# Designing a Concrete Slab to Eurocode

### Introduction

The subject of this post is the design of one way spanning concrete slabs to BS EN 1992-1-1 – Eurocode 2: Design of Concrete Structures – Part 1-1: General Rules for Buildings. The design of such elements is very simple to carry out and thus acts as a good introduction to the concept of reinforced concrete to the Eurocode.

### Principles of Concrete Design

Reinforced concrete is a composite material. The strengths of both the concrete and the steel reinforcement cast within it are what make it work as a structural element. Concrete is excellent in compression while steel’s strength lies in its ability to withstand tension. In very simple terms, if either component were removed then the structural performance of the remaining component would be significantly reduced. It is this basic tenet that must be remembered and enforced whenever designing a reinforced concrete element.

Figure 3.5 of BS EN 1992-1-1 Eurocode 2: Design of Concrete Structures – Part 1-1: General Rules for Buildings defines the area of reinforcement As required in a concrete element that is to resist tension due to bending. From this figure it is possible to derive the following equation that can be used to calculate the area of reinforcement, which will resist tension:

{ A }_{ s }=\frac { { M }_{ ED } }{ 0.87{ f }_{ yk }z }
• MED is the applied design bending moment
•  fyk is the tensile capacity of the steel reinforcement
• z is the lever arm distance between the tension and compression stresses within the element

The lever arm can be calculated using the following quadratic equation

z=d\left[ 0.5+\sqrt { \left( 0.25-\frac { K }{ 1.134 } \right) } \right] \le 0.95d\\ K=\frac { M_{ ED } }{ b{ d }^{ 2 }{ f }_{ ck } }
• fck is the cylindrical strength of concrete at 28 days
• d is the effective depth of the slab
• b is the width of the slab

#### Analysis of Concrete Slabs

While single span slabs are relatively straight forward to analyse, the same cannot be said for continuous slabs. They are, by their very nature, statically indeterminate and therefore more complex analysis techniques such as moment distribution are needed.

Thankfully there are a set of coefficients that can be used instead to determine their shear and bending moments in a slab. Provided the geometry of the slab spans are within 15% of each other and the dead and imposed loads are similar on all spans, then the coefficients described in Table 1 can be used. In addition, the imposed load must be less than or equal to the dead load in order for these coefficients to remain valid. Generally, the following conditions should be satisfied

• The area of each bay must be greater than 30 square meters
• The ratio of the characteristic imposed load to the dead loads must be less or equal 1.25
• The characteristic imposed load must be less or equal to 5kN/m2
• The span must be approximately equal (i.e. the spans should be within 15% of each other).

#### Shear in concrete slabs

A concrete slab is defined as an element whose width is more than 5 times its depth. In all other instances, the element is a beam and therefore must be treated as such. Slabs typically have lower magnitudes of shear applied to them than beams with the exception of flat slabs without drops where punching shear stress is sometimes high. As a matter of fact, shear reinforcement is not desired in concrete slabs, most designers, therefore, strive to control the shear stress such that no shear reinforcement is required.

Shear verification will be presented in the next post on beam design as it is not a critical check for concrete slabs as long as the rules on preliminary sizing is adhered to.

#### Serviceability

##### Deflection

As a composite material, it is quite difficult to ascertain the magnitude of deflection for reinforced concrete. It is for this reason that the concept of allowable span/depth ratios has been developed, which reduces the need to carry out complex calculations.

To verify deflection the actual span/effective depth ratio must be less than the permissible l/d ratio:

\frac { l }{ d } permissble=N\times K\times F1\times F2\times F3
• ρ is the ratio of tension reinforcement at mid-span and supports.
• ρ’ is the ratio of compression reinforcement at mid-span and supports.
• ρ0 is the reference reinforcement ratio
##### Minimum Area of Reinforcing

BS EN 1992-1-1 postulated that the minimum area of steel in a reinforced concrete member subjected to flexure should not be less than: As,min

{ A }_{ s,min }=\frac { 0.26{ f }_{ ctm } }{ { f }_{ yk } } { b }_{ t }d\ge 0.013{ b }_{ t }d\\ \\ { f }_{ ctm }=0.3{ f }_{ ck }^{ 2/3 }

The maximum spacing between bars is explained in Clause 7.3.3 of BS EN 1992-1-1. It is dependent upon the stress in the reinforcement, with the trend being the higher the magnitude of stress, the closer the spacing between them

#### Corrosion and Protection

Before any design can be carried out, some parameters concerning the exposure conditions of the concrete element must be established. This is controlled by establishing the amount of concrete cover that is assumed to be in place with respect to the reinforcement within the element. Clause 4.4.1 of BS EN 1992-1-1 explains how the minimum concrete cover is determined

The nominal cover to reinforcement cnom is defined as

{ C }_{ nom }={ C }_{ min }+\triangle { C }_{ dev }
• cmin is the minimum cover to the reinforcement that allows bond forces to be transmitted, prevents the reinforcement from corroding and provides adequate ﬁre resistance
• ΔCdev is an allowance for deviation from the design due to tolerance and is taken to be 10mm.

