文章目录
- 3. 格林应变与阿尔曼西应变
3. 格林应变与阿尔曼西应变
变形体在变形前的线元 O A → \overrightarrow{OA} OA，在变形后变成 o a → \overrightarrow{oa} oa，那么应变应该度量这种线元变形前后的差别。
∣ o a → ∣ 2 − ∣ O A → ∣ 2 = d x i d x i − d x i ′ d x i ′ = ∂ x i ∂ x j ′ d x j ′ ⋅ ∂ x i ∂ x k ′ d x k ′ − d x i ′ d x i ′ = ∂ x i ∂ x j ′ ∂ x i ∂ x k ′ d x j ′ d x k ′ − d x i ′ d x i ′ = ( ∂ x i ∂ x j ′ ∂ x i ∂ x k ′ − δ j k ) d x j ′ d x k ′ (3.1) \begin{aligned} |\overrightarrow{oa}|^2-|\overrightarrow{OA}|^2&=dx_idx_i-dx^{'}_idx^{'}_i\\ &=\frac{\partial x_i}{\partial x^{'}_j}dx^{'}_j\cdot\frac{\partial x_i}{\partial x^{'}_k}dx^{'}_k-dx^{'}_idx^{'}_i\\ &=\frac{\partial x_i}{\partial x^{'}_j}\frac{\partial x_i}{\partial x^{'}_k}dx^{'}_jdx^{'}_k-dx^{'}_idx^{'}_i\\ &=(\frac{\partial x_i}{\partial x^{'}_j}\frac{\partial x_i}{\partial x^{'}_k}-\delta_{jk}) dx^{'}_jdx^{'}_k \end{aligned}\tag{3.1} ∣oa∣2−∣OA∣2=dxidxi−dxi′dxi′=∂xj′∂xidxj′⋅∂xk′∂xidxk′−dxi′dxi′=∂xj′∂xi∂xk′∂xidxj′dxk′−dxi′dxi′=(∂xj′∂xi∂xk′∂xi−δjk)dxj′dxk′(3.1)
上式中括号中为度量线元变化的无量纲度量，将此取为应变度量，即格林应变
E i j = 1 2 ( ∂ x k ∂ x i ′ ∂ x k ∂ x j ′ − δ i j ) (3.2) E_{ij}=\frac{1}{2}(\frac{\partial x_k}{\partial x^{'}_i}\frac{\partial x_k}{\partial x^{'}_j}-\delta_{ij}) \tag{3.2} Eij=21(∂xi′∂xk∂xj′∂xk−δij)(3.2)
将其写成矩阵形式，如下所示
E = 1 2 ( F T F − I ) (3.3) E=\frac{1}{2}(F^TF-I) \tag{3.3} E=21(FTF−I)(3.3)
格林应变以变形前的位形作为参考状态，同样可以用变形后的位形作为参考状态，那么线元前后的变化如下式所示
∣ o a → ∣ 2 − ∣ O A → ∣ 2 = d x i d x i − d x i ′ d x i ′ = δ i j d x i d x j − ∂ x i ′ ∂ x j d x j ⋅ ∂ x i ′ ∂ x k d x k = δ i j d x i d x j − ∂ x i ′ ∂ x j ∂ x i ′ ∂ x k d x j d x k = ( δ j k − ∂ x i ′ ∂ x j ∂ x i ′ ∂ x k ) d x j d x k (3.4) \begin{aligned} |\overrightarrow{oa}|^2-|\overrightarrow{OA}|^2&=dx_idx_i-dx^{'}_idx^{'}_i\\ &=\delta_{ij} dx_idx_j-\frac{\partial x^{'}_i}{\partial x_j}dx_j\cdot\frac{\partial x^{'}_i}{\partial x_k}dx_k\\ &=\delta_{ij} dx_idx_j-\frac{\partial x^{'}_i}{\partial x_j}\frac{\partial x^{'}_i}{\partial x_k}dx_jdx_k\\ &=(\delta_{jk}-\frac{\partial x^{'}_i}{\partial x_j}\frac{\partial x^{'}_i}{\partial x_k}) dx_jdx_k \end{aligned}\tag{3.4} ∣oa∣2−∣OA∣2=dxidxi−dxi′dxi′=δijdxidxj−∂xj∂xi′dxj⋅∂xk∂xi′dxk=δijdxidxj−∂xj∂xi′∂xk∂xi′dxjdxk=(δjk−∂xj∂xi′∂xk∂xi′)dxjdxk(3.4)
