CH7 - Design for Shear Loads

8/21/2019 CH7 - Design for Shear Loads

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7-1

7 Chapter 7Design for shear loads

7.1 IntroductionThe aim of chapter 4 was to show how to calculate the main reinforcement of a beamsubjected to bending moments and axial forces which give rise to distributions ofnormal stresses in cross-sections. The main principle was that steel reinforcement is

provided in those areas where the concrete cracks due to large tensile stresses.

Figure 7.1-1 shows a beam in a four point bending disposition. The central part of thebeam is loaded in pure bending; the longitudinal reinforcement is calculated asindicated in chapter 4. The two parts of the beam between the concentrated forces andthe supports are subjected to a more complex loading because of the combination of the

bending moment and the shear force. Yet, in the early days of reinforced concrete,

people tried out the behaviour of beams with only longitudinal reinforcement andobserved for increasing loads the appearance of inclined cracks in the zones with shearloads. Without special reinforcement to bridge the inclined cracks, it is even observedthat failure of the beam is determined by shear: one crack is prolonged suddenly up tothe upper side of the beam which causes the total collapse of the structural element (asshown in figure 7.1-2). This type of failure happens in a sudden (brittle) way and hasthus absolutely to be avoided. The logical solution is to provide inclined reinforcement,

perpendicular to the cracks (figure 7.1-3), but a valuable alternative is to use verticallinks (or stirrups) which bridge the crack at a certain angle.

Shear force

Bending moment

Figure 7.1-1

Four point bending test, applied on a beam with only longitudinal reinforcement


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7-2

Figure 7.1-2

Typical failure mode for a beam with only longitudinal reinforcement: a shear crackleads to total collapse in a sudden way

(b)(a)

Figure 7.1-3Two possible solutions to bridge shear cracks; (a) inclined shear reinforcement:

longitudinal bars (main reinforcement on the bottom side) may be bent up to the upperside of the beam instead of being simply curtailed; (b) vertical shear reinforcement:

links or stirrups

7.2 Members not requiring shear reinforcement7.2.1 IntroductionProviding shear reinforcement leads to a substantial cost; it is thus useful to analyse theconditions which may allow omitting this type of reinforcement. This paragraph focuseson the determination of the shear resistance of members without shear reinforcement.

7.2.2 A starting point: overview of results from theory of elasticity for beams withcontinuous, homogeneous, isotropic and elastic materials

The following paragraph presents an overview of main notions and formulas concerningshear forces and shear stresses in a beam loaded in bending, taken from theory ofelasticity and strength of materials courses. The formulas are valid for homogeneous,

isotropic, continuous and elastic materials. The setting is defined in figure 7.2.2-1.


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7-3

M

V+

-y

zG -

x

P

Vy> 0 Mz> 0

x

NA

N' N'+dN'

dx

Figure 7.2.2-1

Principle figure for the elaboration of the formulas for shear stresses in beams loaded in

bending

Main results are:

longitudinal shear force dxbdN b ..' =

rotation equilibrium in a cross-section leads to (figure 7.2.2-2):

zNM '.=

zdNdM '.=

zNAM

N'

N

Figure 7.2.2-2

Rotation equilibrium in a cross-section of a beam loaded in bending

the relationship between VyandMz(in absolute values):

dx

dMV zy =

the shear stress on the level of the NA:

zb

V

dxzb

dM

dxb

dNNL

....

'=== (7.2.2-1)

the formula of JOURAWSKI for the shear stress in a certain point (or on a certainlevel) of the cross-section:


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z

zxy

Ib

SV

.

.= (7.2.2-2)

with

xy they-component of the shear stress on an elementary surface perpendicular to

thex-axis;V the shear load in the cross-section;

b the width of the cross-section at the level where the shear stress isdetermined;

zI the moment of inertia of the full cross-section with respect to thez-axis (axispassing through the centre of gravity G);

zS the static moment of the part of the cross-section situated above the levelwhere the stress is determined, with respect to thez-axis.

xy is characterized by a parabolic distribution for a rectangular cross-section,

with maximum value at the level of the NA equal tohb

V

..

2

3.

Figure 7.2.2-3 presents a beam loaded by a uniformly distributed load. In uncrackedsituation, and assuming continuous, homogeneous, isotropic and elastic material, oneobtains the set of trajectories of the principal stresses. The orientation and magnitude ofthe stresses are determined in each point with a theory of elasticity approach; MOHRscircle can be used for graphical representation. Figure 7.2.2-4 presents in a schematicway the reasoning that permits to determine the principal orientations, the principal

elementary areas and principal stresses in a point A on the NA (x= 0; xy max) and inpoint B in the cross-section. It is observed in point A that the principal tensile stress hasthe same magnitude as the shear stress and is oriented with an angle of 45 with respectto the axis of the beam. In punt B, the principal elementary area with the largest

principal tensile stress is much more horizontally oriented. These results help tounderstand the crack pattern due to shear load in a beam in reinforced concrete withonly main reinforcement and in which, from a macroscopic point of view, the concretemay be considered as a homogeneous material: see figure 7.2.2-5.

Figure 7.2.2-3

Principal stress trajectories in uncracked situation (continuous, homogeneous, isotropicand elastic material)


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7-5

x

y +

+

V> 0

Ib

SV

.

.=

A

B

XA

YA= poleA

1A

YBpoleB

1B

XB

A

B

Tensile zonecracks

Figure 7.2.2-4

Application of MOHRs circle for the identification of the principle tensile stress at theNA (axis of the beam); deduction of the crack pattern influenced by the presence of

shear


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Figure 7.2.2-5

Figure (a) presents the trajectories of the principle compression stresses in an uncrackedbeam; figure (b) presents the experimentally observed crack pattern obtained by a four

points bending test on a beam in reinforced concrete without shear reinforcement(WIGHT, 2009)

7.2.3 Effect of the cracking in reinforced concrete (beam without shearreinforcement)

The appearance of cracks has an important influence on the further distribution ofinternal forces. As cracks develop in the lower part of the beam, the NA is shifted

upwards which leads to the vertical elongation of the cracks; these cracks only deviatetowards the 45 orientation on the level of the new NA. This explains why the crack

pattern shown in figure 7.2.3-1 is characterized by much more vertical cracks than the45 disposition in uncracked material. When load intensity increases, crackingcontinues until one crack becomes instable: that means that the crack develops in a

brittle way over the whole depth of the beam. Internal equilibrium is not possible anymore and failure is reached.

Figure 7.2.3-1


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7-7

Crack development with increasing load, in a beam in reinforced concrete with mainreinforcement and without shear reinforcement (WALRAVEN, 1995)

Another consequence of the cracking is that equations (7.2.2-1) and (7.2.2-2) are strictlynot valid anymore. Moreover, the stress distribution in the cross-section, in ULS, is non-linear and is thus highly different from the distribution in uncracked state.

Note:

WALRAVEN (1995) assumes that the following formula still allowsdetermining a reasonable estimation of the mean" shear stress in a section inreinforced concrete:

zb

V

.= (7.2.3-1)

with

b= the width of the cross-section or the minimum width of the web of I- or T-beams;

z= the lever arm, which in first approximation can be taken as 0,9.d.

7.2.4 Mechanisms of the transfer of shear loads in a cracked beam in reinforcedconcrete

Figure 7.2.4-1 gives an overview of the different mechanisms which explain the transferof the shear load in a beam in reinforced concrete without shear reinforcement.

(a)

VV

(b)

(c)

(d)

Figure 7.2.4-1

Mechanisms for shear load transfer in cracked reinforced concrete:

(a) uncracked concrete in compression; (b) tensile stresses at the tip of the crack;

(c) granulate interlocking; (d) dowel action

The following mechanisms are identified:

the uncracked compression concrete in the upper part of the beam (above theshear crack) is able to transfer high shear loads;


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tensile contact stresses are present at the crack tip as long as both sides are notseparated more thanw 0,15 mm (WALRAVEN, 1995). In order to further openthe crack tip, an additional tensile force has to be developed;

the shear displacement of one part of the beam with respect to the other part ishindered by the mechanical friction resistance provided by the sliding of two

irregular crack surfaces. This is called the aggregate interlocking effect; the shear displacement of one part of the beam with respect to the other part is

also hindered by the dowel action of the main reinforcement bars. On top of thelocal shear resistance of the steel bars, one may also take account of the resistanceto local crushing of the concrete adjacent to the bars: figure 7.2.4-2.

V

V

Figure 7.2.4-2

Dowel action of the main reinforcement and resistance to local crushing of theconcrete adjacent to the bars

It can thus be concluded that the following factors determine the shear load bearingcapacity of beams without shear reinforcement:

the concrete class;

the main reinforcement ratio (a larger ratio also leads to smaller crack widths);

the width of the cross-section;

the depth of the cross-section. An important observation is that shear load bearingcapacity indeed increases with depth but less than proportional. This is a wellknown phenomenon in the course on failure mechanics: a large crack is moresensitive for instable elongation than a short crack (small sections are moreefficient to bear shear loads);

an eventual axial force, which may influence the crack width.


