lecture 3 mohr s circle and theory of failure
TRANSCRIPT
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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
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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
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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
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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.
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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
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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
Normal stress on oblique plane
length ED = Tangential stress on Oblique plane
length AE = Resultant stress on Oblique plane=
t
= t
2+
n
2O
+_
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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
Normal stress on oblique plane
length ED = Tangential stress on Oblique plane
length AE = Resultant stress on Oblique plane=
t
= t
2+
n
2L
M
O
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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)
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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
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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
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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
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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
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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
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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.
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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