Designing a Concrete Beam to Eurocode

Introduction

The subject of this post is the design of reinforced concrete beams to BS EN 1992-1-1- Eurocode 2: Design of Concrete Structures (part1-1): General Rules for Buildings. It covers the design of multi-span beams that have both ‘L’ and ‘T’ cross-section profiles.

The principles of reinforced concrete design have been explained in the last post on concrete slabs design. You are directed to the post prior to reading this post in-order to appreciate the concepts that are lay down within it.

Principles of Concrete Design

A concrete beam is defined as an element whose width is less than 5 times its depth. In all other instances, the element is a slab and therefore must be treated as such. The general procedure to be adopted in designing a concrete beam according to the Eurocode 2 are:

check that the section complies with the requirements for fire resistance
check that cover and concrete quality comply with durability requirements
calculate bending moments and shear forces

calculate reinforcement required for bending and shear
check serviceability requirement

Analysis of Concrete Beams

The moment and shear forces to be used for design can be obtained by considering the beam alone or as been part of a sub-frame and using traditional methods of analysis such as three-moment Clapeyron equations, moment distribution and matrix methods etc.

The Beams should be analyzed for the most unfavorable arrangements of design loads. For beams subjected to predominantly uniformly distributed loading it will be sufficient to consider following arrangements of loads only for ultimate limit state verification.

All span carrying the design permanent and variable actions (1.35Gk + 1.5Qk)
Alternate spans carrying the maximum permanent and variable actions, other spans carrying the design permanent actions 1.35Gk

The above arrangement yields 3 load-cases in a typical continuous beam, which must be analyzed individually in order to obtain the critical moment at sections. The critical moments obtained, however, can be redistributed up to 30% provided that the resulting distribution of moments remains in equilibrium with the applied load.

Where the designer for some reason could not consider above load-cases, the beam can be analyzed for a single load-case of all span loaded with the design permanent and variable actions, the resulting support moments except those at the supports of cantilevers should then be reduced by 20% with a consequential increase in the span moments.

Like concrete slabs, there are a set of coefficients that can be used to determine the shear and bending moments in a beam. Provided the geometry of the beam spans are within 15% of each other and the dead and imposed loads are similar on all spans, 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. Where the above conditions are not satisfied, the beam must be analyzed using any chosen method of analysis.

Design of Concrete Beams

There are three forms of reinforced concrete beam: rectangular, ‘T’ and ‘L’. These are defined in Clause 5.3.2.1 of BS EN 1992-11. In continuous beams it is necessary to calculate the width of the tension flange in a ‘T’ and ‘L’ beam, as defined in Figure 5.2 of BS EN 1992-1-1 and Figure 1. The ‘T’ beam is the most common as it forms part of a downstand to a reinforced concrete slab. The width of the compression flange b_eff is derived using the following equation:

{ b }_{ eff }={ b }_{ w }+\sum { { b }_{ eff,i } } \le b\\

b_w= width of beam web
b_eff,i=0.2b_i+0.1l₀<0.2l_o

b & l_o is as defined in figure 1 & 2

Analysis of the beam in flexure is the same as a one-way slab, however, there are some instances where compression reinforcement is required in beams due to geometric and load application design constraints. In order to determine whether or not compression reinforcement is required in a reinforced concrete beam, the following equation can be employed

{ M }_{ ED }>0.168{ f }_{ ck }b{ d }^{ 2 }\\

The following equation is used to determine the quantity of steel needed in the compression zone

{ A }_{ s2 }=\frac { { M }_{ ED }-0.168{ f }_{ ck }b{ d }^{ 2 } }{ 0.87{ f }_{ yk }(d-{ d }_{ 2 }) }

The reinforcement in the tension zone is given as

{ A }_{ s1 }=\frac { 0.168{ f }_{ ck }b{ d }^{ 2 } }{ 0.87{ f }_{ yk }z } +{ A }_{ s2 }

Where:
M_ed is the applied ultimate bending moment
f_ck is the cylinder compressive strength of concrete
b is the width of the beam
d is the effective depth of the beam f_ykis the characteristic yield strength of steel d₂ is the effective depth to compression steel z is the lever arm

Shear in Concrete

Shear is an important component in concrete beam design as the vast majority of beams require some form of shear reinforcement to be placed within them. Small diameter reinforcement bars (no greater than 16mm in diameter) are bent around the primary reinforcement bars to form a cage. These are called ‘shear links’ and provide the tension component to the concrete beam as it resists shear. BS EN 1992-1-1 uses the ‘variable strut inclination method’ to determine the amount of shear reinforcement required in a concrete beam.

Procedure for Shear Design

Obtain the shear force V_ED and calculate the concrete shear resistance V_RD,using eqn 3.1 If V_Ed<V_RD no shear reinforcement required but provide minimum shear reinforcement else design shear reinforcement.
If shear reinforcement is required, calculate cot θ from eqn 3.2
If cotθ <1 then the beam should be resized, f_ck increased or the load reduced.

