弹塑性力学作业及答案

发表于 更新于 分类于 Waline： 阅读次数：

弹性力学第一次作业

弹塑性力学第二次作业

  1. 证明\(\varepsilon_{ijk}\)是三阶张量 \[ \begin{align} &\because\varepsilon_{ijk}=\varepsilon_{ijs}\delta_{sk}=\varepsilon_{ijs}\vec{e_s}\vec{e_k}\\ &\therefore\varepsilon_{i'j'k'}=(e_{i'}\times e_{j'})\cdot\vec{e_{k'}}=(C_{i'i}\vec{e_i}\times C_{j'j}\vec{e_j})\cdot C_{k'k}\vec{e_k}\\ &=C_{i'i}C_{j'j}C_{k'k}\varepsilon_{ijk}\\ &故\varepsilon_{ijk}为三阶张量 \end{align}\\ \]

  2. 证明\(\vec{a}\mathcal{A}=\mathcal{A}^T\vec{a}\) \[ \begin{align} \vec{a}\mathcal{A}&=a_k\vec{e_k}A_{ij}\vec{e_i}\vec{e_j}\\ &=A_{ij}a_k\delta_{ki}\vec{e_j}\\ &=A_{ij}a_i\vec{a_j}\\ \mathcal{A}\vec{a}&=A_{ji}\vec{e_i}\vec{e_j}a_k\vec{e_k}\\ &=A_{ji}a_k\delta_{jk}\vec{e_i}\\ &=A_{ji}a_j\vec{e_i}\\ 故\vec{a}&\mathcal{A}=\mathcal{A}^T\vec{a}得证 \end{align} \]

  3. 若\(\vec{a}\mathcal{A}=\mathcal{A}\vec{a}\),则\(\mathcal{A}\)是什么张量？ \[ \begin{align} \vec{a}\mathcal{A}&=A_{ij}a_i\vec{e_j}\\ \mathcal{A}\vec{a}&=A_{ij}a_j\vec{e_i}=A_{ji}a_i\vec{a_j}\\ \because \vec{a}\mathcal{A}&=\mathcal{A}\vec{a}\\ \therefore A&=A^T,A为对称张量 \end{align} \]

  4. \(\mathcal{A}\times\vec{a}\overset{?}{=}\vec{a}\times\mathcal{A}\) \[ \begin{align} \mathcal{A}\times\vec{a}&=A_{ij}\vec{e_i}\vec{e_j}\times a_k\vec{e_k}=A_{ij}a_k\varepsilon_{jks}\vec{e_i}\vec{e_s} \\ \vec{a}\times\mathcal{A}&=a_k\vec{e_k}\times A_{ij}\vec{e_i}\vec{e_j}=a_k A_{ij}\varepsilon_{kis}\vec{e_s}\vec{e_j}\\ \therefore \mathcal{A}\times\vec{a}&\neq\vec{a}\times\mathcal{A} \end{align} \]

  5. \(如果A^T=-A,则\mathcal{A}\vec{a}=-\varepsilon_{ijk}\omega_k\vec{e_i}\vec{e_j}\cdot a_s\vec{e_s}=-\varepsilon_{ijk}\omega_{k}a_j\vec{e_i}=\vec{w}\times\vec{a}\),证明最后一个等号 \[ \begin{align} \vec{\omega}\times\vec{a}&=\omega_k\vec{e_k}\times a_j\vec{e_j}\\ &=\omega_ka_j\varepsilon_{kjs}\vec{e_s}\\ &=-\varepsilon_{ijk}\omega_{k}a_j\vec{e_s}\\ &得证 \end{align} \]

  6. \[ \begin{align} &1) \vec{a}\cdot\mathcal{A}\cdot\vec{b}=a_i\vec{e_i}\cdot A_{jk}\vec{e_j}\vec{e_k}\cdot b_s\vec{e_s}=a_iA_{jk}b_s(\vec{e_i}\cdot\vec{e_j})(\vec{e_k}\cdot\vec{e_s})=a_iA_{ik}b_k\\ &2) \vec{a}\times\mathcal{A}\times\vec{b}=a_i\vec{e_i}\times A_{jk}\vec{e_j}\vec{e_k}\times b_s\vec{e_s}=a_iA_{jk}b_s\varepsilon_{ijp}\varepsilon_{ksq}\vec{e_p}\vec{e_q}\\ &3) \vec{a}\cdot\mathcal{A}\times\vec{b}=a_i\vec{e_i}\cdot A_{jk}\vec{e_j}\vec{e_k}\times b_s\vec{e_s}=a_iA_{ik}b_s\varepsilon_{ksp}\vec{e_p}\\ &4) \vec{a}\times\mathcal{A}\cdot\vec{b}=a_i\vec{e_i}\times A_{jk}\vec{e_j}\vec{e_k}\cdot b_s\vec{e_s}=a_iA_{jk}b_k\varepsilon_{ijp}\vec{e_p}\\ \end{align} \]

  7. 证明\(\mathcal{I}\underset{\times}{\times}\mathcal{A}=J(A)I-A\) \[ \begin{align} \mathcal{I}\underset{\times}{\times}\mathcal{A}&=\vec{e_i}\vec{e_i}\underset{\times}{\times}A_{jk}\vec{e_j}\vec{e_k}\\ &=A_{jk}(\vec{e_i}\times\vec{e_k})(\vec{e_i}\times\vec{e_j})\\ &=A_{jk}\varepsilon_{ikp}\varepsilon_{ijq}\vec{e_q}\vec{e_p}\\ &=(\delta_{kj}\delta_{pq}-\delta_{kq}\delta_{pj})A_{jk}\vec{e_p}\vec{e_q}\\ &=A_{jj}\vec{e_p}\vec{e_p}-A_{pq}\vec{e_p}\vec{e_q}\\ &=J(\mathcal{A})\mathcal{I}-\mathcal{A} \end{align} \]

  8. 证明\(\vec{a}\times\mathcal{A}=-(\mathcal{A}^T\times\vec{a})^T\) \[ \begin{align} \vec{a}\times\mathcal{A}&=a_i\vec{e_i}\times A_{jk}\vec{e_j}\vec{e_k}=a_i A_{jk}\varepsilon_{ijs}\vec{e_s}\vec{e_k} \\ -(A^T\times\vec{a})^T&=-(A_{kj}\vec{e_j}\vec{e_k}\times a_i\vec{e_i})^T\\ &=-(a_i A_{kj}\varepsilon_{kis}\vec{e_s}\vec{e_j})^T\\ &=-a_i A_{ks}\varepsilon_{kij}\vec{e_s}\vec{e_j}\\ &=-a_kA_{kj}\varepsilon_{jis}\vec{e_s}\vec{e_k}\\ &=a_iA_{kj}\varepsilon_{ijs}\vec{e_s}\vec{e_j}\\ &得证 \end{align} \]

  9. 证明\(\vec{a_i}\vec{a_j}\)是二阶张量 \[ \begin{align} &\vec{a_i}\vec{a_j}=C_{ik}\vec{e_k}C_{js}\vec{e_s}=C_{ik}C_{js}\vec{e_k}\vec{e_s}=C_{ik}C_{js}\delta_{ks}\\ &得证 \end{align} \]

  10. 证明\(J(\mathcal{A})=\mathcal{I}\underset{\cdot}{\cdot}\mathcal{A}\) \[ \begin{align} \mathcal{I}\underset{\cdot}{\cdot}\mathcal{A}&=\vec{e_i}\vec{e_i}\underset{\cdot}{\cdot}A_{jk}\vec{e_j}\vec{e_k}\\ &=A_{jk}(\vec{e_i}\cdot\vec{e_k})(\vec{e_i}\cdot\vec{e_j})\\ &=A_{jk}\delta_{ik}\delta_{ij}\\ &=A_{ij}\delta_{ij}\\ &=J(A)\\ &得证 \end{align} \]

  11. 证明\(\mathcal{A}^T\underset{\cdot}{\times}\mathcal{B}^T=\mathcal{A}\underset{\times}{\cdot}\mathcal{B}\) \[ \begin{align} \mathcal{A}^T\underset{\cdot}{\times}\mathcal{B}^T&=A_{ji}\vec{e_i}\vec{e_j}\underset{\cdot}{\times}B_{sk}\vec{e_k}\vec{e_s}\\ &=A_{ji}B_{sk}(\vec{e_i}\cdot\vec{e_s})(\vec{e_j}\times\vec{e_k})\\ &=A_{ji}B_{sk}\delta_{is}\varepsilon_{jkp}\vec{e_p}\\ \\ \mathcal{A}\overset{\cdot}{\times}\mathcal{B}&=A_{ij}\vec{e_i}\vec{e_j}\overset{\cdot}{\times}B_{ks}\vec{e_k}\vec{e_s}\\ &=A_{ij}B_{ks}(\vec{e_i}\times\vec{e_s})(\vec{e_j}\cdot\vec{e_k})\\ &=A_{ij}B_{ks}\varepsilon_{isp}\delta_{jk}\vec{e_p}\\ &=A_{ji}B_{sk}\varepsilon_{jkp}\delta_{is}\vec{e_p}\\ 得证 \end{align} \]

弹塑性力学第三次作业

\[ \begin{aligned} J(\nabla\times A)=J(A\times\nabla)=0证明最后一个等号\\ \nabla\times\mathcal{A}=\begin{pmatrix} \begin{array}{ll} A_{31,2}-A_{21,3}&A_{32,2}-A_{22,3}&A_{33,2}-A_{23,3}\\ A_{11,3}-A_{31,1}&A_{12,3}-A_{32,1}&A_{13,3}-A{33,1}\\ A_{21,1}-A_{11,2}&A_{22,1}-A_{12,2}&A_{23,1}-A_{13,2} \end{array} \end{pmatrix}\\ J(\nabla\times\mathcal{A})=A_{31,2}-A_{21,3}+A_{12,3}-A_{32,1}+A_{23,1}-A_{13,2}\\ \because \mathcal{A}为对称矩阵，\therefore J(\nabla\times\mathcal{A})=0 \end{aligned} \]

