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Quick Design: CSA A23.3:24 Column

Overview

This calculator is for the design of reinforced non-prestressed concrete columns as per CSA A23.3:24. The following calculations are performed:

  • Column slenderness is determined in each direction depending on user inputs.
  • Design bending moments are calculated based on slenderness.
  • Moment capacities are derived from interaction diagrams, which are drawn based on the section properties of the column.
  • Shear capacities of the section are determined using the inclined strut truss model.
  • Detailing checks are performed on the section.

Assumptions

  • When calculating shear, the angle of the shear reinforcement is assumed to be perpendicular to the longitudinal reinforcement and the longitudinal reinforcement is parallel with all faces of the concrete.
  • Aggregate size is assumed to be 20mm for calculations.
  • λ = 1 as normal density concrete is assumed (8.6.5).
  • No transverse load is assumed to act between the column supports.
  • bw is taken as the diameter when calculating shear capacity for circular sections as per 11.2.10.3.

Section Analysis

Interaction Diagrams

The bending capacity of the concrete column is derived from interaction diagrams drawn for each axis. The interaction line is plotted by calculating values of M and P at varying depths of the neutral axis. This is done by determining the strains in the concrete and reinforcement bars for a given depth of the neutral axis. These strains can be converted into forces, which can then be used to calculate the maximum axial load and bending moment that the column can resist for that neutral axis depth.

Notes
  • Moments are taken about the plastic centroid of the section.
  • The resultant concrete force is assumed to act at the centroid of the rectangular stress block.
    • For rectangular sections, this is taken as half the height of the block.
    • For circular sections, this is calculated by finding the centroid of the segment in compression, where the height of the segment is equal to β1.c.

Column Design

Axial Capacity

The greatest compression strength of a column is defined as it’s factored axial resistance with zero eccentricity (Pro). This is calculated assuming the section’s reinforcement has reached it’s full yield stress (Φsfy) in compression and the concrete is under a uniform compressive stress of Φcα1f’c (Equation 10.11). The maximum axial load resistance (Pr,max) comprises a reduced value of Pro, where the reduction relates to the confinement of the column’s core as per equations 10.8 and 10.9.
The tensile capacity of the section is calculated as the strength of the reinforcement at yield and ignores any contribution from the concrete.

Core Confinement

The confinement of the column core, as referenced in section 10.10.4, is calculated by comparing the ratio of the confining spiral reinforcement to the limiting ratio, ρs, as per section 10.9.4. Note that only spiral reinforcement is considered to confine the column’s core and the maximum confinement pitch is 100mm (cl. 7.6.4.3).

Equation 10.7
ρs=0.5(AgAc−1)1.4(f′cfyh) \rho _s = 0.5 {\left(\frac{A_g}{A_c} – 1 \right)}^{1.4} \left(\frac{f’_c}{f_{yh}}\right)
Equation 10.7
ρactual=VspiralVcore≥ρs \rho _{actual} = \frac{V_{spiral}}{V_{core}} \geq \rho _s

Design Moments

Sway and Non-Sway

The calculation of column end moments will depend on whether it acts as a sway or non-sway element, which can be determined as per clause 10.14.4. For non-sway columns, the end moments can be calculated directly using first-order linear analysis and then converted to the design end moments M1 and M2.

If the column is acting as a sway element, then the end moments must be calculated using second-order analysis as stated in clause 10.16.3.1. The design end moments, M1 and M2, can then be derived using section 10.16.2. The calculation of magnified moments to this clause is beyond the scope of this calculator, however users may specify the directions in which the column acts as a sway element and provide magnified moments directly.

Following the calculation of M1 and M2 moments, the column can be checked for slenderness and designed for Mc, regardless of its sway status (10.16.4).

M1=max[M2,minmin[MtopMbottom]] M_1 = \max \left[ M_{2,min} \;\; \min \left[ M_{top} \;\; M_{bottom} \right] \right]
M2=max[M2,minmax[MtopMbottom]] M_2 = \max \left[ M_{2,min} \;\; \max \left[ M_{top} \;\; M_{bottom} \right] \right]
M1=max[M2,minM1ns+δsM1s] M_1 = \max \left[ M_{2,min} \;\; M_{1ns} \;+\; \delta _{s}M_{1s}\right]
M2=max[M2,minM2ns+δsM2s] M_2 = \max \left[ M_{2,min} \;\; M_{2ns} \;+\; \delta _{s}M_{2s}\right]
Notes
  • M1 / M2 is taken as negative when the column is bent in single curvature. In this calculator, this is when the moments have the same sign.
  • When M2,min is greater than M2, the column is assumed to bent in single curvature.
  • M2,min is calculated as Pf(15 + 0.03h).

Calculate Slenderness

A column is deemed to not be slender if it is within the limits outlined below. A column’s slenderness ratio is calculated as klu / r, where radius of gyration is calculated as (I / A)0.5.

