Question

In a tension test, a metallic rod of diameter 16mm produces an elongation of 48 mm whensubjected to 90 kN load. The length of the baris 150 mm. Find modulus of elasticity.​

Answer 1

Answer:

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Answer 2

Given: In a tension test, a metallic rod of diameter 16mm produces an elongation of 48 mm when subjected to 90 kN load. The length of the bar is 150 mm.

To find: Modulus of elasticity of the bar

Explanation: Let the modulus of elasticity be y.

Tension(F)= 90 kN

= 90000 N( 1 kN = 1000N)

Length of the bar(l)

= 150 mm

= 0.15 m( 1 mm = 0.001 m)

Elongation of bar(dl)

= 48 mm

= 0.048 m(1 mm= 0.001 m)

Diameter of bar = 16 mm

Radius(r)= Diameter/2

= 16/2

= 8 mm

= 0.008 m

Cross-sectional area of the bar

=$\pi {r}^{2}$

=$3.14 \times ({0.008)}^{2}$

=$200.96 \times {10}^{ - 6}$

Strain of bar

= Elongation/ Length

= dl / l

= 0.048/0.150

=0.32

Stress in bar

= Tension/ Area

=$\frac{90000}{200.96 \times {10}^{ - 6} }$

=$448 \times {10}^{6}$

The formula relating y, strain and stress is:

y = Stress / Strain

=$\frac{448 \times {10}^{6} }{0.32}$

=$1400 \times {10}^{6}$

=$14 \times {10}^{8} \: Pa$

Therefore, the modulus of elasticity is

$14 \times {10}^{8}$ Pa.