# Comparison Between British Standard and Euro Code in Term of Shear Design

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Comparison between British standard and Euro code in term of shear design

Table of Contents

1.1 Introduction: GENERAL BACKGROUND

1.9 outline of the dissertation:

2.2.1Ultimate limit State (ULS):

2.2.2 Serviceability Limit State (SLS):

2.3 Behaviours in limits states state:

2.3.1Behaviour in service state:

2.3.2Behaviour in ultimate state

2.6Failure in slabs due to shear:

2.7 Intermediate shear failure:

# CHAPTER 1: INTRODUCTION

## 1.1 Introduction: GENERAL BACKGROUND

The structural design of most civil engineering objects (buildings, bridges, etc.) is based on European or international codes of preparation, these codes guide and assist the engineers in the general appraisal of the overall structural scheme, detailed and design.

In many, various countries the British standard 8110 has been used completely with the exception of alteration of nationally determined the limits such as UAE and Kurdistan. An alternative set of codes been assembled to replace the British standard and other European engineering codes in the last decades. Combination of civil engineering codes has been developed to make the Euro code which is been used recently in various and many different countries in the three continents (Africa, Asia and Europe).

BS 8110 is a British code of practice for construction of concrete structures, it was formed and strategized by different technical groups and institutions including the Institution of Civil Engineers, the Steel Reinforcement Commission and the Institution of Structural Engineers. This design scheme based on limit state design. In 2010, it was outdated by the new set of codes of practice Eurocode 2. Eurocode 2.

Eurocode 2 became the chief of all engineering codes of practice for concrete structures in all European Union member countries. Public experts have to apply Eurocode 2 as a lawful method of design on major works. It been considered as one of the most progressive codes of practise in the world.

The Euro codes are new set of developed European structural codes for civil engineering objects and purposes. The Euro codes has presented as united language of European harmonization process of civil engineering department. Eurocode2: design of concrete being employed in the design of concrete structure (Singh, 2015)

The aim of BS8110 and Eurocode2 are to provide common design standards and different procedures to achieve the specified requirements for mechanical resistance and stability including aspects of durability and economy. Furthermore, they provide a general understanding regarding the design of structures between contractors, operators, owners, designers, workers and manufacturing company of the construction material. (nowfor, 2015)

At the present time, the Eurocode2 and BS8110 are being the most used and applied code for design a concrete structures as many countries, worldwide are using the BS8110 and EC2 for designing civil engineering objects, however the Eurocode2 is aiming and intending of becoming the standards and mandatory for all European structures works as it more developed than BS8110 and also more economic.

The steps carried about by the developments of Euro codes have important impacts on British Standard users as considerations must be made in keeping well-informed with developments and technologies in current practices. This paper aims to provide some Shear clauses related to Euro code 2 and BS 8110 code, and some introductory aspects particularly on design principles and the differences

## 1.2 historical background:

During the history, cementing materials have played an important and vital role and used extensively in the ancient world. The old Egyptians (pharaohs) used the gypsum as a cement and the Greeks and Romans used lime which had been made by heating limestone and added sand to make cement, with coarser stones for concrete.

The Romans figured out that a cement created, which set under water and this mixture used for the construction of ports. This special cement made by mixing crushed volcanic ash with lime and later called a “pozzolanic” cement, named after the village of Pozzuoli near Vesuvius. (LAMBERT, 2002)

In places where volcanic ash was rare, such as Britain, crushed brick or tile used instead. The Romans were most probably the first to operate systematically the properties of cementitious materials for specific applications and circumstances.

After the Romans engineers, there was a worldwide loss in structure skills in Europe, mainly with regard to cement. Mortars hardened mainly by carbonation of lime, a slow process. The use of pozzolana revived in the late middle Ages. (LAMBERT, 2002)

The great mediaeval cathedrals, such as Dorham, Lincoln and Rochster in England and Chartres and Rheims in France obviously built by extremely skilled masons. Despite this, it would probably be reasonable to say they did not have the technology to control the properties of cementitious materials in the way the Romans had done a thousand years earlier. (cement, 2012)

The Renaissance and development era carried a new ways of thinking which led to the engineering revolution. In eighteenth century Britaina, the interests of industry and empire matched, with the need to build lighthouses on bare rocks to avoid shipping losses. The continuous loss of trading ships and warships pushed cement technology forwards. (cement, 2012)

John Smeaton, while constructing the third Eddy stone lighthouse (1759) off the coast of Cornwall in South-western England, found that a mix of lime, clay and crushed slag from iron-making formed a mortar which hardened under water. (cement, 2012)

A Portland cement were discovered by joseph Aspdin in 1824. The Portland cement was a combination of finely grounded clay and limestone, he produced it by firing the finely grounded-clay with limestone till it was coincided

Joseph Aspdin was always regards as the inventor of the Portland cement throughout the history, but aspdin cement was not completed, as it was not produced at high-enough temperature to be real forerunner of the current Portland cement

The first modern Portland cement was discovered and created by Isaac Johnson in 1845 a few years later after, the mixture was made by firing a mixture of chalk and clay at much higher temperature than what joseph aspdin used similar to those used today. At these temperature (1400C-1500C), clinkering occurs and minerals form which are very reactive and more strongly cementitious (cement, 2012), and then the reinforced concrete were discovered.

## 1.3 Reinforced concrete:

Concrete is a composite material, containing mainly of Portland cement, water and aggregate (gravel, sand or rock). When these materials mixed, they form an effective paste, which then slowly hardens over time (ICE, 2016).

