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Concrete Column Designer to AS3600's banner

Concrete Column Designer to AS3600

Verified by the CalcTree engineering team on August 6, 2024

This calculator allows the user to assess the structural integrity of concrete columns to ensure compliance with the Australian Standard AS 3600. The calculation will identify the design capacities of concrete columns to meet axial, flexural and shear design requirements to Ultimate Limit State (ULS) methods.
All calculations are performed in accordance with AS3600:2018.

Results Summary

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📃 List of symbols used in this calculator


Ag=bd=gross cross-sectional area(mm2)\footnotesize{A_g = bd = \text{gross cross-sectional area}\hspace{0.1cm}(mm^2)}

Ast=area of tensile longitudinal reinforcement(mm2)\small{A_{st} = \footnotesize\text{area of tensile longitudinal reinforcement}\hspace{0.1cm}(mm^2)}

Asc=area of compressive longitudinal reinforcement(mm2)\footnotesize{A_{sc} = \text{area of compressive longitudinal reinforcement}\hspace{0.1cm}(mm^2)}

Ast=area of transverse/shear reinforcement(mm2)\small{A_{st} = \footnotesize\text{area of transverse/shear reinforcement}\hspace{0.1cm}(mm^2)}

Es=Young’s modulus of concrete(MPa)\footnotesize{E_s = \text{Young's modulus of concrete}\hspace{0.1cm}(MPa)}

Ec=Young’s modulus of steel(MPa)\footnotesize{E_c = \text{Young's modulus of steel}\hspace{0.1cm}(MPa)}

fc′=characteristic strength of concrete(MPa)\footnotesize{f'_c = \text{characteristic strength of concrete}\hspace{0.1cm}(MPa)}

fsy=yield strength of steel(MPa)\footnotesize{f_{sy} = \text{yield strength of steel}\hspace{0.1cm}(MPa)}

D=depth of the cross-section(mm)\footnotesize{D = \text{depth of the cross-section}\hspace{0.1cm}(mm)}

Cc=compressive force in the concrete section(kN)\footnotesize{C_c = \text{compressive force in the concrete section}\hspace{0.1cm}(kN)}

Cs=compressive force in the compression reinforcement(kN)\footnotesize{C_s = \text{compressive force in the compression reinforcement}\hspace{0.1cm}(kN)}

Ts=tensile force in the tensile reinforcement(kN)\footnotesize{T_s = \text{tensile force in the tensile reinforcement}\hspace{0.1cm}(kN)}

dsc=distance from the extreme compressive fibre to the centroid of compression reinforcement(mm)\footnotesize{d_{sc} = \text{distance from the extreme compressive fibre to the centroid of compression reinforcement}\hspace{0.1cm}(mm)}

dst=d=distance from the extreme compressive fibre to the centroid of tension reinforcement(mm)\footnotesize{d_{st} = d= \text{distance from the extreme compressive fibre to the centroid of tension reinforcement}\hspace{0.1cm}(mm)}

kud=distance from the extreme compressive fibre to the neutral axis(mm)\footnotesize{k_ud = \text{distance from the extreme compressive fibre to the neutral axis}\hspace{0.1cm}(mm)}

α1=1.0−0.003fc′0.72≤α1≤0.85\footnotesize{\alpha_1 = 1.0-0.003f'_c \hspace{1cm}0.72\le \alpha_1 \le 0.85}

α2=0.85−0.0015fc′α2≥0.67\footnotesize{\alpha_2 = 0.85-0.0015f'_c \hspace{1cm}\alpha_2 \ge 0.67}

γ=0.97−0.0027fc′α2≥0.67\footnotesize{\gamma = 0.97 - 0.0027f'_c \hspace{1cm}\alpha_2 \ge 0.67}


Calculation



Technical notes

  1. The calculator does not calculate second order effects of slender columns, it assumes the input design bending moment (M*) takes into account any second order effects.
  1. Currently you can only input rectangular sections in the calculator.

Inputs

Material Properties



γc
:24 kN / m3​



f'c
:40MPa​



Ec
:32.8 GPa​



Es
:200 GPa​



fsy
:500 MPa​



Loads



N*
:1 MN​



V*
:50 kN​



M*
:150 kN m​



Section and Reinforcement Geometry



Lu
:3 m​



D
:425 mm​



b
:300 mm​


Tensile Reinforcement, Ast:


Tensile reinforcement diameter
:24 mm​



Number of tensile reinforcement bars
:3​


Compressive Reinforcement, Asc:


Compressive reinforcement diameter
:24 mm​



Number of compressive reinforcement bars
:3​


Shear Reinforcement:


Stirrup diameter
:10 mm​



Spacing of legs
:300 mm​



Number of legs
:3​

Cover:


Exposure classification
:A1​



Formwork and compaction method (1)
:Self-compacting concrete​



Design life
:60 years​

Using a design life of 120 years requires additional 5mm cover


cst
:25 mm​



csc
:25 mm​

This cover distance is to the surface of the first steel reinforcement, either longitudinal or shear, from top and bottom concrete fibres.


