Question

A displacement field is given in terms of scalar components by: u = 2 xż, uz = az, Uz = 23X2X3, where az, az, az are positive constants. a) Find the infinitesimal strain tensor.

Answer 1

Answer:

The infinitesimal strain tensor is

$\left[\begin{array}{ccc}0&a_1x_2&0\\a_1x_2&0&\frac{a_3x_3}{2}\\0&\frac{a_3x_3}{2}&a_3x_2\end{array}\right]$

Explanation:

As the scalar components of the displacement field are not clear thus estimating their values (from the question attached).

Here the values are given as

$u_1=a_1x_2^2\\u_2=a_2\\u_3=a_3x_2x_3$

The infinitesimal strain tensor is given as

$\epsilon=\left[\begin{array}{ccc}\epsilon_{11}&\epsilon_{12}&\epsilon_{13}\\\epsilon_{21}&\epsilon_{22}&\epsilon_{23}\\\epsilon_{31}&\epsilon_{32}&\epsilon_{33}\end{array}\right]$

Here

$\epsilon_{11}=\frac{\partial u_1}{\partial x_1}$

$\epsilon_{12}=\frac{1}{2}(\frac{\partial u_1}{\partial x_2} +\frac{\partial u_2}{\partial x_1} )$

$\epsilon_{13}=\frac{1}{2}(\frac{\partial u_1}{\partial x_3} +\frac{\partial u_3}{\partial x_1} )$

$\epsilon_{21}=\frac{1}{2}(\frac{\partial u_1}{\partial x_2} +\frac{\partial u_2}{\partial x_1} )$

$\epsilon_{22}=\frac{\partial u_2}{\partial x_2}$

$\epsilon_{23}=\frac{1}{2}(\frac{\partial u_2}{\partial x_3} +\frac{\partial u_3}{\partial x_2} )$

$\epsilon_{31}=\frac{1}{2}(\frac{\partial u_1}{\partial x_3} +\frac{\partial u_3}{\partial x_1} )$

$\epsilon_{32}=\frac{1}{2}(\frac{\partial u_2}{\partial x_3} +\frac{\partial u_3}{\partial x_2} )$

$\epsilon_{33}=\frac{\partial u_3}{\partial x_3}$

There values are calculated as

$\epsilon_{11}=\frac{\partial u_1}{\partial x_1}=\frac{\partial (a_1x_2^2)}{\partial x_1}=0$

$\epsilon_{12}=\frac{1}{2}(\frac{\partial u_1}{\partial x_2} +\frac{\partial u_2}{\partial x_1} )=\frac{1}{2}(\frac{\partial (a_1x_2^2)}{\partial x_2} +\frac{\partial (a_2)}{\partial x_1} )=\frac{1}{2}((2a_1x_2)+0 )=a_1x_2$

$\epsilon_{13}=\frac{1}{2}(\frac{\partial u_1}{\partial x_3} +\frac{\partial u_3}{\partial x_1} )=\frac{1}{2}(\frac{\partial (a_1x_2^2)}{\partial x_3} +\frac{\partial (a_3x_2x_3}{\partial x_1} )=\frac{1}{2}(0+0 )=0$

$\epsilon_{21}=\frac{1}{2}(\frac{\partial u_1}{\partial x_2} +\frac{\partial u_2}{\partial x_1} )==\frac{1}{2}(\frac{\partial (a_1x_2^2)}{\partial x_2} +\frac{\partial (a_2)}{\partial x_1} )=\frac{1}{2}((2a_1x_2)+0 )=a_1x_2$

$\epsilon_{22}=\frac{\partial u_2}{\partial x_2}=\frac{\partial (a_2)}{\partial x_2}=0$

$\epsilon_{23}=\frac{1}{2}(\frac{\partial u_2}{\partial x_3} +\frac{\partial u_3}{\partial x_2} )=\frac{1}{2}(\frac{\partial (a_2)}{\partial x_3} +\frac{\partial a_3x_2x_3}{\partial x_2} )=\frac{1}{2}(0 +a_3x_3 )=\frac{a_3x_3}{2}$

$\epsilon_{31}=\frac{1}{2}(\frac{\partial u_1}{\partial x_3} +\frac{\partial u_3}{\partial x_1} )=\frac{1}{2}(\frac{\partial (a_1x_2^2)}{\partial x_3} +\frac{\partial (a_3x_2x_3}{\partial x_1} )=\frac{1}{2}(0+0 )=0$

$\epsilon_{32}=\frac{1}{2}(\frac{\partial u_2}{\partial x_3} +\frac{\partial u_3}{\partial x_2} )=\frac{1}{2}(\frac{\partial (a_2)}{\partial x_3} +\frac{\partial a_3x_2x_3}{\partial x_2} )=\frac{1}{2}(0 +a_3x_3 )=\frac{a_3x_3}{2}$

$\epsilon_{33}=\frac{\partial u_3}{\partial x_3}=\frac{\partial (a_3x_2x_3)}{\partial x_3}=a_3x_2$

So the values are put in the tensor as

$\epsilon=\left[\begin{array}{ccc}0&a_1x_2&0\\a_1x_2&0&\frac{a_3x_3}{2}\\0&\frac{a_3x_3}{2}&a_3x_2\end{array}\right]$

So the infinitesimal strain tensor is $\left[\begin{array}{ccc}0&a_1x_2&0\\a_1x_2&0&\frac{a_3x_3}{2}\\0&\frac{a_3x_3}{2}&a_3x_2\end{array}\right]$