Concrete is certainly the most important and widely used material in structural designs, concrete is solid durable material which can be formed into various shapes and sizes, its ability of moulding into any shapes from simple rectangular beam or column to slender curved dome or shell makes it one of the most important material in civil engineering (ICE, 2016)

What make the concrete one of the most important material in the civil engineering structure design? the concrete behave almost as an elastic material with nearly full recovery of displacement if the load is removed and also one of the main reason why concrete is the most used material in structure design is that concrete generally increase its strength with age as it shown below in figure 1

Reinforced concrete been used in a varied range of applications such as beam, slab, column, frames and foundations. Mostly the reinforced concrete is used in the areas of concrete that are most likely to be subjected to tension as concrete is weak under tension and significantly stronger under compression , so unreinforced concrete is unsuitable for many structures as it is relativity poor at bearing stress created by vibrations, wind (ICE, 2016)

Figure 1ghetr3http://3.bp.blogspot.com/-IdcRjLGX0Y4/VE_sGtXqY8I/AAAAAAAAAHc/tvaHLb2dBMg/s1600/Achieve-Design-Objectives.jpg

## 1.4 Shear on concrete:

Only a strict and wide knowledge of the structural materials and their behaviour up to failure in the individual structural members allows the engineer involved with reinforced concrete design to become accustomed to the conventional and academic calculations methods correctly for the particular problem.

Accepting the responsibility of designing and executing complex reinforced concrete structures, which fulfil the required safety regulations in all parts without being uneconomic. This idea should also apply to the design of reinforced concrete slabs. Reinforced concrete slabs without shear reinforcement can fail in shear, reinforced concrete slabs without shear reinforcement that are subjected to various loadings can fail in shear distribution. It is important to know the type of behaviour the structure will exhibit (ductile, brittle) and at what load it will fail.

The failure types of reinforced concrete slabs can be categorized as follows:

- Flexural failure: This failure type is related with a ductile behaviour of slabs with sensible reinforcement ratios subjected to uniformly distributed loads. Shear stresses in the slab are usually low and the structure can freely feel the plastic strains without a restriction of its capacity.
- Punching shear failure: is a type of failure of reinforced concrete slabs exposed to high localized forces. In flat slab structures this type of failure occurs at column support points. The failure is due to shear. This type of failure is tragic because no visible signs are shown prior to failure. Punching shear failure disasters have occurred several times in this past decade. This failure mode is linked with the local introduction of focused loads such as columns or wheel loads. It is stiff and therefore unwanted. (punching faliure , n.d.)
- One-way shear failure: This failure kind is associated with line loads and linear supports with distributed loads. It is also stiff and unwanted. The actual behaviour of slabs is more complex however, because of the following aspects:
- A shear or punching shear failures may occur either before or after the yielding of flexural reinforcement. In the latter case, this means that a brittle failure type occurs

The slabs which are supported on a beam or wall don’t require a shear reinforcement because the depth is small and the span therefore equally slender so bending and deflection will nearly govern the design.

The developed shear stresses are small because most of the slabs are subjected to uniformly distributed loads.

Slabs are mostly safe in shear that is why most pf slabs are not designed for shear. The reasons are simple-

- Slabs have way smaller depth than Beams.
- Slabs are more governed on the basis of Deflection and Bending & if they are safe in these they are ultimately result in safe in shear.

Slabs generally designed for deflection, to attain the thickness, thickness based on deflection and re-bars provided for flexure is sufficient to be safe against shear, by which the depth of slab decided. Once the slab is safe, in deflection, it satisfies all the other checks and therefore the slab considered safe for the depth and reinforcement provided in shear. So shear reinforcement not provided

On the other hand,

shear reinforcements can be provided In the form of bent-up bars and also shear links, the first one placed near the support and the second one along the slab on the bottom as the tension happen on the bottom for concrete but most of the time the slab don’t need shear links as the slabs has smaller depth than beam. This bent-up bars and links may have seen everywhere now. Because these shear resistance enforcements will take any slight shear in the member, mostly on beam as the depth is big. (Concrete properties , 2009)

(Concrete properties , 2009)

But, Flat slabs are unlike the other slab types, they are supported on columns only that are nearly point supports, which means that the there is a high concentrated shear force and a small boundary through which the column can punch. Punching shear reinforcement is frequently necessary in such slabs and it can take several forms. One may use links (stirrups), bent up bars and so on to support the slab. That what I am going to show on some of my hand calculations.

## 1.5 Shear reinforcement:

Shear reinforcement or lateral ties are the reinforcement given in the form of hoops or ties. Their main purpose is to keep the main longitudinal reinforcement in place and avoid the shear cracks from mitigating.

## 1.6 Design process:

Design process is a series of steps that engineers follow to come up with strategy to finish the project the design process involves three step:

- conceptual design:

This is the first stage of design process and requires many considerations beyond the calculations associated with the later stages using the principles and procedure

Conceptual design is the first step of the object design process, where drawings and other graphics or models are used. It helps to provide a description of the planned project, in relations of a set of integrated ideas and thoughts about what shall be done, behave and look like in a way that is clear for users. It is the design of interactions, experiences, paths and strategies and is the point at which people, knowledge, ideas, services, processes, and productivity meet vision and limitless possibilities, each acting as a different colour on the canvass of the designer. (Conceptual Design Vs. Detailed Design in Product Development, 2015)

- Preliminary design:

This early calculations stage, as mentioned above, will help create the viability of potential conceptual solutions, and improve their progress and refinement. Initial procedures usually be based on the hand calculations, and may lead into subsequent computer analysis for complex structure (Mosley, et al., 2012)

