Practice Problem 7-2 | Back | Index | Next |
Given: 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. |
The axial stress in this section the force divided by the cross-sectional area:
s_{A} = F/t^{2}
The maximum bending stress is s_{B} = Mc/I, which for a square beam is:
s_{B} = 6M/t^{3}
Here, the bending stress is compressive at the top, and tensile at the bottom. Combining the axial stress and the tensile bending stress gives:
s_{max} = F/t^{2} + 6Fb/t^{3}
s_{max} = 100 lb/(1 in.)^{2} + 6(100 lb)(5 in.)/(1 in.)^{3} = 3100 psi
s_{max} = 3.1 ksi
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