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第二章：应变分析

About 1621 wordsAbout 5 min

2026-01-28

1 位移描述

物体内部任意点 P 在变形前的坐标为 a=aiei\bold{a}=a_i\bold{e}_i，变形后坐标为 x=xiei\bold{x}=x_i\bold{e}_i。定义位移矢量为 u=x−a\bold{u}=\bold{x}-\bold{a}，即 ui=xi−aiu_i=x_i-a_i，则位移场 ui=ui(a1,a2,a3)u_i=u_i(a_1,a_2,a_3) 描述了物体内部每一点的位移情况。

2 几何方程

Note

本节讨论均基于小变形假设。

正应变 εi\varepsilon_{i} 表示物体沿坐标轴方向的长度的单位改变量；剪应变 γij=2εij\gamma_{ij}=2\varepsilon_{ij} 表示物体两个正交线元形变之后偏移的角度。它们与位移场满足以下关系，称其为几何方程。

εx=∂u∂x,εy=∂v∂y,εz=∂w∂zγxy=∂u∂y+∂v∂x,γyz=∂v∂z+∂w∂y,γzx=∂w∂x+∂u∂z\varepsilon_x=\frac{\partial u}{\partial x},\quad\varepsilon_y=\frac{\partial v}{\partial y},\quad\varepsilon_z=\frac{\partial w}{\partial z}\\ \gamma_{xy}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x},\quad\gamma_{yz}=\frac{\partial v}{\partial z}+\frac{\partial w}{\partial y},\quad\gamma_{zx}=\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z}

也即

εij=12(∂ui∂xj+∂uj∂xi)\varepsilon_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)

Note

位移梯度可以分解为对称部分和反对称部分：

∂ui∂xj=εij−ωij\frac{\partial u_i}{\partial x_j}=\varepsilon_{ij}-\omega_{ij}

  • 应变张量：εij=12(∂ui∂xj+∂uj∂xi)\varepsilon_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)
  • 转动张量：ωij=12(∂uj∂xi−∂ui∂xj)\omega_{ij}=\frac{1}{2}\left(\frac{\partial u_j}{\partial x_i}-\frac{\partial u_i}{\partial x_j}\right)

3 应变协调方程

  • 物理意义：协调方程保证物体变形后保持连续，既不开裂，又不重叠。

  • 数学意义：协调方程是位移场单值连续的必要条件。

    Note

    对于单连通区域，应变协调方程是保证存在单值连续位移场的充要条件；但对于多连通区域，仅是必要条件。

εij,kl+εkl,ij−εik,jl−εjl,ik=0\varepsilon_{ij,kl}+\varepsilon_{kl,ij}-\varepsilon_{ik,jl}-\varepsilon_{jl,ik}=0

上式含有四个自由指标，共表示 81 个方程，其中除去恒等式和等价方程，可简化为 6 个方程：

εx,yy+εy,xx=γxy,xyεy,zz+εz,yy=γyz,yzεz,xx+εx,zz=γzx,zx(−γyz,x+γzx,y+γxy,z),x=2εx,yz(γyz,x−γzx,y+γxy,z),y=2εy,zx(γyz,x+γzx,y−γxy,z),z=2εz,xy\varepsilon_{x,yy} + \varepsilon_{y,xx} = \gamma_{xy,xy} \\ \varepsilon_{y,zz} + \varepsilon_{z,yy} = \gamma_{yz,yz} \\ \varepsilon_{z,xx} + \varepsilon_{x,zz} = \gamma_{zx,zx} \\ (-\gamma_{yz,x} + \gamma_{zx,y} + \gamma_{xy,z})_{,x} = 2\varepsilon_{x,yz} \\ (\gamma_{yz,x} - \gamma_{zx,y} + \gamma_{xy,z})_{,y} = 2\varepsilon_{y,zx} \\ (\gamma_{yz,x} + \gamma_{zx,y} - \gamma_{xy,z})_{,z} = 2\varepsilon_{z,xy}