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7.2.5 The shear resistance of a beam in reinforced concrete without shearreinforcement

Reference: EN 1992-1-1:2004; 6.2.2

The design value for the shear resistance VRd,cin a beam without shear reinforcement isdetermined by means of the following empirical formula:

dbkfkCV wcpcklcRdcRd ...)..100.(. 13

1

,,

+= (7.2.5-1)

with a minimum of:

[ ] dbkvVwcpminimumcRd

...1min,,

+= (7.2.5-2)

where:

VRd,cis expressed in N;

fckis expressed in MPa;

d

k200

1+= with din mm and k 2;

d= effective depth determining the distance between the centre of gravity of themain reinforcement to the most compressed concrete fibres (top layer of beam);

wb = smallest width of the cross-section in the tensile area;

c

Edcp

A

N= < 0,2.fcd

EdN (expressed in N): axial force in the cross-section due to loading or

prestressing (positive sign for compressive load);

cA (expressed in mm2): area of concrete cross-section;

02,0. =

db

A

w

sll ;

with =slA area of the tensile reinforcement which extends at the least over thedistance d+lbdbeyond the section considered (see figure 7.2.5-1). Note: lbdis therequired anchorage length, discussed in chapter 6 in these course notes;

c

cRdC

18,0, = ; assuming c= 1,5 leads to CRd,c= 0,12;

k1= 0,15;

vmin= 0,035.k3/2.fck

1/2;

The introduction of the recommended values k1 = 0,15 , CRd,c = 0,12 and vmin in

equations (7.2.5-1) and (7.2.5-2) leads to the following equations for the design value ofthe shear resistance of a beam without shear reinforcement:


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dbfkV wcpcklcRd ...15,0)..100.(.12,03

1

,

+= (7.2.5-3)

with a mimimum of:

dbfkV wcpckimumcRd ...15,0..035,02/12/3

min,, += (7.2.5-4)

Figure 7.2.5-1

Definition ofAslin the formula for the calculation of the shear resistance of a beamwithout shear reinforcement: one can only take account of those bars which are

adequately anchored; (a) end support; (b) intermediate support (Figure 6.3 in EN 1992-1-1:2004)

The verification of the shear load bearing capacity of a structural member without shearreinforcement is thus performed by the comparison, in the cross-section to beconsidered, of the design value of the imposed shear load VEdwith VRdc.

7.3 Members requiring design shear reinforcement7.3.1 IntroductionIf preliminary calculation shows that the shear load bearing capacity of the memberwithout shear reinforcement, is not large enough to withstand the imposed shear force

(thus if VEd > VRd,c), an adequate shear reinforcement is necessary. The shearreinforcement provides replacement of the shear load bearing capacity which disappearsgradually with growing cracks, the reduction of the thickness of the compressedconcrete arch and the increased crack width which reduces the granulate interlockingresistance. The presence of shear reinforcement allows to further increase loads whileavoiding catastrophic beam shear failure before the full exploitation of the bendingcapacity.

Throughout the years, it was not easy to find an international agreement on a shearreinforcement calculation model. The models proposed in the CEB-FIP Model Code(precursor of EC2) and later on in the EC2, have been reworked several times.

The shear reinforcement calculation model has been developed on the basis ofremarkable experimental results.


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7.3.2 Remarquable experimental results7.3.2.1Result 1: beams in reinforced concrete may be analyzed by means of an

analoguous truss

Experiments on beams with shear reinforcement (links for example) reveal that thecrack pattern is determined by the presence of the links: figure 7.3.2-1. The cracks in thezone loaded by shear, show a regular pattern and are even somewhat parallel in long

beams. In between the cracks, compressive concrete struts are identified. The strutsguide the loads applied on the upper side of the beam towards the lower side of the

beam; from there on, the loads are back again transferred towards the upper side bymeans of the links; this is a regular process all along the length of the beam.

These experimental observations are the basis of the papers written independently bythe Swiss engineer RITTER in 1899 and the German engineer MRSCH in 1902, inwhich they both proposed to describe the shear load transfer in reinforced concrete

beams by means of an analogous truss (WIGHT, 2009).

z.cotg

z

V

Asw

s

Figure 7.3.2-1

Schematic representation of the regular crack pattern in a beam in reinforced concretewith shear reinforcement: identification of an analogous truss.Aswrepresents the cross-

section of 1 link with two legs

The truss system is composed of fours types of members:

- the non-cracked arch with compressed concrete at the upper side of thebeam, acts as the top compression member of the truss;

- the horizontal tension steel (main reinforcement) acts as bottom chord of thetruss; the distance between top and bottom member is the lever armz;


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- the diagonal compression members inclined at an angle , represent theconcrete compression struts between the (parallel) shear cracks;

- the transverse tension members in the truss, characterized by the distancez.cotg between them, represent the shear reinforcement (in this examplecomposed of vertical links).

7.3.2.2Result 2: the relationship between imposed shear load and the necessary shearreinforcement

The areaAsvof a vertical member in the truss in figure 7.3.2-1, is equal to:

s

zAA sws

cotg..= (7.3.2-1)

with

Asw the cross-section of 1 link (2 vertical legs);s the distance between adjacent links.

The force that has to be resisted by the vertical member is indeed the shear load V.Consequently, the tensile stress svin the vertical member is:

cotg.

.z

s

A

V

A

V

sws

s == (7.3.2-2)

The steel stress (in the links) has to be limited to the design strength fywd. This

reasoning, fully based on the truss analogy, leads to the value of the maximum shearload that can be supported:

ywdsw

u fzs

AV .cotg.. = (7.3.2-3)

However, experimental results (WALRAVEN, 1995) show that the real behaviour doesnot fully coincide with the one suggested by the truss analogy. Figure 7.3.2-2 shows, ina schematic way, the experimentally measured relationship between the steel stress svin the shear reinforcement and the applied shear load V; the solid line shows the

experimental relationship while the dashed line shows the relationship according to thetruss analogy via expression (7.3.2-2).

Vc

Applied shear

loadV

Effective stress

measured in thevertical legs of the

linksstress in linksaccording to truss

analogy model

fywd

sv

Vc


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Figure 7.3.2-2

The steel stress svin the vertical links in function of the imposed shear load V: solidline = experimental measurement; dashed line = theoretical relationship according to the

truss analogy

Figure 7.3.2-2 shows that the shear reinforcement is practically not working whensmall values of shear loads are applied. Shear reinforcement is only activated (increaseof steel stress sv in the links) from the moment on that a shear crack appears. It islearned from the experiments that in cracked situation, the imposed shear load V istransferred by two mechanisms:

- partly by the truss mechanism;- partly by an extra bearing mechanism, which can be explained by:

the fact that the hinges in the idealized truss system are nothinges at all in reality; the nodes of the truss transfer alsomoments;

crack surfaces are not smooth and straight, but are very irregularin shape;

a part of the load is transferred by the uncracked compressionarch to the supports and by the dowel action of the mainreinforcement.

The sum of all non-truss mechanisms can be called Vc; it is as if this part of the loadtransfer is taken care off by the concrete (c< concrete). It is observed that

- Vc is practically constant during loading, on the condition that themechanisms which explain the concrete part Vcare not too much destroyed

by too large crack widths;

- Vcis practically equal to the shear load that causes inclined cracks to appear.This leads to the assumption that this shear load is nothing else than theshear load bearing capacity VRd,c of the same beam but without shearreinforcement.

The experimental result mentioned above, has been confirmed for cross-sections withvarious shapes and reinforcement ratios. It is an important result which has lead to therule in earlier versions of EC2 (1995, 1998) that shear reinforcement in beams could becalculated for the shear load (V-VRd,c) only. The actual version of EC2 (2004) adoptsanother point of vue (see further).

7.3.3 Analogous truss models


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hd

s

Fcd

Ftd

Figure 7.3.3-1

Basic model of analogous truss system for the development of the formulas for shearload verification of a beam in reinforced concrete

Figure 7.3.3-1 presents the general truss model that is used for the development of theformulas for shear load verification of a beam in reinforced concrete. The inclinationangle of the shear reinforcement with the beams axis is called . For inclined bars:< 90 (typical 45); for vertical links: = 90. The inclination angle of the cracks, andthus also the inclination angle of the concrete compression strut, is called . The limitvalues for the angle are fixed in the standard:

EN 1992-1-1:2004; 6.2.3(2):

1 cotg 2,5 (7.3.3-1)

which corresponds to:

45,0 21,8 (7.3.3-2)

The Belgian ANB takes account of the effect of an eventual axial force or prestressingforce which lead to less inclined cracks; this is illustrated by the principle reasoning bymeans of MOHRs circle in figure 7.3.3-2. The ANB defines the limit values for asfollows:

1,0 cotg cotg max (7.3.3-3)

with

3..

