by

subject through these values is

)8(.2

3.

2

1.

2

1.

2

1

, 321321

Uv

73

(5)3

)()21(

3

)(.

2

3 321321

E

uU v

2321 )(

6

)21(

E

uU v (9)

Strain energy due to distortion

U=Uy + Ud

Ud = U- Uy

2321323121

23

22

21 6

212

2

1 E

uu

EU d

(10)

2321323121

23

22

21 21)(6)(3

6

1 uuE

U d74

When the third force in the

2113

23

2332

22

2221

21 222

6

1

E

uU d

213

232

2216

1

E

uU d

(12)

(13)

For biaxial stress system, σ3 = 0

2221

21 2

3

1

E

uU d

(14)

For uniaxial stress system

213

1 E

uU d

(15)

75

For Distortion theory the failure occurs when distortion energy of the

member however equal to the strain energy of distortion at failure

(yielding) in uniaxial torsion (or equilibrium). So substituting σ1=σy in

equ (15)

226

1yd E

uU

(16)

And equating it with (13)

213

232

221

2

6

1)2(

6

1

E

u

E

uy

213

232

221

22 y (17) 76

For biaxial stress system

2221

21

2 2 y (18)

77