Manual for Advanced Design

MAD-11 First Volume, First Edition

MANUAL for ADVANCED DESIGN Flexure and Shear

Assist. Lecturer Eng. Dumitru MOLDOVAN Eng. Horia CONSTANTINESCU

Manual for Advanced Design

Foreword 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

[2]

[overall current page]


Content List Foreword

Page [2]

Content List

Page [3]

[Chapter One] Theoretical background

Page [6]

[Part (One) A] Introduction

Page [6]

[Part (One) B] Milestone’s 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] Milestone’s 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 [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)

[3]

Page [47]

Page [50]

[Section (Three-B) a] Sizing of the cross section

Page [50]

[Section (Three-B) b] Static analysis

Page [52]



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] Longitudinal reinforcement layout

Page [65]

[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] Longitudinal reinforcement layout

Page [74]

[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 background

Page [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] Longitudinal reinforcement layout

Page [89]

[Part (Four) D] Area of reinforcement for flexure – checked by exact calculus [Section (Four-D) a] Longitudinal reinforcement layout [Part (Four) E] Concluding remarks

Page [89] Page [90] Page [90]

Appendix

Page [91]

[Chapter One] DURABILITY

Page [91]

[Part (One) A] CEMENT [4]


Page [91]


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]

[5]



Basics in Reinforced Concrete Design [1]

[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 and reinforcement); < 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, it’s relevant to discuss the designing process from a theoretical point of view, by explaining the model’s milestone’s in flexure and shear.

[Part (One) B] Milestone’s 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) [6]


[pages in chapter]: [20]



[Section (One-B) a] Neutral axis In these authors’ opinion the design process in flexure is very interconnected with the position of the neutral axis

( x)

though the actual design seems to provide no direct link to this, meaning this

variable isn’t as highlighted as others, for most formulas are pairing the neutral axis with other variables. In order to explain, the following imaginary exercise is proposed in which the reader is invited to assume the position of a lab technician on the point of testing a beam in flexure by gravitational loading (perpendicular to its geometric longitudinal axis). Also assume that: < a > You have marked with vertical lines the lateral faces of your beam on its full length; Let’s pretend that each line/slice (a cross section) can be extracted from the beam itself without affecting the structural behaviour of the element (similar to the way in which you may take from a book set your “favourite book” to read). The present discussion will focus on “your favourite book” in view of the stresses that appear (compression is pushing on the book while tension is pulling on the book, each action occurring on the opposite cover; i.e. let’s pretend compression occurs on the front cover and tension on the back one). At first, prior to the loading, there is null stress in each and every point on the height of the cross section. Since the neutral axis is only a limited area in the cross section, it’s safe to asses that the neutral axis is outside of the cross section. Since we assumed the loading to be gravitational in nature, that would mean that the neutral axis is somewhere below the extreme lower fibre of the cross section. As the loading begins, the extreme lower part of the cross section will develop tension stresses while the extreme upper part of the cross section will develop compression stresses, in other words the neutral axis starts to move from the extreme lower edge to the upper edge of the cross section. This movement will continue as long as the section holds. Failure will occur, theoretically, when the entire cross section will be in tension (a „ripping apart” effect). In fact, for normal strength concrete, there will be a small area at the top of the cross section that will crush under compression, therefore causing the collapse. From a mechanical (static) point of view, that is the same with the element becoming a mechanism with a hinge located at the top of the cross section. In this case, the collapse is instantaneous (even under own weight).

[Section (One-B) b] Assumptions Let’s convert the beam described previously to the material we call reinforced concrete. The present design model assumes that cracked concrete does not contribute to the bearing capacity, although it is well known that between cracks, the bond of concrete to steel leads to a reduction in the tensile stresses in the reinforcement. As the concrete grade increases so does its tensile strength making it logical to assume that, in some degree, the previous assumption may deflect the model from the actual behaviour. Since this is still under debate, no further details will be provided herein.

(Space intentionally left blank)

[7]





Other assumptions used in the model are: < 1 > Concrete is considered to be on the brink of failure (the so called 3rd state); < 2 > Bernoulli’s hypothesis of plane sections and the compatibility of strains in concrete and steel for the same fibre in the cross section; < 3 > Hooke’s Law allows for strains and stresses to be considered a ratio of the elastic properties of the material; < 4 > The real stress block for the compressed area of concrete is replaced by a simplified rectangular stress block; < 5 > The reinforcement yields prior to the crushing of concrete.

[Section (One-B) c] Stress block parameters In comparison with a uniaxial loading, whether it’s 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 also as 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 tendency to 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,

Fc ;

< 2 > A tension force in the reinforcement, < 3 > A lever arm in-between,

[8]

Ft ;

z.





The resistive flexural capacity for a given cross section is written as: [1-1]

M Rd = Fc ⋅ z = Ft ⋅ z

[1-2]

Fc = Ft

with

Fc is the compressive resultant; z is the lever arm;

Ft is 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: ε cu

Fc = b ⋅ ∫ σ (ε ) ⋅ ∂ε

[1-3]

with

0

ε cu is the ultimate compressive strain; b is the width of the cross section;

σ 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. NOTE

The cross section is considered to be rectangular both for the shape’s simplicity and for the fact that this particular shape is the most common in constructions.






Figure 1 Design model parameters

Cross section

Strain Distribution

< 1 > The notation for the height is < 2 > The beam has longitudinal

Real Stress Block

Simplified Stress Block

( h ) and for the width ( b ) ; (l )

steel

(s)

reinforcement

( Asl ,1 )

in tension (T , t ) with a stress

( f yd ) equal with the design ( d ) strength at yielding ( y ) in the lower part (1) of the cross section, at position/axis distance ( d1 ) from the extreme lower fibre, and reinforcement ( Asl ,2 ) in compression ( C , c ) with a stress ( f yd ) in the upper part ( 2 ) of it, at the position/axis distance ( d2 ) from the extreme top fibre;

(

)

< 3 > The position of the reinforcement in tension Asl ,1 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 will be named

( Asl ,1 )

in tension (T , t ) will be less than the one at yielding it

( f st ) ;

< 5 > If the stress in the reinforcement yielding it will be named

( Asl ,2 )

in compression

(C, c)

will be less than the one at

( f sc ) ;

< 6 > The maximum stress

( fcd )

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

strength due to a reduction factor (η ) given by:

[1-4]

 1.00 if f ck ≤ 50 MPa  f ck − 50  1 − 200 if 50 ≤ f ck ≤ 90 MPa

η =

< 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 ( λ ) given by:

[1-5]

