Question

A 30-mm-diameter copper rod is 1 m long with a yield strength of 70 MPa. Determine the axial force necessary to cause the diameter of the rod to reduce by 0.01 percent, assuming elastic deformation. Check that the elastic deformation assumption is valid by comparing the axial stress to the yield strength.

Answer 1

Explanation:

Given data:

d = 30 mm = 0.03 m

L = 1m

S$_{y}$ = 70 Mpa

Δd = -0.0001d

Axial force = ?

validity of elastic deformation assumption.

Solution:

O'₂ = Δd/d = (-0.0001d)/d = -0.0001

For copper,

v = 0.326      E = 119×10³ Mpa

O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶

∵δ = F.L/E.A    and σ = F/A so,

σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa

F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN

S$_{y}$ = 70 MPa > σ = 36.5 MPa

∵ elastic deformation assumption is valid.

so the answer is

F = 25800 K N            and     S$_{y}$ > σ