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Concrete One-Way Slab Designer to EC2's banner
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Concrete One-Way Slab Designer to EC2

Verified by the CalcTree engineering team on August 13, 2024

This calculator performs the structural analysis and design of a one-way spanning slab, by performing flexural, shear and deflection checks.
All calculations are performed in accordance with:
  1. BS EN 1992-1-1 (2004) Eurocode 2: Design of Concrete Structures Part 1-1: General Rules for Structures. This code is typically referred to as "EC2".
  2. BS EN 1991-1-1 (2002) Eurocode 1: Actions on Structures Part 1-1 General actions - Densities, self-weight, and imposed loads for buildings. This code is typically referred to as "EC1".
Section view of a one-way spanning slab


Calculation

📃Assumptions

This calculator:
  1. Only designs a single span of a one-way slab which is simply supported
  2. Only designs for a sagging bending moment, therefore only considers bottom reinforcement. Calculation is per meter strip
  3. Performs the flexural, shear, and deflection checks as prescribed by the code. Refer to Explanation section below for more details. Crack width is not checked.
  1. Assumes no redistribution of bending moments
  2. Applicable for
    
    , so that
    
    in the stress distribution for flexural design
  3. UK National Annex values are used for concrete and steel material factors:
    
    

Inputs

Concrete Properties



fck
:40 MPa​



Concrete unit weight
:25 kN / m^3​


Slab Geometry



L
:5 m​



t
:200 mm​


Reinforcement Data



fyk
:500 MPa​



c
:25 mm​



db
:16 mm​



Bar spacing
:200 mm​


Slab Loads

ULS Load Factors:


Permanent load factor
:1.35​



Imposed load factor
:1.50​

Permanent Actions:


Partition
:1.00 kPa​



Floor finishes
:0.50 kPa​



Self Weight
:5.00kN/m2​

Imposed Actions:


Imposed load
:1.50 kPa​



Movable partitions
:0.80 kPa​


Outputs

Design (ULS) Loads

Ultimate Load Calculations:


Gk
:6.50kN/m2​



QK
:2.30kN/m^2​



Total ULS loads
:12.22kN/m^2​

Design Actions:


Med
:38.19kN m​



Ved
:30.55kN​



Bending (ULS) Check

Flexural Design Cl. 6.1:


d
:167mm​



k
:0.034​



Compression status
:OK, k<0.167​



Ast, required
:553.34mm2/m​



Ast, provided
:1,005.00mm2/m​



Slab Design Status
:PASS, there is sufficient bottom reinforcement​

Minimum Area of Steel Cl 9.2.1.1:


Ast, min
:305.00mm2/m​



Minimum steel design status
:PASS, minimum reinforcement has been provided​



Shear (ULS) Check

Shear Design Cl. 6.2.1(3) and 6.2.2:


vrdc
:0.69N/mm2​



Vrdc
:115.73kN​



Shear utilisation ratio
:26%​



Shear design status
:PASS​



Deflection (SLS) Check



N
:21.20​



F1
:1.00​



F2
:1.00​



F3
:1.50​



δ, actual
:29.94mm​



δ, limit
:31.80mm​



Deflection utilisation ratio
:94%​



Deflection check
:PASS​


Explanation

One-way slabs are a form of solid slab that spans mainly in one direction and transfers its loads to the supports by bending. They are commonly used in buildings, bridges, and other structures.
Difference in bending behaviour between a one-way and two-way slab

Usually, if the length-to-breadth ratio of the slab is equal to or greater than 2, the slab can be considered a one-way slab.
A slab is considered a one-way slab when it's length is at least double it's breadth


Actions on Solid Slabs

According to the Eurocodes, loads on solid slabs can be classified into three types:
  1. Permanent, loads that remain constant during the lifetime of the structure, such as the self-weight of the slab or any fixed equipment
  2. Variable, also called imposed loads, vary in magnitude and/or position over time, such as loads from people, furniture, vehicles, etc.
  3. Accidental, loads that result from exceptional or unforeseen events, such as explosions, impacts, earthquakes, etc.
❗Note, this calculator does not consider accidental loads

Flexural Design

The flexural (also referred to as 'bending' or 'moment') capacity of a slab cross-section is determined using the Rectangular Stress Block method, as per EC2. The stress distribution in concrete under bending is curved in reality, however, it can be converted to an equivalent rectangular stress block by the use of reduction factors shown in the following image.
Rectangular stress distribution used for flexural design, adapted from EC2 Figure 3.5

The procedure below is outlined in the Design of Structural Elements 3rd Edition by Chanakya Arya.
The area of reinforcement,

that will withstand tension can be determined using the following equation:

As=MEdfyd×z{ A }_{ s }=\dfrac { M_{Ed}}{f_{yd}\times z }
Where:
  1. 
    
    is the design ULS moment in the major axis
    
    
  2. 
    
    is the design yield strength of reinforcement
    
    
  3. 
    
    is the lever arm between the force from the compression stress block and force from the steel reinforcement
    
    
The lever arm,

can be calculated using the following equation:

z=d[0.5+(0.25−K1.134)]≤0.95dz=d\left[ 0.5+\sqrt { \left( 0.25-\frac { K }{ 1.134 } \right) } \right] \le 0.95d
The parameter

is used to check the position of the neutral axis and is limited by a value of 0.167 to avoid compression reinforcement.

is calculated from:

K=MEdb×d2×fck≤0.167K=\dfrac { M_{ Ed } }{ b \times{ d }^{ 2 }\times{ f }_{ ck } }\le 0.167
Where:
  1. 
    
    is the characteristic compressive cylinder strength of concrete at 28 days
    
    
  2. 
    