....2cotg 1 +=

ywdsw

wcp

maxfzA

sdbk (7.3.3-4)

where: k1, cp, bw, d: defined in paragraph 7.2.5;


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Asw = cross-sectional area of one shear reinforcement: one inclined bar or twovertical legs of one link;

s = spacing of the adjacent shear reinforcement; z = lever arm between the compression and tensile members of the truss;

ANB acceptsz= 0,9.dif cp= 0; fywd = design yield strength of shear reinforcement.

With cp non 0, one may even adopt cotg = 3; this assumption corresponds to veryslightly inclined cracks, with an inclination angle of only 18,4.

If cp= 0, application of ANB leads to the following limit values:

1 cotg 2 (7.3.3-5)

which corresponds to:

45,0 26,6 (7.3.3-6)


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y +

+

V> 0

Ib

SV

.

.=

A

XA

Y'A=YA= poolA

1A

y +

+

Ib

SV

.

.=

A'

(druk)

X'A

poolA'

1A'

XA

YA

X'A

Y'A

Figure 7.3.3-2

Auxiliary reasoning by means of MOHRs circle to show that the presence ofaxial compression stresses leads to a less inclined crack angle


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Note:

It is thus observed that the actual standard accepts the choice of rather smallvalues of the crack inclination angle and thus of the concrete compressionstruts in the truss model. The justification for this choice and the discussion of itsconsequences is presented further in this chapter.

7.3.4 Design of the truss members7.3.4.1IntroductionThe truss model in figure 7.3.3-1 contains four components:

the vertical or inclined tension reinforcement which represent the shearreinforcement (links or stirrups or inclined bars);

the concrete compression struts, with an inclination angle ;

the compression member on top;

the tension member at the bottom (the bottom chord member).

Design for shear means that each of all four members of the truss is designed strongenough in order to make the beam able resisting the imposed shear load.

7.3.4.2The shear reinforcement1. The force in the truss member

The assumed truss model is once again presented in figure 7.3.4-1. The method ofsections (method of RITTER) may be applied to determine the force T in theinclined truss member; vertical translation equilibrium leads to:

sin

VT= (7.3.4-1)

V

T

Figure 7.3.4-1

Application of the method of sections (RITTER) to determine the force in the

inclined truss member


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2. The maximum shear load VRd,sthat can be resisted by the shear reinforcement

A schematic representation of the truss is shown in figure 7.3.4-2; the figureshows clearly that each single stirrup or inclined bar that is represented asinclined truss member (associated with each inclined strut), represents in fact aseries of stirrups or bars distributed along each crack with spacings.

.cotg .cotg.cotg

dstrut

s

Figure 7.3.4-2

Auxiliary figure for the determination of the shear reinforcement

The maximum value of the shear load VRd,s

that may be resisted by the inclinedtensile truss member is:

sin)...(, ywdswsRd fAnV = (7.3.4-2)

with:

swA the cross-sectional area of 1 stirrup (2 legs!) or of 1 inclined bar;

ywdf the design yield strength of the shear reinforcement;

n the number of links or bars that is distributed along the distancez(cotg + cotg). The number is equal to:

s

zn

)cotg.(cotg +=

withs= the spacing of the stirrups or bars;

sin for the vertical projection.

The formula for the maximum shear load that may be resisted by the shearreinforcement is thus:


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sin).cotg.(cotg.., += ywdsw

sRd fzs

AV (7.3.4-3)

For vertical links with )1sin;0(cotg90 === , the formula is:

cotg..., ywdsw

sRd fzs

AV = (7.3.4-4)

Note 1:

With cp= 0, z = 0,9.dmay be assumed.

Note 2:

The earlier versions of the standard (1995, 1998) proposed to apply theso called standard method in which the inclination angle of allcompression strut was = 45. With this assumption, the formulas are:

- with inclined shear reinforcement:

sin).cotg1.(.., += ywdsw

sRd fzs

AV (7.3.4-5)

- with vertical stirrups:

ywdsw

sRd fzs

AV .., = (7.3.4-6)

3. The necessary shear reinforcement to resist the imposed shear load VEd

The necessary shear reinforcement per unit length (along the beams axis) can bededuced from expression (7.3.4-3):

- for inclined shear reinforcement:

sin).cotg.(cotg. +=

ywd

Edsw

fz

V

s

A (7.3.4-7)

- for vertical stirrups:

cotg.. ywd

Edsw

fz

V

s

A= (7.3.4-8)

Note:

The last formula allows to conclude that the choice of a smaller value ofthe angle leads to a smaller cross-sectional area of shear reinforcement(smaller cotg larger). The adoption of less inclined cracks in thetruss model thus leads to savings in shear reinforcement.


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7-20

This conclusion can also be explained in another way: if cracks are lessinclined, the principal tensile stress (which is perpendicular to the crack)is oriented more vertically; this means that vertical stirrups are usedmore efficiently which leads to the reduction of the number of stirrupsneeded.

7.3.4.3The concrete compression struts1. The force in the truss member

The truss model is shown in figure 7.3.4-3. Vertical translation equilibrium leadsto the identification of the forceDin the inclined compression member:

sin

VD= (7.3.4-9)

V

D

Figure 7.3.4-3

Application of the method of sections (RITTER) to determine the force in theinclined concrete compression member

2. The maximum shear load VRd,maxthat can be resisted by the concrete compressionmember

The maximum value of the compression force D that may be resisted by theinclined concrete strut is equal to the product of the maximum concretecompression strength with the cross-sectional area of the strut; the last one isdeduced from figure 7.3.4-2: cross-sectional area of the strut = strutdb. .

The maximum concrete compression strength to be used for the strut calculation,is defined in EN 1992-1-1:2004; 6.2.3(3) and is limited to v.fcd

with fcd= fck/ 1,5 (and notfcd= 0,85 .fck/ 1,5 !)

Note:It should be remembered here that EN 1992-1-1:2004; 3.1.6 definesfcdas


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fcd= cc.fck/ c

with cca factor for which the value 1 is recommended.

In Belgium, the National Annex (NBN EN 1992-1-1 ANB) recommendsthe use of the value cc= 0,85 for verification in ULS for axial loads,

bending and combined axial force with bending; for other loading types(shear and torsion), cc = 1 should be used. This means in practice thatfor calculations in accordance with NBN EN 1992-1-1 ANB, thefollowing design values have to be used for the compressive strength ofconcrete:

- for ULS design of the main reinforcement (thus for normalstresses due to axial loads and bending moments):fcd= 0,85 .fck/1,5

- for ULS design of shear reinforcement (necessary to take up shearloads and torsion):fcd= fck/ 1,5

and with= strength reduction factor, defined by:

( ) 5,0250

1.6,0 = ckf

v (7.3.4-10)

in whichfckis expressed in N/mm2.

The additional strength reduction factorhas to be applied to the concrete designstrength for the calculation of the struts in order to take account of the complex,two-dimensional stress situation in the struts. Indeed, the struts are intersected bylinks or by inclined bars which are loaded in tension; due to the adherence

between steel and concrete, the transverse tensile stresses cause the weakening ofthe compressive struts. Formula (7.3.4-10) is the result of experimental tests.

Note:

EN 1992-1-1:2004; 6.2.3(3) stipulates that when the design stress in theshear reinforcement is less than 80% offywk, the following values may beadopted for the reduction factor:

- = 0,6 for fck60 MPa;

- = 0,9 fck/200 > 0,5 forfck> 60 MPa

The maximum compression force D that can be resisted by the strut is thus:v.fcd.b.dstrut. The vertical projection of this force is designated in EN 1992-1-1:2004 with the symbol VRd,max; this force has to be compared to the imposedshear load VEd.

VRd,maxmay be further expressed as (see figure 7.3.4-2):

sin....max, strutcdRd dbfvV = (7.3.4-11)

with


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)cotg.(cotgsin

+=

z

dstrut

and thus, with inclined (angle ) shear reinforcement:

2

2, cotg1

cotgcotg....)cot.(cotsin....

+

+=+= zbfggzbfvV cdcdmaxRd (7.3.4-12)

With vertical links )1sin;0(cotg90 === , the formula is:

tgcotg

1....cotg..sin... 2,

+== zbfzbfvV cdcdmaxRd

or also: szbfvV cdmaxRd co.sin...., = (7.3.4-13)

Note 1:

If cp= 0, z= 0,9.dmay be assumed.

Note 2:

The earlier versions of the standard (1995, 1998) proposed to apply theso called standard method in which the inclination angle of all

compression strut was = 45.)

2

2sin;1(cot45 === g

With this assumption, the formulas are:

- with inclined shear reinforcement:

)cotg1.(2

1...., += zbfvV cdmaxRd (7.3.4-14)

- with vertical stirrups:

zbfvV cdmaxRd ....2

1, = (7.3.4-15)

3. Stress control in the concrete compression strut

The stress is deduced from expression (7.3.4-12):

cotgcotg

cotg1.

.)cotg.(cotgsin

1.

.