定义应变度量，即阿尔曼西应变
e i j = 1 2 ( δ i j − ∂ x k ′ ∂ x i ∂ x k ′ ∂ x j ) (3.5) e_{ij}=\frac{1}{2}(\delta_{ij}-\frac{\partial x^{'}_k}{\partial x_i}\frac{\partial x^{'}_k}{\partial x_j}) \tag{3.5} eij=21(δij−∂xi∂xk′∂xj∂xk′)(3.5)
将其写成矩阵形式，如下所示
e = 1 2 ( I − F − T F − 1 ) (3.6) e=\frac{1}{2}(I-F^{-T}F^{-1}) \tag{3.6} e=21(I−F−TF−1)(3.6)
欧拉描述和拉格朗日描述的位移如下所示
u i = x i ( x j ′ , t ) − x i ′ u i = x i − x i ′ ( x j , t ) (3.7) u_i=x_i(x^{'}_j,t)-x^{'}_i\\ u_i=x_i-x^{'}_i(x_j,t)\tag{3.7} ui=xi(xj′,t)−xi′ui=xi−xi′(xj,t)(3.7)
如果用位移来表示上述两种应变，那么有
∂ u i ∂ x j ′ = ∂ x i ∂ x j ′ − δ i j ∂ u i ∂ x j = δ i j − ∂ x i ′ ∂ x j (3.8) \frac{\partial u_i}{\partial x^{'}_j}=\frac{\partial x_i}{\partial x^{'}_j}-\delta_{ij}\\ \frac{\partial u_i}{\partial x_j}=\delta_{ij}-\frac{\partial x^{'}_i}{\partial x_j}\tag{3.8} ∂xj′∂ui=∂xj′∂xi−δij∂xj∂ui=δij−∂xj∂xi′(3.8)
将上式代入（3.2）式，可得
E i j = 1 2 ( ∂ x k ∂ x i ′ ∂ x k ∂ x j ′ − δ i j ) = 1 2 [ ( ∂ u k ∂ x i ′ + δ k i ) ( ∂ u k ∂ x j ′ + δ k j ) − δ i j ] = 1 2 ( ∂ u k ∂ x i ′ ∂ u k ∂ x j ′ + δ k i ∂ u k ∂ x j ′ + δ k j ∂ u k ∂ x i ′ + δ k i δ k j − δ i j ) = 1 2 ( ∂ u k ∂ x i ′ ∂ u k ∂ x j ′ + ∂ u i ∂ x j ′ + ∂ u j ∂ x i ′ ) (3.9) \begin{aligned} E_{ij}&=\frac{1}{2}(\frac{\partial x_k}{\partial x^{'}_i}\frac{\partial x_k}{\partial x^{'}_j}-\delta_{ij}) \\ &=\frac{1}{2}[(\frac{\partial u_k}{\partial x^{'}_i}+\delta_{ki})(\frac{\partial u_k}{\partial x^{'}_j}+\delta_{kj})-\delta_{ij}]\\ &=\frac{1}{2}(\frac{\partial u_k}{\partial x^{'}_i}\frac{\partial u_k}{\partial x^{'}_j}+\delta_{ki}\frac{\partial u_k}{\partial x^{'}_j}+\delta_{kj}\frac{\partial u_k}{\partial x^{'}_i}+\delta_{ki}\delta_{kj}-\delta_{ij})\\ &=\frac{1}{2}(\frac{\partial u_k}{\partial x^{'}_i}\frac{\partial u_k}{\partial x^{'}_j}+\frac{\partial u_i}{\partial x^{'}_j}+\frac{\partial u_j}{\partial x^{'}_i})\tag{3.9} \end{aligned} Eij=21(∂xi′∂xk∂xj′∂xk−δij)=21[(∂xi′∂uk+δki)(∂xj′∂uk+δkj)−δij]=21(∂xi′∂uk∂xj′∂uk+δki∂xj′∂uk+δkj∂xi′∂uk+δkiδkj−δij)=21(∂xi′∂uk∂xj′∂uk+∂xj′∂ui+∂xi′∂uj)(3.9)
将上式写成矩阵形式，此时需要引入哈密顿算子 ∇ \nabla ∇和张量积
