第一题

$$\begin{array}{rl}{\int }_{-\infty }^{+\infty }x{e}^{x-e^x}\text{d}x& ={\int }_0^{+\infty }\ln t\cdot e^{\ln t-t}\cdot \frac{1}{t}\text{d}t\\ & ={\int }_0^{+\infty }\ln t\cdot e^{-t}\cdot \frac{1}{t}\cdot t\text{d}t\\ & ={\int }_0^{+\infty }\ln t\cdot e^{-t}\text{d}t=\psi (1)\\ & =-\gamma \end{array}$$

第二题

$$\begin{array}{rl}{\int }_0^{+\infty }& \frac{x^4{e}^x}{(e^x-1{)}^2}dx\\ =& -{\int }_0^{+\infty }{x}^{4}d\frac{1}{e^x-1}\\ =& 4{\int }_0^{+\infty }\frac{x^3}{e^x-1}dx-\frac{x^4}{e^x-1}{|}_0^{+\infty }\\ =& 4{\int }_0^{+\infty }\frac{x^3{e}^{-x}}{1-e^{-x}}dx\\ =& 4{\int }_0^{+\infty }\left(x^3\sum _{n=1}^{+\infty }{e}^{-nx}\right)dx\\ =& 4\sum _{n=1}^{+\infty }\left({\int }_0^{+\infty }{x}^3{e}^{-nx}dx\right)\\ =& 4\sum _{n=1}^{+\infty }\left[-\left(\frac{1}{n}{x}^3+\frac{3}{n^2}{x}^2+\frac{6}{n^3}x+\frac{6}{n^4}\right)e^{-nx}{|}_0^{+\infty }\right]\\ =& 24\sum _{n=1}^{+\infty }\frac{1}{n^4}\\ =& \frac{4{\pi }^4}{15}\end{array}$$

第三题

对于直角应变花，试证明主应变的大小及方向可用以下公式计算：
$$\begin{array}{l}{\epsilon }_{\max }\\ {\epsilon }_{\min }\end{array}\}=\frac{{\epsilon }_{0^{\circ }}+{\epsilon }_{90^{\circ }}}{2}\pm \frac{\sqrt{2}}{2}\sqrt{({\epsilon }_{0^{\circ }}-{\epsilon }_{45^{\circ }}{)}^2+({\epsilon }_{45^{\circ }}-{\epsilon }_{90^{\circ }}{)}^2}$$
$$\tan 2{\alpha }_0=\frac{2{\epsilon }_{45^{\circ }}-{\epsilon }_{0^{\circ }}-{\epsilon }_{90^{\circ }}}{{\epsilon }_{0^{\circ }}-{\epsilon }_{90^{\circ }}}$$

$60^{\circ }$应变花,三个应变片的角度分别为：${\alpha }_1=0^{\circ },{\alpha }_2=60^{\circ },{\alpha }_3=120^{\circ }$。求证主应变的数值及方向由以下公式计算：
$$\begin{array}{l}{\epsilon }_{\max }\\ {\epsilon }_{\min }\end{array}\}=\frac{{\epsilon }_{0^{\circ }}+{\epsilon }_{60^{\circ }}+{\epsilon }_{120^{\circ }}}{3}\pm \frac{\sqrt{2}}{3}\sqrt{({\epsilon }_{0^{\circ }}-{\epsilon }_{60^{\circ }}{)}^2+({\epsilon }_{60^{\circ }}-{\epsilon }_{120^{\circ }}{)}^2+({\epsilon }_{120^{\circ }}-{\epsilon }_{0^{\circ }}{)}^2}$$
$$\tan 2{\alpha }_0=\frac{\sqrt{3}({\epsilon }_{60^{\circ }}-{\epsilon }_{120^{\circ }})}{2{\epsilon }_{0^{\circ }}-{\epsilon }_{60^{\circ }}-{\epsilon }_{120^{\circ }}}$$

证明

对于以上问题我们有
$$\begin{array}{l}{\epsilon }_{{\alpha }_1}=\frac{{\epsilon }_x+{\epsilon }_y}{2}+\frac{{\epsilon }_x-{\epsilon }_y}{2}\cos 2{\alpha }_1+\frac{{\gamma }_{xy}}{2}\sin 2{\alpha }_1\\ {\epsilon }_{{\alpha }_2}=\frac{{\epsilon }_x+{\epsilon }_y}{2}+\frac{{\epsilon }_x-{\epsilon }_y}{2}\cos 2{\alpha }_2+\frac{{\gamma }_{xy}}{2}\sin 2{\alpha }_2\\ {\epsilon }_{{\alpha }_3}=\frac{{\epsilon }_x+{\epsilon }_y}{2}+\frac{{\epsilon }_x-{\epsilon }_y}{2}\cos 2{\alpha }_3+\frac{{\gamma }_{xy}}{2}\sin 2{\alpha }_3\end{array}\}$$

