written 6.4 years ago by
teamques10
★ 69k
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modified 6.4 years ago
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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
ΔW=0.0081mm
ez=Deltatt
0.00039=Δt60
Δt=0.0234mm