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
- thickness of the slab, h is 125 mm
- concrete strength, fck = 25 N/mm2
- yield strength of reinforcement , fyk = 500 N/mm2
- Unit weight of concrete ,γ= 25 kN/m3
- diameter of the reinforcement,Φ is 10 mm
- Cover to the steel reinforcement, c is 25 mm
- for 1 metre width of the slab, the bread (b) is 1000 mm
Loading on the slab:
- self-weight of the slab = (125/1000)*25 = 3.125 kN/m2
- Dead load on the slab, Gk = 3.125 + 1.2 = 4.325 kN/m2
- Live load on the slab, Qk = 1.5 kN/m2
- 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’ = 0.6q -0.16q2 -0.21 where q <= 1.0
Assuming q =1
Since K < K’ , compression reinforcement is not necessary.
lever arm, z is given by:
z= 118.4 mm >0.95d
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 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.
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
Actual span to depth ratio, l/d = (2300/120)= 19.2
Since 208 > 19.2 , the slab passes in deflection.
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 )
|Moments||Coefficients for short span||Coefficients for long span|
|For interior panels|
|Negative moment at continuous edge||0.031||0.044||0.053||0.059||0.063||0.032|
|Positive moment at midspan||0.024||0.034||0.04||0.044||0.048||0.024|
|Two adjacent edges discontinuous|
|Negative moment at continuous edge||0.047||0.066||0.078||0.087||0.093||0.045|
|Positive moment at midspan||0.036||0.049||0.059||0.065||0.07||0.034|
|One short edge discontinuous|
|Negative moment at continuous edge||0.039||0.05||0.058||0.063||0.067||0.037|
|Positive moment at midspan||0.029||0.038||0.043||0.047||0.05||0.028|
|One long edge discontinuous|
|Negative moment at continuous edge||0.039||0.059||0.073||0.083||0.089||0.037|
|Positive moment at midspan||0.03||0.045||0.055||0.062||0.067||0.028|
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.
For more updates like our Facebook Page.
Please don’t forget to subscribe to our newsletter!
YOU MAY ALSO LIKE TO READ
- GREEN STORMWATER SOLUTIONS: HOW TO EFFECTIVELY MANAGE STORMWATERIn the face of climate change threats, constructing green infrastructure is crucial in today’s world.. In a world with global warming and pollution, it is important to seek out infrastructural solutions to that are clean and energy efficient to ensure sustainability. As many cities all over the world have adopted… Read more: GREEN STORMWATER SOLUTIONS: HOW TO EFFECTIVELY MANAGE STORMWATER
- VERTICAL VS HORIZONTAL EXPANSIONCivil engineering involves the art of shaping our world. The choices that engineers make in designing and constructing buildings and infrastructure have a profound impact on communities. In this article, we are going to look to fundamental approaches in the design and construction of buildings. These approaches are vertical expansion… Read more: VERTICAL VS HORIZONTAL EXPANSION
- FIVE MUST-HAVE GADGETS FOR ENGINEERSAs an engineer what sets you apart is what you keep in your toolbox. While knowledge and experience are key to succeeding in engineering, the tools you keep on you can determine your efficiency and effectiveness. Engineers are prone to travelling and working remotely, and hence It is important to… Read more: FIVE MUST-HAVE GADGETS FOR ENGINEERS