Theories of Failure The material properties are usually determined by simple tension or compression tests.

The mechanical members are subjected to biaxial or triaxial stresses.

To determine whether a component will fail or not, some failure theories are proposed which are related to the properties of materials obtained from uniaxial tension or compression tests.

Initially we will consider failure of a mechanical member subjected to biaxial stresses

The Theories of Failures which are applicable for this situation are:

• Max principal or normal stress theory (Rankine’s theory)

• Maximum shear stress theory (Guest’s or Tresca’s theory)

• Max. Distortion energy theory (Von Mises & Hencky’s theory)

• Max. strain energy theory• Max. principal strain theory

Ductile materials usually fail by yielding and hence the limiting strength is the yield strength of material as determined from simple tension test which is assumed the same in compression also.

For brittle materials limiting strength of material is ultimate tensile strength in tension or compression.

Max. Principal or Normal stress theory (Rankine’s Theory):

• It is assumed that the failure or yield occurs at a point in a member when the max. principal or normal stress in the biaxial stress system reaches the limiting strength of the material in a simple tension test.

• In this case max. principal stress is calculated in a biaxial stress case and is equated to limiting strength of the material.

Maximum principal stress

2

2

1 22 xyyxyx

Minimum principal stress

2

2

2 22 xyyxyx

•For ductile materials

1 should not exceed in tension,

FOS

S yt

•For brittle materials

1 should not exceed in tension

FOS

Sut

FOS=Factor of safety

This theory is basically applicable for brittle materials which are relatively stronger in shear and not applicable to ductile materials which are relatively weak in shear.

+σ2

-σ1

-σ2

+σ1

Syc

Syt

Syc

Syto

Boundary for maximum – normal – stress theory under bi – axial stresses

•The failure or yielding is assumed to take place at a point in a member where the max shear stress in a biaxial stress system reaches a value equal to shear strength of the material obtained from simple tension test.

•In a biaxial stress case max shear stress developed is given by

2.Maximum Shear Stress theory (Guest’s or Tresca’s theory):

FOSyt

max

where max = FOS2

Syt

This theory is mostly used for ductile materials.

22

max 2 xyyx

oσ2= σ3 =0

max

σ1

σ

Mohr’s circle for uni – axial tension

max

σσ3 =0

+σ1-σ2

max

o

Mohr’s circle for bi– axial stress condition

2

..max

stressdirectMinstressdirectMax

CASE – 1 (First quadrant )σ1 and σ2 are +ve

yt

yt

Sei

S

1

1131max

..222

0

2

CASE – 2 (Second quadrant)σ1 is -ve and σ2 is +ve ,Then

2

2

2

22

)(

1max

11212max

ytSThen

2maxytS

CASE – 3 (Third quadrant)σ1 is -ve and σ2 is more -ve ,Then

yc

yc

Sei

SThen

.22

2

0

2

)(

max

223max

CASE – 4 (Fourth quadrant)σ1 is +ve and σ2 is -ve ,Then

2

2

2

22

)(

max

2121max

ytSThen

Assuming that σ1> σ2> σ3 and σ3 =0

According to the Maximum shear stress theory,

And also

+σ1

σ1=Syt

+σ2

-σ1

-σ2

Syc

Syt

Syc

Syto

σ1=Syc

•It is assumed that failure or yielding occurs at a point the member where the distortion strain energy (also called shear strain energy) per unit volume in a biaxial stress system reaches the limiting distortion energy (distortion energy at yield point) per unit volume as determined from a simple tension test.

•The maximum distortion energy is the difference between the total strain energy and the strain energy due to uniform stress.

3.Max. Distortion energy theory (Von Mises & Hencky’s theory):

3.Max. Distortion energy theory (Von Mises & Hencky’s theory):

• The criteria of failure for the distortion – energy

theory is expressed as

• Considering the factor of safety

• For bi – axial stresses (σ3=0),

3.Max. Distortion energy theory (Von Mises & Hencky’s theory):

212

22

1 FOS

S yt

213

232

2212

1 FOS

S yt

213

232

2212

1 ytS

3.Max. Distortion energy theory (Von Mises & Hencky’s theory):

• A component subjected to pure shear stresses and the corresponding Mohr’s circle diagram is

Y

X

Element subjected to pure shear stresses

o σ1-σ2

σ

Mohr’s circle for pure shear stresses

In the biaxial stress case, principal stress 1, 2 are calculated based on x ,y & xy which in turn are used to determine whether the left hand side is more than right hand side, which indicates failure of the component.

