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Theories of Failure
Ramadas Chennamsetti
Summary
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Maximum principal stress theory
Maximum principal strain theory
Maximum strain energy theory
Distortion energy theory
Maximum shear stress theory
Octahedral stress theory
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Introduction
Failure occurs when material starts exhibiting
inelastic behavior
Brittle and ductile materials – different modes
of failures – mode of failure – depends on
loading
Ductile materials – exhibit yielding – plastic
deformation before failure
Yield stress – material property
Brittle materials – no yielding – sudden failure
Factor of safety (FS)
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Introduction
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Ductile and brittle materials
σ
σY
εe
σF
εp
Ductile material
σF
σY
ε
0.2% ε
ε
Brittle material
Well – defined yield point in ductile materials – FS on yielding
No yield point in brittle materials sudden failure – FS on failure
load
4
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σ
Introduction
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Stress developed in the material < yield stress
Simple axial load
σx
σx
If σx = σY => yielding starts – failure
σx
σx
Similarly in pure shear – only shear
stress.
If τmax = τY => Yielding in shear
Multi-axial stress state ??
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Yielding is governed by single stress
component, σx
Introduction
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Various types of loads acting at the same time
T
N
M
p
Internal pressure and external UDL
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Axial, moment and torque
Introduction
Multiaxial stress state – six stress components – one
representative value
Define effective / equivalent stress – combination of
components of multiaxial stress state
Equivalents stress reaching a limiting value – property
of material – yielding occurs – Yield criteria
Yield criteria define conditions under which yielding
occurs
Single yield criteria – doesn’t cater for all materials
Selection of yield criteria
Material yielding depends on rate of loading – static &
dynamic
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Introduction
Yield criteria expressed in terms of quantities
like stress state, strain state, strain energy etc.
Yield function => f(σij, Y), σij = stress state
If f(σij, Y)<0 => No yielding takes place – no
failure of the material
If f(σij, Y) = 0 – starts yielding – onset of yield
If f(σij, Y) > 0 - ??
Yield function developed by combining stress
components into a single quantity – effective
stress => σe
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Introduction
Equivalent stress depends on stress state and
yield criteria – not a property
Compare σe with yield stress of material
Yield surface – graphical representation of
yield function, f(σij, Y) = 0
Yield surface is plotted in principal stress
space – Haiagh – Westergaard stress space
Yield surface – closed curve
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Parameters in uniaxial tension
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Maximum principal stress
Applied stress => Y
Y
Y
σ1 = Y, σ2 = 0, σ3 = 0
Maximum shear stress
2
Y
=
2
Maximum principal strain
σ1 = Y, σ2 = 0, σ3 = 0
σ1
υ
Y
ε Y = − (σ 2 + σ 3 ) =
E E
E
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τ max =
σ1 −σ 3
Parameters in uniaxial tension
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Total strain energy density
1
1Y2
U = Yε Y =
2
2 E
Linear elastic material
Y
[σ ] =  0
 0
0
0
0
0 Y − p
0 =  0
0  0
0
−p
0
0  p
0  +  0
− p  0
0
p
0
0
0 
p
First invariant = 0 for deviatoric part => p = Y/3
U = UD + UV
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Distortional energy
Parameters in uniaxial tension
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Volumetric strain energy density, UV = p2/2K
UD
Y 2 (1 − 2υ )Y 2
