Chapter 3
Elements of the theory of plasticity Subjects of interest • Introduction/objectives • The flow curve • True stress and true strain • Yielding criteria for ductile materials • Combined stress tests • The yield locus • Anisotropy in yielding
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Objective • This chapter chapter provide provides s a basic theory theory of plasticity plasticity for the understanding of the flow curve. • Dif Differ ferenc ences es betwe between en the the true true stres stress s – tru true e strain strain curve curv e and the engineer engineering ing stress stress – engi engineer neering ing strain curves will be highlighted. • Finally the understandin understanding g of the yielding yielding criteria criteria for ductile materials will be made.
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Introduction • Plastic deformation is a non reversible process where Hooke’s law is no longer valid.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.
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s s e r t s
Elastic energy
Plastic energy
strain Plastic and elastic deformation in uniaxial tension
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The flow curve • True stress-strain curve for typical ductile materials, i.e., aluminium, show that the stress strain relationship follows up the Hooke’s law up to the yield point , σ σo . • Beyond σ σo , the metal deforms plastically with strain-hardening . This cannot be related by any simple constant of proportionality. • If the load is released from straining up to point A, the total strain will immediately decrease from ε ε1 to ε ε2 . by an amount of σ σ /E .
A
σ σ
σ σο ο
ε ε3 ε ε2 3 2
ε ε1 1
ε ε
Typical true stress-strain curves for a ductile metal.
• The strain ε is the recoverable elastic strain. Also there will be ε1 - ε ε2 a small amount of the plastic strain ε known as anelastic ε2 - ε ε3 behaviour which will disappear by time. (neglected in plasticity theories.) Suranaree University of Technology
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The flow curve • Usually the stress-strain curve on unloading from a plastic strain will not be exactly linear and parallel to the elastic portion of the curve.
σ σ
• On reloading the curve will generally bend over as the stress pass through the original value from which it was unloaded. • With this little effect of unloading and loading from a plastic strain, the stressstrain curve becomes a continuation of the hysteresis behaviour. (but generally neglected in plasticity theories.)
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A
ε ε
Stress-strain curve on unloading from a plastic strain.
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The flow curve • If specimen is deformed plastically beyond the yield stress in tension (+), and then in compression (-), it is found that the yield stress on reloading in compression is less than the original yield stress.
+ σ σ
σ σa
0
σ σb
ε ε
-
• The dependence of the yield stress on loading path and direction is called the Bauschinger effect . (however it is neglected in plasticity theories and it is assumed that the yield stress in tension and compression are the same).
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σ σa
> σ σb
Bauschinger effect
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The flow curve • A true stress – strain curve provides the stress required to cause the metal to flow plastically at any strain it is often called a ‘flow curve’. • A mathematical equation that fit to this curve from the beginning of the plastic flow to the maximum load before necking is a power expression of the type σ
= K ε n
…Eq.1
Where K is the stress at ε ε = 1.0 n is the strain – hardening exponent (slope of a log-log plot of Eq.1)
A
σ σ
σ σο ο
ε ε1 1
ε ε
Typical true stress-strain curves for a ductile metal.
Note: higher σ σo greater elastic region, but less ductility (less plastic region). Suranaree University of Technology
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Idealised flow curves Due to considerable mathematical complexity concerning the theory of plasticity, the idealised flow curves are therefore utilised to simplify the mathematics. 1) Rigid ideal plastic material : no elastic strain, no strain hardening. 2) Perfectly plastic material with an elastic region, i.e., plain carbon steel. 3) Piecewise linear (strain-hardening material) : with elastic region and strain hardening region more realistic approach but complicated mathematics. σ σ
σ σ
σ σο ο
σ σ
σ σο ο ε ε
ε ε
ε ε
(a) Rigid ideal plastic (b) Ideal plastic material (c) Piecewise linear (stain- hardening) material. with elastic region. material. Suranaree University of Technology
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True stress and true strain • The engineering stress – strain curve is based entirely on the original dimensions of the specimen This cannot represent true deformation characteristic of the material.
Stress True stress-strain curve
engineering stress-strain curve
• The true stress – strain curve is based on the instantaneous specimen dimensions.
Strain Engineering stress-strain and true stress-strain curves.
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The true strain According to the concept of unit linear strain,
e=
∆ L Lo
=
1
Lo
∫
L
Lo
…Eq.2
dL
This satisfies for elastic strain where ∆L is very small, but not for plastic strain.
