Question 11.8: The stress in the column found in Example Problem 11–7 seems...

Objective Redesign the eccentrically loaded circular column of Example Problem 11–7 to reduce the stress and achieve a design factor of at least 3.

Given Data from Example Problems 11–6 and 11–7.

Analysis    Use a larger diameter. Use Equation (11–23) to compute the required strength. Then compare that with the strength of SAE 1040 hot-rolled steel. Iterate until the stress is satisfactory.

Results     Appendix A–3 gives the value for the yield strength of SAE 1040 HR to be 414 MPa.

Since we want to retain the same material, the cross-sectional dimensions of the column must be increased to decrease the stress. Equation (11–23) 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} = 4780 N, N = 3, L_{e} = 0.8 m, and e = 19 mm, and the value of the entire right side of the equation is less than 414 MPa. The original design had a circular cross section with a diameter of 18 mm. Let’s try increasing the diameter to D = 25 mm. Then

A = πD² /4 = π ( 25 mm)² /4 = 491 mm²

r = D/4 = (25 mm)/4 = 6.3 mm

r² = (6.3 mm)² = 39.0 mm²

c = D/2 = (25 mm)/2 = 12.5 mm

Now let us call the right side of Equation (11–23) s'_{y} . Then

s'_{y} = \frac{3(4780)}{491} \left[1+ \frac{(0.75)(0.0125)}{(39)}sec\left\lgroup\frac{8}{2(0.0063)} \sqrt{\frac{(3)(4780)}{(491)(207 \times10^{9})}}\right\rgroup \right]

s'_{y} = 258.7 MPa = required valuee of s_{y}

This is well below the value of s_{y} = 414 MPa for the given steel, and it gives the desired design factor of 3.0 or greater.

Now we can evaluate the expected maximum deflection with the new design using Equation (11–24):

y_{max} = 0.75\left[sec\left\lgroup\frac{0.8}{2(0.0063)} \sqrt{\frac{4780}{(491(207 \times10^{9}))}}\right\rgroup -1 \right]

y_{max} = 1.87 mm

Comments   The diameter of 25.0 mm is satisfactory. The maximum deflection for the column is 1.9 mm.