A column has a solid circular cross section, 1.25 inches in diameter; it has a length of 4.50 ft and is pinned at both ends. If it is made from SAE 1020 cold-drawn steel, what would be a safe column loading?

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Accepted Answer

Objective: Specify a safe loading for the column.
Given: Solid circular cross section: diameter = [latex]D[/latex] = 1.25 in; length = [latex]L[/latex] = 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 [latex]K[/latex] = 1.0. The effective length equals the actual length; [latex]KL[/latex] = 4.50 ft = 54.0 in = [latex]L_e[/latex].
Step 2. From Appendix 1, for a solid round section,
[latex]r = D/4 = 1.25/4 = 0.3125[/latex] in
Step 3. Compute the slenderness ratio:

[latex]\frac{K L}{r}=\frac{1.0(54)}{0.3125}=173[/latex]
Step 4. Compute the column constant from Equation (6-8) ([latex]\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}}}[/latex]. For SAE 1020 cold-drawn steel, the yield strength is [latex]51000 \mathrm{psi}[/latex], and the modulus of elasticity is [latex]30 imes 10^{6} \mathrm{psi}[/latex]. Then
[latex]C_{c}=\sqrt{\frac{2 \pi^{2} E}{s_{y}}}=\sqrt{\frac{2 \pi^{2}\left(30 imes 10^{6} ight)}{51000}}=108[/latex]
Step 5. Because [latex]K L / r[/latex] is greater than [latex]C_{c}[/latex], the column is long, and Euler's formula should be used. The area is
[latex]I=\frac{\pi D^{4}}{64}=\frac{\pi(1.25)^{4}}{64}=0.1198 \mathrm{in}^{4}[/latex]
Then the critical load is
[latex]P_{C r}=\frac{\pi^{2} E I}{\left(L_{e} ight)^{2}}=\frac{\pi^{2}\left(30 imes 10^{6} ight)(0.1198)}{(54)^{2}}=12169 \mathrm{lb}[/latex]
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 [latex]N=3[/latex] to compute the allowable load, [latex]P_{a}=P_{c r} / N:[/latex]
[latex]P_{a}=(12169 \mathrm{lb}) / 3=4056 \mathrm{lb}[/latex]
Comment: The safe load on the column is [latex]4056 \mathrm{lb}[/latex].

APPENDIX 1 Properties of Areas
[latex]\begin{array}{ll}A= ext { area } & r= ext { radius of gyration }=\sqrt{I / A} \I= ext {moment of inertia } & J= ext { polar moment of inertia } \S= ext { section modulus } &Z_{p}= ext { polar section modulus }\end{array}[/latex]
	[latex]\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}[/latex]
	[latex]\begin{array}{ll}A=\pi\left(D^{2}-d^{2} ight) / 4 & r=\frac{\sqrt{\left(D^{2}+d^{2} ight)}}{4} \I=\pi\left(D^{4}-d^{4} ight) / 64 & J=\pi\left(D^{4}-d^{4} ight) / 32 \S=\pi\left(D^{4}-d^{4} ight) / 32 D & Z_{p}=\pi\left(D^{4}-d^{4} ight) / 16 D\end{array}[/latex]
	[latex]\begin{aligned}&A=H^{2} \quad\quad\quad r=H / \sqrt{12} \&I=H^{4} / 12 \&S=H^{3} / 6\end{aligned}[/latex]
	[latex]\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}[/latex]
	[latex]\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}[/latex]
	[latex]\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}[/latex]
	[latex]\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}[/latex]
	[latex]\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}[/latex]
	[latex]\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} ight)}{12(a+2 B)} \I_{x} &=\frac{H^{3}\left(a^{2}+4a B+B^{2} ight)}{36(a+B)} \quad\quad r=\frac{H^{2}\left(a^{2}+4 a B+B^{2} ight)}{18(a+B)^{2}} \y &=\underset{ ext { Maximum distance from } x ext {-axis to }}{ ext { outer surface of section }}\end{aligned}[/latex]
	[latex]\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}[/latex]

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

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