03 Elements of The Theory of Plasticity
03 Elements of The Theory of Plasticity
03 Elements of The Theory of Plasticity
Tapany Udomphol
May-Aug 2007
Objective
This chapter provides a basic theory of plasticity for the understanding of the flow curve. Differences between the true stress true strain curve and the engineering stress engineering strain curves will be highlighted. Finally the understanding of the yielding criteria for ductile materials will be made.
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Introduction
Plastic deformation is a non reversible process where Hookes law is no longer valid. One aspect of plasticity in the viewpoint of structural design is that it is concerned with predicting the maximum load, which can be applied to a body without causing excessive yielding. Another aspect of plasticity is about the plastic forming of metals where large plastic deformation is required to change metals into desired shapes.
stress
Elastic energy
Plastic energy
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3 2 1
The strain 1-2 is the recoverable elastic strain. Also there will be a small amount of the plastic strain 2-3 known as anelastic behaviour which will disappear by time. (neglected in plasticity theories.)
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a
0
-
b a > b
Bauschinger effect
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= K n
Eq.1
Where K is the stress at = 1.0 n is the strain hardening exponent (slope of a log-log plot of Eq.1) Note: higher o
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(a) Rigid ideal plastic (b) Ideal plastic material (c) Piecewise linear (stainhardening) material. with elastic region. material.
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L 1 e= = Lo Lo
Lo
dL
Eq.2
This satisfies for elastic strain where L is very small, but not for plastic strain.
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In true strain, the same amount of strain (but the opposite sign) is produced in tension and compression respectively.
= ln(2 Lo / Lo ) = ln 2
Eq.5
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= ln[( Lo / 2) Lo ] = ln 2
Eq.6
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The total conventional strain e The total conventional strain e0-3 is not equal to e0-1 + e1-2 + e2-3.
Eq.7
The total true strain = the summation of the incremental true strains.
01 + 1 2 + 23 = ln
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During plastic deformation, it is considered that the volume of a solid remain constant ( = 0)
+ 1 = (1 + e x )(1 + e y )(1 + e z ) ln 1 = 0 = ln(1 + e x ) + ln(1 + e y ) + ln(1 + e z )
But x = ln(1+ex ) x + y + z = 1 + 2 + 3 = 0 , hence Due to the constant volume AoLo = AL, therefore
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Eq.10
= ln
A L = ln o Lo A
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Eq.11
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P = A
Engineering stress
s=
P Ao
Relationship between the true stress and the engineering stress Since
P P Ao = = A Ao A
But
Ao L = = e +1 A Lo
Hence,
P (e + 1) = s (e + 1) Ao
Eq.12
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[(
2) 2 + ( 2 3 ) 2 + ( 3 1 ) 2 = 6k 2 1
2 2 o = 6k 2 , then k =
Eq.13
o
3
Eq.14
Substituting k from Eq.14 in Eq.13, we then have the von Mises yield criterion
[(
2) + ( 2 3 ) + ( 3 1 )
2 2
1 2 2
= 2 o
Eq.15
In pure shear, to evaluate the constant k, note 1 = 3 = y , 2 = 0, where o is the yield stress; when yields: y2+y2+4y2 = 6k2 then k = y By comparing with Eq 14, we then have
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y = 0.577 o
*** Eq.16
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Example: Stress analysis of a spacecraft structural member gives the state of stress shown below. If the part is made from 7075-T6 aluminium alloy with o = 500 MPa, will it exhibit yielding? If not, what is the safety factor?
z = 50 MPa
From Eq.16
y = 100 MPa
o =
1 2
[(
2 2 2 y ) 2 + ( y z ) 2 + ( z x ) 2 + 6( xy + yz + xz )
12
x = 200 MPa
xy = 30 MPa
o =
1 2
[(200 100)
12
o = 224 MPa
The calculated o = 224 MPa < the yield stress (500 MPa), therefore yielding will not occur. Safety factor = 500/224 = 2.2.
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max =
1 3
2
Eq.17
Where 1 is the algebraically largest and 3 is the algebraically smallest principal stress.
max = 1 3
2 =o =
For uniaxial tension, 1 = o, 2 = 3 = 0, and the shearing yield stress o = o/2. Therefore the maximum - shear stress criterion is given by In pure shear, 1 = -3 = k , 2 = 0, max = y
o
2
Eq.18
1 3 = o
Eq.19
y = 0.5 o
***
Eq.20
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max =
y = 100 MPa
x z
2 2 200 (50) = o
o = 250 MPa
x = 200 MPa xy = 30 MPa
The calculated value of o is less then the yield stress (500 MPa), therefore yielding will not occur.
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Summation
1) Von Mises yield criterion Yielding is based on differences of normal stress, but independent of hydrostatic stress. Complicated mathematical equations. Used in most theoretical work.
y = 0.577 o
***
2) Tresca yield criterion Less complicated mathematical equation used in engineering design.
y = 0.5 o
***
Note: the difference between the two criteria are approximately 1-15%.
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MT
xy
x
MT
x o
+ 4 xy = 1 o
Eq.21
x o
xy + 3 o
=1
Eq.22
Comparison between maximum-shear-stress theory and distortion-energy (von Mises) theory. May-Aug 2007
12 + 32 1 3 = o2
The equation is an ellipse type with -major semiaxis
2 o
Eq.23
- minor semiaxis
2 3
Yield locus
The yield locus for the maximum shear stress criterion falls inside the von Mises yield ellipse. The yield stress predicted by the von Mises criterion is 15.5% > than the yield stress predicted by the maximum-shear-stress criterion.
Comparison of yield criteria for plane stress
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References
Dieter, G.E., Mechanical metallurgy, 1988, SI metric edition, McGraw-Hill, ISBN 0-07-100406-8. Hibbeler, R.C. Mechanics of materials, 2005, SI second edition, Person Prentice Hall, ISBN 0-13-186-638-9.
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