How to Design Axially Loaded RC Short Column as per ACI 318-19? Example Included

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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 (P_u) on the column i.e. transfer loads from slabs and beams to the columns.

2. Assume a reinforcement ratio (p_g) 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.

p_g= A_st/A_g Equation 1

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

5. Plug (A_g) 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:

p_g: gross reinforcement ratio of column

A_st: longitudinal reinforcement ratio of column, mm²

A_g: gross area of column, mm²

P_u: 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

Number of bars	1	2	3	4	5	6	7	8	9	10	11	12
Bar Size, IS	-	-	-	-	-	-	-	-	-	-	-	-
10	71	142	213	284	355	426	497	568	639	710	781	852
13	129	258	387	516	645	774	903	1032	1161	1290	1419	1548
16	199	398	597	796	995	1194	1393	1592	1791	1990	2189	2388
19	284	568	852	1136	1420	1704	1988	2272	2556	2840	3124	3408
22	387	774	1161	1548	1935	2322	2709	3096	3483	3870	4257	4644
25	510	1020	1530	2040	2550	3060	3570	4080	4590	5100	5610	6120
29	645	1290	1935	2580	3225	3870	4515	5160	5805	6450	7095	7740
32	819	1638	2457	3276	4095	4914	5733	6552	7371	8190	9009	9828
36	1006	2012	3018	4024	5030	6036	7042	8048	9054	10060	11066	12072
43	1452	2904	4356	5808	7260	8712	10164	11616	13068	14520	15972	17424
57	2581	5162	7743	10324	12905	15486	18087	20648	23229	25810	28391	30972

Example:

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

Solution:

1- The load on the column is already computed, P_u=2600 KN

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

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

A_st=0.02*A_g

4. Estimate column dimension using Equation-2:

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

A_g= 157609.38 mm²

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

Compute a new gross area of column from rounded dimension:

A_g= 160000 mm²

5. Compute A_st using Equation-2:

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

A_st= 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:

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

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

p_g= A_st/A_g= 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 1.5d_b=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

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