- Detailed design:

It is at this step that a chosen potential solution will be fully analysed using computer program as appropriate and refined to produce detailed calculations, sketches and other documentation necessary for estimating and construction. (Mosley, et al., 2012)

## 1.7 Aims:

The aim of this project is to compare two different codes of practice, BS 8110 and Eurocode 2, for shear design in slabs by carrying out design calculations and analyse the results to find out the most economical design

## 1.8 objects:

- Show the difference between Euro code 2 and BS8110 in shear check process
- Analysis the result obtained from the calculations
- Explain the main rule of shear force in designing a beam or a slab
- compare BS8110 and EC2 for the design of reinforced concrete slabs and beams for shear
- set up an Excel spread sheet to obtain the area of steel for shear calculated to BS8110 and Eurocode 2
- provide a brief explaining how to use Autodesk Revit and Autodesk Robot to compare the two codes

## 1.9 outline of the dissertation:

This dissertation divided into six chapters and an appendix section. The content of each chapter shown briefly below:

- Chapter 2 introduces literature reviews on the research topic that other authors have published about shear design on slab such as shear on one-way slab, two-way slab and flat slab
- Chapter 3 methodology is the method which been used to compare the both codes in order to provide the best economic method of designing a concrete slab and beam
- Chapter 4 results, it shows the method of hand calculating and it outline the method of setting up excel spread sheet
- Chapter 5 discussion, this chapter will compare the different results obtained from the chapter 4 to provide the best way of designing concrete slab and beam
- Chapter 6 gives conclusion and recommendation based on the outcome of the research results in Chapter 5

# CHAPTER 2: LITERAL REIVEW:

** **

## 2.1 introduction:

Shear strength is the strength of a material or an element, against the type of force or structural failure where the material fails in shear. A shear load is a force that tends to produce a sliding failure on a material along a beam that is parallel to the direction of the force. When a concrete beam get cracked with a load, the beam fails in shear as it show on figure 2 (Assleby, 2002).

Figure 2

(wai pay , 2011)

A shear stress between two planes occurs when a force pushes the plane along the same plane as the face of the object against another object that being pushed in the opposite direction

A shear stress inside an object will occur when a force parallel to the plane causes one plane of the material to want to slide against another. Shown in figure 3 below

Figure 3

Shear force is the force in the beam-acting vertical to its longitudinal (x) axis. For design purposes, the beam’s capability to resist shear force is more important than its ability to resist an axial as the shear failure is the one thing that cannot be happening in the beam (Singh, 2015)

(Singh, 2015)

## 2.2 Limit state design:

According to Mosely researches on his book (reinforced concrete) the design method used in structural engineering. The method is in fact a modernization and rationalization of engineering knowledge, which was well established prior to the adoption of LSD. Beyond the concept of a limit state, LSD simply entails the application of statistics to determine the level of safety required by or during the design process. (Mosley, et al., 2012)

### 2.2.1Ultimate limit State (ULS):

Professor k Beyer investigated about the limits states and punching shear on slabs on his article to accomplish the ultimate limit state, the structure must not collapse or fail when subjected to the highest design load for which it was designed. A structure is believed to please the ultimate limit state criteria if all factored bending, shear and tensile or compressive stresses are beneath the factored resistance designed for the section under consideration. (BEYER, 2012)

Whereas Magnification Factor is used for the loads, and Reduction Factor for the resistance of members. The limit state criteria can also be fixed in terms of stress rather than load. Thus the structural element being examined (e.g. abeam or a column or other load bearing element, such as walls) is shown to be safe when the factored “Magnified “loads are less than their factored “Reduced” resistance.

### 2.2.2Serviceability Limit State (SLS):

Dr R Hules get to the point on his book (reinforced concrete): To please the serviceability limit state criteria, a structure must remain useful and functional for its intended use subject to routine loading, and as such the structure must not cause tenant discomfort under routine circumstances. A structure is believed to satisfy the serviceability limit state when the essential elements do not deflect by more than certain limits placed down in the building codes, the floors fall within scheduled vibration criteria, in addition to other possible requirements as required by the applicable building code (Mosley, et al., 2012)

## 2.3 Behaviours in limits states state:

There are two limits state effecting the shear design, which may cause cracking on the slab:

### 2.3.1Behaviour in service state:

The behaviour through the service life simply resembles the moment-curvature relation of state 1 and 2 in Figure 4. The reaction is linear during the un-cracked state and the reinforcement has low impact on the performance. Compared to state 2, the sections have high stiffness which is specified by the slope of the graph. The first crack increase when the cracking moment, *Mcr*, is reached. This direct to a sudden damage of stiffness in the cracked area, which can be inferred by the decreased inclination of the moment-curvature graph. The sudden change of stiffness due to cracking beside the member will provide a raise to stress redistributions, load is transferred to stiffer un-cracked areas and the linearity is lost

*Figure 4 Moment-curvature relationship for a small reinforced concrete region. Adapted from Engström et al. (2008).*

### 2.3.2Behaviour in ultimate state

Dr Robert and poja shams day done some researches on how the limit state can behave in concrete structures he get to point which: The ultimate state begins when one or both of the materials start to behave non-linearly. The illustration in Figure 5 shows that the ultimate state starts when the graph begins to deviate from the linear dotted line. In the sections where yielding takes place stiffness is lost significantly. A similar process of stress redistribution as in the service state takes place. Load transferred from yielded sections to stiffer areas, mostly those with high reinforcement amount. This process called plastic redistribution and requires that the member can withstand the deformations associated with it. These deformations referred to as plastic- rotations or deformations.