[10]

 0.80 if f ck ≤ 50 MPa  f ck − 50 if 50 ≤ f ck ≤ 90 MPa  0.8 − 400

λ=





< 8 > (η ) and ( λ ) 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]

Fc = C = ( λ ⋅ x ) ⋅ (η ⋅ fcd ) ⋅ b

[1-7]

Fs = T = Asl ⋅ f yd

or

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]

T =C → x=

Asl ⋅ f yd

λ ⋅η ⋅ b ⋅ f cd

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

There 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 present manual. 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







Ideal design is for M Rd = M Ed . Based on this [eq. 1-1] will be re-written as: [1-9]

M Ed = T ⋅ z → T =

[1-10]

Asl ,1 =

M Ed z

1 M Ed ⋅ f yd z

with

f yd 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]

M Ed = C ⋅ z = ( λ ⋅ x ) ⋅ (η ⋅ f cd ) ⋅ b  ⋅ z

[1-12]

z = d − 0,5 ⋅ λ ⋅ x

with

[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]

M Ed = ( λ ⋅ x ) ⋅ (η ⋅ f cd ) ⋅ b  ⋅ d − ( λ ⋅ x ) ⋅ (η ⋅ f cd ) ⋅ b  ⋅ ( 0,5 ⋅ λ ⋅ 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 d 2 d 2 , with some additional grouping the above becomes:

[1-14]

M Ed

2  x   x =   ⋅ ( λ ⋅η ) −   ⋅ 0,5 ⋅ λ 2 ⋅η  ⋅ b ⋅ d 2 ⋅ f cd d    d 

(

) (

)

The above form may be further organised as: 2

[1-15]

µ = (ξ ) ⋅ λ ⋅η − (ξ ) ⋅ 0, 5 ⋅ λ 2 ⋅η

[1-16]

µ=

M Ed b ⋅ d 2 ⋅ f cd

[1-17]

ξ=

x d

with and

where

µ is the relative bending moment; ξ is the relative height in compression; f cd is the design compressive strength of concrete in compression.

[12]





The above [eq. 1-15] is a form for ( µ ) that is ONLY dependent on ( ξ ) 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 ( µ ) that is dependent on: < a > ( M Ed ) is given by the loading (case, pattern and values);

( )

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

( fcd ) 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 CONCRETE’s 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 ( ε cu ) and the other in reinforcement ( ε s ) as a consequence of Bernoulli’s Law. By writing proportions for similar triangles (upper and lower), it can be written:

ε 1  x d  = cu → ε s = ε cu ⋅  − 1 = ε cu ⋅  − 1 d − x εs x  ξ 

[1-18]

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

( f yd ) while the strain is higher than the one at yielding (ε yd ) : 1

[1-19]



f

ε s ≥ ε yd → ε cu ⋅  − 1 ≥ yd  ξ  Es

with

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

Es = 200 000 [MPa] 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 ( ε cu ) and implicitly for ( Es ⋅ ε cu ) if the concrete grade ( ≥ C 50 / 60 ) .

[13]





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 ( ξ ) will be extracted as:

1 [1-20]

ξ

≥

f yd Es ⋅ ε cu

1

+1 → ξ ≤

=

f yd Es ⋅ ε cu

+1

Es ⋅ ε cu 700 = f yd + Es ⋅ ε cu f yd + 700

The above sets a maximum limit ( ξ lim ) for ( ξ ) as well as a maximum value ( µlim ) for ( µ ) , both dependent on the steel grade

( f yd ) .

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]

M Rd = µlim ⋅ b ⋅ d 2 ⋅ f cd Should

(µ)

as per [eq. 1-16] be higher than

(µ)

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]

T = C → Asl =

( λ ⋅ x ) ⋅ (η ⋅ f cd ) ⋅ b =  λ ⋅η ⋅ x  ⋅ b ⋅ d ⋅ f yd

 

 d

f cd f = ω ⋅ b ⋅ d ⋅ cd f yd f yd

with

ω is a tabulated coefficient of mechanical reinforcement dependent on the value of ( ξ ) . The above is better understood should one fill in the corresponding values:

Asl ⋅ f yd [1-23]

ω = λ ⋅η ⋅ ξ = λ ⋅η ⋅

Asl ⋅ f yd λ ⋅η ⋅ b ⋅ f cd x = λ ⋅η ⋅ = d d b ⋅ d ⋅ f cd

(Space intentionally left blank)

[14]





[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]

µ=

M Ed 2 > µlim = (ξlim ) ⋅ λ ⋅η − (ξlim ) ⋅ 0,5 ⋅ λ 2 ⋅η 2 b ⋅ d ⋅ fcd

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


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

(

)

< 1 > A singly reinforced section for which ( µ = µlim ) with the reinforcement in tension Asl ,1 to be in equilibrium with the concrete in compression. This section has a capacity in flexure of: [1-25]

M Rd ,1 = µlim ⋅ b ⋅ d 2 ⋅ f cd

< 2 > An area of reinforcement in tension

( Asl ,2 )

to be in equilibrium ONLY with another area of

reinforcement in compression. Since the stress is assumed to be

( f yd ) for both tension and compression

due to the actual properties of steel, the area in tension will be equal with the area in compression.

[15]





This section has a capacity in flexure of.

M Rd ,2 = M Ed − M Rd ,1 = M Ed − µlim ⋅ b ⋅ d 2 ⋅ f cd

[1-26]

(

The reinforcement to be placed in the cross section is Asl ,2

)

(

for compression and Asl ,1 + Asl ,2

)

for tension. The formulas to calculate those areas of reinforcement are easily deduced as:

1 M Ed ,2 1 M Ed − µlim ⋅ b ⋅ d 2 ⋅ fcd ⋅ = ⋅ f yd z2 f yd d − d2

[1-27]

Asl ,2 =

[1-28]

Asl ,1 = ωlim ⋅ b ⋅ d ⋅

and

f cd f yd

[Section (One-B) f] Tee/flanged section (FS) Tee or flanged section refers to cross sections of the same shape that can be in fact the result of: < a > A choice of the designer based on engineering experience to favour this shape (i.e. tee section because of heavy loading); A composite section consisting of a beam (web) aided in withstanding loads by a slab (flange) in compression (flanged section). Definitions specific to this type of reinforcement are presented in [Figure 4] with the same meaning as per [Figure 1] and the subsidiary list: Figure 4 Singly reinforced tee/flanged section

Cross section

Operated Separation

Strain Distribution


< 1 > The web is considered to be rectangular, though it may assume any shape;

< 2 > The notation for its height is ( h ) and for the width ( bw ) ;

( )

( )

< 3 > The height of the flange is h f and the width is beff .






