14 ft. column is subjected to total compressive load of 550 kips.
Design it as reinforced concrete square tied column. Use concrete fc’= 5 ksi, steel fy = 60 ksi.
Solution:
Given:
Height of column (L) = 14 ft.
Total compressive load (P) = 550 kips
Concrete strength (fc') = 5 ksi
Steel strength (fy) = 60 ksi
Assumptions:
1. The column cross-section is square.
2. The steel reinforcement is placed in tied configuration.
3. The effective length of the column is equal to L.
Design Procedure:
Step 1: Determine the design axial load on the column.
The design axial load (Pn) is obtained by applying the appropriate load factor to the total compressive load. For this column, the load factor is taken as 0.65, which corresponds to a strength design method.
Pn = 0.65 x P
Pn = 0.65 x 550 kips
Pn = 357.5 kips
Step 2: Select the column dimensions.
Assuming a square column section, let the width and depth of the column be equal (b = d). From the ACI code, the minimum dimension of a tied column should be 12 inches. Therefore, the column dimensions are assumed as follows:
b = d = 18 inches
Step 3: Determine the effective length factor (K).
For a column with pinned ends, the effective length factor is given by:
K = (effective length)/(least lateral dimension)
For a square column, the least lateral dimension is equal to the width of the column.
K = L/b
K = 14 ft./(18 in./12 in./ft.)
K = 9.33
Step 4: Compute the slenderness ratio (λ).
The slenderness ratio is defined as the ratio of the effective length to the radius of gyration of the column cross-section.
λ = K x √(Ag/Ig)
where,
Ag = gross area of the column cross-section = b x d
Ig = moment of inertia of the column cross-section about its centroid
For a square cross-section,
Ag = b x d = 18 in. x 18 in. = 324 sq. in.
Ig = (b x d^3)/12 = (18 in. x 18 in.^3)/12 = 5832 in.^4
λ = K x √(Ag/Ig)
λ = 9.33 x √(324 sq. in./5832 in.^4)
λ = 0.82
Step 5: Check for slenderness.
The slenderness ratio should not exceed the maximum permissible value given by the ACI code. For a square tied column, the maximum slenderness ratio is 50.
λ = 0.82 < 50
Therefore, no further checks are required for slenderness.
Step 6: Determine the required longitudinal steel reinforcement.
The design axial load capacity of the column (Pn) can be expressed as:
Pn = φPc + Asfy
where,
φ = strength reduction factor = 0.65
Pc = compressive strength of the column = 0.85fc'Ag
As = area of longitudinal steel reinforcement
fy = yield strength of steel
Substituting the values,
Pn = φPc + Asfy
357.5 kips = 0.65 x 0.85 x 5 ksi x 324 sq. in. + As x 60 ksi
As = 2.694 sq. in.
Assuming four No. 8 longitudinal bars, the area of steel provided is:
As provided = 4 x (π/4 x 0.79 in.^2) = 7.854 sq. in.
Therefore, the required longitudinal steel is satisfied.
Step 7: Check for minimum reinforcement.
The area of longitudinal steel reinforcement should not be less than the minimum required by the ACI code. For columns, the minimum reinforcement is 1% of the gross area for tied columns.
Minimum area of steel = 1% x Ag = 0.01 x 324 sq. in. = 3.24 sq. in.
The area of steel provided (7.854 sq. in.) is greater than the minimum required (3.24 sq. in.). Therefore, the minimum reinforcement requirement is satisfied.
Step 8: Check for maximum reinforcement.
The area of longitudinal steel reinforcement should not exceed the maximum percentage specified by the ACI code. For tied columns, the maximum reinforcement is 6% of the gross area.
Maximum area of steel = 6% x Ag = 0.06 x 324 sq. in. = 19.44 sq. in.
The area of steel provided (7.854 sq. in.) is less than the maximum permitted (19.44 sq. in.). Therefore, the maximum reinforcement requirement is satisfied.
Step 9: Check for maximum bar spacing.
The maximum spacing between adjacent longitudinal bars should not exceed the smaller of 3 inches and one-fifth of the least dimension of the column.
Maximum spacing = min(3 in., b/5) = min(3 in., 18 in./5) = 3 in.
The actual spacing of No. 8 bars is less than the maximum permitted (3 in.). Therefore, the maximum bar spacing requirement is satisfied.
Step 10: Check for development length.
The development length of the longitudinal steel reinforcement should be checked to ensure adequate bond strength between steel and concrete. For this column, the longitudinal bars are assumed to be fully developed, and hence, no checks are required.
Conclusion:
The design of the 14 ft. square tied column with a compressive load of 550 kips is complete, and the cross-section dimensions, steel reinforcement configuration, and the required amount of steel reinforcement have been determined. The column satisfies all ACI code requirements for minimum reinforcement, maximum reinforcement, maximum bar spacing, and development length.