For the column of Example Problem 6_4, compute the maximum stress and deflection if a load of 1075 lb is applied with an eccentricity of 0.75 in. The column is initially straight.
For the column of Example Problem 6_4, compute the maximum stress and deflection if a load of 1075 lb is applied with an eccentricity of 0.75 in. The column is initially straight.
Objective Compute the stress and the deflection for the eccentrically loaded column.
Given Data from Example Problem 6_4 , but eccentricity = e = 0.75 in. Solid circular cross section: D = 0.75 in; L = 32 in; Initially straight Both ends are pinned; KL = 32 in; r = 0.188 in; c = DI2 = 0.375 in. Material: AISI 1040 hot-rolled steel; E =30×10^{6} psi.
Analysis Use Equation (6_12) \sigma_{L / 2}=\frac{P}{A}\left[1+\frac{e c}{r^{2}} \sec \left(\frac{K L}{2 r} \sqrt{\frac{P}{A E}}\right)\right] to compute maximum stress. Then use Equation (6_14) y_{\max }=e\left[\sec \left(\frac{K L}{2 r} \sqrt{\frac{P}{A E}}\right)-1\right] to compute maximum deflection.
Results All terms have been evaluated before. Then the maximum stress is found from Equation (6_12): \sigma_{L / 2}=\frac{P}{A}\left[1+\frac{e c}{r^{2}} \sec \left(\frac{K L}{2 r} \sqrt{\frac{P}{A E}}\right)\right]
\sigma_{L / 2}=\frac{1075}{0.422}\left[1+\frac{(0.75)(0.375)}{(0.188)^{2}} \sec \left(\frac{32}{2(0.188)} \sqrt{\frac{1075}{(0.442)\left(30 \times 10^{6}\right)}}\right)\right]
\sigma_{L / 2}=29300 \mathrm{psi} The maximum deflection is found from Equation (6_14): y_{\max }=e\left[\sec \left(\frac{K L}{2 r} \sqrt{\frac{P}{A E}}\right)-1\right]
y_{\max }=0.75\left[\sec \left(\frac{32}{2(0.188)} \sqrt{\frac{1075}{(0.442)\left(30 \times 10^{6}\right)}}\right)-1\right]=0.293 \mathrm{in}
Comments The maximum stress is 29 300 psi at the midlength of the column. The deflection there is 0.293 in from the original straight central axis of the column.