Question

A steel bar 35 mm x 35 mm in section and 100 mm in length is acted upon by a loud of 180 KN along its longitudinal axis and 400 KN and 300 KN along the axes of the lateral surface Determine

i) Change in the dimension of the bar

ii) Change in volume Take E = 205 Gpa and poissons ratio 0.3​

Answer 1

• When a bar is subjected to a normal load, the length of the bar changes. Thus, the ratio of change in length to the original length is known as linear strain. ... Thus, the ratio of change in the lateral dimension to the original dimension is known as lateral strain.

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Answer 2

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