Equation 10.16
klur≤min[4015+10(M1/M2)Pf(f′cAg)] \frac{kl_u}{r} \leq \min \left[ 40 \;\; \frac{15 + 10(M_1/M_2)}{\sqrt{P_f(f’_cA_g)}} \right]
Equation 10.26
lur≤35Pf(f′cAg) \frac{l_u}{r} \leq \frac{35}{\sqrt{P_f(f’_cA_g)}}

Magnified Moment For Curvature, Mc

If a column is found to be slender, then the design moment needs to be factored to account for curvature. Note that in equation 10.19 and 10.20, βd is the ratio of the sustained load, which must be provided by the user in this calculation.

Equation 10.17
Mc=CmM21−PfϕmPc≥M2 M_c = \frac{C_mM_2}{1 – \frac{P_f}{ \phi _mP_c}} \geq M_2
Equation 10.17
ϕm=0.75 \phi _m = 0.75
Equation 10.18
Pc=π2(EI)eff(klu)2 P_c = \frac{\pi ^2(EI)_{eff}}{(kl_u)^2}
Equation 10.19 & 10.20
(EI)eff=max[0.2EcIg+EsIst1+βd0.4EcIg1+βd] (EI)_{eff} = \max \left[ \frac{0.2E_cI_g + E_sI_{st}}{1 + \beta_d} \;\; \frac{0.4E_cI_g}{1 + \beta_d} \right]
Equation 10.21
Cm=0.6−0.4M1M2 C_m = 0.6 – 0.4\frac{M_1}{M_2}

Bending Capacity

The bending moment capacity is calculated in each axis by plotting the coordinates of the design forces (axial load and bending moment) on the interaction diagrams. A line is drawn through the origin, which intersects the design coordinate. The moment capacity (Mr,axis) and axial capacity (Pr,axis) are taken at the point where this line intersects the interaction line. The Bresler equation below considers the axial capacity of each axis (Prz and Pry) with the maximum axial capacity of the section (Pr0) to conclude a resistance for the column (Pr). This value is compared to the factored applied axial load (Pf) to determine a biaxial utilization.

Equation 10.21
1Pr=1Prz+1Pry−1Pr0 \frac{1}{P_r} = \frac{1}{P_{rz}} + \frac{1}{P_{ry}} – \frac{1}{P_{r0}}

Shear Capacity

The total shear capacity of the column (Vr) is calculated as the capacity of the concrete (Vc) combined with the capacity of the shear reinforcement (Vs). This value is limited to the crushing strength of the shear web (Vr,max).

Equation 11.4 and 11.5
Vr=min[Vc+VsVr,max] V_r = \min \left[ V_{c} + V_{s} \;\; V_{r,max} \right]
Equation 11.5
Vr,max=0.25ϕcf′cbwdv V_{r,max} = 0.25\phi _cf’_cb_wd_v
Equation 11.6
Vc=ϕcλβf′cbwdv V_{c} = \phi _c\lambda\beta\sqrt{f’_c}b_wd_v
Equation 11.7
Vs=ϕsAvfytdvcotθs V_{s} = \frac{\phi _sA_vf_{yt}d_v\cot\theta}{s}
Equation 11.11
β=0.41+1500ϵx×13001000+sze≥0.05 \beta = \frac{0.4}{1 + 1500\epsilon_x} \times \frac{1300}{1000 + s_{ze}} \geq0.05
Equation 11.12
θ=29+7000ϵx \theta = 29 + 7000\epsilon_x
Equation 11.10
sze=35sz15+ag⇒szforag=20 s_{ze} = \frac{35s_z}{15 + a_g} \Rightarrow s_z \; for \; a_g=20
Equation 11.13
εx=|Mfdv|+|Vf|+0.5Nf2(EsAs)≤3.0×10−3 \varepsilon_x = \dfrac{ \left|\dfrac{M_f}{d_v}\right| + \left|V_f\right|+ 0.5N_f }{ 2\left(E_s A_{s}\right) } \;\leq\; 3.0 \times 10^{-3}
Mf=max[MdesignVf.dv] M_f = \max \left[ M_{design} \;\; V_f.d_v\right]
Notes
  • Reinforcement bars are assumed to be fully developed in this calculator.
  • Mid-height strain calculation excludes the contribution of steel on the mid-height line.
  • The area of tension steel is calculated as the area of steel in the tension side of the column,
    with the exception of the steel described in the point above. The tension side is the area on the tension side of the mid-height line.
  • Similarly, the area of flexural concrete on the tension side of the member (Act) is calculated as the area on the tension side of the mid-height line (A/2).
  • The value of εx is doubled when flexural compression flange is in tension. Flange force calculated as follows:
    • −Mfdv+0.5Nf+0.5Vfcotθ \dfrac{-M_f}{d_v} + 0.5N_f + 0.5V_f \cot\theta
  • If εx is found to be negative (in compression), then the value should be further modified using the equation below:
εx=−2.0×10−3≤|Mfdv|+|Vf|+0.5Nf2(EsAs+EcAct)≤3.0×10−3 \varepsilon_x = -2.0 \times 10^{-3} \;\leq\; \dfrac{ \left|\dfrac{M_f}{d_v}\right| + \left|V_f\right|+ 0.5N_f }{ 2\left(E_s A_{s} + E_c A_{ct}\right) } \;\leq\; 3.0 \times 10^{-3}

Change Log

  1. Initial deployment July 2026.

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Updated November 2, 2026

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