In the ultimate state, behavioural differences can be observed between structures with high versus low reinforcement amounts. If the member is over-reinforced, it will ex-habit a brittle behaviour, while an under-reinforced member shows high ductility. The same applies for sections within the same member with different reinforcement amounts. Brittle failures occur very sudden and sometimes in an explosion like man-near and should therefore be avoided. On the contrary, ductile failures happen a

Large amounts of deformation, which is a desired property since the deformation act as a warning.

In an over-reinforced section, the concrete in compression will crush before any yielding starts in the flexural reinforcement. It is crushing of the concrete that gives the brittle property. In an under-reinforced section, the flexural reinforcement will start to yield before the concretes compressive strength reached. This gives the characteristic plateau in the right diagram in Figure 5. The length of the plateau de-scribes the plastic rotation capacity (robert, 2012)

*Figure 5 Sectional response of an over-reinforced section (brittle) to the left and under-reinforced section (ductile) to the right. Adapted from Engström et al. (2008).*

## 2.4 Shear in concrete beams:

The deformation of an elastic beam with constant stiffness along its length commonly described by the differential equation according to Bernoulli’s beam theory in equation (1). The bending stiffness EI, times the forth order derivative of the vertical deflection w, equals the distributed load q, see Figure 4. Based on this equation, the bending moment and shear force in a beam are generally expressed as in equations (2) and (3).

EId4wdx4=q……….. (1)

M=-EId2wdx2………..(2)

V=-EId3wdx3………..(3)

Figure 6 Deflected beam according to Bernoulli theory.

It can be seen that the shear force is the initial order derivative of the bending moment. In a situation when a simply supported beam subjected to a uniformly distributed load, the moment and shear force will differ according to Figure 7.

Figure 7 bending and shear force in a simply supported beam

The moment and shear force above cause a crack form in a reinforced concrete beam as shown in Figure 8. The cracks in the beam mid-span caused by the moment while the surface cracks mostly caused by the shear. The inclination of the cracks directs that they subjected to shear. This is because shear gives rise to rotation of major stresses. (Cause of cracking , 2009)

Figure 8 Cracking in a concrete beam due to moment and shear.

## 2.5 Shear failure in concrete beams:

Beams must have a suitable safety margin against other types of failure, some of which May be more risky than flexural failure. Shear failure of reinforced concrete, more properly called “diagonal tension failure” is one example

If a beam without well-designed shear reinforcement is loaded to failure, shear failure is most likely to occur suddenly with no advance warnings, which called (brittle failure).

Therefore, concrete have to be provided by “the right shear reinforcement” to insure flexural failure would occur before shear failure. In different way, we want to make sure that beam will fail in a ductile manner and in flexure not in shear manners.

Shear failure of reinforced concrete beam, more properly called “diagonal tension failure “is difficult to predict accurately. In spite of many years of experimental researches and the use of highly sophisticated computational tools, it is not fully understood

If a beam without properly designed for shear reinforcement is over loaded to failure, shear failure is likely to occur suddenly. (Jmes, 2005)

A lot of experiments been done on beam, a concrete beam with no shear reinforcements provided will most likely fail and crack near the highest shear region which is near the support. (wai pay , 2011)

Cracked Beam without any shear reinforcement

- Force resulting from aggregate interlock at crack.
- Concrete shear stress in compression zone
- Dowel shear from longitudinal flexural reinforcement

Shear failure can be stop if the right steps been taken by doing right calculation for the shear reinforcement. Shear reinforcements comes in couple of configurations:

- Vertical stirrups, also called “ties” or “hoops”:

Stirrups are provided to resist the shear stresses at the edges of the concrete beam. They set vertically for ease of installation.

Stirrups means the rings that used to tie and connect the top and bottom reinforcements in a beam. Stirrups plays important rule of holding the top and bottom reinforcement in a beam in position before concreting and later provide resistance to shear after the concrete beam is set.

The stirrups would have been somehow more efficient of they were installed perpendicular to the angle of shear cracks that spread at the ends of the beam

- Inclined stirrups :

Inclined stirrups are vertical stirrups but been slightly inclined to resist principal tensile force which act on inclined direction, stirrups generally provided in beams because the principal tensile force acts in an inclined direction. Therefore, if you incline the vertical shear reinforcement along the direction of the principal tensile force, it is the effective method to resist the tensile force. Even though, cracks will easily develop on the beam.

- Bend up bars :

Bending tension occur in the bottom of the beam for this reason bent up bar used in the beam, bent up bar resist the tension which happen in the bottom of the beam. Bending moment is small near the span end this why beam require fewer bars near the end of the span.

However, diagonal tension in the beam caused by the shear forces which is large at the span ends ,This area is where the inclined portion of the bent-up bar is placed to resist the diagonal tension due to shear. The bent-up portion of reinforcing bars for continuous beams continues across the midway supports to resist top tension in the support area. When the bent-up bars cannot resist all of the diagonal tension, stirrups (called U Shape bars) added. Because of the tensile stress on the stirrups, they pass under the bottom bar and are inclined or perpendicular to it to prevent slippage.

### 2.4.1 Shear in concrete slab:

according to blaawenddraad usually slabs lay under the category of thin plates since the plate is used to carry out of plane loads and the extent to thickness ratio is larger than 5,0. Skinny plates expected not to have any shear twists.

