The square aluminum (E = 10 Msi) bar at right is subjected to a tensile force of F = 100 lb. Let t = 1 in., a = 24 in. and b = 5 in.
Req'd: Determine the maximum normal stress in the bar.
Sol'n: Take a vertical cut through the middle segment of the bar. At that interior cross-section, both axial force F and bending moment M = Fb act. The maximum normal stress occurs in the middle segment of the bar due to an axial stress and a bending stress.
The axial stress in this section the force divided by the cross-sectional area:
sA = F/t2
The maximum bending stress is sB = Mc/I, which for a square beam is:
sB = 6M/t3
Here, the bending stress is compressive at the top, and tensile at the bottom. Combining the axial stress and the tensile bending stress gives: