with a minimum value of:
VRd.c = the design shear resistance of the section without shear reinforcement
< 2.0 with d expressed in mm
/lsi - the area of tensile reinforcement that extends beyond the section being considered by at least a full anchorage length plus one effective depth id) bw = the smallest width of the section in the tensile area (mm)
Some typical values of the corresponding shear stress capacities (rRd.c = Vri,c/bvd) arc given in chapter 8 (table 8.2).
5.1.2 The variable strut inclination method for sections that do require shear reinforcement
In order to derive the design equations the action of a reinforced concrete beam in shear is represented by an analogous truss as shown in figure 5.2. The concrete acts as the top
Assumed truss model for t variable strut Inclination method compression member and as the diagonal compression members inclined at an angle 0 to the horizontal. The bottom chord is the horizontal tension steel and the vertical links are the transverse tension members. It should be noted that in this model of shear behaviour all shear will be resisted by the provision of links with no direct contribution from the shear capacity of the concrete itself.
The angle 0 increases with the magnitude of the maximum shear force on the beam and hence the compressive forces in the diagonal concrete members. It is set by EC2 to have a value between 22 and 45 degrees. For most cases of predominately uniformly distributed loading the angle 0 will be 22 degrees but for heavy and concentrated loads it can be higher in order to resist crushing of the concrete diagonal members.
The analysis of the truss to derive the design equations will be carried out in the following order:
1. Consideration of the compressive strength of the diagonal concrete strut and its angle 0\
2. Calculation of the required shear reinforcement AiW/s for the vertical tics:
3. Calculation of the additional tension steel /\s| required in the bottom chord member.
The following notation is used in the equations for the shear design
Asw = the cross-sectional area of the two legs of the link ,v = the spacing of the links c the lever arm between the upper and lower chord members of the analogous truss fyw| the design yield strength of the link reinforcement
/yk the characteristic strength of the link reinforcement
VKt| = the shear force due to the actions at the ultimate limit state
VEf = the ultimate shear force at the face of the support
VW(| = the shear force in the link
VWs the shear resistance of the links
VKl( mux the maximum design value of the shear which can be resisted by the
(1) The diagonal compressive strut and the angle 0
The shear force applied to the section must be limited so that excessive compressive stresses do not occur in the diagonal compressive struts, leading to compressive failure of the concrete. Thus the maximum design shear force Vrj max is limited by the ultimate crushing strength of the diagonal concrete member in the analogous truss and its vertical component.
With reference to figure 5.2, the effective cross sectional area of concrete acting as the diagonal strut is taken as bv x jcosW and the design concrete stress/al -/ck/l.5.
The ultimate strength of the strut = ultimate design stress x cross-sectional area concrete strut and its vertical component so that
= C/ck/1-5) x x zcos6) - R/ck/1-5) x (¿w x ccos 6»)] x sin0
which by conversion of the trigometrical functions can also be expressed as
In EC2 this equation is modified by the inclusion of a strength reduction factor (V[) for concrete cracked in shear. Thus v _ UKzv i x
where the strength reduction factor takes the value of V| 0.6(1 -/ck/250) and, pulling z = 0.9(/. equation 5.3 becomes
_ 0.9</ x x 0.6( 1 -/ck/2501/ck VRdm"" 1,5(cot0 + tan 8) 0.36£»wÎ/( 1 -/ck/250)/ck
and to ensure that there is no crushing of the diagonal compressive strut:
This must be checked for the maximum value of shear on the beam, which is usually taken as the shear force. Vec, at the face of the beam's supports so that
As previously noted EC2 limits 0 to a value between 22 and 45 degrees.
(i) With 0 = 22 degrees (this is the usual case for uniformly distributed loads) From equation 5.4:
If mux(22> < Vu lhen a larger value of the angle 0 must be used so that the diagonal concrete strut has a larger vertical component to balance VEd.
(ii) With 0 = 45 degrees (the maximum value of 0 as allowed by EC2) From equation 5.4:
which is the upper limit on the compressive strength of the concrete diagonal member in the analogous truss. When V|.;r > VRd.max(45)> from equation 5.7 the diagonal strut will be over stressed and the beam's dimensions must be increased or a higher class of concrete be used.
(Hi) With 0 between 22 degrees and 45 degrees
The required value for 0 can be obtained by equaling V|.j to Vr^™« and solving for i9 in equation 5.4 as follows:
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