2

2+

+=

+=

zb

V

zb

V EdEdc (7.3.4-16)

This formula allows to observe that the choice of a smaller value of inclinationangle leads to larger compression stresses in the concrete strut. This result isobvious when looking at figure 7.3.4-4: a less inclined strut has to transfer a


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larger compression forceDin order to generate the same resisting shear force. Itis observed that this does not cause problems in most practical normal cases,

because the stress is in general quite smaller than the acceptable stress v.fcd(seeapplications). Yet, problems may arise when small inclination angles are chosen.

Expression (7.3.4-16) also shows that the stress in the concrete compression strut

gets smaller with the use of inclined bars (< 90).

Ved

VstrutD

Ved

VstrutD

Figure 7.3.4-4

A less inclined strut has to transfer a larger compression forceDin order togenerate the same resisting shear force

Note:

EN 1992-1-1:2004; 6.2.3(3) stipulates that when prestressing is applied,

the value of VRd,maxmay be increased, in order to take account of thefact that cracks are closed.

7.3.4.4The upper chord and the bottom chord of the truss model1. The forces in the truss members

The forces in the upper and bottom chord can be determined by expressing theequilibrium of the forces applied to the part of the beam shown in figure 7.3.4-5.

At the level of the considered cross-section, a whole series of compression strutsare cut; all these compression forces have a resultant force which isDand which

is applied at half depth; the magnitude of its vertical componentDyis equal to V.In the same cross-section, a whole series of tensile reinforcement bars (stirrups orbars) are cut; the resultant force of all these tensile forces is T, which is applied athalf depth; the magnitude of its vertical component Tyhas to be equal to V.

Consequently, the horizontal components are:

cotg.

cotg.

VT

VD

x

x

=

=


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R

V

ys

cC

/2

D

Dx

TxT

q

S

/2

D

Dx=V.cotg

Dx

Dy

Dy=V

T

Tx

Ty

Tx=V.cotg

Ty=V

Figure 7.3.4-5

Auxiliary figure for the determination of the member forces in the upper trussmember and in the bottom chord

The other forces that are applied to the isolated left part of the beam, are:

the imposed uniformly distributed load q;

the support reaction forceR;

the forceNcin the arch of compressed non-cracked concrete;

the tensile forceNsin the main reinforcement.

Rotation equilibrium written around point S(figure 7.3.4-5) leads to:

2.

2..

2

..

2z

Tz

DzNxq

xR xxcs

s +=

The first member of this equation is nothing else than the bending moment Mzinthe considered cross-section, and thus:

2).cotg.(cotg.

zVzNM cz +=

The force in the upper truss member is thus:


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)cotg.(cotg2

=V

z

MN zc (7.3.4-17)

In an analogous way, the rotation equilibrium around point C (figure 7.3.4-5)leads to:

)cotg.(cotg2

+=V

z

MN zs (7.3.4-18)

2. Discussion

The result of expression (7.3.4-18) is important because this shows that, in zoneswith shear loads, the force to be transmitted by the main reinforcement does not

only depend on the bending moment Mz; indeed: z

MN zs . The main

reinforcement is loaded by an additional tensile force which increases withdecreasing value of (a disadvantage of choosing less inclined cracks and thusless inclined concrete struts). The consequences of this observation are illustratedin a visual way for the particular case with the choices: )1(cot45 == g and

vertical stirrups with = 90 (cotg = 0); expression (7.3.4-18) is than written:

2

V

z

MN zs += (7.3.4-19)

Expression (7.3.4-19) is represented in a schematic way in figure 7.3.4-6, for auniformly distributed load and for a concentrated load.

V

Q

Q

Q/2V/2

|V/2|

V/2s

/z s

q

qL/2

V/2

|V/2|

s

V

V/2

/z

s

(a) (b)

Figure 7.3.4-6


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Schematic representation of the increase in tensile force in the mainreinforcement due to the shear load, for two beams (a) and (b), and with the

assumptions = 45 and = 90 (vertical stirrups)

In the case of the concentrated load (figure 7.3.4-6(b)), one notes Mz=V.x.Substitution in expression (7.3.4-19) leads to:

z

zxVNs

)..( 21+

= (7.3.4-20)

Expression (7.3.4-20) shows that in order to calculate the forceNsin the sectionx,one may not use the bending moment at the distance x from the support, but

instead of that, has to use the bending moment at the distance2

zx + from the

support. The bending moment diagram has thus to be shifted over the

distance z.2

1, in unfavourable direction. In the more general case with arbitrary

values of and , the distance over which the bending moment diagram has to be

shifted is )cotg.(cotg.2

1z .

Note:

The additional tensile force in the main reinforcement disappears wheninclined bars with )1(cotg45 == are used in combination with the

assumption = 45.

3. Prescriptions EN 1992-1-1:2004

The main reinforcement has to be designed for a supplementary force which isdue to the shear load; the problem is due to the fact that the orientation of thereinforcement does not coincide with the orientation of the principal tensile stress.EN 1992-1-1:2004; 6.2.3(7) stipulates that the main reinforcement should be

calculated for an additional tensile force Ftdwhich is due to the imposed shearforce VEdand which may be determined by means of expression (7.3.4-18):

Ftd= 0,5.VEd.(cotg cotg ) (7.3.4-21)

and with the condition that (see figure 7.3.4-7):

(MEd/z) + FtdMEd,max/z (7.3.4-22)

in whichMEd,maxis the maximum moment along the beam.


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In practice, the rule just mentioned above is translated into the much morepractical alternative which is called the shift rule: the bending moment diagramis shifted over the distance al, which also means that the length of the mainreinforcement bars is increased with al at each end. The shifting has thusessentially consequences for the curtailment of the longitudinal tension

reinforcement. The prescriptions in EN 1992-1-1:2004; 9.2.1.3 can besummarized as follows:

for structural members without shear reinforcement, the moment curve maybe shifted over the distance al= d;

for structural members with shear reinforcement, the moment curve may beshifted over the distance al= z/2.(cotg - cotg ). As already said before,in the absence of axial compression loads,zmay be assumed equal to 0,9.d;

the curtailed bars should be anchored with lbdfrom the point on where thebars are not useful anymore. The diagram of the resisting tensile forces

should engulf the envelope diagram of the imposed tensile forces, afterapplication of the shift rule: see figure 7.3.4-7;

the anchorage length of a bent-up bar which contributes to the resistance toshear, should not be less than 1,3.lbd in the tension zone and 0,7.lbd in thecompression zone.

Figure 7.3.4-7

Envelope diagram for the calculation of structural members subjected to bending,with indication of the anchorage lengths to be applied

(figure 9.2 in EN 1992-1-1:2004)


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7.3.4.5Discussion of the variable strut inclination methodThe previous versions of the standard (1995,1998) recommended the so called standardmethod, which is characterized by the choice of the inclination angle of all concretecompression struts equal to 45. This assumption leads to a relatively simpleverification of successive cracks with a constant inclination angle, starting with the firstcrack for the largest imposed shear load: see figure 7.3.4-8. However, the variable strutinclination method was also mentioned in parallel to the standard method. Theadoption of smaller strut inclination angles leads to the following effects:

- less shear reinforcement per unit length along the beam;- more important loading of the concrete compression struts, with higher stress

levels;- larger shift length aland thus longer main reinforcement bars.

The 2004 version of EC2 does not mention the standard method anymore. Moreover, itis observed that the standard accepts rather small values for the inclination angle :

EN 1992-1-1: 2004: 45,0 21,8ANB:

- met cp= 0: 45,0 26,6

- met cp0: 45,0 18,4

With the new prescriptions, the standard wants to take account of the observation that inthe case of shear failure of the structural member, the most important crack close to thesupport and with the highest imposed shear load, is indeed characterized by a smallerinclination angle; see figure 7.3.4-9. Adopting = 45 in this zone of the structuralmember, leads to over-estimation of the necessary shear reinforcement.

Note:

Low inclination cracks only appear with lack of shear resistance. Figure7.3.4-10 shows the crack pattern in a beam with sufficient shearreinforcement; the beam has failed in bending and only nearly verticalcracks are observed in the zone with shear loads.

Finally, it should be stressed that the design of shear reinforcement by means of a trussmodel with variable strut inclination, is in full accordance with the principles of plastic

design which occupies a prominent place in the present version of the standard. Designon the basis of the assumption of a strut inclination which does not fully coincide withthe real inclination, leads to a slightly different failure mechanism: the beam fails in away that is determined by the designer. With the adoption of a too small inclinationangle and thus the provision of less shear reinforcement, the designer asks to the

beam for a rearrangement of tasks with a heavier loading of the struts and of the mainreinforcement. Practically speaking, this rearrangement will be visible by a moreexpressive development of cracks, because the design now asks for heavier loading ofthe struts. The principles of plastic design are discussed in chapter 11 in these coursenotes.


