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
