随笔：hhhhh
第一题
∫ − ∞ + ∞ x e x − e x d x = ∫ 0 + ∞ ln t ⋅ e ln t − t ⋅ 1 t d t = ∫ 0 + ∞ ln t ⋅ e − t ⋅ 1 t ⋅ t d t = ∫ 0 + ∞ ln t ⋅ e − t d t = ψ ( 1 ) = − γ \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*} ∫−∞+∞xex−exdx=∫0+∞lnt⋅elnt−t⋅t1dt=∫0+∞lnt⋅e−t⋅t1⋅tdt=∫0+∞lnt⋅e−tdt=ψ(1)=−γ
第二题
∫ 0 + ∞ x 4 e x ( e x − 1 ) 2 d x = − ∫ 0 + ∞ x 4 d 1 e x − 1 = 4 ∫ 0 + ∞ x 3 e x − 1 d x − x 4 e x − 1 ∣ 0 + ∞ = 4 ∫ 0 + ∞ x 3 e − x 1 − e − x d x = 4 ∫ 0 + ∞ ( x 3 ∑ n = 1 + ∞ e − n x ) d x = 4 ∑ n = 1 + ∞ ( ∫ 0 + ∞ x 3 e − n x d x ) = 4 ∑ n = 1 + ∞ [ − ( 1 n x 3 + 3 n 2 x 2 + 6 n 3 x + 6 n 4 ) e − n x ∣ 0 + ∞ ] = 24 ∑ n = 1 + ∞ 1 n 4 = 4 π 4 15 \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*} ∫0+∞========(ex−1)2x4exdx−∫0+∞x4dex−114∫0+∞ex−1x3dx−ex−1x4 0+∞4∫0+∞1−e−xx3e−xdx4∫0+∞(x3n=1∑+∞e−nx)dx4n=1∑+∞(∫0+∞x3e−nxdx)4n=1∑+∞[−(n1x3+n23x2+n36x+n46)e−nx 0+∞]24n=1∑+∞n41154π4
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：
ε max ε min } = ε 0 ∘ + ε 9 0 ∘ 2 ± 2 2 ( ε 0 ∘ − ε 4 5 ∘ ) 2 + ( ε 4 5 ∘ − ε 9 0 ∘ ) 2 \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} εmaxεmin}=2ε0∘+ε90∘±22(ε0∘−ε45∘)2+(ε45∘−ε90∘)2
tan 2 α 0 = 2 ε 4 5 ∘ − ε 0 ∘ − ε 9 0 ∘ ε 0 ∘ − ε 9 0 ∘ \tan 2\alpha_{0} = \frac{2\varepsilon_{45^{\circ}} - \varepsilon_{0^{\circ}} - \varepsilon_{90^{\circ}}}{\varepsilon_{0^{\circ}} - \varepsilon_{90^{\circ}}} tan2α0=ε0∘−ε90∘2ε45∘−ε0∘−ε90∘
6 0 ∘ 60^{\circ} 60∘应变花,三个应变片的角度分别为： α 1 = 0 ∘ , α 2 = 6 0 ∘ , α 3 = 12 0 ∘ \alpha_{1}=0^{\circ}, \alpha_{2}=60^{\circ}, \alpha_{3}=120^{\circ} α1=0∘,α2=60∘,α3=120∘。求证主应变的数值及方向由以下公式计算：
ε max ε min } = ε 0 ∘ + ε 6 0 ∘ + ε 12 0 ∘ 3 ± 2 3 ( ε 0 ∘ − ε 6 0 ∘ ) 2 + ( ε 6 0 ∘ − ε 12 0 ∘ ) 2 + ( ε 12 0 ∘ − ε 0 ∘ ) 2 \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} εmaxεmin}=3ε0∘+ε60∘+ε120∘±32(ε0∘−ε60∘)2+(ε60∘−ε120∘)2+(ε120∘−ε0∘)2
tan 2 α 0 = 3 ( ε 6 0 ∘ − ε 12 0 ∘ ) 2 ε 0 ∘ − ε 6 0 ∘ − ε 12 0 ∘ \tan 2\alpha_{0} = \frac{\sqrt{3}(\varepsilon_{60^{\circ}} - \varepsilon_{120^{\circ}})}{2\varepsilon_{0^{\circ}} - \varepsilon_{60^{\circ}} - \varepsilon_{120^{\circ}}} tan2α0=2ε0∘−ε60∘−ε120∘3(ε60∘−ε120∘)
证明
对于以上问题我们有
ε α 1 = ε x + ε y 2 + ε x − ε y 2 cos 2 α 1 + γ x y 2 sin 2 α 1 ε α 2 = ε x + ε y 2 + ε x − ε y 2 cos 2 α 2 + γ x y 2 sin 2 α 2 ε α 3 = ε x + ε y 2 + ε x − ε y 2 cos 2 α 3 + γ x y 2 sin 2 α 3 } \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\} εα1=2εx+εy+2εx−εycos2α1+2γxysin2α1εα2=2εx+εy+2εx−εycos2α2+2γxysin2α2εα3=2εx+εy+2εx−εycos2α3+2γxysin2α3⎭ ⎬ ⎫
将 ε x + ε y 2 \frac{\varepsilon_x+\varepsilon_y}{2} 2εx+εy、 ε x − ε y 2 \frac{\varepsilon_x-\varepsilon_y}{2} 2εx−εy和 γ x y \gamma_{xy} γxy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：
{ ε α 1 = A + B cos 2 α 1 + C sin 2 α 1 ε α 2 = A + B cos 2 α 2 + C sin 2 α 2 ε α 3 = A + B cos 2 α 3 + C sin 2 α 3 \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} ⎩ ⎨ ⎧εα1=A+Bcos2α1+Csin2α1εα2=A+Bcos2α2+Csin2α2εα3=A+Bcos2α3+Csin2α3
消去 $ A$ 后利用三角恒等式化简，通过线性方程组解得：
B = ε α 2 − ε α 1 2 sin ( α 2 − α 1 ) cos ( α 1 + α 3 ) − ε α 3 − ε α 1 2 sin ( α 3 − α 1 ) cos ( α 1 + α 2 ) sin ( α 3 − α 2 ) , C = ε α 3 − ε α 1 2 sin ( α 3 − α 1 ) sin ( α 1 + α 2 ) − ε α 2 − ε α 1 2 sin ( α 2 − α 1 ) sin ( α 1 + α 3 ) sin ( α 3 − α 2 ) , A = ε α 1 − B cos 2 α 1 − C sin 2 α 1 . \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*} BCA=sin(α3−α2)2sin(α2−α1)εα2−εα1cos(α1+α3)−2sin(α3−α1)εα3−εα1cos(α1+α2),=sin(α3−α2)2sin(α3−α1)εα3−εα1sin(α1+α2)−2sin(α2−α1)εα2−εα1sin(α1+α3),=εα1−Bcos2α1−Csin2α1.
