Loading
/custom-emojis/emojis/contour-map.png
Templates
📚
Articles & Resources
📖
Guides & Support
🌵
CalcTree
Bust Common Myths About Java Programming
Loading
/custom-emojis/emojis/calculator.png
Tensile Strength and Capacity Control of the W-Shape Sections According to AISC 360-16
Loading
/custom-emojis/emojis/calculator.png
Concrete Cylinder Strength Vs Cube Strength
Loading
/custom-emojis/emojis/calculator.png
Earthquake Design Action Calculation
EM Wave Propagation Calculator
İKSA YAPILARI PROJELENDİRME HİZMET BEDELİ (2024)
section properties with units
GEOTEKNİK RAPOR (EK-B) ASGARİ HİZMET BEDELİ (2024)
Forward Kinematics of Robotic Arm with 6 Degrees of Freedom
ZEMİN İYİLEŞTİRME/DERİN TEMEL PROJELENDİRME ASGARİ HİZMET BEDELİ (2024)
İKSA YAPILARI PROJELENDİRME HİZMET BEDELİ (2023)
Loading
/custom-emojis/emojis/bending-moment.png
Dezi et. al (2010)
🤾
Projectile motion
Sitemap
Neutral Axis Calculator: RC Section to AS3600's banner
/custom-emojis/emojis/rc-beam.png

Neutral Axis Calculator: RC Section to AS3600

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:
  1. Uncracked, linear-elastic section
  2. Cracked, linear-elastic section
  3. Cracked, non-linear elastic section
RC section and the strain distribution in sagging


Calculation

Inputs

Material Properties

Concrete
Grade,

f'c
:32 MPa​

Young's modulus,

Ec
:33 GPa​


Reinforcement
Yield stress,

fsy
:500 MPa​

Young's modulus,

Es
:200 GPa​


Geometry

Width,

B
:200 mm​

Depth,

D
:400 mm​


Reinforcement Layout

Bottom Reinforcement:
Cover,

c_st
:30 mm​

Bar diameter,

d_bst
:16 mm​

Number of bars,

n_st
:2​

Total area,

A_st
:402.12 mm^2​

Top Reinforcement:
Cover,

c_sc
:50 mm​

Bar diameter,

d_bsc
:12 mm​

Number of bars,

n_sc
:0​

Total area,

A_sc
:0.00 mm^2​
Symbols used in this calculator


Output

Geometric Properties

Gross cross-section area,

A_g
:800 cm^2​

Section modulus about x-axis,

Z_x
:5333 cm^3​

Section modulus about y-axis,

Z_y
:2667 cm^3​

Moment of inertia about x-axis,

I_x
:106667 cm^4​

Moment of inertia about y-axis,

I_y
:26667 cm^4​


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​


Explanation

What 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.
Normal force in section

Shear force in section

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.
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.
Beam sagging

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,

is an input parameter to the beam's ultimate flexural capacity,

and the second moment of inertia equation,

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.

Derivation

The 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:

where

and

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

and

terms in the below equations, where

is the modular ratio, transforms the steel areas into equivalent concrete areas.

1) Uncracked and Linear Elastic


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.

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_s
Rearranging the above, the depth of the neutral axis,

is given by:

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}}

2) Cracked and Linear Elastic

The section is cracked, but its behaviour is linear-elastic therefore the transformed area and force equilibrium method can be used.

At this stage, the concrete section below the neutral axis does not contribute to tensile capacity. Hence:

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_s
The force equilibrium can be presented as a quadratic equation with respect to the depth of the neutral axis

and can be calculated as:

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}]=0

3) Cracked and Inelastic

The 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.

At this stage, the concrete section below the neutral axis also does not contribute to tensile capacity. Hence:

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_s
The force equilibrium can be presented as a quadratic equation with respect to the depth of the neutral axis

and can be calculated as:

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} = 0