  1. 等式证明
  1. \[ \begin{aligned} \nabla\cdot(\mathcal{A}\cdot\vec{a})&=\partial_i \vec{e_i}\cdot(A_{jk}\vec{e_j}\vec{e_k}\cdot a_s\vec{e_s})\\ &=(A_{jk,i}a_s+A_{jk}a_{s,i})(\vec{e_i}\cdot\vec{e_j})(\vec{e_k}\vec{e_s})\\ &=(\partial_i\vec{e_i}\cdot A_{jk}\vec{e_j}\vec{e_k})\cdot a_s\vec{e_s}+(A_{jk}\vec{e_j}\vec{e_k})\underset{\cdot}{\cdot}(a_s\vec{e_s}\partial_i \vec{e_i})\\ &=(\nabla\cdot\mathcal{A})\cdot a+\mathcal{A}\underset{\cdot}{\cdot}(\vec{a}\nabla) \end{aligned} \]
  2. \[ \begin{aligned} \nabla\cdot(\mathcal{A}\times\vec{a})&=\partial_i \vec{e_i}\cdot(A_{jk}\vec{e_j}\vec{e_k}\times a_s\vec{e_s})\\ &=(A_{jk,i}a_s+A_{jk}a_{s,i})(\vec{e_i}\vec{e_j}\vec{e_k})\times(\vec{e_s})\\ &=(\partial_i\vec{e_i}\cdot A_{jk}\vec{e_j}\vec{e_k})\times a_s\vec{e_s}+A_{jk}\vec{e_j}\vec{e_k}\underset{\cdot}{\times}(\partial_i\vec{e_i}a_s\vec{e_s})\\ &=(\nabla\cdot\mathcal{A})\times\vec{a}+\mathcal{A}\underset{\cdot}{\times}(\vec{a}\nabla)\\ \end{aligned} \]
  3. \[ \begin{aligned} \nabla\times(\mathcal{A}\cdot\vec{a})&=\partial_i\vec{e_i}\times(A_{jk}\vec{e_j}\vec{e_k}\cdot a_s\vec{e_s})\\ &=(A_{jk,i}a_s+A_{jk}a_{s,i})(\vec{e_i}\times\vec{e_j})(\vec{e_k}\cdot\vec{e_s})\\ &=(\partial_i\vec{e_i}\times A_{jk}\vec{e_j}\vec{e_k})-(A_{jk}\vec{e_j}\vec{e_k})(a_s\vec{e_s}\partial_i \vec{e_i})\\ &=(\nabla\times\mathcal{A})\cdot\vec{a}-\mathcal{A}\underset{\times}{\cdot}(\vec{a}\nabla) \end{aligned} \]
  4. \[ \begin{aligned} \nabla\times(\mathcal{A}\times\vec{a})&=\partial_i\vec{e_i}\times(A_{jk}\vec{e_j}\vec{e_k}\times a_s\vec{e_s})\\ &=A_{jk,i}(\vec{e_i}\times\vec{e_j})(\vec{e_k}\times\vec{e_s})\\ &=(\partial_i\vec{e_i}\times A_{jk}\vec{e_j}\vec{e_k})\times(a_s\vec{a})-A_{jk}\vec{e_j}\vec{e_k}\underset{\times}{\times}a_s\vec{e_s}\partial_i\vec{e_i}\\ &=(\nabla\times\mathcal{A})\times\vec{a}-\mathcal{A}\underset{\times}{\times}(\vec{a}\nabla) \end{aligned} \]
  1. \[ \begin{aligned} \nabla\cdot(\mathcal{I}\underset{\times}{\times}\vec{a})\times \nabla&=\partial_i\vec{e_i}\times(\vec{e_j}\vec{e_j}\underset{\times}{\times}A_{ks}\vec{e_k}\vec{e_s})\times\partial_p\vec{e_p}\\ &=A_{ks,ip}[\vec{e_i}\times(\vec{e_j}\times\vec{e_s})][\vec{e_j}\times(\vec{e_k}\times\vec{e_p})]\\ &=A_{ks,ip}(\delta_{is}\vec{e_j}-\delta_{ij}\vec{e_s})(\delta_{jp}\vec{e_k}-\delta_{kp}\vec{e_j})\\ &=A_{ks,ip}(\delta_{is}\delta_{jp}\vec{e_j}\vec{e_k}+\delta_{ij}\delta_{kp}\vec{e_s}\vec{e_j}-\delta_{is}\delta_{kp}\vec{e_j}\vec{e_j}-\delta_{ij}\delta_{jp}\vec{e_s}\vec{e_k})\\ &=A_{ki,ij}\vec{e_j}\vec{e_k}+A_{ks,jk}\vec{e_s}\vec{e_j}-A_{ks,sk}\vec{e_j}\vec{e_j}-A_{ks,ii}\vec{e_s}\vec{e_k}\\ &=\nabla(\mathcal{A}\cdot\nabla)+(\nabla\cdot\mathcal{A})\nabla-\nabla\cdot(\nabla\cdot\mathcal{A})\mathcal{I}-\nabla^2\mathcal{A}^T\\ \end{aligned} \]
  2. \[ \begin{aligned} \mathcal{I}\underset{\times}{\times}(\nabla\times\mathcal{A}\times\nabla)&=\vec{e_i}\vec{e_i}\underset{\times}{\times}(\partial_j \vec{e_j}\times A_{ks}\vec{e_k}\vec{e_s}\times\partial_p \vec{e_p})\\ &=A_{ks,jp}\vec{e_i}\vec{e_i}\underset{\times}{\times}[(\vec{e_j}\times\vec{e_k})(\vec{e_s}\times\vec{e_p})]\\ &=A_{ks,jp}[\vec{e_i}\times(\vec{e_s}\times\vec{e_p})][\vec{e_i}\times(\vec{e_j}\times\vec{e_k})]\\ &=A_{ks,jp}(\delta_{ip}\vec{e_s}-\delta_{is}\vec{e_p})(\delta_{ik}\vec{e_s}-\delta_{jk}\vec{e_i})\\ &=A_{ks,jp}(\delta_{ip}\delta_{ik}\vec{e_s}\vec{e_s}-\delta_{ip}\delta_{jk}\vec{e_s}\vec{e_i}-\delta_{is}\delta_{ik}\vec{e_p}\vec{e_s}+\delta_{is}\delta_{jk}\vec{e_p}\vec{e_i})\\ &=A_{ki,ij}\vec{e_j}\vec{e_k}+A_{ks,jk}\vec{e_s}\vec{e_j}-A_{ks,ii}\vec{e_s}\vec{e_j}-A_{ii,jp}\vec{e_p}\vec{e_j}\\ &=\nabla(\mathcal{A}\cdot\nabla)+(\nabla\cdot\mathcal{A})\nabla-\nabla\nabla J(A)-\nabla^2A^T\\ \end{aligned} \]

  1. \[ \begin{aligned} \iiint_{\Omega}\mathcal{A}\cdot\nabla\mathrm{d}\tau=\vec{e_i}\{\iiint_\Omega\vec{b_i}\cdot\nabla\mathrm{d}\tau\}=\vec{e_i}\{\oiint_{\partial\Omega}\vec{b_i}\cdot\mathrm{d}\vec{s}\}=\oiint_{\partial\Omega}(\vec{e_i} \vec{b_i})\cdot\mathrm{d}\vec{s}=\oiint_{\partial\Omega}\mathcal{A}\cdot\mathrm{d}\vec{s} \end{aligned} \]

  2. \[ \begin{aligned} 设b_i=A_{ij}\vec{e_j},A&=\vec{e_i}\vec{b_i}\\ \iint_S\mathrm{d}\vec{s}\cdot(\nabla\times\mathcal{A})&=(\iint_s \mathrm{d}\vec{s}\cdot\nabla\times\vec{b_i})\vec{e_i}\\ &=(\iint_s(\nabla\times\vec{b_i})\mathrm{d}\vec{s})\vec{e_i}\\ &=(\oint_{\partial S}\vec{b_i}\mathrm{d}\vec{r})\vec{e_j}\\ &=\oint_{\partial S}\mathcal{A}\cdot\mathrm{d}\vec{r}\\ &=\oint_{\partial S}\mathrm{d}\vec{r}\cdot\mathcal{A} \end{aligned} \]

  3. \[ \begin{aligned} 设b_i=A_{ij}\vec{e_j},A&=\vec{e_i}\vec{b_i}\\ \iint_S(\mathcal{A}\times\nabla)\cdot\mathrm{d}\vec{s}&=\vec{e_i}[\iint_S(\vec{b_i}\nabla)\cdot\mathrm{d}\vec{s}]\\ &=-\vec{e_i}[\iint_S(\nabla\times \vec{b_i})]\cdot \mathrm{d}\vec{s}\\ &=-\vec{e_i}\oint_{\partial S}\vec{b_i}\mathrm{d}\vec{r}\\ &=-\oint_{\partial S}\mathcal{A}\cdot\mathrm{d}\vec{s} \end{aligned} \]

  4. \[ \begin{aligned} \underset{\sim}{\sigma}\underset{\cdot}{\times}\underset{\sim}{I}=0\\ \Downarrow\qquad\\ \sigma_{ij}\vec{e_i}\vec{e_j}\underset{\cdot}{\times}\vec{e_k}\vec{e_k}=0\\ \Downarrow\qquad\\ \sigma_{ij}(\vec{e_i}\vec{e_k})(\vec{e_j}\times\vec{e_k})=0\\ \Downarrow\qquad\\ \sigma_{ij}\delta_{ik}\varepsilon_{jkt}\vec{e_t}=0\\ \Downarrow\qquad\\ \sigma_{kj}\varepsilon_{jkt}\vec{e_t}=0\\ \Downarrow\qquad\\ (\sigma_{13}-\sigma_{31})\vec{e_2}+(\sigma_{21}-\sigma_{12})\vec{e_3}+(\sigma_{32}-\sigma_{23})\vec{e_1}=0\\ \sigma^T=\sigma得证 \end{aligned} \]

  5. 柱坐标系下的平衡方程

\[ \begin{aligned} &x=r\cos\theta\qquad x_2=r\sin\theta\qquad x_3=z\\ &考虑在r方向上力的平衡\\ &(\sigma_{r}+\frac{\partial \sigma_r}{\partial r}\mathrm{d}r)(r+\mathrm{d}r)\mathrm{d}\theta\mathrm{d}z-\sigma_rr\mathrm{d}\theta\mathrm{d}z+(\sigma_{zr}+\frac{\partial \sigma_{zr}}{\partial z})\mathrm{d}r(r+\frac{1}{2}\mathrm{d}r)\mathrm{d}\theta-\sigma_{zr}\mathrm{d}r(r+\frac{1}{2}\mathrm{d}r)\mathrm{d}\theta+\\ &(\sigma_{\theta r}+\frac{\partial \sigma_{\theta r}}{\partial \theta}\mathrm{d}\theta)\mathrm{d}r\mathrm{d}z\cos\frac{\mathrm{d}\theta}{2}-(\sigma_{\theta}+\frac{\partial \sigma_{\theta}}{\partial \theta}\mathrm{d}\theta)\mathrm{d}r\mathrm{d}z\sin\frac{\theta}{2}-\sigma_\theta\mathrm{d}r\mathrm{d}z\sin\frac{\mathrm{d}\theta}{2}+f_r(r+\frac{\mathrm{d}r}{2})\mathrm{d}\theta\mathrm{d}r\mathrm{d}z=0\\ &整理可得:\frac{\partial \sigma_{rr}}{\partial r}+\frac{\partial \sigma_{\theta r}}{r\partial\theta}+\frac{\partial \sigma_{zr}}{\partial r}+\frac{\sigma_{rr}-\sigma_{\theta\theta}}{r}+f_r=0\\ 同理可得： &\frac{\partial \sigma_{r\theta}}{\partial r}+\frac{\partial\sigma_{\theta\theta}}{r\partial\theta}+\frac{\partial \sigma_{z\theta}}{\partial z}+2\frac{\sigma_{r\varphi}}{r}+f_\theta=0\\ &\frac{\partial \sigma_{rz}}{\partial r}+\frac{\partial\sigma_{\theta z}}{r\partial\theta}+\frac{\partial \sigma_{zz}}{\partial z}+\frac{\sigma_{rz}}{r}+f_z=0\\ \end{aligned} \]