Y2
Y2
(3 − 1 + 2υ ) =
(1 + υ )
=
−
=
2E
6E
6E
3E
Y2
UD =
6G
Similarly for pure shear also
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UV
(
p2
Y2
1 − 2υ ) 2
=
=
=
Y
2 K 18 K
6E
U D = U −UV
Failure theories
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Failure mode –
Mild steel (M. S) subjected to pure tension
M. S subjected to pure torsion
Cast iron subjected to pure tension
Cast iron subjected to pure torsion
Max. principal stress theory – Rankine
Max. principal strain theory – St. Venants
Max. strain energy – Beltrami
Distortional energy – von Mises
Max. shear stress theory – Tresca
Octahedral shear stress theory
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Theories of failure
Max. principal stress theory
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Maximum principal stress reaches tensile yield
stress (Y)
For a given stress state, calculate principle
stresses, σ1, σ2 and σ3
Yield function
If,
f < 0
2
,σ
3
)− Y
no yielding
f = 0 onset of yielding
f > 0
not defined
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f = max (σ 1 , σ
Max. principal stress theory
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Yield surface –
Represent six surfaces
σ2
Yield surface
Y
Yield strength – same in
tension and compression
Y
σ1
Y
σ3
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σ 1 = ± Y => σ 1 + Y = 0, σ 1 − Y = 0
σ 2 = ± Y => σ 2 + Y = 0, σ 2 − Y = 0
σ 3 = ± Y => σ 3 + Y = 0, σ 3 − Y = 0
Max. principal stress theory
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In 2D case, σ3 = 0 – equations become
σ 1 = ± Y => σ 1 + Y = 0, σ 1 − Y = 0
σ 2 = ± Y => σ 2 + Y = 0, σ 2 − Y = 0
Y
σ2
-Y
Closed curve
Y
σ1
Stress state inside – elastic, outside => Yielding
Pure shear test => σ1 = + τY, σ2 = - τY
-Y
From the above => σY = τY
τY
τy
Experimental results – Yield stress in shear
is less than yield stress in tension
Predicts well, if all principal stresses are tensile
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Pure shear
τy
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For tension => σ1 = + σY
Max. principal strain theory
“Failure occurs at a point in a body when the
maximum strain at that point exceeds the value
of the maximum strain in a uniaxial test of the
material at yield point”
‘Y’ – yield stress in uniaxial tension, yield
strain, εy = Y/E
The maximum strain developed in the body
due to external loading should be less than this
Principal stresses => σ1, σ2 and σ3 strains
corresponding to these stress => ε1, ε2 and ε3
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Max. principal strain theory
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Strains corresponding to principal stresses -
ε2 =
ε3 =
E
σ2
E
σ3
E
−
−
−
υ
E
υ
E
υ
E
(σ 2 + σ 3 )
(σ 1 + σ 3 )
(σ 2 + σ 1 )
Maximum of this should be
less than εy
For onset of yielding
Y
=> σ1 −υ(σ 2 + σ 2 ) = ±Y
E
Y
ε 2 = => σ 2 −υ(σ 3 + σ1 ) = ±Y
E
Y
ε 3 = => σ 3 −υ(σ1 + σ 2 ) = ±Y
E
ε1 =
There are six equations –
each equation represents a
plane
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ε1 =
σ1
Max. principal strain theory
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Yield function
f = max σ i − υσ j − υσ k − Y , i, j, k = 1, 2, 3
i≠ j≠k
f = σe −Y
σ e = max σ i − υσ j − υσ k
For 2D case
i≠ j ≠k
σ 2 − υσ 1 = Y => σ 2 − υσ 1 = ±Y
There are four equations, each equation represents a
straight line in 2D stress space
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σ 1 − υσ 2 = Y => σ 1 − υσ 2 = ±Y
Max. principal strain theory
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Equations –
σ 1 − υσ 2 = Y , σ 1 − υσ 2 = −Y
σ 2 − υσ 1 = Y , σ 2 − υσ 1 = −Y
Plotting in stress space
σ2 σ 2 −νσ 1 = σ Y
σ 1 −νσ 2 = −σ Y
σy
σ1
σ 1 −νσ 2 = σ Y
Failure – equivalent
stress falls outside yield
surface
σ 2 −νσ 1 = −σ Y
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σy
Max. principal strain theory
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Biaxial loading
-σ2
For onset of yielding –
2
= σ (1 + υ
Y = σ (1 + υ
)
