Answer:
Given
[tex]u_{x}(x,y)=3x^{2}+14y^{2}-8xy[/tex]
The normal strain in x -direction is defined as
[tex]\epsilon _{xx}=\frac{\partial u_{x}}{\partial x}=\frac{\partial }{\partial x}(3x^{2}+14y^{2}-8xy)\\\\\therefore \epsilon _{xx}=6x-8y[/tex]
Similarly the normal strain in y-direction is defined as
[tex]\epsilon _{yy}=\frac{\partial u_{y}}{\partial y}[/tex]
Now since there is no displacement field along y direction thus we have
[tex]\epsilon _{yy}=0[/tex]
Similarly shear strain in xy plane is given by
[tex]\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)[/tex]