    is the effective depth of the slab
    
    
  3. 
    
    is the width of the slab
    
    
The above equations for

and

are derived using the following equations:

MRd=FcczFcc=ηfcd λX bz=d−1/2λXM_{Rd}=F_{cc}z \\F_{cc}= \eta f_{cd} \space \lambda X \space b \\ z = d-1/2 \lambda X
The

is defined assuming:

αcc=0.85,γc=1.5η=1,λ=0.8,X≤0.45dfor ≤50MPa\alpha_{cc}=0.85, \gamma_c=1.5 \\ \eta=1, \lambda=0.8,X \leq0.45d \hspace{1cm}\text{for } \leq50MPa
Cl 5.6.3 of EC2 limits the depth of the neutral axis in order to provide a ductile section.
The depth of the neutral axis,

at ULS is found using equilibrium of the stress distribution:

Fcc=Fstηfcd λX b=Astfyd→X=Astfydηfcd λF_{cc}=F_{st} \\ \eta f_{cd} \space \lambda X \space b=A_{st}f_{yd} \\\rightarrow X= \dfrac{A_{st}f_{yd}}{\eta f_{cd} \space \lambda}

Minimum Area of Steel

According to EC2 Cl 9.2.1.1, a reinforced concrete member that is susceptible to flexure should have a minimum area of steel given by:

As,min=0.26fctmfykbtd≥0.013btd{ A }_{ s,min }=0.26\dfrac {{ f }_{ ctm } }{ { f }_{ yk } } { b }_{ t }d\ge 0.013{ b }_{ t }d
Where:
  1. 
    
    is the characteristic yield strength of reinforcement
  2. 
    
    is the mean value of axial tensile strength of concrete, taken from Table 3.1 in EC2 or using the equation
    
    

Shear Design

Shear reinforcement will rarely be necessary in slabs subjected to uniformly distributed loads since shear stresses in these slabs are often minimal. The slab is typically designed so that the ultimate shear force,

is smaller than the shear strength of an unreinforced section,

and therefore no shear reinforcement is required. See Clause 6.2.2 of EC2.
❗Note: In the presence of significant point loads on the slab, such as those from columns, the engineer should verify punching shear.

Deflection Checks

The deflection of one-way solid slabs is evaluated to ensure it meets serviceability requirements, that is, to avoid excessive deflections. The deflection of a one-way slab depends on factors such as the span length, load, material properties, and support conditions.
EC2 Clause 7.4.2 provides guidelines for calculating deflections in reinforced concrete structures. EC2 has two alternative methods of checking deflection, by:
  1. Assessing the theoretical deflection using the expressions given in the code
  2. Limiting the span-to-depth ratio, provided in EC2 Table 7.4N for common reinforced concrete members
❗Method 2 is used in this calculator
For method 2, the values in Table 7.4N can be derived from the following expressions:
The permissible span-to-depth ratio,

is given by:

ld=N×K×F1×F2×F3\dfrac { l }{ d } =N\times K\times F1\times F2\times F3
For slabs:
  1. 
    
    
  2. 
    
    
  3. 
    
    
  4. 
    
    

The parameter

can be calculated from the following expressions:

Ifρo≥ρ  N={11+1.5fckρoρ+3.2fck(ρoρ−1)3/2} \text{If}\quad { \rho }_{ o }\ge \rho\ \ \hspace{0.7cm} N=\left\{ 11+\dfrac { 1.5\sqrt { { f }_{ ck } } { \rho }_{ o } }{ \rho } +3.2\sqrt { { f }_{ ck } } \left( \dfrac { { \rho }_{ o } }{ \rho } -1 \right) ^{ 3/2 } \right\} \quad

If  ρ0<ρ N={11+1.5fckρ0(ρ−ρ′)+fck12ρ′ρ0}\text{If }\ \rho_0 < \rho\ \hspace{0.7cm} N = \left\{ 11 + \dfrac{1.5 \sqrt{f_{ck}} \rho_0}{(\rho - \rho')} + \dfrac{\sqrt{f_{ck}}}{12} \sqrt{\dfrac{\rho'}{\rho_0}} \right\} \quad
Where:
  1. 
    
    is the ratio of tension reinforcement at mid-span and supports
  2. 
    
    is the ratio of compression reinforcement at mid-span and supports
  3. 
    
    is the reference reinforcement ratio

The steel stress,

(used in the equation for

) can then be calculated from the following equation:

σs=fykγs[gk+ψqkULS design action]As,reqAs,prov1δ{ \sigma }_{ s }=\dfrac { { f }_{ yk } }{ { \gamma }_{ s } } \left[ \dfrac { { g }_{ k }+\psi { q }_{ k } }{ \text{ULS design action} } \right] \dfrac { { A }_{ s,req } }{ { A }_{ s,prov } } \dfrac { 1 }{ \delta }
Where:
  1. 
    
    = material factor of safety for steel, taken as 1.15 in this calculator
  2. 
    
    = permanent actions
    
    
  3. 
    
    = variable actions
    
    
  4. 
    
    = factor on the live load, taken as 0.6 in this calculator
  1. 
    
    = area of steel required
    
    
  2. 
    
    = Area of steel provided
    
    
  3. 
    
    = redistribution ratio, taken as 1 for one-way slabs

Acknowledgements

This calculation was built in collaboration with Kunle Yusuf. Learn more.

Related Resources

  1. RC one way slab design to ACI318
  2. Slab Thickness Calculator to ACI 360R-10
  3. Concrete Slab-on-Grade Designer to AS3600
  4. Concrete Beam Design Calculator to AS3600