The cross section is momentarily divided for calculus, see [Figure 4], into the following pieces: < 1 > A singly reinforced web section for which the reinforcement in tension

( Asl ,1 )

can be easily

calculated (similarly to SRRS);

(

< 2 > An area of reinforcement in tension Asl ,2

) 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 Asl ,1 + Asl ,2 for tension. Before presenting the actual reinforcement formula, it is necessary to understand the calculation of

( ) a parameter ONLY for flanged sections (for T sections it is simply named

the effective slab width beff top 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 ( lo ) for the calculation of the effective slab width beff Inner support

End Span

Inner support

Inner Span

Cantilever

Definitions specific as per [Figure 5] are: < 1 > The drawing must be read as referring either to an end span

( l1 ) ,

an interior span

( l2 )

or a

cantilever ( l3 ) ; < 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 ( lo ) , whose values are well established from Statics for each type of span, load pattern and type of accompanying supports.






Figure 6

( ) parameters

Effective slab width beff

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 ( 2 ⋅ b1 ) to the left and ( 2 ⋅ b2 ) to the right;

(

)

(

)

< 2 > The effective slab width is beff ,1 to the left and beff ,2 to the right; < 3 > If the web assumes any other shape other than a rectangular one, the width ( bw ) to be taken into consideration is the minimum value over the height of the cross section.

( ) to be taken into consideration

Based on all the previous definitions, the effective slab width beff as contributing to withstand compression together with the web ( bw ) is: [1-29]

2

beff = bw + ∑ beff ,i ≤ b

and

1

[1-30]

beff ,i = 0, 2 ⋅ b + 0,1⋅ lo ≤ min ( 0, 2 ⋅ lo ; 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 doesn’t matter at what point in time the summation occurs since the [18]





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]

[1-32]

[1-33]

(

 h f ⋅ (η ⋅ f cd ) ⋅ b3  ⋅ z3 h f ⋅ (η ⋅ f cd ) ⋅ beff ,1 + beff ,2 Asl ,3 =  = f yd ⋅ (d − 0,5 ⋅ h f ) f yd

µ1 =

M Ed − M Rd ,3 2

bw ⋅ d ⋅ f cd

Asl ,1 = ωlim ⋅ b ⋅ d ⋅

=

(

)

)

M Ed −  h f ⋅ (η ⋅ f cd ) ⋅ beff ,1 + beff ,2  ⋅ ( d − 0,5 ⋅ h f ) bw ⋅ d 2 ⋅ f cd

and

f cd f yd

In the previous relations the lever arm for the flanges has been written both in its short form ( z3 ) as well as in its full form (d − 0,5 ⋅ h f ) . Likewise, the width of the flanges is written as ( b3 ) and as

( beff ,1 + beff ,2 ) respectively, although in most cases ( beff ,1 = beff ,2 ) . In the above [eq. 1-27] the web will have to withstand the remaining flexural moment which isn’t resisted by the flanges (therefore the subtraction indicated). In the above [eq. 1-28] ( ω1 ) is the coefficient of mechanical reinforcement usually extracted from Tables for the corresponding value of ( µ1 ) . 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


The cross section is momentarily divided for calculus, see [Figure 7], into the following pieces: < 1 > A singly reinforced web section for which the reinforcement in tension

( Asl ,1 )

can be easily

calculated (similarly to SRRS); (Continued on next page)

[19]





< 2 > An area of reinforcement in tension

( Asl ,2 )

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 Asl ,3 to be in equilibrium ONLY with flange in compression.

(

)

(

The reinforcement to be placed in the cross section is Asl ,1 + Asl ,2 + Asl ,3 for tension and Asl ,2

)

for compression. The formulas to calculate the areas of reinforcement are easily deduced as: [1-34]

[1-35]

(

 h f ⋅ (η ⋅ f cd ) ⋅ b3  ⋅ z3 h f ⋅ (η ⋅ f cd ) ⋅ beff ,1 + beff ,2 = Asl ,3 =  f yd ⋅ (d − 0,5 ⋅ h f ) f yd

µw =

M Ed − M Rd ,3 2

bw ⋅ d ⋅ f cd

=

(

)

)

M Ed −  h f ⋅ (η ⋅ f cd ) ⋅ beff ,1 + beff ,2  ⋅ ( d − 0,5 ⋅ h f ) bw ⋅ d 2 ⋅ f cd

2 1 M Ed ,2 1 M Ed − M Rd ,3 − µlim ⋅ bw ⋅ d ⋅ f cd ⋅ = ⋅ f yd z2 f yd d − d2

[1-36]

Asl ,2 =

[1-37]

Asl ,1 = ωlim ⋅ b ⋅ d ⋅

and

f cd f yd

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 isn’t 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 ( µ w ) for the web and checking if ( µ w > µlim ) . [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 cross section; < 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. [20]





[Part (One) C] Milestone’s 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 of books – 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)

[21]


Section through strut

Section through tie




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 (α ) ; < 2 > The angle between the strut (modelling concrete chord between inclined cracks) and the axis of the longitudinal area of reinforcement (horizontal direction) is (θ ) . Based on experiments it has been

(

)

concurred that the domain for this angle is 21,8o ≤ θ ≤ 45o . Since (as it will be demonstrated) the angle is used with its trigonometric function ( ctg ) , previous relation turns into (1 ≤ ctg θ ≤ 2,5 ) ; < 3 > The distance in-between chords in compression is ( a ) ; < 4 > The compression force in the strut is ( Fcw ) for which the stress (σ cw ) may reach a maximum of

( fcd ) , the design strength of concrete; < 5 > The tension force in the tie is ( Fsw ) for which the stress (σ sw ) may reach a maximum of

( f ywd ) ,

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 ( Asw ) . 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]

VEd = Fcw ⋅ sin θ Since ( Fcw ) is a force resultant of the stresses (σ cw ) acting on the area ( a ⋅ bw ) and furthermore

( a ) may be fully written as a function of the geometry of the truss, the previous becomes: [1-39]

{

}

VEd = ( a ⋅ bw ⋅ σ cw ) ⋅ sin θ =  z ⋅ ( ctg θ + ctg α ) ⋅ sin θ  ⋅ bw ⋅ σ cw ⋅ sin θ = ctg θ + ctg α = bw ⋅ z ⋅ σ cw ⋅ 1 + ctg 2 θ






The above may be further transformed based on the following:

( fcd ) ;

