## 由一张A4纸引发的思考1)

*南京航空航天大学,南京 210016

## INSPIRATION OF A SHEET OF A4 PAPER1)

HUANG Ke*, GUO Yujun*, HAO Yang*, LI Dongdong,*,2), FAN Qinshan*,

*Nanjing University of Aeronautics and Astronautics, Nanjing 210016, China

Department of Engineering Mechanics, School of Aerospace Engineering, Tsinghua University, Beijing 100084, China

 基金资助: 1)江苏高校优势学科建设工程资助项目

Abstract

In order to explain the phenomenon that "a sheet of printing paper cannot be erected, while a certain degree curved one can" in terms of mechanics, the geometric properties of the cross section under two different states are analyzed to form a formula for the principal centroidal moments of inertia, which is verified by numerical simulations. The stability of the sheet of paper or a curved one is reduced into the stability of a column under its own weight. It is found that the curvature of the arc has a great influence on the stability. When the arc is 2.52, the critical value of the evenly distributed axial load of the cylindrical A4 paper is about $3.7\times10^5$ times of that of the planar one.

Keywords： printing paper; moment of inertia; stability analysis

HUANG Ke, GUO Yujun, HAO Yang, LI Dongdong, FAN Qinshan. INSPIRATION OF A SHEET OF A4 PAPER1). MECHANICS IN ENGINEERING, 2021, 43(5): 801-805 DOI:10.6052/1000-0879-20-310

### 图1

A4打印纸(以下简称A4纸)是生活中十分常见的物品,在很多中小学生的趣味活动中,也有将A4纸立起来的项目。虽然说将A4纸弯曲就能将其立起来是一件非常容易的事情,但是这背后的力学知识可不简单。

## 1 截面几何性质的理论计算

### 图2

$\left. {\begin{array}{l} I_{z} =\int_A {y^{2}{\rm d}A} \\ I_{y} =\int_A {z^{2}{\rm d}A} \\ \end{array}} \right\}$

$\left. {\begin{array}{l} I_{z} =\dfrac{bh^{3}}{12} \\ I_{y} =\dfrac{hb^{3}}{12} \\ \end{array}} \right\}$

### 1.2 圆柱面状A4纸

$\rho =\dfrac{b}{2\alpha }$

### 图3

$\left. {\begin{array}{l} z_{C} =\dfrac{\int_A {z{\rm d}A} }{A} \\ y_{C} =\dfrac{\int_A {y{\rm d}A} }{A} \\ \end{array}} \right\}$

$\left. {\begin{array}{l} z=\rho \cos \theta \\ y=\rho \sin \theta \\ \end{array}} \right\}$

${\rm d}A=\rho {\rm d}\theta {\rm d}\rho$

### 图4

$\left. {\begin{array}{l} z_{C} =0 \\ y_{C} =\dfrac{\int_{{\pi}/{2}-\alpha }^{{\pi}/{2}+\alpha }{\int_{\rho -{h}/{2}}^{\rho +{h}/{2}} {(\rho \sin \theta )\rho{\rm d}\theta {\rm d}\rho } } }{hb}=\\[3mm]\qquad \dfrac{2\sin \alpha }{b}\left( {\rho^{2}+\dfrac{1}{12}h^{2}} \right) \\ \end{array}} \right\}$

$a=y_{C} =\dfrac{2\sin \alpha }{b}\left( {\rho^{2}+\dfrac{1}{12}h^{2}}\right)$

$\left. {\begin{array}{l} I_{z} =\left( {\alpha +\dfrac{1}{2}\sin 2\alpha } \right)\left( {\rho^{3}h+\dfrac{1}{4}\rho h^{3}} \right) \\ I_{y} =\left( {\alpha -\dfrac{1}{2}\sin 2\alpha } \right)\left( {\rho^{3}h+\dfrac{1}{4}\rho h^{3}} \right) \\ \end{array}} \right\}$

