﻿ 梯度型非局部高阶梁理论与非局部弯曲新解法
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 力学学报  2016, Vol. 48 Issue (1): 127-134  DOI: 10.6052/0459-1879-15-170 0

### 引用本文 [复制中英文]

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Chen Ling, Shen Jiping, Li Cheng, Liu Xinpei. GRADIENT TYPE OF NONLOCAL HIGHER-ORDER BEAM THEORY AND NEW SOLUTION METHODOLOGY OF NONLOCAL BENDING DEFLECTION[J]. Chinese Journal of Ship Research, 2016, 48(1): 127-134. DOI: 10.6052/0459-1879-15-170.
[复制英文]

### 文章历史

2015-05-12收稿
2015-08-17录用
2015-08-27网络版发表

1 梯度型非局部梁的高阶弯曲模型

 ${\sigma _{ij}}\left( r \right) = \int {_V} \alpha \left( {\left| {r' - r} \right|,\tau } \right){{\sigma '}_{ij}}\left( {r'} \right)\;{\rm{d}}V\left( {r'} \right)$ (1)
 $\sigma _{ij}-\left( {e_0 a} \right)^2\nabla ^2\sigma _{ij} = {\sigma }'_{ij}$ (2)

 $\sigma - {\mkern 1mu} {\left( {{e_0}a} \right)^2}\frac{{{{\rm{d}}^2}\sigma }}{{{\rm{ d}}{x^2}}} = \sigma '$ (3)

 $\sigma ' = - Ey\frac{{{{\rm{d}}^2}w}}{{{\rm{d}}{x^2}}}$ (4)

 ${\sigma _n} - {\left( {{e_0}a} \right)^2}\frac{{{d^2}{\sigma _{n - 1}}}}{{{\rm{d}}{x^2}}} = \sigma '$ (5)

 ${\sigma _n} = - Ey\sum\limits_{m = 0}^n {{{\left( {{e_0}a} \right)}^{2m}}} \frac{{{d^{2m + 2}}w}}{{d{x^{2m + 2}}}}$ (6)

 $\begin{array}{l} \sigma = - Ey\sum\limits_{m = 0}^\infty {{{\left( {{e_0}a} \right)}^{2m}}} \frac{{{{\rm{d}}^{2m + 2}}w}}{{{\rm{d}}{x^{2m + 2}}}} = \\ \qquad \sigma ' - Ey\sum\limits_{m = 1}^\infty {{{\left( {{e_0}a} \right)}^{2m}}} \frac{{{{\rm{d}}^{2m + 2}}w}}{{{\rm{d}}{x^{2m + 2}}}} \end{array}$ (7)

 $\begin{array}{l} \sigma = \frac{1}{{\left[{1 - {{\left( {{e_0}a} \right)}^2}\frac{{{{\rm{d}}^2}}}{{{\rm{d}}{x^2}}}} \right]\frac{1}{{\sigma '}}}} = \\ \qquad \sigma ' + {\left( {{e_0}a} \right)^2}\frac{{{{\rm{d}}^2}\sigma '}}{{{\rm{d}}{x^2}}} + {\left( {{e_0}a} \right)^4}\frac{{{{\rm{d}}^4}\sigma '}}{{{\rm{d}}{x^4}}} + \cdots = \\ \qquad \sum\limits_{m = 0}^\infty {{{\left( {{e_0}a} \right)}^{2m}}} \frac{{{{\rm{d}}^{2m}}\sigma '}}{{{\rm{d}}{x^{2m}}}} = - Ey\sum\limits_{m = 0}^\infty {{{\left( {{e_0}a} \right)}^{2m}}} \frac{{{{\rm{d}}^{2m + 2}}w}}{{{\rm{d}}{x^{2m + 2}}}} \end{array}$ (8)

 $\frac{M}{{E{I_z}}} = \frac{{{{\rm{d}}^2}w}}{{{\rm{d}}{x^2}}} + {\rm{ }}\sum\limits_{m = 1}^\infty {{{\left( {{e_0}a} \right)}^{2m}}} \frac{{{{\rm{d}}^{2m + 2}}w}}{{{\rm{d}}{x^{2m + 2}}}}$ (9)

