# How to Design Axially Loaded RC Short Column as per ACI 318-19? Example Included Reading time: 1 minute

The design of an axially loaded reinforced concrete (RC) short column is quite simple and straightforward. It is governed by the strength of materials and the cross-section of the member. If vertical axial loads act on the center of gravity of the column's cross-section, it is termed as an axially loaded column.

Axially loaded columns rarely exist in practice because of factors like inaccuracy in the layout of the column, unsymmetrical loading due to differences in the thickness of slabs in adjacent spans, and imperfections in the alignment that can easily shift the point of vertical load action away from the center of the column cross-section and, consequently, create eccentricities.

The axially loaded columns are those with relatively small eccentricity, e, of about 0.1 h or less, where h is the total depth of the column and e is the eccentric distance from the center of the column. The interior column of multistory buildings with symmetrical loads from floor slabs from all sides is an example of an axially loaded column. ACI 318-19 equips designers with several specifications and requirements to produce an economical and safe design.

## Procedure for Design of Axially Loaded RC Short Column

1- Calculate the total axial load (Pu) on the column i.e. transfer loads from slabs and beams to the columns.

2. Assume a reinforcement ratio (pg) that should be equal to or greater than the minimum reinforcement ratio (0.01) and equal to or smaller than the maximum reinforcement ratio (0.08) provided by ACI 318-19.

3. Express steel area in terms of reinforcement ratio times the gross area of the column cross-section.

pg= Ast/Ag Equation 1

4. Select column dimensions using Equation-1 and then round the dimensions.

5. Plug (Ag) from Step-4 into Equation-4 and then calculate longitudinal reinforcement area.

6. Assume a bar size from Table-1 and then determine the number of bars for the column. The minimum number of bars for a square column is four and for a circular column is six.

7. Calculate the column reinforcement ratio and check whether it is within the minimum and maximum reinforcement ratio or not.

8. Design ties; determine the size of the ties and estimate the spacing.

9. Check for code requirements; clear spacing between longitudinal bars, a minimum number of bars, minimum tie diameter, and arrangement of ties.

Where:

pg: gross reinforcement ratio of column

Ast: longitudinal reinforcement ratio of column, mm2

Ag: gross area of column, mm2

Pu: ultimate axial load on the column, KN

phi: strength reduction factor, which 0.65 for tied column and 0.75 for spiral column.

fc': concrete compressive strength, MPa

fy: yield strength of steel, MPa

Table-1: Area of Groups of Bars, mm2

## Example:

Design an axially loaded short square tied column to support a maximum factored load of  Pu=2600 KN. Material strength: fc'= 28 MPa and fy= 420 MPa.

### Solution:

2. Assume pg= 0.02, which is between 0.01 and 0.02.

3. Express Ast in terms of gross area of column using Equation-1:

Ast=0.02*Ag

4. Estimate column dimension using Equation-2:

2600*103=0.85*0.65*[0.85*28*(Ag-0.02*Ag)+0.02*Ag*420

Ag= 157609.38 mm2

Ag=h2 , h=(157609.38)1/2= 397 mm= 400 mm.

Compute a new gross area of column from rounded dimension:

Ag= 160000 mm2

5. Compute Ast using Equation-2:

2600*103=0.85*0.65*[0.85*28*(160000-Ast)+Ast*420

Ast= 2838.09 mm2

6. Assume a bar size, select No. 32 from Table-1

Take bar of No. 32 and go to the right side to select the bars area that is close to the estimate reinforcement area, after that go to the top of the column to determine the number of bars:

Ast,provided= 3276 mm2, and number of bars are 4

7. Compute reinforcement ratio using bars area from Step-6.

pg= Ast/Ag= 3276/160000= 0.0204 (OK), since it is greater than minimum reinforcement ratio and less than maximum reinforcement ratio

8. Design Ties:

Assume Tie bar of No. 10

The spacing between ties is the smallest of the following:

48* tie diameter= 48*10= 480 mm

16* longitudinal bar diameter= 16*32= 512 mm

least dimension of the column= 400 mm

So, the spacing between ties is 400 mm

9. Checks:

a- clear spacing between longitudinal bar= (400-2*40-2*10-2*32)=236 mm< 40 mm or 1.5db=32*1.5= 48 mm OK.

b- The minimum number of bars for the square columns is 4, and 4 bars have been assigned for the column under consideration, OK.

c- Minimum tie diameter 10 for 32 longitudinal bar diameters, ok

d- Tie arrangement, sice all longitudinal bars are placed in a corner, check for tie arrangement is not needed.

## FAQs

What is axially loaded reinforced concrete short columns?

If vertical axial loads act on the center of gravity of the column's cross-section, it is termed an axially loaded column. The axially loaded columns are those with relatively small eccentricity, e, of about 0.1 h or less, where h is the total depth of the column and e is the eccentric distance from the center of the column.

What are the factors that govern the design of RCC short columns?

The design of axially loaded short column is governed by the strength of materials and the cross-section of the member.

What are the minimum number of longitudinal bars for square and circular columns?

ACI 318-19 specifies a minimum of four bars for square columns and six bars for circular columns.

What are column stirrups?

Stirrups are closed loop of reinforcement bars that prevent buckling of longitudinal bars in column and hold them at their position.