Note

六个应变协调方程并不完全独立，不能由它们独立解出六个应变分量。

4 由应变求位移

4.1 线积分法

  • 求 u1u_1 ：

u1=∫cdu1+u10=∫c(∂u1∂x1dx1+∂u1∂x2dx2+∂u1∂x3dx3)+u10u_1 = \int_c du_1 + u_{10} = \int_c \left( \frac{\partial u_1}{\partial x_1} \text{d}x_1 + \frac{\partial u_1}{\partial x_2} \text{d}x_2 + \frac{\partial u_1}{\partial x_3} \text{d}x_3 \right) + u_{10}

∂∂x1(∂u1∂x2)=∂ε11∂x2∂∂x2(∂u1∂x2)=∂γ12∂x2−∂ε22∂x1∂∂x3(∂u1∂x2)=12(∂γ12∂x3+∂γ31∂x2−∂γ23∂x1)}∂u1∂x2[C1]\quad \left. \begin{array}{l} \displaystyle \frac{\partial}{\partial x_1} \left( \frac{\partial u_1}{\partial x_2} \right) = \frac{\partial \varepsilon_{11}}{\partial x_2} \\ \displaystyle \frac{\partial}{\partial x_2} \left( \frac{\partial u_1}{\partial x_2} \right) = \frac{\partial \gamma_{12}}{\partial x_2} - \frac{\partial \varepsilon_{22}}{\partial x_1} \\ \displaystyle \frac{\partial}{\partial x_3} \left( \frac{\partial u_1}{\partial x_2} \right) = \frac{1}{2} \left( \frac{\partial \gamma_{12}}{\partial x_3} + \frac{\partial \gamma_{31}}{\partial x_2} - \frac{\partial \gamma_{23}}{\partial x_1} \right) \end{array} \right\} \dfrac{\partial u_1}{\partial x_2} [C_1] \\

∂∂x1(∂u1∂x3)=∂ε11∂x3∂∂x2(∂u1∂x3)=∂∂x3(∂u1∂x2)∂∂x3(∂u1∂x3)=∂γ31∂x3−∂ε33∂x1}∂u1∂x3[C2]\quad \left. \begin{array}{l} \displaystyle \frac{\partial}{\partial x_1} \left( \frac{\partial u_1}{\partial x_3} \right) = \frac{\partial \varepsilon_{11}}{\partial x_3} \\ \displaystyle \frac{\partial}{\partial x_2} \left( \frac{\partial u_1}{\partial x_3} \right) = \frac{\partial}{\partial x_3} \left( \frac{\partial u_1}{\partial x_2} \right) \\ \displaystyle \frac{\partial}{\partial x_3} \left( \frac{\partial u_1}{\partial x_3} \right) = \frac{\partial \gamma_{31}}{\partial x_3} - \frac{\partial \varepsilon_{33}}{\partial x_1} \end{array} \right\} \dfrac{\partial u_1}{\partial x_3} [C_2] \\

∂u1∂x1=ε11∂u1∂x2[C1]∂u1∂x3[C2]}u1[u10,C1,C2]\quad \left. \begin{array}{l} \displaystyle \frac{\partial u_1}{\partial x_1} = \varepsilon_{11} \\ \displaystyle \frac{\partial u_1}{\partial x_2} [C_1] \\ \displaystyle \frac{\partial u_1}{\partial x_3} [C_2] \end{array} \right\} u_1[u_{10}, C_1, C_2]\\

  • 求 u2u_2 ：

u2=∫cdu2+u20=∫c(∂u2∂x1dx1+∂u2∂x2dx2+∂u2∂x3dx3)+u20u_2 = \int_c du_2 + u_{20} = \int_c \left( \frac{\partial u_2}{\partial x_1} \text{d}x_1 + \frac{\partial u_2}{\partial x_2} \text{d}x_2 + \frac{\partial u_2}{\partial x_3} \text{d}x_3 \right) + u_{20}

