WORLD FOR CIVIL

Topic 7.2c: Structural Steel Column Selection - Example 3

Select the best (safe & lightest) I-beam to be used as a 16 foot vertical column, with one end fixed and the other end pinned, and which is to carry an axial load of 120,000 lb. Young’s Modulus for the steel is 30 x 10⁶ psi., and the yield stress for the steel is 34,000 psi. (Note that since this is a fixed-pinned column, the effective length is equal to .7 * L.) Use the Table of I-Beams to select from.

One method of determining the best I-beam to use is an iterative type process. We first find the minimum cross sectional area of the beam by assuming the slenderness ratio, L_e/r =0. Then the factor of safety, FS = 5/3 = 1.667, and the allowable stress = (yield stress/safety factor) = 34,000 lb/in²/1.667 = 20,400 lb/in². Since the allowable stress is also Force/Area, we can solve for the minimum area: Area = 120,000 lb/20,400 lb/in² = 5.88 in².

Next we choose (guess) a beam with an area greater than the minimum area found in step 1. From our table we will try a W10 x 21, which has an area of 6.2 in² and a minimum radius of gyration of 1.32 in. The Slenderness ratio for this column = L_e/r = (.7*16' * 12"/ft)/ 1.32 in. = 101.8. We also calculate the critical slenderness ratio C_c² = ( 2 * 3.14² * 30 x 10⁶ lb/in² /34,000 lb/in²) = 17417; and C = 132 Since the slenderness ratio of the column is less than the critical slenderness ratio, we use formula for intermediate columns to find the allowable stress.

Before we can determine the allowable stress, we first calculate the factor of safety. FS = (5/3) + (3/8)(101.8/132) - (1/8)(101.8/132)³ = 1.9. Then the Allowable Stress = (34,000/1.9)[1 - (1/2)(101.8/132)²] = 10,994 lb/in²

Finally the Allowable Load = Stress * Area = 10,994 lb/in² * 6.2 in² = 68,165 lb.

We notice that the allowable load is about half of the load we would like to apply. This means we need an I-beam with a larger area and/or a larger radius of gyration. So we now select a larger cross section beam, and repeat steps 2, 3, and 4.

From our table we will try a W10 x 33, which has an area of 9.71 in² and a minimum radius of gyration of 1.94 in. The Slenderness ratio for this column = L_e/r = (.7*16' * 12"/ft)/ 1.94 in. = 69.3. The critical slenderness ratio remains the same [C_c² = ( 2 * 3.14² * 30 x 10⁶ lb/in² /34,000 lb/in²) = 17417; and C = 132] 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.

Before we can determine the allowable stress, we first calculate the factor of safety. FS = (5/3) + (3/8)(69.3/132) - (1/8)(69.3/132)³ = 1.85. Then the Allowable Stress = (34,000/1.85)[1 - (1/2)(69.3/132)²] = 15,845 lb/in².

Finally the Allowable Load = Stress * Area = 15,845 lb/in² * 9.71 in² = 153,855 lb. We notice that the allowable load is somewhat higher than the load we would like to apply. This means, of course, that this I-beam would work, but it is probably not the best (lightest). So we make one more guess/estimate. We need a beam with a slightly lower area and/or radius of gyration.

From our table we will try a W12 x 31, which has area of 9.13 in² and a minimum radius of gyration of 1.54 in. The Slenderness ratio for this column = L_e/r = (.7*16' * 12"/ft)/ 1.54 in. = 87.3. The critical slenderness ratio remains the same [C_c² = ( 2 * 3.14² * 30 x 10⁶ lb/in² /34,000 lb/in²) = 17417; and C = 132] 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.

Before we can determine the allowable stress, we first calculate the factor of safety. FS = (5/3) + (3/8)(87.3/132) - (1/8)(87.3/132)³ = 1.88. Then the Allowable Stress = (34,000/1.88)[1 - (1/2)(87.3/132)²] = 14,130 lb/in²

Finally the Allowable Load = Stress * Area = 14,130 lb/in² * 9.13 in² = 129,000 lb. Here we see our maximum allowable load is just slightly above the load we would like to apply, (120,000 lb.). This beam is a very good candidate for the best beam. It might be possible to find a slightly better beam, but we will end here.

There are other design procedures for selecting columns than the one above. Most of these also utilize a trial and error type of procedure. However, with computers and spreadsheets today, it is not a hard process to develop a program to try a number of different columns and arrive at the best one in a short time.