Design of Columns | RCC & Prestressed Concrete - Civil Engineering (CE) PDF Download

Slenderness ratio (λ)
λ = effective length/least lateral dimension
If λ > 12 then the column is long.

Load carrying capacity for short column
P = σ_scA_sc + σ_ccA_c 
where, AC = Area of concrete, A_c = A_g - A_SC
σ_SC Stress in compression steel
σ_CC Stress in concrete
A_g Total gross cross-sectional area
A_SC Area of compression steel

Load carrying capacity for long column
P = C_r(σ_scA_sc + σ_ccA_c)
where, Cr = Reduction factor

where, l_eff = Effective length of the column
B = Least lateral dimension
i_min = Least radius of gyration and
where, l = Moment of inertia and A = Cross-sectional area

Effective length of column

Effective length of Compression Members

Column with helical reinforcement
Strength of the column is increased by 5%
P = 1.05(σ_scA_sc + σ_ccA_c) for short column
P = 1.05C_r(σ_scA_sc + σ_ccA_c) for long column

Longitudinal reinforcement

1. Minimum area of steel = 0.8% of the gross area of the column
2. Maximum area of steel
   (i) When bars are not lapped Amax = 6% of the gross area of the column
   (ii) When bars are lapped Amax = 4% of the gross area of the column

Minimum number of bars for reinforcement
For rectangular column  4
For circular column  6
Minimum diameter of bar = 12 mm
Maximum distance between longitudinal bar = 300 mm
Pedestal: It is a short length whose effective length is not more than 3 times of lest lateral dimension.

Transverse reinforcement (Ties)

where ϕ_min dia of the main longitudinal bar
φ = dia of the bar for transverse reinforcement
Pitch (p)

where, φ_min = minimum dia of main longitudinal bar

Helical reinforcement

1. Diameters of helical reinforcement is selected such that
   
2. Pitch of helical reinforcement: (p)
   
   where, d_C = Core diameter = d_g – 2 × clear cover to helical reinforcement
   A_G = Gross area = π/4(d_g)²
   d_g = Gross diameter
   V_h = Volume of helical reinforcement in a unit length of the column
   φ_h = Diameter of steel bar forming the helix
   
   d_h = centre to centre dia of the helix
   = d_g – 2 clear cover - φ_h
   φ_h = diameter of the steel bar forming the helix

Some others IS recommendations

1. Slenderness limit
   (i) Unsupported length between end restrains 60 times least lateral dimension.
   (ii) If in any given plane one end of the column is unrestrained than its unsupported length
2. All column should be designed for a minimum eccentricity of

Limit state method

• Slenderness ratio (λ) if

 λ<12 Short column

1. Eccentricity
   
   If e_min ≤0.05D then it is a short axially loaded column.
   where, P_u = axial load on the column.
2. Short axially loaded column with helical reinforcement
   P_u = 0.4 f_ckA_c + 0.67 f_yA_sc
3. Some others IS code Recommendations
   P_u = 1.05(0.4 f_ckA_c + 0.67 f_yA_sc)
   (i) Slenderness limit
   (a) Unsupported length between end restrains  60 times least lateral dimension
   (b) If in any given plane one end of the column is unrestrained than its unsupported length
   (ii) All column should be designed for a minimum eccentricity of

Concentrically Loaded Columns
Where e = 0, i.e., the column is truly axially loaded.
P_u = 0.45 f_ckA_c + 0.75 f_yA_sc
This formula is also used for member subjected to combined axial load and bi-axial bending and also used when e > 0.05 D.