Slabs are horizontal structural elements that form the floor system of a building. The main purpose of the slabs is to support live load and dead loads which can result from people using the building, furniture, equipment and self weight of the building.

Types of Slabs

There are different types of slabs depending on their shape, support conditions, materials and the way they transfer loads. Regard to shape, slabs can be either solid or ribbed. Solid slabs have greater load carrying capacity compared to ribbed slabs of the same depth, and they are suitable for areas with wide spanning panels and heavy loads. On the other hand, ribbed slabs are more economical as they reduce on the amount of material used in construction.

Depending on the beam layout, slabs can be considered to transfer loads as either one-way or two-way slabs. One-way slabs transfer loads to the beam in one direction, while two-way slabs transfer loads to the beam in two directions. To determine whether a slab is either one-way or two-way, one needs to consider the ratio of the longer span (ly) to the shorter span (lx). One-way slabs have a ratio greater than 2 while two-ways slabs have a ratio less than or equal to 2.

The nature of supports also influences the type of the slab ,and hence slabs can either be continuous or simply supported. Simply supported slabs have only one span with two supports, and may carry more load while continuous slabs have a series of supports and carry less load within span due to load redistribution to other spans.

Slabs can also be made of either a single material or a mixture of two different materials. Slabs made of a number of materials are considered to be “composite slabs”. Typical composite slabs consist of a concrete slab with a steel decking ,and this kind of slab is common for the steel framed buildings.

Planning floor layout

The choice of floor layout determines both the type of the slab as well as the design approach. The floor layout consists of the configuration of the slab panels and beams. When designing a floor layout, it is important to use economical slab types like ribbed slabs and short spanning panels. One way slabs are also generally economical, however, these slab panels should be kept small during the layout design.

How to design reinforced concrete slabs (Eurocode Design)

When designing slabs there are important aspects to considers such as; the slab loading, design moment, shear and deflections. The typical loads on a slab are dead loads (Gk) and live loads (Qk). The dead loads on slabs include the self-weight of the slab, finishes and fixed equipment within the building. The live loads mainly consist of the imposed loads from people using the building ,and these depend of the usage of the building. For domestic and residential building, the minimum live loads for floors, stairs and balconies are 1.5kNm2, 2.0kNm2 and 2.5 kNm2 respectively. More examples of imposed floor loads can be found in Table 6.2 (Eurocode 1)

In order to determine the design moment, shear and deflections, the characteristic loads (dead and live loads) have to be combined using factors of safety. In Eurocode, the dead and live loads are given factors of 1.35 and 1.5 respectively. The following formula shows the design load (Wk) computed from the load combination.

Wk = 1.34 Gk+ 1.5 Qk

In Eurocode, slabs are designed for both ultimate limit state and serviceability state. In the ultimate limit state, slab is designed for flexural behaviour as well as shear. For the serviceability state, the deflection of the slab is checked to ensure it does not exceed the serviceability limit.

Depending the nature of the slab, the design moment (MEd) and shear (VEd) can be determined from the analysis of the slab. For simply supported one-way slabs, the design moment and shear from analysis can be obtained using the formulas below.

MEd = W L2/8

VEd = W L/2

where W is the load per 1 metre width of the slab and L is the span of the slab.

Example (Simply supported one way slab design according to Eurocode):

Consider a floor slab panel for residential building with the following conditions:

Load due to finishes = 1.2 kN/m2

Live load = 1.5 kN/m2

span of the slab = 2300 mm



  1. thickness of the slab, h is 125 mm
  2. concrete strength, fck = 25 N/mm2
  3. yield strength of reinforcement , fyk = 500 N/mm2
  4. Unit weight of concrete ,γ= 25 kN/m3
  5. diameter of the reinforcement,Φ is 10 mm
  6. Cover to the steel reinforcement, c is 25 mm
  7. for 1 metre width of the slab, the bread (b) is 1000 mm

Loading on the slab:

  1. self-weight of the slab = (125/1000)*25 = 3.125 kN/m2
  2. Dead load on the slab, Gk = 3.125 + 1.2 = 4.325 kN/m2
  3. Live load on the slab, Qk = 1.5 kN/m2
  4. Design load per 1 metre width , W = (1.35 * Gk + 1.5 * Qk ) = 8.089 kN/m