< a > The maximum stress to be attained by concrete is its design stress,

 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

( fcd ) and the stress due to prestressing (σ cp ) which, if present, has a positive effect: for reinforced concrete

1,00 1+

[1-40]

α cw =

for prestressed concrete

σ cp fcd

1, 25 if

0 < σ cp ≤ 0, 25 ⋅ fcd

if

f cd < σ cp ≤ 0,5 ⋅ f cd

 σ cp  2,5 ⋅ 1 −  if 0,5 ⋅ f cd < σ cp < fcd f cd   < c > 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 (σ cw ) develop:

f   0, 6 ⋅  1 − ck  if f ywd > 0,8 ⋅ f ywk  200  [1-41]

υ1 = υ = 0, 6 if 0,9 −

f ck ≤ 60 MPa f ywd ≤ 0,8 ⋅ f ywk

f ck >0,5 200

if

f ck > 60 MPa f ywd ≤ 0,8 ⋅ f ywk

(

)

< d > By considering the real positioning of stirrups along the member, α = 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]

VRd ,max =

α cw ⋅ bw ⋅ z ⋅υ1 ⋅ f cd ctg θ + tg θ

with

VRd ,max is the maximum capacity in shear of the concrete in compression.

(

)

Should the case of loading lead to an inequality such that VEd > VRd ,max , 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.

[23]





For the 2nd section (through the tie) the following apply: [1-43]

VEd = Fsw ⋅ sin α

( Asw s ) uniformly distributed along the member. Since (σ sw ) acting on the longitudinal length z ⋅ ( ctg θ + ctg α )

The stirrups give an equivalent area of

( Fsw )

is a force resultant of the stresses

between a strut and a tie, the previous becomes: [1-44]

A  VEd =  sw ⋅ σ sw ⋅ z ⋅ ( ctg θ + ctg α )  ⋅ sin α  s  The above may be further transformed based on the following: < a > The maximum stress to develop in the tie is the design strength,

( f ywd ) ;

(

)

 By considering the real positioning of stirrups along the member, α = 90o ; < c > The right member of [eq. 1-39] represents the capacity in shear of the inclined tie that is reached upon failure by yielding of the reinforcement in tension. The full form for all the above supplementary definitions incorporated in [eq. 1-39] is: [1-45]

VRd ,s =

Asw ⋅ f ywd ⋅ z ⋅ ctg θ s

(

)

Should the case of loading lead to an inequality such that VEd > VRd , s , the ties would fail by yielding of the reinforcement, making the cross section unsuited to uphold the loading and therefore is a situation to be avoided at all times. Apart of insuring the cross section will uphold the loading acting on it (avoiding the failure of both the strut and the tie), there is an additional condition to be considered. It’s about the type of failure which must be a warned one, making it mandatory that the tie fails in tension prior to the strut failing in

(

)

compression, VRd , s ≤ VRd ,max , which may be written in full form as: [1-46]

Asw α cw ⋅ bw ⋅ z ⋅υ1 ⋅ f cd VEd ≤ = s f ywd ⋅ z ⋅ ctg θ ⋅ ( ctg θ + tg θ ) f ywd ⋅ z ⋅ ctg θ

(

)

Assuming a particular value as ctg θ = 1 : [1-47]

α cw ⋅ bw ⋅υ1 ⋅ f cd  Asw    ≤ 2 ⋅ f ywd  s  max

The above are presented as such to emphasize the upcoming of the relationships to be used further in design. Apart from using vertical stirrups to uphold shear, there is always the possibility to use bent up bars, as part of the bars used primarily for upholding sagging a.k.a. positive bending, which decreases as [24]





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 Advice Up 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 one’s good judgement in deciding whether or not all the steps are necessary in the provided order depending on the design problem at hand.

[25]




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 ( M Ed ) and the maximum shear

(VEd )

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 a “domino 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 it’s 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 it’s 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 produced worldwide). In order to achieve SUSTAINABILITY (by its generalised meaning) it’s 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. [26]




Steps to Design [2]

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 one’s best judgement, the first provided answer in the previously proposed endeavour is to correctly calculate concrete cover.

[Part (Two) C] Concrete cover Any 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 Appendix Default values (recommended)

Pages [43-48] [Table 2 & 3] 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 Appendix Default values (recommended)

Pages [24-42 & 190-191] [Table 6, 7 & 8] C30/37 for concrete & S500 for steel

< 3 > Choosing the steel bar size (herein “diameter” will be referred to by “size”) øsl ,max = 6 [ mm] for slabs

Default values (recommended)

øsl ,max = 25 [ mm] for beams øsl ,max = 28 [ mm] for columns øsw = 8 [ mm ] {6 ÷ 12(14) [ mm]} for stirrups

(

)

(

)

< 4 > Calculate nominal concrete cover for both stirrups, cnom, sw , and longitudinal bars, cnom, sl : [1] [2] [2-1]

[2-2]

cnom = cmin + ∆cdev

cmin

cmin,b = max cmin, dur + ∆cdur ,γ − ∆cdur , st − ∆cdur ,add 10 mm

with

cmin,b [mm] is the concrete cover based on bond conditions (Appendix, [Table 5a]); cmin,dur [mm] is the concrete cover based on exposure conditions (Appendix, [Table 4, 2 & 5b]) ∆cdur ,γ [mm] is the safety cover, ∆cdur ,γ = 0 mm ; ∆cdur , st [mm] is the reduction due to using stainless steel, ∆cdur ,st = 0 mm ; ∆cdur ,add [mm] is the reduction due to additional concrete protection, ∆cdur ,add = 0 mm .

[27]




Steps to Design [2]

∆cdev [mm] is the deviation (tolerances) in actual pouring ( 10 mm ≥ ∆cdev ≥ 5 mm , see [Figure 9] where ( d1 ) 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]

cv = max

cnom , sw cnom , sl − φsw

< 6 > Check if design concrete cover is at least the minimum cover after pouring: [2-4]

cv ≥ cmin,dur + ∆cdev

[Part (Two) D] Sizing of the cross section It’s well known that a board set flat over two supports will bend downward when pressed upon in direct 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)


[28]


60 [mm] for roofs 70 [mm] for civil structures 80 [mm] for industrial structures 100 [mm] for pavements



Steps to Design [2]

The previous will lead to Establishing the height of the cross section ( h ) as a multiple of [50 mm] if the result is less than [800 mm] or as a multiple of [100 mm] otherwise. The width of the cross section ( b ) is (also as a multiple of [50/100 mm] accordingly):

1,5 ≤ h b ≤ 3 for rectangular cross sections

2 ≤ h b ≤ 4 for T sections


bmin ≥ 200 mm

The above are true for members with no connections to other structural/non-structural members that may prevent the member to fall over prior to its fixing in the final position in the structure (mainly precast elements). This is ALWAYS a transitory state for precast members; more details are available in Kiss & Onet (2010) Proiectarea... [9], Pages 332-342. For all other members a thinner web is recommended ( h b ≅ 3) as for reinforced concrete members the height is more relevant to designing than the width. Still, a maximum ratio should be ( h b ≤ 5 ) because a higher ratio may lead to a cross section too thin for withstanding shear. This should be avoided as the best solution (section & corresponding reinforcement) should be reached in one step (with the first chosen section). After calculating the concrete cover and Establishing the dimensions for the cross section, designing may proceed to the main step, the calculation of the reinforcement depending on its shape (rectangular/tee/flanged) and type (singly/doubly reinforced).

