Question

Question 9. 8. A piece of copper having a rectangular cross-section of 15.2 mm x 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain? Shear modulus of elasticity of copper is 42 x 109 N/m2.​

Answer 1

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length = 19.1 mm = 1.9 x 10⁻³ m

breadth = 15.2 mm = 15.2 x 10⁻³m

Area = length x breath

= 19.1 x 10⁻³ x 15.2 x 10⁻³

= 2.9 x 10 ⁻⁴ m²

Tension on copper , T = 44500 N

Modulus of elasticity of copper , η = 42 x 10 ⁹ N / m²

Modulus of elasticity = stress / strain

η = (F/A) / Strain

Strain = F/Aη

= 44500 / 2.9 X 10⁻⁴ x 42 x 10⁹

= 3.65 x 10⁻³

Answer 2

Given ,

C.SA of copper piece (A) = 15.2mm × 19.1mm

=( 15.2 × 10^-3)×(19.1×10^-3) m²

= 15.2 × 19.1 × 10^-6 m²

Force applied (F) = 44500 N

Young's modulus (Y) = 0.42× 10¹¹ N/m²

We know,

Young's modulus = stress/strain

= F/A.strain

1.1 × 10¹¹ = 44500/(15.2×19.1×10^-6)×strain

Strain = 44500/(15.2×19.1×10^-6×0.42 ×10¹¹)

= 364.95 × 10^-5

= 3.65 mm