)
Maximum principal stress theory –
Y=σ
σ1
σ1 = | σ2 |= σ
Max. principal strain theory predicts smaller value of stress
than max. principal stress theory
Conservative design
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Y = σ 1 − υσ
Max. principal strain theory
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Pure shear
Principal stresses corresponding to
shear yield stress
τy
σ1 = +τy , σ2 = -τy
For onset of yielding – max. principal
strain theory
Relation between yield stress in tension and shear
τy = Y/ (1 + υ) for υ = 0.25
τy = 0.8Y Not supported by experiments
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Y = τy + υ τy = τy (1 + υ)
Strain energy theory
“Failure at any point in a body subjected to a
state of stress begins only when the energy
density absorbed at that point is equal to the
energy density absorbed by the material when
subjected to elastic limit in a uniaxial stress
state”
In uniaxial stress (yielding)
ε
σ = Eε => Hooke’s law
Strain energy density,
U =
∫σ
ij
d ε ij => U =
1 Y 2
U =
2 E
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y
∫ σdε
0
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Strain energy theory
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Body subjected to external loads => principal
stresses
σ
2
Strain energy associated with principal
stresses
1
U = (σ 1ε 1 + σ 2 ε 2 + σ 3 ε 3 )
2
ε2 =
ε3 =
σ1
E
σ2
E
σ2
E
−
−
−
υ
E
υ
E
υ
E
σ3
(σ 2 + σ 3 )
U=
(σ 3 + σ 1 )
For onset of yielding,
(σ 1 + σ 2 )
Y2
1 2
=
σ1 + σ 22 + σ 32 − 2υ(σ1σ 2 + σ 3σ1 + σ 2σ 3 )
2E 2E
[
]
1 2
σ1 + σ 22 + σ 32 − 2υ(σ1σ 2 + σ 3σ1 + σ 2σ 3 )
2E
[
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ε1 =
σ1
]
24
Strain energy theory
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Yield function –
f = σ 12 + σ 22 + σ 32 − υ (σ 1σ 2 + σ 2σ 3 + σ 3σ 1 ) − Y 2
f = σ e2 − Y 2
Equivalent stress => σ e2 = σ 12 + σ 22 + σ 32 − υ (σ 1σ 2 + σ 2σ 3 + σ 3σ 1 )
For 2D stress state => σ3 = 0 – Yield function
becomes
f = σ 2 + σ 2 − υσ σ − Y 2
1
For onset of yielding => f = 0
2
1
2
σ 12 + σ 22 − υσ 1σ 2 − Y 2 = 0
Plotting this in principal stress space
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Yielding => f = 0, safe f < 0
Strain energy theory
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Rearrange the terms –
σ2
2
3
ζ
Y
This represents an ellipse –
Transform to ζ-η csys
45o
Y
Equivalent stress
inside – no failure
σ1
1
(ζ − η )
σ 1 = ζ cos 45 − η sin 45 =
2
1
(ζ + η )
σ 2 = ζ sin 45 + η cos 45 =
2
Substitute these in the above
expression
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η
 σ1   σ 2 
 σ1 σ 2 
  +   − 2υ 
 =1
Y  Y 
Y Y 
Strain energy theory
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ζ2
2
Y
(1 − υ )
+
Semi major axis – OA => a =
Semi minor axis – OB => b =
η2
2
Y
(1 + υ )
= 1 =>
Y
(1 − υ )
Y
(1 + υ )
ζ2
a
2
η
+
η2
b
2
=1
σ2
Y
A
ζ
B
45o
o
σ1
Y
Higher Poisson ratio – bigger major axis, smaller minor axis
If υ = 0 => circle of radius ‘Y’
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Simplifying,
Strain energy theory
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Pure shear
τy
Principal stresses corresponding to
shear yield stress
σ1 = +τy , σ2 = -τy
E
τy
(1 + υ ), ε 2 = − (1 + υ )
Strain energy, U τ =
E
(1 + υ ) 2τ 2
2E
y
=
1
Y
2E
2
=> Y =
2 (1 + υ )τ y
τy = 0.632 Y
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ε1 =
τy
Distortional energy theory (von-Mises)
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applying uniform stress from all the
directions on a body
Large amount of strain
energy can be stored
Experimentally verified
Pressure ‘p’ applied from all sides
Pressures beyond yield
stress – no failure of material
Hydrostatic loading – change in size – volume
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Hydrostatic loading
von-Mises theory
Energy associated with volumetric change –
volumetric strain energy
Volumetric strain energy – no failure of
material
Strain energy causing material failure –
distortion energy – associated with shear –
