A 14 ft. column is subjected to total compressive load of 550 kips.

Question

Design it as reinforced concrete square tied column. Use concrete fc’= 5 ksi, steel fy = 60 ksi.
**Design it as steel column. Use K=1, Fy= 35 ksi.

Answer 1

Reinforced Concrete Column Design:

1. Determine the effective length factor (K) based on the column's end conditions. For a column fixed at both ends, K = 0.5.

K = 0.5

2. Determine the minimum required column dimension (b and h) based on the axial load and the allowable stress in the concrete and steel.

b = h = sqrt(550 kips / (0.4 * 5 ksi + 0.8 * 60 ksi * K * sqrt(fc'))) = 15.05 inches

Assuming a square column, use 15 inches for both b and h.

3. Calculate the area of steel required to resist the axial load and check for minimum steel requirements.

Assuming #8 bars spaced evenly on all sides of the column:

Ag = b * h = 15 inches * 15 inches = 225 sq. inches
Pn = 0.85 * fc' * Ag + As * fy = 0.85 * 5 ksi * 225 sq. inches + (#8 bars * 0.79 sq. inches/sq. ft. * 4 sides) * 60 ksi
Pn = 1275 kips + 151 kips = 1426 kips
As = (550 kips - 0.85 * fc' * Ag) / (fy * 0.79 sq. inches/sq. ft. * 4 sides) = 5.50 sq. inches

The minimum steel required is determined by As,min = 0.01 * Ag = 2.25 sq. inches. Therefore, the design is adequate.

4. Check for column stability and minimum reinforcement spacing requirements.

Since the column is tied, it should be sufficient to prevent buckling. The minimum spacing requirement is 1 inch for #8 bars.

Steel Column Design:

1. Determine the slenderness ratio (L/r) of the column based on its geometry.

Assuming a W10x15 shape with a length of 14 feet:

r = sqrt(I/A) = sqrt(2.63 in^4 / 2.53 in^2) = 1.63 inches
L = 14 feet = 168 inches
L/r = 103

2. Determine the allowable axial load based on the slenderness ratio and the column's yield strength.

Assuming K = 1:

Pu = (0.658^(Fy/ksi) / (L/r)^2) * Ag * Fy = (0.658^(35 ksi/ksi) / (103)^2) * 8.29 sq. inches * 35 ksi = 253 kips

3. Compare the allowable axial load to the applied load and check for buckling.

Pu > 550 kips, therefore the column will buckle and cannot be designed as a steel column. A larger column size or higher strength steel would be required.