Question

Consider a cylindrical titanium wire 3.0 mm (0.12 in.) in diameter and 2.5 Ã 104 mm (1000 in.) long. calculate its elongation when a load of 500 n (112 lbf) is applied. assume that the deformation is totally elastic.

Answer 1

Given:

Diameter,d of the titanium wire = 3.0 mm = 0.003 m

Length, L = 104 mm = 0.0104 m

Force applied F = 500 N

To determine:

The deformation, i.e.change in length ΔL

Explanation:

Stress on the wire = Force/Area

F = 500 N

A = π(d/2)² = π(0.003/2)² = 7.069*10⁻⁶ m²

Stress = 500 N/7.069*10⁻⁶ m² = 70.731*10⁶ N/m2 = 0.0707 GPa

Now, the Young's modulus for Ti = 107 GPa

Young's modulus = Stress/Strain

Strain = 0.0707/107 = 0.000660

Now,

Strain = ΔL/L

ΔL = 0.000660*0.0104 = 6.87 * 10⁻⁶ m

Ans: The elongation in the Ti wire would be 6.87 * 10⁻⁶ m