Design moment, MEd = 8.089 * (2300/1000)2 / 8 = 5.35 kNm

Design shear at the supports, VEd = 8.089 * (2300/1000) / 2 = 9.3 kN

Effective depth of the slab, d = h – c – Φ/2 = 150 – 25 – (10/2) = 120 mm

The breadth, b = 1000 mm

K =0.015

K’ = 0.6q -0.16q2 -0.21 where q <= 1.0

Assuming q =1

K’= 0.21

Since K < K’ , compression reinforcement is not necessary.

lever arm, z is given by:

z= 118.4 mm >0.95d

Hence z=114

The required tension reinforcement is given by:

fyd = (fyk/1.15)=(500/1.15) =434.8

As = 107.9 mm2

Minimum area of reinforcement required is by :

for fck = 25 N/mm2 , fctm = 2.6 N/mm2 as obtained from the table below.
for fyk = 500 N/mm2

As,min = 162.2 mm2

Maximum area of reinforcement required is by:

Ac = 1000 x 150 mm2

As,max = 6000 mm2

Since As < As,min, the required area of steel, Areq = 162.2 mm2

Size and spacing of reinforcement

 Spacing of Bars        

From the table above, for an area of steel of 162.2 mm2 , the required reinforcement is T10 – 300 for the main bars.

Basing on the same minimum area of steel, the distribution reinforcement is T10 – 300.

Deflection Check

Span of the slab, l = 2300

Basic span to depth ratio, l/d is given by

The value of K can be obtained based the span conditions:

  • K = 1.5 for interior span condition
  • K = 1.3 for end span condition
  • K = 0.4 for cantilevers

In this case, we can assume K =1.5

The compression reinforcement ,ρ’ = As2/bd

where As2 is the area of compression reinforcement

For this case, ρ’ can also be taken as 0

The reference reinforcement ratio, ρ0 = (√fck)/1000 = 0.005

ρ = As/bd = (107.9)/(1000×120) =0.0008991

Since ρ < ρ0 , then basic span to depth ratio, l/d = 208


Since the span is less than 7m, F2 =1.0

F3 can also be assumed as 1.0

Actual span to depth ratio, l/d = (2300/120)= 19.2

Since 208 > 19.2 , the slab passes in deflection.

Shear Check

Slabs are normally not provided with shear reinforcement, therefore the concrete alone is first checked for adequacy under shear.

The applied shear stress, vEd= VEd/(bd) = 0.0775 N/mm2

The concrete shear stress capacity without shear reinforcement, vRd,c is given by:


k = 1+ √(200/120) = 2.29 > 2

hence k =2

ρ1 = (107.9)/(1000×120) = 0.0008991 < 0.02

hence ρ1 = 0.0008991

0.035k1.5fck0.5 = 0.495

vrd,c = 0.12 x 2 (100×0.0008991×25)1/3 = 0.314 < 0.495

hence vRd,c = 0.495 N/mm2

Since vEd < vRd,c the slab is adequate is shear, and hence no shear reinforcement is required.

Designing Two way slab

In the case of two-way slabs, the procedure above is repeated for each span using the respective design moment. The design moments for the different spans can be computed from:

Msx = βsx w lx2 for the shorter span
Msy = βsy w lx2 for the longer span

where βsx and βsy are bending moment coefficients which can be got from the table below, and lx is the shorter span.

Bending moment coefficients ( Extract from: How to Design Concrete Structures using Eurocode 2 )

MomentsCoefficients for short span    Coefficients for long span
For interior panels
Negative moment at continuous edge0.0310.0440.0530.0590.0630.032
Positive moment at midspan0.0240.0340.040.0440.0480.024
Two adjacent edges discontinuous
Negative moment at continuous edge0.0470.0660.0780.0870.0930.045
Positive moment at midspan0.0360.0490.0590.0650.070.034
One short edge discontinuous
Negative moment at continuous edge0.0390.050.0580.0630.0670.037
Positive moment at midspan0.0290.0380.0430.0470.050.028
One long edge discontinuous
Negative moment at continuous edge0.0390.0590.0730.0830.0890.037
Positive moment at midspan0.030.0450.0550.0620.0670.028


Cook, N. J. (2007). Designers’ guide to EN 1991-1-4: Eurocode 1: Actions on structures, general actions. Wind actions. Thomas Telford Publishing..

Institution, B. S. (2004). Eurocode 2: Design of concrete structures : Part 1-1: General rules and rules for buildings.

Bond, A. (2006). How to design concrete structures using Eurocode 2.

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