Definition: true strain or natural strain (first proposed by Ludwik ) is the change in length referred to the instantaneous gauge length.
ε
=∑
Hence the relationship between the true strain and the conventional linear L1 − Lo L2 − L1 L3 − L2 + + + ... strain becomes ∆ L L − Lo L Lo L1 L2 e=
ε
L
dL
Lo
L
=∫
= ln
L
Lo
Lo
e +1 = …Eq.3 …Eq.4
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ε
= ln
=
Lo
=
Lo
−1
L Lo L
Lo
= ln(e + 1) May-Aug 2007
Comparison of true strain and conventional linear strain True stain
ε ε
Conventional strain e
0.01
0.10
0.20
0.50
1.0
4.0
0.01
0.105
0.22
0.65
1.72
53.6
• In true strain, the same amount of strain (but the opposite sign) is produced in tension and compression respectively.
Ex: Expanding the cylinder to twice its length.
Ex: Compression to half the original length. Compression
Tension ε
= ln(2 Lo / Lo ) = ln 2
ε
= ln[( Lo / 2) Lo ] = − ln 2 …Eq.6
…Eq.5 Suranaree University of Technology
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Total true strain and conventional strain Increment
Length of rod
0
50
1
55
2
60.5
e1-2 = 5.5/55=0.1
3
66.5
e2-3 = 6.05/60.5= 0.1
eo-1 = 5/50=0.1
The total conventional strain e • The total conventional strain e0-3 is not equal to e0-1 + e1-2 + e2-3.
e0−1 + e1− 2 + e2−3 = 0.3 ≠ e0−3 = 16.55 / 50 = 0.331
…Eq.7
The total true strain ε ε • The total true strain = the summation of the incremental true strains. ε 0 −1
+ ε 1− 2 + ε 2−3 = ln
55 50
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+ ln
60.5 55
+ ln
66.55 60.5
= ln
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66.55 50
= ε 0−3 = 0.286
…Eq.8
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The volume strain According to the volume strain ∆
∆=
∆V V
=
(1 + e x )(1 + e y )(1 + e z )d x d y d z − d x d y d z
d x d y d z
∆ = (1 + e x )(1 + e y )(1 + e z ) − 1
…Eq.9
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 ε ε x = ln(1+e x ) ε x , hence
+ ε y + ε z = ε 1 + ε 2 + ε 3 = 0
Due to the constant volume AoLo = AL, therefore Suranaree University of Technology
ε
= ln
L Lo
= ln
…Eq.10
Ao A
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…Eq.11
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The true stress Definition: the true stress is the load divided by the instantaneous area. True stress
σ
=
P A
Engineering stress
s =
P Ao
Relationship between the true stress and the engineering stress Since
σ
=
P A
=
P Ao Ao A
Ao
But
A
Hence,
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σ
=
=
L Lo
P Ao
= e +1
(e + 1) = s (e + 1)
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…Eq.12
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Example: A tensile specimen with a 12 mm initial diameter and 50 mm gauge length reaches maximum load at 90 kN and fractures at 70 kN. The maximum diameter at fracture is 10 mm. Determine engineering stress at maximum load (the ultimate tensile strength), true fracture stress, true strain at fracture and engineering strain at fracture Engineering stress at maximum load True fracture stress True strain at fracture Engineering strain at fracture Suranaree University of Technology
P max Amax
P f A f
=
=
90 × 10 3 π (12 × 10
−3 2
) /4
70 × 10 3 π (10 × 10
−3
)/4
= 796 MPa
= 891 MPa
2
12 = ln = 2 ln 1.2 = 0.365 ε f = ln A f 10 Ao
e f = exp(ε ) − 1 = exp(0.365) − 1 = 0.44
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Yielding criteria for ductile metals • Plastic yielding of the material subjected to any external forces is of considerable importance in the field of plasticity. • For predicting the onset of yielding in ductile material, there are at present two generally accepted criteria, 1) Von Mises’ or Distortion-energy criterion
2) Tresca or Maximum shear stress criterion
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Von Mises’ criterion Von Mises proposed that yielding occur when the second invariant of the stress deviator J 2 > critical value k 2 . (σ 1 −σ 2) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2 = 6k 2 For yielding in uniaxial tension, σ σ1 = σ σo , σ σ2 = σ σ3 = 0