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zone bending + shear zone pure bending

d

zone with calculated shear

reinforcement

zone with technological

shear reinforcement

zone bending + shear zone pure bending

d

zone with calculated shear

reinforcement

zone with technological

shear reinforcement

= 45

Figure 7.3.4-8

Comparison of truss models with the standard model (= 45) on one hand andthe variable strut inclination method on the other hand

Figure 7.3.4-9

Shear failure of a beam without sufficient shear reinforcement: the mostimportant crack, associated with the maximum shear load, is characterized by an

inclination angle smaller than 45


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Figure 7.3.4-10

Crack pattern in a beam subjected to a four point bending test; failure is inbending and not in shear; shear reinforcement has been well designed

7.4 Design for shear7.4.1 IntroductionThis paragraph focuses on the ULS design calculation of shear reinforcement accordingto EN 1992-1-1:2004; 6.2 and the complementary Belgian ANB prescriptions.

7.4.2 DefinitionsThe verification of shear resistance is based on three design values of resisting shearforces:

VRd,c design shear resistance of the member in a section without shearreinforcement. VRd,cis calculated by means of expressions (7.2.5-3) and (7.2.5-4)in these course notes;

VRd,s design value of the shear force which can be sustained by the yieldingshear reinforcement. VRd,s is calculated by means of expressions (7.3.4-3) and(7.3.4-4) in these course notes;

VRd,max design value of the maximum shear force which can be sustained by themember, limited by crushing of the compression struts. VRd,max is calculated bymeans of expressions (7.3.4-12) and (7.3.4-13) in these course notes.

VEd is the imposed design shear force in the section to be verified, resulting fromexternal loading on the structural member.


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7.4.3 The principles of the shear verification procedure- In the regions of the member where VEdVRd,cno calculated shear reinforcement

is necessary. Yet, when on the basis of the design calculation, no shearreinforcement is required, minimum (technological) shear reinforcement should

be provided. The minimum shear reinforcement may be omitted in certain special

cases such as:- slabs where transverse redistribution of loads is possible;- members of minor importance which do not contribute significantly to the

overall resistance and stability of the structure; example: lintels with span 2 m.

- In regions where VEd> VRd,csufficient shear reinforcement should be provided inorder that VEdVRd.. VRdis the resisting shear force and is equal to the smallestof the two values VRd,sand VRd,max.

Important note:It was already mentioned in figure 7.3.2-2, that there is experimentalevidence for the fact that the shear reinforcement only starts to workas a member in a truss sytem, for a reduced value of the imposed shearload. In the previous versions (1995, 1998) of EC2, it was accepted thatthe reduction could be taken equal to the shear force VRd,cwhich is infact the shear load resisted without shear reinforcement, by thefollowing mechanisms:

- the shear resistance of the non-cracked compression concrete arch;

- the granulate interlocking effect along the shear crack;

- the dowel action of the main reinforcement.It was thus accepted in the previous versions of EC2, in which thestandard method was used for shear verification (struts with constantinclination angle of 45), to design shear reinforcement for the forceVEd- VRd,c.

This is not the case anymore in the present version (2004) of EC2, inspite of the experimental evidence shown in figure 7.3.2-2; shearreinforcement has now to be calculated for the full imposed shear forceVEd. The reason for this is that the present standard does not want toaccumulate too much favourable effects. Indeed, the present EC2

allows adopting small inclination angles in the regions where high shearloads are applied; this leads already to smaller shear reinforcement(while causing more severe loading of the concrete struts). EC2 doesnot want to accumulate this positive effect on the shear reinforcementwith a second one generated by the reduction with VRd,cof the imposedshear force VEd.

- The longitudinal tension reinforcement should be able to resist the additionaltensile force caused by shear; in practice, the shift rule is used.

- For members subjected to predominantly uniform distributed loading, the designshear force need not to be checked at a distance less than dfrom the face of thesupport; see figure 7.4.3-1. This rule takes into account that the loads applied


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close to the support, are directly transmitted to the support without causingbending and shear of the beam itself. The shear verification thus starts with thefirst crack which is initiated at the tensile side of the beam and which developsupwards with a certain inclination angle. Any shear reinforcement required in thefirst verified section, should continue to the support. On top of that, it should

always be verified that the imposed shear force at the support is not larger thanVRd,max.

d

d

Figure 7.4.3-1

In the case of uniformly distributed loads, direct transmission to the support of the loadsapplied close to the support may be assumed

Note 1:

For members with inclined chords (upper side and lower side), the valueof VRd should be increased with two additional resisting shear loadcomponents (see figure 7.4.3-2):

VRd= minimum(VRd,s; VRd,max) + Vccd+ Vtd

with:

- Vccd = the design value of the shear component of the force in thecompression area, in the case of an inclined compression chord (upperside);

- Vtd = the design value of the shear component of the force in the

tensile reinforcement, in the case of an inclined tensile chord (lowerside).


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Vccd

D

T

Vtd

actionResisting

reaction

Figure 7.4.3-2

Additional shear resistance due to the presence of inclined chords in structuralmembers

(adaption of figure 6.2 in EN 1992-1-1:2004)

Note 2:

When loads are applied to the lower side of a structural member,sufficient vertical reinforcement, in addition to the shear reinforcement,is needed in order to transfer the loads to the upper side: see figure 7.4.3-3.

Figure 7.4.3-3

Additional reinforcement is needed to transfer loads applied to the lower side ofthe beam to the upper side

7.4.4

Members not requiring design shear reinforcement: VEdVRd,cSome prescriptions:- in the regions where VEdVRd,cno calculated shear reinforcement is necessary,

but minimum (technological) shear reinforcement should be provided: see furtherin the paragraph on technological prescriptions;

- for the design of the longitudinal reinforcement, the MEd-diagram should beshifted over a distance al= din the unfavourable direction;

- for members with concentrated loads applied on the upper side and rather close tothe support, it may be assumed that a part of the load is transferred directly to thesupport (without interaction of the beam itself), which gives a reduction of theimposed shear force VEd. The prescriptions stipulate that when the concentrated

load is applied on the upper side within a distance 0,5dav2d(see figure 7.4.4-1) from the edge of the support, the contribution of this load to the shear force VEd


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may be reduced by multiplying the load by the factor = av/2d. This reduction isonly valid provided that the longitudinal reinforcement is fully anchored at thesupport. For av0,5dthe value av= 0,5dshould be used.

Important remark: the imposed shear force VEd, calculated without reduction bythe factor, should always satisfy the condition:

VEd0,5.bw.d..fcd (7.4.4-1)

where = the strength reduction factor for concrete cracked in shear:

( ) 5,0250

1.6,0 = ckf

v (7.4.4-2)

in which fckis expressed in N/mm2.

This condition corresponds in fact to the verification of a fictive concretecompression strut right above the support. Expression (7.4.4-1) is deduced fromexpression (7.3.4-13) which defines the maximum shear force that can be resistedfrom the point of view of strut failure:

szbfvV cdmaxRd co.sin...., =

By introduction of the lever armztaken equal to dand = 45, expression (7.4.4-1) is obtained, which corresponds to a sort of upper limit for VEd. Indeed, takinginto account the following evolution of the term sin.cos:

sin.cos45 0,5

40 0,49

35 0,47

30 0,43

one finds the condition VEd...sin.cosmore severe for VEdin order to take intoaccount that the strut is more heavily loaded in compression when it is lessinclined.


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Figure 7.4.4-1

It may be assumed that a fraction of the loads applied near supports, istransmitted directly to the support and does not give a contribution to theimposed shear force VEd, on the condition that the main reinforcement is

sufficiently anchored (figure 6.4 in EN 1992-1-1:2004)

7.4.5 Structural members requiring design shear reinforcement: VEd> VRd,cIn regions where VEd> VRd,c, sufficient shear reinforcement should be provided in orderthat VEdVRd. VRdis the resisting shear force and is equal to the minimum of the two

values VRd,sand VRd,max. The formulas for VRd,sand VRd,maxare developed on the basis ofan analogous truss model which is once again represented in figure 7.4.5-1.


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Figure 7.4.5-1

Truss model used for the calculation of the shear reinforcement in structural members(figure 6.5 in EN 1992-1-1:2004)

7.4.5.1When vertical shear reinforcement is usedcotg..., ywd

swsRd fz

s

AV = (7.4.5-1)

tgcotg

1....,

+= cdwmaxRd fzbV (7.4.5-2)

with the additional condition that defines the maximum effective cross-sectional area of

shear reinforcement for = 45 ( cotg = 1 ):

sbffA wcdywdmaxsw ....2

1., (7.4.5-3)

7.4.5.2When inclined shear reinforcement is used

sin).cotg.(cotg.., += ywdsw

sRd fzs

AV (7.4.5-4)

2, cotg1

cotgcotg....

+

+= cdwmaxRd fzbV (7.4.5-5)

with the additional condition that defines the maximum effective cross-sectional area ofshear reinforcement for = 45 ( cotg = 1 ):

sin

1.....