[Part (Two) E] Singly reinforced rectangular sections (SRRS) As discussed in the previous chapter, singly reinforcement is the most common type of layout for the bars subjected to flexure. Since all design cases refer to a moment in time just prior to final collapse several conditions are assumed to be true (almost in all situations they are). The most important is that reinforcement yields (warned failure) and therefore the stress developed is the maximum possible value

( )

as given by f yd , the design strength of steel. One should also keep in mind that one-way slabs are designed as strips bearing transversely on the supports with a width of [1 m] (rectangular cross section having the width equal with the unity). < 1 > Calculate the design strengths of the materials (Appendix, [Table 6 to 9]):

[2-5]

fcd = α cc ⋅

fck

γc

and

f yd =

f yk

γs

< 2 > Assume the chosen reinforcement is to be placed in one layer and therefore calculate its axis distance/position in the cross section: [2-6]

d1 = cnom,sl + 0,5 ⋅ φsl

< 3 > Calculate the depth of the cross section: [2-7]

[29]

d = h − d1




Steps to Design [2]

< 4 > Calculate the relative bending moment and check the section to be SR (Appendix, [Table 15]): [2-8]

µ=

M Ed ≤ µlim b ⋅ d 2 ⋅ f cd

< 5 > Extract (Appendix, [Table 16]) or calculate (chapter 1)

(ω )

and calculate the tension

reinforcement: [2-9]

Asl ,1 = ωlim ⋅ b ⋅ d ⋅

(

f cd f yd

< 6 > Choose bar size and no A

sl , eff

)

≥ Asl ,1 (Appendix, [Table 10 & 11]).

Pay attention should one choose a no. of bars which would imply placing on more than one row as assumed. In this case, don’t forget to check in the end the section’s capacity in flexure (it must be at least equal to the bending as given by the loading). Any change in the layout of the reinforcement (placing on multiple rows) affects the position of the tension centroid which is given by the centre of masses of the bars in tension in the cross section thus in turn reducing the depth ( d ) . Since this value is taken into consideration in the formula of

(µ)

at the power of 2 that means any

reduction in depth drastically reduces flexure capacity. That is why at one point (predesign dimensions for cross section) we specified that out of the width ( b ) and the height ( h ) for a given section the most important factor in flexure is having an increased height. The bars should be chosen as having the same size or a maximum combination of two consecutive sizes. It is possible to jump over one size BUT no more than one. As a recommended practice choose bigger sizes and fewer bars as this will lead to less steel consumption in the end. < 7 > Choose the best placing and arrangement for bars (layout) (Appendix, [Table 18]); < 8 > Check the effective reinforcement to fulfil minimum/maximum criterion:

[2-10]

Asl ,min

 0, 26 ⋅ fctm ⋅ bt ⋅ d  f yk = max  ≤ Asl ,eff ≤ Asl ,max = 0, 04 ⋅ b ⋅ d  0, 0013 ⋅ b ⋅ d  t

with

bt is the 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 ) .

The above [eq. 2-10] is limited at the lower end by two conditions that take into consideration the minimum area of reinforcement to be placed in a cross section to control crack opening and crack spacing, respectively. The upper limit is imposed by the assumed collapse mechanism which must involve the crushing of the concrete after reinforcement yields (warned failure).

[30]




 

The ratio  ρ =

Steps to Design [2]

Asl  is named coefficient of longitudinal reinforcement and is the measure of steel b ⋅ d 

consumption for a cross section having values (in most common situations) ( ρ = 0, 005 ÷ 0, 010 ) for slabs and ( ρ = 0, 015 ÷ 0, 020 ) for beams. Figure 10 Strains for tension reinforcement

ε x = c d − x εs

→ ε s = εc ⋅

Strain Distribution

d−x d  = ε c ⋅  − 1 x x 

Demonstration

< 9 > Check that the reinforcement yields (as assumed): [2-11]

f yd d  − 1 ≥ ε yd = Es x 

ε ud ≥ ε s = ε c ⋅ 

with

ε ud = 0, 9 ⋅ ε 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 ( µ > µlim ) 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]

Asl ,2

M Ed − µlim ⋅ b ⋅ d 2 ⋅ f cd = = f yd ⋅ ( d − d 2 ) f yd ⋅ ( d − d 2 ) M Ed ,2

< 2 > Extract (Appendix, [Table 16]) or calculate (chapter 1)

(ωlim ) and

calculate the area of

reinforcement (maximum) in the singly reinforced web: [2-13]

Asl ,1 = ωlim ⋅ b ⋅ d ⋅

f cd f yd

< 3 > Calculate the area of reinforcement to be placed in the cross section: [2-14]

for tension (bottom)

Asl ,1 + Asl ,2

for compression (top) Asl ,2

(

)

< 4 > Choose bar size and no Asl ,eff ≥ Asl ,calc for both of the previous (Appendix, [Table 10 &11]);

[31]




Steps to Design [2]

< 5 > Choose the best placing and arrangement for bars (layout) (Appendix, [Table 18]); Figure 11 Strains for compression reinforcement

x − d 2 ε sc = x ε cu

→ ε sc = ε cu ⋅

Strain Distribution

x − d2  d  = ε cu ⋅ 1 − 2  x x  

Demonstration

< 6 > Check that the reinforcement in compression yields as assumed: [2-15]

 

ε ud ≥ ε s ,2 = ε cu ⋅  1 −

f yd d2   ≥ ε yd = x  Es

< 7 > Should ε s ,2 < 0 ↔ x < d 2 there is no need for supplementary compression reinforcement and: [2-16]

