Topic 7.2b: Structural Steel - Example 2
A structural steel T-beam (WT 15 x 66) is used as a column. The beam is 12 ft. long and is fixed on one end and free on the other. Young’s Modulus for the steel is 30 x 106 psi., and the yield stress for the steel is 38,000 psi. Determine the maximum allowable axial load. (Note that since this is a fixed-free column, the effective length is equal to 2 * L.). Repeat the problem if the column is 20 ft long.
| - | - | Depth | Flange | Flange | Stem | - | Cross | Section | Info. | - |
| Designation | Area | of T | Width | thick | thick | - | x-x axis | x-x axis | x-x axis | x-x axis |
| - | A-in2 | d - in | wf - in | tf - in | tw - in | d/tw | I - in4 | S -in3 | r - in | y - in |
| WT15x66 | 19.40 | 15.15 | 10.551 | 1.000 | 0.615 | 24.60 | 421.00 | 37.400 | 4.650 | 3.900 |
We will first do this problem assuming that radius of gyration is lowest about the X-X Axis, and buckling will occur first with respect to that axis. Calculating the slenderness ratio = Le/r = (2*12ft * 12"/ft)/4.65 = 61.94
Next we calculate the critical slenderness ratio Cc2 = ( 2 * 3.142 * 30 x 106 lb/in2 /38,000 lb/in2) = 15568; and C = 124.8. Since the slenderness ratio of the column is less than the critical slenderness ratio, we use the intermediate formula to find the allowable stress.
Before we can determine the allowable stress, we first calculate the factor of safety.
FS = (5/3) + (3/8)(61.94/124.8) - (1/8)(61.94/124.8)3 = 1.84.
Then Allowable Stress = (38,000/1.84)[1 - (1/2)(61.94/124.8)2] = 18,110 lb/in2.
Finally the allowable load = Stress * Area = 18,110 lb/in2 * 19.4 in2 = 351,334 lb.
PART 2: Repeat if the column is 20 ft long.
We again assume that radius of gyration is lowest about the X-X Axis, and buckling will occur first with respect to that axis. Calculating the slenderness ratio = Le/r = (2*20ft * 12"/ft)/4.65 in = 103.2
Next we calculate the critical slenderness ratio Cc2 = ( 2 * 3.142 * 30 x 106 lb/in2 /38,000 lb/in2) = 15568; and C = 124.8. Since the slenderness ratio of the column is less than the critical slenderness ratio, we use the formula for intermediate columns to find the allowable stress. However, before we can determine the allowable stress, we must first calculate the factor of safety.
FS = (5/3) + (3/8)(103.2/124.8) - (1/8)(103.2/124.8)3 = 1.91
Then Allowable Stress = (38,000/1.91)[1 -(1/2)(103.2/124.8)2] = 13,090 lb/in2
Finally the Allowable Load = Stress * Area = 13,090 lb/in2 * 19.4 in2 = 253,946 lb.
Notice how much the change in length has reduced the allowable load.
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