  1. 球坐标系下的平衡方程

\[ \begin{aligned} &x=r\sin\theta\cos\varphi\qquad y=r\sin\theta\sin\varphi\qquad z=r\cos\theta\\ &取r方向上力的平衡\\ &(\sigma_r+\frac{\partial\sigma_r}{\partial r}\mathrm{d}r)\mathrm{d}S_2-\sigma_r\mathrm{d}S_1-(\sigma_\theta+\frac{\partial\sigma_\theta}{\partial\theta}\mathrm{d}\theta)\mathrm{d}S_3\sin\frac{\mathrm{d}\theta}{2}+(\sigma_{\theta r}+\frac{\partial \sigma_{\theta r}}{\partial \theta}\mathrm{d}\theta)\mathrm{d}S_3\cos\frac{\mathrm{d}\theta}{2}-\\ &\sigma_{\theta r}\mathrm{d}{S_4}\cos\frac{\mathrm{d}\theta}{2}-(\sigma_{\varphi}+\frac{\partial \sigma_{\varphi}}{\partial\varphi}\mathrm{d}\varphi)\mathrm{d}S_5\sin(\frac{\sin\theta\mathrm{d}\varphi}{2})-\sigma_\varphi\mathrm{d}S_5\sin(\frac{\sin\theta\mathrm{d}\varphi}{2})+\\ &(\sigma_{\varphi r}+\frac{\partial \sigma_{\varphi r}}{\partial \varphi})\mathrm{d}S_5\sin(\frac{\sin\theta\mathrm{d}\varphi}{2})-\sigma_{\varphi r}\mathrm{d}S_5\cos(\frac{\sin\theta\mathrm{d}\varphi}{2})+f_r\mathrm{d}V=0\\ &整理可得:\frac{\partial \sigma_{rr}}{\partial r}+\frac{\partial \sigma_{\theta r}}{r\partial\theta}+\frac{\partial \sigma_{\varphi r}}{r\sin\theta\partial\varphi}+\frac{2\sigma_{rr}-\sigma_{\theta\theta}-\sigma_{\varphi\varphi}}{r}+\cot\theta\frac{\sigma_{r\theta}}{r}+f_r=0\\ 同理可得： &\frac{\partial \sigma_{r\theta}}{\partial r}+\frac{\partial\sigma_{\theta\theta}}{r\partial\theta}+\frac{\partial \sigma_{\varphi\theta}}{r\sin\theta\partial\varphi}+\frac{3\sigma_{r\theta}}{r}+\cot\theta\frac{\sigma_{\theta\theta}-\sigma_{\varphi\varphi}}{r}+f_\theta=0\\ &\frac{\partial \sigma_{r\varphi}}{\partial r}+\frac{\partial\sigma_{\theta\varphi}}{r\partial\theta}+\frac{\partial \sigma_{\varphi\varphi}}{r\sin\theta\partial\varphi}+\frac{3\sigma_{\varphi r}}{r}+2\cot\theta\frac{\sigma_{\theta\varphi}}{r}+f_\varphi=0\\ \end{aligned} \]

弹塑性力学第四次作业

  1. 证明\(J_2^\prime=(\sigma_1-\sigma_0)(\sigma_2-\sigma_0)+(\sigma_2-\sigma_0)+(\sigma_3-\sigma_0)(\sigma_1-\sigma_0)\),\(J_3^\prime=(\sigma_1-\sigma_0)(\sigma_2-\sigma_0)(\sigma_3-\sigma_0)\) \[ \begin{aligned} &\sigma_0=\sigma_{ii}/3\\ &\begin{pmatrix} \sigma_{11}-\sigma_{0}&\sigma_{12}&\sigma_{13}\\ \sigma_{21}&\sigma_{22}-\sigma_0&\sigma_{23}\\ \sigma_{31}&\sigma_{32}&\sigma_{33}-\sigma_0 \end{pmatrix}\\ &1)证明：\\ &J_2^\prime=\begin{vmatrix} \sigma_{11}-\sigma_{0}&\sigma_{12}\\ \sigma_{21}&\sigma_{22}-\sigma_0 \end{vmatrix}+\begin{vmatrix} \sigma_{22}-\sigma_{0}&\sigma_{23}\\ \sigma_{32}&\sigma_{33}-\sigma_{0} \end{vmatrix}+\begin{vmatrix} \sigma_{33}-\sigma_{0}&\sigma_{31}\\ \sigma_{13}&\sigma_{11}-\sigma_0 \end{vmatrix}\\ &=(\sigma_1-\sigma_0)(\sigma_2-\sigma_0)+(\sigma_2-\sigma_0)+(\sigma_3-\sigma_0)(\sigma_1-\sigma_0)-\sigma_{12}\sigma_{21}-\sigma_{23}\sigma_{32}-\sigma_{13}\sigma_{31}\\ &主坐标系下，\sigma_{12}=\sigma_{21}=0;\sigma_{13}=\sigma_{31}=0;\sigma_{32}=\sigma_{23}=0\\ &2) 证明：在主坐标系下，主方向间的剪应力为0，\\ &J_3^\prime=\begin{vmatrix} \sigma_{11}-\sigma_{0}&\sigma_{12}&\sigma_{13}\\ \sigma_{21}&\sigma_{22}-\sigma_0&\sigma_{23}\\ \sigma_{31}&\sigma_{32}&\sigma_{33}-\sigma_0 \end{vmatrix}=(\sigma_1-\sigma_0)(\sigma_2-\sigma_0)(\sigma_3-\sigma_0) \end{aligned} \]

  2. 证明\(J_2=-\frac{3}{2}\tau_0^2\)

\[ \begin{aligned} &设a=\sigma_{1}-\sigma_{2},b=\sigma_2-\sigma_3,c-\sigma_3-\sigma_1\\ &J_2^\prime=\frac{1}{9}[(a-b)(b-c)+(b-c)(c-a)+(c-a)(a-b)]\\ &=\frac{1}{9}[-a^2-b^2-c^2+ab+bc+ca]\\ &=\frac{1}{9}[-\frac{3}{2}(a^2+b^2+c^2)+\frac{1}{2}(a+b+c)^2]\\ &=-(a^2+b^2+c^2)/6\\ &=-\frac{3}{2}\tau^2 \end{aligned} \]

  1. 一点应力张量值\(\underset{\sim}{\sigma}\)

\[ \begin{pmatrix} 3&1&1\\ 1&0&2\\ 1&2&0\\ \end{pmatrix}\times10^5Mpa \]

求改点主应力和相应主方向 \[ \begin{aligned} &|\lambda E-\sigma|=\sigma_{ij}=\begin{vmatrix} \lambda-3&-1&-1\\ -1&\lambda&-2\\ -1&-2&\lambda \end{vmatrix}=0\\ &\Downarrow\\ &(\lambda-1)(\lambda-4)(\lambda+2)=0\\ &\Downarrow\\ &\lambda_1=4,\lambda_2=1,\lambda_3=-2\\ &\Downarrow\\ &\sigma_1=4\times10^5Mpa,\sigma_2=1\times 10^5Mpa,\sigma_3=-2\times10^5Mpa\\ &\Downarrow\\ &\left\{ \begin{array}{ll} l\sigma=\sigma_{11}l+\sigma_{21}m+\sigma_{31}n\\ m\sigma=\sigma_{12}l+\sigma_{22}m+\sigma_{32}n\\ n\sigma=\sigma_{13}l+\sigma_{23}m+\sigma_{33}n \end{array} \right.\\ &\Downarrow\\ &把\sigma_1代入\left\{ \begin{array}{lr} \frac{m_1}{l_1}+\frac{n_1}{l_1}=1\\ -\frac{4m_1}{l_1}+\frac{2n_1}{l_1}=-1 \end{array} \right. 故：\frac{m_1}{l_1}=\frac{3}{5},\frac{n_1}{l_1}=\frac{2}{5}\\ &l_1=\frac{1}{\sqrt{1+(\frac{m_1}{l_1})^2+(\frac{n_1}{l_1})^2}}=\frac{\sqrt{6}}{3},m_1=\frac{\sqrt{6}}{6},n_1=\frac{\sqrt{6}}{6}\\ &同理可得l_2=\frac{\sqrt{3}}{3},m_2=-\frac{\sqrt{3}}{3},n_2=-\frac{\sqrt{3}}{3},l_3=0,m_3=\frac{\sqrt{2}}{2},n_3= \frac{\sqrt{2}}{2} \end{aligned} \]

  1. 一点应力状态：\(\sigma_{11}=\sigma_{22}\sigma_{33}=0,\sigma_{12}=\tau,\sigma_{21}=\tau,\sigma_{23}=\sigma_{32}=0,\sigma_{13}=\sigma_{31}=0\)

    求\(x_1+2x_2+3x_3=0\)截面上正应力和剪应力 \[ \begin{aligned} &平面x_1+2x_2+3x_3=0的法线的方向余弦为：\\ &l=\frac{1}{\sqrt{1^2+2^2+3^2}}=\frac{1}{\sqrt{14}},m=\frac{2}{1^2+2^2+3^2}=\frac{2}{\sqrt{14}},n=\frac{3}{\sqrt{1^2+2^2+3^2}}=\frac{3}{\sqrt{14}}\\ &\left\{ \begin{array}{ll} p_{v1}=\sigma_1l+\sigma_{21}m+\sigma_{31}n\\ p_{v2}=\sigma_{12}l+\sigma_{2}m+\sigma_{32}n\\ p_{v3}=\sigma_{13}l+\sigma_{23}m+\sigma_{3}n \end{array} \right.\\ &1)\sigma_{22}=\sigma_{33}=0\\ &\sigma_v=p_{v1}l+p_{v2}m+p_{vn}n=\frac{2\tau}{7}\\ &\tau_v=\sqrt{(p_{v1}^2+p_{v2}^2+p_{v3}^2)-\sigma_v^2}=\frac{3\sqrt{6}\tau}{14}\\ &2)\sigma_{22}\neq\sigma_{33},\sigma_{22}=\sigma,\sigma_{33}=0\\ &可得，\sigma_v=\frac{2(t+\sigma)}{7} ,\tau_v=\frac{3\sqrt{6}\tau}{14}+\frac{\sqrt{10}\sigma}{14}+\frac{\sqrt{6}\tau\sigma}{7}\\ &3)\sigma_{22}=0,\sigma_{33}=\sigma\\ &可得，\sigma_v=\frac{2\tau}{14}+\frac{9\sigma}{14} ,\tau_v=\frac{3\sqrt{6}\tau}{14}+\frac{3\sqrt{5}\sigma}{14}+\frac{9\tau\sigma}{14} \\ \end{aligned} \]