2 高阶弯曲方程的正则摄动解法

 $\begin{array}{l} \frac{{{q_0}L}}{{2E{I_z}}}x - \frac{{{q_0}}}{{2E{I_z}}}{x^2} + \frac{{{M_{\rm{e}}}x}}{{E{I_z}L}} = \\ \qquad \frac{{{{\rm{d}}^2}w}}{{{\rm{d}}{x^2}}} + \sum\limits_{m = 1}^\infty {{{\left( {{e_0}a} \right)}^{2m}}} \frac{{{{\rm{d}}^{2m + 2}}w}}{{{\rm{d}}{x^{2m + 2}}}} \end{array}$ (10)
 图 1 简支梁受均布载荷及力偶矩作用示意图 Fig.1 Sketch of a simply supported beam subjected to uniformly distributed loads and moment of couple

 $\begin{array}{l} \frac{{{{\bar q}_0}}}{2}\bar x - \frac{{{{\bar q}_0}}}{2}{{\bar x}^2} + {{\bar M}_{\rm{e}}}\bar x = \\ \qquad \frac{{{{\rm{d}}^2}\bar w}}{{{\rm{d}}{{\bar x}^2}}} + \sum\limits_{m = 1}^\infty {{\tau ^{2m}}} \frac{{{{\rm{d}}^{2m + 2}}\bar w}}{{{\rm{d}}{{\bar x}^{2m + 2}}}} \end{array}$ (11)

 $\bar {w}\left( 0 \right) = \bar {w}\left( 1 \right) = 0$ (12)

 $\begin{array}{l} \frac{{{{\bar q}_0}}}{2}\bar x - \frac{{{{\bar q}_0}}}{2}{{\bar x}^{{\kern 1pt} 2}} + {{\bar M}_{\rm{e}}}\bar x = \\ \qquad {\tau ^0}\left( {\frac{{{{\rm{d}}^2}{{\bar w}_0}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 2}}}} + \tau \frac{{{{\rm{d}}^2}{{\bar w}_1}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 2}}}} + {\tau ^2}\frac{{{{\rm{d}}^2}{{\bar w}_2}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 2}}}} + {\tau ^3}\frac{{{{\rm{d}}^2}{{\bar w}_3}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 2}}}} + \cdots } \right) + \\ \qquad {\tau ^2}\left( {\frac{{{{\rm{d}}^4}{{\bar w}_0}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 4}}}} + \tau \frac{{{{\rm{d}}^4}{{\bar w}_1}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 4}}}} + {\tau ^2}\frac{{{{\rm{d}}^4}{{\bar w}_2}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 4}}}} + {\tau ^3}\frac{{{{\rm{d}}^4}{{\bar w}_3}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 4}}}} + \cdots } \right) + \\ \qquad {\tau ^4}\left( {\frac{{{{\rm{d}}^6}{{\bar w}_0}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 6}}}} + \tau \frac{{{{\rm{d}}^6}{{\bar w}_1}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 6}}}} + {\tau ^2}\frac{{{{\rm{d}}^6}{{\bar w}_2}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 6}}}} + {\tau ^3}\frac{{{{\rm{d}}^6}{{\bar w}_3}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 6}}}} + \cdots } \right) + \\ \qquad {\tau ^6}\left( {\frac{{{{\rm{d}}^8}{{\bar w}_0}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 8}}}} + \tau \frac{{{{\rm{d}}^8}{{\bar w}_1}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 8}}}} + {\tau ^2}\frac{{{{\rm{d}}^8}{{\bar w}_2}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 8}}}} + {\tau ^3}\frac{{{{\rm{d}}^8}{{\bar w}_3}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 8}}}} + \cdots } \right) + \cdots = \\ \qquad \sum\limits_{m = 0}^\infty {{\tau ^{2m}}} \left( {\sum\limits_{n = 0}^\infty {{\tau ^n}} \frac{{{{\rm{d}}^{2m + 2}}{{\bar w}_n}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 2m + 2}}}}} \right) \end{array}$ (13)

