Question 11.6: A column has both ends pinned and has a length of 80 cm. It ...
A column has both ends pinned and has a length of 80 cm. It has a circular cross section with a diameter of 18.0 mm and an initial crookedness of 3.0 mm. The material is SAE 1040 hot-rolled steel. Compute the allowable load for a design factor of 3.
Objective Specify the allowable load for the column.
Given Solid circular cross section: D = 18 mm; L = 800 mm; use N = 3.
Both ends are pinned; K = 1.0. Initial crookedness = a = 3.0 mm
Material: SAE 1040 hot-rolled steel
Analysis Use Equation (11-21). First, evaluate C_{1} and C_{2} . Then solve the quadratic equation for P_{a} .
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 (11–21)
Results
s_{y} = 414 MPa, E = 207 GPa
A = πD² /4 = (π)(18)² /4 = 255 mm²
r = D/4 = 18/4 = 4.5 mm
c = D/2 = 18/2 = 9.0 mm
KL/r = [(1.0)(800)]/4.5 = 178
P_{cr} = \frac{π²EA}{(KL/r)²} = \frac{π²(207 \times 10^{9})(255)}{178²(1000²)} = 16 442 N
C_{1} = \frac{-1}{N} [S_{y}A+\left\lgroup 1+ \frac{ac}{r²}\right\rgroup P_{cr} ] = -42 497
C_{2} = \frac{s_{y}AP_{cr}}{N^{2}} = 1.93 \times 10^{8}
The quadratic is therefore
P_{a}^{2} – 42 497 + (1.93 \times 10^{8}) = 0
Solving this quadratic equation gives P_{a} = 5170 N, the allowable load.