| Practice Problem 10-4 | Back | Index | Next |
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Given: An aluminum column (E = 70 GPa, Sy = 270 MPa) of length L = 4.0 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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scr = (p2EI) / (Le)2
For this member:
E = 70 GPa
Le = L = 4.0 m
I = (p/4)[(ro)4 – (ri)4] = (p/4)[(0.045 m)4 – (0.040 m)4] = 1.21·10–6 m2
Combining theses we get:
scr = [p2(70 GPa )(1.21·10–6 m2)] / (4 0m)2 = 52.25 kN
If a factor of safety of 1.5 is used, the allowable load is then:
Pallow = Pcr/1.5 = 52.25 kN/1.5 = 34.7 kN
Pallow = 34.7 kN
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