将$\frac{{\epsilon }_x+{\epsilon }_y}{2}$、$\frac{{\epsilon }_x-{\epsilon }_y}{2}$和${\gamma }_{xy}$视为变量，设 $ A = \frac{\varepsilon_x + \varepsilon_y}{2}$，$B = \frac{\varepsilon_x - \varepsilon_y}{2}$，$C = \frac{\gamma_{xy}}{2}$，原方程组化为：
$$\{\begin{array}{l}{\epsilon }_{{\alpha }_1}=A+B\cos 2{\alpha }_1+C\sin 2{\alpha }_1\\ {\epsilon }_{{\alpha }_2}=A+B\cos 2{\alpha }_2+C\sin 2{\alpha }_2\\ {\epsilon }_{{\alpha }_3}=A+B\cos 2{\alpha }_3+C\sin 2{\alpha }_3\end{array}$$
消去 $ A$ 后利用三角恒等式化简，通过线性方程组解得：
$$\begin{array}{rl}B& =\frac{\frac{{\epsilon }_{{\alpha }_2}-{\epsilon }_{{\alpha }_1}}{2\sin ({\alpha }_2-{\alpha }_1)}\cos ({\alpha }_1+{\alpha }_3)-\frac{{\epsilon }_{{\alpha }_3}-{\epsilon }_{{\alpha }_1}}{2\sin ({\alpha }_3-{\alpha }_1)}\cos ({\alpha }_1+{\alpha }_2)}{\sin ({\alpha }_3-{\alpha }_2)},\\ C& =\frac{\frac{{\epsilon }_{{\alpha }_3}-{\epsilon }_{{\alpha }_1}}{2\sin ({\alpha }_3-{\alpha }_1)}\sin ({\alpha }_1+{\alpha }_2)-\frac{{\epsilon }_{{\alpha }_2}-{\epsilon }_{{\alpha }_1}}{2\sin ({\alpha }_2-{\alpha }_1)}\sin ({\alpha }_1+{\alpha }_3)}{\sin ({\alpha }_3-{\alpha }_2)},\\ A& ={\epsilon }_{{\alpha }_1}-B\cos 2{\alpha }_1-C\sin 2{\alpha }_1.\end{array}$$
还原变量得：
$${\epsilon }_x=A+B,{\epsilon }_y=A-B,{\gamma }_{xy}=2C.$$

由此可以带入应变圆得到上述题目各个表达式。

第四题

使用广义胡克定律证明弹性常数之间的关系。$$G=\frac{E}{2(1+\mu )}$$$$K=\frac{E}{3(1-2\mu )}$$

证明

广义胡克定律

$$\{\begin{array}{l}{\epsilon }_x=\frac{1}{E}[{\sigma }_x-\mu ({\sigma }_y+{\sigma }_z)]\\ {\epsilon }_x=\frac{1}{E}[{\sigma }_x-\mu ({\sigma }_y+{\sigma }_z)]\\ {\epsilon }_x=\frac{1}{E}[{\sigma }_x-\mu ({\sigma }_y+{\sigma }_z)]\\ {\gamma }_{xy}=\frac{{\tau }_{xy}}{G}\\ {\gamma }_{yz}=\frac{{\tau }_{yz}}{G}\\ {\gamma }_{zx}=\frac{{\tau }_{zx}}{G}\end{array}$$

(a)

首先，证明$$\boxed{G=\frac{E}{2(1+\mu )}}$$

在纯剪切状态下：
$${\sigma }_x={\sigma }_y={\sigma }_z=0,{\tau }_{xy}\ne 0$$
由剪切本构方程：
$${\gamma }_{xy}=\frac{{\tau }_{xy}}{G}(1)$$
旋转坐标系 $$45^{\circ }$$ 后，主应力为：
$${\sigma }_1={\tau }_{xy},{\sigma }_2=-{\tau }_{xy}$$
对应正应变为：
$${\epsilon }_1=\frac{{\tau }_{xy}}{E}(1+\mu ),{\epsilon }_2=-\frac{{\tau }_{xy}}{E}(1+\mu )$$
几何关系：
$${\gamma }_{xy}={\epsilon }_1-{\epsilon }_2=\frac{2{\tau }_{xy}(1+\mu )}{E}(2)$$
联立 $(1)$ 和 $(2)$：
$$\frac{{\tau }_{xy}}{G}=\frac{2{\tau }_{xy}(1+\mu )}{E}⟹G=\frac{E}{2(1+\mu )}$$

(b)

其次，证明$$\boxed{K=\frac{E}{3(1-2\mu )}}$$

对于六面体其变形前的体积

$$V=\text{d}x\text{d}y\text{d}z$$

变形后体积

$$V_1=(1+{\epsilon }_1+{\epsilon }_2+{\epsilon }_3)V$$

体应变

$$\theta ={\epsilon }_1+{\epsilon }_2+{\epsilon }_3=\frac{1-2\mu }{E}({\sigma }_1+{\sigma }_2+{\sigma }_3)$$

对比

$$\theta =\frac{{\sigma }_m}{K}$$

其中，${\sigma }_m=\frac{{\sigma }_1+{\sigma }_2+{\sigma }_3}{3}$

得到
$$K=\frac{E}{3(1-2\mu )}$$