 $\left. \begin{array}{l} \frac{{{{\bar q}_0}}}{2}\bar x - \frac{{{{\bar q}_0}}}{2}{{\bar x}^{{\kern 1pt} 2}} + {{\bar M}_{\rm{e}}}\bar x = \frac{{{{\rm{d}}^2}{{\bar w}_0}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 2}}}}\\ \frac{{{{\rm{d}}^2}{{\bar w}_1}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 2}}}} = 0\\ \frac{{{{\rm{d}}^2}{{\bar w}_2}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 2}}}} + \frac{{{{\rm{d}}^4}{{\bar w}_0}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 4}}}} = 0\\ \frac{{{{\rm{d}}^2}{{\bar w}_3}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 2}}}} + \frac{{{{\rm{d}}^4}{{\bar w}_1}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 4}}}} = 0\\ \frac{{{{\rm{d}}^2}{{\bar w}_4}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 2}}}} + \frac{{{{\rm{d}}^4}{{\bar w}_2}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 4}}}} + \frac{{{{\rm{d}}^6}{{\bar w}_0}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 6}}}} = 0\\ \frac{{{{\rm{d}}^2}{{\bar w}_5}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 2}}}} + \frac{{{{\rm{d}}^4}{{\bar w}_3}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 4}}}} + \frac{{{{\rm{d}}^6}{{\bar w}_1}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 6}}}} = 0\\ \quad \quad \quad \quad \quad \vdots \\ \frac{{{{\rm{d}}^2}{{\bar w}_{2j - 1}}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 2}}}} + \frac{{{{\rm{d}}^4}{{\bar w}_{2j - 3}}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 4}}}} + \frac{{{{\rm{d}}^6}{{\bar w}_{2j - 5}}}}{{{\rm{d}}{{\bar x}^6}}} + \cdots + \\ \qquad \frac{{{{\rm{d}}^{2j}}{{\bar w}_1}}}{{{\rm{d}}{{\bar x}^{2j}}}} = 0\frac{{{{\rm{d}}^2}{{\bar w}_{2j}}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 2}}}} + \frac{{{{\rm{d}}^4}{{\bar w}_{2j - 2}}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 4}}}} + \frac{{{{\rm{d}}^6}{{\bar w}_{2j - 4}}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 6}}}} + \cdots + \\ \qquad \frac{{{{\rm{d}}^{2j + 2}}{{\bar w}_0}}}{{{\rm{d}}{{\bar x}^{{\kern 1pt} 2j + 2}}}} = 0 \end{array} \right\}$ (14)

 $\bar {w}_0 = \dfrac{\bar {q}_0 \bar {x}^{3}}{12} - \dfrac{\bar {q}_0 \bar {x}^{4}}{24} + \dfrac{\bar {M}_{\rm e} \bar {x}^{3}}{6} + C_0 \bar {x} + D_0$ (15a)
 $\bar {w}_2 = \dfrac{\bar {q}_0 \bar {x}^{2}}{2} + C_2 \bar {x} + D_2$ (15b)
 $\bar {w}_j = C_j \bar {x} + D_j ,\ \ \left( {j = 1,3,4,5,\cdots } \right)$ (15c)

 $\bar {w}_0 = \dfrac{\bar {q}_0 \bar {x}^{3}}{12} - \dfrac{\bar {q}_0 \bar {x}^{4}}{24} + \dfrac{\bar {M}_{\rm e} \bar {x}^{3}}{6} - \dfrac{\bar {q}_0 + 4\bar {M}_{\rm e} }{24}\bar {x}$ (16a)
 $\bar {w}_2 = \dfrac{\bar {q}_0 \bar {x}^{2}}{2} - \dfrac{\bar {q}_0 \bar {x}}{2}$ (16b)
 $\bar {w}_j = 0 \ \ \ \left( {j = 1,3,4,5,\cdots } \right)$ (16c)

 $\begin{array}{l} \bar w = \frac{{{{\bar q}_0}{{\bar x}^{{\kern 1pt} 3}}}}{{12}} - \frac{{{{\bar q}_0}{{\bar x}^{{\kern 1pt} 4}}}}{{24}} + \frac{{{{\bar M}_{\rm{e}}}{{\bar x}^{{\kern 1pt} 3}}}}{6} - \\ \qquad \frac{{{{\bar q}_0} + 4{{\bar M}_{\rm{e}}}}}{{24}}\bar x + {\tau ^2}\left( {\frac{{{{\bar q}_0}{{\bar x}^{{\kern 1pt} 2}}}}{2} - \frac{{{{\bar q}_0}\bar x}}{2}} \right) \end{array}$ (17)

 $\begin{array}{l} \bar w = \frac{{{{\bar q}_0}{{\bar x}^3}}}{{12}} - \frac{{{{\bar q}_0}{{\bar x}^4}}}{{24}} + \frac{{{{\bar M}_{\rm{e}}}{{\bar x}^3}}}{6} + {C_0}\bar x + \\ \qquad {D_0} + {\tau ^2}\left( {\frac{{{{\bar q}_0}{{\bar x}^2}}}{2} + {C_2}\bar x + {D_2}} \right) + \\ \qquad \tau \left( {{C_1}\bar x + {D_1}} \right) + \sum\limits_{n = 3}^\infty {{\tau ^n}} \left( {{C_n}\bar x + {D_n}} \right) \end{array}$ (18)

