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Bending Deformation of Hinged-Fixed Beams at Both Ends under Nonlinear Constitutive Relations
DOI: 10.12677/IJM.2021.101001, PDF, HTML, XML, 下载: 284  浏览: 530  国家自然科学基金支持

Abstract: Based on the classical beam theory, this paper assumes that the elastic modulus and the strain become a linear relationship (non-linear constitutive relationship), and derives the basic equation for the bending of a fixed hinge beam at both ends under a vertical uniform load. And the basic equations and boundary conditions are dimensionless to make them general. Then, the numerical method is used to solve the numerical results of the non-dimensional basic equation of the problem. The influence of the nonlinear constitutive relationship and the vertical distributed load on the bending deformation of the beam and the position of the neutral layer in the two cases of large deflection and small deflection are analyzed and compared.

1. 引言

2. 基本方程

2.1. 梁的模型

2.2. 几何方程

${\epsilon }_{\text{0}}=\frac{\text{d}u}{\text{d}x}+a\cdot \frac{1}{2}{\left(\frac{\text{d}w}{\text{d}x}\right)}^{2}$ (1)

$\epsilon =\frac{\text{d}u}{\text{d}x}-z\cdot \frac{{\text{d}}^{2}w}{\text{d}{x}^{2}}+a\cdot \frac{1}{2}{\left(\frac{\text{d}w}{\text{d}x}\right)}^{2}$ (2)

Figure 1. Schematic diagram of beam size and coordinate system

2.3. 物理方程

$E=E\left(\epsilon \right)={E}_{0}+{E}_{1}\epsilon ={E}_{0}\left(1+{E}_{r}\epsilon \right)$ (3)

$\sigma =E\epsilon =E\left(\epsilon \right)\cdot \epsilon ={E}_{0}\left(1+{E}_{r}\epsilon \right)\epsilon ={E}_{0}\left(\epsilon +{E}_{r}{\epsilon }^{2}\right)$ (4)

${z}_{0}=\left[\frac{\text{d}u}{\text{d}x}+a\cdot \frac{1}{2}{\left(\frac{\text{d}w}{\text{d}x}\right)}^{2}\right]/\frac{{\text{d}}^{2}w}{\text{d}{x}^{2}}$ (5)

$\begin{array}{l}{F}_{N}={\int }_{-h/2}^{h/2}\sigma \cdot b\text{d}z=\frac{1}{4}{E}_{0}b\left\{\frac{1}{3}{E}_{r}{\left(\frac{{\text{d}}^{2}w}{\text{d}{x}^{2}}\right)}^{2}{h}^{3}+\left[{E}_{r}{\left(\frac{\text{d}w}{\text{d}x}\right)}^{2}a+2{E}_{r}\frac{\text{d}u}{\text{d}x}+2\right]\left[{\left(\frac{\text{d}w}{\text{d}x}\right)}^{2}a+2\frac{\text{d}u}{\text{d}x}\right]h\right\}\\ M={\int }_{-h/2}^{h/2}-\sigma z\cdot b\text{d}z=\frac{1}{12}{E}_{0}b\frac{{\text{d}}^{2}w}{\text{d}{x}^{2}}\left[{E}_{r}{\left(\frac{\text{d}w}{\text{d}x}\right)}^{2}a+2{E}_{r}\frac{\text{d}u}{\text{d}x}+1\right]{h}^{3}\end{array}$ (6)

2.4. 平衡方程

$\frac{\text{d}{F}_{N}}{\text{d}x}=0$ (7)

$\frac{{\text{d}}^{2}M}{\text{d}{x}^{2}}-{F}_{N}\frac{{\text{d}}^{2}w}{\text{d}{x}^{2}}-q=0$ (8)

$u=0$, $w=0$, $M=0$ (9)

3. 对基本方程和边界条件进行无量纲处理

$W=\frac{w}{h}$, $X=\frac{x}{l}$, $U=\frac{ul}{{h}^{2}}$, $\delta =\frac{h}{l}$, $Q=\frac{12q}{{E}_{0}b{\delta }^{4}}$ (10)

$\begin{array}{l}\stackrel{¯}{{F}_{N}}=\frac{12{F}_{N}}{{E}_{0}bl{\delta }^{3}}={E}_{r}{\delta }^{2}{\left(\frac{{\text{d}}^{2}W}{\text{d}{X}^{2}}\right)}^{2}+3\left[{E}_{r}{\delta }^{2}a{\left(\frac{\text{d}W}{\text{d}X}\right)}^{2}+2{E}_{r}{\delta }^{2}\frac{\text{d}U}{\text{d}X}+2\right]\left[{\left(\frac{\text{d}W}{\text{d}X}\right)}^{2}a+2\frac{\text{d}U}{\text{d}X}\right]\\ \stackrel{¯}{M}=\frac{12M}{{E}_{0}b{l}^{2}{\delta }^{4}}=\frac{{\text{d}}^{2}W}{\text{d}{X}^{2}}\left[{E}_{r}{\delta }^{2}{\left(\frac{\text{d}W}{\text{d}X}\right)}^{2}a+2{E}_{r}{\delta }^{2}\frac{\text{d}U}{\text{d}X}+1\right]\end{array}$ (11)

$\frac{\text{d}\stackrel{¯}{{F}_{N}}}{\text{d}X}=0$, $\frac{{\text{d}}^{2}\stackrel{¯}{M}}{\text{d}{X}^{2}}-Q-\stackrel{¯}{{F}_{N}}\frac{{\text{d}}^{2}W}{\text{d}{X}^{2}}=0$ (12)

4. 数值求解与结果分析

Figure 2. Relationship between beam midpoint deflection W(0.5) and load Q at small deflection

(a) 弯曲挠度 (b) 中性轴位置

Figure 3. Bending deflection and neutral axis position of small deflection beam when Q = 50

(a) 中点挠度W(0.5)与载荷Q关系 (b) 中性轴位置

Figure 4. The relationship between the midpoint deflection W(0.5) of the large deflection beam and the load Q and the position of the neutral axis when Q = 90

5. 结论

1) Er影响梁的弯曲刚度，且同一载荷下Er增大时梁的弯曲刚度增大。小挠度情况下Er取正值和负值时梁的中性层位置相反，大挠度情况下中性层位置在同一侧。

2) 两端固定铰支使梁的几何中面轴向变形受到约束，因而 ${E}_{r}\ne 0$ 时梁的横截面产生了轴向力，所以中性层位置改变了且随着Er的变化而改变。

3) 若研究梁的弯曲问题时考虑了本构关系的非线性，采用大挠度问题的方程求解较为合适。

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