Manual for Advanced Design

8/10/2019 Manual for Advanced Design

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Manual for Advanced Design

Foreword

[2] [overall current page]

It is the authors great pleasure to introduce their first handbook on reinforced concrete design

according to SR EN 1992-1, Eurocode 2: Design of concrete structures (2004), as part of a more complex

project on designing safe, fast casting and cost attractive members and structures.

Above all, this book is for the use of civil engineering students, especially those at their first

contact with structural concrete and its design. Nonetheless, this does not exclude other interested parties

to read, comment, refer to the present work or address suggestions to the authors to better organise the

information provided or improve specific points.

Concrete is, in these authors opinion, one of the oldest and greatest inventions. Still, there are

some aspects regarding this material that may be approached differently, in order to unlock its full

potential, or, in a more general definition, to achieve sustainability.

The first step towards that goal is to proper understand the limits of the material, plain and

(especially) reinforced. By doing so, it is possible to design members and structures that are safe and cost

attractive, avoiding waste of energy, materials and manpower.

That being said, the authors invite the reader to address together the designing process. Simply, get

MAD!

Cluj-Napoca, 12th March 2011


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Content List


Foreword Page [2]

Content List Page [3]

[Chapter One]Theoretical background Page [6]

[Part (One) A]Introduction Page [6]

[Part (One) B]

Milestones for designing in flexure Page [6]

[Section (One-B) a]

Neutral axis Page [7]

[Section (One-B) b]Assumptions Page [7]

[Section (One-B) c]

Stress block parameters Page [8]

[Section (One-B) d]

Singly Reinforced Rectangular Section (SRRS) Page [11]

[Section (One-B) e]Doubly reinforced rectangular section (DRRS) Page [15]

[Section (One-B) f]Tee/flanged section (FS) Page [16]

[Part (One) C]

Milestones for shear design Page [21]

[Section (One-C) a]

Variable angle truss model Page [21]

[Section (One-C) b]Final Advice Page [25]

[Chapter Two]Steps to Design Page [26]

[Part (Two) A]

Citations used Page [26]

[Part (Two) B]Introduction Page [26]

[Part (Two) C]Concrete cover Page [27]

[Part (Two) D]

Sizing of the cross section Page [28]

[Part (Two) E]

Singly reinforced rectangular sections (SRRS) Page [29]

[Part (Two) F]

Doubly reinforced rectangular sections (DRRS) Page [31][Part (Two) G]

Singly reinforced tee/flanged sections (SRFS) Page [33]

[Part (Two) H]

Doubly reinforced tee/flanged sections (DRFS) Page [35]

[Part (Two) I]

Shear design Page [36]

[Chapter Three]Worked Examples Page [46]

[Part (Three) A]

Parameters independent of the cross section Page [47]

[Section (Three-A) a]

Bar size Page [47]

[Section (Three-A) b]Concrete cover Page [47]

[Section (Three-A) c]

Axis distance Page [48]

[Section (Three-A) d]

Supports Page [48]

[Section (Three-A) e]

Effective span Page [49]

[Section (Three-A) f]Partial load calculus (A set) Page [49]

[Part (Three) B]

Singly reinforced rectangular section (SRRS) Page [50]

[Section (Three-B) a]

Sizing of the cross section Page [50]

[Section (Three-B) b]Static analysis Page [52]


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Content List


[Section (Three-B) c]

Area of reinforcement for flexure Page [52]

[Section (Three-B) d]

Longitudinal reinforcement layout Page [52]

[Section (Three-B) e]Area of reinforcement for shear Page [56]

[Section (Three-B) f]Transverse reinforcement layout Page [58]

[Section (Three-B) g]

Anchorage length Page [61]

[Part (Three) C]

Doubly reinforced rectangular section Page [64]

[Section (Three-C) a]Area of reinforcement for flexure Page [64]

[Section (Three-C) b]


[Section (Three-C) c]

Area of reinforcement for shear Page [69]

[Section (Three-C) d]Transverse reinforcement layout Page [71]

[Section (Three-C) e]Anchorage length Page [71]

[Part (Three) D]

Tee singly reinforced section Page [72]

[Section (Three-D) a]

Load calculus (B set) Page [72]

[Section (Three-D) b]Static analysis Page [73]

[Section (Three-D) c]Area of reinforcement for flexure Page [73]

[Section (Three-D) d]


[Section (Three-D) e]Area of reinforcement for shear Page [75]

[Section (Three-D) f]Transverse reinforcement layout Page [77]

[Section (Three-D) g]

Anchorage length Page [80]

[Section (Three-D) h]

Check shear between web and flange theoretical

backgroundPage [82]

[Section (Three-D) i]Check shear between web and flange calculus Page [84]

[Chapter Four]Fast track to design (FTD) Page [86]

[Part (Four) A]

Introduction Page [86]

[Part (Four) B]

Deflection control by calculus Page [86]

[Part (Four) C]Singly reinforced rectangular section (SRRS) Page [87]

[Section (Four-C) a]Sizing of the cross section Page [88]

[Section (Four-C) b]

Static analysis Page [88]

[Section (Four-C) c]

Area of reinforcement for flexure Page [89]

[Section (Four-C) d]


[Part (Four) D]

Area of reinforcement for flexure checked by exact calculus Page [89]

[Section (Four-D) a]


[Part (Four) E]Concluding remarks Page [90]

Appendix Page [91]

[Chapter One]

DURABILITY Page [91]

[Part (One) A]

CEMENT Page [91]


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Content List


[Part (One) B]

EXPOSURE CLASSES Page [94]

[Part (One) C]

CONCRETE COVER Page [98]

[Chapter Two]MATERIALS PROPERTIES Page [100]

[Part (Two) A]CONCRETE Page [100]

[Part (Two) B]

STEEL Page [102]

[Chapter Three]

REINFORCEMENT Page [104]

[Part (Three) A]WELDED WIRE Page [104]

[Part (Three) B]

BARS Page [106]

[Chapter Four]

FIRE RESISTANCE Page [107]

[Part (Four) A]SLABS Page [108]

[Part (Four) B]BEAMS Page [111]

[Chapter Five]

DEPTH-to-SPAN RATIO Page [114]

[Chapter Six]

DESIGN TABLES Page [115]

[Part (Six) A]REINFORCEMENT LAYOUT Page [118]

[Chapter Seven]SHEAR REDUCTIONS Page [119]

References Page [120]

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Manual for Advanced Design Basics in Reinforced Concrete Design [1]

[6] [overall current page] [pages in chapter]: [20]

[Chapter One]

Theoretical Background

[Part (One) A]

Introduction

Knowledge in general and engineering in particular works with models that may be considered to

showcase the following levels of understanding:

< 1 >

The global level, at which one perceives everything as one unitary object (in constructions, a

structure);

< 2 >The system level, at which one perceives the object to be composed of different groups of elements

(in constructions, all the members with similar functions, i.e. the slabs, the beams or the columns);

< 3 >

The element level, at which one perceives a particular component (in constructions, a slab, a beam,

or a column);

< 4 >The macroscopic level, at which one perceives the major structure of that particular component (in

constructions, in the case of reinforced concrete, the cross section of an element as a mix of concrete andreinforcement);

< 5 >

The microscopic level, at which one perceives the properties and interaction of the different

constituents (in constructions, in the case of concrete, the cement, aggregates, water, etc. that form it);

< 6 >The atomic level, at which one perceives the properties and interaction of atoms (in constructions,

in the case of concrete, yet to be established).

By only addressing the macroscopic level (4th) it is possible for the engineer to predict the

behaviour of the superior levels of knowledge (3rd, 2nd and 1st) in limits deemed satisfactory. Of course,

as knowledge progresses it is possible to minimize those limits by use of advanced computer calculations

or a more fundamental approach.

Reinforced concrete is subject to the previous as well, that is why, before all, its relevant to discuss

the designing process from a theoretical point of view, by explaining the models milestones in flexure

and shear.

[Part (One) B]Milestones for designing in flexure

At hand is a very important task, that of providing a needed background for the reader to

comprehend the behaviour of an element under flexure. That is why it is mandatory to start by explaining

what flexure is all about.

In very general terms, flexure is a state of loading in which the same cross section will have

opposite stresses, of tension in one part and in compression for the rest. The transition area in-between is

called the neutral axis (null stress, or better said very close to null stress). Its position on the height of the

cross section is variable, depending on the values opposing stresses reach.