∂∂x1(∂u2∂x3)=12(∂γ23∂x1+∂γ12∂x3−∂γ31∂x2)∂∂x2(∂u2∂x3)=∂ε22∂x3∂∂x3(∂u2∂x3)=∂γ23∂x3−∂ε33∂x2}∂u2∂x3[C3]\quad \left. \begin{array}{l} \displaystyle \frac{\partial}{\partial x_1} \left( \frac{\partial u_2}{\partial x_3} \right) = \frac{1}{2} \left( \frac{\partial \gamma_{23}}{\partial x_1} + \frac{\partial \gamma_{12}}{\partial x_3} - \frac{\partial \gamma_{31}}{\partial x_2} \right) \\ \displaystyle \frac{\partial}{\partial x_2} \left( \frac{\partial u_2}{\partial x_3} \right) = \frac{\partial \varepsilon_{22}}{\partial x_3}\\ \displaystyle \frac{\partial}{\partial x_3} \left( \frac{\partial u_2}{\partial x_3} \right) = \frac{\partial \gamma_{23}}{\partial x_3} - \frac{\partial \varepsilon_{33}}{\partial x_2} \end{array} \right\} \dfrac{\partial u_2}{\partial x_3} [C_3] \\

∂u2∂x1=γ12−∂u1∂x2[C1]∂u2∂x2=ε22∂u2∂x3[C3]}u2[u20,C1,C3]\quad \left. \begin{array}{l} \displaystyle \frac{\partial u_2}{\partial x_1} = \gamma_{12} - \frac{\partial u_1}{\partial x_2} [C_1] \\ \displaystyle \frac{\partial u_2}{\partial x_2} = \varepsilon_{22} \\ \displaystyle \frac{\partial u_2}{\partial x_3} [C_3] \end{array} \right\} u_2[u_{20}, C_1, C_3]\\

  • 求 u3u_3 ：

u3=∫cdu3+u30=∫c(∂u3∂x1dx1+∂u3∂x2dx2+∂u3∂x3dx3)+u30u_3 = \int_c du_3 + u_{30} = \int_c \left( \frac{\partial u_3}{\partial x_1} \text{d}x_1 + \frac{\partial u_3}{\partial x_2} \text{d}x_2 + \frac{\partial u_3}{\partial x_3} \text{d}x_3 \right) + u_{30}

∂u3∂x1=γ31−∂u1∂x3[C2]∂u3∂x2=γ23−∂u2∂x3[C3]∂u3∂x3=ε33}u3[u30,C2,C3]\quad \left. \begin{array}{l} \displaystyle \frac{\partial u_3}{\partial x_1} = \gamma_{31} - \frac{\partial u_1}{\partial x_3} [C_2] \\ \displaystyle \frac{\partial u_3}{\partial x_2} = \gamma_{23} - \frac{\partial u_2}{\partial x_3} [C_3] \\ \displaystyle \frac{\partial u_3}{\partial x_3} = \varepsilon_{33} \end{array} \right\} u_3[u_{30}, C_2, C_3]

只要应变满足协调方程，以上各式中的积分均与路径无关，一般取与坐标轴平行的折线为积分路径，故称线积分法。

上面式子中出现的 6 个积分常数 u10,u20,u30u_{10},u_{20},u_{30} 和 C1,C2,C3C_1,C_2,C_3 分别对应于刚体平移和刚体转动的 6 个自由度，须由外部约束条件来决定。

4.2 直接积分法

对某些应变分量表达式较为简单的情况，也可采用直接积分法。下面以无应变状态 εij=0\varepsilon_{ij}=0 为例展示基本步骤。

由正应变表达式

ε11=∂u1∂x1=0\varepsilon_{11}=\frac{\partial u_1}{\partial x_1}=0

ε22=∂u2∂x2=0\varepsilon_{22}=\frac{\partial u_2}{\partial x_2}=0

ε33=∂u3∂x3=0\varepsilon_{33}=\frac{\partial u_3}{\partial x_3}=0

分别对 x1,x2,x3x_1,x_2,x_3 积分一次得

u1=f1(x2,x3)u_1=f_1(x_2,x_3)

u2=f2(x3,x1)u_2=f_2(x_3,x_1)

u3=f3(x1,x2)u_3=f_3(x_1,x_2)

代入剪应变表达式

γ12=∂u1∂x2+∂u2∂x1=0\gamma_{12}=\frac{\partial u_1}{\partial x_2}+\frac{\partial u_2}{\partial x_1}=0

