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)

216 Shear, bond and torsion
v (minimum links) = Ve + 0.4 = 1.10 N/mm 2
V (minimum links) = (1.10) (350) (800) (10- 3 ) = 308 kN
Beyond 3 m from B:
Therefore
As before,
As (two size 32 bars) = 1608 mm 2
100 As/bvd = 0.57; d> 400 mm
Ve (Table 6.4-1) = 0.61 N/mm 2 (by interpolation)
V (minimum links) = Ve + 0.4 = 1.01 N/mm 2
V (minimum links) = (1.01) (350) (800) (10- 3 ) = 282 kN
The shear resistance of 308 and 282 kN are drawn as the chain-dot lines
in Fig. 6.4-2(a); minimum links only are required where the shear force
envelope lies inside these limits.
From Example 6.4-1(b):
Minimum link requirements are met by grade 250 (mild steel) links of
0.18% bvSv.
so that
A sv = 0.0018 bvSv
Asv/sv = (0.0018) (350) = 0.63 mm
308kN limil
(a) Shear force envelope
7R12-110150
16R12-110300
23R12-110100
A~~----------------------B
1..900J 4800 • .1. 2300 .. 1
(b) Arrangement of links
Fig. 6.4-2