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 SAE 1040 hot-rolled steel. Compute the allowable load for a design factor of 3.

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Accepted Answer

Objective: Specify the allowable load for the column.
Given: Solid circular cross section: [latex]D[/latex] = 0.75 in; [latex]L[/latex] = 32 in; use [latex]N[/latex] = 3.
Both are ends pinned. Initial crookedness = a = 0.125 in.
Material: SAE 1040 hot-rolled steel.
Analysis: Use Equation (6–13).

[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] (6-13)

First evaluate [latex]C_1[/latex] and [latex]C_2[/latex]. Then solve the quadratic equation for [latex]P_a[/latex].
Results:

[latex]\begin{aligned}s_{y} &=42000 \mathrm{psi} \A &=\pi D^{2} / 4=(\pi)(0.75)^{2} / 4=0.442\mathrm{in}^{2} \r &=D / 4=0.75 / 4=0.188 \mathrm{in} \c &=D / 2=0.75 / 2=0.375\mathrm{in} \K L / r &=[(1.0)(32)] / 0.188=171 \P_{c r} &=\frac{\pi^{2} E A}{(KL/ r)^{2}}=\frac{\pi^{2}(30000000)(0.442)}{(171)^{2}}=4476 \mathrm{lb} \C_{1}&=\frac{-1}{N}\left[s_{y} A+\left(1+\frac{a c}{r^{2}} ight) P_{c r} ight]=-9649 \C_{2}&=\frac{s_{y} A P_{c r}}{N^{2}}=9.232 imes 10^{6}\end{aligned}[/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}=1076 \mathrm{lb}[/latex] is the allowable load.

Question 6.4: A column has both ends pinned and has a length of 32 in. It ...

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