Asl ,calc = Asl ,1

< 8 > Should ε s ,2 ≤ ε yd it means that the compression reinforcement has not reached yielding, therefore:

[2-17]

Asc =

M Ed ,2 f sc ⋅ ( d − d 2 )

= Asl ,2 ⋅

f sc f yd

 d  700 ⋅ 1 − 2  x   = Asl ,2 ⋅ f yd

with

  d   d  f sc = ε sc ⋅ Es → f sc = ε cu ⋅ 1 − 2   ⋅ Es = 700 ⋅  1 − 2  is the stress in the x  x    

[2-18]

compression reinforcement prior to yielding, calculated as of Hooke’s law Since

( Asl ,2 )

is the calculated compression area when it was assumed that reinforcement yields, it’s

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]

f yd d  − 1 ≥ ε yd = Es x 

ε ud ≥ ε s = ε cu ⋅ 

< 10 > Should ε s ≤ ε yd it means that the tension reinforcement has not reached yielding, therefore: [2-20]

Ast =

M Ed ,1 f st ⋅ ( d − 0,5 ⋅ λ ⋅ x )

= Asl ,1 ⋅

700 ⋅ ( d x − 1) f st = Asl ,1 ⋅ f yd f yd

with

(Continued on next page) [32]




Steps to Design [2]

  d  d  f s = ε s ⋅ Es → f s = ε cu ⋅  − 1  ⋅ Es = 700 ⋅  − 1 is the stress in the tension  x  x  

[2-21]

reinforcement prior to yielding, calculated as of Hooke’s law < 11 > Choose bar size and no

( Asl ,eff ≥ Asl ,calc ) for

both tension and compression reinforcement

(Appendix, [Table 10 & 11]); < 12 > Choose the best placing and arrangement for bars (layout) (Appendix, [Table 18]); < 13 > Check the effective reinforcement to fulfil minimum/maximum criterion:

[2-22]

Asl ,min


with

bt 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

h f < 0, 05 ⋅ 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 beff : 2

beff = bw + ∑ beff ,i ≤ b

[2-23]

1

beff ,i = 0, 2 ⋅ bi + 0,1 ⋅ lo ≤ min ( 0.2 ⋅ lo ; bi )

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:

(

)

(

M Rd , f =  h f ⋅ beff ⋅ (η ⋅ f cd )  ⋅ d − 0, 5 ⋅ λ ⋅ h f

[2-24]

)

< 4 > Should M Rd , f ≥ M Ed ↔ the neutral axis x lies in the flange and designing is as for a rectangular with beff ⋅ h ;

[33]




Steps to Design [2]

< 5 > Should M Rd , f < M Ed ↔ 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]

(

) (

M Rd ,3 = (η ⋅ f cd ) ⋅ h f ⋅ beff ,1 + beff ,2  ⋅ d − 0, 5 ⋅ h f

)

< 7 > Calculate the area of reinforcement to be in equilibrium with the compression in the flanges: [2-26]

Asl ,3 =

(η ⋅ f cd ) ⋅ h f ⋅ ( beff

− bw

f yd

)

the form for the flanges of the general Asl =

1 M Ed ⋅ f yd z

< 8 > Calculate the relative bending moment and check the section to be SR (Appendix, [Table 15]): [2-27]

µw =

M Ed − M Rd ,3 bw ⋅ d 2 ⋅ f cd

≤ µlim

< 9 > Extract (Appendix, [Table 16]) or calculate (ω ) (Chapter 1) and Establish the area of tension reinforcement in equilibrium with the compression in the web: [2-28]

f cd f yd

Asl ,1 = ω1 ⋅ bw ⋅ d ⋅

< 10 > Calculate the area of reinforcement to be placed in tension: [2-29]

  beff  h f  bw ⋅ d ⋅ fcd Asl ,tot = Asl ,1 + Asl ,3 = ω + η ⋅  − 1 ⋅ ⋅ f yd  bw  d  

(

)

< 11 > Choose bar size and no Asl ,eff ≥ Asl ,tot for the previous (Appendix, [Table 10 & 11]); < 12 > Choose the best placing and arrangement for bars (layout) (Appendix, [Table 18]); < 13 > Check the effective reinforcement to fulfil minimum/maximum criterion:

[2-30]

Asl ,min


with


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





Steps to Design [2]

[Part (Two) H] Doubly reinforced tee/flanged sections (DRFS) The proposed steps presented in this subchapter are true for that particular situation for which

( µ > µlim ) . 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]

Asl ,2 =

M Ed ,2 f yd ⋅ ( d − d 2 )

=

M Ed

− M Rd ,3 − µ lim

2

⋅ bw ⋅ d ⋅ fcd

f yd ⋅ ( d − d 2 )

< 2 > Extract (Appendix, [Table 16]) or calculate (chapter 1) (ω ) and calculate the area of tension reinforcement in equilibrium with the compression in the web: [2-32]

Asl ,1 = ωlim ⋅ bw ⋅ d ⋅

f cd f yd

< 3 > Calculate the area of reinforcement to be placed in the cross section: [2-33]


Asl ,1 + Asl ,2 + Asl ,3

for compression (top) Asl ,2

< 4 > Choose bar size and no.

( Asl ,eff ≥ Asl ,calc ) for tension and compression reinforcement as above

(Appendix, [Table 10 & 11]); < 5 > Choose the best placing and arrangement for bars (layout) (Appendix, [Table 18]); < 6 > Check that the reinforcement in compression yields as assumed: [2-34]

 

ε ud ≥ ε s ,2 = ε cu ⋅  1 −

f yd d2   ≥ ε yd = x  Es

< 7 > Should ε s ,2 < 0 ↔ x < d 2 there is no need for supplementary compression reinforcement and: [2-35]


Asl ,1 + Asl ,3

< 8 > Should ε s ,2 ≤ ε yd it means that the compression reinforcement has not reached yielding and therefore: [2-36]

Asc =

M Ed ,2 f sc ⋅ ( d − d 2 )

= Asl ,2 ⋅

700 ⋅ (1 − d 2 x ) f sc = Asl ,2 ⋅ f yd f yd

< 9 > Check that the reinforcement in tension yields as assumed: [2-37]

[35]

f yd d  − 1 ≥ ε yd = Es x 

ε ud ≥ ε s = ε cu ⋅ 




Steps to Design [2]

< 10 > Should ε s ≤ ε yd it means that the tension reinforcement has not reached yielding and therefore: [2-38]

Ast =

M Ed ,1 f st ⋅ ( d − 0,5 ⋅ λ ⋅ x )

< 11 > Choose bar size and no

= Asl ,1 ⋅

700 ⋅ ( d x − 1) f st = Asl ,1 ⋅ f yd f yd

( Asl ,eff ≥ Asl ,calc ) for

both tension and compression reinforcement

(Appendix, [Table 10 & 11]); < 12 > Choose the best placing and arrangement for bars (layout) (Appendix, [Table 18]); < 13 > Check the effective reinforcement to fulfil minimum/maximum criterion:

[2-39]

Asl ,min


with


( 0, 26 )

is replaced by ( 0,50 ) .

