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# Design of short concrete columns

## Strength of column subjected to axial load only

Ideally, if a column is subjected the pure axial load, concrete and reinforcing steel will have the same amount of shortening.  Concrete reaches its maximum strength at 0.85fc' first.  Then, concrete continues to yield until steel reaches its yield strength, fy, when the column fails. The strength contributed by concrete is 0.85f’c(Ag-Ast), where fc' is compressive strength of concreter, Ag is gross area of column, Ast is areas of reinforcing steel.  The strength provided by reinforcing steel is Astfy.  Therefore, the nominal strength of a reinforced concrete column,  is

Pn = 0.85f’c(Ag-Ast)+Astfy                                                                                            

For design purpose, ACI specify column strength as follows

For a spiral column, the design strength is

fPn = 0.85f[0.85f’c(Ag-Ast)+Astfy]                                                             

For a regular tie column, the design strength is

fPn = 0.80f[0.85f’c(Ag-Ast)+Astfy]                                                             

where f is strength reduction factor.  For spiral column f = 0.75 (ACI 318-99), f = 0.7 (ACI 318-02, 05). For spiral column f = 0.7 (ACI 318-99), f = 0.65 (ACI 318-02, 05)

The factors 0.85f and 0.8f are considering the effect of confinement of column ties and strength reduction due to failure mode.  Nevertheless, column loads are never purely axial.  There is always bending along with axial load.

## Strength of column subjected to axial load and bending

Consider a column subjected to axial load, P and bending moment, M.  Axial load, P produces an uniform stress distribution across the section while bending moment produces tensile stress on one side and compressive stress on the other.

### Strain and stress distributions of short concrete column at failure and interactive diagram

Assumption:

1. A plan section remains a plan at failure.  Strain distributes linearly across section

2. Concrete fails at a strain of 0.003.

3. Reinforcing steel fails at a strain of 0.005.

 Axial load only (moment = 0) Failure occurs when concrete strain reaches 0.003 Moment only (Axial load = 0) Failure occurs when steel strain reaches 0.005 first. Large axial load with small momentFailure occurs when concrete strain reaches 0.003 Small axial load with large momentFailure occurs when steel strain reaches 0.005 Balanced conditionFailure occurs when concrete strain reaches 0.003 and steel strain reaches 0.005 at the same time. Interaction diagram for Pn and Mn Design aid:

The interaction diagrams of concrete column with strength reduction factor is available on ACI design handbook.  The vertical axis is fPn /Ag and the horizontal axis is fMn /Agh, where h is the dimension of column in the direction of moment.  The chart is arranged based on the ratio, g which is the ratio of the distance between center of longitudinal reinforcements to h. Column strength interaction diagram for rectangular column with g =0.6 (Coursey of American Concrete Institute)

ACI  design handbook can be purchase from ACI book store.  The title is "SP-17: Design Handbook: Beams, One-Way Slabs, Brackets, Footings, Pile Caps, Columns, Two-Way Slabs, and Seismic Design in accordance with the Strength Design Method of 318-95"

### Design of short concrete column

Design requirements:

1. Design strength:fPn ³ Pu   and fMn ³ Mu

2. Minimum eccentricity, e = Mu/Pu ³ 0.1.

Design procedure:

1. Calculate factored axial load, Pu and factored moment, Mu.

2. Select a trial column column with b and column depth, h in the direction of moment.

3. Calculate gross area, Ag and ratio, g = distance between rebar/h.

4. Calculate ratio, Pu/Ag and Mu/Agh.

5. Select reinforcement ratio,r, from PCA design chart based on concrete strength, fc', steel yield strength, fy, and the ratio, g.

6. Calculate area of column reinforcement, As, and select rebar number and size.

7. Design column ties.

### Design example:

Example: A 12"x12" interior reinforced concreter short column is supporting a factored axial load of 350 kips and a factored moment of 35 kip-ft.

Desogn data:

Factored axial: Pu = 350 kips

Factored moment: Mu = 35 ft-kips

Compressive strength of concrete: fc'= 4000 psi

Yield strength of steel: fy = 60 ksi

Requirement: design column reinforcements.

Column size: b = 12 in, h = 12 in

Gross area, Ag = 144 in2.

Concrete cover: dc = 1.5 in

Assume #4 ties, dt = 0.5 in and #6 bars, ds = 0.75 in

Calculate  g = (h - 2 dc-2 dt - ds)/h = 0.6

Calculate,

Pu/Ag = 300/144 = 2.43 ksi,

Mu/Agh = 35/[(144)(12)] = 0.243

From ACI design handbook, reinforcement ratio, r = 0.018

Area of reinforcement, As = (0.0018)(144) = 2.6 in2.

Use 6#6, area of reinforcement,  As = (6)(0.44) = 2.64 in2.

Check Bar spacing, s = (h - 2 dc - 2 dt - ds)/2 = 3.625 in  (O.K.)

Calculate minimum spacing of column ties:

48 times of tie bar diameter = (48)(0.5) = 24 in

16 times of longitudinal bar diameter = (16)(0.75) = 12 in

Minimum diameter of column = 12 in

Use #4 ties at 12 inches on center.