F. K. Kong MA, MSc, PhD, CEng, FICE, FIStructE, R. H. Evans CBE, DSc, D ès Sc, DTech, PhD, CEng, FICE, FIMechE, FIStructE (auth.)-Reinforced and Prestressed Concrete-Springer US (1987)

206 Shear, bond and torsion
V must be carried by Vez, Va and V d • As the diagonal crack widens, the
aggregate interlock becomes less effective and Va decreases, forcing V ez
and V d to increase rapidly. Failure of the beam finally occurs either by
dowel splitting of the concrete along the longitudinal reinforcement or
by crushing of the concrete compression zone resulting from the combined
shear and direct stresses.
In current design practice, the stresse~ in the shear reinforcement are
analysed by the truss analogy, iIIustrated in Fig. 6.3-4, in which the web
bars are assumed to form the tension members of an imaginary truss, while
the thrusts in the concrete constitute the compression members (shown
dotted in the figure). The figure shows a general case of Iinks at a longitudinal
spacing SV. The links and the concrete 'struts' are shown inclined at
the general angles a and p, respectively, to the beam axis. To derive design
equations using the truss analysis, draw a line A-A in Fig. 6.3-4, parallel
to the concrete 'struts'. Consider the vertical. equilibrium of the free body
to the left of the line A-A. The web resistance V s is contributed by the
vertical components of the tensions Asvfyv in the individual links that are
crossed by A-A:
v: A . [Number of links crossed ]
s = svfyv sm a by A-A (see Example 6.3-1)
_ A f, . [(d - d')cot a + (d - d')cot P]
- sv yv sm a
Sv
= Asvfyv [cos a + sin a cot P] [d --s:- - d']
(6.3-2)
where A sv is the area of both legs of each link and fyv is the characteristic
strength of the Iinks.
In the particular case of verticallinks, a = 90° and eqn (6.3-2) becomes
[d- d']
Vs = Asvfys -s-v- cot P
'* Asvfyv [~] cot P (6.3-3)
Tests [2] have led to the recommendation that pin eqn (6.3-3) could be
taken as 45°, so that
(6.3-4)
From eqn (6.3-1), V s = V - Ve; and if we write V = vbd and Ve = vebd,
the above equation for vertical Iinks becomes
(v - ve) b = A;!yv (6.3-5)
When the web reinforcement consists of a system of bent-up bars, it would
be reasonable to use eqn (6.3-2) as it stands:
[ shear carried by ] _ [ . P] d - d'
b t b t - Asvfyv cos a + sm a cot
en -up ar sys em
Sv
(6.3-6)