2σ o2 = 6k 2 , then k =
σ o
3
…Eq.13
…Eq.14
Substituting k from Eq.14 in Eq.13, we then have the von Mises’ yield criterion
[(σ −σ ) 1
2
2
+ (σ 2 − σ 3 ) + (σ 3 − σ 1 ) 2
1 2 2
]
= 2σ o
…Eq.15
In pure shear , to evaluate the constant k , note σ σ1 = σ σ3 = τ τy , σ σ2 = 0 , where σ σo is the yield stress; when yields: τ τy 2 +τ τy 2 +4τ τy 2 = 6k 2 then k = τ τy By comparing with Eq 14, we then have Suranaree University of Technology
τ y Tapany Udomphol
= 0.577σ o
***
…Eq.16 May-Aug 2007
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 σ σo = 500 MPa, will it exhibit yielding? If not, what is the safety factor? σ σz =
50 MPa
From Eq.16 σ σy =
σ σ x =
200 MPa
τ τ xy =
100 MPa
σ o
=
1 2
[(σ
x
2 2 )] − σ y ) 2 + (σ y − σ z ) 2 + (σ z − σ x ) 2 + 6(τ xy2 + τ yz + τ xz
12
30 MPa
1
[(200 − 100)
σ o
=
σ o
= 224 MPa
2
2
+ (100 − (−50)) 2 + (−50 − 200) 2 + 6(30) 2 ]
12
The calculated σ σo = 224 MPa < the yield stress (500 MPa), therefore yielding will not occur. Safety factor = 500/224 = 2.2. Suranaree University of Technology
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Tresca yield criterion Yielding occurs when the maximum shear stress τ reaches τmax the value of the shear stress in the uniaxial-tension test, τ τo .
τ max
=
σ 1
− σ 3 2
…Eq.17
Where σ σ1 is the algebraically largest and σ σ3 is the algebraically smallest principal stress.
For uniaxial tension, σ σ1 = σ σo , σ σ2 = σ σ3 = 0 , and the shearing yield stress τ τo = σ σo /2 . Therefore the maximum - shear stress criterion is given by = k , σ In pure shear , σ σ1 = - σ σ3 σ2 = 0 , τ = τ τmax τy
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τ max
=
σ 1
− σ 3 2
σ 1
− σ 3 = σ o
τ y
= 0.5σ o
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= τ o =
σ o
2
…Eq.18
…Eq.19
***
…Eq.20
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Example: Use the maximum-shear-stress criterion to establish whether yielding will occur for the stress state shown in the previous example. σ σz =
50 MPa
τ max σ σy =
100 MPa
200 MPa
τ τ xy =
σ x
− σ z
=
σ o
2 2 200 − (−50) = σ o
σ o σ σ x =
=
= 250 MPa
30 MPa
The calculated value of σ σo 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.
τ y
= 0.577σ o
***
τ y
= 0.5σ o
***
2) Tresca yield criterion • Less complicated mathematical equation used in engineering design.
Note: the difference between the two criteria are approximately 1-15%. Suranaree University of Technology
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Combined stress tests In a thin-wall tube, states of stress are various combinations of uniaxial and torsion with maybe a hydrostatic pressure being introduced to produce a circumferential hoop stress in the tube. y
In a thin wall, σ , σ σ1 = - σ σ3 σ2 = 0
M T
The maximum shear-stress criterion of yielding in the thin wall tube is given by 2
P
τ τ xy σ σ x
y
2
τ σ x + 4 xy = 1 σ o σ o
σ σ x
P
x
M T
Combined tension and torsion in a thin-wall tube.
…Eq.21
The distortion-energy theory of yielding is expressed by 2
2
τ σ x + 3 xy = 1 σ o σ o Suranaree University of Technology
…Eq.22
Comparison between maximum-shear-stress theory and distortion-energy (von Mise’s) theory. May-Aug 2007
The yield locus For a biaxial plane-stress condition ( σ = 0) the von-Mise’s yield σ2 criterion can be expressed as 2
σ 1
+ σ 32 − σ 1σ 3 = σ o2
The equation is an ellipse type with -major semiaxis
2σ o
- minor semiaxis
2 σ 3 o
…Eq.23
Yield locus
• The yield locus for the maximum shear stress criterion falls inside the von Mise’s yield ellipse. • The yield stress predicted by the von Mise’s criterion is 15.5% > than the yield stress predicted by the maximum-shear-stress criterion. Comparison of yield criteria for plane stress Suranaree University of Technology
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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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