Question 6.EP.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: AISI 1020 cold-drawn steel.

Analysis      Use the procedure in Figure 6_4.

Results        Step 1. For the pinned-end column, the end-fixity factor is AT = 1.0. The effective length equals the actual length; KL = 4.50 ft = 54.0 in.

Step 2. From Appendix 1, for a solid round section,r=D/4=1.25/4=0.3125in

Step 3. Compute the slenderness ratio:\frac{KL}{r}=\frac{1.0\left(45\right) }{0.3125}=173

Step 4. Compute the column constant from Equation (6_4) C_{c}=\sqrt{\frac{2\pi ^{2}E }{s_{y} } } . For AISI 1020 cold-drawn steel, the yield strength is 51 000 psi. and the modulus of elasticity is 30 × 10^{6} psi.ThenC_{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 KL/r is greater thanC_{c},the column is long, and Euler’s formula should be used. The area is A=\frac{\pi D^{2}}{4}=\frac{\pi(1.25)^{2}}{4}=1.23 \mathrm{in}^{2} Then the critical load is P_{c r}=\frac{\pi^{2} E A}{(K L / r)^{2}}=\frac{\pi^{2}\left(30 \times 10^{6}\right)(1.23)}{(173)^{2}}=12200 \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_{cr}/N: P_{a}=\left(12200\right)/3=4067 Ib

Comment       The safe load on the column is 4067 lb.