First invariant of deviatoric stress = 0
For a given stress state estimate distortion
energy – this should be less than distortion
energy due to uniaxial tensile – safe
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von-Mises theory
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Given stress state referred to principal coordinate system –
σ1
[σ ] =  0
 0
0
σ2
0
0  σ1 − p
0
0  p
0  =  0
σ2 − p
0  +  0
σ 3   0
0
σ 3 − p  0
Firstinvariant, J1 = 0
0
p
0
0
0 
p
1
=> p = σ ii
3
Principal strains => ε1, ε2, ε3
Volumetric strain => εV = ε1+ ε2 + ε3
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(σ1 − p) + (σ 2 − p) + (σ 3 − p) = 0
von-Mises theory
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1
ε V = ε 1 + ε 2 + ε 3 = {(σ 1 + σ 2 + σ 3 ) − 2υ (σ 1 + σ 2 + σ 3 )}
E
(
1 − 2υ )
3(1 − 2υ )
(σ 1 + σ 2 + σ 3 ) =
εV =
p
E
E
1
Volumetric strain energy, U V = p ε V
2
1 3(1 − 2υ )
3(1 − 2υ ) 2 (1 − 2υ )
(σ 1 + σ 2 + σ 3 )2
UV = p
p=
p =
2
E
2E
6E
U = strain energy due to principal stresses & strains
υ
1
2
2
2
U =
σ 1 + σ 2 + σ 3 − (σ 1σ 2 + σ 2σ 3 + σ 3σ 1 )
2E
E
32
(
)
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This gives –
von-Mises theory
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Distortional energy –
σ2 - σm
σm
σm
=
σ1 - σm
+
σm
σ3 - σm
1
 1 − 2υ 
2
2
2
2
UD =
σ 1 + σ 2 + σ 3 − 2υ (σ 2σ 1 + σ 2σ 3 + σ 3σ 1 ) − 
(σ 1 + σ 2 + σ 3 )
2E
 6E 
[(
]
)
Simplifying this
[
1
2
2
2
(σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 )
UD =
12G
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UD = U - UV
von-Mises theory
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Compare this with distortion in uniaxial tensile
stress
[
Y2
1
(σ 1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ 1 )2
U D ==
=
6G 12G
2
2
2
=> 2Y 2 = (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 )
]
f = σ e2 − Y 2
[
1
2
2
2
Equivalent stress, σ = (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 )
2
2
e
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Yield function,
von-Mises theory
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Principal stresses of deviatoric shear stress, Sii
0  σ1 − p
σ2
0  =  0
σ 3   0
0
0
S1
[σ ] =  0
S2
 0
0
0
0
σ2 − p
0
0  p
0  +  0
S3   0
0  p
0  +  0
σ 3 − p  0
0
p
0
0
p
0
0
0 
p
0
0 
p
Sii = σii – p => σii = Sii + p
2Y 2 = (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 )
2
2
2
2Y 2 = ((S1 + p ) − (S 2 + p )) + ((S 2 + p ) − (S3 + p )) + ((S3 + p ) − (S1 + p ))
2
2
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σ1
[σ ] =  0
 0
von-Mises theory
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Simplifying this expression –
2Y 2 = (S1 − S 2 ) + (S 2 − S3 ) + (S 3 − S1 )
2
2
2
von-Mises criteria has square terms – result
independent of signs of individual stress
components
Von-Mises equivalent stress => +ve stress
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Hydrostatic pressure does not appear in the expression
von-Mises theory
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2D stress state => σ3 = 0
f = σ 12 + σ 22 − σ 1σ 2 − Y 2
Onset of yielding,
σ 12 + σ 22 − σ 1σ 2 = Y 2
σ2
Re-arrange the terms –
 σ 1   σ 2   σ 1σ 2 
  +  − 2  =1
Y  Y   Y 
This represents an ellipse
2
2
η
Y
A
ζ
B
45o
o
σ1
Y
Semi - major axis, OA = 2Y
2
Semi - minor axis, OB =
Y
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Yield function,
von-Mises theory
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Pure shear –
τy
Principal stresses corresponding to
shear yield stress
σ1 = +τy , σ2 = -τy
Y 2 = σ 12 + σ 22 − σ 1σ 2 = 3τ y2 => τ y = 0.577Y
Suitable for ductile materials
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Shear yield = 0.577 * Tensile yield
von-Mises theory
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Plot yield function in 3D principal stress space
σ1= σ2= σ3
σ1= σ2= σ3
σ2
n
σ2
~
A
σ1 + σ 2 + σ 3 = 0
B
Deviatoric plane
o
σ3
f = (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 ) − 2Y 2 = 0
2
2
2
Cylinder, with hydrostatic stress as axis
σ3
1 