If 2.5<cotθ<1 then check reinforcement required at zcotθ from end of beam with shear force at that point.
If cotθ>2.5 then assume cotθ= 2.5 for the calculation of shear reinforcement
Calculate shear reinforcement from equation 3.3

Check minimum shear reinforcement with equation 3.4

{ V }_{ RD }=0.12k(100\rho { f }_{ ck })^{ (1/3) }{ b }_{ w }d\ge 0.035{ k }^{ 2/3 }{ b }_{ w }d——(3.1)\\ \quad \quad \quad where:\quad k=1+\sqrt { \frac { 200 }{ d } } \le 2\quad ;\quad \rho =\frac { { A }_{ s } }{ { b }d } \\

\theta =0.5{ sin }^{ -1 }\left[ \frac { { 5.56V }_{ ED } }{ { b }_{ w\quad }d(1-{ f }_{ ck }/250){ f }_{ ck } } \right] ———(3.2)

\frac { { A }_{ sv } }{ { s }_{ v } } \ge \frac { { V }_{ ED } }{ { f }_{ ywd }zcot\theta } ——————(3.3)

\frac { { A }_{ sw,min } }{ { s }_{ v } } =\frac { 0.08\sqrt { { f }_{ ck } } }{ { f }_{ yk } } —————–(3.4)

The maximum spacing between shear links should be limited to 0.75d

Serviceability and Detailing Requirement

Clause 8.8 of BS EN 1992-1-1 highlights special rules when using 32mm bars and it is not permitted to use 6mm bars as a reinforcement element within concrete as they do not comply with Annex C of BS EN 1992-1-1.

The minimum spacing between bars is one of the following:

The maximum bar size in the element
The maximum aggregate size +5mm

20mm

The maximum spacing is dependent upon the stress σs that is being applied to the reinforcement bar. This can be determined using the following equation:

{ \sigma }_{ s }=\frac { { f }_{ yk } }{ { \gamma }_{ s } } \left[ \frac { { G }_{ k }+{ \psi }_{ 2 }{ Q }_{ k } }{ { 1.35G }_{ k }+1.5Q_{ K } } \right] \left( \frac { { A }_{ s,req } }{ { A }_{ s,prov } } \right) \left( \frac { 1 }{ \delta } \right)

The stress is read against the bar size in Table 2, which describes the maximum distance between bars.

In beams that have an overall depth of more than 750mm, side lacer bars need to be installed in order to prevent cracking as the concrete sets. These bars need to be evenly distributed from the neutral axis to the bottom of the beam. The size of bars should be based on the minimum area of steel reinforcement

N:B Deflection checks as well as the minimum reinforcing areas is the same as with concrete slabs

Worked Example

A continuous central band-beam 7m long, double-equal span is placed at 6m intervals. The beam supports an ultimate line load of 100kN/m2. The beam is to have a fire rating of 1hour, the grade of concrete is C30/37 and the beam is not directly exposed to water. Determine the tension and shear reinforcement required in the beam.

Element size estimation

Beam is mult-span; span=7m ; udl=100kN/m2 therefore d=350mm

Analysis

Since span is approximately equal and beam supports a uniformly distributed loading. Coefficients from Table 1 can be used

Moment\\ \qquad End-span\quad =0.09FL=0.09(100\times 7)\times 7=441kN.m\\ \qquad Interior\quad Support\quad =0.11FL=0.11(100\times 7)\times 7=539kN.m\\ Shear\\ \qquad Interior\quad Support\quad =0.6F=0.6\times (100\times 7)=420kN\\ \qquad Outer\quad Support\quad =0.4F=0.4(100\times 7)=280kN

Cover

{ c }_{ nom }={ c }_{ min }{ +\triangle c }_{ dev }\quad ;\quad { c }_{ min }=max({ c }_{ min,b };{ c }_{ min,dur })\\ { c }_{ min,b }=20mm\quad (assuming\quad 20mm\quad bars)\\ { c }_{ min,dur }=25mm\quad (for\quad class\quad XCI);\quad \triangle { c }_{ dev }=10mm\\ { c }_{ nom }=\quad 25+10=35mm

Flexural Design

End-Span\\ { M }_{ ED }=441kN.m\\ d=h-(\frac { \emptyset }{ 2 } +{ c }_{ nom })=350-(20/2+35)=305mm\\ { b }_{ eff }={ b }_{ w }+{ b }_{ eff,1 }+{ b }_{ eff,2 }<b

