EGM 5653
CHAPTER 4
Inelastic Material Behavior
EGM 5653
Advanced Mechanics of Materials
Namas Chandra
Advanced Mechanics of Materials
Chapter 4-1
EGM 5653
Objectives
Nonlinear material behavior
Yield criteria
Yielding in ductile materials
Sections
4.1 Limitations of Uniaxial Stress- Strain data
4.2 Nonlinear Material Response
4.3 Yield Criteria : General Concepts
4.4 Yielding of Ductile Materials
4.5 Alternative Yield Criteria
4.6 General Yielding
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Advanced Mechanics of Materials
Chapter 4-2
EGM 5653
Introduction
When a material is elastic, it returns to the same state (at
macroscopic, microscopic and atomistic levels) upon removal of all
external load
Any material is not elastic can be assumed to be inelastic
E.g.. Viscoelastic, Viscoplastic, and plastic
 To use the measured quantities like yield strength etc. we need
some criteria
The criterias are mathematical concepts motivated by strong
experimental observations
E.g. Ductile materials fail by shear stress on planes of maximum
shear stress
Brittle materials by direct tensile loading without much yielding
 Other factors affecting material behavior
- Temperature
- Rate of loading
- Loading/ Unloading cycles
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Chapter 4-3
EGM 5653
Types of Loading
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Chapter 4-4
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4.2.1 Models for Uniaxial stress-strain
All constitutive equations are models that are supposed to represent
the physical behavior as described by experimental stress-strain
response
Experimental Stress strain curves
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Idealized stress strain curves
Elastic- perfectly plastic response
Chapter 4-5
EGM 5653
4.2.1 Models for Uniaxial stress-strain contd.
.
Linear elastic response
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Elastic strain hardening response
Chapter 4-6
EGM 5653
4.2.1 Models for Uniaxial stress-strain contd.
Rigid models
Rigid- perfectly plastic
response
.
Rigid- strain hardening plastic
response
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Chapter 4-7
EGM 5653
Ideal Stress Strain Curves
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Chapter 4-8
EGM 5653
4.2.1 Models for Uniaxial stress-strain contd.
4.4 The members AD and CF are made of
elastic- perfectly plastic structural steel, and member BE
is made of 7075 –T6 Aluminum alloy. The members
each have a cross-sectional area of 100 mm2.Determine
the load P= PY that initiates yield of the structure and the
fully plastic load PP for which all the members yield.
Soln:
.
Contd..
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Advanced Mechanics of Materials
Chapter 4-9
EGM 5653
4.2.1 Models for Uniaxial stress-strain contd.
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Advanced Mechanics of Materials
Chapter 4-10
EGM 5653
4.3 The Yield Criteria : General concepts
General Theory of Plasticity defines
Yield criteria : predicts material yield under multi-axial state of stress
Flow rule : relation between plastic strain increment and stress increment
Hardening rule: Evolution of yield surface with strain
Yield Criterion is a mathematical postulate and is defined by a yield
function f  f ( ij , Y )
where Y is the yield strength in uniaxial load, and is correlated with the
history of stress state.
Some Yield criteria developed over the years are:
Maximum Principal Stress Criterion:Maximum Principal Strain Criterion:Strain energy density criterion:Maximum shear stress criterion (a.k.a Tresca):Von Mises or Distortional energy criterion:Namas Chandra
Advanced Mechanics of Materials
used for brittle materials
sometimes used for brittle materials
ellipse in the principal stress plane
popularly used for ductile materials
most popular for ductile materials
Chapter 4-11
EGM 5653
4.3.1 Maximum Principal Stress Criterion
Originally proposed by Rankine
2c cos 
YC 
1  sin 
f  max   1 ,  2 ,  3   Y
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Yield surface is:
1  Y
 2  Y
 3  Y
Chapter 4-12
EGM 5653
4.3.2 Maximum Principal Strain
This was originally proposed by St. Venant
f1  1  2  3  Y  0 or
1  2  3  Y
f2  1  2  3  Y  0 or  2 1  3  Y
f3   3 1  2  Y  0 or  3 1  2  Y
Hence the effective stress may be defined as
 e  max  i   j   k
i  j k
The yield function may be defined as
f  e Y
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Chapter 4-13
EGM 5653
4.3.2 Strain Energy Density Criterion
This was originally proposed by Beltrami
Strain energy density is found as
U0 
1
 12   12   12  2  1 2   1 3   2 3    0
2E
Strain energy density at yield in uniaxial tension test
U 0Y
Y2

