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

Given

`u_(x)(x,y)=3x^(2)+14y^(2)-8xy`

The normal strain in x -direction is defined as

`\epsilon _(xx)=(\partial u_(x))/(\partial x)=(\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)=(\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)=(1)/(2)((\partial u_(x))/(\partial y)+(\partial u_(y))/(\partial x))\\\\\therefore \epsilon _(xy)=(1)/(2)((\partial )/(\partial y)(3x^(2)+14y^(2)-8xy)+(\partial 0)/(\partial x))\\\\\epsilon _(xy)=(1)/(2)(28y-8x)`