Given:
A titanium bar with dimensions L = 50 mm and D = 10 mm is subjected to axial force P = 20 kN. Titanium has a Young's Modulus of 115 GPa and a Poisson's Ratio of 0.3.

Req'd: Determine the normal stress s.

Sol'n: The normal or axial stress is the force divided by the cross-sectional area, thus:

s = P/(pD^{2}/4) = (20,000 N) / [p(0.010 m)^{2}/4] = 254.6 x 10^{6} N/m^{2}