A bar of steel is (60 mm x 60 mm) in section and 180 mm along. It is subjected to a tensile load of 320 kN along the longitudinal axis and tensile loads of 760 kN and 600 kN on the lateral faces. Find the change in the dimensions of the bar and the change in volume. Take E=200 GN/m2 and Poisson’s ratio =0.3

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,

$e_y=\frac{P_y}{E}-\frac{\mu P_z}{E}-\frac{\mu P_x}{E}$

$e_y=\frac{70.37}{200*10^3}-\frac{0.3*55.56}{200*10^3}-$ $\frac{0.3*88.89}{200*10^3}=0.000135$

$e_z=\frac{P_z}{E}-\frac{\mu P_x}{E}-\frac{\mu P_y}{E}$

$e_z=\frac{55.56}{200*10^3}-\frac{0.3*88.89}{200*10^3}-$ $\frac{0.3*70.37}{200*10^3}=0.00039$

Volumetric sign$=e_x+e_y+e_z$

$=0.000256+0.000135+0.00039$

$e_v=0.000781$

Also, $volumetric strain=\frac{change in volume}{original volume}$

$0.000781=\frac{change in volume}{180*60*60}$

$Change in volume=506.088mm^3$

$Now, e_x=\frac{Delta L}{L}$

$0.000256=\frac{\Delta}{180}$

$\Delta L=0.046mm$

$e_y=\frac{Delta W}{W}$

$0.000135=\frac{\Delta W}{60}$

$\Delta W=0.0081mm$

$e_z=\frac{Delta t}{t}$

$0.00039=\frac{\Delta t}{60}$

$\Delta t=0.0234mm$