Question 11.4: Compute the allowable compressive strength, Pa , for a colum...
Compute the allowable compressive strength, P_{a} , for a column made from metric steel rectangular structural tubing, 50×100×4. The material is ASTM A500, Grade B, structural steel. The column length is 3050 mm and its ends are pinned.
We will use Equations (11–8) through (11–13) with E = 200 GPa = 200 000 MPa and s_{y} = 290 MPa. The tube is expected to buckle about the Y–Y axis so r_{min} = r_{y} = 20.4 mm [Appendix A–8(c)] and A = 1136 mm²:
\text{Transition slenderness ratio } = SR_t = 4.71 \sqrt{E/s_y} \\ ( SR_t \text{ is approximately } 6\% \text{ larger than } C_c ) (11-8)
\text{ Elastic critical buckling stress } = s_e = π^2 E /(SR)^2 \\ ( \text{sometimes called Eulerstress }) (11-9)
S_{cr} = [0.658^d]s_y \quad \quad \text{ and the exponent } d = s_y/s_e (11-10)
s_{cr} = 0.877 s_e (11-11)
\text{ Nominal buckling strength } = P_n = s_{cr}A_g (11-12)
\text{ Allowable compressive strength } = P_a = P_n / 1.67 (11-13)
Actual slenderness ratio = SR = KL/r = 1.00 (3050 mm/ 20.4 mm) = 150
Transition slenderness ratio = SR_{t} = 4.71 \sqrt{E/s_{y}} = 4.71 \sqrt{(200 000/290)} = 123.7
Elastic critical buckling stress = s_{e} = π² E/(SR)² = π²(200 000 MPa)/(150)²
s_{e} = 87.73 MPa
Because SR > SR_{t} , the column is long, and we use Equation (11–11):
s_{cr} = 0.877 s_{e} = 0.877(87.73 MPa) = 76.9 MPa
Now, using Equation (11–12),
Nominal buckling strength = P_{n} = s_{cr} A_{g} = (76.9 N/mm²)(1136 mm²) = 87.40 kN
Using Equation (11–13),
Allowable compressive strength = P_{a} = P_{n} /1.67 = 87.40 kN/1.67 = 52.3 kN