(Space intentionally left blank)


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Other assumptions used in the model are:

Concrete is considered to be on the brink of failure (the so called 3rd state);< 1 >

Bernoullis hypothesis of plane sections and the compatibility of strains in concrete and steel for the< 2 >

same fibre in the cross section;

Hookes Law allows for strains and stresses to be considered a ratio of the elastic properties of the< 3 >

material;

The real stress block for the compressed area of concrete is replaced by a simplified rectangular< 4 >

stress block;

The reinforcement yields prior to the crushing of concrete.< 5 >

[Section (One-B) c]Stress block parameters

In comparison with a uniaxial loading, whether its compression or tension, flexure determines

different fibres on the height of the cross section to be under different stress, not only as per value but alsoas per nature (compression on top of the cross section and tension at the bottom in the case of

gravitational loads). That is why, although flexure can be considered an eccentric compression, the

compressive stress in concrete subjected to flexure is not the same as in pure compression.

First, in pure compression, all fibres are under about the same stress.

NOTE

Testing has shown that there is a reduction in the stress value with the increase of the

distance from the centre of gravity of the sample to the edges (further details are available

in literature).

This is not the case for flexure where eccentricity introduces variation in values, some fibres being

subjected to higher stresses than others. Therefore, different longitudinal layers of fibres have a tendencyto slip from each other.

This is in these authors opinion a positive effect as it will lead to:

< a >An increase in the strains an element can develop due to a decrease in the speed of strain

development over time;



A delayed failure of concrete due to a roll-over mechanism which transmits the stress from

the fibres under the maximum effort to the less loaded fibres closer to the neutral axis.

Second, the longitudinal splitting effect is in opposition with the compression stress which will lead

to a reduction in the amount of stress the most compressed fibres will bear.

NOTE Further details are available in literature.

Calculus model in flexure consists of two ideal forces in perfect equilibrium (see [Figure 1]):

< 1 >A compression force in concrete, cF ;

< 2 >A tension force in the reinforcement, tF;

< 3 >A lever arm in-between, z .

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The resistive flexural capacity for a given cross section is written as:

[1-1]Rd c tM F z F z= = with

[1-2]

c tF F=

cF is the compressive resultant;

z is the lever arm;

tFis the tension resultant.

Since the final form is to be a formula for the area of the reinforcement, one should:

< a >Evaluate the compressive resultant:

Establish a function for the stress variation based on strain values (due to their easiness of

measuring in tests);

< ii >

Establish/find the limit for integration;



Establish its position over the height of the cross section;

< c >Calculate the lever arm.

The function needed for the first operation has the general form:

[1-3]

0

( )cu

cF b

= with

cu is the ultimate compressive strain;

b is the width of the cross section;

c is the stress function of strains.

Since the above is difficult to evaluate precisely, a simplified stress block replacing the real

distribution of stresses while being easier to evaluate has been deemed necessary. The substitution of one

with the other is based on two conditions:

< a >The volume of stresses must the correctly evaluated (very close to equal);

The position of the compressive resultant in the real and simplified diagram must be the same

(this will insure the correct estimation of the resistive capacity in flexure).

The notations used in general with reinforced concrete and their meaning is presented in [Figure 1]and the subsequent list.

NOTEThe cross section is considered to be rectangular both for the shapes simplicity and for

the fact that this particular shape is the most common in constructions.

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Figure 1

Design model parameters

Cross section Strain Distribution Real Stress Block Simplified Stress Block

< 1 >The notation for the height is ( )h and for the width ( )b ;

< 2 >The beam has longitudinal ( )l steel ( )s reinforcement ( ),1slA in tension ( ),T t with a stress

( )ydf equal with the design ( )d strength at yielding ( )y in the lower part ( )1 of the cross section, at

position/axis distance ( )1d from the extreme lower fibre, and reinforcement ( ),2slA in compression

( ),C c with a stress ( )ydf in the upper part ( )2 of it, at the position/axis distance ( )2d from theextreme top fibre;

< 3 >

The position of the reinforcement in tension ( ),1slA from the extreme fibre in compression (i.e., the

top of the section) is the (effective) depth ( )d ;

< 4 >If the stress in the reinforcement ( ),1slA in tension ( ),T t will be less than the one at yielding it

will be named

( )stf ;

< 5 >

If the stress in the reinforcement ( ),2slA in compression ( ), cC will be less than the one at

yielding it will be named ( )scf ;

< 6 >

The maximum stress ( )cdf to develop in every fibre of concrete ( )c is less than the design ( )d

strength due to a reduction factor ( ) given by:

[1-4]

1.00 50

50

1 50 90200

ck

ckck

if f MPa

f

if f MPa

=

< 7 >The height of the simplified stress block is given by the height in compression (position of the

neutral axis) ( )x reduced by a factor ( ) given by:

[1-5]

0.80 50

500.8 50 90

400

ck

ckck

if f MPa

fif f MPa

=

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< 8 > ( ) and ( ) are determined based on the two fore mentioned conditions (the same volume of

stresses and the same position for the resultant).

With this model in mind, eq. [1-2] will be re-written as:

[1-6] ( ) ( )c cdF C x f b = = or

[1-7]

s sl yd F T A f = =

In [eq. 1-6] terms have been grouped to outline the simplified stress block while in [eq. 1-7]

subscripts dependent on the part of the cross section (lower or upper) have been omitted to outline the

formula for the resultant in the reinforcement (whether in tension or compression). In addition, by writing

equilibrium for the horizontal forces, the above [eq. 1-6 &7] give the height in compression as:

[1-8] sl yd

cd

A fT C x

b f

= =

This may be used to calculate the position of the neutral axis ( )x ONLY after the reinforcementhas been calculated.

NOTEThere is NO need to actually calculate the height in compression (further guidance will be

provided on the matter as described in subsequent chapters).

[Section (One-B) d]

Singly Reinforced Rectangular Section (SRRS)

A singly reinforced cross section is the one for which only tension reinforcement is provided and

therefore calculated (herein, the bottom of the cross section).

Other types of reinforcement are possible and are presented in the subsequent parts of the presentmanual. Those are in fact extrapolations of singly reinforcement.

Definitions specific to this type of reinforcement are presented in [Figure 2] with the same meaning

as per [Figure 1].

Figure 2

Singly reinforced rectangular section

Cross section Strain Distribution Real Stress Block Simplified Stress Block


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Ideal design is for Rd EdM M= . Based on this [eq. 1-1] will be re-written as:

[1-9] EdEd

MM T z T

z= =

[1-10],1

1 Edsl yd

MA

f z=

with

ydf is the design tensile (yielding) strength of steel;

z is the lever arm.

The above [eq. 1-9] is a simplified form for designing and checking the actual reinforcement if a

correct value is chosen for the lever arm ( )z . The limitations for this form will be explained in a later

part. Still, the authors present this form here due to its simplicity and invite the reader to keep it in mind.

Another form of the above is:

[1-11] ( ) ( )Ed cdM C z x f b z = = with

[1-12] 0,5z d x=

[Eq. 1-11] may be further re-written should ( )z be substituted as a function of ( )x or vice-versa.

By choosing the first, one can write:

[1-13] ( ) ( ) ( ) ( ) ( )0,5Ed cd cdM x f b d x f b x =

By multiplying each term in the right member with ( )1 written for the first term as ( )d d and for

the second one as ( )2 2d d , with some additional grouping the above becomes:

[1-14] ( ) ( ) ( )

2

2 20,5Ed cdx x

M b d fd d

=

The above form may be further organised as:

[1-15] ( ) ( )2 20,5 = with

[1-16]

2Ed

cd

M

b d f=

and

[1-17]

x

d =

where

is the relative bending moment;

is the relative height in compression;

cdf is the design compressive strength of concrete in compression.

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The above [eq. 1-15] is a form for ( ) that is ONLY dependent on ( ) and therefore ONLY on

the height in compression ( )x (in turn x depends on the quantity of reinforcement and the steel grade as

well as on the sizing of the cross section and the concrete grade, see above [1-8]).

The above [eq. 1-16] is a form for ( ) that is dependent on:

< a > ( )EdM is given by the loading (case, pattern and values);

 ( )b , ( )2d are given by the dimensions of the cross section;

< c > ( )cdf is given by the concrete grade.