i
+
j
+
k


~
3 ~ ~ ~
OA = σ 1 i + σ 2 j + σ 3 k
n =
~
Axis makes equal DCs with all axes
σ1
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~
~
~
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σ1
von-Mises theory
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Projection of OA on hydrostatic axis
~
~
~
~
n
= OB
 σ i + σ j + σ k . i + j + k 
 1~ 2

3 
~
~
~
~
 ~ 
OB = 
3
1
(σ 1 + σ 2 + σ 3 )
OB =
3
1
1
(
=> OB = OB n =
σ 1 + σ 2 + σ 3 )  i + j + k 
~
~
3
3 ~ ~ ~ 
(
σ1 + σ 2 + σ 3 ) 
 = p i + j + k 
+
+
OB =
i
j
k
~

~

~
3
 ~ ~
 ~ ~
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OA = OB + BA
~
~
~
BA = OA − OB
~
~
~
BA = r = radius
of cylinder
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OA . n = OA n cosθ => OA cosθ =
OA . n
von-Mises theory
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Radius of cylinder
BA = R = OA − OB =  σ 1 i + σ 2 j + σ 3 k  − p  i + j + k 
~
~
~
~
~
~
 ~
~ ~ ~ 
 2σ − σ 2 − σ 3   2σ 2 − σ 3 − σ 1   2σ 3 − σ 1 − σ 2 
R= 1
 i+ 
 j+ 
k
~
3
3
3