弹塑性力学第五次作业

1、 已知：平面应变状态\(\gamma_{33}=\gamma_{13}=\gamma_{23}=0\)，且知道\(30^\circ,60^\circ,120^\circ\)方向上线应变，分别为\(\varepsilon_{30},\varepsilon_{60},\varepsilon_{120}\)，求\(\gamma_{11},\gamma_{22},\gamma_{12}\) \[ \begin{aligned} &由题意知\boldsymbol{\Gamma}=\begin{bmatrix} \gamma_{11}&\gamma_{12}&0\\ \gamma_{21}&\gamma_{22}&0\\ 0&0&0 \end{bmatrix}\\ &设\boldsymbol{\xi_1}=(\frac{\sqrt{3}}{2},\frac{1}{2},0)为30^\circ方向单位矢量,\boldsymbol{\xi_2}=(\frac{1}{2},\frac{\sqrt{3}}{2},0)为60^\circ方向上单位矢量\\ &\boldsymbol{\xi_3}=(-\frac{1}{2},\frac{\sqrt{3}}{2},0)为120^\circ方向上单位矢量\\ &由\varepsilon=\boldsymbol{\xi}\cdot\boldsymbol{\Gamma}\cdot\boldsymbol{\xi}可知\varepsilon_{30^\circ}=\xi_1\boldsymbol{\Gamma}\xi_1=\frac{3}{4}\gamma_{11}+\frac{\sqrt{3}}{2}\gamma_{12}+\frac{1}{4}\gamma_{22}\\ &\varepsilon_{60^\circ}=\xi_2\boldsymbol{\Gamma}\xi_2=\frac{1}{4}\gamma_{11}+\frac{\sqrt{3}}{2}\gamma_{12}+\frac{3}{4}\gamma_22\\ &\varepsilon_{30^\circ}=\xi_1\boldsymbol{\Gamma}\xi_1=\frac{1}{4}\gamma_{11}-\frac{\sqrt{3}}{2}\gamma_{12}+\frac{3}{4}\gamma_22\\ &解得\left\{ \begin{array}{ll} \gamma_{11}=\frac{3}{2}\varepsilon_{30}-\varepsilon_{60}+\frac{1}{2}\varepsilon_{120}\\ \gamma_{12}=\frac{\sqrt{3}}{3}\varepsilon_{60}-\frac{\sqrt{3}}{3}\varepsilon_{120}\\ \gamma_{22}=-\frac{1}{2}\varepsilon_{30}+\varepsilon_{60}+\frac{1}{2}\varepsilon_{120} \end{array} \right. \end{aligned} \] 2、已知位移场 \[ \left\{ \begin{array}{ll} u_1=a(3x_1^2+x_2+4x_3)\\ u_2=a(3x_1+2x_2+3x_3)\\ u_3=a(2x_1^2+x_2+4x_3^2) \end{array} \right. \] a为常数，对于\((3,3,2)\)点，求（1）沿\((1,0,1)\)方向线应变；（2）沿\((1,1,1)\)和\((1,-1,0)\)方向剪应变及夹角的改变 \[ \begin{aligned} &\frac{\partial u_1}{\partial x_1}=6ax_1\qquad\frac{\partial u_1}{\partial x_2}=a\qquad\frac{\partial u_1}{\partial x_3}=4a\\ &\frac{\partial u_2}{\partial x_1}=3a\qquad\frac{\partial u_2}{\partial x_2}=2a\qquad \frac{\partial u_2}{\partial x_3}=3a\\ &\frac{\partial u_3}{\partial x_1}=2ax_1\qquad\frac{\partial u_3}{\partial x_2}=a\qquad \frac{\partial u_2}{\partial x_3}=4ax_3\\ &可知在点(3,3,2)处应变张量为\boldsymbol{\Gamma}=\begin{bmatrix} 18a&2a&8a\\ 2a&2a&2a\\ 8a&2a&16a \end{bmatrix}\\ &(1)沿着(1,0,1)方向的单位矢量为\boldsymbol{\xi}=(\frac{\sqrt{2}}{2},0,\frac{\sqrt{2}}{2})\\ &线应变\varepsilon=\xi_i v_{ij}\xi_j=25a\\ &(2)(1,1,1)方向的单位矢量为\boldsymbol{\xi}=(\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3})\\ &(1,-1,0)方向的单位矢量为\boldsymbol{\eta}=(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2},0)\\ &可知\boldsymbol{\xi}与\boldsymbol{\eta}正交，则沿(1,1,1)和(1,-1,0)方向剪应变为\\ &\gamma=\xi\cdot\boldsymbol{\Gamma}\cdot\eta=\frac{11}{3}\sqrt{6}a\\ &夹角改变为2\gamma=\frac{22}{3}\sqrt{6}a \end{aligned} \] 3、 \[ \Gamma=\begin{bmatrix} 0&1&-1\\ 1&0&1\\ -1&1&0 \end{bmatrix}\times10^{-5} \] 求（1）主应变（2）主方向 \[ \begin{aligned} &由题意可知\boldsymbol{\Gamma}的第一不变量、第二不变量和第三步变量分别为：\\ &I_1=r_{ii}=0\\ &I_2=\begin{vmatrix} 0&1\\ 1&0 \end{vmatrix}+\begin{vmatrix} 0&1\\ 1&0 \end{vmatrix}+\begin{vmatrix} 0&-1\\ -1&0 \end{vmatrix}=-3\\ &I_3=|r|=\begin{vmatrix} 0&1&-1\\ 1&0&1\\ -1&1&0 \end{vmatrix}=-2\\ &特征方程为\lambda^3-I_1\lambda^2+I_2\lambda-I_3=0,可得\lambda^3-3\lambda+2=0\\ &可得\lambda_1=\lambda_2=1,\lambda_3=-2\\ &(1)主应变为\varepsilon_x=\varepsilon_y=10^{-5},\varepsilon_z=-2\times10^{-5}\\ &(2)将\lambda_3=-2代入\left\{ \begin{array}{ll} n_1(\gamma_{11}-\lambda)+n_2\gamma_{12}+n_3\gamma_{13}=0\\ n_2\gamma_{21}+n_2(\gamma_{22}-\lambda)+n_3\lambda_{23}=0\\ n_3\gamma_{31}+n_2\gamma_{32}+n_3(\gamma_{33}-\lambda)=0\\ n_1^2+n_2^2+n_3^2=1 \end{array} \right.\\ &得\left\{ \begin{array}{ll} 2n_{31}+n_{32}-n_{33}=0\\ n_{31}+2n_{32}+n_{33}=0\\ -n_{31}+n_{32}+2n_{33}=0\\ n_{31}^2+n_{32}^2+n_{33}^2=1 \end{array} \right.\\ &解得(\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}),(-\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3})\\ &故主方向\xi_1=(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0),\xi_2=(-\frac{\sqrt{6}}{6},\frac{\sqrt{6}}{6},\frac{\sqrt{6}}{3}),\xi_3=(-\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3}) \end{aligned} \]

4、证明线性位移场 \[ \left\{ \begin{array}{ll} u_1=u_0+a_1x_1+b_1x_2+c_1x_3\\ u_2=v_0+a_2x_1+b_2x_2+c_2x_3\\ u_1=w_0+a_3x_1+b_3x_2+c_3x_3\\ \end{array} \right. \] (1)直线变成直线；(2)平面变成平面;(3)球面变成椭球

  1. \[ \begin{aligned} &设平面内直线上一点P(x_1,x_2,x_3),直线参数方程为\left\{ \begin{array}{ll} x_1=x_1^0+a_0t\\ x_2=x_2^0+b_0t\\ x_3=x_3^0+c_0t\\ \end{array} \right.\\ &经线位移场P点移动到\overset{\sim}{P}(\overset{\sim}{x_1},\overset{\sim}{x_2},\overset{\sim}{x_3})点\\ &\left\{ \begin{array}{ll} \overset{\sim}{x_1}=x_1+u_1\\ \overset{\sim}{x_2}=x_2+u_2\\ \overset{\sim}{x_3}=x_3+u_3\\ \end{array} \right.\\ &\left\{ \begin{array}{ll} u_1=u_0+a_1(x_1^0+a_0t)+b_1(x_2^0+b_0t)+c_1(x_3^0+c_0t)\\ u_2=v_0+a_2(x_1^0+a_0t)+b_2(x_2^0+b_0t)+c_2(x_3^0+c_0t)\\ u_3=w_0+a_3(x_1^0+a_0t)+b_3(x_2^0+b_0t)+c_1(x_3^0+c_0t)\\ \end{array} \right.\\ &代入得\left\{ \begin{array}{ll} \overset{\sim}{x_1}=u_0+(a_1+1)x_1^0+b_1x_2^0+c_1x_3^0+(a_0+a_0a_1+b_0b_1+c_0c_1)t\\ \overset{\sim}{x_2}=v_0+a_2x_1^0+(b_2+1)x_2^0+c_2x_3^0+(b_0+a_0a_2+b_0b_2+c_0c_2)t\\ \overset{\sim}{x_1}=w_0+a_3x_1^0+b_3x_2^0+(c_3+1)x_3^0+(c_0+a_0a_3+b_0b_3+c_0c_3)t\\ \end{array} \right.\\ &也为空间直线，所以直线进过线性位移变换仍为直线 \end{aligned} \]
  2. \[ \begin{aligned} &设平面上一点P(x_1,x_2,x_3),直线参数方程为\left\{ \begin{array}{ll} x_1=x_1^0+m_1k+n_1t\\ x_2=x_2^0+m_2k+n_2t\\ x_3=x_3^0+m_3k+n_3t\\ \end{array} \right.\\ &经线位移场P点移动到\overset{\sim}{P}(\overset{\sim}{x_1},\overset{\sim}{x_2},\overset{\sim}{x_3})点\\ &\left\{ \begin{array}{ll} \overset{\sim}{x_1}=x_1+u_1\\ \overset{\sim}{x_2}=x_2+u_2\\ \overset{\sim}{x_3}=x_3+u_3\\ \end{array} \right.\\ &\left\{ \begin{array}{ll} u_1=u_0+a_1(x_1^0+m_1k+n_1t)+b_1(x_2^0+m_2k+n_2t)+c_1(x_3^0+m_3k+n_3t)\\ u_2=v_0+a_2(x_1^0+m_1k+n_1t)+b_2(x_2^0+m_2k+n_2t)+c_2(x_3^0+m_3k+n_3t)\\ u_3=w_0+a_3(x_1^0+m_1k+n_1t)+b_3(x_2^0+m_2k+n_2t)+c_3(x_3^0+m_3k+n_3t)\\ \end{array} \right.\\ &代入得\left\{ \begin{array}{ll} \overset{\sim}{x_1}=u_0+x_1^0+a_1x_1^0+b_1x_2^0+c_1x_3^0+(m_1+a_1m_1+b_1m_2+c_1m_3)k+(n_1+a_1n_1+b_1n_2+c_1n_3)t\\ \overset{\sim}{x_2}=v_0+x_2^0+a_2x_1^0+b_2x_2^0+c_2x_3^0+(m_2+a_2m_1+b_2m_2+c_2m_3)k+(n_2+a_2n_1+b_2n_2+c_2n_3)t\\ \overset{\sim}{x_3}=w_0+x_3^0+a_3x_1^0+b_3x_2^0+c_3x_3^0+(m_3+a_3m_1+b_3m_2+c_3m_3)k+(n_3+a_3n_1+b_3n_2+c_3n_3)t\\ \end{array} \right.\\ &仍为空间平面，故平面经过线性位移场变换仍为平面 \end{aligned} \]
  3. \[ \begin{align} &设空间内球上一点P(x_1,x_2,x_3),进过线位移场P点变形为点\overset{\sim}{P}(\overset{\sim}{x_1},\overset{\sim}{x_2},\overset{\sim}{x_3})\\ &即\left\{ \begin{array}{ll} x_1=x_1^0+r\sin\alpha\cos\beta\\ x_2=x_2^0+r\sin\alpha\sin\beta\\ x_3=x_3^0+r\cos\alpha \end{array} \right.\qquad\left\{ \begin{array}{ll} \overset{\sim}{x_1}=x_1+u_1\\ \overset{\sim}{x_2}=x_2+u_2\\ \overset{\sim}{x_3}=x_3+u_3\\ \end{array} \right.\\ &\left\{ \begin{array}{ll} u_1=u_0+a_1(x_1^0+r\sin\alpha\cos\beta)+b_1(x_2^0+r\sin\alpha\sin\beta)+c_1(x_3^0+r\cos\alpha)\\ u_2=v_0+a_2(x_1^0+r\sin\alpha\cos\beta)+b_2(x_2^0+r\sin\alpha\sin\beta)+c_2(x_3^0+r\cos\alpha)\\ u_3=w_0+a_3(x_1^0+r\sin\alpha\cos\beta)+b_3(x_2^0+r\sin\alpha\sin\beta)+c_3(x_3^0+r\cos\alpha)\\ \end{array} \right.\\ &\therefore \left\{ \begin{array}{ll} \overset{\sim}{x_1}=x_1^0+u_0+a_1x_1^0+b_1x_2^0+c_1x_3^0+r\sin\alpha\cos\beta+a_1r\sin\alpha\cos\beta+b_1r\sin\alpha\sin\beta+c_1r\cos\alpha\\ \overset{\sim}{x_2}=x_2^0+v_0+a_2x_1^0+b_2x_2^0+c_2x_3^0+r\sin\alpha\sin\beta+a_2r\sin\alpha\cos\beta+b_2r\sin\alpha\sin\beta+c_2r\cos\alpha\\ \overset{\sim}{x_3}=x_3^0+2_0+a_3x_1^0+b_3x_2^0+c_3x_3^0+r\cos\alpha+a_3r\sin\alpha\cos\beta+b_3r\sin\alpha\sin\beta+c_1r\cos\alpha\\ \end{array} \right.\\ &令\overset{\sim}{x^0_1}=x_1^0+u_0+a_1x_1^0+b_1x_2^0\qquad\overset{\sim}{x^1_0}=x_2^0+v_0+a_2x_1^0+b_2x_2^0+c_2x_3^0\qquad\overset{\sim}{x_3^0}=x_3^0+2_0+a_3x_1^0+b_3x_2^0+c_3x_3^0\\ &可得\left\{ \begin{array}{ll} \overset{\sim}{x_1}=\overset{\sim}{x_1^0}+kr\sin\alpha_1\cos\beta_1\\ \overset{\sim}{x_1}=\overset{\sim}{x_1^0}+mr\sin\alpha_1\sin\beta_1\\ \overset{\sim}{x_1}=\overset{\sim}{x_1^0}+nr\cos\alpha_1\\ \end{array} \right.\\ &\therefore \frac{(\overset{\sim}{x_1}-\overset{\sim}{x_1^0})^2}{k^2}+\frac{(\overset{\sim}{x_1}-\overset{\sim}{x_1^0})^2}{k^2}+\frac{(\overset{\sim}{x_1}-\overset{\sim}{x_1^0})^2}{k^2}=r^2\\ &所以球经过线性位移场变为椭球 \end{align} \]