2

1., sbffA wcdywdmaxsw (7.4.5-6)

7.4.5.3Additional prescriptionsJust as in the case where shear reinforcement is not necessary, the rule for directtransmission to the support of a fraction of the loads that are applied near supports, can

be applied here too. The prescriptions stipulate that when the load is applied on theupper side within a distance 0,5dav2d (see figure 7.4.4-1) from the edge of thesupport, the contribution of the load to the imposed shear force VEdmay be reduced bymultiplying the load by the factor = av/2d. This reduction may only be applied if themain reinforcement is fully anchored above the support. For av0,5d, the value

av= 0,5dmay be adopted.


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Important: the imposed shear force VEd, calculated in this way (thus with application ofthe reduction factor), should satisfy the condition:

VEdAsw.fywd. sin

where Asw.fywd= the resistance of the shear reinforcement crossing the inclined shearcrack between the loaded areas (see figure 7.4.5-2); only the shear reinforcement withinthe central 0,75.avshould be taken into account.

Note:

The reduction with the factor may only be applied

- for the calculation of shear reinforcement and not for the strut verification;- provided that the longitudinal reinforcement is fully anchored at the

support.

Figure 7.4.5-2

Auxiliary figure for the calculation of shear reinforcement with direct strut action closeto the support (figure 6.6 in EN 1992-1-1:2004)

7.5 Overview of the shear reinforcement calculation for a beam withuniformly distributed load and concentrated load

7.5.1 The first value of VEdto be considered: in principleIn principle, the largest value of VEd should be considered for the first shear designcalculation. To remember, design of shear reinforcement is performed in ULS; thedesign values of the loads have thus to be considered (use of partial safety factors). Seeexample in figure 7.5.1-1.


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VQd

Q.Q g.g

Vgd

B

(Vg+VQ)d

(Vg+Q)dA

Figure 7.5.1-1

In principle, the largest value of VEdshould be considered for the first shear designcalculation; that is the value at the support

7.5.2 Reduction of VEdto take account of direct load transfer to the supports7.5.2.1Uniformly distributed loadThe shear calculation is performed by considering successive shear cracks, starting withthe first crack in the region with the largest value of the imposed shear force. The crackis initiated on the tensile side (at the bottom of the beam in figure 7.5.1-1) at the edge ofthe support; the crack is assumed to be vertical in the concrete cover, but starting fromthe main reinforcement, develops upwards with an inclination angle. In reality, thecrack gets stuck in the concrete compression arch, but in the simplified truss model, thecrack is extended up to the upper side of the beam with a constant inclination angle.

When the first crack is considered with the inclination angle , the assumption isadopted that the uniformly distributed load to the left of the upper end of the crack is

transferred directly to the support: see figure 7.5.2-1. The first value of VEd to becalculated is thus the imposed shear force in the cross-section at the distance d.cotg(dwhen = 45) from the edge of the support (cross-section C in figure 7.5.2-1). Thevalue of (VEd)C is equal to the maximum shear force in support A (due to g and Q)reduced with the portion of shear load due togover the distance a + d.cotg:

)cot..(.)()( gdagVV gdAQgCEd += +

with

a: see determination of the design span for different support conditions in paragraph

2.5.2.2.


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Q.Q

g.g

B

d

da

Q

C

(VEd)C= (Vg+Q)dA- g.g.(a+d)

45

d

d.cotga

Q

C'

(VEd)C'= (Vg+Q)dA- g.g.(a+d.cotg)

Figure 7.5.2-1

First shear verification in the cross-section with the largest imposed shear force VEd,taking account of the reduction due to the direct transfer ofgto the support

Note:

The adoption of smaller values of (than 45) leads to smaller values of VEd(see figure 7.5.2-1). But with this assumption, it is very important to verify theconcrete compression strut above the support with the formula:

VEd(with reduction ofgonly)bw.z. .fcd. sin. cos

If this condition is OK, then further comparison of VEd with VRd,c can beperformed.

7.5.2.2Concentrated loads close to supportsFor concentrated loads which are applied at the distance avfrom the edge of the support,with 0,5.dav2.d, a reduction may be adopted of the imposed shear force. Thecontribution of the concentrated force to the shear force VEd may be reduced bymultiplying the concentrated force with the reduction factor = av/2d(1). This rule isindependent from the choice of the inclination angle .

Figure 7.5.2-2 shows the example of the concentrated force Q applied within thedistance av= 2.dfrom the edge of the support. The imposed shear load (VQ)dAin cross-


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sectionAmay be reduced to .(VQ)dA; this means that the difference (VQ)dA- .(VQ)dAisdirectly transferred to the support. The shear force (VEd)C to be considered in cross-section C, at the distance d.cotgfrom the edge of the support is thus:

[ ]dAQdAQgdAQgCSd VVgdagVV

)()()cot..(.)()(

+=+

Q.Q

B

d

2d

Q

C

VSd

45

av

(VQ)dA

Q.Q

d

VSd

C'

Figure 7.5.2-2

Reduction of the imposed shear force due to direct transfer of a portion of theconcentrated force to the support

Important note:

The reduction by means of the factor is not allowed for the verification of theconcrete compression struts. The reduction is only considered for:

- the comparison between VEdand VRd,c;

- the calculation of the shear reinforcement.

It is thus recommended to calculate two distinct values of VEdfor further use:

- VEdgin cross-section C(with only the reduction of the contribution ofg);

- VEdg+Qin cross-section C(inclusive the reduction ofgand Q).

7.5.3 The verifications to be performed for the first crackThe following verifications have to be performed.


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7.5.3.1VRd,cCalculation of the resisting shear force VRd,c. The crack that has to be considered with(VEd)C in cross-section C, starts from the tensile side at the edge of the support and

develops with an inclination angle up to the cross-section C. This crack is determiningfor the area of reinforcement Asl that has to be considered for the calculation of VRd,c.This value of VRd,cwhich is calculated in order to be compared with (VEd)C in cross-section C, has to take account of the main reinforcement that contributes to the shearresistance by means of the dowel action in the crack. Again, only that reinforcement can

be considered that is sufficiently anchored with lbdbeyond the section where the dowelaction takes place; this is thusAslwhich continues over the length d+lbdsuch as pointedout in figure 7.2.5-1(a) and in the definition ofAslin expression (7.2.5-1).

7.5.3.2VEdVRd,cIs VEd(with reduction ofgand Q)VRd,c, then:- the technological shear reinforcement is sufficient (see further);

- the strut above the support should be verified by means of expression (7.4.4-1):

VEd(with only reduction ofg)0,5.bw.d..fcd

This expression takes account of = 90, as the technological reinforcement iscomposed of stirrups.

7.5.3.3VEd> VRd,cFirst, the strut-condition has to be verified. At this stage, if necessary, one can still makea new choice of the inclination angle :

- with stirrups

tgcotg

1....)ofreduction(

+ cdwgonlyEd fzbV (7.5.3-1)

- with bent-up bars

2)ofreduction( cotg1

cotgcotg.... +

+

cdwgonlyEd fzbV (7.5.3-2)

Next, the necessary shear reinforcement per unit length can be determined by means ofthe following formulas:

- with stirrups

cotg..)andofreductionwith(

ywd

QgEdsw

fz

V

s

A (7.5.3-3)

- with bent-up bars


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sin).cotg.(cotg.)andofreductionwith(

+

ywd

QgEdsw

fz

V

s

A (7.5.3-3)

Once Asw/s calculated, one has to verify if this reinforcement is at least equal to the

minimum reinforcement ration (see further in detailing of reinforcement). Thepractical translation of Asw/s into a suitable diameter and spacing, can be realized bymeans of table 7.5.3-1.

Table 7.5.3-1

Values ofAswands

Asw for stirrups with two vertical legs

(mm) 6 8 10 12 14 16

Asw(mm2) 56,5 100 157 226 308 402

s(mm)s

Asw (mm2/mm)

50 1,131 2,011 3,141 4,524 6,158 8,042

60 0,942 1,676 2,618 3,770 5,131 6,702

70 0,808 1,436 2,244 3,231 4,398 5,745

80 0,707 1,257 1,963 2,827 3,848 5,027

90 0,628 1,117 1,745 2,513 3,421 4,468

100 0,565 1,005 1,571 2,262 3,079 4,021

120 0,471 0,838 1,309 1,885 2,566 3,351

140 0,404 0,718 1,122 1,616 2,199 2,872150 0,377 0,670 1,047 1,508 2,053 2,681

160 0,353 0,628 0,982 1,414 1,924 2,513

180 0,314 0,559 0,873 1,257 1,710 2,234

200 0,283 0,503 0,785 1,131 1,539 2,011

250 0,226 0,402 0,628 0,905 1,232 1,608

300 0,188 0,335 0,524 0,754 1,026 1,340

Note:

The additional condition related to the use of the -factor for concentrated loadsshould not be forgotten: the imposed shear force VEd, calculated with applicationof the -factor, should always respect the following condition:

VEdAsw.fywd. sin (7.5.3-5)

where Asw.fywd= the force that is generated by the shear reinforcement in thisarea (at the support); only the reinforcement in the central part 0,75.avshould be

taken into account (see figure 7.4.5-2).


