Question

Consider a cylindrical nickel wire 1.8 mm in diameter and 2.6 × 104 mm long. Calculate its elongation when a load of 290 N is applied. Assume that the deformation is totally elastic and that the elastic modulus for nickel is 207 GPa (or 207 × 109 N/m2).

Answer 1

Answer:

e = 3.97*10^-4

Explanation:

1.8 mm = 0.0018 m

2.6*10^4 mm = 26 m

Elongation is The ratio between the stretched length and the original length.

e = L/L0

This is calculated with Hooke's law:

e = σ/E

Where

σ: normal stress

E: elastic constant

σ = P/A

Where

P: normal load

A: cross section

A = π/4 * d^2

Therefore:

e = P / (A * E)

e = 4 * P / (π * d^2 * E)

e = 4 * 290 / (π * 0.0018^2 * 207*10^9) = 3.97*10^-4