~ 
~ 
~
R = S1 i + S 2 j + S 3 k
Radius =>
~
~
~
R=
S12 + S 22 + S 32
First invariant of deviatoris stress tensor, J 1 = 0 => S1 + S 2 + S 3 = 0
(S1 + S 2 + S 3 )2 = 0 = S12 + S 22 + S 32 = −2(S1 S 2 + S 2 S 3 + S 3 S1 )
2
2
2
Yield criteria => (S1 − S 2 ) + (S 2 − S 3 ) + (S 3 − S1 ) = 2Y 2
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~
von-Mises theory
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Yield criteria,
Y 2 = S12 + S 22 + S 32 − S1 S 2 − S 2 S 3 − S 3 S1
Use, S12 + S 22 + S 32 = − 2(S1 S 2 + S 2 S 3 + S 3 S1 )
(
1 2
Y = S + S + S + S1 + S 22 + S 32
2
3 2
3 2
2
2
2
Y = S1 + S 2 + S 3 = R
2
2
3
Y =
R
2
2
1
2
2
(
2
3
)
)
Yielding depends on deviatoric stresses
Hydrostatic stress has no role in yielding
Ramadas Chennamsetti
42
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2
von-Mises theory
R&DE (Engineers), DRDO
Second invariant of deviatoric stress
0
S2
0
0
S 0 S2
0  => J 2 = 1
+
0
0 S2
S3 
J 2 = S1S 2 + S 2 S3 + S 3 S1
(
0 S1
+
S3 0
0
S3
)
1 2
S1S 2 + S 2 S3 + S3 S1 = − S1 + S 22 + S32
2
1 2
1 2
R2
2
2
2
2
J 2 = − S1 + S 2 + S3 => J 2 = S1 + S 2 + S3 =
2
2
2
3
Y 2 = R 2 => Y 2 = 3 J 2
2
Redfining yield function => f = 3 J 2 − Y 2
(
)
J2 Materials
(
Ramadas Chennamsetti
)
43
rd_mech@yahoo.co.in
 S1
[S ] =  0
 0
Max. shear stress theory (Tresca)
R&DE (Engineers), DRDO
“Yielding begins when the maximum shear
stress at a point equals the maximum shear
stress at yield in a uniaxial tension”
τ max = KT =
σ1 − σ 2
2
=> τ max =
Y
= KT
2
Y
Y
If maximum shear stress < Y/2 => No failure occurs
τ max = KT =
σ1 − σ 2
2
τy
=> τ max = τ y = KT
Shear yield = 0.5 Tensile yield
Ramadas Chennamsetti
44
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For pure shear, σ1 = +τy , σ2 = -τy
Tresca theory
R&DE (Engineers), DRDO
In 3D stress state – principal stresses => σ1, σ2
and σ3
Maximum shear stress
 σ1 − σ 2 σ 2 − σ 3 σ 3 − σ1 
max .

