A cylindrical specimen of some metal alloy having an elastic modulus of 126 GPa and an original cross-sectional diameter of 3.4 mm will experience only elastic deformation when a tensile load of 2130 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.45 mm.

The maximum length of the specimen before deformation is 240.64 mm

Explanation:

Strain = stress ÷ elastic modulus

stress = load ÷ area

load = 2130 N

diameter = 3.4 mm = 3.4×10^-3 m

area = πd^2/4 = 3.142 × (3.4×10^-3)^2/4 = 9.08038×10^-6 m^2

stress = 2130 N ÷ 9.08038×10^-6 m^2 = 2.35×10^8 N/m^2

elastic modulus = 126 GPa = 126×10^9 Pa

Strain = 2.35×10^8 ÷ 126×10^9 = 0.00187

Length = extension ÷ strain = 0.45 mm ÷ 0.00187 = 240.64 mm

the answer is 0.8m.

by: nicholas