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news 来源：原创 2025/6/20 1:41:13

第一题

$\begin{align*} \int_{-\infty}^{+\infty}xe^{x-e^x}\text{d}x&=\int_{0}^{+\infty} \ln t \cdot e^{\ln t - t} \cdot \frac{1}{t} \text{d}t \\ &= \int_{0}^{+\infty} \ln t \cdot e^{ - t} \cdot \frac{1}{t} \cdot t \text{d}t\\ & = \int_{0}^{+\infty} \ln t \cdot e^{-t} \text{d}t=\psi(1)\\&=-\gamma \end{align*}$

第二题

$\begin{align*} \int_{0}^{+\infty}&\frac{x^{4}e^{x}}{(e^{x}-1)^{2}}dx\\ =&-\int_{0}^{+\infty}x^{4}d\frac{1}{e^{x}-1}\\ =&4\int_{0}^{+\infty}\frac{x^{3}}{e^{x}-1}dx-\frac{x^{4}}{e^{x}-1}\big|_{0}^{+\infty}\\ =&4\int_{0}^{+\infty}\frac{x^{3}e^{-x}}{1 - e^{-x}}dx\\ =&4\int_{0}^{+\infty}\left(x^{3}\sum_{n = 1}^{+\infty}e^{-nx}\right)dx\\ =&4\sum_{n = 1}^{+\infty}\left(\int_{0}^{+\infty}x^{3}e^{-nx}dx\right)\\ =&4\sum_{n = 1}^{+\infty}\left[-\left(\frac{1}{n}x^{3}+\frac{3}{n^{2}}x^{2}+\frac{6}{n^{3}}x+\frac{6}{n^{4}}\right)e^{-nx}\big|_{0}^{+\infty}\right]\\ =&24\sum_{n = 1}^{+\infty}\frac{1}{n^{4}}\\ =&\frac{4\pi^{4}}{15} \end{align*}$

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：
$\left. \begin{array}{l} \varepsilon_{\max} \\ \varepsilon_{\min} \end{array} \right\} = \frac{\varepsilon_{0^{\circ}} + \varepsilon_{90^{\circ}}}{2} \pm \frac{\sqrt{2}}{2} \sqrt{(\varepsilon_{0^{\circ}} - \varepsilon_{45^{\circ}})^2 + (\varepsilon_{45^{\circ}} - \varepsilon_{90^{\circ}})^2}$
$\tan 2\alpha_{0} = \frac{2\varepsilon_{45^{\circ}} - \varepsilon_{0^{\circ}} - \varepsilon_{90^{\circ}}}{\varepsilon_{0^{\circ}} - \varepsilon_{90^{\circ}}}$

$60^{\circ}$ 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：
$\left. \begin{array}{l} \varepsilon_{\max} \\ \varepsilon_{\min} \end{array} \right\} = \frac{\varepsilon_{0^{\circ}} + \varepsilon_{60^{\circ}} + \varepsilon_{120^{\circ}}}{3} \pm \frac{\sqrt{2}}{3} \sqrt{(\varepsilon_{0^{\circ}} - \varepsilon_{60^{\circ}})^2 + (\varepsilon_{60^{\circ}} - \varepsilon_{120^{\circ}})^2 + (\varepsilon_{120^{\circ}} - \varepsilon_{0^{\circ}})^2}$
$\tan 2\alpha_{0} = \frac{\sqrt{3}(\varepsilon_{60^{\circ}} - \varepsilon_{120^{\circ}})}{2\varepsilon_{0^{\circ}} - \varepsilon_{60^{\circ}} - \varepsilon_{120^{\circ}}}$

证明

对于以上问题我们有
$\left. \begin{array}{l} \varepsilon_{\alpha_1} = \frac{\varepsilon_x + \varepsilon_y}{2} + \frac{\varepsilon_x - \varepsilon_y}{2} \cos 2\alpha_1 + \frac{\gamma_{xy}}{2} \sin 2\alpha_1 \\ \varepsilon_{\alpha_2} = \frac{\varepsilon_x + \varepsilon_y}{2} + \frac{\varepsilon_x - \varepsilon_y}{2} \cos 2\alpha_2 + \frac{\gamma_{xy}}{2} \sin 2\alpha_2 \\ \varepsilon_{\alpha_3} = \frac{\varepsilon_x + \varepsilon_y}{2} + \frac{\varepsilon_x - \varepsilon_y}{2} \cos 2\alpha_3 + \frac{\gamma_{xy}}{2} \sin 2\alpha_3 \end{array} \right\}$

