Question 6.5: For the column of Example Problem 6–4, compute the maximum s...

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 = D/2 = 0.375 in.
Material: SAE 1040 hot-rolled steel; E=30 \times 10^{6} \mathrm{psi}, \mathrm{s}_{y}=42 \mathrm{psi}

Analysis: Use Equation (6–14) to compute maximum stress. Then use Equation (6–16) to compute maximum deflection.

\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]                        (6-14)
Results: All terms have been evaluated before. Then the maximum stress is found from Equation (6–14):

\begin{aligned}\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 \times10^{6}\right)}}\right)\right] \\\sigma_{L / 2} &=29300 \mathrm{psi}\end{aligned}
The maximum deflection is found from Equation (6-16):

y_{\max }=e\left[\sec \left(\frac{K L}{2 r} \sqrt{\frac{P}{A E}}\right)-1\right]                     (6-16)

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 \text { in }
Comments: The maximum stress is 29300 psi at the midlength of the column. The deflection there is 0.293 in from the original straight central axis of the column.