### 2.4.2 One-way and two-way slab:

When it comes to designing concrete slabs it is necessary to differentiate between one- and two-way acting slabs, show in Figure 9. This depends on the support circumstances and indicate to the direction that the forces and moments mostly transferred in. Slabs with two-way action need a bending reinforcement in two directions, while slabs with one-way action act similar to beams and require bending reinforcement in essentially one direction

Figure 9 One-way slab on the left and two-way slab on the right

### 2.4.3: Failure in slabs or beams due to shear:

Usually, a failure mode produced by shear considered as one of the two following types. The first failure mode called shear failure and the second failure mode called punching shear failure. These two forms of failure can be label as mode 1 and mode 2 in Figure 10.

- Mode one: One-way shear failures are mostly associated with distributed or line loads and linear supports such as (walls or the webs of a bridge girder). This failure includes a straight crack that will occur parallel to the direction of the support. Simply a slab without shear reinforcement need to be tested against one-way shear failure if this form of flow of inner forces is present, this failure mode is usually brittle and can occur without any signs of a coming collapse

- Mr Norman investigated about Mode 2:

Figure 10 The modes of failure due to shear. 1)Mode one on the left Shear failure caused by one-way shear

2) mode two on the right Punching shear failure caused by two-way shear

Design codes usually separate these two modes from each other and give different ways of calculating the capacities. In Euro code 2, no exact capacity given for one-way shear in slabs. Instead, the slab is supposed to be designed and calculated in the same way as a beam. In reality the shape of the failure mode is complex and may for that reason become something in between a straight and a circular crack, causing a shear capacity that does not settle with the capacities from design codes.

Regularly when designing concrete slabs, the designer needs to avoid diagonal reinforcement. This is specifically true when designing bridge decks where the loads are always in motion and the use of stirrups in the entire slab is unreasonable. A common solution to increase the shear capacity and avoid using stirrups is to increase the thickness of the slab.

## 2.5 Failure on slab and beams due to punching shear:

XUESONG ZHANG investigated about the punching shear that mostly affect the concrete beams and it might result in collapsing he got the following explanations: Shear failure of reinforced concrete beams is usually sudden, occur without sufficient advanced warning. This type of shear failure considered high-risk type of failure. The cost and safety of shear reinforcement in reinforced concrete beams led to study and absorb other alternatives. (ZHANG, 2002)

A concrete failure happens due to lack in shear resistance, is the most serious type of failure because shear failures are preceded by slight, if any bend or cracking to give early warning. Simply, the main problem here is that most of engineers do not know whether- and to what range- reinforcement acts together and simultaneously with the concrete in resisting shear. Neither do they know how the torsional stress being produced by non-symmetrical loading influences the maximum shearing stress to be (Feld, 1964)resisted. However, mostly the main cause to that all shear failures result from:

- lack of knowledge of what requires resistance;
- not properly applying all available data to the problem
- not providing the correct concrete strength or dimension
- Not providing the right amount of reinforcement or positioning it properly. (Feld, 1964)

## 2.6 BS8110 and EC2:

- Designing a concrete beam and slab to BS8110 is depended on limit-state design. This method provides suitable possibilities that they will not reach or pass the limit state, which makes the structure unsuitable for the purpose it designed for. BS8110 is based on cubic strength of concrete (35MPA) so its covers less concrete grades than EC2.The partial safety factor for the design of concrete structure is a bit different from the partial safety factor of EC2 as 1.4 for dead load and 1.6 for imposed load (1.4GK +1.6QK). In addition, partial safety factor for material given as 1.05. The code recommends using 460 N/mm
^{2}for yield strength of steel (fy). (slabe, 2104) - According to Mosely who done many researches about concrete beam, the designing of concrete beam and slab to Euro code 2 is similar to BS8110. It based on limit-state design. One of the two differences is that EC2 based on characteristic cylinder strength and it covers higher concrete grades than Bs8110, which is up to C90/105, which makes the EC2 more reliable than Bs8110. The partial factor for steel reinforcement is 1.15. However, the characteristic yield strength of steel (fy) that meets the requirements will be 500N/mm2, so overall the effect is insignificant. The partial safety factor is given as 1.35 for dead load and 1.5 for imposed load (1.35GK+1.5QK) (Mosley, et al., 2012)

### 2.6.1 Shear to EC2:

The design of shear elements to EC2 is very similar to that of BS8110. EC2 does not give a specific guidance, but more the general principles to be applied. The general procedure of designing a concrete beam or slab to EC2 include number of steps that need to be followed carefully, the design load (w) need to be found using (1.35GK+1.5QK).1m and also bending moment, shear force, k, As etc. The only differences are that in EC2 the design load will be different as the partial safety factor is different, also the steel check will be different as the yield strength of the steel is different from BS8110 and concrete resistance will be different too as EC2 use Fck=30 N/MM2

### 2.6.2 Shear to BS8110:

Designing a concrete slab to BS8110 is similar to EC2, only a few small differences occur during the procedure. The general procedure is the same as EC2, again the design load (w) need to be found using (1.4GK+1.6QK).1m and bending moment, shear force, k, As etc. The only differences are that in Bs8110, the design load will be different as the partial safety factor is different, also the steel check will be different as the yield strength of the steel is different from EC2 and concrete resistance will be different too as Bs8110 use Fck=38 N/MM2

# CHAPTER 3: Methodology

## 3.0 INTRODUCTION:

This project compares shear member design for BS 8110 and EC2 based on 30 different cases generated by Microsoft Excel’s spreadsheets. The goal is to discover differences of these two codes, also to investigate which code carries more benefits in designing and which code is economic and efficient, and to provide more knowledge about EC 2 and BS8110 to engineers.