还原变量得：
ε x = A + B , ε y = A − B , γ x y = 2 C . \varepsilon_x = A+B, \quad \varepsilon_y = A-B, \quad \gamma_{xy} = 2C. εx=A+B,εy=A−B,γxy=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 = E 2 ( 1 + μ ) G=\frac{E}{2(1+\mu)} G=2(1+μ)E K = E 3 ( 1 − 2 μ ) K=\frac{E}{3(1-2\mu)} K=3(1−2μ)E
证明
广义胡克定律
{ ε x = 1 E [ σ x − μ ( σ y + σ z ) ] ε x = 1 E [ σ x − μ ( σ y + σ z ) ] ε x = 1 E [ σ x − μ ( σ y + σ z ) ] γ x y = τ x y G γ y z = τ y z G γ z x = τ z x G \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} ⎩ ⎨ ⎧εx=E1[σx−μ(σy+σz)]εx=E1[σx−μ(σy+σz)]εx=E1[σx−μ(σy+σz)]γxy=Gτxyγyz=Gτyzγzx=Gτzx
(a)
首先，证明 G = E 2 ( 1 + μ ) \boxed{G=\frac{E}{2(1+\mu)}} G=2(1+μ)E
在纯剪切状态下：
σ x = σ y = σ z = 0 , τ x y ≠ 0 \sigma_x = \sigma_y = \sigma_z = 0, \quad \tau_{xy} \neq 0 σx=σy=σz=0,τxy=0
由剪切本构方程：
γ x y = τ x y G ( 1 ) \gamma_{xy} = \frac{\tau_{xy}}{G} \quad (1) γxy=Gτxy(1)
旋转坐标系 4 5 ∘ 45^\circ 45∘ 后，主应力为：
σ 1 = τ x y , σ 2 = − τ x y \sigma_{1} = \tau_{xy}, \quad \sigma_{2} = -\tau_{xy} σ1=τxy,σ2=−τxy
对应正应变为：
ε 1 = τ x y E ( 1 + μ ) , ε 2 = − τ x y E ( 1 + μ ) \varepsilon_1 = \frac{\tau_{xy}}{E}(1+\mu), \quad \varepsilon_2 = -\frac{\tau_{xy}}{E}(1+\mu) ε1=Eτxy(1+μ),ε2=−Eτxy(1+μ)
几何关系：
γ x y = ε 1 − ε 2 = 2 τ x y ( 1 + μ ) E ( 2 ) \gamma_{xy} = \varepsilon_1 - \varepsilon_2 = \frac{2\tau_{xy}(1+\mu)}{E} \quad (2) γxy=ε1−ε2=E2τxy(1+μ)(2)
联立 ( 1 ) (1) (1) 和 ( 2 ) (2) (2)：
τ x y G = 2 τ x y ( 1 + μ ) E ⟹ G = E 2 ( 1 + μ ) \frac{\tau_{xy}}{G} = \frac{2\tau_{xy}(1+\mu)}{E} \implies G = \frac{E}{2(1+\mu)} Gτxy=E2τxy(1+μ)⟹G=2(1+μ)E
(b)
其次，证明 K = E 3 ( 1 − 2 μ ) \boxed{K=\frac{E}{3(1-2\mu)}} K=3(1−2μ)E
对于六面体其变形前的体积
V = d x d y d z V=\text{d}x\text{d}y\text{d}z V=dxdydz
变形后体积
V 1 = ( 1 + ε 1 + ε 2 + ε 3 ) V V_1=(1+\varepsilon_1+\varepsilon_2+\varepsilon_3)V V1=(1+ε1+ε2+ε3)V
体应变
θ = ε 1 + ε 2 + ε 3 = 1 − 2 μ E ( σ 1 + σ 2 + σ 3 ) \theta=\varepsilon_1+\varepsilon_2+\varepsilon_3=\frac{1-2\mu}{E}(\sigma_1+\sigma_2+\sigma_3) θ=ε1+ε2+ε3=E1−2μ(σ1+σ2+σ3)
对比
θ = σ m K \theta=\frac{\sigma_m}{K} θ=Kσm
其中， σ m = σ 1 + σ 2 + σ 3 3 \sigma_m=\frac{\sigma_1+\sigma_2+\sigma_3}{3} σm=3σ1+σ2+σ3
得到
K = E 3 ( 1 − 2 μ ) K=\frac{E}{3(1-2\mu)} K=3(1−2μ)E