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7.5.4 Verification of further cracksAfter the control of the first crack, the second crack has to be verified: as long asVEd> VRd,c(and a calculated shear reinforcement is necessary), a next crack has to beverified. In other words: if it is found in the considered cross-section that the calculatedshear reinforcement is larger than the minimum reinforcement, a next crack has to beverified.

Figure 7.5.4-1 presents the example of an end support, with indication of the cross-sections C1, C2, C3... that have to be verified successively. The cracks start in the tensileregion at the bottom side of the beam; this determines the areas of the mainreinforcementAslto be considered in the calculation of VRd,c; see figure 7.2.5-1(a) andthe definition of Asl in expression (7.2.5-1): Asl1, Asl2, Asl3 are the areas of mainreinforcement that continue over the distance d+lbdto the left beyond the cross-sectionsthat are considered.

C1

VEd1

d

VEd2 VEd3

C2 C3

d d d

Asl1 Asl2 Asl3

Figure 7.5.4-1

The cross-sections to be considered successively in shear verification, in a beam close toan end support

Figure 7.5.4-2 presents the example of an intermediate support in a continuous beam.The first crack is easy to identify on the basis of the cone with direct load transfer to thesupport. The first cross-section to be calculated is C1 with (VEd)C1. The crack that

corresponds to this (VEd)C1starts in the tensile region, at the upper side in C2; this is theplace where the dowel action of the main reinforcement is activated. In accordance withfigure 7.2.5-1(b) and the definition of Asl in expression (7.2.5-1), the mainreinforcementAsl1should continue over the distance d+lbdbeyond C2, in order to assurefor 100% the dowel resistance in C2.


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C1

VEd1VEd2

VEd3

C2C3

Asl1Asl2Asl3

Figure 7.5.4-2

The cross-sections to be considered successively in shear verification, in a beam close toan intermediate support

7.6 Detailing of shear reinforcementReference: EN 1992-1-1:2004; 9.2.2

7.6.1 Shape and nature of shear reinforcement- The shear reinforcement should form an angle between 45 and 90 to the

longitudinal axis of the structural element.- The shear reinforcement may consist of a combination of:

links enclosing the longitudinal tension reinforcement and the concretecompression zone: see figure 7.6.1-1;

bent-up bars;

cages, ladders, etc. which are cast in without enclosing the longitudinalreinforcement but are properly anchored in the compression and tensionzones.

Figure 7.6.1-1

Examples of shear reinforcement (figure 9.5 in EN 1992-1-1:2004)

- Links should be effectively anchored. Stirrups should form a closed rectangle; alap joint on the vertical leg is permitted provided that the link is not required toresist torsion (see chapter on torsion).

- At least 50 % of the necessary shear reinforcement should be in the form of links.


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7.6.2 Minimum shear reinforcementThe geometric ratio of shear reinforcement is defined as:

sin.. w

sww

bs

A= (7.6.2-1)

where:

w : geometric ratio of shear reinforcement, with ww,min;

swA : area of shear reinforcement within lengths;

s : spacing of the shear reinforcement measured along the longitudinal axis ofthe structural member;

wb : width (or breadth) of the web of the structural member;

: angle between the shear reinforcement and the longitudinal axis.

The value of w,minis identified by the following expression:

ywk

ck

wf

f.08,0min, = (7.6.2-2)

withfckandfywkin MPa.

w,mindepends of the steel grade of the shear reinforcement and of the concrete class.Table 7.6.2-1 presents the values for w,minfor different combinations.


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Table 7.6.2-1

Values of w,minfor different concrete-steel combinationsSteel grade of shear reinforcement

Concrete class

S220

(fywk= 220 MPa)

S400

(fywk= 400 MPa)

S500

(fywk= 500 MPa)

S600

(fywk= 600 MPa)C12/15 (fck= 12 MPa) 1,26E-03 6,93E-04 5,54E-04 4,62E-04

C16/20 (fck= 16 MPa) 1,45E-03 8,00E-04 6,40E-04 5,33E-04

C20/25 (fck= 20 MPa) 1,63E-03 8,94E-04 7,16E-04 5,96E-04

C25/30 (fck= 25 MPa) 1,82E-03 1,00E-03 8,00E-04 6,67E-04

C30/37 (fck= 30 MPa) 1,99E-03 1,10E-03 8,76E-04 7,30E-04

C35/45 (fck= 35 MPa) 2,15E-03 1,18E-03 9,47E-04 7,89E-04

C40/50 (fck= 40 MPa) 2,30E-03 1,26E-03 1,01E-03 8,43E-04

C45/55 (fck= 45 MPa) 2,44E-03 1,34E-03 1,07E-03 8,94E-04

C50/60 (fck= 50 MPa) 2,57E-03 1,41E-03 1,13E-03 9,43E-04

C55/67 (fck= 55 MPa) 2,70E-03 1,48E-03 1,19E-03 9,89E-04

C60/75 (fck= 60 MPa) 2,82E-03 1,55E-03 1,24E-03 1,03E-03

C70/85 (fck= 70 MPa) 3,04E-03 1,67E-03 1,34E-03 1,12E-03

C80/95 (fck= 80 MPa) 3,25E-03 1,79E-03 1,43E-03 1,19E-03

C90/105 (fck= 90 MPa) 3,45E-03 1,90E-03 1,52E-03 1,26E-03

7.6.3 Spacing7.6.3.1In the longitudinal directionThe recommended value for the maximum longitudinal spacing sl,maxbetween adjacentstirrups, is determined by the formula:

sl,max= 0,75.d.(1 + cotg ) (7.6.3-1)

The recommended value for the maximum longitudinal spacing sb,maxbetween adjacentbent-up bars, is determined by the formula:

sb,max= 0,60.d.(1 + cotg ) (7.6.3-2)

Table 7.6.3-1 presents some values for bent-up bars with = 45 and vertical stirrups (= 90).

Note:

The longitudinal spacing between stirrups may be quite large. The earliereditions of EC2 were more severe at this point, with a maximum value of 300mm. In practice, it is observed that the spacing criterion of 30 mm is stilladopted in workshops, because of practical considerations in the realization of

the reinforcement cages.


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Table 7.6.3-1

Values of recommended maximum longitudinal spacingsl,maxbetween adjacent bent-upbars (with = 45) and vertical stirrups (= 90)

Bent-up bars with = 45( cotg = 1 )

Stirrups with= 90

( cotg = 1 )

sb,max= 0,6.d.(1+cotg ) sl,max= 0,75.d.(1+cotg )

d

(mm)sb,max= 1,2.d

(mm)

sl,max= 0,75.d

(mm)

400 480 300

600 720 450

800 960 600

1000 1200 750

7.6.3.2In the transverse directionThe transverse spacing of the legs in a series of shear links should not exceed st,max,determined by the formula:

st,max= 0,75.d600 mm (7.6.3-3)

7.7 Shear between web and flanges of T-sections7.7.1 Analysis of the compression flange7.7.1.1Identification of the problem and the truss analogyFigure 7.7.1-1 shows a T-beam subjected to simple bending. The NA is situated in

principle in the upper part of the cross-section; the assumption may be adopted that theflange corresponds to the compression part of the cross-section. The force N' in the

compression zone (distributed over the whole width of the flange) changes over thedistance xwith the value:

z

MN

= '

The indication N'f ( f< flange) corresponds to the compression force in 1 part of theflange left or right of the web (abstraction is made of an eventually compressed small

part of the web);N'fis thus a fraction ofN' as expressed by the formula:


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f

wf

fb

bb

NN 2'.'

= (7.7.1-1)

The forceN'fin 1 part of the flange left or right from the web, changes over the distancexwith the value:

f

wf

xfb

bb

z

MN 2.)'(

=

(7.7.1-2)

N'

x

N'+N' hf

bw bf

RA

N'

N'

M/z

N' N'+N'

N' N'+N'

I

I

I

I

hf

bw

bf

bf -bw

2

bf -bw

2

Figure 7.7.1-1

T-beam loaded in simple bending

One of the two hatched pieces of flange in figure 7.7.1-1 is now isolated: see detail infigure 7.7.1-2. The difference in compression force (N'f)xhas to be equilibrated by theshear force distributed over the contact surfaceI-Iwith area: hf.x.


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x

I I

I I

(N'f)x(N'f)x

.hf.x=(N'f)x

Figure 7.7.1-2

Detail: equilibrium of a part of the flange, left or right of the web

The equilibrium equation leads to the expression for the mean shear stress in thecontact surfaceI-I, in the uncracked elastic phase:

f

wf

ff

xf

b

bb

x

M

hzxh

N 2...

1

.

)( '

=

=

(7.7.1-3)

And asx

MV

= , one can thus also write:

f

wf

f b

bb

hz

V 2..