∇ u = [ ∂ ∂ x ∂ ∂ y ∂ ∂ z ] [ u x u y u z ] = [ ∂ u x ∂ x ∂ u y ∂ x ∂ u z ∂ x ∂ u x ∂ y ∂ u y ∂ y ∂ u z ∂ y ∂ u x ∂ z ∂ u y ∂ z ∂ u z ∂ z ] u ∇ = [ u x u y u z ] [ ∂ ∂ x ∂ ∂ y ∂ ∂ z ] = [ ∂ u x ∂ x ∂ u x ∂ y ∂ u x ∂ z ∂ u y ∂ x ∂ u y ∂ y ∂ u y ∂ z ∂ u z ∂ x ∂ u z ∂ y ∂ u z ∂ z ] (3.10) \nabla u=\begin{bmatrix} \frac{\partial}{\partial x}\\ \frac{\partial}{\partial y}\\ \frac{\partial}{\partial z} \end{bmatrix}\begin{bmatrix} u_x& u_y& u_z \end{bmatrix}=\begin{bmatrix} \frac{\partial u_x}{\partial x}& \frac{\partial u_y}{\partial x}& \frac{\partial u_z}{\partial x}\\ \frac{\partial u_x}{\partial y}& \frac{\partial u_y}{\partial y}& \frac{\partial u_z}{\partial y}\\ \frac{\partial u_x}{\partial z}& \frac{\partial u_y}{\partial z}& \frac{\partial u_z}{\partial z} \end{bmatrix}\\\quad\\ u\nabla =\begin{bmatrix}u_x\\ u_y\\ u_z \end{bmatrix}\begin{bmatrix} \frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z} \end{bmatrix}=\begin{bmatrix} \frac{\partial u_x}{\partial x}& \frac{\partial u_x}{\partial y}& \frac{\partial u_x}{\partial z}\\ \frac{\partial u_y}{\partial x}& \frac{\partial u_y}{\partial y}& \frac{\partial u_y}{\partial z}\\ \frac{\partial u_z}{\partial x}& \frac{\partial u_z}{\partial y}& \frac{\partial u_z}{\partial z} \end{bmatrix}\tag{3.10} ∇u= ∂x∂∂y∂∂z∂ [uxuyuz]= ∂x∂ux∂y∂ux∂z∂ux∂x∂uy∂y∂uy∂z∂uy∂x∂uz∂y∂uz∂z∂uz u∇= uxuyuz [∂x∂∂y∂∂z∂]= ∂x∂ux∂x∂uy∂x∂uz∂y∂ux∂y∂uy∂y∂uz∂z∂ux∂z∂uy∂z∂uz (3.10)
那么式（3.9）矩阵形式为
E = 1 2 ( ∇ 0 u + u ∇ 0 + ∇ 0 u ⋅ u ∇ 0 ) (3.11) E=\frac{1}{2}(\nabla_0 u+u\nabla_0 +\nabla_0 u\cdot u\nabla_0)\tag{3.11} E=21(∇0u+u∇0+∇0u⋅u∇0)(3.11)
其中下标表示对初始坐标求微分。
将式（3.8）代入（3.5）式，可得
e i j = 1 2 ( δ i j − ∂ x k ′ ∂ x i ∂ x k ′ ∂ x j ) = 1 2 [ δ i j − ( δ k i − ∂ u k ∂ x i ) ( δ k j − ∂ u k ∂ x j ) ] = 1 2 ( δ i j − δ k i δ k j + δ k j ∂ u k ∂ x i + δ k i ∂ u k ∂ x j − ∂ u k ∂ x i ∂ u k ∂ x j ) = 1 2 ( ∂ u j ∂ x i + ∂ u i ∂ x j − ∂ u k ∂ x i ∂ u k ∂ x j ) (3.12) \begin{aligned} e_{ij}&=\frac{1}{2}(\delta_{ij}-\frac{\partial x^{'}_k}{\partial x_i}\frac{\partial x^{'}_k}{\partial x_j}) \\ &=\frac{1}{2}[\delta_{ij}-(\delta_{ki}-\frac{\partial u_k}{\partial