Question 6.6: The stress in the column found in Example Problem 6–5 seems ...

Objective: Redesign the eccentrically loaded column of Example Problem 6–5 to reduce the stress and achieve a design factor of at least 3 .
Given: Data from Example Problems 6-4 and 6-5.
Analysis: Use a larger diameter. Use Equation (6-15) to compute the required strength. Then compare that with the strength of SAE 1040 hot-rolled steel. Iterate until the stress is satisfactory.

$s_{y}=\frac{N P_{a}}{A}\left[1+\frac{e c}{r^{2}} \sec \left(\frac{K L}{2 r} \sqrt{\frac{N P_{a}}{A E}}\right)\right]$                (6-15)

Results: Appendix 3 gives the value for the yield strength of SAE $1040 \mathrm{HR}$ to be 42000 psi. If we choose to retain the same material, the cross-sectional dimensions of the column must be increased to decrease the stress. Equation (6-15) can be used to evaluate a design alternative.

The objective is to find suitable values for $A, c$, and $r$ for the cross section such that $P_{a}=1075 \mathrm{lb} ; N=3 ; L_{e}=32 \mathrm{in} ; e=0.75 \mathrm{in}$; and the value of the entire right side of the equation is less than 42000 psi. The original design had a circular cross section with a diameter of 0.75 in. Let’s try increasing the diameter to $D=1.00$ in. Then
$\begin{aligned}A &=\pi D^{2} / 4=\pi(1.00 \mathrm{in})^{2} / 4=0.785 \mathrm{in}^{2} \\r &=D / 4=(1.00 \mathrm{in}) / 4=0.250 \mathrm{in} \\r^{2} &=(0.250 \mathrm{in})^{2}=0.0625 \mathrm{in}^{2} \\c &=D / 2=(1.00 \mathrm{in}) / 2=0.50 \mathrm{in}\end{aligned}$
Now let’s call the right side of Equation (6-15) $s_{y}^{\prime}$. Then
$\begin{aligned}&s_{y}^{\prime}=\frac{3(1075)}{0.785}\left[1+\frac{(0.75)(0.50)}{(0.0625)} \sec \left(\frac{32}{2(0.250)} \sqrt{\frac{(3)(1075)}{(0.785)\left(30 \times10^{6}\right)}}\right)\right] \\&s_{y}^{\prime}=37740 \mathrm{psi}=\text { requiredvalue of } s_{y}\end{aligned}$
This is a satisfactory result because it is just slightly less than the value of $s_{y}$ of 42000 psi for the steel.

Now we can evaluate the expected maximum deflection with the new design using Equation (6-16);

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

$\begin{aligned}&y_{\max }=0.75\left[\sec \left(\frac{32}{2(0.250)} \sqrt{\left.\frac{1075}{(0.785)\left(30 \times 10^{6}\right.}\right)-1}\right]\right. \\&y_{\max }=0.076 \text { in }\end{aligned}$
Comments: The diameter of 1.00 in is satisfactory. The maximum deflection for the column is 0.076 in.