Question

A displacement field in the +ve x direction is given as u=3x^2+14y^2-8xy. Determine the strains in the x and y directions.

Answer 1

Answer:

Given

$u_{x}(x,y)=3x^{2}+14y^{2}-8xy$

The normal strain in x -direction is defined as

$\epsilon _{xx}=\frac{\partial u_{x}}{\partial x}=\frac{\partial }{\partial x}(3x^{2}+14y^{2}-8xy)\\\\\therefore \epsilon _{xx}=6x-8y$

Similarly the normal strain in y-direction is defined as

$\epsilon _{yy}=\frac{\partial u_{y}}{\partial y}$

Now since there is no displacement field along y direction thus we have

$\epsilon _{yy}=0$

Similarly shear strain in xy plane is given by

$\epsilon _{xy}=\frac{1}{2}(\frac{\partial u_{x}}{\partial y}+\frac{\partial u_{y}}{\partial x})\\\\\therefore \epsilon _{xy}=\frac{1}{2}(\frac{\partial }{\partial y}(3x^{2}+14y^{2}-8xy)+\frac{\partial 0}{\partial x})\\\\\epsilon _{xy}=\frac{1}{2}(28y-8x)$