= (7.7.1-4)

If this shear stress is too large, causing the principal tensile stress to reach the tensilestrength of the concrete, cracks will appear with an inclination angle with respect to thelongitudinal axis. Shear reinforcement is necessary in that case, in order to allow the

build-up of the longitudinal forces in the flange. The design of this shear reinforcementcan again be done by means of an analogous truss system in which shear is resisted bythe combined action of struts and tensile rods. Figure 7.7.1-3 shows the decompositionof the shear force into two forces: a compression force which has to be resisted by theconcrete struts and a tensile force perpendicular to the longitudinal axis of the beam.


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x f

(N'f)x(N'f)x

f

Fwapening

Fschoor

Figure 7.7.1-3

Decomposition of the shear force at the level of the contact surface flange-web, into two

forces: a compression force to be resisted by concrete struts and a tensile forceperpendicular to the longitudinal axis of the beam, which has to be resisted by steel

reinforcement

7.7.1.2Transverse reinforcementIt is assumed that the orientation of the compression diagonals in the flange ischaracterized by the inclination angle f. As:

xff N

entreinforcemtransverseinforcetensile

=

)(

tg

'

(7.7.1-5)

one finds that the transverse reinforcement should be able to transfer a force (indicatedbyFreinforcementin figure 7.7.1-3) equal to:

cotg

)( ' xfentreinforcem

NF

=

It is assumed that the transverse reinforcement is composed of rods with sectional areaAsfand spacing sf; this means that the ratio Asf/sfrepresents the transverse reinforcementper unit length. The necessary reinforcement is thus:

cotg

)'(.. xfyd

f

sf Nfx

s

A

=

This leads to the transverse reinforcement in the flange per unit length of the beamf

sf

s

A:

cotg..

)( '

yd

xf

f

sf

fx

N

s

A

=

(7.7.1-7)


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or, when expression (7.7.1-2) is used:

f

wf

yd

x

f

sf

b

bb

fxzM

sA 2.cotg. 1.. )(

=

(7.7.1-8)

or

f

wf

ydf

sf

b

bb

fz

V

s

A 2.cotg.

1.

=

(7.7.1-9)

7.7.1.3The concrete strutsIt is also necessary to verify if the concrete struts are able to withstand the imposedcompression force (indicated by Fstrut in figure 7.7.1-3). This means that thecompression stress in the strut should be limited to .fcd.

From figure 7.7.1-3, it appears that:

f

xf

strut

NF

cos

)'(

= (7.7.1-10)

zodat

strutf

f

xf

strutstrut

bh

N

F

.

cos

)'(

sectionstrut

==

with bstrut= x.sin f (see figure 7.7.1-4).

The condition thus becomes:

cd

fff

xf

strut fxh

N

.sin...cos

)'(

=

(7.7.1-11)

As (N'f)x= .hf.x, expression (7.7.1-11) also allows to develop a condition for :

cd

fff

ff

xh

xh.

sin.cos..

..

or

ffcdf sin.cos.. (7.7.1-12)


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x bschoor= x.sinff

Figure 7.7.1-4

Auxiliary figure for the determination of the width of the compression strut

7.7.1.4Alternative formulation of the shear problem between web and flanges by meansof the strut and tie method

The following reasoning is based on the philosophy of the strut and tie method, which isdiscussed in a separate chapter in these course notes (see further); this method presents afurther generalization of the truss system analogy that was already introduced for theshear verification of beams loaded in bending.

The analysis of the analogous truss system model, presented in figure 7.7.1.5, revealsthat the upwards inclined compression diagonal in the web makes equilibrium with thetensile force in the vertical member and the compression force in the flange. In order to

install equilibrium in the flange, the force has to be spread out over the whole width ofthe flange (effective width!). A new analogous truss model appears in the flange: theforce spreads out via two compression diagonals aband ab', which on their turn have tomake equilibrium with the compression forcesN'fin the flange. This equilibrium needsthe tensile member bb'. The further development of this model leads to the sameformulas as developed before.

a

N'f

N'f

b

b'

ab b'

(a) (b)


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Figure 7.7.1-5

Transfer of forces in the flange of a T-beam; (a) truss model for beam withcompression flange; (b) top vue of the beam (WALRAVEN, 1995)

7.7.2 The tension flangeThe developments presented for the compression flange are also applicable to thetension flange; see figure 7.7.2-1.

(Asl)II

Asl

hf2

II

II

Figure 7.7.2-1The shear problem in the tension flange

In analogy with expression (7.7.1-4), the shear stress in the contact sectionII-IIin figure7.7.2-1, is defined by:

=

sl

IIsl

f A

A

hz

V )(.

. 2 (7.7.2-1)

with

slA = the total area of longitudinal reinforcement in the whole tensionflange;

IIslA )( = the area of the longitudinal reinforcement in the isolated part of the

tension flange.

In analogy with expression (7.7.1-9), the necessary transverse reinforcement per unitlength of the beam, is determined by:


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=

sl

IIsl

sf

sf

A

A

fz

V

s

A )(.

cotg.

1.

(7.7.2-2)

Figure 7.7.2-2 presents a model of force transfer in the tension flange and web

(philosophy of the strut and tie method).

side view

Projection of

bottom flange sl st

Figure 7.7.2-2

Model of force transfer from web to tension flange, based on the philosophy of the strutand tie method (WALRAVEN, 1995)

7.7.3 Prescriptions concerning the passage from the web to the flanges in T-beams7.7.3.1PrinciplesReference: EN 1992-1-1:2004; 6.2.4

The shear strength of the flange may be calculated by considering the flange as a systemof compressive struts combined with ties in the form of tensile reinforcement. Thenotations in figure 7.7.3-1 have to be applied:

Fd= the variation of the normal force in one part of the flange, over the length x

(notation (Nf)xwas used in the text before);hf= the heigth (or thickness) of the flange at the contact surface between web and

flange;

vEd= the longitudinal shear stress (notation was used in the text before) at the contactsurface between web and flange. Consequently: vEd= Fd/ (hf. x);

Asf= the cross-sectional area of one transverse reinforcement bar;

sf= the spacing between the transverse reinforcement bars.

Note:


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The bar which is indicated with the letter B in figure (7.7.3-1) corresponds tolongitudinal reinforcement bars which are eventually present (for example totake up occasional fixing moments 25% of the moment in the span).

Figure 7.7.3-1

Notations used in the standard for the analysis of the force transfer between weband flange via compression struts and transverse reinforcement (figure 6.7 in EN

1992-1-1:2004)

For the length xto be used, the standard presents the following recommendations:- x the half of the distance between the cross-sections with bending

momentsM= 0 andMmax. Example: see figure 7.7.3-2;- where point loads are applied, the length xthe distance between the point

loads.

MMmax

M=0 M=0

x x x x


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Figure 7.7.3-2

Indication of the length xto be used for the calculation of the transversereinforcement to assure the force transfer web-flange: in the particular case of asimply supported beam with uniformly distributed load, there are at least four

regions to be considered

The value for the inclination angle f which determines the orientation of the struts,should be in accordance with the following conditions:

- for compression flanges

1 cotg f2 or 45 f26,5

- for tension flanges

1 cotg f1,25 or 45 f38,6

7.7.3.2The transverse reinforcementThe transverse reinforcement per unit length Asf/sf is determined by means of thefollowing expression (see also figure 7.7.3-3):

resisting tensile force tensile force due to the shear force

and thus:

f

fEdyd

f

sf xhvfxsA

cotg....

or finally:

fyd

fEd

f

sf

f

hv

s

A

cotg.

. (7.7.3-1)

7.7.3.3The verification of the compression strutsThe verification of the compression struts is performed in accordance with figure 7.7.3-3, which leads to the following formula:

strutf

f

d

strutstrut

bh

F

F

.

cos

strutsection

==

)sin..(.cos

..

fff

fE

strutxh

xhv

=


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As: cdstrut f.

one finds:

ffcdE fv sin.cos.. (7.3.3-2)

x

Fd

Fd+Fd=Fd+vE.hf.x

fFd=vE.hf.x

bstrut=x.sinf

Compression force

to be resisted bythe strut

f

dF

cos

=

Tensile force to

be resisted by

transverse

reinforcement

= Fd.tg

Figure 7.7.3-3

Auxiliary figure for the development of the formulas for the calculation of thetransverse reinforcement to assure the force transfer web-flange and for the

verification of the compression struts

7.7.3.4Combined actionsIn the case of combined

- shear between the flange and the web, and

- transverse bending (bending in the horizontal plane),

two possible solutions are available:

- the area of steel should be greater than the one given by expression (7.7.3-1);

- half of the area of steel given by expression (7.7.3-1) in combination with thereinforcement needed to resist transverse bending.

7.7.3.5RemarkExtra transverse reinforcement (in addition to the normal one for bending) is notnecessary when the following conditions are met:

- vEd0,4.fctdaccording to EN 1992-1-1:2004; 6.2.4(6)

CH7 - Design for Shear Loads

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