Question 11.3: Refer to the results of Example Problem 11–1 in which a 25 m...
Refer to the results of Example Problem 11–1 in which a 25 mm diameter steel column with a length of 950 mm was analyzed to determine the critical buckling load. The column was made from SAE 1020 cold-drawn steel. It was found that P_{cr} = 43.4 kN. Now consider a redesign of the structure for which this column is a part. It is decided to provide lateral bracing in all directions for the column at its midlength. Determine the critical buckling load for the redesigned column.
Objective Compute the critical buckling load for the braced column.
Given Data from Example Problem 11–1; circular cross section, D = 25 mm, L = 950 mm; pinned ends; SAE 1020 CD steel; E = 207 GPa; s_{y} = 441 MPa; column braced at 450 mm from each end
Analysis Use the Method of Analyzing Columns.
Results Step 1. End-fixity factor, K = 1.0 for pinned ends
Step 2. Effective length: The unbraced length is now 450 mm. Then L_{e} = KL = 1.0(450 mm) = 450 mm
Step 3. Radius of gyration = r = 6.25 mm [from Example Problem 11–1]
Step 4. Slenderness ratio = SR = L_{e} /r = (450 mm)/(6.25 mm) = 72
Step 5. Column constant = C_{c} = 96.2 [from Example Problem 11–1]
Step 6. Because SR < C_{c} , use the Johnson formula, A = 491 mm²:
P_{cr} = As_{y}[1- \frac{s_{y}(SR)^{2}}{4 \pi^{2}E} ]
P_{cr} =\left\lgroup491 mm^{2}\right\rgroup \left\lgroup\frac{414 N}{mm^{2}}\right\rgroup [1- \frac{(414 \times 10^{6} N/m^{2})(72)^{2}}{4 \pi^{2}(207 \times10^{9} N/m^{2})} ] = 156 kN
Comment The critical buckling load has been increased from 43.4 to 156 kN, over 3 \frac{1}{2} times. That is a significant improvement. The column behaves as if it were only one-half as long.