A column has both ends pinned and has a length of 32 in. It has a circular cross section with a diameter of 0.75 in and an initial crookedness of 0.125 in. The material is AISI 1040 hot-rolled steel. Compute the allowable load for a design factor of 3.

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Objective Specify the allowable load for the column.

Given Solid circular cross section: D = 0.75 in; L = 32 in: use N = 3. Both are ends pinned. Initial crookedness = a = 0.125 in. Material: AISI 1040 hot-rolled steel.

Analysis Use Equation (6_11) [latex]P_{a}^{2}-\frac{1}{N}\left[s_{y} A+\left(1+\frac{a c}{r^{2}} ight) P_{c r} ight] P_{a}+\frac{s_{y} A P_{c r}}{N^{2}}=0[/latex]. First evaluate [latex]C_{1}[/latex] and [latex]C_{2}[/latex]. Then solve the quadratic equation for [latex]P_{a}[/latex].

Results [latex]s_{y}=42000 psi [/latex]

[latex]A=\pi D^{2} / 4=(\pi)(0.75)^{2} / 4=0.442 \mathrm{in}^{2}[/latex]

[latex]r=D/4=0.75/4=0.188 in[/latex]

[latex]c=D/2=0.75/2=0.375 in [/latex]

[latex]K L / r=[(1.0)(32)] / 0.188=171[/latex]

[latex]P_{cr}=\frac{\pi^{2} E A}{(K L / r)^{2}}=\frac{\pi^{2}(30000000)(0.442)}{(171)^{2}}=4476 \mathrm{lb}[/latex]

[latex]C_{1}=\frac{-1}{N}\left[s_{y} A+\left(1+\frac{a c}{r^{2}} ight) P_{c r} ight]=-9649[/latex]

[latex]C_{2}=\frac{s_{y} A P_{c r}}{N^{2}}=9.232 imes 10^{6}[/latex]

The quadratic is therefore [latex]P_{a}^{2}-9649 P_{a}+9.232 imes 10^{6}=0[/latex]

Comment From this, [latex]P_{a}[/latex] = 1077 Ib is the allowable load.

Question 6.EP.4: A column has both ends pinned and has a length of 32 in. It ...

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