{ b }_{ eff,10 }={ b }_{ eff,2 }=0.2b+0.1{ l }_{ 0 }\le 0.2{ l }_{ 0 }\\ b=(6000-1200)/2=2400mm\\ { l }_{ o }=0.85l=0.85\times 7000=5950mm\\ { b }_{ eff,1 }=0.2(2400)+0.1(5950)\le 0.2(5950)={ b }_{ eff,2 }\\ \qquad \quad =1075mm\\ { b }_{ eff }=2400+1075+1075=4550\le 2400\\ \qquad =2400mm

k=\frac { { M }_{ ED } }{ { bd }^{ 2 }{ f }_{ ck } } =\frac { 441\times { 10 }^{ 6 } }{ 2400\times { 305 }^{ 2 }\times 30 } =0.066<0.168\\ z=d\left[ 0.5+\sqrt { 0.25-(\frac { k }{ 1.134 } ) } \right] \le 0.95d\\ z=0.94d=0.94\times 305=286.7mm\\ { A }_{ s }=\frac { { M }_{ ED } }{ 0.87{ f }_{ yk }z } =\frac { 441\times { 10 }^{ 6 } }{ 0.87\times 500\times 286,7 } =3536.08{ mm }^{ 2 }\\ \qquad Try\quad 12H20\quad bars\quad ({ A }_{ s,prov }=3768{ mm }^{ 2 })\\

Interior-Support\\ { M }_{ ED }=539kN.m\\ d=h-(\frac { \emptyset }{ 2 } +{ c }_{ nom })=350-(20/2+35)=305mm\\ k=\frac { { M }_{ ED } }{ { bd }^{ 2 }{ f }_{ ck } } =\frac { 539\times { 10 }^{ 6 } }{ 2400\times { 305 }^{ 2 }\times 30 } =0.080<0.168\\ z=d\left[ 0.5+\sqrt { 0.25-(\frac { k }{ 1.134 } ) } \right] \le 0.95d\\ z=0.92d=0.92\times 305=280.6mm\\ { A }_{ s }=\frac { { M }_{ ED } }{ 0.87{ f }_{ yk }z } =\frac { 441\times { 10 }^{ 6 } }{ 0.87\times 500\times 280.6 } =4415.83{ mm }^{ 2 }\\ \qquad Try\quad 16H20\quad bars\quad ({ A }_{ s,prov }=5024{ mm }^{ 2 })\\

Shear Design

Shear\quad is\quad critical\quad @\quad Interior\quad support\\ { V }_{ ED }=420kN\\ { V }_{ RD }=0.12k(100{ f }_{ ck })^{ \frac { 1 }{ 3 } }{ b }_{ w }d>0.035{ k }^{ \frac { 3 }{ 2 } }\sqrt { { f }_{ ck } } { b }_{ w }d\\ \rho =\frac { { A }_{ s } }{ bd } =\frac { 3768 }{ 2400\times 305 } =5.15\times 10^{ -3 }\\ k=1+\sqrt { \frac { 200 }{ d } } =1+\sqrt { \frac { 200 }{ 305 } } =1.8<2\\ { V }_{ RD }=0.12\cdot 1.8\cdot (100\times 5.15\times { 10 }^{ -3 }\times 30)^{ \frac { 1 }{ 3 } }\cdot 2400\times 305\\ \qquad \quad >0.035\cdot 1.8^{ \frac { 3 }{ 2 } }\cdot \sqrt { 30 } \times 2400\times 305\\ \qquad =393.8kN(shear\quad reinforcement\quad required)\\ \theta =0.5{ sin }^{ -1 })\frac { 5.56{ V }_{ ED } }{ { b }_{ w }d(1-{ { f }_{ ck } }/_{ 250 }){ f }_{ ck } } =0.5{ sin }^{ -1 }\frac { 5.56\cdot 420\times { 10 }^{ 3 } }{ 2400\cdot 305(1-30/250)30 } =3.4\\ cot\theta =16.83>2.5\quad therefore\quad take\quad cot\theta =2.5\\ \frac { { A }_{ sw } }{ { s }_{ v } } \ge \frac { { V }_{ ED } }{ z{ f }_{ ywd }cot\theta } =\frac { 420\times { 10 }^{ 3 } }{ 0.9\times 305\times 500\times 2.5 } =1.224\\ \frac { { A }_{ sw,min } }{ { s }_{ v } } =\frac { 0.08\sqrt { { f }_{ ck } } { b }_{ w } }{ { f }_{ yk } } =\frac { 0.08\times \sqrt { 30 } \times 2400 }{ 500 } =2.10\\ maximum\quad spacing\quad =0.75d=0.75\times 305=228.75mm\\ Try\quad 12H8legs-200mm\quad both\quad ways\quad ({ A }_{ sv }/{ s }_{ v }=3.02\\

Deflection is done the same way as concrete slab, also maximum spacing between bars can be checked using Table 2. if need be, its obviously not critical for this design.