212

22

1 FOS

S yt

From the figure, σ1 = -σ2 = and σ3=0

Substituting the values in the equation

We get

Replacing by Ssy, we get

3ytS

ytyt

sy SS

S 577.03

+σ1

+σ2

-σ1

-σ2

Syc

Syt

Syc

Syto

Boundary for distortion – energy theory under bi – axial stresses

Case 1 (First quadrant)

σ1 and σ2 are +ve and equal to σ, then

FOS

SFOS

S

yt

yt

212

22

1

Case 4 (Fourth quadrant)

σ1 is +ve and σ2 is -ve and equal to σ, then

FOS

SFOS

SFOS

SFOS

S

yt

yt

yt

yt

577.0

33 2

212

22

1

212

22

1

Case 2 (Second quadrant)

σ1 is -ve and σ2 is +ve and equal to σ, then

FOS

SFOS

SFOS

SFOS

S

yt

yt

yt

yt

577.0

33 2

212

22

1

212

22

1

Case 3 (Third quadrant)

σ1 is -ve and σ2 is +ve and equal to σ, then

FOS

SFOS

S

yt

yt

212

22

1

MPaMPa

MPa

•Failure is assumed to take place at a point in a member where strain energy per unit volume in a biaxial stress system reaches the limiting strain energy that is strain energy at yield point per unit volume as determined from a simple tension test.

• Strain energy per unit volume in a biaxial system is

• The limiting strain energy per unit volume for yielding as determined from simple tension test is

mEU 212

22

11

2

2

1

4. Max. Strain energy theory (Heigh’s Thoery):

2

2 2

1

FOS

S

EU yt

Equating the above two equations then we get

In a biaxial case 1, 2 are calculated based as x, y & xy

2

2122

21

2

FOS

S

myt

It will be checked whether the Left Hand Side of Equation is less than Right Hand Side of Equation or not. This theory is used for ductile materials.

EFOS

S

mE

σ

E

σE yt21

max

•It is assumed that the failure or yielding occurs at a point in a member where the maximum principal (normal) strain in a biaxial stress exceeds limiting value of strain (strain at yield port) as obtained from simple tension test.

• In a biaxial stress case

•One can calculate 1 & 2 given x , y & xy and check whether the material fails or not, this theory is not used in general as reliable results could not be detained in variety of materials.

5.Max. Principal Strain theory (Saint Venant’s Theory):

Example :1

• The load on a bolt consists of an axial pull of 10kN together with a transverse shear force of 5kN. Find the diameter of bolt required according to

1. Maximum principal stress theory 2. Maximum shear stress theory 3. Maximum principal strain theory4. Maximum strain energy theory5. Maximum distortion energy theory Permissible tensile stress at elastic limit =100MPa and

Poisson’s ratio =0.3

Solution 1

• Cross – sectional area of the bolt,

• Axial stress,

• And transverse shear stress,

22 7854.04

ddA

2221 /73.12

7854.0

10mmkN

ddA

P

22

/365.67854.0

5mmkN

dA

Ps

According to maximum principal stress theory• Maximum principal stress,

• According to maximum principal stress theory, Syt = σ1

2

2

1 22 xyyxyx

22

1 22 xyxx

221

2

2

2

221

/15365

365.6

2

73.12

2

73.12

mmNd

ddd

mmdd

4.1215365

1002

According to maximum shear stress theory • Maximum shear stress,

• According to maximum shear stress,

2

2

max 2 xyyx

22

22

2

2

2

2

22

max

/9000

/9365.673.12

2

mmNd

mmkNddd

xyx

mmdd

S yt

42.132

1009000

2 2max

According to maximum principal strain theory • The maximum principal stress,

• And minimum principal stress,

2

2

1 22 xyyxyx

2

2

2 22 xyyxyx

22

2

1

15365

22 dxyxx

222

2

2

2

222

2

2

/2635

365.6

2

73.1273.12

22

mmNd

dddxyxx

• And according to maximum principal strain theory,

mmddd

7.12

1003.0263515365

Sm

σσ

E

S

mE

σ

E

σ

22yt2

1

yt21

• According to maximum strain energy theory

• According to maximum distortion theory mmd

dddd

Sm yt

78.12

1003.0263515365

2263515365

2

222

2

2

2

2

22122

21

mmd

dddd

S yt

4.13

263515365263515365100

22

2

2

2

2

212

22

1