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The design of axially loaded circular columns is mainly governed by the column’s cross-section and its material properties. This is because the load acts at the center of the column, and the entire cross-sectional area of the column is in compression.
An entire column subjected to purely compression rarely exists in practice as the perfect vertical alignment of a column is difficult to achieve. As a result, the eccentricity of loads relative to the center of the column is inevitable.
According to ACI 318-19, a column is considered to be axially loaded if the eccentricity is not greater than 1% of the column's total depth. The nominal capacity of such a column is equal to the internal concrete compression force and internal steel force.
Several specifications and practical considerations are presented in ACI 318-19 that help designers develop adequate, safe, and practical column designs.
Nominal Load Capacity of RC Circular Short Column
The nominal load capacity of the reinforced concrete column can be expressed as follows:
Where Ac is the gross cross-sectional area of the column, and Ast is the tensile steel area.
Therefore, Equation-1 can be modified as follows:
The design strength of axially loaded columns can be computed by multiplying Equation-2 by a specific strength reduction factor provided by ACI 318-19. The strength reduction factor for spirally reinforced concrete columns is 0.75.
The ACI code has considered an upper limit on column strength to allow for accidental eccentricities of loading not accounted for in the analysis. This upper limit is taken as 0.85 times the design strength. Therefore, Equation-2 can be rewritten for a spirally reinforced concrete column as follows:
Pu: ultimate load on the column, KN
Ø: strength reduction factor, which is 0.75
fc’: concrete compressive strength, MPa
Ag: gross area of column cross-section, mm2
Ast: reinforcement area, mm2
fy: yield strength of steel, MPa
- Longitudinal reinforcement is the main bar in the reinforced concrete column.
- The minimum concrete cover over longitudinal reinforcement in columns is 40 mm.
- The minimum number of longitudinal bars should be six for spiral reinforcement as per ACI 318-19, Section 10.7.3.
- The ratio of longitudinal steel area to concrete cross-section area ranges from 0.01 to 0.08.
- Spiral reinforcements are closely-spaced continuous spirals that hold the longitudinal bars in the forms while the concrete is being placed.
- The spiral reinforcement prevents the sudden crushing of concrete and the buckling of longitudinal steel bars. It has the advantage of producing a tough column that undergoes gradual and ductile failure.
- The minimum diameter of spirals is 10 mm, and their clear spacing should not be more than 75 mm nor less than 25 mm, according to the ACI Code 7.10.4.
- 1.5 extra turns of the spiral bar shall be provided as the anchorage of spiral reinforcement.
- The minimum volumetric spiral reinforcement ratio is defined as the ratio of the volume of spiral steel to the volume of core concrete, and it is calculated according to Equation-4.
- The minimum spiral ratio required by the ACI Code is meant to provide an additional compressive capacity to compensate for the spalling of the column shell.
Ag: gross area of the section
Ach: area of the core of the spirally reinforced column measured to the outside diameter of spiral
ρs: yield strength of spiral reinforcement
The spiral may be selected with an expression in which (ρs) is written in terms of the volume of the steel in one loop:
Dch: diameter of the core out to out of the spiral
as: cross-sectional area of the spiral bar
ds: diameter of the spiral bar
See Figure-1 for the detail of the parameters of Equation-5
Design Procedure for Axially Loaded Circular RC Column
- Assume a reinforcement ratio between the minimum value of (0.01) and the maximum value of 0.08, desirably not greater than 0.04.
- Express steel area in terms of reinforcement ratio times the gross area of the column cross-section using the expression (ρg=As/As).
- Calculate the ultimate or applied load on the column; use a suitable load combination.
- Use Equation-3 to calculate the gross area of the column cross-section. After that, estimate the diameter of the column and round them.
- Compute column gross cross-sectional area from the rounded dimension value of the column.
- Estimate the longitudinal reinforcement ratio by using Equation-3.
- Assume the bar's size and determine the number of bars by dividing the estimated steel area by the column's gross area.
- Calculate the column reinforcement ratio and check whether it is within the minimum and maximum reinforcement ratio or not.
Design spiral reinforcement
- Assume the size of the spiral bar.
- Calculate the minimum volumetric spiral reinforcement ratio by using Equation-4.
- Plug the volumetric reinforcement ratio from Step-4 to calculate the spacing of spirals by using Equation 5.
- Check for ACI Code requirements which include clear spacing between longitudinal bars, the minimum number of bars, minimum spiral diameter, and clear spacing for one loop.
Design an axially loaded short rounded spiral column to support a dead load of 850 KN and a live load of 1600 KN. Material strength: fc’= 35 MPa, and fy= 420 MPa.
Estimate Column dimension and required steel area
Step-1: Assume reinforcement ratio:
Step-2: Express reinforcement area in terms of assumed reinforcement ratio (Step 1) and gross cross-sectional area:
Step-3: Calculate ultimate load using most critical load combinations:
Pu =1.2DL+1.6LL= 1.2*850+1.6*1600= 3580 KN
Step-4: Select column dimensions using Equation 3:
3580*103= 0.85*0.75[0.85*35(Ag-0.02Ag )+(0.02Ag*420]=149532.31 mm2
Ag=π/4 D2 → D =√((4*149532.31)/π) =436 mm≈450 mm
Step-5: Compute Ag from new D= 450 mm
Ag=π/4*4502= 159043.12 mm2
Step-6: Estimate area and number of longitudinal steel bars:
3580*103= 0.85*0.75[0.85*28(159043.12-Ast )+Ast*420]= 2265.61 mm2
Step-7: Estimate number of longitudinal bars, Assume ∅22:
No. of bars = 2265.61/(π/4(22)2)= 5.96 ≈ 6
As,provided = π/4(22)2*6 = 2280.79 mm2
Step-8: Estimate and check column reinforcement ratio:
ρg= Ast/Ag =(2280.79)/159043.12 = 0.0201, since 0.01<ρg= 0.01434<0.08, ok
Design Spiral Reinforcement
Step-1: Assume ∅10.
Step-2: Calculate minimum volumetric spiral reinforcement:
Dch= D-2*concrete cover = 450-2*40=370 mm
Ach= π/4(370)2 = 107521.008 mm2
ρs= 0.45(159043.12/107521.008-1)*(35/420) = 0.017969
Step-3: Calculate spacing from one loop to another, s:
as= π/4*102 = 78.5398 mm2
0.017969 = (4*78.5398*(370-10))/(s*3702 )→ s = 45.97 ≈ 40 mm
Step-4: Check for ACI Code requirements:
- Diameter of centroidal circle of bars =450-2*40-10*2-22 = 328 mm,
- Clear spacing =(π*328-6*22)/6 =149.74 mm
- No. of bars, 6 =6 for column enclosed by spirals, ok
- The minimum spiral diameter is ∅10, ok
- Clear spacing for one loop = s-ds =40-10 =30mm, 25mm<30mm<40 mm, Ok
A reinforced concrete column is considered to be axially loaded if the point of load action coincides with the center of the column cross-section. A column is said to be axially loaded if the eccentricity of the load from the center of the column cross-section is not greater than 0.1 times the maximum depth of the column.
Eccentricity is the distance between the center of the column cross-section and the action point of the applied load.
Spirally reinforced concrete columns are the ones in which longitudinal bars are arranged in a circular manner supported by closely spaced circular spirals.
Spiral reinforcement holds the longitudinal bars in the forms while the concrete is being placed, and prevents the sudden crushing of concrete and the buckling of longitudinal steel bars.
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