同济大学弹塑性力学Ch5Constitutive

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Strain energy 应变能
Note that
δ W = σ ijδε ij
The function of strain energy is a unique function of strain 应变能（应变能密度）是状态变量应变的单值函数 Therefore, we have the following relation that
These are not yet sufficient to solve substantial problems without knowing the constitutive equations (本构关系)of the material!
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Foreword
General Hooke’s law 广义虎克定律
σ x = C11ε x + C12ε y + C13ε z + C14γ xy + C15γ yz + C16γ zx σ y = C21ε x + C22ε y + C23ε z + C24γ xy + C25γ yz + C26γ zx σ z = C31ε x + C32ε y + C33ε z + C34γ xy + C35γ yz + C36γ zx τ xy = C41ε x + C42ε y + C43ε z + C44γ xy + C45γ yz + C46γ zx τ yz = C51ε x + C52ε y + C53ε z + C54γ xy + C55γ yz + C56γ zx τ zx = C61ε x + C62ε y + C63ε z + C64γ xy + C65γ yz + C66γ zx
∂W σ ij = ∂ε ij
If we can work out W ( ε ) , then we can find the relation between σ ij and ij .
Green function 格林公式
ε
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Strain energy 应变能
V
During the loading process, the external work is positive.(外力做正功)
δ WdV > 0 ∫
Since the relation is valid for any small volume dV, then W ≥ 0 (上式对于任意小体积dV均成立,因此W ≥0)
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(1)Elastic body with a single symmetric plane
具有对称面的弹性体
Suppose there is a symmetric plane of material, stresses and strains perpendicular to this plane are symmetric. (方向垂直于对称平面的应力和应变分别是对称的)
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§ 5.1 Strain energy and constitutive relations 应变能和本构关系
Strain energy 应变能计算
1 (δ ui , j + δ u j ,i ) = δε ij 2
Strain and displacement relation External work (外力做功)is calculated as ∫ Tiδ ui ds + ∫ fiδ ui dV = ∫ σ ij n jδ ui ds + ∫ fiδ ui dV
General Hooke’s law 广义虎克定律
In the initial state, we assume the absence of strain(初始状态无应变), it requires bij = 0 and c = 0
σ ij = bij + Eijklε kl = Eijklε kl
1 W= c + bijε ij + Eijklε ijε kl 2 1 1 = Eijklε ijε kl σ ijε ij 2 2
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Linear relation between stress and strain 应力与应变呈线性关系
15
There were originally 81 possible independent components of Eijkl
s V s V
V V V V
= ∫ (σ ijδ ui ), j dV + ∫ fiδ ui dV = ∫ (σ ij , j + fi )δ ui dV + ∫ σ ijδ ui , j dV
V
σ δε dV ∫ δ WdV ∫=
ij ij V
Equilibrium(平衡方程)
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The variation is applied to strain “功”的变分对应于应变的变分
dW = σ ij d ε ij
ε ij
0
W = ∫ σ ij d ε ij
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Strain energy 应变能
ε ij
0
From W = σ ij d ε ij we can get ∫
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§ 5.2 General Hooke’s law 广义虎克定律
General Hooke’s law 广义虎克定律

c =W
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1 W= c + bij ε ij + Eijkl ε ij ε kl 2 where
ε ij = 0
(for ε ij = 0)
bij =
∂W
∂ε ij
Eijkl
ε ij = 0
∂W = ∂ε ij ∂ε kl
2
ε ij = 0
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General Hooke’s law 广义虎克定律
Since we have the commutation law
2 2 2
∂W ∂W ∂W ∂W = = = ∂ε ij ∂ε kl ∂ε ji ∂ε kl ∂ε ij ∂ε lk ∂ε kl ∂ε ij
5 Constitutive Equations 本构关系
Xiaoying Zhuang
Elasticity
Foreword
In previous lectures, we have learned strain analysis 应变分析 stress analysis 应力分析 equilibrium equations 平衡方程 compatibility equations 几何（协调）方程
= σ= σz, σ x′ σ= x , σ y′ y , σ z′ τ xy , τ y′z′ = −τ yz , τ z′x′ = −τ zx τ x′y′ =
For example if such plane is x-y plane then we will get the mapping between stresses and strains as
Let’s write the constitutive relation σ ij = Eijkl ε kl
C11 = E1111 , C12 = E1122 , C14 = E1112
Eijkl = Eklij
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Cij = C ji (Cij is not a 2nd order tensor)
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In this chapter, our goal is to build the stress and strain relations! 建立应力应变关系！
σ ij
ε ij
stress
strain
Outline
5.1 Strain energy and constitutive relations 5.2 General Hooke’s law 5.3 Anisotropic elastic body 5.4 Isotropic elastic body 5.5 Complementary energy density
= ε= εz, ε x′ ε= x , ε y′ y , ε z′ γ xy , γ y′z′ = −γ yz , γ z′x′ = −γ zx γ x′y′ =
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(1) Elastic body with a single symmetric plane
E = E = E = Eklij ijkl jikl ijlk
How comes the number of
parameters are reduced from 81 to 36 then to 21?
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General Hooke’s law 广义虎克定律
A large variety of constitutive equations for different materials Materials are too COMPLICATED… Here we only look at LINEAR ELASTIC MATERIAL （我们只关注线弹性材料）
Cijkl is the flexibility coefficient
ε ij = Cijklσ kl
柔度系数
σ ij = Eijkl ε kl
弹性系数
Flexibility coefficient
Elasticity coefficient
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§ 5.3 Anisotropic elastic body 各向异性弹性体
1 W= c + bij ε ij + Eijkl ε ij ε kl 2
E = E = E = Eklij ijkl jikl ijlk
1 ∂W bij + ( Eijkl ε kl + Eklij ε kl ) = bij + Eijkl ε kl σ ij == 2 ∂ε ij
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2
Recall that Eijkl
∂W = ∂ε ij ∂ε kl
2
ε ij = 0
E = E = E = Eklij ijkl jikl ijlk
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therefore
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Fra Baidu bibliotek
General Hooke’s law 广义虎克定律
This
∂W σ ij = ∂ε ij
Substitute the relation of stress and strain into constitutive relations on P20 we will get
σ x′ = C11ε x′ + C12ε y′ + C13ε z′ + C14γ x′y′ − C15γ y′z′ − C16γ z′x′ σ y′ = C21ε x′ + C22ε y′ + C23ε z′ + C24γ x′y′ − C25γ y′z′ − C26γ z′x′ σ z′ = C31ε x′ + C32ε y′ + C33ε z′ + C34γ x′y′ − C35γ y′z′ − C36γ z′x′ τ x′y′ = C41ε x′ + C42ε y′ + C43ε z′ + C44γ x′y′ − C45γ y′z′ − C46γ z′x′ −τ y′z′= C51ε x′ + C52ε y′ + C53ε z′ + C54γ x′y′ − C55γ y′z′ − C56γ z′x′ C61ε x′ + C62ε y′ + C63ε z′ + C64γ x′y′ − C65γ y′z′ − C66γ z′x′ ′ −τ z′x=
Reduced components
= E = E = Eklij Then E ijkl jikl ijlk
(3×3)×(3×3)=81
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?
21
6×6
16
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Question time!