Question

A bar of steel is 60 mm x 60 mm in section and 180 mm long. It is subjected to a tensile load of 300 kN along the longitudinal axis and tensile loads of 750 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 1/m = 0.3.

Answer 1

Answer:

The stresses in the direction of X, Y and Z axes,

Along X-axes, Px=FxA=320∗10360∗60(N/mm∗mm)=88.89N/mm2

Along Y axis, Py=FyA=760∗103180∗60(N/mm∗mm)=70.37N/mm2

Along Z axis, Pz=FzA=600∗103180∗60(N/mm∗mm)=55.56N/mm2

Now, the strain along the three principal directions are, due to stresses, Px,Py,Pz,

ex=PxE−μPYE−μPzE

ex=88.89200∗103−0.3∗70.37200∗103− 0.3∗55.56200∗103=0.000256

Now,

ey=PyE−μPzE−μPxE

ey=70.37200∗103−0.3∗55.56200∗103− 0.3∗88.89200∗103=0.000135

ez=PzE−μPxE−μPyE

ez=55.56200∗103−0.3∗88.89200∗103− 0.3∗70.37200∗103=0.00039

Volumetric sign=ex+ey+ez

=0.000256+0.000135+0.00039

ev=0.000781

Also, volumetricstrain=changeinvolumeoriginalvolume

0.000781=changeinvolume180∗60∗60

Changeinvolume=506.088mm3

Now,ex=DeltaLL

0.000256=Δ180

ΔL=0.046mm

ey=DeltaWW

0.000135=ΔW60