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

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: SAE 1040 hot-rolled steel.
Analysis: Use Equation (6–13).

P_{a}^{2}-\frac{1}{N}\left[s_{y} A+\left(1+\frac{a c}{r^{2}}\right) P_{c r}\right] P_{a}+\frac{s_{y} A P_{c r}}{N^{2}}=0                       (6-13)

First evaluate C_1 and C_2. Then solve the quadratic equation for P_a.
Results:

\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}}\right) P_{c r}\right]=-9649 \\C_{2}&=\frac{s_{y} A P_{c r}}{N^{2}}=9.232 \times 10^{6}\end{aligned}
The quadratic is therefore
P_{a}^{2}-9649 P_{a}+9.232 \times 10^{6}=0

Comment: From this, P_{a}=1076 \mathrm{lb} is the allowable load.