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