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)

212 Shear, bond and torsion
reduce the shear strength [1]. Therefore, in using Table 6.4-1,
any longitudinal bars which are terminated within a distance d
of the section concerned cannot be considered; indeed, near
such termination, it is desirable (though not mandatory) to put in
additional links locally [1].
(b) The use of Ve in shear design was explained in Section 6.3 in relation
to eqn (6.3-7). Earlier, it was explained in Section 6.2 that the
nominal shear stress at the collapse of a beam increases with the
concrete strength feu and the tension steel ratio e (= Asi bd). The Ve
values in Table 6.4-1, which already allow for the partial safety
factor Ym, are plotted against experimental data in Fig. 6.2-6. This
figure shows that some of the experimental values fall below the
BS 8110 design values.
(c) Equation (6.4-2) gives the requirements for minimum links. The
technical background to eqn (6.4-2) was given earlier in eqn (6.3-8):
(d)
evfyv ;::: 0.4
That is,
A sv f, > 04
b"sv yv - .
which becomes eqn (6.4-2), when the partial safety factor 0.87 is
used for fyv.
The requirement that the link spacing sv should not exceed 0.75d
follows from eqn (6.3-9). At right angles to the span, the horizontal
link spacing must also be kept sufficiently small to prevent the
pressing down of any longitudinal reinforcement bar and hence
maintain the dowel capacity.
Comments on Step 4
When v > 0.5ve , BS 8110 allows for the possibility of diagonal cracking; if
a diagonal crack does occur (because of accidental overloading, for
example) there should be sufficient shear reinforcement to arrest its
growth, and the minimum links specified by eqn (6.4-2) provide such a
safeguard.
Comments on Step 5
(a)
Equation (6.4-3) follows directly from eqn (6.3-5), if fyv is
multiplied by the coefficient 0.87 (= 1/1.15 where 1.15 is the partial
safety factor Y m).
(b) Note that irrespective of the amount of shear reinforcement used, the
shear stress v from eqn (6.4-1) must noI exceed O.8Heu or 5 Nlmm 2 •
See Step 2 again, if necessary.
(c)
BS 8110 does not permit shear reinforcement to be provided entirely
in the form of bent-up bars, because there is insufficient evidence to
show that beot-up bars 00 their own are satisfactory_ Bent-up bars,
however, may be used in combination with links (Fig. 6.3-2) but the
links must contribute at least 50% of the total resistance of the shear
reinforcement. The combined resistance of the links and bent-up bars
is to tie taken as the sum of their separate resistance; this point was