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)

(b) From Table 6.11-1,
Viu = 5 N/mm 2
From eqn (6.11-3),
VI + V = 5
Design and detailing-illustrative example 243
where V = 0.57 N/mm 2 from Example 6.11-2(b). Therefore
VI = 5 - 0.57 = 4.43 N/mm 2
Using eqn (6.11-1),
T _ (4.43) (3502 ) [800 - 350/3J
- 2
= 185 kNm
6.12 Design and detailing-illustrative example
The following example is a continuation of Example 4.11-1.
Example 6.12-1
Complete Step 7 of the solution to Example 4.11-1.
SOLUTION
From Example 4.11-1 (Solution: Step 6),
V = 86.15 kN
V (86.15) (10 3 ) 2
V = bvd = (325) (320) = 0.83 N/mm
< (i) 5 N/mm 2 and (ii) 0.8 V/eu (= 5.1 N/mm 2 )
Therefore the overall dimensions are adequate for shear.
From Example 4.11-1 (Solution: Step 4),
As provided at interior support = 628 mm 2
~ - 628 _ O 6001
bvd - (325) (320) - . 10
From Table 6.4-1,
Hence
Ve = 0.66 N/mm 2 by interpolation
0.5vc = 0.33 N/mm 2 ; Ve + 0.4 = 1.06 N/mm 2
0.5vc < V (= 0.83 N/mm 2 ) < Ve + 0.4
Provide minimum links, using eqn (6.4-2) (see also Example 6.4-1(b»:
= O.4bvSv = (0.4) (325)sv
A
sv 0.87/yv (0.87) (250)