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)

Design and detailing-illustrative examples 439
M = 40% of 307 kNm (Fig. 11.5-5: Case 2)
= 123 kNm
M (123)(10 6 )
fcubd 2 = (40)(350)(500Z) = 0.03S
From Table 4.7-2,
z
d = 0.94
X
d = 0.13
< K' of Table 4.7-1
From eqn (4.7-9),
M (123)(10 6 )
As = 0.87/yz = (0.87)(460)(0.94)(500)
= 654 mm 2 2T25 U-bars (982 mm 2 )
From Table 4.10-2
ultimate anchorage bond length = 32(j) = 800 mm
Using eqns (6.6-3(a) (b)),
required anchorage length = ~~i (800) = 533 mm 2
The curtailment diagram for the longitudinal reinforcement is shown in
Fig. 11.5-10, which should be read in conjunction with the reinforcement
details in Fig. 11.5-12 (seep. 444).
Step 6 Shear reinforcement
For further explanations of the calculations, see Comments on Step 6 at
the end.
(a)
From Fig. 11.5-12, the minimum tension reinforcement along the
beam may reasonably be taken as 2T25, i.e. As= 982 mm 2 •
100As _ (100)(982) = O 56
bvd - (350)(500) ·
From Table 6.4-1,
Vc = 0.61 N/mm 2
(b) Minimum links (see Section 6.4: 'Shear resistance in design
calculations: Step 4') will be provided where v :s: (vc + 0.4), i.e.
where V :S: (vc + 0.4)bvd.
V :s: (0.61 + 0.4)(350)(500) N
= 177 kN