Column Design for Concentric Loading Problem: A beachfront house is to be jacked up 10 ft above grade and placed on a set of steel columns. The weight to be supported by each column is estimated to be 200 000 lb. Two designs are to be considered, one using square steel tubes and the other using

Question

Column Design for Concentric Loading

Problem: A beachfront house is to be jacked up 10 ft above grade and placed on a set of steel columns. The weight to be supported by each column is estimated to be 200 000 lb. Two designs are to be considered, one using square steel tubes and the other using round steel tubes.

Given: Design the columns using a safety factor of 4. Determine the columns’ outer dimensions for each shape, assuming a 0.5-in-thick tube wall in each case. The steel alloy has a compressive yield stress [latex]S_{yc}[/latex] = 60 kpsi.

Assumptions: The loading is concentric and the columns are vertical. Their bases are set in concrete and their tops are free, creating a fixed-free end-constraint condition. Use AISC recommended end-condition factors.

Accepted Answer

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.

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

[latex] \begin{aligned} \frac{S_{y c}}{2} &=\frac{\pi^2 E}{S_r^2} \ \left(S_r ight)_D &=\pi \sqrt{\frac{2 E}{S_{y c}}} \end{aligned} [/latex] (4.42)

[latex] \frac{P_{c r}}{A}=S_{y c}-\frac{1}{E}\left\lgroup\frac{S_{y c} S_r}{2 \pi} ight group ^2 [/latex] (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 [latex]A[/latex], second moment of area [latex]I[/latex], radius of gyration [latex]k[/latex], and the slenderness ratio [latex]l_{eff}/k[/latex], 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 [latex](S_{r})_{D}[/latex] at the tangent point should be found from Eq 4.42 and compared to the column’s actual [latex]S_{r}[/latex] 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 [latex]A[/latex], second moment of area [latex]I[/latex], and radius of gyration [latex]k[/latex] for a 0.5-in wall-thickness round tube of that outside diameter are then

[latex] \begin{aligned} A &=\frac{\pi\left(d_o^2-d_i^2 ight)}{4}=\frac{\pi(64-49)}{4}=11.781 in ^2 \ I &=\frac{\pi\left(d_o^4-d_i^4 ight)}{64}=\frac{\pi(4096-2401)}{64}=83.203 in ^4 \ k &=\sqrt{\frac{I}{A}}=\sqrt{\frac{83.203}{11.781}}=2.658 in \end{aligned} [/latex] (a)

4 Calculate the slenderness ratio [latex]S_{r}[/latex] for this column and compare it to the value of [latex](S_{r})_{D}[/latex] 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 [latex]l_{eff} = 2.1 l[/latex].

[latex] \begin{aligned} S_r &=\frac{l_{e f f}}{k}=\frac{120(2.1)}{2.658}=94.825 \ \left(S_r ight)_D &=\pi \sqrt{\frac{2 E}{S_y}}=\pi \sqrt{\frac{2(30 E 6)}{60000}}=99.346 \end{aligned} [/latex] (b)

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 [latex]P_{cr}[/latex] and apply the safety factor to determine the allowable load [latex]P_{allow}[/latex].

[latex] \begin{array}{l} P_{c r}=A\left[S_y-\frac{1}{E}\left\lgroup\frac{S_y S_r}{2 \pi} ight group^2 ight]=11.8\left\{6 E 4-\frac{1}{3 E 7}\left[\frac{6 E 4(94.83)}{2 \pi} ight]^2 ight\}=384866 lb \ P_{ ext {allow }}=\frac{P_{c r}}{S F}=\frac{384866}{4}=96217 lb \end{array} [/latex] (c)

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 ([latex]a[/latex]) 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 ([latex]D_{out}[/latex]) 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.

Table 4-5 Example 4-11 – Column Design Part 1 of 2 Round Column Design
Input	Variable	Output	Unit	Comment
				Input Data
circle	Shape			column shape 'square or 'circle
120	L		in	length of column
0.5	Wall		in	column wall thickness
2.1	end			AISC end condition factor
30E6	E		psi	Young's modulus
60 000	Sy		psi	compressive yield stress
4	FS			safety factor
200 000	Allow		lb	allowable load desired
				Output Data
G[latex]^{ *}[/latex]	Dout	11.35	in	outside dia of column
	Leff	252	in	effective length of column
	Sr	65.60		slenderness ratio
	Srd	99.35		tangency point in [latex]S_{r}[/latex]
	Load	46 921	lb	critical unit load
	Johnson	46 921	lb	Johnson unit load
	Euler	68 811	lb	Euler unit load
	Din	10.35	in	inside dia of column
	k	3.84	in	radius of gyration
	I	251.63	in^4	second moment of area
	A	17.05	in^2	area of cross section
* Indicates that a guess value is required to start the iteration.

Table 4-5 Example 4-11 – Column Design Part 2 of 2 Square Column Design
Input	Variable	Output	Unit	Comment
				Input Data
circle	Shape			column shape 'square or 'circle
120	L		in	length of column
0.5	Wall		in	column wall thickness
2.1	end			AISC end condition factor
30E6	E		psi	Young's modulus
60 000	Sy		psi	compressive yield stress
4	FS			safety factor
200 000	Allow		lb	allowable load desired
				Output Data
G[latex]^{ *}[/latex]	Dout	9.34	in	outside dia of column
	Leff	252	in	effective length of column
	Sr	69.69		slenderness ratio
	Srd	99.35		tangency point in [latex]S_{r}[/latex]
	Load	45 235	lb	critical unit load
	Johnson	45 235	lb	Johnson unit load
	Euler	60 956	lb	Euler unit load
	Din	8.34	in	inside dia of column
	k	3.62	in	radius of gyration
	I	231.21	in^4	second moment of area
	A	17.69	in^2	area of cross section
* Indicates that a guess value is required to start the iteration.

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

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