Acylindrical specimen of a nickel alloy having
an elastic modulus of 207 gpa (
psi) and an original diameter of 10.2 mm
(0.40 in.) will experience only elastic deformation
when a tensile load of 8900 n (2000 lb )
is applied. compute the maximum length
of the specimen before deformation if the
maximum allowable elongation is 0.25 mm
(0.010
6.5 an aluminum

L = 0.475 m = 475 mm = 18.7 inches

Explanation:

A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa and an original diameter of 10.2 mm (0.40 in.) will experience only elastic deformation when a tensile load of 8900 N (2000 lb ) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.25 mm (0.010 in).

E = 207 GPa = 207*10⁹ Pa

D = 10.2 mm = 0.0102 m

P = 8900 N

ΔL = 0.25 mm = 2.5*10⁻⁴ m

L = ?

We can use the Equation of the Hooke's Law

ΔL = P*L / (A*E)    ⇒   L = ΔL*A*E / P

⇒   L = (2.5*10⁻⁴ m)*(π*(0.0102 m)²*0.25)*(207*10⁹ Pa) / (8900 N)

⇒   L = 0.475 m = 475 mm = 18.7 inches

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