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(2) The shear resistance of a member with shear reinforcement is equal to:

VRd= VRd,s+ Vccd+ Vtd (6.1)

(3) In regions of the member where VEdVRd,cno calculated shear reinforcement is necessary.

VEdis the design shear force in the section considered resulting from external loading andprestressing (bonded or unbonded).

(4) When, on the basis of the design shear calculation, no shear reinforcement is required,minimum shear reinforcement should nevertheless be provided according to 9.2.2. Theminimum shear reinforcement may be omitted in members such as slabs (solid, ribbed orhollow core slabs) where transverse redistribution of loads is possible. Minimum reinforcement

may also be omitted in members of minor importance (e.g. lintels with span 2 m) which do notcontribute significantly to the overall resistance and stability of the structure.

(5) In regions where VEd

> VRd,c

according to Expression (6.2), sufficient shear reinforcement

should be provided in order that VEd VRd(see Expression (6.8)).

(6) The sum of the design shear force and the contributions of the flanges, VEd- Vccd-Vtd,should not exceed the permitted maximum value VRd,max(see 6.2.3), anywhere in the member.

(7) The longitudinal tension reinforcement should be able to resist the additional tensile forcecaused by shear (see 6.2.3 (7)).

(8) For members subject to predominantly uniformly distributed loading the design shear forceneed not to be checked at a distance less than dfrom the face of the support. Any shear

reinforcement required should continue to the support. In addition it should be verified that theshear at the support does not exceed VRd,max (see also 6.2.2 (6) and 6.2.3 (8).

(9) Where a load is applied near the bottom of a section, sufficient vertical reinforcement tocarry the load to the top of the section should be provided in addition to any reinforcementrequired to resist shear.

6.2.2 Members not requir ing design shear reinforcement

(1) The design value for the shear resistance VRd,cis given by:

VRd,c= [CRd,ck(100 l fck)1/3

+ k1cp] bwd (6.2.a)

with a minimum of

VRd,c= (vmin +k1cp) bwd (6.2.b)

where:fckisin MPa

k = 0,2200

1 +d

with din mm

l = 02,0w

sl

db

A

Asl is the area of the tensile reinforcement, which extends (lbd+ d) beyond thesection considered (see Figure 6.3).

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bw is the smallest width of the cross-section in the tensile area [mm]

cp = NEd/Ac < 0,2 fcd [MPa]NEd is theaxial force in the cross-section due to loading or prestressing [in N] (NEd>0

for compression). The influence of imposed deformations on NEmay be ignored.A

C is the area of concrete cross section [mm

2

]VRd,c is [N]

Note: The values of CRd,c, vminand k1for use in a Country may be found in its National Annex. The

recommended value for CRd,cis 0,18/c, that for vminis given by Expression (6.3N) and that for k1is 0,15.

vmin=0,035k3/2

fck1/2

(6.3N)

A - section considered

Figure 6.3: Defin ition of Aslin Expression (6.2)

(2) In prestressed single span members without shear reinforcement, the shear resistance ofthe regions cracked in bending may be calculated using Expression (6.2a). In regions

uncracked in bending (where the flexural tensile stress is smaller than fctk,0,05/c) the shear

resistance should be limited by the tensile strength of the concrete. In these regions the shearresistance is given by:

( ) ctdcp2

ctdw

Rd,c ffS

bV

l

+

= (6.4)

where

is the second moment of areabw is the width of the cross-section at the centroidal axis, allowing for the presence of

ducts in accordance with Expressions (6.16) and (6.17)S is the first moment of area above and about the centroidal axis

I =lx/lpt2 1,0 for pretensioned tendons

= 1,0 for other types of prestressinglx is the distance of section considered from the starting point of the transmission

lengthlpt2 is the upper bound value of the transmission length of the prestressing element

according to Expression (8.18).

cp is the concrete compressive stress at the centroidal axis due to axial loading

and/or prestressing (cp= NEd /Acin MPa, NEd> 0 in compression)

For cross-sections where the width varies over the height, the maximum principal stress mayoccur on an axis other than the centroidal axis. In such a case the minimum value of the shearresistance should be found by calculating VRd,cat various axes in the cross-section.

45o45o

VEd

lbd

45o

Asl

dd

VEd

VEdAslAsl

lbd

lbd A

AA

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(3) The calculation of the shear resistance according to Expression (6.4) is not required forcross-sections that are nearer to the support than the point which is the intersection of theelastic centroidal axis and a line inclined from the inner edge of the support at an angle of 45

o.

(4) For the general case of members subjected to a bending moment and an axial force, whichcan be shown to be uncracked in flexure at the ULS, reference is made to 12.6.3.

(5) For the design of the longitudinal reinforcement, in the region cracked in flexure, the MEd-line should be shifted over a distance al= d in the unfavourable direction (see 9.2.1.3 (2)).

(6) For members with loads applied on the upper side within a distance 0,5dav2dfromthe edge of a support (or centre of bearing where flexible bearings are used), the contribution

of this load to the shear force VEdmay be multiplied by = av/2d. This reduction may beapplied for checking VRd,cin Expression (6.2.a). This is only valid provided that the

longitudinal reinforcement is fully anchored at the support. For av0,5dthe value av= 0,5dshould be used.

The shear force VEd, calculated without reduction by , should however always satisfy thecondition

VEd 0,5 bwd fcd (6.5)

where is a strength reduction factor for concrete cracked in shear

Note:The value for use in a Country may be found in its National Annex. The recommended value followsfrom:

=

25016,0 ckf (fckin MPa) (6.6N)

av

d

av

d

(a) Beam with direct support (b) Corbel

Figure 6.4: Loads near supports

(7) Beams with loads near to supports and corbels may alternatively be designed with strut andtie models. For this alternative, reference is made to 6.5.