\[ \begin{aligned} &(\mathrm{d}\overset{\sim}{r})^2=(\mathrm{d}r)^2(1+2\xi\cdot\boldsymbol{\Gamma}\cdot\xi)\\ &左右同乘(1-2\xi\cdot\boldsymbol{\Gamma}\cdot\xi),略去无穷小量\\ &得(1-2\xi\cdot\boldsymbol{\Gamma}\cdot\xi)(\mathrm{d}\overset{\sim}{r})^2=(\mathrm{d}r)^2\\ &\mathrm{d}\overset{\sim}{\boldsymbol{r}}\cdot(\boldsymbol{I}-2\boldsymbol{\Gamma})\cdot\mathrm{d}\overset{\sim}{\boldsymbol{r}}=(\mathrm{d}r)^2\\ &将坐标旋转到主坐标系\\ &(\mathrm{d}\overset{\sim}{x_1},\mathrm{d}\overset{\sim}{x_2},\mathrm{d}\overset{\sim}{x_3})\begin{pmatrix} 1-2\gamma_1&0&0\\ 0&1-2\gamma_2&0\\ 0&0&1-2\gamma_3 \end{pmatrix}\begin{pmatrix} \mathrm{d}\overset{\sim}{x_1}\\\mathrm{d}\overset{\sim}{x_2}\\\mathrm{d}\overset{\sim}{x_3} \end{pmatrix}=(\mathrm{d}r)^2\\ &(1-2\lambda_1)(\mathrm{d}\overset{\sim}{x_1})^2+(1-2\lambda_2)(\mathrm{d}\overset{\sim}{x_2})^2+(1-2\lambda_3)(\mathrm{d}\overset{\sim}{x_3})^2=(\mathrm{d}r)^2\\ &令1-2\lambda_1\triangleq(1+\lambda_1)^{-1},1-2\lambda_2\triangleq(1+\lambda_2)^{-1},1-2\lambda_3\triangleq(1+\lambda_3)^{-1}\\ &\frac{\mathrm{d}\overset{\sim}{x_1}}{(1+\lambda_1)^2}+\frac{\mathrm{d}\overset{\sim}{x_2}}{(1+\lambda_2)^2}+\frac{\mathrm{d}\overset{\sim}{x_3}}{(1+\lambda_3)^2}=0\\ &所以球面经过线位移场的作用变为椭球 \end{aligned} \]

5、已知弹性体内一点M附近，两个不同向单位线元\(\vec{\xi}\)和\(\vec{\eta}\)间夹角为\(\theta\)(\(\theta\)不一定为直角)，变形后（即小变形，去Cauchy应变张量\(\boldsymbol{\Gamma}\)）,夹角为\(\theta-2\gamma\),确定平均剪应变\(\gamma\)与\(\boldsymbol{\Gamma}\)及\(\theta\)的关系，并证明之。 \[ \begin{aligned} &取\mathrm{d}\boldsymbol{r}=(\mathrm{d}r)\boldsymbol{\xi}\qquad\delta\boldsymbol{r}=(\delta r)\boldsymbol{\eta}\qquad \boldsymbol{\xi}=(\xi_1,\xi_2,\xi_3)\qquad\boldsymbol{\eta}=(\eta_1,\eta_2,\eta_3)\\ &从P,A,B变形为\overset{\sim}{P},\overset{\sim}{A},\overset{\sim}{B}位移分别为\boldsymbol{u},\boldsymbol{u}+\mathrm{d}\boldsymbol{u},\boldsymbol{u}+\delta\boldsymbol{u}\\ &可得\boldsymbol{\overset{\sim}{P}\overset{\sim}{A}}=\mathrm{d}\boldsymbol{\overset{\sim}{r}}=\mathrm{d}\boldsymbol{r}+\mathrm{d}\boldsymbol{u}\qquad \boldsymbol{\overset{\sim}{P}\overset{\sim}{B}}=\delta\boldsymbol{\overset{\sim}{r}}=\delta\boldsymbol{r}+\mathrm{d}\boldsymbol{u}\\ &\mathrm{d}\boldsymbol{\overset{\sim}{r}}\cdot\mathrm{d}\boldsymbol{\overset{\sim}{r}}=\mathrm{d}\boldsymbol{r}\cdot\delta\boldsymbol{r}+\mathrm{d}\boldsymbol{r}\cdot\delta\boldsymbol{u}+\mathrm{d}\boldsymbol{u}\cdot\delta\boldsymbol{r}+\mathrm{d}\boldsymbol{u}\cdot\delta\boldsymbol{u}\qquad \mathrm{d}\boldsymbol{r}\cdot\delta\boldsymbol{r}=\mathrm{d}r\delta r\cos\theta \\ &\because \mathrm{d}\boldsymbol{u}=\mathrm{d}\boldsymbol{r}(\nabla\boldsymbol{u})\qquad \delta\boldsymbol{u}=(\boldsymbol{u}\nabla)\cdot\delta\boldsymbol{r}\\ &\therefore\mathrm{d}\boldsymbol{\overset{\sim}{r}}\cdot\delta\boldsymbol{\overset{\sim}{r}}=\mathrm{d}\boldsymbol{r}(\boldsymbol{\omega}\times\delta\boldsymbol{r}+\boldsymbol{\Gamma}\cdot\delta\boldsymbol{r})+\delta\boldsymbol{r}\cdot(\boldsymbol{\omega}\times\mathrm{d}\boldsymbol{r}+\boldsymbol{\Gamma}\cdot\mathrm{d}\boldsymbol{r})+\mathrm{d}\boldsymbol{r}\cdot(\nabla\boldsymbol{u})\cdot(\boldsymbol{u}\nabla)\cdot\delta\boldsymbol{r}\\ &=\mathrm{d}r\cdot\delta r(\cos\theta+2\xi\cdot\boldsymbol{G}\cdot\eta)\\ &\mathrm{d}\boldsymbol{\overset{\sim}{r}}与\delta\boldsymbol{r}的夹角为\theta-2\gamma\\ &设\mathrm{d}\boldsymbol{\overset{\sim}{r}}=(1+\varepsilon_1)\mathrm{d}r\qquad\delta\boldsymbol{\overset{\sim}{r}}=(1+\varepsilon_2)\delta r\\ &则有\mathrm{d}\boldsymbol{\overset{\sim}{r}}\cdot\delta\boldsymbol{\overset{\sim}{r}}=\mathrm{d}\overset{\sim}{r}\cdot\delta\overset{\sim}{r}\cos(\theta-2\gamma)=\mathrm{d}r\delta[\cos\theta+2\boldsymbol{\xi\cdot\Gamma}\cdot\eta]\\ &\therefore(1+\varepsilon_1)(1+\varepsilon_2)\cos(\theta-2\gamma)=\cos\theta+2\xi\cdot\boldsymbol{\Gamma}\cdot\eta\\ &(1+\varepsilon_1+\varepsilon_2+\varepsilon_1\varepsilon_2)[\cos\theta\cos2\gamma+\sin\theta\sin2\gamma]=\cos\theta+2\xi\cdot\boldsymbol{\Gamma}\cdot\eta\\ &小变形情况下(1+\varepsilon_1+\varepsilon_2)(\cos\theta+2\gamma\sin\theta)=\cos\theta+2\xi\cdot\boldsymbol{\Gamma}\cdot\eta\\ &(1+\varepsilon_1+\varepsilon_2)(1+2\gamma\tan\theta)=1+\frac{2}{\cos\theta}\xi\cdot\boldsymbol{\Gamma}\cdot\eta\\ &省略无穷小量2\gamma\tan\theta=\frac{2}{\cos\theta}\xi\cdot\boldsymbol{\Gamma}\cdot\eta-(\varepsilon_1+\varepsilon_2)\\ &\varepsilon_1=\xi\cdot\boldsymbol{\Gamma}\cdot\xi\qquad\varepsilon=\eta\cdot\boldsymbol{\Gamma}\cdot\eta\\ &r=\frac{1}{\sin\theta}\xi\cdot\boldsymbol{\Gamma}\cdot\eta-\frac{\cot\theta}{2}(\xi\cdot\boldsymbol{\Gamma}\cdot\xi+\eta\cdot\boldsymbol{\Gamma}\cdot\eta) \end{aligned} \]

6、 \[ \boldsymbol{\Gamma}=\begin{bmatrix} 0&1&-1\\ 1&0&1\\ -1&1&0\\ \end{bmatrix}\times 10^{-5} \] 求（1）球应变（2）偏应力张量（3）偏应力张量主值、主方向（4）八面体\(\varepsilon_0,\gamma_0\) \[ \begin{aligned} &根据题意可以求出第一不变量、第二不变量、第三步变量为\\ &I_1=r_{ii}=0\\ &I_2=\begin{vmatrix} 0&1\\ 1&0 \end{vmatrix}+\begin{vmatrix} 0&1\\ 1&0 \end{vmatrix}+\begin{vmatrix} 0&-1\\ -1&0 \end{vmatrix}=-3\\ &I_3=\begin{vmatrix} 0&1&-1\\ 1&0&1\\ -1&1&0\\ \end{vmatrix}=-2\\ &特征方程为\lambda^3-I_1\lambda^2+I_2\lambda-I_3=0,即\lambda^3-3\lambda+2=0\\ &解得\lambda_1=\lambda_2=1,\lambda_3=-2\\ &应变张量\boldsymbol{\Gamma}的主应变为\varepsilon_x=\varepsilon_y=10^{-5},\varepsilon_z=-2\times10^{-5}\\ &(1)球应力张量\varepsilon=\frac{1}{3}I_1=0\\ &(2)偏应力张量e_{ij}=\lambda_{ij}-\varepsilon_0\delta_{ij}=r_{ij},记为矩阵E=\Gamma\\ &(3)主值e_x=e_y=10^{-5},e_z=-2\times10^{-5}\\ &主方向\eta_1=(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0),\eta_2=(-\frac{\sqrt{6}}{6},\frac{\sqrt{6}}{6},\frac{\sqrt{6}}{3}),\eta_3=(-\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3})\\ &(4)\varepsilon_0=\frac{1}{3}I_1=0\\ &\gamma_0=\frac{2}{3}\sqrt{(\lambda_1-\lambda_2)^2+(\lambda_2-\lambda_3)^2+(\lambda_3-\lambda_1)^2}=0 \end{aligned} \] 7、等倾面法线与等倾面内任一方向线元剪应力最大值\(\gamma_0=\frac{2}{3}\sqrt{(\lambda_1-\lambda_2)^2+(\lambda_2-\lambda_3)^2+(\lambda_3-\lambda_1)^2}\) \[ \begin{aligned} &考虑主坐标系中第一卦限的等倾面，其法向矢量\boldsymbol{v}=\frac{1}{\sqrt{3}}(\boldsymbol{e_1}+\boldsymbol{e_2}+\boldsymbol{e_3})\\ &设\boldsymbol{t}为等倾面上任一单位矢量，\boldsymbol{v}\cdot\boldsymbol{t}=0\\ &可得\gamma_{vt}=2\boldsymbol{v}\cdot\boldsymbol{\epsilon}\cdot\boldsymbol{t}=2(v_1\epsilon_1 t_1+v_2\epsilon_2 t_2+v_3\epsilon_3 t_3)=\frac{2}{\sqrt{3}}(\epsilon_1 t_1+\epsilon_2 t_2)\\ &当\epsilon给定时，\gamma_{vt}是\boldsymbol{t}的函数。\\ &令\gamma_{vt}=2\boldsymbol{d}\cdot\boldsymbol{t}\\ &其中\boldsymbol{d}=\boldsymbol{v}\cdot\boldsymbol{\epsilon}=\frac{1}{\sqrt{3}}(\epsilon_1\boldsymbol{e_1}+\boldsymbol{\epsilon_2}\boldsymbol{e_2}+\boldsymbol{\epsilon_3 e_3})\\ &显然矢量\boldsymbol{d},\boldsymbol{v},\boldsymbol{t}共面时，\boldsymbol{d}在\boldsymbol{t}上的投影\boldsymbol{d\cdot t}(\frac{1}{2}\gamma_{vt})取得最大值。\\ &故(\max_t \boldsymbol{d\cdot t})^2=|\boldsymbol{d}|^2-(\boldsymbol{d\cdot t})^2=\frac{1}{3}(\varepsilon_1^2+\varepsilon_2^2+\varepsilon_3^2)-\frac{1}{9}(\varepsilon_1+\varepsilon_2+\varepsilon_3)^2\\ &\max\gamma_{vt}=\frac{2}{3}\sqrt{3(\varepsilon_1^2+\varepsilon_2^2+\varepsilon_3^2)-(\varepsilon_1+\varepsilon_2+\varepsilon_3)^2}\\ &=\frac{2}{3}\sqrt{(\epsilon_1-\varepsilon_2)^2+(\varepsilon_2-\varepsilon_3)^2+(\varepsilon_3-\varepsilon_1)^2}\\ &\gamma_0=\frac{2}{3}\sqrt{(\lambda_1-\lambda_2)^2+(\lambda_2-\lambda_3)^2+(\lambda_3-\lambda_1)^2}得证 \end{aligned} \]