Since this form DOES NOT depend on other parameters, it may be asserted that it is an evaluation

of the BENDING CAPACITY of the cross section BASED ONLY ON ITS SIZE and the CONCRETEs

PROPERTIES irrespectively of the area of reinforcement in the cross section.

One short discussion is needed to fully exploit the above. As known, should a reinforced concrete

member fail it MUST FAIL WITH WARNING, visible in the form of large deflections. In other words,

the reinforcement must YIELD (very large strains and relatively no increase in stress).

As per [Figure 2], the variation of strains over the height of the cross section is considered to be

linear between two extremes, one in concrete ( )cu and the other in reinforcement ( )s as a consequence

of Bernoullis Law. By writing proportions for similar triangles (upper and lower), it can be written:

[1-18]1

1 1cu s cu cus

x d

d x x

= = =

A basic assumption is yielding of the reinforcement; this implies that the stress is the design

strength at yielding ( )ydf while the strain is higher than the one at yielding ( )yd :

[1-19]1

1 yd

s yd cu

s

f

E

with

[ ]3.5 cu = for concrete grade ( )50/60C and any stress-strain code curve is the ultimate strain in

concrete;

200 000 [MPa]sE = is the modulus of elasticity for steel.

SR EN 1992-1-1 presents three possible stress-strain curves:

< a >Non-linear, referenced by subscript 1;



Parabolic-rectangular, referenced by subscript 2 and;

< c >

Trapezoid, referenced by subscript 3.

with different values for ( )cu and implicitly for ( )s cuE if the concrete grade ( )50/60C .

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It is acknowledged that the most favourable variation in terms of safety in design is the trapezoid

stress-strain curve. Therefore all values as set here-in are corresponding to this type of stress-strain curve.

From the above ( ) will be extracted as:

[1-20]

1 1 700

1 7001

yd s cu

yds cu yd s cu yd

s cu

f E

fE f E f

E

+ = = + ++

The above sets a maximum limit ( )lim for () as well as a maximum value ( )lim for ( ) , both

dependent on the steel grade ( )ydf .

DO REMEMBER that the above EXTREME VALUES are calculated on the assumption that the

cross section is SRRS. Should one need to calculate the maximum capacity in flexure of a given member

the following applies:

[1-21]

2limRd cdM b d f=

Should ( ) as per [eq. 1-16] be higher than ( ) as per [eq. 1-15] the assumption of singly

reinforcement is void because the cross section for a given load case and pattern cannot provide the

necessary bearing capacity.

Therefore, it is necessary either to:

< a >Change the dimensions of the cross section, or



Provide additional reinforcement in compression and thus design a doubly reinforced

member.

The area of the reinforcement may be calculated as:

[1-22]( ) ( )

cd cd cd

sl

yd yd yd

x f b f fxT C A b d b d

f d f f

= = = =

with

is a tabulated coefficient of mechanical reinforcement dependent on the value of( ) .

The above is better understood should one fill in the corresponding values:

[1-23]sl yd

sl yd cd

cd

A f

A fb fx

d d b d f

= = = =

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[Section (One-B) e]Doubly reinforced rectangular section (DRRS)

If by some reason, generally height restrictions or functionality requirements, the dimensions of a

cross section (particularly the height) must comply with a specified maximum, providing singly

reinforcement may be insufficient in terms of necessary bearing capacity.

One way of providing additional bearing capacity is to place reinforcement in compression.

Still, one should keep in mind that doubly reinforcement is not always the panacea and that

changing (at least) the shape of the cross section, if not the dimensions, is sometimes a better approach.

Changing the shape of the cross section will be presented in the subsequent part, when referring to

tee/flanged sections.

Introducing a change in the placing of the reinforcement, by arranging it both in tension and in

compression, is a valid solution ONLY after using to its full the reinforcement in tension (singly

reinforcement, as presented by [eq. 1-21]).

In other words:

[1-24] ( ) ( )

2 2lim lim lim2

0,5Ed

cd

M

b d f = > =

Definitions specific to this type of reinforcement are presented in [Figure 3] with the same meaning

as per [Figure 1].

Figure 3

Doubly reinforced rectangular section

Cross section Operated Separation Strain Distribution Simplified Stress Block

The cross section is momentarily divided for calculus, see [Figure 3], into the following pieces:

< 1 >A singly reinforced section for which ( )lim = with the reinforcement in tension ( ),1slA to be in

equilibrium with the concrete in compression. This section has a capacity in flexure of:

[1-25] 2,1 limRd cdM b d f=

< 2 >

An area of reinforcement in tension ( ),2slA to be in equilibrium ONLY with another area of

reinforcement in compression. Since the stress is assumed to be ( )ydf for both tension and compressiondue to the actual properties of steel, the area in tension will be equal with the area in compression.


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< 1 >A singly reinforced web section for which the reinforcement in tension ( ),1slA can be easily

calculated (similarly to SRRS);

< 2 >

An area of reinforcement in tension ( ),2slA to be in equilibrium ONLY with the two flanges (to the

left and the right of the web) in compression.

The area of reinforcement to be placed in the cross section is ( ),1 ,2sl slA A+ for tension.

Before presenting the actual reinforcement formula, it is necessary to understand the calculation of

the effective slab width ( )effb a parameter ONLY for flanged sections (for T sections it is simply namedtop width) dependent on:

< a >

The static scheme for the beam to be designed;



The position of this beam in the slab-beam system in place.

Figure 5

Definition of for the calculation of the effective slab width

Inner support Inner support

End Span Inner Span Cantilever

Definitions specific as per [Figure 5] are:

< 1 >

The drawing must be read as referring either to an end span ( )1l , an interior span ( )2l or a

cantilever ( )3l ;

< 2 >

The distance in between the sections where the flexural moment is null (where the part of the

section in tension turns from bottom to top or vice versa) is ( )ol , whose values are well established from

Statics for each type of span, load pattern and type of accompanying supports.


( )ol ( )effb

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Figure 6

Effective slab width ( )effb parameters

Beam to the left Beam to be designed Beam to the right

Definitions specific as per [Figure 6] are:

< 1 >

The distance in-between beams is to the left and to the right;

< 2 >

The effective slab width is to the left and to the right;

< 3 >

If the web assumes any other shape other than a rectangular one, the width to be taken into

consideration is the minimum value over the height of the cross section.

Based on all the previous definitions, the effective slab width ( )effb to be taken into consideration

as contributing to withstand compression together with the web ( )wb is:

[1-29]2

,1

eff w eff ib b b b= + and

[1-30]

( ), 0,2 0,1 min 0,2 ;eff i o ob b l l b= +

In order to design the reinforcement as for a tee/flanged section, it is imperative for the division as

presented in [Figure 4], to hold true in the case studied. In other words, the neutral axis ( )x must lie in

the web. The other case, the neutral axis ( )x lying in the flange, is a particular case of a SRRS.

This can be easily understood if one takes into account that for tee/flanged section there are two

areas of concrete in compression, each having its own height:

< a >The flanges are fully compressed;



The web is partially compressed but in a higher degree that the flanges.

Another explanation to the above result is that since ONLY the area of concrete in compression is

contributing to the capacity in flexure of a given member, the shape of the cross section below the neutral

axis HAS NO RELEVANCE. Therefore, two cross sections with the same area of concrete in

compression will require identical areas of reinforcement in tension no matter what the general shape is.

The area of the reinforcement to be placed in the cross section is, as explained, the sum of the areas

in equilibrium with the above two (flanges and web). Should in both cases the height in compression be

the same, mathematically speaking, it doesnt matter at what point in time the summation occurs since the

( )12 b ( )22 b

( ),1effb

( ),2effb

( )wb

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end result is the same. It may occur at the end (after designing as for a tee/flanged section) or, even better,

at the very beginning by summing up the areas of concrete in compression and AFTERWARDS

calculating the area of reinforcement.