$\left. {\begin{array}{l} I_{z_{C} } =I_{z} -a^{2}A \\ I_{y_{C} } =I_{y} \\ \end{array}} \right\}$

$\left. {\begin{array}{l} I_{z_{C} } =\left( {\alpha +\dfrac{1}{2}\sin 2\alpha } \right)\left({\dfrac{b^{3}h}{8\alpha^{3}}+\dfrac{bh^{3}}{8\alpha }} \right)-\\[3mm]\qquad \dfrac{h\sin^{2}\alpha }{4b}\left( {\dfrac{b^{2}}{\alpha^{2}}+\dfrac{1}{3}h^{2}}\right)^{2} \\ I_{y_{C} } =\left( {\alpha -\dfrac{1}{2}\sin 2\alpha } \right)\left({\dfrac{b^{3}h}{8\alpha^{3}}+\dfrac{bh^{3}}{8\alpha }} \right) \\ \end{array}} \right\}$

## 2 计算机仿真模拟与验证

$I_{z_{C}} =6 466.79~{\rm mm}^{4},\ \ I_{y_{C}} =55 934.93~{\rm mm}^{4}$

$\left. {\begin{array}{l} I_{z_{C} } =\rho^{3}h\left( {\dfrac{b}{2\rho }+\dfrac{1}{2}\sin\dfrac{b}{\rho }} \right)-\\[3mm]\qquad \dfrac{4h\rho^{4}\sin^{2}\left( {{b/{2\rho }}}\right)}{b}=6 491.68~ {\rm mm}^{4} \\[2mm] I_{y_{C} } =\rho^{3}h\left( {\dfrac{b}{2\rho }-\dfrac{1}{2}\sin\dfrac{b}{\rho }} \right)=56 095.60~{\rm mm}^{4} \\ \end{array}} \right\}$

$\left. {\begin{array}{l} I_{z_{C} } \to \dfrac{bh^{3}}{12} \\[1.5mm] I_{y_{C} } \to \dfrac{hb^{3}}{12} \\ \end{array}} \right\}$

$I_{z_{C} } -I_{y_{C} } =\frac{2\rho^{3}h\sin [b/(2\rho)]}{b}\cdot \ \left( {b\cos [b/(2\rho)]-2\rho \sin [b/(2\rho)]}\right)$

## 3 两种不同初始形状A4纸截面惯性矩的对比

A4纸抵抗绕$z$轴发生失稳的能力,与截面对于$z$轴的形心主惯性矩有直接关系。由上面的计算我们可以看到：当曲率半径$\rho = 83.4$ mm时,圆柱面状A4纸截面对于$z$轴的形心主惯性矩$I_{z_{C}} =6 491.68$ mm$^{4}$,而平面状A4纸截面对于$z$轴的形心主惯性矩$I_{z_{C}} =0.017 5$ mm$^{4}$。前者约为后者的$3.7\times10^{5}$倍,也就是说圆柱面状A4纸抵抗绕$z$轴发生失稳的能力远远强于平面状A4纸。

## 4 对于形心主惯性矩极值的探讨

$I_{z_{C} } =\left( {\alpha +\dfrac{1}{2}\sin 2\alpha } \right)\left({\dfrac{b^{3}h}{8\alpha^{3}}+\dfrac{bh^{3}}{8\alpha }} \right)-\ \frac{h\sin^{2}\alpha }{4b}\left( {\dfrac{b^{2}}{\alpha^{2}}+\frac{1}{3}h^{2}}\right)^{2}$

## 5 对于此问题本质的讨论

### 5.2 利用能量法确定均布轴向载荷的临界值

$w\left( x \right)=a_{1} \left( {1-\cos \dfrac{\pi x}{2l}} \right)$

### 图6

$\varDelta V_{{\rm t}} =V_{{\rm P}} +V_{\varepsilon}$

$\left. {\begin{array}{l} V_{{\rm P}} =-\int_l { {q{\rm d}x} } \varDelta \left( x \right) \\[3mm] V_{\varepsilon} =\int_0^l {\dfrac{M^{2}}{2EI}} {\rm d}x \\ \end{array}} \right\}$