8、\(\boldsymbol{\nabla}\times\boldsymbol{\Gamma}\times\boldsymbol{\nabla}=0\)化为指标记法 \[ \begin{aligned} &\boldsymbol{\nabla}\times\boldsymbol{\Gamma}\times\boldsymbol{\nabla}=\partial_i\boldsymbol{e_i}\times\gamma_{jk}\boldsymbol{e_je_k}\times\partial_i \boldsymbol{e}_s=\varepsilon_{pij}\varepsilon_{qks}\gamma_{jk,is}\boldsymbol{e}_p\boldsymbol{e}_q=\boldsymbol{0}\\ &可得\varepsilon_{pij}\varepsilon_{qks}\gamma_{jk,is}=0\qquad(p,q=1,2,3)\\ &若p=q=1,则\varepsilon_{123}\varepsilon_{123}\gamma_{32,23}+\varepsilon_{132}\varepsilon_{123}\gamma_{23,23}+\varepsilon_{123}\varepsilon_{132}\gamma_{33,22}+\varepsilon_{132}\varepsilon_{123}\gamma_{22,33}=0\\ &即\gamma_{22,33}+\gamma_{33,22}=2\gamma_{23,23}\\ &若p=q=2,则\varepsilon_{213}\varepsilon_{213}\gamma_{31,13}+\varepsilon_{231}\varepsilon_{231}\gamma_{13,31}+\varepsilon_{213}\varepsilon_{231}\gamma_{33,11}+\varepsilon_{231}\varepsilon_{213}\gamma_{11,33}=0\\ &即\gamma_{33,11}+\gamma_{11,33}=2\gamma_{31,13}\\ &若p=q=3,则\varepsilon_{312}\varepsilon_{312}\gamma_{21,12}+\varepsilon_{321}\varepsilon_{321}\gamma_{12,21}+\varepsilon_{312}\varepsilon_{321}\gamma_{22,11}+\varepsilon_{321}\varepsilon_{312}\gamma_{11,22}=0\\ &即\gamma_{11,22}+\gamma_{22,11}=2\gamma_{12,12}\\ &若p=3,q=2,则\\ &\varepsilon_{123}\varepsilon_{132}\gamma_{21,13}+\varepsilon_{132}\varepsilon_{312}\gamma_{23,11}+\varepsilon_{213}\varepsilon_{132}\gamma_{11,23}+\varepsilon_{213}\varepsilon_{312}\gamma_{13,21}=0\\ &即\gamma_{31,32}-\gamma_{12,33}+\gamma_{23,31}=\gamma_{33,12}\\ &若p=3,q=1,则\\ &\varepsilon_{123}\varepsilon_{231}\gamma_{21,13}+\varepsilon_{123}\varepsilon_{321}\gamma_{23,12}+\varepsilon_{213}\varepsilon_{231}\gamma_{12,23}+\varepsilon_{213}\varepsilon_{321}\gamma_{13,22}=0\\ &即\gamma_{23，21}-\gamma_{31，22}+\gamma_{12，23}=\gamma_{23，31}\\ &若p=1,q=2,则\varepsilon_{123}\varepsilon_{231}\gamma_{33,21}+\varepsilon_{132}\varepsilon_{213}\gamma_{21,23}+\varepsilon_{123}\varepsilon_{213}\gamma_{31,23}+\varepsilon_{132}\varepsilon_{231}\gamma_{23,31}=0\\ &即\gamma_{31,32}-\gamma_{12,33}+\gamma_{23,31}=\gamma_{33,12}\\ &得证. \end{aligned} \]

9、\(\varepsilon_x=k(x^2+y^2),\varepsilon_y=k(y^2+z^2),\gamma_{xy}=k^\prime xyz\),其余为0，验证是否为真实的应变场（\(k,k^\prime\)为const） \[ \begin{aligned} &证明是否为真实应变场即为证明是否满足变形协调方程\\ &\frac{\partial^2 \varepsilon_x}{\partial y^2}+\frac{\partial^2 \varepsilon_y}{\partial x^2}=2k\qquad 2\frac{\partial^2 \gamma_{xy}}{\partial x\partial y}=2k^\prime z\\ &欲使得\frac{\partial^2 \varepsilon_x}{\partial y^2}+\frac{\partial^2 \varepsilon_y}{\partial x^2}=2\frac{\partial^2 \gamma_{xy}}{\partial x\partial y}\\ &只需要k=k^\prime z\\ &\frac{\partial^2 \varepsilon_y}{\partial z^2}+\frac{\partial^2 \varepsilon_z}{\partial y^2}=2k\qquad 2\frac{\partial^2 \gamma_{yz}}{\partial y\partial z}=0\\ &欲使得\frac{\partial^2 \varepsilon_y}{\partial z^2}+\frac{\partial^2 \varepsilon_z}{\partial y^2}=2\frac{\partial^2 \gamma_{yz}}{\partial y\partial z}\\ &故k=0\\ &则k^\prime z=k=0\\ &所以\boldsymbol{\Gamma}为\boldsymbol{0}，不是真实的应变场 \end{aligned} \]

弹塑性力学第六次作业

1、 证明\(E_{ijkl}^\prime=C_{im}C_{jn}C_{ko}C_{lp}E_{mnop}\)

\[ \begin{aligned} &\sigma_{mn}^\prime=E_{mnop}^\prime\gamma_{op}^\prime\\ &\sigma_{ij}=C_{ri}C_{tj}\sigma_{rt}^\prime,\gamma_{kl}=C_{pk}C_{ql}r_{pq}^\prime\\ &可得C_{ri}C_{tj}\sigma_{tr}^\prime=E_{ijks}C_{pk}C_{qs}\gamma_{pq}^\prime\\ &\sigma_{mn}^\prime=C_{mi}C_{nj}C_{ok}C_{ps}E_{ijks}\gamma_{op}^\prime\\ &(E_{mnop}^\prime-C_{mi}C_{nj}C_{ok}C_{ps}E_{ijks})\gamma_{pq}^\prime=0\\ &E_{mnop}^\prime=C_{mi}C_{nj}C_{ok}C_{ps}E_{ijks}\\ &故E_{mnop}^\prime 为四阶张量 \end{aligned} \]

2、Cauchy,\(\boldsymbol{\sigma}\sim\boldsymbol{\Gamma}\)，是否能证明\(E_{ijkl}=E_{klij}\) \[ \begin{aligned} &\sigma_{ij}=F_{ij}(\varepsilon_{kl})\\ &\sigma_{ij}=D_{ij}+E_{ijkl}\varepsilon_{kl}\\ &假设初始状态无应变状态，对应无应力状态\\ &\sigma_{ij}=E_{ijkl}\varepsilon_{kl}\\ &E_{ijkl}=\frac{\partial \sigma_{ij}}{\partial \gamma_{kl}}=\frac{\partial \sigma_{ji}}{\partial \gamma_{kl}}=\frac{\partial \sigma_{ij}}{\partial \gamma_{lk}}=\frac{\partial \sigma_{ji}}{\partial \gamma_{lk}}\\ &即E_{ijkl}=E_{jikl}=E_{ijlk}=E_{jilk},无法推出E_{ijkl}=E_{klij} \end{aligned} \]

3、证明\(\boldsymbol{\sigma^\prime}=\lambda J(\boldsymbol{\Gamma^\prime})\boldsymbol{I}+2\mu\boldsymbol{\Gamma^\prime}\)坐标变化下形式是不变的 \[ \begin{aligned} &\boldsymbol{\sigma^\prime}=\lambda J(\boldsymbol{\Gamma^\prime})\boldsymbol{I}+2\mu\boldsymbol{\Gamma^\prime}\\ &经坐标变换后，\sigma=C\boldsymbol{\sigma^\prime} C^T\\ &\sigma=C(\lambda J(\boldsymbol{\Gamma^\prime})\boldsymbol{I}+2\mu\boldsymbol{\Gamma^\prime})C^T=\lambda J(\boldsymbol{\Gamma^\prime})\boldsymbol{I}+2\mu C\boldsymbol{\Gamma^\prime}C^T=\lambda J(\boldsymbol{\Gamma})\boldsymbol{I}+2\mu\boldsymbol{\Gamma}\\ \end{aligned} \] 4、各向同性，应变主轴和应力主轴一致 \[ \begin{aligned} &设1，2，3轴为物体内某点的应变主轴，对应剪应变\gamma_{12}=\gamma_{13}=\gamma_{23}=0\\ &根据广义胡克定律\tau_{23}=C_{41}\varepsilon_{1}+C_{42}\varepsilon_{2}+C_{43}\varepsilon_{3}\\ &\varepsilon_i(i=1,2,3)为该点的主应变\\ &将此坐标系沿着轴2旋转180^\circ,得新的坐标轴1',2',3',\\ &用(l_1,m_1,n_1),(l_2,m_2,n_2),(l_3,m_3,n_3)分别表示1',2',3'轴对原坐标1，2，3各轴的方向余弦\\ &即(l_1,m_1,n_1)=(-1,0,-1),(l_2,m_2,n_2)=(0,1,0),(l_3,m_3,n_3)=(0,0,-1)\\ &因为新坐标轴也指向应变主轴方向，剪应变也等于0，且因为各向同性时，弹性系数C_{41},C_{42}和C_{43}不随方向面改变\\ &\tau_{2'3'}=C_{41}\varepsilon_1^\prime+C_{42}\varepsilon_2^\prime+C_{43}\varepsilon_3^\prime\\ &\tau_{2'3'}=n_3m_2\tau_{23}=-\tau_{23}(转轴应力分量变换公式)\\ &\left\{ \begin{array}{ll} \varepsilon_1^\prime=l_1^2\varepsilon_1=\varepsilon_1\\ \varepsilon_2^\prime=l_2^2\varepsilon_2=\varepsilon_2\\ \varepsilon_3^\prime=l_3^2\varepsilon_3=\varepsilon_3\\ \end{array} \right.\Rightarrow -\tau_{23}=C_{41}\varepsilon_1+C_{42}\varepsilon_2+C_{43}\varepsilon_3\\ &所以\tau_{23}=-\tau_{23}=0\\ &同理可证\tau_{12}=\tau_{13}=0,故应变主轴和应变主轴重合 \end{aligned} \] 5、主应力与主应变成比例 \[ \begin{aligned} &在主坐标系内\sigma_{11}=\lambda(\gamma_{11}+\gamma_{22}+\gamma_{33})+2\mu\lambda_{22}\\ &E_{1122}=E_{2211}=E_{2233}=E_{3322}=E_{3311}=E_{1133}=\lambda\\ &\because \lambda\gamma_{11}=E_{2211}\gamma_{11}=\sigma_{22},\lambda\gamma_{11}=E_{3311}\gamma_{11}=\sigma_{33}\\ &\therefore \sigma_{22}=\sigma_{33}\\ &\because \lambda\gamma_{22}=E_{3322}\gamma_{22}=\sigma_{33},\lambda\gamma_{22}=E_{1122}\gamma_{22}=\sigma_{11}\\ &\therefore\sigma_{11}=\sigma_{33}=\sigma_{22}\\ &\sigma_{11}=\lambda(\gamma_{11}+\gamma_{22}+\gamma_{33})+2\mu\lambda_{22}=3\sigma_{11}+2\mu\gamma_{11}\\ &\therefore\sigma_{11}=-\mu\gamma_{11},\sigma_2=-\mu\gamma_{22},\sigma_3=-\mu\gamma_{33}\\ &\therefore\sigma_r=-\mu\gamma_i\\ &故主应力与主应变成比例 \end{aligned} \]

