Question 6.1: A column has a solid circular cross section, 1.25 inches in ...

Objective: Specify a safe loading for the column.
Given: Solid circular cross section: diameter = D = 1.25 in; length = L = 4.50 ft.
Both ends of the column are pinned.
Material: SAE 1020 cold-drawn steel.

Analysis: Use the procedure in Figure 6–7.
Results:

Step 1. For the pinned-end column, the end-fixity factor is K = 1.0. The effective length equals the actual length; KL = 4.50 ft = 54.0 in = L_e.
Step 2. From Appendix 1, for a solid round section,
r = D/4 = 1.25/4 = 0.3125 in
Step 3. Compute the slenderness ratio:

\frac{K L}{r}=\frac{1.0(54)}{0.3125}=173
Step 4. Compute the column constant from Equation (6-8) (\frac{s_{y}}{2}=\frac{\pi^{2} E}{S_{r T}^{2}} \Rightarrow C_{c}=S_{r T}=\sqrt{\frac{2 \pi^{2} E}{s_{y}}}. For SAE 1020 cold-drawn steel, the yield strength is 51000 \mathrm{psi}, and the modulus of elasticity is 30 \times 10^{6} \mathrm{psi}. Then
C_{c}=\sqrt{\frac{2 \pi^{2} E}{s_{y}}}=\sqrt{\frac{2 \pi^{2}\left(30 \times 10^{6}\right)}{51000}}=108
Step 5. Because K L / r is greater than C_{c}, the column is long, and Euler’s formula should be used. The area is
I=\frac{\pi D^{4}}{64}=\frac{\pi(1.25)^{4}}{64}=0.1198 \mathrm{in}^{4}
Then the critical load is
P_{C r}=\frac{\pi^{2} E I}{\left(L_{e}\right)^{2}}=\frac{\pi^{2}\left(30 \times 10^{6}\right)(0.1198)}{(54)^{2}}=12169 \mathrm{lb}
At this load, the column should just begin to buckle. A safe load would be a reduced value, found by applying the design factor to the critical load. Let’s use N=3 to compute the allowable load, P_{a}=P_{c r} / N:
P_{a}=(12169 \mathrm{lb}) / 3=4056 \mathrm{lb}
Comment: The safe load on the column is 4056 \mathrm{lb}.

APPENDIX 1 Properties of Areas
\begin{array}{ll}A=\text { area } & r=\text { radius of gyration }=\sqrt{I / A} \\I=\text {moment of inertia } & J=\text { polar moment of inertia } \\S=\text { section modulus } &Z_{p}=\text { polar section modulus }\end{array}
	\begin{array}{ll}A=\pi D^{2} / 4 & r=D / 4 \\I=\pi D^{4} / 64 & J=\pi D^{4} / 32 \\S=\pi D^{3} / 32 & Z_{p}=\pi D^{3} / 16\end{array}
	\begin{array}{ll}A=\pi\left(D^{2}-d^{2}\right) / 4 & r=\frac{\sqrt{\left(D^{2}+d^{2}\right)}}{4} \\I=\pi\left(D^{4}-d^{4}\right) / 64 & J=\pi\left(D^{4}-d^{4}\right) / 32 \\S=\pi\left(D^{4}-d^{4}\right) / 32 D & Z_{p}=\pi\left(D^{4}-d^{4}\right) / 16 D\end{array}
	\begin{aligned}&A=H^{2} \quad\quad\quad r=H / \sqrt{12} \\&I=H^{4} / 12 \\&S=H^{3} / 6\end{aligned}
	\begin{array}{ll}A=B H & r_{x}=H / \sqrt{12} \\I_{x}=B H^{3} / 12 & r_{y}=B / \sqrt{12} \\I_{y}=H B^{3} / 12 & \\S_{x}=B H^{2} / 6 & \\S_{y}=H B^{2} / 6 &\end{array}
	\begin{aligned}&A=B H-b h \\&I_{x}=\frac{B H^{3}-b h^{3}}{12} \quad S_{x}=\frac{B H^{3}-b h^{3}}{6 H}\quad r_{x}=0.289 \sqrt{\frac{B H^{3}-b h^{3}}{B H-b h}} \\&I_{y}=\frac{H B^{3}-h b^{3}}{12} \quad S_{y}=\frac{H B^{3}-h b^{3}}{6 B} \quad r_{y}=0.289 \sqrt{\frac{H B^{3}-h b^{3}}{H B-h b}}\end{aligned}
	\begin{aligned}&A=B H / 2 \quad\quad\quad r=H / \sqrt{18} \\&I=B H^{3} / 36 \\&S=B H^{2} /24\end{aligned}
	\begin{aligned}&A=\pi D^{2} / 8 \quad\quad\quad r=0.132 D \\&I=0.007 D^{4} \\&S=0.024D^{3}\end{aligned}
	\begin{aligned}&A=0.866 D^{2} \quad\quad\quad r=0.264 D \\&I=0.06 D^{4} \\&S=0.12 D^{3}\end{aligned}
	\begin{aligned}A &=H(a+B) / 2 \\y &=\frac{H(a+2 B)}{3(a+B)} \quad\quad\quad\quad\quad\quad S=\frac{H^{2}\left(a^{2}+4 a B+B^{2}\right)}{12(a+2 B)} \\I_{x} &=\frac{H^{3}\left(a^{2}+4a B+B^{2}\right)}{36(a+B)} \quad\quad r=\frac{H^{2}\left(a^{2}+4 a B+B^{2}\right)}{18(a+B)^{2}} \\y &=\underset{\text { Maximum distance from } x \text {-axis to }}{\text { outer surface of section }}\end{aligned}
	\begin{aligned}&A=\pi b h \\&I=\frac{\pi h^{3} b}{4} \\&S=\frac{\pi h^{2} b}{4} \\&r=h / 2\end{aligned}