RC Section to AS3600

Verified by the CalcTree engineering team on August 5, 2024
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The neutral axis (N.A) is an axis in the cross-section of a beam where there are no stresses or strain. In a symmetrical cross-section the neutral axis is at the centroidal depth. How to calculate the depth of the neutral axis and how the neutral axis affects a member's performance is important to understand when designing RC sections.
All calculations are performed in accordance with AS3600-2018.
You can use this calculator to determine the depth to the neutral axis of an RC section in 3 different stages of concrete cracking: 
Uncracked, linear-elastic section
Cracked, linear-elastic section
Cracked, non-linear elastic section
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RC section and the strain distribution in sagging
CalculationInputsMaterial Properties
Concrete
Grade,﻿f'c
:32 MPa​
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Young's modulus,﻿Ec
:33 GPa​
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Reinforcement
Yield stress,﻿fsy
:500 MPa​
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Young's modulus,﻿Es
:200 GPa​
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Geometry
Width,﻿B
:200 mm​
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Depth,﻿D
:400 mm​
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Reinforcement Layout
Bottom Reinforcement:
Cover,﻿c_st
:30 mm​
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Bar diameter,﻿d_bst
:16 mm​
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Number of bars,﻿n_st
:2​
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Total area,﻿A_st
:402.12 mm^2​
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Top Reinforcement:
Cover,﻿c_sc
:50 mm​
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Bar diameter,﻿d_bsc
:12 mm​
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Number of bars,﻿n_sc
:0​
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Total area,﻿A_sc
:0.00 mm^2​
 
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Symbols used in this calculator
OutputGeometric Properties
Gross cross-section area, ﻿A_g
:800 cm^2​
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Section modulus about x-axis, ﻿Z_x
:5333 cm^3​
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Section modulus about y-axis, ﻿Z_y
:2667 cm^3​
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Moment of inertia about x-axis, ﻿I_x
:106667 cm^4​
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Moment of inertia about y-axis, ﻿I_y
:26667 cm^4​
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Depth of the N.A.
Uncracked, linear-elastic section
Cracked, linear-elastic section
Cracked, inelastic section
﻿﻿dne
:204.05 mm​
 