x_i})(\delta_{kj}-\frac{\partial u_k}{\partial x_j})]\\ &=\frac{1}{2}(\delta_{ij}-\delta_{ki}\delta_{kj}+\delta_{kj}\frac{\partial u_k}{\partial x_i}+\delta_{ki}\frac{\partial u_k}{\partial x_j}-\frac{\partial u_k}{\partial x_i}\frac{\partial u_k}{\partial x_j})\\ &=\frac{1}{2}(\frac{\partial u_j}{\partial x_i}+\frac{\partial u_i}{\partial x_j}-\frac{\partial u_k}{\partial x_i}\frac{\partial u_k}{\partial x_j}) \tag{3.12} \end{aligned} eij=21(δij−∂xi∂xk′∂xj∂xk′)=21[δij−(δki−∂xi∂uk)(δkj−∂xj∂uk)]=21(δij−δkiδkj+δkj∂xi∂uk+δki∂xj∂uk−∂xi∂uk∂xj∂uk)=21(∂xi∂uj+∂xj∂ui−∂xi∂uk∂xj∂uk)(3.12)
那么式（3.12）矩阵形式为
e = 1 2 ( ∇ u + u ∇ − ∇ u ⋅ u ∇ ) (3.13) e=\frac{1}{2}(\nabla u+u\nabla -\nabla u\cdot u\nabla)\tag{3.13} e=21(∇u+u∇−∇u⋅u∇)(3.13)
下面对格林应变和阿尔曼西应变进行一些讨论。
将式（3.1）式进行一些变化，原式
∣ o a → ∣ 2 − ∣ O A → ∣ 2 = 2 E i j d x i ′ d x j ′ (3.14) \begin{aligned} |\overrightarrow{oa}|^2-|\overrightarrow{OA}|^2&=2E_{ij}dx^{'}_idx^{'}_j \end{aligned}\tag{3.14} ∣oa∣2−∣OA∣2=2Eijdxi′dxj′(3.14)
令 d s = ∣ o a → ∣ ds=|\overrightarrow{oa}| ds=∣oa∣， d S = ∣ O A → ∣ dS=|\overrightarrow{OA}| dS=∣OA∣，那么
d s 2 − d S 2 = 2 E i j d x i ′ d x j ′ (3.15) ds^2-dS^2=2E_{ij}dx^{'}_idx^{'}_j\tag{3.15} ds2−dS2=2Eijdxi′dxj′(3.15)
将上式左右各除 d S 2 dS^2 dS2，那么有
d s 2 d S 2 − 1 = 2 E i j d x i ′ d S d x j ′ d S = 2 E i j α i ′ α j ′ (3.16) \begin{aligned} \frac{ds^2}{dS^2}-1&=2E_{ij}\frac{dx^{'}_i}{dS}\frac{dx^{'}_j}{dS}\\ &=2E_{ij}\alpha_i^{'}\alpha_j^{'} \end{aligned}\tag{3.16} dS2ds2−1=2EijdSdxi′dSdxj′=2Eijαi′αj′(3.16)
其中 α i ′ \alpha_i^{'} αi′为 O A → \overrightarrow{OA} OA单位化后的各分量。
在一维情况下，小应变为 ϵ = d s − d S d S \epsilon=\frac{ds-dS}{dS} ϵ=dSds−dS，代入可得
( ϵ + 1 ) 2 = 1 + 2 E 11 → 2 ϵ + ϵ 2 = 2 E 11 (3.17) (\epsilon+1)^2=1+2E_{11}\rightarrow2\epsilon+\epsilon^2=2E_{11}\tag{3.17} (ϵ+1)2=1+2E11→2ϵ+ϵ2=2E11(3.17)
上述都是针对线应变，现在来讨论角应变，在讨论角应变时，式（3.1）就应该改为