The 30 cases different from each other by span lengths and imposed load. Generally, the imposed load stays within the range of 5 to 10 kN/m2, while the span length of beams varies from 4 to 7 meters. Both codes been used in the spreadsheet. The spread sheet contains 8 tables (2 tables for EC2 and 2 tables for BS8110) each table contain 4 cases

- Table 1 same load but different lengths
- Table 2 same length but different loads

Figure 11 example of excel spread sheet of 8 case for BS8110 cases

After obtaining the concrete beam and slab design results of 4 tables (30 cases), graphs will be produced to compares the results of BS 8110 and EC2. The value of bending moment and shear force are always lower for EC2 due to the partial safety factors, which are lower than the one used by BS 8110. Material safety factor for Bs8110 is lower as well.

## 3.1 Excel spread sheet process:

The process of setting up Excel spreadsheet to calculate and design concrete slab in term of shear check. Two codes of practice were used during this process.

### 3.1.2 The benefit of using Excel:

Excel process of calculations is the finest method of obtaining accurate, efficient and quick calculations compared to past where are the calculations were done manually without the present technology. For the purpose of saving time and accuracy Microsoft excel was used as spread sheet to calculate shear design checks values using different load and slab lengths. It is the most suitable method to present, arrange and analyse the data by means of tables and graphs. Thanks Dan brickling (Microsoft excel spread sheet inventor)

### 3.1.3 Excel formulas calculator set up:

Formulas calculator is two sided swords could be helpful by making the calculations procedure easier and provide accurate results and it could mess up the whole calculation when making a tiny mistake during the procedure of entering the equation

The user should be cautious when entering the formula as a small mistake could provide wrong data, which will differently affect the project outcomes

Formulas calculator allows a user to make calculation. The flow chart show the formulas entered into Excel spreadsheet for the calculation shear check to BS8110 and EC2.

It is advisable to calculate the terms twice and calculate one of the bracts manually using hand calculator. By doing this step, area of error and miscalculation can be avoided. Figure 12 shows solution proposal by Excel when error occurs during the process of entering the formula. Before accepting the proposed correction, it should be carefully checked if the correction is the same as what the user want it to be.

Figure 12 Microsoft excel formula correction proposal

Example: The way of entering the formula

Normal calculator |
Excel calculator |

2×2=4 |
2*2=4 |

2/2=1 |
(2)/(2)=1 |

2^{2}=4 |
2^2=4 |

(2×2)2^{2}/2 |
(2*2)2^2/2 |

The full Excel spreadsheet results for this project shown in the appendix. The results will be presented on graphs and will be discussed in chapter 6

## 3.2 Hand calculations:

In order to compare the BS8110 and EC2 more accurately and to correct the answers on the excel spreadsheet if there is a mistake, set of hand calculations been provided to ensure the correction of the excel spread sheet answers as a tiny mistake could happen so easily while using excel calculator.

Designing a concrete slab or beam in term of shear, calculated using both BS8110 and EC2. The process of designing concrete slab using EC2 and Bs8110 was divided into set of steps which made the calculation process so understandable and useful according to Stephen vary very useful book (the design of concrete structures) (Vary, 2010) the steps shown below:

Slab and beam both have the same process but mostly the slab do not need shear links if it does then it is a poor design.

- Find imposed load and dead load
- Calculated the design load (W)
- Calculate Shear force and MED
- Calculate concrete resistance (flexural check) Mrd)
- Calculate As required (area of steel) as it will be used to find Ved
- Calculate Ѵrd and Ved
- Compare Vrd and Ved
- If Vrd<Ved then shear links need to be designed
- Design minimum links Vmin
- Design maximum links Vmax

Two important factors every civil engineer should know about are Imposed load, dead load shear force and bending as any tiny miscalculation of any of the factors above could cause a tragic failure of a structural member, which might cause death.

### 3.2.1 Imposed and dead load:

- Dead load: The dead load includes loads that cannot be moved, including the weight of the structure itself (concrete weight), and fixed fixtures such as walls, plasterboard, carpet, kitchen fittings etc. The roof also considers as dead load. Dead loads are also known as permanent or static loads. Building materials are not dead loads until assembled in position, which it cannot move. (Lauri, 2012)

- Imposed load: refer to loads that do, or can, change it position over time, such as people moving through the building (occupancy or tenants) or movable stuffs such as a pizza on the kitchen deck which can disappear. In addition to live loads, what is known, as environmental loads are loads that are created naturally by the environment and include wind, snow, seismic and lateral soil pressures? (Lauri, 2012)

### 3.2.2 Shear force and bending moment:

- Shear force is a force that acts on a structural member such as (beam, slab) in a direction perpendicular to the extension of the object. For example, a person pushing down on a table. Shear stress occurs when there is a shear force on the member, which might eventually crack or break the member. Shear stress should be taken into account during the stage of designing the structural members; shear forces tend to result in terrible failure before a structure shows any signs of bending or twisting.

- Bending moment is a reaction produced by external force or a moment applied to the member causing bend or twist. Beam and slab are the most and simplest member that maybe be subjected to bending moment.