将 $\frac{\varepsilon_x+\varepsilon_y}{2}$ 、 $\frac{\varepsilon_x-\varepsilon_y}{2}$ 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：
$\begin{cases} \varepsilon_{\alpha_1} = A + B \cos 2\alpha_1 + C \sin 2\alpha_1 \\ \varepsilon_{\alpha_2} = A + B \cos 2\alpha_2 + C \sin 2\alpha_2 \\ \varepsilon_{\alpha_3} = A + B \cos 2\alpha_3 + C \sin 2\alpha_3 \end{cases}$
消去 $ A$ 后利用三角恒等式化简，通过线性方程组解得：
$\begin{align*} B &= \frac{\frac{\varepsilon_{\alpha_2}-\varepsilon_{\alpha_1}}{2\sin(\alpha_2-\alpha_1)}\cos(\alpha_1+\alpha_3) - \frac{\varepsilon_{\alpha_3}-\varepsilon_{\alpha_1}}{2\sin(\alpha_3-\alpha_1)}\cos(\alpha_1+\alpha_2)}{\sin(\alpha_3-\alpha_2)}, \\ C &= \frac{\frac{\varepsilon_{\alpha_3}-\varepsilon_{\alpha_1}}{2\sin(\alpha_3-\alpha_1)}\sin(\alpha_1+\alpha_2) - \frac{\varepsilon_{\alpha_2}-\varepsilon_{\alpha_1}}{2\sin(\alpha_2-\alpha_1)}\sin(\alpha_1+\alpha_3)}{\sin(\alpha_3-\alpha_2)}, \\ A &= \varepsilon_{\alpha_1} - B\cos2\alpha_1 - C\sin2\alpha_1. \end{align*}$
还原变量得：
$\varepsilon_x = A+B, \quad \varepsilon_y = A-B, \quad \gamma_{xy} = 2C.$

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

$\begin{cases} \varepsilon_{x}=\frac{1}{E}[\sigma_x-\mu(\sigma_y+\sigma_{z})]\\ \varepsilon_{x}=\frac{1}{E}[\sigma_x-\mu(\sigma_y+\sigma_{z})]\\ \varepsilon_{x}=\frac{1}{E}[\sigma_x-\mu(\sigma_y+\sigma_{z})]\\ \gamma_{xy}=\frac{\tau_{xy}}{G}\\ \gamma_{yz}=\frac{\tau_{yz}}{G}\\ \gamma_{zx}=\frac{\tau_{zx}}{G} \end{cases}$

(a)

首先，证明 $\boxed{G=\frac{E}{2(1+\mu)}}$

在纯剪切状态下：
$\sigma_x = \sigma_y = \sigma_z = 0, \quad \tau_{xy} \neq 0$
由剪切本构方程：
$\gamma_{xy} = \frac{\tau_{xy}}{G} \quad (1)$
旋转坐标系 $45^\circ$ 后，主应力为：
$\sigma_{1} = \tau_{xy}, \quad \sigma_{2} = -\tau_{xy}$
对应正应变为：
$\varepsilon_1 = \frac{\tau_{xy}}{E}(1+\mu), \quad \varepsilon_2 = -\frac{\tau_{xy}}{E}(1+\mu)$
几何关系：
$\gamma_{xy} = \varepsilon_1 - \varepsilon_2 = \frac{2\tau_{xy}(1+\mu)}{E} \quad (2)$
联立 $(1)$ 和 $(2)$ ：
$\frac{\tau_{xy}}{G} = \frac{2\tau_{xy}(1+\mu)}{E} \implies G = \frac{E}{2(1+\mu)}$

(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

$V_1=(1+\varepsilon_1+\varepsilon_2+\varepsilon_3)V$

体应变

$\theta=\varepsilon_1+\varepsilon_2+\varepsilon_3=\frac{1-2\mu}{E}(\sigma_1+\sigma_2+\sigma_3)$

对比

$\theta=\frac{\sigma_m}{K}$

其中， $\sigma_m=\frac{\sigma_1+\sigma_2+\sigma_3}{3}$

得到
$K=\frac{E}{3(1-2\mu)}$

第一题

第二题

第三题

第四题

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