γ23=∂u2∂x3+∂u3∂x2=0\gamma_{23}=\frac{\partial u_2}{\partial x_3}+\frac{\partial u_3}{\partial x_2}=0

γ31=∂u3∂x1+∂u1∂x3=0\gamma_{31}=\frac{\partial u_3}{\partial x_1}+\frac{\partial u_1}{\partial x_3}=0

可得

∂f1(x2,x3)∂x2+∂f2(x3,x1)∂x1=0(a)\frac{\partial f_1(x_2,x_3)}{\partial x_2}+\frac{\partial f_2(x_3,x_1)}{\partial x_1}=0 \tag{a}

∂f2(x3,x1)∂x3+∂f3(x1,x2)∂x2=0(b)\frac{\partial f_2(x_3,x_1)}{\partial x_3}+\frac{\partial f_3(x_1,x_2)}{\partial x_2}=0 \tag{b}

∂f3(x1,x2)∂x1+∂f1(x2,x3)∂x3=0(c)\frac{\partial f_3(x_1,x_2)}{\partial x_1}+\frac{\partial f_1(x_2,x_3)}{\partial x_3}=0 \tag{c}

由 (a) 式对 x2x_2 求导得

∂2f1(x2,x3)∂x22=0\frac{\partial^2 f_1(x_2,x_3)}{\partial x_2^2}=0

因此

∂f1∂x2=g1(x3),f1=g1(x3)x2+g0(x3)\frac{\partial f_1}{\partial x_2}=g_1(x_3),\quad f_1=g_1(x_3)x_2+g_0(x_3)

同理由 (c) 式有

∂2f1(x2,x3)∂x32=0\frac{\partial^2 f_1(x_2,x_3)}{\partial x_3^2}=0

将 f1f_1 的表达式代入后有

g1′′(x3)x2+g0′′(x3)=0g_1^{''}(x_3)x_2+g_0^{''}(x_3)=0

因此

g1′′(x3)=0,g0′′(x3)=0g_1^{''}(x_3)=0,\quad g_0^{''}(x_3)=0

故

g1(x3)=a3x3+a1,g0(x3)=a2x3+a0g_1(x_3)=a_3x_3+a_1,\quad g_0(x_3)=a_2x_3+a_0

综上

u1=f1(x2,x3)=a0+a1x2+a2x3+a3x2x3(d)u_1=f_1(x_2,x_3)=a_0+a_1x_2+a_2x_3+a_3x_2x_3 \tag{d}

同理可由

∂2f2∂x12=0,∂2f2∂x32=0\frac{\partial^2 f_2}{\partial x_1^2}=0,\quad \frac{\partial^2 f_2}{\partial x_3^2}=0

得

u2=f2(x3,x1)=b0+b1x3+b2x1+b3x3x1(e)u_2=f_2(x_3,x_1)=b_0+b_1x_3+b_2x_1+b_3x_3x_1 \tag{e}

由

∂2f3∂x12=0,∂2f3∂x22=0\frac{\partial^2 f_3}{\partial x_1^2}=0,\quad \frac{\partial^2 f_3}{\partial x_2^2}=0

得

u3=f3(x1,x2)=c0+c1x1+c2x2+c3x1x2(f)u_3=f_3(x_1,x_2)=c_0+c_1x_1+c_2x_2+c_3x_1x_2 \tag{f}

将 (d) (e) (f) 代入 (a) (b) (c) 中得

a1=−b2,b1=−c2,c1=−a2,a3=b3=c3=0a_1=-b_2,\quad b_1=-c_2,\quad c_1=-a_2,\quad a_3=b_3=c_3=0

于是待定常数降为 6 个，(d) (e) (f) 化简为

u1=a0−b2x2+a2x3u2=b0−c2x3+b2x1u3=c0−a2x1+c2x2u_1=a_0-b_2x_2+a_2x_3\\ u_2=b_0-c_2x_3+b_2x_1\\ u_3=c_0-a_2x_1+c_2x_2

其中常数由刚体平动和转动确定。

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