## 3.3 Revit and robot analysis:

Robot and Revit been used during the method of calculations as those programs consider as some of the most professional and useful programs for Civil , architects and mechanical engineers in the field of structural objects analysis such as ( analyzing beams , columns and other structural members ) and also those software’s provide the possibility of designing accurate 3D building

### 3.3.1 Autodesk Revit:

Autodesk software developers invented a program might replace Autodesk AutoCAD in the near future. Autodesk Revit provide the structural, civil and designers with a software that could manage a number of several engineering tasks such as (3D modelling building which is useful for Civil and architects engineers, it’s also a way easier and affective than AutoCAD and the Autodesk Revit could complete other types of task ( plumbing , electrical works and piping ),

In addition, Revit is 4D BIM capable with tools to plan and track several periods in the building’s lifecycle, from thoughts to construction and later destruction.

### 3.3.2 Autodesk Robot:

Robot Structural Analysis Professional software provides Civil engineers with advanced and developed BIM-integrated analysis and design tools to help engineers understand the behaviour of any structure type and objects , all the forces on the members and verify code compliance. Robot analysis examine the behaviour of the member when it subjected to forces and provide shear, moment and deflection etc. graphs

### 3.3.3 Setting up the model instructions:

Numbers of steps been followed in order to accomplish the first stage of designing a concrete slab, columns, beam , levels and grid lines been created in order to build a slab for analysing :

The steps as follows:

- Setting the grid lines to get accurate structural lines to organise and place the beams, column’s and slabs: go to structure window and then chose grid lines.

- Setting the levels as the columns and beam will not be create unless the level are set: go to structure window and then select level.

- Place the columns and beams using the structure window :

- Apply the load on the slab using the boundary and analysis window : firstly the view should be on analytical 3D view , secondly the boundary condition should be applied as it shown below :

- The last step of the Revit software is to apply Udl on the slab: dead and imposed load should be applied using the area load

- Transfer the Revit drawing to robot for structural member’s analysis : go to analysis option and then chose robot analysis on the right as it shown below

note* robot software should be lunched while transferring |*

- The design will transfer to robot within seconds depending on the size of the profile, the only step left is to apply the load combination for BS8110 and Eurcode 2 : firstly, go to option window and select loads after, a number of options will appear on the right hand side , chose load combination as it shown below :

## 3.4 summary:

This chapter presented the process of setting up Microsoft Excel spreadsheet, how to benefit from Autodesk robot and Autodesk Revit to obtain result for shear in beam or slab. The steps of setting up Excel spread sheet and Autodesk programs mentioned briefly.

# Chapter 4 the method of hand calculation and Result

## 4.1 introduction:

This chapter compares the formulas for designing reinforced concrete beam and slabs in term of shear to BS8110 and EC2. This study or chapter base on comparing the both codes and is design led. The method of hand calculation is following by use of comparative calculation examples.

In this dissertation, shear values calculated to BS8110 and EC2. The equations used for hand calculation of shear presented.

## 4.2 Design of concrete beam or slab to EC2:

Number of steps would be follow carefully during the design of concrete slab in shear to EC2.

### 4.2.1 Imposed and Dead load:

The live load (QK) of a building can be determined using a table, which shows the different type of building imposed load for example three stores residential building hold 2.5 Kn/m2 imposed load tables are given on this website and different other websites

http://www.bd.gov.hk/english/documents/code/DIL2011e.pdf

The dead load or permanent load (GK) could be calculated using: depth of the section * unit load of concrete (25 KN/m^{2})

The depth of the section can be taken as 150 +. In this dissertation 225mm been used and the unit weight of concrete in EC2 is always 25 KN/M^{2}

^{ }

^{ }

### 4.2.2 The Design load (w):

This step includes the one of the two differences between EC 2 and Bs8110:

Calculating w using the following equation 4:

(1.35GK+1.5 QK

)*1m

Where GK is the dead load and Qk is the imposed load

### 4.2.3 Shear Force and bending moment:

Shear force can be measured easily using equation 5

WL2

….5

Bending moment can be calculate using the following equation 6

WL28

…6

Where W is design load and L is the length of the slab

### 4.2.4 Area of steel (As required):

Area of steel can be calculated using equation 7: AS=

MED0.87 fyk z

….7 (Vary, 2010)

Z can be find using the following equation 8: Z=

d0.5+ 0.25- k1.134

… (Vary, 2010)

Alternatively, using the Figure 13:

Figure 13 Z/d vs K to find Z

Where d is the effective depth d=H-cover – assumed diameter of the bar

Where Med is bending moment and Fyk is the yield strength of the steel which can be found using the table in Figure 14 for Fck= 30Mpa therefore Fyk =500Mpa

Figure 14 strength classes for concrete

### 4.2.5 VEd and Vrd:

The equation used to find Ѵed =

VedBw.d

…9 (Vary, 2010)

Vrd can be calculated using equation Ѵrd.c=

100 . Asb.d……..10

and then apply the (Vary, 2010)answers to the table in figure 15 to find Vrd

Figure 15 Ultimate shear stress Vrd

Where Ѵed is the shear stress applied to the member and Ѵrd.c is ultimate shear stress resistance

Where b is the breadth and d is the effective depth of the section

- Other way of investigating the shear resistance using the equation below :

### 4.2.6 Shear links design:

Shear links mostly need on beam as the beam takes more load so links require to connect the reinforcement to resist tension. Links require when Vrd<Ved, there is two type of links maximum links and minimum links the following formula shows how to design shear links:

For minimum links: =0.15 ∗ ∗ ( ) ∗√ (Vary, 2010)

Where D is the effective depth

Bw is the section breath

Designing minimum links using this equation:

ASvSV

= 0.08 * BW.

