Q. 84.2( 39 Votes )

# A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

Answer :

Given Data,

Length of the piece of copper = l = 19.1 mm = 19.1 × 10^{-3}m

Breadth of the piece of copper = b = 15.2 mm = 15.2× 10^{-3}m

Tension force applied on the piece of cooper, F = 44500N

Area of rectangular cross section of copper piece,

Area = l× b

⇒ Area = (19.1 × 10^{-3}m) × (15.2× 10^{-3}m)

⇒ Area = 2.9 × 10^{-4} m^{2}

Modulus of elasticity of copper from standard list, η = 42× 10^{9} N/m^{2}

By definition, Modulus of elasticity, η = stress/strain

⇒

⇒ Strain = F/Aη

⇒

⇒ Strain = 3.65 × 10^{-3}

Hence, the resulting strain is 3.65 × 10^{-3}

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