written 5.5 years ago by
teamques10
★ 65k
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modified 5.5 years ago
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The stresses in the direction of X, Y and Z axes,
Along X-axes, $P_x=\frac{F_x}{A}=\frac{320*10^3}{60*60}(N/mm*mm)=88.89 N/mm^2$
Along Y axis, $P_y=\frac{F_y}{A}=\frac{760*10^3}{180*60}(N/mm*mm)=70.37 N/mm^2$
Along Z axis, $P_z=\frac{F_z}{A}=\frac{600*10^3}{180*60}(N/mm*mm)=55.56 N/mm^2$
Now, the strain along the three principal directions are, due to stresses, $P_x, P_y, P_z,$
$e_x=\frac{P_x}{E}-\frac{\mu P_Y}{E}-\frac{\mu P_z}{E}$
$e_x=\frac{88.89}{200*10^3}-\frac{0.3*70.37}{200*10^3}-$
$\frac{0.3*55.56}{200*10^3}=0.000256$
Now,
$e_y=\frac{P_y}{E}-\frac{\mu P_z}{E}-\frac{\mu P_x}{E}$
$e_y=\frac{70.37}{200*10^3}-\frac{0.3*55.56}{200*10^3}-$ …
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