Question

A steel bar 400mm long, 60mm wide and 15 mm thick is sub- jected to an axial tension of 100KN. Calculate the final dimension and change in volume of the bar. Take E= 2 x 105 N/mm2 and 1/m=0.3
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Answer 1

Answer:

Change in length (δl)=P∗LAE

Given, P=160∗103N,L=2000mm

W=40mm,t=20mm

E=200GPa,E=200∗103N/mm2

μ=0.3

δL=160∗103∗200040∗20∗2000)103

δL=2mm

Poisson′sRatio=LateralstrainLongitudinalstrain

Calculations of changes in width : -

Lateralstrain=δwworδtt

Longitudinalstrain=δLL

μ=δwwδLL

0.3=δw4022000

δW=0.012mm

Calculation of change in thickness:

μ=δttδLL

0.3=δt2022000

δt = 0.006mm.

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