Question 4.11: Column Design for Concentric Loading Problem: A beachfront h...

See Table 4-5, parts 1 and 2.

1    This problem, as stated, requires an iterative solution because the allowable load is specified and the column cross-sectional dimensions are requested. If the reverse were desired, equations 4.38c (p. 195), 4.42 (p. 197), and 4.43 (p. 198) could be solved directly to determine the allowable load for any chosen geometry.

\frac{P_{c r}}{A}=\frac{\pi^2 E}{S_r^2}      (4.38c)

\frac{P_{c r}}{A}=S_{y c}-\frac{1}{E}\left\lgroup\frac{S_{y c} S_r}{2 \pi}\right\rgroup ^2      (4.43)

2    To solve this problem using only a calculator requires choosing a trial cross-section dimension, such as the outside diameter, calculating its cross-sectional properties of area A, second moment of area I, radius of gyration k, and the slenderness ratio l_{eff}/k, then using these values in equations 4.38c, 4.42, and 4.43 to determine the allowable load after applying a safety factor. It is not known at the outset whether the column will turn out to be a Johnson or Euler one, so the slenderness ratio (S_{r})_{D} at the tangent point should be found from Eq 4.42 and compared to the column’s actual S_{r} to decide whether Euler’s or Johnson’s equation is required.

3    Assume an 8-in outside diameter for a first-trial, round column. The area A, second moment of area I, and radius of gyration k for a 0.5-in wall-thickness round tube of that outside diameter are then

4    Calculate the slenderness ratio S_{r} for this column and compare it to the value of (S_{r})_{D} corresponding to the tangent point between the Euler and Johnson curves (Eq 4.42). Use the AISC recommended value (Table 4-4, p. 196) for a fixed-free column of l_{eff} = 2.1 l.

5    This column’s slenderness ratio is to the left of the tangent point and thus is in the Johnson region of Figure 4-42 (p. 197), so use equation 4.43 (p. 198) to find the critical load P_{cr} and apply the safety factor to determine the allowable load P_{allow}.

6    This load is substantially below the required 200 000 lb, so we must repeat the calculations in steps 3–5 using larger outside diameters (or wall thicknesses) until we obtain a suitable allowable load. The problem also requests the design of a squaresection column, which will only change the equations (a) in step 3 above.

7    This is clearly a tedious solution process when using only a calculator, and it cries out for a better approach. An equation solver or spreadsheet package can provide such a tool. An iterative-solver is needed for this problem that lets us specify the desired allowable load and have the program iterate until it converges to a value for the outside diameter (D_{out}) that will support the desired load, given an assumed wall thickness. Guess values for one or more of the unknown parameters are needed to start the iteration. The files EX04-10 are on the CD-ROM.

8    These programs allow for the solution of either a square or round cross section design and choose whether to use the Euler or Johnson equations based on the relative values of the slenderness ratios calculated in step 4 above. The complete solution takes only a few seconds. Table 4-5 part 1 shows the solution for the circular column and Table 4-5 part 2 shows the solution for the square column.

9    A round column of 11.3-in diameter and 0.5-in wall is adequate to carry the specified load. This is a Johnson column with an effective slenderness ratio of 65.6 and a weight of 579 lb. The Euler formula predicts a critical load nearly 1.5 times that of the Johnson formula, so this column would be in the “danger region” labeled ABDA in Figure 4-42 (p. 197) if the Euler formula were used. If a 0.5-in wall, square column cross section is chosen, its outside dimension will be 9.3 in for the same slenderness ratio and allowable load, but the column will weigh more at 600 lb. A square column will always be stronger than a round one of the same outside dimension and wall thickness because its area, second moment of area, and radius of gyration are larger due to the material in the corners being at a larger radius. The additional weight of material also makes it more expensive than a round column of the same strength.