There is a generally accepted model for the prediction of the effects of shear reinforcement. This is the ‘truss model’, illustrated in Fig. 6.1 for the commonest case where vertical links are used. In this model, the top and bottom compression and tension members are, respectively, the concrete in the compression zone and the tension steel. The members connecting the top and bottom members are represented by steel tension members and virtual concrete ‘struts’. The truss in Fig. 6.1 can be analysed to give the forces in the various members as indicated below:
F1= N/2 + V(a - v/z - 0.5 cotθ) (D6.3)
F3= N/2 - V(a - v/z + 0.5 cotθ) (D6.4)
F2= V/sinθ (D6.5)
Two factors can be observed from this analysis. Firstly, the forces in the vertical tension member 4 and the compression strut are both independent of the axial force, N. Secondly, the forces in the compression and tension chords both differ from that calculated from the
moment alone by 0.5V cotθ. (Note that Vav= M, and the forces in the upper and lower
chords are given by bending theory as ±M/z.)
In developing the truss under discussion into a design approach for a reinforced or prestressed concrete beam, it is necessary to consider it as a ‘smeared’ truss, as illustrated in Fig. 6.2. In this idealized truss, the vertical stirrups are represented by a uniform vertical
tensile stress ofρwfykor a vertical force ofρwfykbwper unit length of the beam. This assumes
that, at failure, the stirrups yield. The virtual compressive strut is replaced by a uniform
uniaxial compressive stress ofσcacting parallel to the line of action of the strut over the
concrete between the centroid of the tension reinforcement and the centre of compression. A complete analysis of this smeared truss may be carried out by considering the equilibrium of two sections:
• Vertical equlibrium across a section parallel to the virtual compression strut. In this
section, all the shear is supported by the shear reinforcement (section A-A in Fig. 6.2). This gives
V =ρwfykbwz cotθ (D6.7)
Fig. 6.1. Idealized truss
• Vertical equilibrium across a section perpendicular to the strut (section B-B in Fig. 6.2). Across this section, some of the shear is carried by the strut and some by the stirrups. The total strut force is
Fc=σcbwz/cosθ (D6.8)
The horizontal projection of the section is given by z tanθ, and hence equilibrium is
given by
V =ρwfykbwz tanθ + Fccosθ Since, from equation (D6.7),
ρwfykbwz = V tanθ
we can rewrite this as
V = V tan2θ + σ
cbwz tanθ
This can be rearranged to give
V =σcbwz/(cotθ + tan θ) (D6.9)
These equations can be converted to design values by simply replacing the characteristic
material properties by the appropriate design values.σcis replaced byαcwν1fcd. The parameter
ν1is an efficiency factor which allows for the actual distribution of the stress within the strut
and the effects of cracking.αcwis a coefficient taking account of any applied compression
force. Both these factors are Nationally Defined Parameters, and the recommended value
forν1isν,given by
ν = 0.6(1 - fck/250) (D6.10)
A higher value is permitted where the design stress in the shear reinforcement is less than
0.8fyk.
αcwtakes a value of 1 for non-prestressed structures or structures subjected to tension. For
cases where an axial compression is applied, equations (6.11.aN), (6.11.bN) and (6.11.cN) in EN 1992-1-1 provide values.
Equation (D6.9) then gives the maximum shear that can be carried by a section before failure by crushing of the notional compression struts.
Equations for inclined shear reinforcement can be derived in a similar way by replacing the vertical tension by a uniform tension inclined at an angle to the horizontal.
The design forms of equations (D6.7) and (D6.9) appear in EN 1992-1-1 as equations (6.8) and (6.9).
There are two areas where methods in various codes differ. The first is in the choice of a
value for the truss angle,θ. The second is whether the truss should carry all the shear force or
whether part of the shear can be considered to be carried by the concrete.
Internationally, by far the most common approach is to assume that cotθ is 1 (i.e. the
truss is assumed to form at 45oto the axis of the beam). When this assumption is made, it
is found that the shear capacity of a beam with shear reinforcement is underestimated, and it is found that a generally consistent prediction of strength is achieved if the shear strength is assumed to be equal to the shear capacity of the shear reinforcement calculated according to equation (D6.9) plus the capacity of the concrete given by equation (D6.1) (or its equivalent in whatever code is being considered). The shear response is thus assumed to be plastic. Both BS 8110 and the ACI code follow this approach.
EN 1992-1-1 adopts a second approach. In this approach, all the shear is assumed to be
carried by the shear reinforcement, but the truss angle, θ, can take any value between
cot-10.4 and cot-12.5. This variable strut angle approach is considered to be the more
rigorous of the two methods, and is also slightly more economical in some cases. However, as written in EN 1992-1-1, it is open to a misunderstanding. The code implies that the designer may select any strut angle between the specified limits. This concept of free choice does not,
however, reflect the behaviour of a beam. Beams will fail in a manner corresponding to a
strut angle of roughly cot-12.5 unless constrained by the detailing or the geometry of the system
to fail at some steeper angle. A steeper-angled failure could be induced either by the way in which the tension steel was curtailed or where the load is so close to the support that only a steeper failure can occur or where the shear strength is limited by the crushing strength of the strut. This last consideration will be considered further after the crushing strength of the strut has been discussed. Figure 6.3 illustrates this aspect of behaviour.