[Part (Two) I] Shear design It’s 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

[36]




Steps to Design [2]

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 θ + ctg α ) , code provisions allow to further reduce 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 T =

M Ed . The vertical z

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 node where 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]

FE =

z ⋅ ctg θ z ⋅ ctg α − VEd ⋅ = ∆M 2 2 VEd ⋅ 0,5 ⋅ z ⋅ ( ctg θ − ctg α ) = FE ⋅ z VEd ⋅

with

VEd ⋅ al is the additional tension in the longitudinal reinforcement due to shear alone; z

al = 0,5 ⋅ z ⋅ ( ctg θ − ctg α ) is the “shift rule” for the bending moment envelope. Figure 13 Variable angle truss model for shear design


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]

[37]

shifted M Ed VEd ⋅ al M Ed T= + = z z z




Steps to Design [2]

Should an axial force ( N Ed ) 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]

VEd ,red = VEd − F ⋅ ( a1 + d )

with

a1 , (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: 13 C ⋅ k ⋅ (100 ⋅ ρl ⋅ f ck ) + k1 ⋅ σ cp  VRd ,c = max  Rd ,c  ⋅ bw ⋅ d υmin + k1 ⋅ σ cp  

[2-43]

CRd ,c = k = 1+

0,18

γc

=

with

0,18 = 0,12 [no units] is the coefficient provided by national code provisions; 1,5

200 ≤ 2, 0 [no units] is the size factor taking into account the influence of the height of the d

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

ρl =

Asl ≤ 0, 02 [no units] is the coefficient of longitudinal reinforcement taking into account the bw ⋅ d

dowelling effect with a maximum value of „0,02”; Asl [mm2] is the area of tension reinforcement which extends beyond a considered section (named „A”)

with a minimum length of d + lbd (see [fig. 14]); Figure 14 Definitions for anchoring reinforcement at the supports

lbd [mm] is the design anchorage length; bw [mm] is the minimum transverse tension width of the cross section;

[38]




σ cp =

Steps to Design [2]

N Ed ≤ 0, 2 ⋅ f cd [MPa] is the mean (medium/average) stress due to an axial force in the cross Ac

section; N Ed [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;

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

υmin = 0, 035 ⋅ k 3 2 ⋅ f ck1 2 [N/mm2] is the minimum shear capacity of a concrete section (NO longitudinal reinforcement). < 3 > Should VEd ,red ≤ VRd ,c 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 > ρw =

0.08 ⋅ f ck Asw ≥ ρ w,min = s ⋅ bw ⋅ sin α f yk

is the coefficient of transverse reinforcement

having a minimum value as indicated; Asw 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

sl ,max = 0, 75 ⋅ d ⋅ (1 + ctg α ) ; < d > The maximum centre to centre distance in-between vertical stirrups’ arms is

st ,max = 0, 75 ⋅ d ≤ 600 mm . < 4 > Should VEd ,red > VRd ,c 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 doesn’t crush in compression: [2-44]

VRd ,max =

α cw ⋅ bw ⋅ z ⋅υ1 ⋅ f cd ≥ VEd , red ctg θ + tg θ

with

for reinforced concrete 1,00

α cw =

1 + σ cp fcd for prestressed concrete

1, 25 if

(

if

0 < σ cp ≤ 0, 25 ⋅ fcd

f cd < σ cp ≤ 0,5 ⋅ fcd

2, 5 ⋅ 1 − σ cp f cd

)

if 0, 5 ⋅ f cd < σ cp < fcd

is the coefficient taking into account the influence of the stress state in the section;

(Continued on next page)

[39]




Steps to Design [2]

is the coefficient taking into account the influence of the crack state in the section; fck is the characteristic compressive strength of concrete; f ywk is the characteristic tensile strength

f   0, 6 ⋅  1 − ck  if f ywd > 0,8 ⋅ f ywk  200 

υ1 = υ = 0, 6 if 0,9 −

f ck ≤ 60 MPa f ywd ≤ 0,8 ⋅ f ywk

f ck >0,5 200

if

of transverse reinforcement;

f ck > 60 MPa

f ywd is the design tensile (yielding)

f ywd ≤ 0,8 ⋅ f ywk

strength of transverse reinforcement;

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

θ = 21,8 → ctg 21,8 = 2,5 ↔ VRd ,max,inf [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 ctg α = 0 for stirrups are vertical ( α = 90o ). Please now focus on

(θ )

and the

corresponding ctg θ and on ( Fcw ) and the corresponding sin θ respectively. < a > On one hand, as (θ ) decreases to a MINIMUM ctg θ increases to a MAXIMUM making the chord width ( a ) larger. On the other hand, as (θ ) decreases to a MINIMUM sin θ decreases also to a MINIMUM. Therefore, the vertical component of the shear capacity of concrete

( Fcw ⋅ sin θ )

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

MINIMUM ( ctg θ = 2, 5 DEFAULT VALUE ONLY in the above); Should VRd ,max,inf ≥ VEd ,red hold true the strut at its lowest capacity can withstand shear without crushing of the concrete; < c > Should VRd ,max,inf < VEd ,red 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 it’s necessary to change the cross section by (recommended practice) increasing its width ( b ) . Figure 15 Variable angle truss model for shear design

Truss model parameters (longitudinal section)

[40]



Section through tie



Steps to Design [2]

< 6 > Calculate the value for ctg θ based on the particular pattern of loading by interpolation: [2-45]

VRd ,c V V ≤ Ed ,red ≤ Rd ,max,inf ctg θ 2,5 1

In Chapter 1 (see [eq. 1-39] and the explanations there) it was concluded that < a > The lower limit

(VRd ,c )

VRd , s Asw . ≤ s f ywd ⋅ z ⋅ ctg θ

is met for members with CONSTRUCTIVE SHEAR

REINFORCEMENT (minimum stirrup area at the maximum space). Therefore the ratio ( Asw s ) is MINIMUM making in turn ctg θ be MAXIMUM ( ctg θ = 2, 5 ).

