几何非线性| 应变张量

2024-04-30 17:39:15 浏览数 (24)

考虑二维空间中的一个连续体,分别是其中的两个物质点，如图3.1所示。在连续体变形前(时刻)引入物质坐标系，另外，在连续体变形之后(时刻)引入空间坐标系。两个坐标系相关的基向量分别为和。

按照描述，位置矢量

begin{split} mathbf X &= X_1 mathbf E_1 X_2 mathbf E_2 \ end{split} tag{1.1}
begin{split} mathbf x &= x_1 mathbf E_1 x_2 mathbf E_2 \ end{split} tag{1.2}

位移矢量

mathbf u = u_1 mathbf E_1 u_2 mathbf E_2 tag{2}

变形前后的位置矢量之间的关系为

begin{split} mathbf x &= mathbf X mathbf u \ dmathbf x &= d mathbf X d mathbf u end{split} tag{3}

使用坐标系,变形后的物体中任意点的位置矢量：

begin{split} x_1 &= x_1(X_1,X_2) \ end{split} tag{4.1}
begin{split} x_2 &= x_2(X_1,X_2) \ end{split} tag{4.2}

变形前的在变形后移动到新的位置,记

begin{split} dS^2 &= d mathbf X cdot d mathbf X = dX_1dX_1 dX_2dX_2\ ds^2 &= d mathbf x cdot d mathbf x = dx_1dx_1 dx_2dx_2 end{split} quad (5)

于是

Delta^2= ds^2-dS^2 = d mathbf x cdot d mathbf x-d mathbf X cdot d mathbf X quad (6)

定义梯度算子

begin{split} nabla(cdot) &= frac{partial(cdot) } {partialmathbf X}\ nabla_x(cdot) &= frac{partial(cdot) } {partialmathbf x} end{split} quad (7)

则

begin{split} dmathbf x &= nabla(mathbf x) d mathbf X = mathbf F cdot d mathbf X = frac{partialmathbf x(mathbf X,t) }{partialmathbf X}d mathbf X \ dmathbf u &= nabla(mathbf u) d mathbf X = mathbf Hcdot d mathbf X end{split} quad (8)

其中，叫做变形梯度，叫做位移梯度。

由(3)可得

begin{split} mathbf H &= nabla(mathbf x -mathbf X) \ &= nabla mathbf x -nabla mathbf X \ &= frac{partialmathbf x}{partialmathbf X}- frac{partialmathbf X}{partialmathbf X} \ &= mathbf F - mathbf I end{split} tag{9}
begin{split} Delta^2 &= ds^2-dS^2\ &=(mathbf F cdot d mathbf X)(mathbf F cdot d mathbf X) - d mathbf X cdot d mathbf X \ &= (d mathbf Xcdot mathbf F^T)(mathbf F cdot d mathbf X)-d mathbf X cdot d mathbf X\ &=dmathbf X(mathbf F^Tmathbf F)dmathbf X - dmathbf X(mathbf I)dmathbf X\ &=dmathbf X(mathbf F^Tmathbf F-mathbf I)dmathbf X end{split} tag{10}

定义应变

mathbf E = frac{1}{2}(mathbf F^Tmathbf F-mathbf I) tag{11}

则

Delta^2 = 2 dmathbf X (mathbf F^Tmathbf F-mathbf I) dmathbf X tag{12}

由(9)可得

mathbf F = mathbf H mathbf I tag{13}

则

mathbf E = frac{1}{2}((mathbf H mathbf I)^T(mathbf H mathbf I)-mathbf I) tag{14}

展开，得

mathbf E = frac{1}{2}(mathbf H mathbf H^T mathbf H^T mathbf H ) tag{15}

忽略高阶量，线性化的拉格朗日应变张量为

hat{mathbf E }= frac{1}{2}(mathbf H mathbf H^T ) tag{16}

[例1] 给出如下的运动

begin{split} x_1 &= X_1 - X_2X_3\ x_2 &= X_2 X_1X_3\ x_3 &= X_3 phi(X_1,X_2)\ end{split}

则由得

begin{split} u_1 &= - X_2X_3\ u_2 &= X_1X_3\ u_3 &= phi(X_1,X_2)\ end{split}

作求导运算

begin{split} frac{partial u_1 }{partial X_1} &= 0 \ frac{partial u_1 }{partial X_2} &= -X_3\ frac{partial u_1 }{partial X_3} &= -X_2\ frac{partial u_2 }{partial X_1} &= X_3 \ frac{partial u_2 }{partial X_2} &= 0\ frac{partial u_2 }{partial X_3} &= X_1\ frac{partial u_3 }{partial X_1} &= frac{partial phi }{partial X_1} \ frac{partial u_3 }{partial X_2} &= frac{partial phi }{partial X_2} \ frac{partial u_3 }{partial X_3} &= 0\ end{split}

位移梯度

begin{split} mathbf H &= frac{partial mathbf u_i }{partial mathbf X_j } \ &= begin{bmatrix} frac{partial u_1 }{partial X_1} & frac{partial u_1 }{partial X_2} & frac{partial u_1 }{partial X_3} \ frac{partial u_2 }{partial X_1} & frac{partial u_2 }{partial X_2} & frac{partial u_2 }{partial X_3} \ frac{partial u_3 }{partial X_1} & frac{partial u_3 }{partial X_2} & frac{partial u_3 }{partial X_3} \ end{bmatrix} \ &= begin{bmatrix} 0 & -X_3 & -X_2 \ X_3 & 0 & X_1\ frac{partial phi }{partial X_1} & frac{partial phi }{partial X_2} & 0\ end{bmatrix} \ end{split}
begin{split} mathbf H &= frac{partial mathbf u_i }{partial mathbf X_j } \ &= begin{bmatrix} frac{partial u_1 }{partial X_1} & frac{partial u_1 }{partial X_2} & frac{partial u_1 }{partial X_3} \ frac{partial u_2 }{partial X_1} & frac{partial u_2 }{partial X_2} & frac{partial u_2 }{partial X_3} \ frac{partial u_3 }{partial X_1} & frac{partial u_3 }{partial X_2} & frac{partial u_3 }{partial X_3} \ end{bmatrix} \ &= begin{bmatrix} 0 & -X_3 & -X_2 \ X_3 & 0 & X_1\ frac{partial phi }{partial X_1} & frac{partial phi }{partial X_2} & 0\ end{bmatrix} \ end{split}
begin{split} mathbf H &= frac{partial mathbf u_i }{partial mathbf X_j } \ &= begin{bmatrix} frac{partial u_1 }{partial X_1} & frac{partial u_1 }{partial X_2} & frac{partial u_1 }{partial X_3} \ frac{partial u_2 }{partial X_1} & frac{partial u_2 }{partial X_2} & frac{partial u_2 }{partial X_3} \ frac{partial u_3 }{partial X_1} & frac{partial u_3 }{partial X_2} & frac{partial u_3 }{partial X_3} \ end{bmatrix} \ &= begin{bmatrix} 0 & -X_3 & -X_2 \ X_3 & 0 & X_1\ frac{partial phi }{partial X_1} & frac{partial phi }{partial X_2} & 0\ end{bmatrix} \ end{split}
partial split

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