| Practice Problem 10-4 |
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Given: An aluminum-alloy compression member (E = 70 GPa, sY = 270 MPa) of length L = 4 m is loaded in a manner that permits free rotation at its ends. Let ro = 45 mm, and ri = 40 mm. Req'd: Determine the maximum compressive load, Pallow that may be applied if a factor of safety against buckling failure of 1.5 is to be applied. |
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Solution: From the Euler formula for buckling, the critical buckling stress for a pinned-pinned element is: scr = (p2EI) / A(Le)2 For this member we have: Combining theses we get: scr = [p2(70·109N/m2)(1.21·10-6 m2)] / [(1.34·10-3 m2)(4 m)2] = 40.0 MPa If a factor of safety of 1.5 is used, the allowable stress is then: sallow = scr/1.5 = 40 MPa/1.5 = 26 MPa The maximum compressive load, Pallow is then: Pallow = sallow·A = (26 MPa)(1.34·10-3 m2) = 34.7 kN |
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