2E
Yield surface is given by
12  12  12  2 1 2  13   2 3   Y 2  0
f   e2  Y 2
.
 e   12   12   12  2  1 2   1 3   2 3 
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Advanced Mechanics of Materials
Chapter 4-14
EGM 5653
4.4.1 Maximum Shear stress (Tresca) Criterion
This was originally proposed by Tresca
Yield function is defined as
f  e 
Y
2
where the effective stress is
 e   max
Magnitude of the extreme values of the stresses
are
2 3
1 
2 
3 
.
2
 3  1
2
1   2
2
Conditions in which yielding
can occur in a
multi-axial stress state
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 2   3  Y
 3  1  Y
1   2  Y
Chapter 4-15
EGM 5653
4.4.2 Distortional Energy Density (von Mises) Criterion
Originally proposed by von Mises & is the most popular for ductile materials
Total strain energy density = SED due to volumetric change +SED due to distortion
U0
   2   3 
 1
18
UD
2
   2    2   3    3  1 
 1
1   2 

2
2
2
12G
2
  2   3    3  1 
12G
2
2
The yield surface is given by
1
J2  Y 2
3
f 
1
1
2
2
2
 1   2     2   3    3   1    Y 2

 3
6
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Advanced Mechanics of Materials
Chapter 4-16
EGM 5653
4.4.2 Distortional Energy Density (von Mises) Criterion contd.
Alternate form of the yield function
f   e2  Y 2
where the effective stress is
e 
1
2
2
2
1   2    2   3    3  1    3 J 2


2
2
2
1
2
2
2
2
e 












3













xx
yy
yy
zz
zz
xx
xy
yz
xz




2
J2 and the octahedral shear stress are related by
3 2
J 2    oct
2
Hence the von Mises yield criterion can be written as
f   oct 
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2
Y
3
Chapter 4-17
EGM 5653
4.4.3 Effect of Hydrostatic stress and the π- plane
Hydrostatic stress has no influence on yielding
Definition of a π- plane
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Chapter 4-18
EGM 5653
4.5 Alternate Yield Criteria
Generally used for non ductile materials like rock, soil, concrete and
other anisotropic materials
4.5.1 Mohr-Coloumb Yield Criterion
 Very useful for rock and concretes
 Yielding depends on the hydrostatic stress
f  1  3  1   3  sin   2c cos
f  max  i   j   i   j  sin    2c cos 
i j
YT 
2c cos 
1  sin 
YC 
2c cos 
1  sin 
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Advanced Mechanics of Materials
Chapter 4-19
EGM 5653
4.5.2 Drucker-Prager Yield Criterion
This is the generalization of von Mises
criteria with the hydrostatic stress effect
included
Yield function can be written as
f   I1 
J2  K
2sin 
6c cos 

