RC Section to AS3600

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Definition📝 This template has been designed in accordance with AS3600-2018
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.
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
:33MPa​
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Young's modulus,﻿Ec
:32,800MPa​
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Reinforcement
Yield stress,﻿fsy
:500MPa​
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Young's modulus,﻿Es
:200,000MPa​
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Geometry
Width,﻿B
:400mm​
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Depth,﻿D
:500mm​
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Reinforcement Layout
Bottom Reinforcement:
Cover,﻿c_st
:50mm​
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Bar diameter,﻿d_bst
:16mm​
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Number of bars,﻿n_st
:5​
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Total area,﻿A_st
:1,005mm2​
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Top Reinforcement:
Cover,﻿c_sc
:50mm​
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Bar diameter,﻿d_bsc
:12mm​
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Number of bars,﻿n_sc
:5​
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Total area,﻿A_sc
:565mm2​
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Symbols used in this calculator
OutputGeometric Properties
Gross cross-section area,﻿A_g
:2.00e+5mm2​
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Section modulus about x-axis,﻿Z_x
:1.67e+7mm3​
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Section modulus about y-axis,﻿Z_y
:1.33e+7mm3​
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Moment of inertia about x-axis,﻿I_x
:4.17e+9mm4​
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Moment of inertia about y-axis,﻿I_y
:2.67e+9mm4​
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Depth of the N.A.
Uncracked, linear-elastic section
Cracked, linear-elastic section
Cracked, inelastic section
Neutral axis,﻿(1) dn
:252mm​
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Neutral axis:﻿(2) dn
:99mm​
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Neutral axis:﻿(3) dn
:399mm​
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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
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.
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​As​cCompression 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=(h/2bh)+d(n−1)Ast+dsc(n−1)Ascbh+(n−1)Ast+(n−1)Ascd_n= \dfrac{(h/2bh) + 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​(h/2bh)+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​As​cCompression in steel, Cs​=Es​εsc​Asc​Tension in concrete, Tc​=0Tension in steel, Ts​=Es​εst​As​t∴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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Compression in concrete, Cc=α2fc′γbdn2Compression in steel, Cs=εcuEsAsc(dn−dsc)Tension 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}^2 \\ \text{Compression in steel, }C_s = \varepsilon_{cu} E_s A_{sc} (d_n - d_{sc}) \\ \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′​γbdn​2Compression in steel, Cs​=εcu​Es​Asc​(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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