12.1 plastic behavior in simple tension and compression （单轴拉伸下材料的塑性行为）
DESCRIPTION
D. s. B. s e. E. A. C. s p. Theory of Elasticity. s s. e. e p. e e. o. P. A 0. l 0. s ' s. subsequent yield stress （后继屈服应力）. P. Chapter. Page. 12.1 Plastic behavior in simple tension and compression （单轴拉伸下材料的塑性行为）. s b. s ' s. s p limit of proportionality( 比例极限 ). - PowerPoint PPT PresentationTRANSCRIPT
12.1 Plastic behavior in simple tension and compression （单轴拉伸下材料的塑性行为）
Chapter Page
Th
eory
of
Ela
sti
cit
y
PA0
l0
P
0A
P
l
ll 0
o
A
B
C
D
Ep
p limit of proportionality( 比例极限 )
e
e elastic limits ( 弹性极限 )
s
s initial yield stress （初始屈服应力）
b
b strength limit （强度极限）
's
''ssubsequent yield stress （后继屈服应力）'s Bauschinger Effect
（ Bauschinger 效应）
sss
12 3
ep
12.1 Plastic behavior in simple tension and compression( 单轴拉伸下材料的塑性行为）
Chapter Page
Th
eory
of
Ela
stic
ity
Loading, unloading and reloading( 加载，卸载，再加载 )
Loading （加载） :
0d
Unloading （卸载） :
0d
12 4
12.1 Plastic behavior in simple tension and compression( 单轴拉伸下材料的塑性行为）
Chapter Page
Th
eory
of
Ela
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ity
Definition of large plastic strains ( 对数、自然应变 Ludwik, 1909)
L
dld )(
00l
lIn
L
dld
l
l
True strainNatural strain( 对数，自然应变 )
Definition of true stress （真实应力）
A
P
_
Assume the materials is incompressible （假设材料不可压缩）
00lAAl el
l
A
P
00
)1()(0
Inl
lIn .....
!32
32
For small deformations,
512
12.1 Plastic behavior in simple tension and compression( 单轴拉伸下材料的塑性行为）
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Th
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of
Ela
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ity
Example 1
A bar o length l0 is stretched to a final length of 2 l0. Values of the engineering and true strain? If the bar is compressed again to its very initial length , compute the engineering and true strains.
0.12
0
00
l
ll
693.0)2
(0
0 l
lIn
5.02
2
0
00
l
ll
693.0)2
(0
0 l
lIn
The natural strain yields the same magnitude while the engineering strain magnitude is different
0
0
2
20.1
l
lL
0L
Physically impossible
真实应变关于原点对称。
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12.1 Plastic behavior in simple tension and compression( 单轴拉伸下材料的塑性行为）
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Example 2 A bar of 100 mm initial length is elongated to a length of 200 mm by drawing in three stages. The length after each stage are 120, 150 and 200mm, respectively: a) Calculate the engineering strain for each stage separately and compare the sum with the total overall value of ε b) Repeat (a) for the true strain
33.0150
50,25.0
120
30,2.0
100
20321 Solution:
78.0321 sum But 0.1100
100200
overall
29.0)150
200(,22.0)
120
150(,18.0)
100
120( 321 InInIn
69.0321 sum 69.0)100/200( InoverallThis illustrates the additive property of true strain.( 对数应变可加 )
712
12.2 Modeling of uniaxial behavior in plasticity
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Idealized stress-strain curves (1) （理想应力－应变曲线）
s
s
E
E1
Elastic linear strain-hardening( 弹、线性强化 )
sss
s
E
E
)(1
812
12.2 Modeling of uniaxial behavior in plasticity
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Th
eory
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Ela
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Idealized stress-strain curves (2) （理想应力－应变曲线）
s
s
Elastic perfectly-plastic( 理想弹塑性 )
ss
sE
912
12.2 Modeling of uniaxial behavior in plasticity
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eory
of
Ela
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Idealized stress-strain curves (3) （理想应力－应变曲线）
Rigid-perfectly plastic(理想刚塑性 )
Rigid-linear strain-hardening( 刚、线性强化 )
Exponential hardening(幂次强化模型 )
1012
12.2 Modeling of uniaxial behavior in plasticity
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eory
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Strain hardening / Working hardening （强化现象）
The effect of the material being able to withstand a greater stress after plastic deformation （塑性变形后，材料的承载能力提高）
Strain softening （软化）
1112
12.2 Modeling of uniaxial behavior in plasticity
Hardening rules(强化模型 )
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Isotropic hardening( 等向强化 )
*||
|||'| BCCB
The reversed compressive yield stress is assumed equal to the tensile yield stress. （反向压缩屈服应力等于拉伸屈服应力）
1212
12.2 Modeling of uniaxial behavior in plasticity
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Hardening rules(强化模型 )
Kinematic hardening( 随动强化 )
The elastic range is assumed to be unchanged during hardening.