∣ o a → ∣ ⋅ ∣ o b → ∣ − ∣ O A → ∣ ⋅ ∣ O B → ∣ = d x i δ x i − d x i ′ δ x i ′ = ∂ x i ∂ x j ′ d x j ′ ⋅ ∂ x i ∂ x k ′ δ x k ′ − d x i ′ δ x i ′ = ∂ x m ∂ x i ′ ∂ x m ∂ x j ′ d x i ′ δ x j ′ − d x i ′ δ x i ′ = ( ∂ x m ∂ x i ′ ∂ x m ∂ x j ′ − δ i j ) d x i ′ δ x j ′ = 2 E i j d x i ′ δ x j ′ (3.1’) \begin{aligned} |\overrightarrow{oa}|\cdot |\overrightarrow{ob}|-|\overrightarrow{OA}|\cdot|\overrightarrow{OB}|&=dx_i\delta x_i-dx^{'}_i\delta x^{'}_i\\ &=\frac{\partial x_i}{\partial x^{'}_j}dx^{'}_j\cdot\frac{\partial x_i}{\partial x^{'}_k}\delta x^{'}_k-dx^{'}_i\delta x^{'}_i\\ &=\frac{\partial x_m}{\partial x^{'}_i}\frac{\partial x_m}{\partial x^{'}_j}dx^{'}_i\delta x^{'}_j-dx^{'}_i \delta x^{'}_i\\ &=(\frac{\partial x_m}{\partial x^{'}_i}\frac{\partial x_m}{\partial x^{'}_j}-\delta_{ij}) dx^{'}_i\delta x^{'}_j\\ &=2E_{ij}dx^{'}_i\delta x^{'}_j \end{aligned}\tag{3.1'} ∣oa∣⋅∣ob∣−∣OA∣⋅∣OB∣=dxiδxi−dxi′δxi′=∂xj′∂xidxj′⋅∂xk′∂xiδxk′−dxi′δxi′=∂xi′∂xm∂xj′∂xmdxi′δxj′−dxi′δxi′=(∂xi′∂xm∂xj′∂xm−δij)dxi′δxj′=2Eijdxi′δxj′(3.1’)
上式进一步改为
∣ o a → ∣ ⋅ ∣ o b → ∣ − ∣ O A → ∣ ⋅ ∣ O B → ∣ = d s δ s cos θ − d S δ S cos θ 0 = 2 E i j d x i ′ δ x j ′ (3.18) \begin{aligned} |\overrightarrow{oa}|\cdot |\overrightarrow{ob}|-|\overrightarrow{OA}|\cdot|\overrightarrow{OB}| &=ds\delta s\cos\theta-dS\delta S\cos\theta_0\\ &=2E_{ij}dx^{'}_i\delta x^{'}_j \end{aligned}\tag{3.18} ∣oa∣⋅∣ob∣−∣OA∣⋅∣OB∣=dsδscosθ−dSδScosθ0=2Eijdxi′δxj′(3.18)
上式两端除以 d S δ S dS\delta S dSδS，同时 θ 0 = π 2 \theta_0=\frac{\pi}{2} θ0=2π， α i = d x i ′ d S \alpha_i=\frac{dx^{'}_i}{dS} αi=dSdxi′， α j = δ x j ′ δ S \alpha_j=\frac{\delta x^{'}_j}{\delta S} αj=δSδxj′，因此上式改为
d s d S δ s δ S cos θ − cos θ 0 = d s d S δ s δ S cos θ = λ a λ b sin θ a b = 2 E i j α i α j (3.19) \begin{aligned} \frac{ds}{dS} \frac{\delta s}{\delta S}\cos\theta-\cos\theta_0 &=\frac{ds}{dS} \frac{\delta s}{\delta S}\cos\theta\\ &=\lambda^{a}\lambda^{b}\sin\theta_{ab}\\ &=2E_{ij}\alpha_i\alpha_j \end{aligned}\tag{3.19} dSdsδSδscosθ−cosθ0=dSdsδSδscosθ=λaλbsinθab=2Eijαiαj(3.19)
其中 sin θ a b = π 2 − θ \sin\theta_{ab}=\frac{\pi}{2}-\theta sinθab=2π−θ即为角应变，由下式确定
sin θ a b = 2 E i j α i α j λ a λ b (3.20) \sin\theta_{ab}=\frac{2E_{ij}\alpha_i\alpha_j}{\lambda^{a}\lambda^{b}}\tag{3.20} sinθab=λaλb2Eijαiαj(3.20)
格林应变张量中的 E i j E_{ij} Eij与角应变的关系中还包含线伸长量，因此不想小应变假设中的 γ i j \gamma_{ij} γij这么简单。
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