Practice Problem 3-2(a)

Practice Problem 3-2 (a)

Given: A steel shaft with dimensions L = 50 mm and D = 10 mm is subjected to a force of P = 20 kN. The shaft has a Young's Modulus of 100 GPa and a Poisson's Ratio of 0.3.

Req'd:
What is the longitudinal stress (parallel to the applied stress) s_L [MPa]?

Solution: The longitudinal or axial stress is simply the force divided by the cross-sectional area, thus:

s_L = P/(pD²/4)

s_L = (20,000 N) / [p(0.010 m²)/4] = 254.6 x 10⁶ N/m² = 255 MPa