2
,
2
,
2


 σ1 − σ 2 σ 2 − σ 3 σ 3 − σ1 
 Y
f = max .
,
,
 − KT  = 
2
2 
 2
 2
f < 0 => No yielding
f = 0 => Onset of yielding
Ramadas Chennamsetti
45
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Yield function
Tresca theory
R&DE (Engineers), DRDO
Following equations are obtained
= K T =>
2
f1 (σ 1 , σ 2 ) = σ 1 − σ 2 − 2 K T ;
σ2 −σ3
= K T =>
2
f 3 (σ 2 , σ 3 ) = σ 2 − σ 3 − 2 K T ;
σ 3 −σ1
= K T =>
2
f 5 (σ 3 , σ 1 ) = σ 3 − σ 1 − 2 K T ;
σ1 −σ 2
2
= ±KT
f 2 (σ 1 , σ 2 ) = σ 1 − σ 2 + 2 K T
σ2 −σ3
2
= ± KT
f 4 (σ 2 , σ 3 ) = σ 2 − σ 3 + 2 K T
σ 3 −σ1
2
= ± KT
f 6 (σ 3 , σ 1 ) = σ 3 − σ 1 + 2 K T
Ramadas Chennamsetti
46
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σ1 −σ 2
Tresca theory
R&DE (Engineers), DRDO
Redefining yield function as,
f (σ 1 , σ 2 , σ 3 ) = f1 (σ 1 , σ 2 ). f 2 (σ 1 , σ 2 ). f 3 (σ 2 , σ 3 ). f 4 (σ 2 , σ 3 ). f1 (σ 1 , σ 2 )
) = (σ 1 − σ 2 − 2 K T )(σ 1 − σ 2 + 2 K T )
(σ 2 − σ 3 − 2 K T )(σ 2 − σ 3 + 2 K T )
(σ 3 − σ 1 − 2 K T )(σ 3 − σ 1 + 2 K T )
3
Each function represents a plane in 3D principal stress space
(
)(
)(
f (σ 1 , σ 2 , σ 3 ) = (σ 1 − σ 2 ) − 4 K T2 (σ 2 − σ 3 ) − 4 K T2 (σ 3 − σ 1 ) − 4 K T2
2
2
2
No effect of hydrostatic pressure in Tresca criteria
Ramadas Chennamsetti
)
47
rd_mech@yahoo.co.in
f (σ 1 , σ 2 , σ
Tresca theory
R&DE (Engineers), DRDO
Yield function in principal stress space
σ2
Hydrostatic axis
σ3
A
σ1
Tresca yield surface
σ1 - σ2 = 2KT
σ1
O
σ2 - σ1 = 2KT
σ2
View ‘A’ – along hydrostatic axis
Ramadas Chennamsetti
48
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σ3
Tresca theory
R&DE (Engineers), DRDO
Yield surface intersects principal axes at 2KT
Wall/plane of hexagon
Hydrostati c axis => σ 1 = σ 2 = σ 3
σ3 - σ1 = 2K
1
cos α =
=> α = 54 .73 0
3
α + θ = 90 0 => θ = 35 .26 0
A
α
θ
OA = 2 K T
C
Hydrostatic axis
O
Deviatoric plane, σ1 + σ2 + σ3 = 0
OB = OA cos θ = 2 K T
2
3
OB – projection of OA on
deviatoric plane
Ramadas Chennamsetti
49
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B
Tresca theory
R&DE (Engineers), DRDO
Tresca hexagon
2KT
2
3
A, B
D
σ3
300
D
C
OC = OD cos 30
=> OC = 2 K T
=> OC =
2 3
3 2
σ1 - σ2 = 2KT
σ1
O
σ2 - σ1 = 2KT
C
σ2
2KT
Ramadas Chennamsetti
50
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O
Tresca theory
R&DE (Engineers), DRDO
2D stress state - σ3 = 0
Each equation represents two
lines in 2D stress space
σ2
σ 1 − σ 2 = ±2 K T
σ 2 = ±2 K T
σ 1 = ±2 K T
- 2KT σ2 = 2KT C
σ1 = - 2KT
σ2 = - 2KT
450
σ1 = 2KT
450
A
2KT
B
σ1 - σ2 = 2KT
A
σ1
B
OB = OA cos 45 = 2 K T
1
= 2KT
2
OA
OC =
= 2 2KT
cos 45
Yield curve – elongated hexagon
Ramadas Chennamsetti
51
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- σ1 + σ2 = 2KT
O
2KT
O
Tresca theory
R&DE (Engineers), DRDO
2D stress state - σ3 = 0
σ2
- 2KT σ2 = 2KT C
- σ1 + σ2 = 2KT
σ1 = 2KT
O
σ1 = - 2KT
450
A
2KT
σ1
σ 1 − σ 2 = ±2 K T
σ 2 = ±2 K T
σ 1 = ±2 K T
B
σ1 - σ2 = 2KT
σ2 = - 2KT
Yield curve – elongated hexagon
Ramadas Chennamsetti
52
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Each equation represents two
lines in 2D stress space
von-Mises – Tresca theories
R&DE (Engineers), DRDO
σ1 = Y, σ2 = 0, σ3 = 0
[
]
1
(σ1 −σ2 )2 + (σ2 −σ3 )2 + (σ3 −σ1 )2
6
 σ1 − σ 2 σ 2 − σ 3 σ 3 − σ1 
K T = max 
,
,

2
2
2


von-Mises criteria => J2 = KM2 =
Tresca’s criteria =>
KM
1
=
Y,
3
1
KT = Y
2
Pure shear => σ1 = +τy, σ2 = -τy => K M = τ y = K T
1
1
KM =
Y = τ y , KT = Y = τ y
2
3
τ y = 0.577 Y (von − Mises), τ y = 0.5Y (Tresca )
von-Mises criteria predicts 15% higher shear stress than Tresca
53
Ramadas Chennamsetti
rd_mech@yahoo.co.in
Pure tension –
von-Mises – Tresca theories
R&DE (Engineers), DRDO
2D stress space – von-Mises and Tresca
σ
σ2
Y
A
ζ
A ζ
2τy
η
B
B
45o
o
45o
σ1
Y
2τy
o
3τ
Y 2 = σ 12 + σ 22 − σ 1σ 2
σ1
y
3τ y2 = σ 12 + σ 22 − σ 1σ 2
Y = max .{σ 1 − σ 2 , σ 1 , σ 2 }
τy
Yielding in uniaxial tension
Yielding in shear
Tresca – conservative
von-Mises – conservative
 σ −σ 2 σ1 σ 2 
= max . 1
,
,

