A 1.5-m-long aluminum rod must not stretch more than 1 mm andthe normal stress must not exceed 40 MPa when the rod is subjectedto a 3-kN axial load. Knowing that E = 70 GPa, determine therequired diameter of the rod.

d = 11.2 mm

Explanation:

L = 1.5 m

σ = 40 * 10^6 Pa

E = 40 * 10^9 Pa

P = 3 * 10^3 N

δ = 1 * 10^-3 m

stress:

σ = P/A

A = P/σ

= 3 * 10^3 N/40 * 10^6 Pa

= 75 mm^2

deformation:

δ = P*L/A*E

A = P*L/E*δ

= 75 mm^2

A = π/4*d^2 (1)

putting all values in eq (1)

d = 11.2 mm

note:

there maybe calculation mistake but method is correct