$\varDelta \left( x \right)=\int_0^x {\left({{\rm d}x-{\rm d}x\cos \theta } \right)} =\int_0^x {\left( {1-\cos \theta } \right)}{\rm d}x=\ \ \frac{1}{2}\int_0^x {\left( {\frac{{\rm d}w}{{\rm d}x}} \right)}^{2}{\rm d}x$

$\varDelta \left( x \right)=\frac{1}{2}\int_0^x{\left( {\frac{{\rm d}w}{{\rm d}x}} \right)}^{2}{\rm d}x=\ \ \frac{1}{2}\left( {\frac{\pi }{2l}} \right)^{2}a_{1}^{2}\left( {-\frac{l}{\pi }\sin \frac{\pi x}{2l}\cos \dfrac{\pi x}{2l}+\frac{x}{2}} \right)$

$V_{{\rm P}} =-\int_0^l {{q{\rm d}x} \varDelta \left( x \right)} =-q\dfrac{\pi^{2}}{32}a_{1}^{2} \times \dfrac{\pi ^{2}-4}{\pi^{2}}$

$M=\int_x^l q (\eta -w){\rm d}\xi$

$\eta =a_{1} \left( {1-\cos \dfrac{\pi \xi }{2l}} \right)$

$M=a_{1} q\left[ {(l-x)\cos \dfrac{\pi x}{2l}-\dfrac{2l}{\pi }\left( {1-\sin \dfrac{\pi x}{2l}} \right)} \right]$

$V_{\varepsilon} =\int_0^l {\dfrac{M^{2}}{2EI}} {\rm d}x=\dfrac{a_{1}^{2}q^{2}l^{3}}{2EI}\left( {\dfrac{1}{6}+\dfrac{9}{\pi^{2}}-\dfrac{32}{\pi ^{3}}} \right)$

$q_{\rm cr} =\dfrac{7.89EI}{l^{3}}$

$q_{\rm cr1} =\dfrac{7.837EI}{l^{3}}$

## 6 对于纸张稳定性的定量分析

### 6.1 纸张在两种状态下的稳定性分析

$q_{\rm cr} =\dfrac{7.837EI}{l^{3}}=6.654\times 10^{-3}~{\rm N/m}$

$q_{纸} =9.8kb=0.093 {\rm N/m} \gg q_{\rm cr}$

$q_{\rm cr} =\dfrac{7.837EI}{l^{3}}=2 468.22~\mbox{N/m}\gg q_{纸}$

### 6.2 对于圆柱面状A4纸在自重作用下发生屈曲的临界弧度

$I_{z} =\dfrac{q_{纸}l^{3}}{7.837E}=2.568 9\times 10^{-4}~{\rm mm}^{4}$

$\rho_{\rm cr} =1 485 964.35~{\rm mm}$

$\theta_{\rm cr} =2\alpha =\dfrac{b}{\rho_{\rm cr} }=1.413\times 10^{-4}~{\rm rad}$

## 7 结论

(1) 用于推导惯性矩时使用的方法,具有一定的普适性及合理性。

(2) 在$\alpha$的定义域范围内,对于$z$轴的形心主惯性矩随着$\alpha$先增大后减小,且在$\alpha = 2.577$ rad时取得最大值,即在此时圆柱面状A4纸最稳定。

(3)本文给出的均布轴向载荷临界值的表达式精度较高,满足一般的工程需求,且推导过程简洁。

(4)平面状A4纸立不起来,其力学本质是自重状态下发生了失稳,而圆柱面状A4纸能够立起来,是因为截面形状的改变使得形心主惯性矩显著增加,从而增强了A4纸抵抗失稳的能力。

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