弹塑性力学第七次作业

1、证明\(\boldsymbol{\Gamma}=\frac{1}{E}[(1+\upsilon)\boldsymbol{\sigma}-\upsilon\Theta\boldsymbol{I}]\) \[ \begin{aligned} &考虑各向同性弹性体，其应力应变关系为\boldsymbol{\sigma}=\lambda\theta\boldsymbol{I}+2\mu\boldsymbol{\Gamma}\\ &\therefore \boldsymbol{\Gamma}=\frac{1}{2\mu}\boldsymbol{\sigma}-\frac{\lambda\theta}{2\mu}\boldsymbol{I}\\ &\theta=J(\boldsymbol{\Gamma})=\varepsilon_x+\varepsilon_y+\varepsilon_z=\frac{\Theta}{3\lambda+2\mu}=\frac{\Theta(1-2\upsilon)}{E}\\ &\mu=\frac{E}{2(1+\upsilon)}\qquad\lambda=\frac{E\upsilon}{(1+\upsilon)(1-2\upsilon)}\\ &\boldsymbol{\Gamma}=\frac{1}{2\times\frac{E}{2(1+\upsilon)}}\boldsymbol{\sigma}+\frac{\frac{E\upsilon}{(1+\upsilon)(1-2\upsilon)}\frac{\Theta(1-2\upsilon)}{E}}{2\times\frac{E}{2(1+\upsilon)}}\boldsymbol{I}=\frac{1+\upsilon}{E}\boldsymbol{\sigma}-\frac{\upsilon}{E}\Theta\boldsymbol{I}\\ &=\frac{1}{E}[(1+\upsilon)\boldsymbol{\sigma}-\upsilon\Theta\boldsymbol{I}] \end{aligned} \] 2、证明\(W=\frac{\Theta^2}{18k}-\frac{J_2^\prime}{2\mu}\) \[ \begin{align} w&=\frac{\Theta^2}{18k}-\frac{J_2^\prime}{2\mu}\\ &=\frac{k\theta^2}{2}-\frac{J_2^\prime}{2\mu}\\ &=\frac{1}{6}(3\lambda+2\mu)\theta^2-\frac{J_2^\prime}{2\mu}\\ &=\frac{1}{2}\lambda\gamma_{kk}\gamma_{jj}+\frac{1}{3}\mu\theta^2+\frac{(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2}{12\mu}\\ &=\frac{1}{2}\lambda\gamma_{kk}\gamma_{jj}+\frac{1}{3}\mu(\gamma_{11}+\gamma_{22}+\gamma_{33})^2+\frac{4\mu^2[(\gamma_{11}-\gamma_{22})^2+(\gamma_{22}-\gamma_{33})^2+(\gamma_{33}-\gamma_{11})^2]}{12\mu}\\ &=\frac{1}{2}\lambda\gamma_{kk}\gamma_{jj}+\frac{\mu(\gamma_{11}^2+\gamma_{22}^2+\gamma_{33}^2+2\gamma_{11}\gamma_{22}+2\gamma_{11}\gamma_{33}+2\gamma_{33}\gamma_{11})}{3}\\ &+\frac{\mu(\gamma_{11}^2+\gamma_{22}^2-2\gamma_{11}\gamma_{22}+\gamma_{22}^2+\gamma_{33}^2-2\gamma_{22}\gamma_{33})}{3}\\ &=\frac{1}{2}\lambda\gamma_{kk}\gamma_{jj}+\mu\gamma_{ij}\gamma_{ij}\\ &考虑各向同性弹性体：W=\frac{1}{2}a\gamma_{kk}\gamma_{ij}+2\mu\gamma_{ij}\gamma_{ij}=\frac{1}{2}\lambda\gamma_{kk}\gamma_{jj}+\mu\gamma_{ij}\gamma_{ij}\\ &故W=\frac{\Theta^2}{18k}-\frac{J_2^\prime}{2\mu}\\ &其中\Theta=\sigma_{ii}=J(\boldsymbol{\sigma})\qquad J_2^\prime=-\frac{1}{6}[(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_2)^2]\\ &k=\frac{1}{3}(3\lambda+2\mu) \end{align} \] 3、物体某点在x，y方向正应力分量：\(\sigma_x=35N/mm^2,\sigma_y=25N/mm^2\)，沿z轴方向被完全限制，求\(\sigma_z,\varepsilon_x,\varepsilon_y\),试求\(\sigma_z,\varepsilon_x,\varepsilon_y。E=2.1\times10^5N/mm^2,\upsilon=0.3\) \[ \begin{aligned} &\left\{ \begin{array}{ll} \varepsilon_x=\frac{1}{E}[\sigma_x-\upsilon(\sigma_y+\sigma_z)]\\ \varepsilon_y=\frac{1}{E}[\sigma_y-\upsilon(\sigma_z+\sigma_x)]\\ \varepsilon_z=\frac{1}{E}[\sigma_z-\upsilon(\sigma_x+\sigma_y)]\\ \end{array} \right.\\ &得到\left\{ \begin{array}{ll} \sigma_z=\upsilon(\sigma_x+\sigma_y)=0.3\times(35+25)=18N/mm^2\\ \varepsilon_y=\frac{1}{2.1\times10^5}\times[25-0.3\times(18+35)]=4.333\times10^{-5}mm\\ \varepsilon_x=\frac{1}{2.1\times10^5}\times[35-0.3\times(25+18)]=1.052\times10^{-5}mm\\ \end{array} \right. \end{aligned} \] 4、\(\nabla^2\vec u+\frac{1}{1-2\nu}\vec\nabla(\vec\nabla\cdot\vec u)=0\)转换为直角坐标系下投影形式和指标记法 \[ \begin{cases} \nabla^2u+\frac{1}{1-2\nu}\frac{\partial}{\partial x}(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z})=0,\\ \nabla^2v+\frac{1}{1-2\nu}\frac{\partial}{\partial y}(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z})=0,\\ \nabla^2w+\frac{1}{1-2\nu}\frac{\partial}{\partial z}(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z})=0,\\ \end{cases}\\ 指标记法：u_{i,jj}+\frac{1}{1-2\nu}u_{j,ji}+=0 \] 5、 \(\nabla^2T+\frac{1}{1+\nu}\vec\nabla\vec\nabla\Theta=0\)转换为直角坐标系下投影形式和指标记法 \[ \begin{cases} \nabla^2\sigma_x+\frac{1}{1+\nu}\Theta_{,xx}=0,\\ \nabla^2\sigma_y+\frac{1}{1+\nu}\Theta_{,yy}=0,\\ \nabla^2\sigma_z+\frac{1}{1+\nu}\Theta_{,zz}=0,\\ \nabla^2\tau_{yz}+\frac{1}{1+\nu}\Theta_{,yz}=0,\\ \nabla^2\tau_{zx}+\frac{1}{1+\nu}\Theta_{,zx}=0,\\ \nabla^2\tau_{xy}+\frac{1}{1+\nu}\Theta_{,xy}=0. \end{cases}\\ 指标形式：\nabla^2\sigma_{ij}+\frac{1}{1+\nu}\Theta_{,ij}=0 \]

弹塑性力学第八次作业

1、 矩形截面管沟刚性壁，填充土比重\(\rho\),求\(\boldsymbol{\sigma},\boldsymbol{u},\boldsymbol{\Gamma}\),设\(\varphi=Ax^2y+By^2+Cy^2+Dx^2\) \[ \begin{align} &设体力f_X=0,f_y=-\varphi\\ &1）相容方程\nabla^4\varphi=0\\ &\frac{\partial^4\varphi}{\partial x^4}=0\qquad \frac{\partial^4\varphi}{\partial y^4}=0\qquad\frac{\partial^4\varphi}{\partial x^2\partial y^2}=0满足相容方程\\ &2)应力分量表达式\\ &\begin{cases} \sigma_x=\frac{\partial^2 \varphi}{\partial y^2}=2B+2C\\ \sigma_y-f=\frac{\partial^2 \varphi}{\partial x^2}=2Ay+2D\\ \tau_{xy}=-\frac{\partial^2 \varphi}{\partial x\partial y}=-2Ax \end{cases}\\ &3)边界条件\\ &I.y=h时\tau_{xy}=0，II,y=h时\sigma_{y}=0,III.\int_{-a}^{a}\varepsilon_x\mathrm{d}x=0\rightarrow\varepsilon_x|_{x=x^*}\cdot 2a=0\rightarrow \varepsilon_x与x无关\\ &故\varepsilon_x=0\\ &4)各应力分量带回应力边界条件\\ &\begin{cases} -Ah=0\\ 2Ah+2D+\rho h=0\\ \varepsilon_x=\frac{1}{E^{'}}(\sigma_x-\upsilon^{'}\sigma_y)=0 \end{cases}\\ &解得:\begin{cases} A=0\\ D=-\frac{1}{2}\rho h\\ \sigma_x=\upsilon^\prime\sigma_y\\ \end{cases}\\ &\sigma_y=\rho y-\rho h\rightarrow \sigma_x=\upsilon^\prime(\rho y-\rho h)=\frac{\upsilon}{1-\upsilon}\rho(y-h),可得B=\frac{\upsilon}{1-\upsilon}\cdot\frac{\rho}{6},C=-\frac{\upsilon}{1-\upsilon}\cdot\frac{h}{2}\\ &\boldsymbol{\sigma}=\frac{\upsilon}{1-\upsilon}\rho(y-h),\sigma_y=\rho(y-h),\tau_{xy}=0,\tau{xz}=\tau_{yz}=0,\sigma_z=\upsilon(\sigma_x+\sigma_y)=\frac{\upsilon}{1-\upsilon}(y-h)\\ &故\varphi=\frac{\rho\upsilon}{6(1-\upsilon)}y^3-\frac{\upsilon h}{2(1-\upsilon)}y^2-\frac{1}{2}\rho hx^2\\ &5)位移\\ &u=0,w=0,v=\int_{0}^y\varepsilon_x\mathrm{d}y=\int_0^y\frac{1}{E^{'}}(\sigma_y-\upsilon^{'}\sigma_x)=\frac{1}{E}\frac{(1+\upsilon)(1-2\upsilon)}{1-\upsilon}\rho(\frac{y^2}{2-hy})\\ &6)\boldsymbol{\Gamma}\\ &\varepsilon_x=0,\varepsilon_y=\frac{1}{E}\frac{(1+\upsilon)(1-2\upsilon)}{1-\upsilon}\rho(\frac{y^2}{y-h}),\gamma_{xy}=\frac{1}{2\mu}\tau_{xy}=0 \end{align} \] 2、单位厚度板，\(\tau=c\),下端固定，体力为重力，比重为\(\rho\),求\(\sigma_{ij}(i,j=1,2)\) \[ \begin{align} &可知纵向纤维不受挤压，故\sigma_x=0\\ &\sigma_x=\frac{\partial^2\varphi}{\partial y^2}=0\\ &积分后得\varphi=yf_1(x)+f_2(x)\\ &\varphi满足相容方程\nabla^4\varphi=0,y\frac{\partial^4 f_1(x)}{\partial x^4}+\frac{\partial^4 f_2(x)}{\partial x^4}=0(\forall y)\\ &故\frac{\partial^4 f_1(x)}{\partial x^4}=0,\frac{\partial^4 f_2(x)}{\partial x^4}=0,可得\begin{cases} f_1(x)=Ax^3+Bx^2+Cx+I\\ f_2(x)=Dx^3+Ex^2+Fx+J\\ \end{cases}\\ &\varphi=(Ax^3+Bx^2+Cx)y+Dx^3+Ex^2\\ &\begin{cases} \sigma_x=\varphi_{,yy}=0\\ \sigma_y=(6Ax+2B)y+6Dx+2E-\rho y\\ \tau_{xy}=-3Ax^2-2Bx-C\\ \end{cases}\\ &边界条件\\ &x=0,\sigma_x=0,\tau_{xy}=0\rightarrow c=0\\ &x=h,\sigma_x=0,\tau_{xy}=\tau\rightarrow-3Ah^2-2Bh=\tau\\ &y=0,\int_0^h\tau_{xy}\mathrm{d}x=0\rightarrow 3Dh^2+2Eh=0\\ &\int_0^h\sigma_{y}\mathrm{d}x=0\rightarrow 3Dh^2+2Eh=0\\ &\int_0^h\sigma_{y}x\mathrm{d}x=0\rightarrow 2Dh^3+Eh^2=0\\ &\varphi=(-\frac{c}{h^2}x^3+\frac{c}{h}x^2)y\\ &\sigma_x=0,\sigma_y=(-\frac{6c}{h^2}+\frac{2c}{h})-\rho y,\tau_{xy}=\frac{3c}{h^2}x^3-\frac{2c}{h}x \end{align} \] 3、逆解法\(\varphi=Ax^3+bx^2y+cxy^2+dy^3\) \[ \begin{cases} \sigma_x=\frac{\partial^2 \varphi}{\partial y^2}=2Cx+6Dy\\ \sigma_y=\frac{\partial^2 \varphi}{\partial x^2}=6Ax+2By\\ \tau_{xy}=-\frac{\partial^2 \varphi}{\partial x\partial y}=-2Bx-2Cy \end{cases} \] 1) \(A\neq 0,B=C=D=0\),\(\sigma_x=0,\sigma_y=6Ax,\tau_{xy}=0\)