Assuming the latter, the formulas to calculate the areas of reinforcement are easily deduced as:

[1-31]

( ) ( ) ( ),1 ,23 3,3( 0,5 )

f cd eff efff cd

sl

yd f yd

h f b bh f b zAf d h f

+ = =

[1-32]

( ) ( ),1 ,2,31 2 2

( 0,5 )Ed f cd eff eff fEd Rd

w cd w cd

M h f b b d hM M

b d f b d f

+ = =

and

[1-33] ,1 lim

cdsl

yd

fA b d

f=

In the previous relations the lever arm for the flanges has been written both in its short form ( )3z

as well as in its full form ( 0,5 )fd h . Likewise, the width of the flanges is written as ( )3b and as

( ),1 ,2eff eff b b+ respectively, although in most cases ( ),1 ,2eff eff b b= .

In the above [eq. 1-27] the web will have to withstand the remaining flexural moment which isnt

resisted by the flanges (therefore the subtraction indicated).

In the above [eq. 1-28] ( )1 is the coefficient of mechanical reinforcement usually extracted from

Tables for the corresponding value of( )1 .

Just as for a rectangular cross section, tee/flanged sections may sometimes be DOUBLY

REINFORCED. In this case, all previous definitions remain the same as per [Figure 1, 4, 5 & 6].

Figure 7

Doubly reinforced tee/flanged section

Cross section Operated Separation Strain Distribution Simplified Stress Block


< 1 >

A singly reinforced web section for which the reinforcement in tension ( ),1slA can be easily

calculated (similarly to SRRS);

(Continued on next page)

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< 2 >An area of reinforcement in tension ( ),2slA to be in equilibrium ONLY with the same area of

reinforcement in compression at the top of the cross section;

< 3 >

An area of reinforcement in tension ( ),3slA to be in equilibrium ONLY with flange in compression.

The reinforcement to be placed in the cross section is ( ),1 ,2 ,3sl sl slA A A+ + for tension and ( ),2slA for compression.

The formulas to calculate the areas of reinforcement are easily deduced as:

[1-34]

( ) ( ) ( ),1 ,23 3,3

( 0,5 )

f cd eff efff cd

sl

yd f yd

h f b bh f b zA

f d h f

+ = =

[1-35]

( ) ( ),1 ,2,32 2

( 0,5 )Ed f cd eff eff fEd Rdw

w cd w cd

M h f b b d hM M

b d f b d f

+ = =

[1-36]2

,2 ,3 lim,2

2 2

1 1Ed Ed Rd w cdsl

yd yd

M M M b d fA

f z f d d

= =

and

[1-37] ,1 lim

cdsl

yd

fA b d

f=

The above [eq. 1-30] is identical to [eq. 1-27]. It is presented as a separate equation for ease of

reference. This result is not surprising since the geometry of the flanges has specific values for a given

section and therefore is independent of the characteristics of the web, which has to withstand the

remaining flexural moment that isnt resisted by the flanges (as indicated next in [eq. 1-31]).

In the above [eq. 1-31] there is a difference in the subscript used to depict the type of reinforcement

as per [eq. 1-27] (w instead of 3) but no difference in meaning. It is presented as a separate equation to

outline the importance of correctly calculating ( )w for the web and checking if ( )limw > .

[Eq. 1-32] is just a reminder that DOUBLY REINFORCEMENT is necessary ONLY after fully

using the potential of a SINGLY REINFORCED cross section, as per [eq. 1-33].

As a conclusion (so far regarding only flexure) the reader is advised to remember that the order in

which to calculate the area of reinforcement is the one used in all previous equations and may be

indicated (in greater details in Chapter 2) as being the following:

< 1 >

Calculate the influence of the flanges in compression (if any) on the bearing capacity of the crosssection;

< 2 >Check if the web is to be designed as a SINGLY or DOUBLY reinforced cross section;

< 3 >

In the case of DOUBLY REINFORCEMENT calculate the area of reinforcement in compression;

< 4 >

Calculate the area of reinforcement in tension ONLY for the web;

< 5 >SUM UP all the partial areas of reinforcement calculated (flanges + compressed + web) and/or

(compressed) and PLACE them correctly in tension and/or compression.

21/45



[Part (One) C]Milestones for shear design

It is well known that elementary stresses pair up depending on the type of loading. In the case of

gravitational loading for horizontal structural members that means a combined action of flexure and

SHEAR, which is actually a combination of two stresses:

< a >

VERTICAL SLIDDING effect (vertical shear) that can be observed when carrying a set ofbooks without squeezing them tight together the books would slip and fall;



HORIZONTAL SLIPPING effect (horizontal shear) that can be observed in the case of a set

of boards which are laid flat over two supports and then loaded each individual board would bend

downward but would also slip along each other horizontally.

Under the combined effect of the two previously described stresses, the cracks that appear along a

beam in flexure, although perpendicular to the axis in the middle third (approximately) of the beam

because of bending (flexure) moment, are rotated to an angle of about 22 to 60 degrees to the axis near

the supports as the bending decreases and shear increases.

That means that a beam under shear has several distinctive sub-elements:

< a >A fibre in compression (the top of the cross section);



A fibre in tension (the bottom of the cross section);

< c >

An inclined chord (the concrete between the cracks) in compression;

< d >

An inclined chord (the stirrups) in tension.

[Section (One-C) a]Variable angle truss model

Based on the previously described behaviour, a variable angle truss model has been proposed to

map the response of any beam under shear. Herein after, the strut will refer to the inclined chord in

compression (the contribution of concrete) and the tie to the inclined chord in tension (the contribution of

stirrups).

Figure 8

Variable angle truss model for shear design

Truss model parameters (longitudinal section) Section through strut Section through tie

22/45



Thus, the model actually used for the design in shear is that of a truss with chords made both of

concrete and of stirrups, as presented in [Figure 8] and the following list:

< 1 >The angle between the tie (modelling stirrups) and the axis of the longitudinal area of reinforcement

(horizontal direction) is ( ) ;

< 2 >

The angle between the strut (modelling concrete chord between inclined cracks) and the axis of the

longitudinal area of reinforcement (horizontal direction) is ( ) . Based on experiments it has been

concurred that the domain for this angle is ( )21,8 45o o . Since (as it will be demonstrated) the

angle is used with its trigonometric function ( )ctg , previous relation turns into ( )1 2,5ctg ;

< 3 >

The distance in-between chords in compression is ( )a ;

< 4 >The compression force in the strut is ( )cwF for which the stress ( )cw may reach a maximum of

( )cdf , the design strength of concrete;

< 5 >

The tension force in the tie is ( )swF for which the stress ( )sw may reach a maximum of ( )ywdf ,the design strength of steel for stirrups;

< 6 >

The spacing (real, as placed along the member) in-between stirrups is ( )s ;

< 7 >

The area of reinforcement to uphold shear (stirrups) is ( )swA .

One observation is a must: all the previous definitions have an additional subscript ( )w to outline

that the referenced source is the web/width of the member for stirrups are placed and arranged in the web

of the member as opposed to the longitudinal reinforcement placed for flexure.

Based on the previous definitions, for each section may be established the equilibrium equations.

For the 1st

section (through the strut) the following applies:

[1-38] sinEd cwV F =

Since ( )cwF is a force resultant of the stresses ( )cw acting on the area ( )wa b and furthermore

( )a may be fully written as a function of the geometry of the truss, the previous becomes:

[1-39]

( ) ( ){ }

2

sin sin sin

1

Ed w cw w cw

w cw

V a b z ctg ctg b

ctg ctgb z

ctg

= = + =

+=

+

23/45



The above may be further transformed based on the following:

< a >

The maximum stress to be attained by concrete is its design stress, ( )cdf ;



The state of stresses in the strut may be expressed by a non-dimensional coefficient taking

into account the type of concrete as well as the relationship in-between the design strength of

concrete ( )cdf and the stress due to prestressing ( )cp which, if present, has a positive effect:

[1-40]

for reinforced concrete

for prestressed concrete

1,00

1 0 0,25

1,25 0,5

2,5 1 0,5

cp

cp cd

cd

cw

cd cp cd

cp

cd cp cd

cd

if ff

if f f

if f f f

+ <

=<

<

The state of cracking may be expressed by a non-dimensional coefficient taking into account

the concrete strength decrease as cracking increases. In turn, the distance in-between successive

struts ( )a decreases, causing a reduction in the area over which stresses of compression ( )cw

develop:

[1-41]

1

0,6 1 0,8200

600,6

0,8

600,9 >0,50,8200

ckywd ywk

ck

ywd ywk

ckck

ywd ywk

fif f f

f MPaif

f f

f MPaf iff f

>

= =

>

< d >

By considering the real positioning of stirrups along the member, ( )90o = ;

< e >

The right member of [eq. 1-34] represents the capacity in shear of the inclined strut that is

reached upon failure by crushing of the concrete in compression.

