# 离心力对轴对称弯道河槽张力作用的研究Study on the Effect of Centrifugal Force on River Channel Tension in Axisymmetric Bend

DOI: 10.12677/APP.2019.911056, PDF, HTML, XML, 下载: 1,132  浏览: 3,943

Abstract: A uniform circular motion system model with symmetrical mirror image is adopted according to the action and reaction properties of centripetal force and centrifugal force. Under the action of constrained centrifugal force, the whole system generates reciprocating vibration. It is applied to the calculation of the effect of the centrifugal force on the tension of the river bed. It can be applied to the design of circular arc water conservancy projects and flood control safety.

1. 引言

2. 圆周运动物体系统的向心力与离心力是作用力与反作用力的依据

3. 匀速圆周运动物体系统在受约束离心力作用下产生周期性运动

3.1. 构建物理实验模型

Figure 1. Mirror symmetrical circular motion

Figure 2. Centrifugal force analysis

3.2. 匀速圆周运动系统离心力作用的解析

3.2.1. 匀速圆周运动系统离心力及在x轴方向上的冲量

${F}_{R}=-{{F}^{\prime }}_{R},$

${F}_{x}={F}_{R}\mathrm{sin}\varphi =-{{F}^{\prime }}_{R}\mathrm{sin}\varphi =-m\frac{{v}^{2}}{R}\mathrm{sin}\varphi$.

$\because {v}_{x}=v\mathrm{cos}\varphi$${v}_{{x}_{1}}=v\mathrm{cos}{\varphi }_{1}$${v}_{{x}_{2}}=v\mathrm{cos}{\varphi }_{2}$$t=\frac{R\varphi }{v}$

$\begin{array}{l}\therefore {\int }_{{t}_{1}}^{{t}_{2}}{F}_{x}\text{d}t={\stackrel{¯}{F}}_{x}\left({t}_{2}-{t}_{1}\right)={\int }_{{t}_{1}}^{{t}_{2}}\left(-m\frac{{v}^{2}}{R}\right)\mathrm{sin}\varphi \text{d}t={\int }_{{\varphi }_{1}}^{{\varphi }_{2}}\left(-m\frac{{v}^{2}}{R}\right)\mathrm{sin}\varphi \text{d}\left(\frac{R\varphi }{v}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=mv\mathrm{cos}{\varphi }_{2}-mv\mathrm{cos}{\varphi }_{1}=M{v}_{2\left(M\right)}-M{v}_{1\left(M\right)}\end{array}$

${\stackrel{¯}{F}}_{x}=\frac{mv\mathrm{cos}{\varphi }_{2}-mv\mathrm{cos}{\varphi }_{1}}{{t}_{2}-{t}_{1}}$.

${v}_{2\left(M\right)}=\frac{mv\mathrm{cos}{\varphi }_{2}-mv\mathrm{cos}{\varphi }_{1}+M{v}_{1\left(M\right)}}{M}$

${\int }_{0}^{\frac{T}{2}}{F}_{x}\text{d}t={\stackrel{¯}{F}}_{x}\frac{T}{2}={\stackrel{¯}{F}}_{x}\frac{\pi R}{v}={mv\mathrm{cos}{\varphi }_{2}-mv\mathrm{cos}{\varphi }_{1}|}_{0}^{\pi }=-2mv$

${\stackrel{¯}{F}}_{x}=\frac{mv\mathrm{cos}{\varphi }_{2}-mv\mathrm{cos}{\varphi }_{1}}{{t}_{2}-{t}_{1}}=\frac{-2m{v}^{2}}{\pi R}$.

${v}_{1\left(M\right)}=0$，且 $M=km$

${v}_{2\left(M\right)}=\frac{mv\mathrm{cos}{\varphi }_{2}-mv\mathrm{cos}{\varphi }_{1}}{km}=\frac{-2mv}{km}=-\frac{2}{k}v.$

$k=3$${v}_{2\left(M\right)}=-\frac{2}{3}v$$k=2$${v}_{2\left(M\right)}=-v$

${\int }_{\frac{T}{2}}^{T}{F}_{x}\text{d}t={\stackrel{¯}{F}}_{x}\frac{T}{2}={\stackrel{¯}{F}}_{x}\frac{\pi R}{v}={mv\mathrm{cos}{\varphi }_{2}-mv\mathrm{cos}{\varphi }_{1}|}_{\pi }^{2\pi }=2mv$

${\stackrel{¯}{F}}_{x}=\frac{mv\mathrm{cos}{\varphi }_{3}-mv\mathrm{cos}{\varphi }_{2}}{{t}_{3}-{t}_{2}}=\frac{2m{v}^{2}}{\pi R}$.