2
2
2
2 

Ramadas Chennamsetti
54
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η
2
von-Mises – Tresca theories
R&DE (Engineers), DRDO
Experiments by Taylor & Quinney**
Thin walled tube subjected to axial and
torsional loads
A
P
τxy
σxx
A
T
σ xx
2
σ xx
2
 σ xx 
2
σ2 =
− 
 + τ xy
2
 2 
σxx
** Taylor and Quinney “Plastic deformation of metals”, Phil. Trans. Roy. Soc.A230, 323-362, 1931
Ramadas Chennamsetti
55
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 σ xx 
2
σ1 =
+ 
+
τ

xy
2
2


P
T
von-Mises – Tresca theories
R&DE (Engineers), DRDO
Tresca criteria
 σ xx 
2
2
2
2
2
Y = σ1 −σ 2 = 2 
 + τ xy => Y = σ xx + 4τ xy
 2 
 σ xx   τ xy 
 < 1

 + 
 Y  Y /2
2
No yielding if,
2
Y 2 = σ 12 + σ 22 − σ 1σ 2
Y 2 = σ xx2 + 3τ xy2
 σ xx   τ xy 
 < 1
No yielding if, 
 + 
 Y  Y / 3 
2
Ramadas Chennamsetti
2
56
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von-Mises criteria
von-Mises – Tresca theories
R&DE (Engineers), DRDO
Plotting these two criteria –
Experimental data shows
good agreement with vonMises theory.
0.6
von-Mises
Tresca
1
Aluminium
Mild steel
Copper
σxx/ Y
Tresca – conservative
von-Mises theory more
accurate – generally used
in design
Experiments show that for
ductile materials yield in
shear is 0.5 to 0.6 times of
yield in tensile
57
Ramadas Chennamsetti
rd_mech@yahoo.co.in
τxy/ Y
Octahedral shear stress theory
R&DE (Engineers), DRDO
Octahedral plane – makes equal angles with all
principal stress axes – direction cosines same
Shear stress acting on this plane – octahedral
shear
]
Body subjected to pure tension, σ1 = Y, σ2 = σ3 = 0
2 2
τ = Y
9
2
2
2
2
2Y = (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 )
2
oct
[
Ramadas Chennamsetti
]
58
rd_mech@yahoo.co.in
[
1
2
2
2
τ = (σ1 −σ2 ) + (σ2 −σ3 ) + (σ3 −σ1)
9
2
oct
Octahedral shear stress theory
R&DE (Engineers), DRDO
Comparing this with von-Mises theory =>
both are same
Pure shear
τy
σ1 = τy, σ2 = - τy, σ3 = 0
]
[
]
theory
Ramadas Chennamsetti
59
rd_mech@yahoo.co.in
[
1
Same as von-Mises
(σ 1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ 1 )2 = 6 τ y2
9
9
theory in pure shear
2 2
2
τ oct = τ y
3
Octahedral shear stress
2
2
2
6τ y2 = (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 )
theory => von-Mises
2
=
τ oct
Tensile & shear yield strengths
R&DE (Engineers), DRDO
Ramadas Chennamsetti
60
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Each failure theory gives a relation between
yielding in tension and shear (υ = 0.25)
Failure theories in a nut shell
Ramadas Chennamsetti
61
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R&DE (Engineers), DRDO
Ramadas Chennamsetti
62
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R&DE (Engineers), DRDO