The full form for all the above supplementary definitions incorporated in [eq. 1-34] is:

[1-42]

1

,max

cw w cd

Rd

b z f

V ctg tg

=

+ with

,maxRdV is the maximum capacity in shear of the concrete in compression.

Should the case of loading lead to an inequality such that ( ),maxEd RdV V> , the struts would fail by

crushing of the concrete, making the cross section unsuited to withstand the loads and therefore is a

situation to be avoided at all times.


24/45


25/45



coming closer to any support, may be bent up so to uphold hogging a.k.a. negative bending, which

increases as closing to any support. An additional effect of bending up the bars is that on their inclined

portion the bars link concrete over cracks and therefore uphold some of the shear in the cross section.

Since this solution is less common because of the increase in workmanship cost it will not be presented.

[Section (One-C) b]

Final AdviceUp to this point the authors have presented the basics on the theoretical background on design, for a

better understanding of the formulas to be used in the actual design work, because only after reaching this

minimum set of knowledge, design can be accomplished easily and swiftly.

The next chapter will present in successive steps a guide to designing reinforced concrete members

under both flexure and shear. The reader is advised to use ones good judgement in deciding whether or

not all the steps are necessary in the provided order depending on the design problem at hand.

26/45

Manual for Advanced Design Steps to Design [2]


[Chapter Two]

Steps to Design

[Part (Two) A]

Citations used

Herein will be presented the full final forms of all the definitions mentioned so far. The reader is

advised to refer to the code provisions, SR EN 1992-1-1:2006[1], SR EN 1992-1-2:2006 [2] and the

Appendix at the end of this book to make use of the Tables and other referenced material as indicated in

this chapter. Also, it is assumed that the maximum flexural moment ( )EdM and the maximum shear

( )EdV are known.

[Part (Two) B]Introduction

Any structure must fulfil two fundamental requirements:

< 1 >

To be designed in such a manner that it does not collapse under normal loading conditions, partially

or totally, and that any partial collapse does not impair the unaffected part of the structure causing adomino effect to bring the structure down (to be therefore redundant). This is the so called Ultimate

Limit State Design (ULSD);

< 2 >

To be designed in such a manner that it does not impair on the intended use of that structure,

partially or totally. This is the so called Service Limit State Design (SLSD).

To address the first limit state its enough to provide for a cross section defined primarily by its

width and height ( )b h the corresponding reinforcement. Sizing of the cross section is subject to certain

conditions, which will be detailed herein.

To address the second limit state its enough to check that the proposed section fulfils additional

requirements which in the case of RC members are deflection and crack width.

For Normal Strength Concrete, by obeying certain limitation as per code provisions, generally

speaking, a section proposed for an ULSD will be checked for SLSD also. That is why SR EN 1992-1-1

states firmly that in all the cases that limitations are respected there is no need to check for SLSD

conditions.

Therefore the first step in any design is to propose a cross section which would best provide

safety by bearing the loads acting upon it and which is also economical and easy to cast. That is why,

today, as mankind struggles to find better management plans for the depleting resources at our disposal,

engineers in general and civil engineers in particular are called upon to find ways unexplored before to

achieve that end (i.e. construction industry consumes about 40% of the overall energy producedworldwide). In order to achieve SUSTAINABILITY (by its generalised meaning) its primordial to insure

DURABILITY for each member and therefore the structure itself.

DURABILITY is nothing more than the response of a member subjected to exposure conditions

due to climatic conditions (rain, snow, etc.) or processes (wanted or accidental) which take place inside

the structure or inside a perimeter around the structure that makes it vulnerable to that specific exposure.

It is achieved by providing a minimum concrete cover to protect the reinforcement from corrosion or the

adverse effects of fire. Of course, the concrete grade is the most important factor to be considered, as it

will be explained herein.

27/45



Since exposure conditions pair up with fire safety conditions to impose a concrete cover and even

minimum dimensions for the cross section and furthermore are general valid conditions with no respect to

a particular cross section that may be chosen by the engineer on ones best judgement, the first provided

answer in the previously proposed endeavour is to correctly calculate concrete cover.

[Part (Two) C]

Concrete coverAny cross section has (generally speaking) two type of reinforcement, for flexure longitudinal to

the axis of the member and for shear transverse to the same axis, which means that there will be two

types of concrete cover to be checked against the required thickness.

< 1 >Exposure conditions are explained in:

[1] SR EN 1992-1-1:2006

Pages [43-48]

Appendix [Table 2 & 3]

Default values (recommended) Exposure conditions XC2 & structural class, S4

< 2 >

Choosing the material (concrete and steel grade), if no values are imposed

[1] SR EN 1992-1-1:2006

Pages [24-42 & 190-191]

Appendix [Table 6, 7 & 8]

Default values (recommended) C30/37 for concrete & S500 for steel

< 3 >

Choosing the steel bar size (herein diameter will be referred to by size)

Default values (recommended)

,max 6 [ ]sl mm= for slabs

,max 25 [ ]sl mm= for beams

,max 28 [ ]sl mm= for columns

{ }8 [ ] 6 12(14) [ ]sw mm mm= for stirrups

< 4 >

Calculate nominal concrete cover for both stirrups, ( ),nom swc , and longitudinal bars, ( ),nom slc :

[2-1]

minnom devc c c= +

[2-2]

min,

min min, , , ,

max

10

b

dur dur dur st dur add

c

c c c c c

mm

= + with

min,bc [mm] is the concrete cover based on bond conditions (Appendix, [Table 5a]);

min,durc [mm] is the concrete cover based on exposure conditions (Appendix, [Table 4, 2 & 5b])

,durc [mm] is the safety cover, , 0durc mm = ;

,dur stc [mm] is the reduction due to using stainless steel, , 0dur stc mm = ;

,dur addc [mm] is the reduction due to additional concrete protection, , 0dur addc mm = .

28/45



devc [mm] is the deviation (tolerances) in actual pouring (10 5devmm c mm , see [Figure 9]

where ( )1d is referred to by ( )a ).

Figure 9

Concrete cover parameters

Concrete cover (notations) Concrete cover (tolerances)

< 5 >

Calculate design concrete cover (based on the actual position of bars in the cross section):

[2-3]

,

,

max

nom sw

vnom sl sw

cc

c =

< 6 >

Check if design concrete cover is at least the minimum cover after pouring:

[2-4]min,v dur devc c c +

[Part (Two) D]

Sizing of the cross section

Its well known that a board set flat over two supports will bend downward when pressed upon indirect ratio to the height of the board. Therefore, to avoid excessive deflections a first condition used to

Establish the dimensions of the cross section are the so called:

< 1 >Rigidity conditions:

Appendix [Table 14]

Other conditions are:

< 2 >

Fire safety conditions:

Appendix [Table 12 to 13]

< 3 >Technological conditions (mainly for the thickness of slabs)

Default values (recommended)

60 [mm] for roofs

70 [mm] for civil structures

80 [mm] for industrial structures

100 [mm] for pavements


29/45


30/45

31/45



The ratio slA

b d

=

is named coefficient of longitudinal reinforcement and is the measure of steel

consumption for a cross section having values (in most common situations) ( )0,005 0,010= for

slabs and ( )0,015 0,020= for beams.

Figure 10

Strains for tension reinforcement

1c s c cs

x d x d

d x x x

= = =

Strain Distribution Demonstration

< 9 >

Check that the reinforcement yields (as assumed):

[2-11]

yd1ud s c yd

s

fd

x E

= =

with

0,9ud uk = is ultimate design strain for steel (dependent on the steel yield class).

[Part (Two) F]Doubly reinforced rectangular sections (DRRS)

The case presented here is ( )lim > and steps up to no. 4 as presented in subchapter [2-E] are

considered true with no change in the assumptions made there.

