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A steel rod 800mm long and 60mm x 20mm in cross section is subjected to an axial push of 89kN.If modulus of elasticity is $2.1 x 10^{5} N/mm^{2}$.

Calculate stress, strain and reduction in length of rod.

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Data: L=800mm, b=60mm, d=20mm, P=89kN, E=2.1x105N/mm2

F$ind \sigma, \mathrm{e}, \delta_{\mathrm{L}}$

Stress induced in the steel rod:

$\sigma=\frac{P}{A}=\frac{89 \times 10^{3}}{60 \times 20}=74.17 \mathrm{N} / \mathrm{mm}^{2}$

Strain induced in the steel rod:

$E=\frac{\sigma}{e}$

$2.1 \times 10^{5}=\frac{74.17}{e}$

$\mathrm{e}=3.53 \times 10^{-4}$

Reduction in the length:

$\delta_{\mathrm{L}}=\frac{P L}{A E}=\frac{89 \times 10^{3} \times 800}{(60 \times 20) \times 2.1 \times 10^{5}}$

$\delta_{\mathrm{L}}=0.2835 \mathrm{mm}$

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