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A mild steel bar of square cross-section 40 mm x 40 mm is 400 mm long. It is subjected to a longitudinal tensile stress of 440 N/mm2 and lateral compressive stress of 200 N/mm2 in perpendicular direction. E = 2 x 105 N/mm2, μ = 0.3. What is the approximate elongation of the bar in the longitudinal direction?

A
0.44 mm
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B
0.88 mm
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C
0.22 mm
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D
1 mm
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Solution

The correct option is D 1 mm
The lateral compressive stress is applied on per
perpendicular direction (not directions). So considering 200 N/mm2lateral stress in one direction. Therefore, strain in longitudinal direction

ϵi=1E(σ1μσ2)

=12×105×(440+0.3×200)
=1400
Elongation of bar in longitudinal direction
=ϵi×400
=1 Mm

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