< 1 >Calculate the area of reinforcement in compression:

[2-12]

( ) ( )

2,2 lim

,2

2 2

Ed Ed cd sl

yd yd

M M b d f A

f d d f d d

= =

< 2 >Extract (Appendix, [Table 16]) or calculate (chapter 1) ( )lim and calculate the area of

reinforcement (maximum) in the singly reinforced web:

[2-13] ,1 lim

cdsl

yd

fA b d

f=

< 3 >

Calculate the area of reinforcement to be placed in the cross section:

[2-14],1 ,2

,2

for tension (bottom)

for compression (top)

sl sl

sl

A A

A

+

< 4 >

Choose bar size and no ( ), ,sl eff sl calcA A for both of the previous (Appendix, [Table 10 &11]);

32/45



< 5 >

Choose the best placing and arrangement for bars (layout) (Appendix, [Table 18]);

Figure 11

Strains for compression reinforcement

2 2 2 1sc sc cu cucu

x d x d d

x x x

= = =

Strain Distribution Demonstration

< 6 >Check that the reinforcement in compression yields as assumed:

[2-15]

2,2 1

yd

ud s cu yd

s

fd

x E

= =

< 7 >

Should ,2 20s x d < < there is no need for supplementary compression reinforcement and:

[2-16]

, ,1sl calc slA A=

< 8 >

Should ,2s yd it means that the compression reinforcement has not reached yielding, therefore:

[2-17]

( )

2

,2,2 ,2

2

700 1Ed sc

sc sl sl

sc yd yd

d

M f xA A A

f d d f f

= = =

with

[2-18]

2 2 1 700 1sc sc s sc cu sd d

f E f Ex x

= = =

is the stress in the

compression reinforcement prior to yielding, calculated as of Hookes law

Since ( ),2slA is the calculated compression area when it was assumed that reinforcement yields, its

easier to calculate the new area of reinforcement as indicated in [eq. 2-17] (the initial value modified by

the ratio of the stresses).

< 9 >Check that the reinforcement in tension yields as assumed:

[2-19]

1

yd

ud s cu yd s

fd

x E

= =

< 10 >

Should s yd it means that the tension reinforcement has not reached yielding, therefore:

[2-20]

( )

( ),1 st,1 ,1

yd

700 1

0,5

Ed

st sl sl

st yd

M d xfA A A

f d x f f

= = =

with

(Continued on next page)

33/45



[2-21]

1 700 1s s s s cu sd d

f E f Ex x

= = =

is the stress in the tension

reinforcement prior to yielding, calculated as of Hookes law

< 11 >

Choose bar size and no ( ), ,sl eff sl calcA A for both tension and compression reinforcement

(Appendix, [Table 10 & 11]);

< 12 >


< 13 >Check the effective reinforcement to fulfil minimum/maximum criterion:

[2-22]

,min , ,max

0,26

max 0,04

0,0013

ctm t

yksl sl eff sl

t

f b d

fA A A b d

b d

= =

with

tb is minimum width in tension.

Whenever seismic load is a CASE LOAD taken into consideration in design, the above coefficient

( )0,26 is replaced by ( )0,50 .

[Part (Two) G]Singly reinforced tee/flanged sections (SRFS)

As previously explained tee/flanged sections may occur on different basis. That is why at first

several checks are performed to determine whether or not the design is a true tee/flanged one. The same

assumption holds true (i.e. the reinforcement in tension yields).

< 1 >Should 0,05fh h< designing is as for a rectangular with b h

Should the influence of flanges be less than 5% they are completely neglected (very thin flanges as

compared to the section therefore very small effect on the flexural capacity).

< 2 >

Should the tee/flanged section be a composite shape, calculate the effective slab width effb :

[2-23]

( )

2

,1

, 0,2 0,1 min 0.2 ;

eff w eff i

eff i i o o i

b b b b

b b l l b

= +

= +

ONLY if the ratio between consecutive

spans is from 2/3 to 3/2 while the span of

the cantilever is a maximum of 1/2 of the

adjacent span

< 3 >

Calculate the flexural capacity of the flange by assuming the neutral axis lies just under the extreme

lower fibre of the flange:

[2-24] ( ) ( ) ( ), 0,5Rd f f eff cd fM h b f d h =

< 4 >

Should ,Rd f EdM M the neutral axis x lies in the flange and designing is as for a rectangular

with effb h ;

34/45



< 5 >

Should ,Rd f EdM M< the neutral axis x lies in the web and designing is as for a true

tee/flanged section;

< 6 >Calculate the flexural capacity of the two flanges (fully compressed):

[2-25] ( ) ( ) ( ),3 ,1 ,2 0,5Rd cd f eff eff fM f h b b d h = +

< 7 >Calculate the area of reinforcement to be in equilibrium with the compression in the flanges:

[2-26]

( ) ( ),3

cd f eff w

sl

yd

f h b bA

f

= the form for the flanges of the general

1 Edsl

yd

MA

f z=

< 8 >

Calculate the relative bending moment and check the section to be SR (Appendix, [Table 15]):

[2-27] ,3lim2

Ed Rd

w

w cd

M M

b d f

=

< 9 >

Extract (Appendix, [Table 16]) or calculate ( ) (Chapter 1) and Establish the area of tension

reinforcement in equilibrium with the compression in the web:

[2-28] ,1 1cd

sl w

yd

fA b d

f=

< 10 >Calculate the area of reinforcement to be placed in tension:

[2-29], ,1 ,3 1

eff f w cdsl tot sl sl

w yd

b d fA A

f

b hA

b d +

+ =

=

< 11 >Choose bar size and no ( ), ,sl eff sl tot A A for the previous (Appendix,[Table 10 & 11]);

< 12 >


< 13 >

Check the effective reinforcement to fulfil minimum/maximum criterion:

[2-30],min , ,max

0,26

max 0,04

0,0013

ctm t

yksl sl eff sl

t

f b d

fA A A b d

b d

= =

with



( )0,26 is replaced by ( )0,50 .

35/45



[Part (Two) H]Doubly reinforced tee/flanged sections (DRFS)

The proposed steps presented in this subchapter are true for that particular situation for which

( )lim > . The steps up to no. 8 as presented in the previous subchapter are considered to be calculated.

The same assumptions hold true (i.e. the reinforcement in tension and compression yields).

< 1 >

Calculate the area of reinforcement in the web in compression:

[2-31]

( ) ( )

2

lim,2 ,3,2

2 2

MEd Ed Rd w cd

sl

yd yd

b d fM MA

f d d f d d

=

=

< 2 >

Extract (Appendix, [Table 16]) or calculate (chapter 1) ( ) and calculate the area of tension

reinforcement in equilibrium with the compression in the web:

[2-32] ,1 lim

cdsl w

yd

fA b d

f=

< 3 >

Calculate the area of reinforcement to be placed in the cross section:

[2-33],1 ,2 ,3

,2

for tension (bottom)

for compression (top)

sl sl sl

sl

A A A

A

+ +

< 4 >

Choose bar size and no. ( ), ,sl eff sl calcA A for tension and compression reinforcement as above(Appendix, [Table 10 & 11]);

< 5 >


< 6 >

Check that the reinforcement in compression yields as assumed:

[2-34]

2,2 1

yd

ud s cu yd

s

fd

x E

= =

< 7 >

Should ,2 20s x d < < there is no need for supplementary compression reinforcement and:

[2-35]

,1 ,3for tension (bottom) sl slA A+

< 8 >Should ,2s yd it means that the compression reinforcement has not reached yielding and

therefore:

[2-36]

( )

( ),2 2,2 ,2

2

700 1Ed scsc sl sl

sc yd yd

M d xfA A A

f d d f f

= = =

< 9 >

Check that the reinforcement in tension yields as assumed:

[2-37] 1

yd

ud s cu yd

s

fd

x E

= =

36/45



< 10 >

Should s yd it means that the tension reinforcement has not reached yielding and therefore:

[2-38]

( )

( ),1 st,1 ,1

yd

700 1

0,5

Ed

st sl sl

st yd

M d xfA A A

f d x f f

= = =

< 11 >

Choose bar size and no ( ), ,sl eff sl calcA A for both tension and compression reinforcement(Appendix, [Table 10 & 11]);

< 12 >


< 13 >

Check the effective reinforcement to fulfil minimum/maximum criterion:

[2-39],min , ,max

0,26

max 0,04

0,0013

ctm t

yksl sl eff sl

t

f b d

fA A A b d

b d

= =

with



( )0,26 is replaced by ( )0,50 .