|'||'| AABB The center of elastic region is moved along the straight line aa’
1312
12.2 Modeling of uniaxial behavior in plasticity
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Hardening rules(强化模型 )
Mixed hardening( 组合强化 )
1412
A combination of kinematic and isotropic hardening (Hodge, 1957)
Kinematic hardening( 随动强化 )
Isotropic hardening 等向强化
Mixed hardening 组合强化
12.3 Basics of Yield Criteria （屈服准则概述）
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O
N
P
Q
平面
H-W stress space(H-W 应力空间 )
Principal stress space （主应力空间）A possible stress state:P （ 1 、 2 、 3 ）
Orientation of the stress state is ignored（忽略主应力方向）
Hydrostatic axis （静水压力轴）Pass through the origin and making the same angle with each of the coordinate axes. （过原点和三个坐标轴夹角相等。）Deviatoric plane
Perpendicular to ON
ON3321
π- plane 0321
12 15
12.3 Basics of Yield Criteria （屈服准则概述）
O
N
P
Q
平面
OQONOP ),,(),,( 000 pppON
pI
nOPON
33
)3
1,
3
1,
3
1(),,(
1
321
),,(
)](),(),[(
),,(),,(
),,(
321
321
321
321
sss
ppp
ppp
ONOQ
822/12
32
22
1 32)( JsssOQ
Division of stress state of a point
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12.3 Basics of Yield Criteria （屈服准则概述）
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Yield criterion （屈服准则）
Defines the elastic limits of a material under combined states of stress.（在复杂应力状态下，材料的弹性极限）
Yield criterion of Simple states of stress （简单应力状态下的屈服准则）
s
s
Uniaxial tension or compression（单轴拉伸或压缩）
Pure shear （纯剪切）
f (ij,k1,K2,K3,….) = 0 Kn are material constants
General Yield criterion （通常情况下的屈服准则）
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12.3 Basics of Yield Criteria （屈服准则概述）
General Yield criterion （通常情况下的屈服准则）
f (ij,k1,K2,K3,….) = 0 Kn are material constants
Isotropic, Orientation of principal stress is immaterial
f (1 、 2 、 3 、 1 、 2 、 3,k1,K2,K3,….) = 0
f (1 ， 2 ， 3, k1,K2,K3,…. )=0
1 ， 2 ， 3, can be expressed in terms of I1,J2,J3
f (I1 ， J2 ， J3 , k1,K2,K3,…. )=0
Experimental results: hydrostatic pressure not appreciable
f(J2 ， J3 , k1,K2,K3,…. )= 0
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12 18
12.3 Basics of Yield Criteria （屈服准则概述）
Respresention of yield criteria( 屈服准则的表达 )
平面
屈服面
Since hydrostastic pressure has no effect, the yield surface in stress space is a cylinder. 静水压力不影响屈服，则屈服面在应力空间为柱形。
f(J2 ， J3 , k1,K2,K3,…. )= 0
3D case
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12 19
12.3 Basics of Yield Criteria （屈服准则概述）
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Tresca yield criterion （ Tresca 屈服准则）
f (ij) = 02
131
k
x= (s1 s3) = (1 3) = k1 22
22
2
π- plane: A straight line
12 = 2k1
13 = 2k1
23 = 2k1
e' e'
e'
平面
屈服面
1864,Tresca, first yield creterion
12 20
12.3 Basics of Yield Criteria （屈服准则概述）
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Determination of the material constant （材料常数的确定）
Simple tension, 1 =s ， 2 =3 =0
f (ij) = 02
131
k k1= s/2
Pure shear, =s 1= s ， 2=0 ， 3= s,
f (ij) = 02
131
k k1= s
s=2s
e' e'
e'
12 21
12.3 Basics of Yield Criteria （屈服准则概述）
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Von-mises yield criterion （ Von-mises 屈服准则）
1913, Von mises
0222 kJf (ij) =
r= k2 =const ， 22 2 J
e' e'
e' 2D case (π- plane)
平面
屈服面
3D case
Yielding begins when the octahedral shearing stress reaches a critical value k
22 3
2
3
2kJoct
J2, an invariant of the stress deviator tensor
22
213
232
221 6k
In terms of principal stresses
12 22
12.3 Basics of Yield Criteria （屈服准则概述）
Chapter Page
Th
eory
of
Ela
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Determination of the material constant （材料常数的确定）
Simple tension, 1 =s ， 2 =3 =0
Pure shear, =s 1= s ， 2=0 ， 3= s,
22
2
23
kJ s
sk 3
12
J2= k2 k2 = s
ss 3
12 23