Column Restraints

AS 3600:2018 Figure 10.5.3(A) - effective length factor (k) for columns with simplified end restraints

End #1


Rotation - #1
:Fixed​



Translation - #1
:Fixed​

End #2


Rotation - #2
:Fixed​



Translation - #2
:Fixed​




Output

Section Properties



Ag
:127500mm2​



Zx
:9.03e+6mm3​



Zy
:6.38e+6mm3​



Ixx
:1.92e+9mm4​



Iyy
:9.56e+8mm4​




Ast
:1357mm2​



Asc
:1357mm2​



Asv
:236mm2​



Total minimum cover
:20mm​



dst
:378mm​



dsc
:47mm​



Distance to centroid of shear reinforcement, dsv
:395mm​



Column and Slenderness Properties



k
:0.7​



Le
:2100mm​



r
:127.5mm​



sr
:16.47​



Column classification
:Braced, Short​


Column Strength Checks

Interaction Curve (combined flexural and axial check)



(M*, N*) < Interaction curve
:PASS​

Squash Load


SL - ϕ
:0.65​



SL - φNuo
:3640kN​

Decompression Point


DP - ϕ
:0.6​



DP - ku
:1​



DP - ϕNu
:2298kN​



DP - ϕMu
:161kN m​

Balanced Point


BP - ϕ
:0.6​



BP - ku
:0.54545​



BP - ϕNu
:990kN​



BP - ϕMu
:255kN m​

Pure Bending


PB - ϕ
:0.85​



PB - kuo
:0.159​



PB - ϕMu
:200kN m​





Flexural Checks



ϕMu / M*
:0.59​



ϕMu > M*
:PASS​


Minimum moment check:


Mu,min
:41.13kN m​



M* > Mu,min
:PASS​


Longitudinal reinf. check:


As,min
:1275mm2​



As,max
:5100mm2​



As > As,min
:PASS​



As ≤ As,max
:PASS​

As,min = 1% of Ag
As,max = 4% of Ag
Ductility check:


kuo
:0.1587​



kuo < 0.36
:PASS​




Shear Checks



θ
:36.0​



bv
:425​



dv
:306mm​



kv
:0.15​



Vuc
:123.376262911469​



Vus
:165.4kN​



ϕVu
:216.6kN​



V*/ϕVu
:0.23​



ϕVu > V*
:PASS​

Minimum shear reinf. check:


Minimum Asv/s
:0.430mm2/mm​



Asv/s
:0.785mm2/mm​



Asv/s > Asv/s,min
:PASS​






Explanation

Columns are typically subject to combined compression and bending load and should be checked using an interaction curve, as per Cl 10.6.2 of AS3600:2018. An interaction curve is a graphical representation of the ultimate strength of a column's cross-section. It is defined by four key points (A, B, C and D on the adjacent figure) which are design capacities that form the boundary of failure modes for a section subject to combined bending and axial load. See the toggle blocks below for further information on the failure modes.
If the design forces N* and M* are within the region bound by the interaction curve, then the column is deemed to be safe.
Interaction curve


Four key points on the Interaction Curve

Note the design capacities are calculated using strain compatibility across the section. The maximum (ultimate) strain of concrete, εcu is 0.003 and the strain at yield for class 500N reinforcing bars is 0.0025.

A - Squash Load

The squash load, Nuo, is the point where a column fails in pure compression. The concrete is at ultimate strain of 0.003 and, due to strain compatibility, the steel therefore has exceeded its yield strain and will be at yield strength.


Nuo=Cc+CswhereCc=α1fc′(Ag−Asc)Cs=Cs1+Cs2=fsyAscN_{uo} = C_c + C_s \\ \text{where}\\ C_c = \alpha_1f'_c(A_g-A_{sc}) \\ C_s = C_{s_{1}}+C_{s_{2}} \\ \hspace{0.5cm} = f_{sy}A_{sc}

B - Decompression Point

The decompression point is where a column fails under combined bending and compression while providing no tensile capacity in the section. At this point, the strain in the tension reinforcement is zero and the extreme compressive fibre of the concrete is at its ultimate strain of 0.003. The concrete section in tension is assumed to provide no resistance against tension.


Nu=Cc+CswhereCc=α2fc′⋅γkud⋅bCs=εsEsAscN_u = C_c + C_s \\ \text{where}\\ C_c = \alpha_2f'_c \cdot \gamma k_ud \cdot b\\C_s = \varepsilon_sE_sA_{sc}

C - Balanced Failure

The balanced failure point is where a column fails under combined bending and compression by simultaneous crushing of the concrete and yielding of the reinforcement. At this point, the concrete is at ultimate strain, 0.003 and the outer steel strain reaches yield, 0.0025 and hence ku is fixed at 0.545. The balanced failure point represents the maximum bending capacity of a column.


Nu=Cc+Cs−TswhereCc=α2fc′⋅γkud⋅bCs=εs1EsAscTs=fsyAstandku=0.545N_u = C_c + C_s - T_s\\ \text{where}\\ C_c = \alpha_2f'_c \cdot \gamma k_ud \cdot b\\C_s = \varepsilon_{s1}E_sA_{sc}\\T_s = f_{sy}A_{st}\\ \text{and}\\k_u=0.545

D - Pure Bending

The pure bending point is where the column fails in bending without an external axial load. The column capacity is calculated in the same way as a doubly reinforced beam, taking moments about any point.


Mu=Cc(γkud2)+Cs(dsc)−Ts(d)whereCc=α2fc′⋅γkud⋅bCs=εs1EsAscTs=fsyAstM_u = C_c(\frac{\gamma k_ud}{2})+C_s(d_{sc})-T_s(d)\\\text{where}\\ C_c = \alpha_2f'_c \cdot \gamma k_ud \cdot b\\C_s = \varepsilon_{s1}E_sA_{sc}\\T_s = f_{sy}A_{st}

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