FCKFyk

(Vary, 2010)

Where Sv is the assumed spacing which should be smaller than 0.75*d (Vary, 2010)

For maximum links:

ASw=S * Ѵed *bw0.87*fyk*cotQ

Where S is assumed spacing and cotQ is the angle, which can be find using FCk table

(Vary, 2010)

## 4.3 Designing concrete beam and slab to BS8110:

Designing concrete slab using British standard 8110 is quite similar to Euro code 2. There are a couple of different steps between the two codes procedures but the rest are the same.

*Only the different steps would be mentioned*

### 4.3.1 The Design load (w):

This step is one of the differences between the two codes:

Calculating w using the following equation 4:

(1.4GK+1.6 QK

)*1m

Where GK is the dead load and Qk is the imposed load

### 4.3.2 Area of steel (As required):

This step is also quite similar to EC2, only one tiny difference occurs when calculation Z, in EC 2 its K/1.134 but in Bs8110 its K/0.9 as it shown below:

Area of steel can be calculated using equation 7: AS=

MED0.87 fyk z

….7

Z can be find using the following equation 8: Z=

d0.5+ 0.25- k0.9

…8

### 4.3.3 VC and VED:

Finding the shear stress resistance (Vc) and the design shear stress (n) is bit similar to EC2 process, n can be find using the following equation from Bs8110 book:

n=vedbd

…9

Vc can be find using the table from the following equation from the Bs8110 book:

100 Asbd

…10

### 4.3.4 Shear links design:

When V or n > Vc then shear links require in order to help the beam or slab resist tension. Shear links can be designed to Bs8110 using the following process:

There are couple of cases for example if 0.5 Vc<n<(0.4+vc) then use the following equation

ASvsv=0.4 bv0.87 fyv–from bs8110 book

In addition, the last step is to choose the suitable bar with spacing, which shall not be bigger, than 0.75*d from the table below:

## 4.4 summary:

This chapter presented the steps been taken in order to obtain the result of the hand calculation. This chapter shows the stages of the hand calculation briefly.

# Chapter 5 Discussion:

This chapter will shows and provide an explaining to a different graphs comparing bs8110 and ec2 and provide an explanation of the differences between BS8110 and EC2

## 5.1 factors difference between Bs8110 and EC2 comparison:

A numbers of difference been noticed during the process of designing a concrete beam and slab, some of the differences are critical differences and some are less important differences on my opinion : the differences are explained bellow :

### Grades of concrete:

One of the most noticeable differences between the two codes is concrete grade as EC2 allows aids to be derived from using high strength concretes, which BS 8110 does not. Concrete grade comes in two forms cylinder strength which is roughly 10-20 % less than the corresponding cubic strength. The maximum characteristic cubic strength FCK allowed is 105 N/mm2, which corresponds to a characteristic cylinder strength of 90 N/mm2 for example if cubic strength FCK = 40 N/mm2, the corresponding cylindrical strength FCK = 40 * 0.75 = 30 N/mm2

__Understanding 1__: EC2 allows benefits to move away from using high concrete strength which is aim to save costs but in the other hand Bs8110 aim to be safer as safety is the main factor in construction department.

## Load combination:

One of the main factors which affect the design of a concrete beam to both codes as BS8110 aim to be more protective and safer as it allows a little bit extra load on the beam so the engineer can provide perfect suitable sections as we call this method (worst case scenario) but in the other hand EC2 aim to be more Economic as its safety factor.

BS8110 = 1.4 (GK) + 1.6 (QK)

Eurocode2 = 1.35(GK) + 1.6 (QK)

__Understanding 2__: EC2 considers lower load than BS8110 as it uses 1.35 for dead load, BS8110 uses 1.4 for dead load so this compartment shows clearly that EC2 aims to be more specific which might save time and money but BS8110 aims to be a little bit more costly than EC2 which is a really little more cost but safer

### Material partial safety factors:

BS 8110 and EC2 both use a basic material partial safety factor for concrete of 1.5. Several years’ ago the design in BS 8110:1997 was originally on the basis of Grade 460 steel with a partial safety factor of 1.05. This was amended in 2005 to Grade 500 steel with a partial safety factor of 1.15, bringing it into line with the changes in the reinforcement standards , the material partial safety factor for reinforcing steel in BS 8110 was reduced from1.15 to 1.05.EC2 uses a value of 1.15 although this is subject to National Annex. This is unlikely to have any practical impact however as steel intended to meet the existing yield strength of 460N/mm2assumed by BS 8110 is likely to be able to meet the500N/mm2assumption made by EC2, so that the design yield strength f yd will be virtually identical

### Design values for loading:

In due course these will be given by EC1.The comparisons made in this paper in general consider only the resistance side of the equation although some mention is made of the partial load factors to be used. It is worth noting that a value of 25kN/m3 is taken for the density of normal weight concrete as opposed to the currently assumed value of 23.6 kN/m3

### Summary:

The advent of EC2 as for the other Euro codes will have a big impact on the design of all types of structures. There will be a learning curve associated with gaining familiarity and using the new code. To make this as painless an exercise is possible, the concrete industry in conjunction with Beware producing design aids and information to assist the profession, and can answer detailed queries, by way of answers to frequently asked ques-tins posted on the above website. In general EC2, used in conjunction with the National Annex, is not wildly different from BS 8110 in terms of the design approach. It gives similar answers and offers scope for more economic structures

Explain the differences and couple of graphs

Bs8110 and ec2 compared

Personal decision : I would chose bs8110 over ec2 as it safer , ec2 is more economic but not by much just by 0.09 % so which make no big saving money difference so I would rather go for safer option than little saving option and less safety

Strand

Difference between cubic strength and cylinder strength

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