,K
3(3  sin  )
3(3  sin  )
2sin 
6c cos 

,K 
3(3  sin  )
3(3  sin  )
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Advanced Mechanics of Materials
Chapter 4-20
EGM 5653
4.5.3 Hill’s Yield Criterion for Orthotropic Materials
This is the criterion is used for non-linear materials
The yield function is given by
f  F  22   33   G  33   11   H ( 11   22 )2
2
2
2
2
 L( 23
  322 )  M ( 132   312 )  N ( 122   21
) 1
1
1
1
2F  2  2  2
Z
Y
X
1
1
1
2G  2  2  2
Z
X
Y
1
1
1
2H  2  2  2
X
Y
Z
1
1
1
2 L  2 , 2M  2 , 2 N  2
S23
S13
S12
For an isotropic material
6 F  6G  6 H  L  M  N
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Chapter 4-21
EGM 5653
General Yielding
The failure of a material is when the structure cannot support the
intended function
For some special cases, the loading will continue to increase even
beyond the initial load
At this point, part of the member will still be in elastic range. When
the entire member reaches the inelastic range, then the general
yielding occurs
bh 2
PY  Ybh, M Y  Y
6
PP  Ybh  PY
bh 2
MP  Y
 1.5M Y
4
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Chapter 4-22
EGM 5653
4.6.1 Elastic Plastic Bending
Consider a beam made up of elastic-perfectly plastic material
subjected to bending. We want to find the maximum bending
moment the beam can sustain
 zz  1  k Y
(a)
where,
Y
Y 
E
h
yY 
2k
 FZ    zz dA  0
(b)
(c )
(d )
yY
h/2
0
yY
 M x  M  2   ZZ ydA  2  Yy dA  0
or
yY
M  M EP  2   zz ydA  2Y
0
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h/2

ydA
(e)
yY
Chapter 4-23
EGM 5653
4.6.1 Elastic Plastic Bending contd.
M EP
Ybh2  3 1 
3 1 

  2   MY   2 
6  2 2k 
 2 2k 
(4.43)
where, MY  Ybh / 6
2
as k becomes large
M EP 
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3
MY  M P
2
Chapter 4-24
EGM 5653
4.6.2 Fully Plastic Bending
Definition: Bending required to
cause yielding either in tension
or compression over the entire
cross section
Equilibrium condition
 F  
z
zz
dA  0
Fully plastic moment is
t b 
M P  Ybt 

2


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Chapter 4-25
EGM 5653
Comparison of failure yield criteria
For a tensile specimen
of ductile steel the
following six quantities
attain their critical
values at the same load PY
1. Maximum principal stress (  P / A) reaches the yield strength Y
2. Maximum principal strain (   / E) reaches the value Y  Y / E
3. Strain energy Uo absorbed by the material per unit volume reaches
the valueU0Y  Y 2 / 2E
4. The maximum shear stress ( max  PY / 2 A) reaches the
tresca shear strength (Y  Y / 2)
5. The distortional energy density UD reaches U DY  Y 2 / 6G
6. The octahedral shear stress  oct  2Y / 3  0.471Y
max
max
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Y
max
Chapter 4-26
EGM 5653
Failure criteria for general yielding
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Chapter 4-27
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Interpretation of failure criteria for general yielding
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Chapter 4-28
EGM 5653
Combined Bending and Loading
According to Maximum shear stress criteria, yielding starts when
Y
 
 
 
2



or

4
 
 
  1
2
2
2
Y 
2
2
2
According to the octahedral shear-stress criterion, yielding starts when
2 2  6 2
2Y
 
 

or    3    1
3
3
Y 
Y 
2
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2
Chapter 4-29
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Interpretation of failure criteria for general yielding
Comparison of von Mises and Tresca criteria
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Chapter 4-30
EGM 5653
Problem 4.24
4.24 A rectangular beam of width b and depth h is subjected to pure
bending with a moment M=1.25My. Subsequently, the moment is released.
Assume the plane sections normal to the neutral axis of the beam remain
plane during deformation.
a. Determine the radius of curvature of the beam under the applied bending
moment M=1.25My
b. Determine the distribution of residual bending stress after the applied
bending moment is released
Solution:
Namas Chandra
Advanced Mechanics of Materials
Chapter 4-31
EGM 5653
Problem 4.24 contd.
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Chapter 4-32
EGM 5653
Problem 4.24 contd.
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Chapter 4-33
EGM 5653
Problem 4.40
4.40 A solid aluminum alloy (Y= 320 Mpa)
shaft extends 200mm from a bearing support
to the center of a 400 mm diameter pulley.
The belt tensions T1and T2 vary in magnitude
with time. Their maximum values of the belt
tensions are applied only a few times during
the life of the shaft, determine the required
diameter of the shaft if the factor of safety is
SF= 2.20
Solution:
Namas Chandra
Advanced Mechanics of Materials
Chapter 4-34