SSC JE Mechanical Engineering 11 Oct 2020 Shift 1 Papers

Question : 122 of 200

Marks: +1, -0

A cylindrical metal bar of 12 mm diameter is located by an axial force of 20 kN results in a change in diameter by 0.003 mm. Poisson’s ratio is given by: (Assume modulus of rigidity = 80 GPa)

0.2923
0.056
0.025
0.56

Validate

Solution:

Given:
δd = 0.003 mm, d = 12 mm, F = 20 kN, G = 80 GPa
Longitudinal strain:

∈_{\text "long "} ={\σ}/{E}={F \/ A}/{2 G(1+\µ)}

\∈_{\text "long "} ={20000 \/ \π6^{2}}/{2 \×80000(1+\µ)}

\∈_{\text "long "} ={176.8388}/{160000(1+\µ)}

........(1)
Lateral strain:

\∈_{\text "late "}={\δ{d}}/{{d}}={0.003}/{12}=2.5 \×10^{-4}

.........(2)
Poisson ratio:
From equation (1) and (2)

\µ={\∈_{\text "lateral "}}/{\∈_{\text "Long "}}={2.5 \×10^{-4}}/{{176.8388}/{160000(1+\µ)}}

176.8388μ = 2.5 × 10^{-4}

× 160000(1 + μ)

μ = 0.2923

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