Question 6.EP.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_13) \text { Required } 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] to compute the required strength. Then compare that with the strength of AISI 1040 hot-rolled steel. Iterate until the stress is satisfactory.

Results       Appendix 3 gives the value for the yield strength of AISI 1040 HR to be 42 000 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_13) \text { Required } 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] 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 lb; N = 3; L_{e} = 32 in; e =0.75 in; and the value of the entire right side of the equation is less than 42 000 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 A=\pi D^{2} / 4=\pi(1.00 \mathrm{in})^{2} / 4=0.785 \mathrm{in}^{2}

c=D / 2=(1.00 \mathrm{in}) / 2=0.50 \mathrm{in}  Now let’s call the right side of Equation (6_13)s^{\prime }_{y}. Then \text { Required } 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]

s_{y}^{\prime}=37740 \mathrm{psi}=\text { required value of } s_{y} This is a satisfactory result because it is just slightly less than the value of s_{y} of 42 000 psi for the steel. Now we can evaluate the expected maximum deflection with the new design using Equation (6_14):

Comments         The diameter of 1.00 in is satisfactory.The maximum deflection for the column is 0.076 in.