Question

A metal rod of 10mm square cross section and length 2 m is loaded in tension with an axial tensile load of 4 kN. If the tensile modulus is 72 GPa, what will be the extension of the rod?

Answer 1

Deformation of bar is given by

$\Delta l=\frac{P\times l}{A\times E}$

Where P=4kN=4x1000N

L=2m=2000mm

A=10x10mm²

E=72GPa=72000MPa=72000N/mm²

Δl=(4000x2000)/(10x10x72000)

=1.11mm

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