Question

7. (20 pts) You want to redo these experiments to calculate Green's strain as a function of displacements that you measure using a digital camera. You perform the experiment by applying three different stresses in the x-direction. You fit the data for each stress and fit a line to determine the mapping from your unloaded configuration to your loaded configuration and assume that you only have displacement in the X-direction: x = AX, y = Y, z= Z where for each stress is shown in the table below. Use the following equation to calculate the linearized and non-linear Green's strain for each case: Linear Strain Non-Linear Strain Stress (MPa) 0.1 1.01 1.1 1.3 For which case is the linear assumption most appropriate?

Answer 1

as Per given data

$\\u_x=\cap X\\\\u_y=Y\\\\u_z=Z \\\\\frac{\partial u_x}{\partial x}= \cap \\\\\frac{\partial u_y}{\partial x}=\frac{\partial u_z}{\partial x}=0$

so ,

$\\u_x=\cap X\\\\u_y=Y\\\\u_z=Z \\\\\frac{\partial u_x}{\partial x}= \cap \\\\\frac{\partial u_y}{\partial x}=\frac{\partial u_z}{\partial x}=0 \\\\Linear \ strain=\frac{\partial u_x}{\partial x}= \cap \\\\non \ linear \ green's \ strain=\frac{\partial u_x}{\partial x} +\frac{1}{2}\left [ \left (\frac{\partial u_x}{\partial x} \right )^2+0+0 \right ] \\\\non \ linear \ green's \ strain=\cap+\frac{\cap^2}{2}$

Stress	Ո	Linear strain=Ո	Non-Linear strain= Ո+(Ո2)/2	Error
0.1	1.01	1.01	1.52005	33.55482
1	1.1	1.1	1.705	35.48387
5	1.3	1.3	2.145	39.39394

so for none case the linear assumpstion model is appropiate

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