[Part (Two) I]

Shear design

Its well known that for a beam under continuous uniformly distributed gravitational loads the

paths of the principal stresses are (see [Figure 12]):

< a >

Arches extending from support to support (compressive principal stress)



Inverted arches (tension principal stress)

This is easy to observe should one remember that the arch is the only structural shape loaded only

in compression which unloads directly to supports. Since cracks develop perpendicularly to the axis in the

middle third and then rotate near the supports to an angle close to 45 degrees, by removing/extracting

the area of cracked concrete, the resulting shape is an arch. This behaviour has been experimentally tested

and used in previous design models and may be expressed as: part of the load near the supports is

unloading directly to them without loading the cross section of the member; therefore it is possible to

design to a lower loading, taking into consideration this unloading mechanism.

Figure 12

Main stresses path for horizontal members

37/45



In the variable angle truss model several conditions are met which must be explained prior to

giving relationships for design:

< 1 >Cracked concrete is assumed NOT to withstanding shear while ONLY stirrups do. This is a rather

gross approximation and that is why, for a particular pattern of loading (CONTINUOUS UNIFORMLY

DISTRIBUTED) which does not disturb the main stresses path, for every elementary distance in-between

the foot of a strut and the foot of a tie given as ( )z ctg ctg + , code provisions allow to furtherreduce the design shear value to the minimum shear on this elementary distance (implicitly the difference

is being carried by concrete). This second reduction is very different in nature from the first one for the

unloading mechanism near the supports which is applicable even for concentrated loads and should

therefore be applied ONLY in specific cases.

< 2 >

The horizontal reinforcement is bridging concrete over cracks and is under tension mainly due to

flexure. In Chapter 1 it was explained the formula to calculate this tension as EdM

Tz

= . The vertical

shear (for which reinforcement is provided in the form of stirrups) is causing a dowel effect on the

horizontal reinforcement which is bent downward in-between two consecutive stirrups. This means the

bars will be loaded by an additional tension due to shear alone. By taking moments around the upper nodewhere the strut and tie meet (remember from Statics of Constructions that sectioning of the strut and tie is

performed at the middle of each chord denoted as O in Figure [13]) the equilibrium may be written as

(exterior moments are in equilibrium and are omitted for clarity):

[2-40]

( )

2 2

0,5

Ed Ed

Ed E

z ctg z ctgV V M

V z ctg ctg F z

=

=

with

Ed lE

V aF

z

= is the additional tension in the longitudinal reinforcement due to shear alone;

( )0,5la z ctg ctg = is the shift rule for the bending moment envelope.

Figure 13


Section through strut Section through tie

The use of [eq. 2-38] will be explained in the end of this subchapter when presenting the anchorage

lengths of the bars at the supports. The above leads in turn to writing the overall tension as:

[2-41]

shiftedEd Ed l EdM V a MT

z z z

= + =

38/45



Should an axial force ( )EdN be present, the above [eq. 2-39] would sum up that value with the

given right member.

< 3 >

Since supports have their own dimensions and joints have various fixity in-between members,

further assuming supports to be pointed as in Statics may work ONLY for one span members whose

design in flexure correctly disregards supports dimensions. Still, EVEN in that case, there is operable a

relevant reduction in shear which depends on dimensions of the supports (see Appendix, [Chapter 7]).

With all of the above in mind, the steps to design for shear are:

< 1 >Calculate the design shear force reduced at the face of the support assuming a continuous uniformly

distributed load pattern (for concentrated loads use Appendix, [Chapter 7]):

[2-42] ( ), 1Ed red EdV V F a d = + with

1a , (subscript one) is half the width of the support taken in consideration.

< 2 >Calculate the maximum shear capacity of the cross section assuming only concrete withstands

shear:

[2-43]

( )1 3

, 1,

min 1

100max Rd c l ck cpRd c w

cp

C k f k V b d

k

+ =

+ with

,

0,18 0,180,12

1,5Rd c

c

C

= = = [no units] is the coefficient provided by national code provisions;

2001 2,0k

d= + [no units] is the size factor taking into account the influence of the height of the

cross section (by use of depth in [mm]) with a maximum value of 2,00;

0,02sllw

A

b d =

[no units] is the coefficient of longitudinal reinforcement taking into account the

dowelling effect with a maximum value of 0,02;

slA [mm2] is the area of tension reinforcement which extends beyond a considered section (named A)

with a minimum length of bdd l+ (see [fig. 14]);

Figure 14

Definitions for anchoring reinforcement at the supports

bdl [mm] is the design anchorage length;

wb [mm] is the minimum transverse tension width of the cross section;

39/45



0,2Edcp cd c

Nf

A = [MPa] is the mean (medium/average) stress due to an axial force in the cross

section;

EdN [kN] is the axial force in the cross section, having POSITIVE values for compression (for it tightens

vertical slides and prevent vertical slipping; tension untightens vertical slides easing cracking therefore is

NEGATIVE) without taking into consideration the influence of imposed deformations;

cA [mm2] is the area of the gross concrete section (for a rectangular section cA b d= );

3 2 1 2min 0,035 ckk f = [N/mm

2] is the minimum shear capacity of a concrete section (NO longitudinal

reinforcement).

< 3 >Should , ,Ed red Rd cV V there is no need for shear reinforcement. Cage fabrication pairs in this case

constructive longitudinal reinforcement placed at the extreme fibre (top for one span member) with

designed reinforcement (bottom for one span member) and stirrups respectively, which are provided as

CONSTRUCTIVE SHEAR REINFORCEMENT (minimum size & maximum spacing) with respect to:

< a > ,min

0.08

sin

cksww w

w yk

fA

s b f

= =

is the coefficient of transverse reinforcement

having a minimum value as indicated;

 swA is the area of transverse reinforcement spaced at ( )s taking into account the no. of

vertical arms that are placed in the cross section (minimum 2 x size of the stirrups; should the cross

section, due to its width necessitate supplementary stirrups or as an option of the designer in the

case of heavy shear loaded members, the number of arms may increase)

< c >

The maximum centre to centre distance for stirrups over the member is

( ),max 0,75 1ls d ctg = + ;

< d >

The maximum centre to centre distance in-between vertical stirrups arms is

,max 0,75 600ts d mm= .

< 4 >

Should , ,Ed red Rd cV V> transverse reinforcement is necessary. The design is based on the variable

angle truss model as previously explained considering concrete not to withstand shear.

< 5 >

Calculate the shear capacity of the strut and check if the concrete doesnt crush in compression:

[2-44]

1

,max ,

cw w cd

Rd Ed red

b z fV V

ctg tg

=

+

with

( )

for reinforced concrete

for prestressed concrete

1,00

1 0 0,25

1,25 0,5

2,5 1 0,5

cp cd cp cd

cw

cd cp cd

cp cd cd cp cd

f if f

if f f

f if f f

+ < =

<

< 0,50,8200

ckywd ywk

ck

ywd ywk

ckck

ywd ywk

fif f f

f MPaif

f f

f MPaf iff f

>

= =

>

is the coefficient taking into account theinfluence of the crack state in the section;

ckf is the characteristic compressive

strength of concrete;

ywkf is the characteristic tensile strength

of transverse reinforcement;

ywdf is the design tensile (yielding)

strength of transverse reinforcement;

0,9z d [mm] is the simplified formula for the lever arm;

,max,inf21,8 21,8 2,5 Rdctg V = =

[no units] is the strut angle (imposed);

Explanation of the previous is at ease should the reader remember the model for design in shear. In Figure

[15] assume 0ctg = for stirrups are vertical ( 90o = ). Please now focus on ( ) and the

corresponding ctg and on ( )cwF and the corresponding sin respectively.

< a >

On one hand, as ( ) decreases to a MINIMUM ctg increases to a MAXIMUM makingthe chord width ( )a larger. On the other hand, as ( ) decreases to a MINIMUM sin decreases

also to a MINIMUM. Therefore, the vertical component of the shear capacity of concrete

( )sincwF decreases accordingly making the overall result as per [eq. 2-42] to reach a

MINIMUM ( 2,5ctg = DEFAULT VALUE ONLY in the above);

Should ,max,inf ,Rd Ed redV V hold true the strut at its lowest capacity can withstand shear

without crushing of the concrete;

< c >

Should ,max,inf ,Rd Ed redV V< hold true the strut at its lowest capacity cannot withstand shear

for concrete will crush in compression causing failure of the member. To avoid that its necessary

to change the cross section by (recommended practice) increasing its width ( )b .

