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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 0.85f_{c}' first. Then, concrete continues to yield until steel reaches its yield strength, f_{y}, when the column fails. The strength contributed by concrete is 0.85f’_{c}(A_{g}A_{st}), where f_{c}' is compressive strength of concreter, A_{g} is gross area of column, A_{st} is areas of reinforcing steel. The strength provided by reinforcing steel is A_{st}f_{y}. Therefore, the nominal strength of a reinforced concrete column, is
P_{n} = 0.85f’_{c}(A_{g}A_{st})+A_{st}f_{y }[1]
For design purpose, ACI specify column strength as follows
For a spiral column, the design strength is
fP_{n} = 0.85f[0.85f’_{c}(A_{g}A_{st})+A_{st}f_{y}] [2]
For a regular tie column, the design strength is
fP_{n} = 0.80f[0.85f’_{c}(A_{g}A_{st})+A_{st}f_{y}] [3]
where f is strength reduction factor. For spiral column f = 0.75 (ACI 31899), f = 0.7 (ACI 31802, 05). For spiral column f = 0.7 (ACI 31899), f = 0.65 (ACI 31802, 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.
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.
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
moment Failure occurs when concrete strain reaches 0.003

Small axial load with large
moment Failure occurs when steel strain reaches 0.005

Balanced condition Failure occurs when concrete strain reaches 0.003 and steel strain reaches 0.005 at the same time.

Interaction diagram for P_{n} and
M_{n}

Design aid:
The interaction diagrams of concrete column with strength reduction factor is available on ACI design handbook. The vertical axis is fP_{n} /A_{g} and the horizontal axis is fM_{n} /A_{g}h, 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 "SP17: Design Handbook: Beams, OneWay Slabs, Brackets, Footings, Pile Caps, Columns, TwoWay Slabs, and Seismic Design in accordance with the Strength Design Method of 31895"
Design requirements:
1. Design strength:fP_{n} ³ P_{u }and_{ }fM_{n} ³ M_{u}
2. Minimum eccentricity, e = M_{u}/P_{u} ³ 0.1.
Design procedure:
1. Calculate factored axial load, P_{u} and factored moment, M_{u}.
2. Select a trial column column with b and column depth, h in the direction of moment.
3. Calculate gross area, A_{g} and ratio, g = distance between rebar/h.
4. Calculate ratio, P_{u}/Ag and M_{u}/A_{g}h.
5. Select reinforcement ratio,r, from PCA design chart based on concrete strength, f_{c}', steel yield strength, f_{y}, and the ratio, g.
6. Calculate area of column reinforcement, A_{s}, and select rebar number and size.
7. Design column ties.
Example: A 12"x12" interior reinforced concreter short column is supporting a factored axial load of 350 kips and a factored moment of 35 kipft.
Desogn data:
Factored axial: P_{u} = 350 kips
Factored moment: M_{u} = 35 ftkips
Compressive strength of concrete: f_{c}'= 4000 psi
Yield strength of steel: f_{y} = 60 ksi
Requirement: design column reinforcements.
Column size: b = 12 in, h = 12 in
Gross area, Ag = 144 in^{2}.
Concrete cover: d_{c} = 1.5 in
Assume #4 ties, d_{t} = 0.5 in and #6 bars, d_{s} = 0.75 in
Calculate g = (h  2 d_{c}2 d_{t } d_{s})/h = 0.6
Calculate,
P_{u}/A_{g }= 300/144 = 2.43 ksi,
M_{u}/A_{g}h = 35/[(144)(12)] = 0.243
From ACI design handbook, reinforcement ratio, r = 0.018
Area of reinforcement, A_{s} = (0.0018)(144) = 2.6 in^{2}.
Use 6#6, area of reinforcement, A_{s} = (6)(0.44) = 2.64 in^{2}.
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.