The value of cmin is taken as the greater of two dimensions. The first the minimum cover due to bond Cmin,b which relates to how the reinforcement is interacting with the concrete. BS-EN 1992-1-1 states that the minimum cover due to bond should not be less than the bar diameter. The second is Cmin,dur is required to protect the reinforcement depending on the projected environmental condition the concrete is going to be exposed to. The exposure class is defined in Table 4.1 of BS EN 1992-1-1 and this is cross-referenced to the structural classification of concrete elements which is found in Table A.4 of BS 8500: 2006 – Concrete Complementary British Standard to BS EN 206-1 – Part 1: Method of specifying and guidance for the specifier. This only applies to the UK where it is recommended within the National Annex to BS EN 1992-1-1, that this table is used.

Fire protection is dependent upon the overall thickness of the element as well as the cover of concrete to the reinforcement. The cover in terms of fire protection is defined as the distance from the centre of the group of bars (or bar) to the outer surface of the concrete element. Table 5.8 of BS EN 1992-1-2 Eurocode 2: Design of concrete structures – Part 1-2: General rules – Structural fire design provides data which can be used to determine the thickness and cover requirement for concrete elements. This is plotted against the overall fire rating of the structure that is being designed.

#### Worked Example

A one way simply supported slab spanning 4.5m is required to support screed, finishes, and an office variable action of 3.0kN/m2 (demountable partitions inclusive). Design the slab completely assuming fck=30mpa and fyk=500mpa.

##### Element size estimation

Single span (span=4.5m). Thickness; h=175mm (conservatively)

##### Flexural Design
{ M }_{ ED }=\frac { { n }_{ s }{ l }^{ 2 } }{ 8 } =\frac { 11.85\times { 4.5 }^{ 2 } }{ 8 } =30kN.m\\ d=h-(\Phi /2+{ c }_{ nom })=175-(12/2+25)=144mm\\ k=\frac { { M }_{ ED } }{ { bd }^{ 2 }{ f }_{ ck } } =\frac { 30\times { 10 }^{ 6 } }{ 1000\times { 144 }^{ 2 }\times 30 } =0.048\\ z=d\left[ 0.5+\sqrt { 0.25-\frac { k }{ 1.134 } } \right] \le \quad 0.95d\\ z=0.95d\quad =0.95\times 144\quad =136.8mm\\ { A }_{ s }=\frac { { M }_{ ED } }{ 0.87{ f }_{ yk }z } =\frac { 30\times { 10 }^{ 6 } }{ 0.87\times 500\times 136.8 } =504{ mm }^{ 2 }\\ Try\quad H12\quad @\quad 200mm\quad c/c\quad ({ A }_{ sprov }=565mm^{ 2 }/m)
##### Deflection
{ \frac { l }{ d } }_{ permissible\quad }=\quad N\times K\times F1\times F2\times F3\\ \rho =\frac { { A }_{ s } }{ bd } =\frac { 565 }{ 1000\times 144 } =3.92\times { 10 }^{ -3 }\\ { \rho }_{ o }=\sqrt { { f }_{ ck } } \cdot { 10 }^{ -3 }=5.48\times { 10 }^{ -3 }\\ { \rho }_{ o }\ge \rho \quad therefore\\ N=\left[ 11+\frac { 1.5\sqrt { { f }_{ ck } } { \rho }_{ o } }{ \rho } +3.2\sqrt { { f }_{ ck } } \left( \frac { { \rho }_{ o } }{ \rho } -1 \right) ^{ \frac { 3 }{ 2 } } \right] \\ \quad =\left[ 11+\frac { 1.5\cdot \sqrt { 30 } \cdot 5.48 }{ 3.92 } +3.2\sqrt { 30 } \left( \frac { 5.48 }{ 3.92 } -1 \right) ^{ \frac { 3 }{ 2 } } \right] \\ \quad \quad =\quad 26.89\\
F1=1.0;\quad F2=1.0\quad since\quad l<7m\\ F3=\frac { 310 }{ { \sigma }_{ s } } <1.5\\ { \sigma }_{ s }=\frac { { f }_{ yk } }{ { \gamma }_{ s } } \left[ \frac { { g }_{ k }+\psi { q }_{ k } }{ 1.25{ g }_{ k }+1.5{ q }_{ k } } \right] \frac { { A }_{ sreq } }{ { A }_{ sprov } } \frac { 1 }{ \delta } \\ \quad \quad =\frac { 500 }{ 1.15 } \left[ \frac { 5.875+0.7(3.0) }{ 11.85 } \right] \frac { 504 }{ 565 } \cdot 1=261mpa\\ F3=\frac { 310 }{ 261 } =1.2<1.5\\ { \frac { l }{ d } }_{ permissible }=26.89\times 1.0\times 1.0\times 1.0\times 1.2=32.07\\ { \frac { l }{ d } }_{ actual\quad }=\frac { 4500 }{ 144 } =31.25\\ since\quad { \frac { l }{ d } }_{ actual }<{ \frac { l }{ d } }_{ permissible }deflection\quad is\quad satisfied
##### Verify maximum spacing

• The Institution of Structural Engineers (2006) Manual for the design of concrete building structures to Eurocode 2 London: The Institution of Structural Engineers
• The Concrete Centre (2009) Worked Examples to Eurocode 2: Volume 1 [Online] Available at: www.concretecentre.com/ pdf/Worked_Example_Extract_Slabs.pdf (Accessed: February 2013)

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