 ${D_0} + \sum\limits_{n = 1}^\infty {{\tau ^n}} \;{D_n} = 0$ (19a)
 $\begin{array}{l} \frac{{{{\bar q}_0} + 4{{\bar M}_{\rm{e}}}}}{{24}} + {C_0} + {D_0} + \\ \qquad \frac{{{\tau ^2}{{\bar q}_0}}}{2} + {\rm{ }}\sum\limits_{n = 1}^\infty {{\tau ^n}} \;\left( {{C_n} + {D_n}} \right) = 0 \end{array}$ (19b)

 $\frac{{{{\bar q}_0} + 4{{\bar M}_{\rm{e}}}}}{{24}} + {C_0} + \frac{{{\tau ^2}{{\bar q}_0}}}{2} + {\rm{ }}\sum\limits_{n = 1}^\infty {{\tau ^n}} \;{C_n} = 0$ (20)

 $\left. \begin{array}{l} {C_0} = - \frac{{{{\bar q}_0} + 4{{\bar M}_{\rm{e}}}}}{{24}}\\ {C_2} = - \frac{{{{\bar q}_0}}}{2}\\ {C_j} = 0\;\;\;\left( {j = 1,3,4,5,\cdots } \right)\\ {D_i} = 0\;\;\;\left( {i = 0,1,2,3,\cdots } \right) \end{array} \right\}$ (21)

 $\begin{array}{l} \frac{{{p_0}L}}{{2E{I_z}}}x - \frac{{{p_0}}}{{6E{I_z}L}}{x^3} - \frac{{{p_0}{L^2}}}{{3E{I_z}}} = \\ \qquad \frac{{{{\rm{d}}^2}w}}{{{\rm{d}}{x^2}}} + \sum\limits_{m = 1}^\infty {{{\left( {{e_0}a} \right)}^{2m}}} \;\frac{{{{\rm{d}}^{2m + 2}}w}}{{{\rm{d}}{x^{2m + 2}}}} \end{array}$ (22)
 图 2 悬臂梁受线性分布载荷示意图 Fig.2 Sketch of a cantilever beam subjected to linearly distributed loads

 $\bar w = \frac{{10{{\bar p}_0}{{\bar x}^{{\kern 1pt} 3}} - {{\bar p}_0}{{\bar x}^{{\kern 1pt} 5}} - 20{{\bar p}_0}{{\bar x}^{{\kern 1pt} 2}} + 20{\tau ^2}{{\bar p}_0}{{\bar x}^{{\kern 1pt} 3}}}}{{120}}{\rm{ }}$ (23)

 $\begin{array}{l} - \frac{{{q_0}}}{{2E{I_z}}}{x^2} + \frac{{{q_0}L}}{{2E{I_z}}}x - \frac{{{q_0}{L^2}}}{{12E{I_z}}} = \\ \qquad \frac{{{{\rm{d}}^2}w}}{{{\rm{d}}{x^2}}} + \sum\limits_{m = 1}^\infty {\;{{\left( {{e_0}a} \right)}^{2m}}} \frac{{{{\rm{d}}^{2m + 2}}w}}{{{\rm{d}}{x^{2m + 2}}}} \end{array}$ (24)
 图 3 固支梁受均布载荷示意图 Fig.3 Sketch of a clamped beam subjected to uniformly distributed loads

 $\bar w = - \frac{{{{\bar q}_0}{{\bar x}^{{\kern 1pt} 4}}}}{{24}} + \frac{{{{\bar q}_0}{{\bar x}^{{\kern 1pt} 3}}}}{{12}} - \frac{{{{\bar q}_0}{{\bar x}^{{\kern 1pt} 2}}}}{{24}}{\rm{ }}$ (25)

3 数值算例与分析 3.1 算例

 图 4 非局部弯曲挠度随着轴坐标及非局部尺度因子的变化趋势 Fig.4 The variations of nonlocal bending deflections with respect to axial coordinate and nonlocal scale factor

 ${\bar w_{{\rm{mid}}}} = - \frac{{5{{\bar q}_0}}}{{384}} - \frac{{{{\bar M}_{\rm{e}}}}}{{16}} - \frac{{{{\bar q}_0}}}{8}{\tau ^2}$ (26)

