lecture 3 mohr s circle and theory of failure

7/30/2019 Lecture 3 Mohr s Circle and Theory of Failure

1/16


2/16

Unit 1- Stress and Strain

Lecture -1 - Introduction, state of plane stress Lecture -2 - Principle Stresses and Strains Lecture -3 - Mohr's Stress Circle and Theory of

Failure

Lecture -4- 3-D stress and strain, Equilibriumequations and impact loading

Lecture -5 - Generalized Hook's law and Castigliono's

Topics Covered


3/16

Mohr Stress Circle

n=

1+

2

2+

1

2

2cos2+sin2

t=

1

2

2sin2 cos2

We derived these two equations- These equations represent the equation of

a circle

n

1+

2

2

2

=

1

2

2

cos2+sin2

2

t

( )2

=

1

2

2sin2 cos2

2


4/16

Mohr Stress Circle

n

1+

2

2

2

=

1

2

2cos2+sin2

2

t( )2

=

1

2

2 sin2 cos2

2

Add above 2 equations. We will equation of circle.

n1+

2

2

2

+t

2=

1

2

2

2

+ ( )2

x a( )2

y2

r2

Equation of circle


5/16

Mohr Stress Circle Graphical method to determine stresses.

Body subjected to two mutually perpendicular principalstresses of unequal magnitude.

Body subjected to two mutually perpendicular principalstresses of unequal magnitude and unlike (one tensile

and other compressive).

Body subjected to two mutually perpendicular principalstresses + simple shear stress.


6/16

Mohr Stress Circle Body subjected to two mutually perpendicularprincipal stresses of unequal magnitude

O 1

2

1

2n

t

A C

B

D

E

(1 - 2 ) length AD =

=n

Normal stress on oblique plane

length ED = Tangential stress on Oblique plane

length AE = Resultant stress on Oblique plane=

t

= t

2+

n

2


7/16

Mohr Stress Circle Body subjected to two mutually perpendicular principal

stresses of unequal magnitude and unlike (one tensile

and other compressive).

2

1

2n

t

A

C B

D

E

(1+2 )

length AD =

=n




t

= t

2+

n

2O

+_


8/16

Mohr Stress Circle

Body subjected to two mutually perpendicular principalstresses + simple shear stress.

1

2

1

2n

t

A C

BD

E

length AD =

=n




t

= t

2+

n

2L

M

O


9/16

Theories of failure Maximum principal stress (Rankine theory) Maximum principal strain (Saint Venant theory) Maximum shear stress (Guest theory) Maximum strain energy (Haigh theory) Maximum shear strain energy (Mises & Henky

theory)


10/16

1. Maximum principal

stress theory

1,

2,

3=principal stresses in 3 perpendicular

directions

*

Maximum principal stress should be less than the max stress (yield stress) that material

can bear in tension or compression.

max(1,2,3) *

= max tensile or compressive strength of material

max principal stress=

*

safety _ factor


11/16

2. Maximum principal

strain theory

1,

2,

3=principal stresses in 3 perpendicular

directions

*

Maximum principal strain should be less than the max strain (yield strain) that material

can bear in tension or compression.

e1=

1

E2

E3

E

= max tensile or compressive strength of material

max principal stress=

*

safety _ factor

e2=

2

E1

E3

E

e3=

3

E1

E2

E

max(e1,e

2,e

3) e

*

e*=

*

E


12/16

3. Maximum shear stress

theory

=

1

21

3( )

t

*

Maximum shear stress should be less than the max shear stress in simple tension (at

elastic limit) that material can bear.

= max tensile of material

allowable stress =t

*

safety _ factor

=

1

2(

t

* 0)

max shear stress =half the difference of max and min principal stresses

To prevent failure max shear stress should be less that shear

stress in simple tension at elastic limit

max shear stress at elastic limit

(

1

3)

t

*

In simple tension the stressis existing in one direction


13/16

4. Maximum strain

energy theoryStrain energy per unit volume should be less than the strain energy per unit volume in

simple tension (at elastic limit) that material can bear.

max allowable stress=t

*

safety _ factor

1

2+

2

2+

3

2 2

1

2+

1

3+

2

3( )[ ] t*( )2


14/16

5. Maximum shear strain

energy theoryShear strain energy per unit volume should be less than the shear strain energy per unit

volume in simple tension (at elastic limit).

max allowable stress=t

*

safety _ factor

1

2( )2

+ 1

3( )2

+ 2

3( )2

2* t

*( )2


15/16

Important points Brittle material -> Max principal stress

Brittle material do not fail in shear

Ductile material -> Max shear stress/max shear strainenergy

Ductile material fail in shear because their yieldstrength is high.


16/16

Failure Theory PROBLEM- The principal stresses at a point in an

elastic material are 200 N/mm2 (tensile), 100 N/mm2 (tensile) and 50 N/mm2 (compressive). If thestresses at the elastic limit in simple tension is 200

N/mm2, determine whether the failure of thematerial will occur according to different failuretheory. (take Poisson's ratio =0.3)

Max principal strain theory

Max shear stress theory Max strain energy theory Max shear strain energy theory

lecture 3 mohr s circle and theory of failure

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