(

 The upper limit VRd ,max,inf

) is met for members with HEAVY SHEAR REINFORCEMENT

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

( Asw s )

is MAXIMUM

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

(

< c > VEd , red

)

slides in-between the above limits dependent on the load. As more shear

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

(

“2.5” to “1.0” while VEd , red

) moves away form (VRd ,c ) and closes (VRd ,max,inf ) .

< 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]

VEd ,red  Asw    =  s  rqd z ⋅ f ywd ⋅ ctg θ

< 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 Asw (use Appendix, [Table 11]):

[2-47]

srqd =

Asw VEd ,red z ⋅ f ywd ⋅ ctg θ

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]

( Asw )rqd

  VEd , red = s ⋅ and then choose the size for stirrups knowing the no. of  z ⋅ f ywd ⋅ ctg θ   

vertical arms Remember that (recommended practice): < a > 100 mm ≤ s ≤ 300 mm , for all structures in seismic zones; 100 mm ≤ s ≤ 400 mm , otherwise; [41]




Steps to Design [2]

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

 Asw   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 it’s 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]

ρw =

0, 08 ⋅ fck Asw ≥ ρ w,min = s ⋅ bw ⋅ sin α f yk

< 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]

 Asw  ≤  Asw  = α cw ⋅ bw ⋅υ1 ⋅ f cd     2 ⋅ f ywd  s eff  s  max

< 11 > Additional checks: < a > The

maximum

centre

to

centre

distance

for

stirrups

over

the

member

is

sl ,max = 0, 75 ⋅ d ⋅ (1 + ctg α ) ; The maximum centre to centre distance in-between vertical stirrups’ arms is

st ,max = 0, 75 ⋅ d ≤ 600 mm . With this final step the design in shear is over. Still, it’s necessary to calculate the anchorage length for the longitudinal reinforcement to insure that indeed the bars won’t 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 T = FE , which may further be written as: [2-51]

FE = σ sd ⋅ Asl = σ sd ⋅ n ⋅

π ⋅φ 2

with

4

σ sd is the stress to develop in a bar; n is the no. of bars to be anchored at the support;

φ is the bar size (it’s assumed for easier calculation that all bars are the same size).

[42]




Steps to Design [2]

The bond that develops on the surface of all the bars in the cross section is: [2-52]

Fbd = Aslsurface ⋅ fbd = n ⋅ [π ⋅ φ ⋅ lb ] ⋅ f bd

with

π ⋅ φ ⋅ lb is the surface of the bar on which bond acts; lb is the anchorage length; fbd is the design bond strength. By setting the above forces equal to each other (equilibrium): [2-53]

n ⋅ [π ⋅ φ ⋅ lb ] ⋅ fbd = σ sd ⋅ n ⋅

π ⋅φ 2 4

→ lb = 0, 25 ⋅ φ ⋅

σ sd fbd

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]

FE =

VEd ⋅ al z

with

al = 0,5 ⋅ z ⋅ ( ctg θ − ctg α ) is the “shift rule” for the bending moment envelope. < 13 > Calculate the stress acting in the bars: [2-55]

σ sd =

FEd Asl ,anchored

< 14 > Calculated the required anchorage length:

σ sd fbd

[2-56]

lb,rqd = 0, 25 ⋅ φ ⋅

[2-57]

fbd = 2, 25 ⋅ h1 ⋅ h2 ⋅ f ctd

and with

fbd [MPa] is the design bond stress (see [Table 1] or Appendix, [Table 19]); h1 = 1, 00 for good bond conditions h1 = 0, 70 for poor bond conditions or stepping formwork (if not proven otherwise)

h2 = 1, 00 for φ ≤ 32 mm

h2 =

132 − φ otherwise 100

f ctd [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]

lbd = α 1⋅α 2 ⋅α 3 ⋅α 4 ⋅α 5 ⋅lb,rqd ≥ lb,min

with

α1 [no units] is the coefficient taking into account the type of the end of the bar when concrete cover is

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

[43]




Steps to Design [2]

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

(stirrups); α 4 [no units] is the coefficient taking into account the case of transverse bars welded to the longitudinal

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

given by a concentrated force ( Ft ) as the result of the loading over the length ( t1 ) and width of the support, ( b ) ;

lb,min , minimum anchorage length in the absence of other provisions, for reinforcement in: [2-59]

for tension

lb,min =

for compression

{ } max {0, 6 ⋅ lb,rqd ; 10φ ; 100 mm} max 0,3 ⋅ lb,rqd ; 10φ ; 100 mm

Table 1 Bond Conditions (“A” is the direction of concrete pouring)

Good bond conditions for all bars

all similar cases whether in tension or compression

45o ≤ α ≤ 90o

h ≤ 250 mm

h > 600 mm

h > 250 mm

Hatched area – poor bond conditions Plain area – good bond conditions

all similar cases whether in tension or compression

Remember that any of the above coefficients α1 ÷ α 5 reduces the anchorage length (all are ≤ 1). Still, don’t forget about the condition: [2-60]

α 2 ⋅ α3 ⋅ α5 ≥ 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 (α1 ÷ α 5 ) please see [Table 2].





Steps to Design [2]

Table 2 Anchorage length parameters

Straight end

Hooked End

U Shape End

 a 2  cd = min  c1  c 

 a 2 cd = min   c1

α 4 = 0, 70 for all similar

cd = c

α1 = 0,70 for all similar cases in tension for which

cases whether in tension or compression

cd > 3 ⋅ φ

cd − φ

α 2 = 1 − 0,15 ⋅

φ

≥ 0, 70 ≤ 1, 00

α1 = 1, 00 otherwise

cd − 3 ⋅ φ

φ

≥ 0, 70 ≤ 1, 00

for all similar cases in tension

cases whether in tension or compression

α 4 = 1,00 otherwise

α1 = 1, 00 for all similar

α 2 = 1 − 0,15 ⋅

Transverse welded bar

α 2 = 1,00 otherwise


λ=∑

Ast − ∑ Ast ,min As

∑ Ast ,min = 0, 25 ⋅ As for beams

∑ Ast ,min = 0 otherwise 0, 70 ≤ α3 = 1 − k ⋅ λ ≤ 1, 00

∑ Ast


is the area of stirrups placed over lbd

∑ Ast ,min is the minimum area of stirrups placed over lbd

α 3 = 1, 00 otherwise

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

0, 70 ≤ α5 = 1 − 0, 04 ⋅ p ≤ 1, 00

[45]


p=

Ft b ⋅ t1

Manual for Advanced Design

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