 $\frac{{{{\bar w}_{{\rm{mid}}}}}}{{{{\bar w}_{{\rm{c - mid}}}}}} = 1 + \frac{{48{{\bar q}_0}{\tau ^2}}}{{5{{\bar q}_0} + 24{{\mathop {\bar M}\limits^\; }_{\rm{e}}}}}$ (27)

 图 5 各种边界及外载下非局部挠度与经典挠度之比随非局部尺度因子的变化 Fig.5 The ratio of nonlocal to classical deflections versus nonlocal scale factor for various boundary conditions and external loads

 $\frac{{{{\bar w}_{{\rm{free}}}}}}{{{{\bar w}_{{\rm{c - free}}}}}} = 1 - \frac{{20{\tau ^2}}}{{11}}$ (28)

3.2 分析与讨论

 $6{\bar x_m} - \bar x_m^3 + 12{\tau ^2}{\bar x_m} - 8 = 0{\rm{ }}$ (29)

 图 6 悬臂梁最大非局部挠度发生位置与非局部尺度因子的关系 Fig.6 Relation between the position of maximum nonlocal deflection and nonlocal scale factor for a cantilever beam

 $\frac{{{F_0}}}{{E{I_z}}}x + \frac{{{F_0}L}}{{E{I_z}}} = \frac{{{{\rm{d}}^2}w}}{{{\rm{d}}{x^2}}} + \sum\limits_{m = 1}^\infty {{{\left( {{e_0}a} \right)}^{2m}}} \;\frac{{{{\rm{d}}^{2m + 2}}w}}{{{\rm{d}}{x^{2m + 2}}}}$ (30)

 $\bar w = \frac{{{{\bar F}_0}{{\bar x}^3}}}{6} - \frac{{{{\bar F}_0}{{\bar x}^2}}}{2}$ (31)

 $\begin{array}{l} \frac{{3{p_0}L}}{{20E{I_z}}}x - \frac{{{p_0}}}{{6E{I_z}L}}{x^3} - \frac{{{p_0}{L^2}}}{{30E{I_z}}} = \\ \qquad \frac{{{{\rm{d}}^2}w}}{{{\rm{d}}{x^2}}} + \sum\limits_{m = 1}^\infty {{{\left( {{e_0}a} \right)}^{2m}}} \;\frac{{{{\rm{d}}^{2m + 2}}w}}{{{\rm{d}}{x^{2m + 2}}}} \end{array}$ (32)
 图 7 固支梁受线性分布载荷示意图 Fig.7 Sketch of a clamped beam subjected to linearly distributed loads

 $\bar w = - \frac{{{{\bar p}_0}{{\bar x}^5}}}{{120}} + \frac{{{{\bar p}_0}{{\bar x}^3}}}{{40}} - \frac{{{{\bar p}_0}{{\bar x}^2}}}{{60}}$ (33)

4 结论

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GRADIENT TYPE OF NONLOCAL HIGHER-ORDER BEAM THEORY AND NEW SOLUTION METHODOLOGY OF NONLOCAL BENDING DEFLECTION
Chen Ling, Shen Jiping, Li Cheng, Liu Xinpei
School of Urban Rail Transportation, Soochow University, Suzhou 215131, Jiangsu, China
Abstract: Di erent predictions were found in di erent literatures about the trend of nonlocal e ects on nanostructural sti ness. By employing an iterative method and the Taylor expansion method, an infinite series for nonlocal high-order stress is achieved based on the gradient-type of nonlocal di erential constitutive model. The nonlocal stress consists of the classical bending stress and each order gradient of nonlocal deflection. Consequently, the di erential equation of the bending deflection curve for nonlocal high-order beam is derived. The nonlocal deflection is determined via the regular perturbation method. Some numerical examples are provided to reveal and quantize the e ects of nonlocal scale factor on bending deflection. It is shown that compared with the corresponding classical results, the nonlocal bending deflections of nanostructure may increase, decrease or remain unchanged. The sti ness of nanostructures can be reduced or enhanced or the same as classical structures by considering the gradient-type of nonlocal high-order stress e ect, and the trend depends on external loads and boundary constraints which are found to play significant roles independently in nonlocal bending of nanostructures. Moreover, it is observed for the first time that the position of maximum nonlocal deflection may be influenced by nonlocal scale factor. The present studies are expected to solve the problems in the application of nonlocal elasticity theory to nanostructures, and further provide supports for the development and optimization of such theory.
Key words: nanostructure    high-order stress    gradient-type of nonlocal theory    regular perturbation method