Figure 15


Truss model parameters (longitudinal section) Section through strut Section through tie

41/45



< 6 >

Calculate the value for ctg based on the particular pattern of loading by interpolation:

[2-45] , , ,max,inf 2,5 1

Rd c Ed red RdV V V

ctg

In Chapter 1 (see [eq. 1-39] and the explanations there) it was concluded that

,

Rd ssw

ywd

VA

s f z ctg

.

< a >

The lower limit ( ),Rd cV is met for members with CONSTRUCTIVE SHEAR

REINFORCEMENT (minimum stirrup area at the maximum space). Therefore the ratio ( )swA s

is MINIMUM making in turn ctg be MAXIMUM ( 2,5ctg = ).



The upper limit ( ),max,infRdV is met for members with HEAVY SHEAR REINFORCEMENT

(maximum stirrup area at the minimum space). Therefore the ratio ( )swA s is MAXIMUM

making in turn ctg be MINIMUM ( 1, 0ctg = ).

< c > ( ),Ed redV slides in-between the above limits dependent on the load. As more shear

reinforcement is needed the ratio ( )swA s must increase thus resulting in ctg decreasing from

2.5 to 1.0 while ( ),Ed redV moves away form ( ),Rd cV and closes ( ),max,infRdV .

< 7 >

Calculate the required area of transverse reinforcement uniformly distributed over the member

(based on the condition for stirrups to yield prior to concrete crushing, see [eq. 1-39] and notes there):

[2-46]

,

Ed redsw

ywdrqd

VA

s z f ctg

=

< 8 >

The above leads either to calculating the spacing by imposing the transverse area of stirrups as size

and no. of vertical arms as given by swA (use Appendix, [Table 11]):

[2-47] ,

swrqd

Ed red

ywd

As

V

z f ctg

=

or to (recommended practice) calculating the transverse area of stirrups by imposing the spacing as given

by s , multiple of [10 mm] (it should be multiple of [50 mm] to ease cage fabrication):

[2-48]

( ) ,

Ed red

sw rqdywd

VA sz f ctg

= and then choose the size for stirrups knowing the no. of

vertical arms

Remember that (recommended practice):

< a >100 300mm s mm , for all structures in seismic zones;

100 400mm s mm , otherwise;

42/45



< c >

Along the member there are intervals on which shear decreases so providing a constant ratio

swA

s

although great for shear bearing capacity is uneconomical and therefore unpractical. Both

the area of reinforcement (size and no. of arms) and spacing may be modified but in terms of

workmanship cost and time its easier to modify ONLY the spacing in a manner that generates

multiples of the same basic length (the spacing chosen for the maximum shear). For example,assume the basic length is 100 mm; another spacing should be 200 mm (2 times the basic length),

then 300 mm (3 times the basic length) and so on.

< 9 >

Calculate the coefficient of transverse reinforcement and check the minimum condition:

[2-49],min

0,08

sin

cksww w

w yk

fA

s b f

= =

< 10 >Check the effective area of transverse reinforcement uniformly distributed over the member (based

on the condition for stirrups NOT to fail under excessive tension, see [eq. 1-40] in conjunction with the

above [eq. 2-43] with the corresponding explanations):

[2-50]

1

max 2sw sw cw w cd

eff ywd

A A

s s

b f

f

=

< 11 >Additional checks:

< a >The maximum centre to centre distance for stirrups over the member is

( ),max 0,75 1ls d ctg = + ;



The maximum centre to centre distance in-between vertical stirrups arms is

,max 0,75 600ts d mm= .

With this final step the design in shear is over.

Still, its necessary to calculate the anchorage length for the longitudinal reinforcement to insure

that indeed the bars wont slip inside the concrete. REMEMBER that any reduction in shear that is used

in design MUST be paired with correctly anchoring the longitudinal bars (continued numbered list as set

out below).

For all members under flexure and shear, bending decreases while shear increases while closing the

supports from midspan. In the case on multiple span members, bending at the supports turns from sagging

to hogging therefore is correct to asses that, according to [eq. 2-39], for bars anchored at the support the

tension developing in the bars is ET F=

, which may further be written as:

[2-51]

2

4E sd sl sdF A n

= = with

sd is the stress to develop in a bar;

n is the no. of bars to be anchored at the support;

is the bar size (its assumed for easier calculation that all bars are the same size).

43/45



The bond that develops on the surface of all the bars in the cross section is:

[2-52] [ ]surfacebd sl bd b bd F A f n l f = = with

bl is the surface of the bar on which bond acts;

bl is the anchorage length;

bdf is the design bond strength.

By setting the above forces equal to each other (equilibrium):

[2-53] [ ]

2

0,254

sdb bd sd b

bd

n l f n lf

= =

So, calculation of the anchorage length starts with:

< 12 >

Calculate the tension for the reinforcement to be anchored as (due to shear alone):

[2-54]

Ed lE

V aF

z

= with

( )0,5la z ctg ctg = is the shift rule for the bending moment envelope.

< 13 >

Calculate the stress acting in the bars:

[2-55]

,

Edsd

sl anchored

F

A =

< 14 >

Calculated the required anchorage length:

[2-56]

, 0,25 sd

b rqd

bd

lf= and

[2-57]

1 22,25bd ctd f h h f=

with

bdf [MPa] is the design bond stress (see [Table 1] orAppendix,[Table 19]);

1 1,00h = for good bond conditions 2 1,00h = for 32 mm

1 0,70h = for poor bond conditions or stepping formwork (if not

proven otherwise)

2

132

100h

= otherwise

ctdf [MPa] is the corresponding design tensile strength but limited to C60/75 if not proven otherwise.

< 15 >Calculate design anchorage length and check minimum criterion:

[2-58]

1 2 3 4 5 , ,minbd b rqd bl l l = with

1 [no units] is the coefficient taking into account the type of the end of the bar when concrete cover is

correctly insured;

2 [no units] is the coefficient taking into account the minimum concrete cover;


44/45



3 [no units] is the coefficient taking into account the confinement effect due to transverse reinforcement

(stirrups);

4 [no units] is the coefficient taking into account the case of transverse bars welded to the longitudinal

reinforcement;

5 [no units] is the coefficient taking into account the confinement effect due to transverse pressure

given by a concentrated force ( )tF as the result of the loading over the length ( )1t and width of the

support, ( )b ;

,minbl , minimum anchorage length in the absence of other provisions, for reinforcement in:

[2-59]

{ }

{ }

,

,min

,

for tension

for compression

max 0,3 ; 10 ; 100

max 0,6 ; 10 ; 100

b rqd

b

b rqd

l mml

l mm

=

Table 1

Bond Conditions (A is the direction of concrete pouring)

Good bond conditions for allbars

all similar cases whether intension or compression 45 90

o o 250h mm

Hatched area poor bondconditions

Plain area good bondconditions

all similar cases whether intension or compression 600h mm> 250h mm>

Remember that any of the above coefficients 1 5 reduces the anchorage length (all are 1).

Still, dont forget about the condition:

[2-60]

2 3 5 0,7

meaning the reduction due to these conditions must not be taken more than [70%] of the initial anchorage

length. For specific values of coefficients ( )1 5 please see [Table 2].



45/45


Table 2

Anchorage length parameters

Straight end Hooked End U Shape End Transverse welded bar

1

2

mind

a

c c

c

=

1

2mind

ac

c

=

dc c=

4 0,70 = for all similar

cases whether in tension orcompression

4 1,00 = otherwise

1 1,00 = for all similar

cases whether in tension orcompression

1 0,70 = for all similar cases in tension for which

3dc > 1 1,00 = otherwise

2 1 0,15

0,70

1,00

dc

=

for all similar cases intension

2

31 0,15

0,70

1,00

dc

=

for all similar cases in tension

2 1,00 = otherwise

,minst st

s

A A

A

=

,min 0,25st sA A=

for

beams

,min 0stA = otherwise

30,70 1 1,00k = for all similar cases in tension

stA is the area of stirrups placed over bdl

,minstA is the minimum area of stirrups placed over bdl

sA is the area of one anchored bar having the biggest size

3 1,00 = otherwise

50,70 1 0,04 1,00p = 1

tFpb t

=

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