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A bar of steel is (60 mm x 60 mm) in section and 18mm 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.

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

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

Along X-axes, Px=FxA=3201036060(N/mmmm)=88.89N/mm2

Along Y axis, Py=FyA=76010318060(N/mmmm)=70.37N/mm2

Along Z axis, Pz=FzA=60010318060(N/mmmm)=55.56N/mm2

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

ex=PxEμPYEμPzE

ex=88.892001030.370.37200103 0.355.56200103=0.000256

Now,

ey=PyEμPzEμPxE

ey=70.372001030.355.56200103 0.388.89200103=0.000135

ez=PzEμPxEμPyE

ez=55.562001030.388.89200103 0.370.37200103=0.00039

Volumetric sign=ex+ey+ez

=0.000256+0.000135+0.00039

ev=0.000781

Also, volumetricstrain=changeinvolumeoriginalvolume

0.000781=changeinvolume1806060

Changeinvolume=506.088mm3

Now,ex=DeltaLL

0.000256=Δ180

ΔL=0.046mm

ey=DeltaWW

0.000135=ΔW60

ΔW=0.0081mm

ez=Deltatt

0.00039=Δt60

Δt=0.0234mm

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