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)

366 Prestressed concrete simple beams
StepS
Check that in no case should Vlbvd exceed 0.8~/cu or 5 N/mm 2 whichever
is the lesser. (These stress limits include an allowance for Ym·)
Step6
If V < 0.5Vc, no shear reinforcement is required.
Step7
If V ~ 0.5Vc but ::5 (Vc + 0.4bvd), provide shear links as follows:
(9.6-5)
where symbols have their usual meanings (see eqn 6.4-2 if necessary).
Step8
If V > (Vc + 0.4bvd), provide shear links as follows:
Asv V- Vc
S: = 0.87/yvdt
(9.6-6)
where Asv' Sv (sv ::5 0.75dt) and/yv have their usual meanings (see eqn
6.4-3 if necessary), and dt is the depth from the extreme compression
fibre to the centroid of the tendons or to the longitudinal corner bars
around which the links pass, whichever is the greater.
Comments
BS 8110's shear design procedure is essentially similar to that of the
previous Code CP 110. Smith [9] has written a useful article on the subject.
Example 9.6-l
Design the shear reinforcement for a symmetrical prestressed 1-section,
given that:
SOLUTION
VL = 400 kN
M = 800 kNm
rib width hw = 200 mm overall depth h = 1000 mm
area A = 310 x 10 3 mm 2 I = 36 x 10 9 mm 4
Aps = 1803 mm 2 /pu = 1750 N/mm 2 /pe = 0.6/pu
es = 290 mm (and tendon is inclined at p = 3° at the section
considered)
feu = 50 N/mm 2 /yv = 250 N/mm 2
Pe = Aps/pe = 1803 X 0.6 X 1750 = 1893 kN
/cp = Pe!A = 1893 x 1W/310000 = 6.10 N/mm 2
ft = 0.24Hcu = 0.24~50 = 1.7 N/mm 2
Stepl
Case 1: From eqn (9.6-3),
V = VL- PesinP