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)

240 Shear, hond and tors ion
Step2
Check whether 0.9x :S h f • From Step 1,
x = (0.13) (750) = 98 mm
0.9x = 88 mm < h f
Hence eqn (4.8-3) is applicable:
_ M (215) (10 6 ) _ 2
As - 0.87fyz = (0.87) (460) (0.94) (750) - 762 mm
where 0.94 is zid from Step 1.
Use two size 25 bars (981 mm 2 ; (! = AJbwd = 0.37%)
(b) Shear. We shall follow the steps in Section 6.4.
Stepl
Step2
V 150000 2
v = byd = (350) (750) = 0.57 N/mm
0.8 Heu = 5.1 N/mm 2 (for feu of 40 N/mm 2 ) > 5 N/mm 2
Hence the upper Iim it is 5 N/mm 2 •
Step3
v of Step 1 < 5 N/mm 2 OK
d = 750 mm and 100AJbyd = 0.37 from (a)
From Table 6.4-1,
Hence
Step4
Hence
Ve = 0.53 N/mm 2 by interpolation
v > 0.5 Ve
ve + 0.4 = 0.93 N/mm 2
0.5ve < v < Ve + 0.4
Provide minimum links from eqn (6.4-2):
A
= 0.4 bvSy = (0.4) (350)sy
Sy 0.87 fyy (0.87) (460)
A sy 035 -= . mm
Sv
(c) TorsioD. We shall follow the steps in Section 6.11.
Stepl
Omitted.