3.2.2. 离心力在x轴上的分量所做的功

$\begin{array}{c}{\int }_{{x}_{1}}^{{x}_{2}}{F}_{x}\text{d}x={\stackrel{¯}{F}}_{x}\left({x}_{2}-{x}_{1}\right)={\int }_{{x}_{1}}^{{x}_{2}}\left(-m\frac{{v}^{2}}{R}\right)\mathrm{sin}\varphi \text{d}x\\ ={\int }_{{x}_{1}}^{{x}_{2}}\left(-m\frac{{v}^{2}}{R}\right)\mathrm{sin}\varphi \text{d}\left(R\mathrm{sin}\varphi \right)\\ =\frac{1}{2}m{v}^{2}{\mathrm{cos}}^{2}{\varphi }_{2}-\frac{1}{2}m{v}^{2}{\mathrm{cos}}^{2}{\varphi }_{1}\\ =\frac{1}{2}M{v}_{2\left(M\right)}^{2}-\frac{1}{2}M{v}_{1\left(M\right)}^{2}\end{array}$

$W={\int }_{a}^{b}|f\left(x\right)\text{d}x|=±{\int }_{a}^{b}f\left(x\right)\text{d}x$

${C}_{1}$ 在0→π区间的转动，相对应的系统A速度为 ${v}_{0\left(M\right)}\to {v}_{M}$。根据参考文献 [3] [4] [5] 和分析研究，离心力在前半周(0→π/2)、(π/2→π)做正功，假设系统机械能守恒：

$\begin{array}{c}{W|}_{0}^{\pi }={\int }_{0}^{\frac{\pi }{2}}|{{F}^{\prime }}_{x}\text{d}x|+{\int }_{\frac{\pi }{2}}^{\pi }|{{F}^{\prime }}_{x}\text{d}x|\\ =\frac{1}{2}m{v}^{2}+\frac{1}{2}m{v}^{2}=\frac{1}{2}M{v}_{M}^{2}-\frac{1}{2}M{v}_{0\left(M\right)}^{2}\end{array}$

$M=km$，假设 ${v}_{0\left(M\right)}=0$

$\begin{array}{c}{\int }_{{x}_{1}}^{{x}_{2}}{F}_{x}\text{d}x={\stackrel{¯}{F}}_{x}\left[{x}_{2\left(M\right)}-{x}_{1\left(M\right)}\right]={\stackrel{¯}{F}}_{x}{X}_{M}=m{v}^{2}\\ =\frac{1}{2}M{v}_{2\left(M\right)}^{2}=\frac{1}{2}km{\left(-\frac{2v}{k}\right)}^{2}=\frac{2}{k}m{v}^{2}，\end{array}$

${X}_{M}=\frac{\frac{1}{2}M{v}_{2\left(M\right)}^{2}}{{\stackrel{¯}{F}}_{x}}=\frac{2}{k}m{v}^{2}\frac{\pi R}{2m{v}^{2}}=\frac{\pi R}{k}$

${X}_{M}=\frac{\pi R}{2}$.

3.2.3. 后半周(π→2π)区间离心力在x轴上的分量做的功和振幅

$\begin{array}{c}{|W||}_{\pi }^{2\pi }={\int }_{\pi }^{\frac{3\pi }{2}}|{{F}^{\prime }}_{x}\text{d}x|+{\int }_{\frac{3\pi }{2}}^{2\pi }|{{F}^{\prime }}_{x}\text{d}x|\\ =\frac{1}{2}m{v}^{2}+\frac{1}{2}m{v}^{2}=-\left[\frac{1}{2}M{v}_{{M}_{3}}^{2}-\frac{1}{2}M{v}_{M}^{2}\right].\end{array}$

${\stackrel{¯}{F}}_{x}=\frac{mv\mathrm{cos}{\varphi }_{3}-mv\mathrm{cos}{\varphi }_{2}}{{t}_{3}-{t}_{2}}=\frac{-2m{v}^{2}}{\pi R}$.

。(同上半周)

3.2.4. 前后半周(0→2π)做功之和

$W={W|}_{0}^{\pi }-{|W||}_{\pi }^{2\pi }=0$.

4. 轴对称 [6] 圆弧弯道水流离心力对河道张力作用的解析算法

Figure 3. Schematic diagram of arc curve river channel

Figure 4. River cross section

$M=Nm=n\mathcal{l}m$

$\begin{array}{c}{\int }_{-\mathcal{l}}^{\mathcal{l}}{F}_{x}n\text{d}\mathcal{l}=n{\int }_{{\varphi }_{1}}^{{\varphi }_{2}}{{F}^{\prime }}_{R}\mathrm{cos}\varphi \text{d}\left(R\varphi \right)=n{\int }_{{\varphi }_{1}}^{{\varphi }_{2}}m\frac{{v}^{2}}{R}\mathrm{cos}\varphi \text{d}\left(R\varphi \right)\\ =\frac{M}{\mathcal{l}}{v}^{2}\left(\mathrm{sin}{\varphi }_{2}-\mathrm{sin}{\varphi }_{1}\right)=2\frac{M}{\mathcal{l}}{v}^{2}\mathrm{sin}{\varphi }_{2}.\end{array}$

2012年7月24日长江三峡发生7万m3洪峰流量的罕见洪水，夹杂着泥沙直扑三峡大坝。如流经圆弧河槽段，平均河宽1000 m，平均河深5 m，圆弧河段有效长1000 m，有效半径5000 m， $\text{2}\varphi ={\text{17}}^{\circ }\text{27'}$，平均流速14 m/s，该段圆弧弯道水流沿x轴方向对河槽的张力：

${\int }_{-\mathcal{l}}^{\mathcal{l}}{F}_{x}n\text{d}\mathcal{l}=n{\int }_{{\varphi }_{1}}^{{\varphi }_{2}}{{F}^{\prime }}_{R}\mathrm{cos}\varphi \text{d}\left(R\varphi \right)=2\frac{M}{\mathcal{l}}{v}^{2}\mathrm{sin}{\varphi }_{2}=195673.5\text{\hspace{0.17em}}\text{N}=\text{19}\text{.57}\left(吨\right)$

5. 结论

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