﻿﻿dnt
:82.75 mm​
 
﻿﻿dn
:1.00 mm​
 
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ExplanationWhat is the Neutral Axis?To visualize the neutral axis in structural engineering applications, picture a beam with an external load applied to it. When a beam is loaded, internal forces develop within it to maintain equilibrium. The internal forces in a beam have two components: shear forces in the vertical direction and normal force in the axis of the beam. 
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Normal force in section
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Shear force in section
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Resultant shear and bending moment
There’s an area in the middle of the beam’s cross-section that is neither stretched nor squashed; this is known as the neutral axis. At this point within the beam's cross-section, internal stresses are zero. The strain of the beam varies linearly from top of the section to bottom and passes through zero at the neutral axis. At this point, compressive and tensile stresses are in equilibrium, that is, no internal stresses exist.
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Loaded beam in sagging
💡 Learn about sagging and hogging
If you apply a load on the topside of a beam, the beam will curve downwards, causing a sagging action. When a beam is sagging, the top of the beam will shorten, and the forces at the top of the beam will be compressive. Simultaneously, the bottom side of the beam will stretch, and the fibres in this part of the beam will be in tension.
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Beam sagging
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Beam hogging
When the load is applied from the underside of the beam or when a beam runs continuously over supports, the beam deforms upwards; this is known as hogging action. The bottom side of the beam is now shortening, the normal forces in the bottom are compressive, and the forces on the top of the beam are in tension because the top half is stretching.
Why is it Important to Determine the Neutral Axis?The concept of the neutral axis is fundamental when understanding and determining the flexural bending stresses and deflection of reinforced concrete beams. The neutral axis, ﻿dn​
 is an input parameter to the beam's ultimate flexural capacity, ﻿ϕMu​
 and the second moment of inertia equation, ﻿I
 for the beam's deflection. 
The neutral axis shows how much of the cross-section is in tension or compression. The further the neutral axis is from the extreme tensile fibre, the larger the area under tension and therefore the more tensile cracks there can be. Cracks begin propagating when the tensile stress reach the characteristic flexural tensile strength of the concrete. Concrete is inherently weaker in tension than compression, so steel reinforcements are used to increase the tensile strength of concrete. Steel is ductile and possesses great tensile and compressive strength; therefore, they are combined with concrete to provide extra structural strength.
DerivationThe location of the neutral axis depends on the geometry and cracking of the RC section. The neutral axis can be found using equilibrium of internal forces: ﻿ΣC−ΣT=0
 where ﻿ΣC
 and ﻿ΣT
 are the sums of compression and tension forces, respectively.
Three different stages can be defined for the determination of the neutral axis, as provided below.
👉To simplify the calculation, we "transform" our section into an equivalent homogenous section, that is, a section with one material (concrete) rather then two materials (concrete and steel). The ﻿(n−1)Ast​
 and ﻿(n−1)Asc​
 terms in the below equations, where ﻿n=Ec​Es​​
 is the modular ratio, transforms the steel areas into equivalent concrete areas. 
1) Uncracked and Linear Elastic
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When the concrete is uncracked, the location of the neutral axis is the centre of the geometric centre of the transformed section. At this stage, the concrete section below the neutral axis demonstrates some tensile capacity.
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Compression in concrete, Cc=12Ecεcompdnb−EcεscAscCompression in steel, Cs=EsεscAscTension in concrete, Tc=12Ecεtens(h−dn)b−EcεstAstTension in steel, Ts=EsεstAst∴Cc+Cs=Tc+Ts\text{Compression in concrete, }C_c = \frac{1}{2} E_c \varepsilon_{comp} d_nb - E_c\varepsilon_{sc}A_{sc}  \\\text{Compression in steel, } C_s = E_s\varepsilon_{sc}A_{sc}  \\ \text{Tension in concrete, }T_c = \frac{1}{2}E_c\varepsilon_{tens}(h-d_n)b - E_c\varepsilon_{st}A_{st} \\ \text{Tension in steel, } T_s=E_s\varepsilon_{st}A_st \\ \therefore C_c + C_s=T_c+T_sCompression in concrete, Cc​=21​Ec​εcomp​dn​b−Ec​εsc​Asc​Compression in steel, Cs​=Es​εsc​Asc​Tension in concrete, Tc​=21​Ec​εtens​(h−dn​)b−Ec​εst​Ast​Tension in steel, Ts​=Es​εst​As​t∴Cc​+Cs​=Tc​+Ts​
Rearranging the above, the depth of the neutral axis, ﻿dn​
 is given by:
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dn=h2bh+d(n−1)Ast+dsc(n−1)Ascbh+(n−1)Ast+(n−1)Ascd_n= \dfrac{\frac{h}{2}bh + d(n-1)A_{st}+d_{sc}(n-1)A_{sc}}{bh+(n-1)A_{st}+(n-1)A_{sc}}dn​=bh+(n−1)Ast​+(n−1)Asc​2h​bh+d(n−1)Ast​+dsc​(n−1)Asc​​
2) Cracked and Linear ElasticThe section is cracked, but its behaviour is linear-elastic therefore the transformed area and force equilibrium method can be used.
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At this stage, the concrete section below the neutral axis does not contribute to tensile capacity. Hence:
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Compression in concrete, Cc=12Ecεcompdnb−EcεscAscCompression in steel, Cs=EsεscAscTension in concrete, Tc=0Tension in steel, Ts=EsεstAst∴Cc+Cs=Ts\text{Compression in concrete, }C_c = \frac{1}{2} E_c\varepsilon_{comp}d_nb - E_c\varepsilon_{sc}A_{sc}  \\\text{Compression in steel, } C_s = E_s\varepsilon_{sc}A_{sc}\\ \text{Tension in concrete, }T_c =0 \\\text{Tension in steel, }T_s=E_s\varepsilon_{st}A_{st} \\ \therefore C_c + C_s=T_sCompression in concrete, Cc​=21​Ec​εcomp​dn​b−Ec​εsc​Asc​Compression in steel, Cs​=Es​εsc​Asc​Tension in concrete, Tc​=0Tension in steel, Ts​=Es​εst​Ast​∴Cc​+Cs​=Ts​
The force equilibrium can be presented as a quadratic equation with respect to the depth of the neutral axis ﻿dn​
 and can be calculated as:
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12bdn2+[(n−1)Asc+nAst]dn−[(n−1)Ascdsc+nAstdst]=0\large \frac{1}{2}bd_n^2+[(n-1)A_{sc}+nA_{st}]d_n-[(n-1)A_{sc}d_{sc}+nA_{st}d_{st}]=021​bdn2​+[(n−1)Asc​+nAst​]dn​−[(n−1)Asc​dsc​+nAst​dst​]=0
3) Cracked and InelasticThe section is cracked and has reached its ultimate strength capacity, therefore the stress distribution is non-linear. At this stage, rectangular stress block and force equilibrium method can be used. 
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At this stage, the concrete section below the neutral axis also does not contribute to tensile capacity. Hence:
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Compression in concrete, Cc=α2fc′γ b dnCompression in steel, Cs=εcuEsAsc(dn−dsc)dnTension in concrete, Tc=0Tension in steel, Ts=fsyAst∴Cc+Cs=Ts\text{Compression in concrete, } C_c = \alpha_2 f'_c \gamma \ b \ d_n \\ \text{Compression in steel, }C_s = \varepsilon_{cu} E_s A_{sc} \frac{(d_n - d_{sc})}{d_n} \\ \text{Tension in concrete, }T_c =0 \\ \text{Tension in steel, } T_s = f_{sy}A_{st}\\ \therefore C_c + C_s=T_sCompression in concrete, Cc​=α2​fc′​γ b dn​Compression in steel, Cs​=εcu​Es​Asc​dn​(dn​−dsc​)​Tension in concrete, Tc​=0Tension in steel, Ts​=fsy​Ast​∴Cc​+Cs​=Ts​
The force equilibrium can be presented as a quadratic equation with respect to the depth of the neutral axis ﻿dn​
 and can be calculated as:
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α2fc′γbdn2+εcuEsAsc(dn−dsc)−fsyAst=0\large \alpha_2 f'_c \gamma b {d_n}^2 + \varepsilon_{cu} E_s A_{sc} (d_n - d_{sc}) - f_{sy} A_{st} = 0α2​fc′​γbdn​2+εcu​Es​Asc​(dn​−dsc​)−fsy​Ast​=0
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Neutral Axis Calculator: RC Section to AS3600

Calculation

Inputs

Material Properties

Geometry

Reinforcement Layout

Output

Geometric Properties

Depth of the N.A.

Explanation

What is the Neutral Axis?

💡 Learn about sagging and hogging

Why is it Important to Determine the Neutral Axis?

Derivation

1) Uncracked and Linear Elastic

2) Cracked and Linear Elastic

3) Cracked and Inelastic
