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 × 10-3m


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


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


Area of rectangular cross section of copper piece,


Area = l× b


Area = (19.1 × 10-3m) × (15.2× 10-3m)


Area = 2.9 × 10-4 m2


Modulus of elasticity of copper from standard list, η = 42× 109 N/m2


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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