image-20201112013612488

2)\(B\neq0,A=C=D=0\),\(\sigma_x=0,\sigma_y=2By,\tau_{xy}=-2Bx\)

image-20201124100006820

3)\(C\neq0,A=B=D=0\),\(\sigma_x=2Cx,\sigma_y=0,\tau_{xy}=-2Cy\)

image-20201124100023520

4)\(D\neq0,A=B=C=0\),\(\sigma_x=BDy,\sigma_y=0,\tau_{xy}=0\)

image-20201112013946319

5)整体

image-20201112014019828

4、逆解法，\(\varphi=Ax^4+Bx^3y+Cx^2y^2+Dxy^3+Ey^4\) \[ \begin{align} &\nabla^4\varphi=0,故e=-(a+\frac{1}{3}c)\\ &\begin{cases} \sigma_x=\frac{\partial^2 \varphi}{\partial y^2}=2Cx^2+6Dxy+12Ey^2\\ \sigma_y=\frac{\partial^2 \varphi}{\partial x^2}=12Ax^2+6Bxy+2Cy^2\\ \tau_{xy}=-\frac{\partial^2 \varphi}{\partial x\partial y}=-3Bx^2-4Cxy-3Dy^2 \end{cases} \end{align} \] 1)\(A=B=C=0,E=0,D\neq0\),\(\sigma_x=6Dxy,\sigma_y=0,\tau_{xy}=-3Dy^2\)

image-20201124100047425

2)\(A=B=D=E=0,B\neq0\),\(\sigma_x=0,\sigma_y=6Bxy,\tau_{xy}=-3Bx^2\)

image-20201112014659215

3)\(B=D=0,A\neq0,C\neq0,E\neq0\),\(\sigma_x=2Cx^2+12ey^2,\sigma_y=12Ax^2+2Cy^2,\tau_{xy}=-4Cxy\)image-20201112014906651

  1. 整体
image-20201112015038753

弹塑性力学第九次作业

1、求\(\varphi,\boldsymbol{\sigma},\boldsymbol{\varepsilon}\)

image-20201118214753453 \[ \begin{align} &在梁的区域0<x<l,-c<y<c中满足相容方程\nabla^4\varphi(x,y)=0\\ &边界条件:y=\pm c.\tau_{yx}=\sigma_y=0\\ &在梁端加载:\int_{-c}^{c}\sigma_x\mathrm{d}y=0,\int_{-c}^c\sigma_{yx}\mathrm{d}y=0,\int_{-c}^{c}y\sigma_{x}\mathrm{d}y=0\\ &设\varphi=f_0(y)+(l-x)f_1(y)满足双调和方程，则f_0^{(4)}(y)+(l-x)f_1^{(4)}(y)=0,f_0^{(4)}(y)=0,f_1^{(4)}(y)=0\\ &\begin{cases} f_0(y)=a_0+a_1y+a_2y^2+a_3y^3\\ f_1(y)=b_0+b_1y+b_2y^2+b_3y^3 \end{cases}\\ &当y=\pm c时,\sigma_y=\frac{\partial^2 \phi}{\partial x^2}=0,\tau_{yx}=-(b_1\pm2b_2c+3b_3c)=0\Rightarrow b_2=0,b_3=-\frac{b_1}{3c^2}\\ &\int_{-c}^c\sigma_x\mathrm{d}y=\int_{-c}^c[2a_2+6a_3y+6b_3y(l-x)]\mathrm{d}y=0\Rightarrow a_2=a\\ &\int_{-c}^c y\sigma_x\mathrm{d}y=\int_{-c}^c [6a_3y^2+6b_3y^2(l-x)]\mathrm{d}y=0\Rightarrow a_3=0\\ &\int_{-c}^c\tau_{xy}\mathrm{d}y=\int_{-c}^{c}b(1-\frac{y^2}{c^2})\mathrm{d}y=\frac{4}{3}b_1h=P\Rightarrow b_1=\frac{3P}{4h},a_2=a_3=0\\ &\varphi=\frac{3P}{4c}(y-\frac{1}{3c^2}y^3)(l-x) \end{align} \]

2、 image-20201118214832147 \[ \begin{align} &在梁区域0<x<l,-c<y<c中，满足相容方程\nabla^4\varphi(x,y)=0\\ &边界条件：y=\pm c，\tau_{xy}=0;y=c,\sigma_y=0;y=-c,\sigma_y=-1\\ &在梁端加载处(x=l),\int_{-c}^c\tau_{xy}\mathrm{d}y=-\frac{1}{2}ql,\int_{-c}^c\sigma_{x}\mathrm{d}y=0,\int_{-c}^c\sigma_{x}y\mathrm{d}y=-\frac{1}{2}ql\\ &令\varphi=f_0(y)+(l^2-x^2)f_1(y)\\ &要满足双调和条件f_0^{(4)}(y)+(l^2-x^2)f_1^{(4)}(y)-4f_1^{\prime\prime}(y)=0\\ &故f_1^{(4)}(y)=0,f_0^{(4)}(y)-4f_1^{\prime\prime}(y)=0\\ &设\begin{cases} f_1(y)=a_0+a_1y+a_2y^2+a_3y^3\\ f_2(y)=b_0+b_1y+b_2y^2+b_3y^3+\frac{1}{3}a_2y^4+\frac{1}{5}a_3y^5 \end{cases}\\ &可知，求导后b_0,b_1被消去，当y=\pm c时，\tau_{xy}=2x(a_1\pm2a_2c+3a_3c^2)=0\\ &故a_2=0,a_3=-\frac{a_1}{3c^2}\\ &\sigma_{y}=\frac{\partial^2 \varphi}{\partial x^2}=-2[a_0+a_1(y-\frac{y^3}{3c^2})]\\ &\therefore y=c,\sigma_y=-2[a_0+a_1(c-\frac{1}{3}c)]=0;y=-c,\sigma_y=-2[a_0-a_1(c-\frac{c}{3})]=-q\\ &得a_0=\frac{q}{4},a_1=-\frac{3}{8c}c,a_3=\frac{1}{8h^3}c\\ &\begin{cases} \int_{-c}^c\sigma_x\mathrm{d}y=\int_{-c}^{c}(4a_3y^3+2b_2+6b_3y)\mathrm{d}y=0\\ \int_{-c}^cy\sigma_x\mathrm{d}y=\int_{-c}^{c}(4a_3y^4+2b_2y+6b_3y^2)\mathrm{d}y=0 \end{cases}\qquad b_2=0,b_3=-\frac{q}{20c}\\ &应力函数:\varphi=\frac{q}{40l^3}y^5-\frac{q}{20l}y^3+\frac{q}{4c}(l^2-x^2)(l-\frac{3}{2}y+\frac{y^3}{2c^2})\\ &应力分量:\begin{cases} \sigma_x=\frac{q}{2c^3}y^3-\frac{3q}{10c}y+\frac{q}{4c}(l^2-x^2)(c+\frac{3y}{c^2})\\ \sigma_y=-\frac{q}{4c^3}(y-c)(y^2+cy-2h^2)\\ \tau_{xy}=-\frac{3q}{3c^3}x(l^2-y^2) \end{cases}\\ &应变分量:\begin{cases} \varepsilon_x=\frac{1-\upsilon^2}{E}(\sigma_x-\frac{\upsilon}{1-\upsilon}\sigma_y)=\frac{1-\upsilon^2}{E}[\frac{q}{2c^3}y^3-\frac{3q}{10c}y+\frac{q}{4c}(l^2-x^2)(c+\frac{3y}{c^2})+\frac{1-\upsilon^2}{E}\frac{q}{4c^3}(y-c)(y^2+cy-2h^2)]\\ \varepsilon_y=\frac{1-\upsilon^2}{E}\sigma_y-\frac{\upsilon}{1-\upsilon}\sigma_x)=\frac{1-\upsilon^2}{E}\{-\frac{q}{4c^3}(y-c)(y^2+cy-2h^2)-\frac{1-\upsilon^2}{E}[\frac{q}{2c^3}y^3-\frac{3q}{10c}y+\frac{q}{4c}(l^2-x^2)(c+\frac{3y}{c^2})]\}\\ \varepsilon_{xy}=-\frac{1+\upsilon}{E}\cdot\frac{3q}{4c^3}x(c^2-y^2) \end{cases} \end{align} \]

3、 以D为起点，求\(\varphi\)

image-20201118214908970 \[ \begin{align} &以D为起点，假定\varphi_0=0,\varphi_{,1}=\varphi_{,2}=0\\ &D\rightarrow C中，\varphi_c=0,\varphi_{,1}|_c=0,\varphi_{,2}|_c=0\\ &C\rightarrow B中,\varphi=\frac{1}{2}q(b-y)^2,\varphi_{,x}=0,\varphi_{,y}=-q(b-y),则\varphi=\frac{1}{2}qb^2,\varphi_{,x}|_B=0,\varphi_{,y}|_B=-qb\\ &B\rightarrow A中，\varphi_{A}=\frac{1}{2}qb^2,\varphi_{,x}|_A=0,\varphi_{,y}|_A=-qb\\ &A\rightarrow D中，\varphi_D=\frac{1}{2}q(b-y)^2,\varphi_{,y}=0,\varphi=\frac{1}{2}q(b-y)^2 \end{align} \]