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 427
Fig.ll.S-2
SOLUTION
(See Fig. 11.5-2).
Step 1 Durability and fire resistance
From Table 2.5-7,
nominal cover for mild exposure condition = 20 mm
From table 8.8-1,
fire resistance of 180 mm slab with 20 mm cover to main bars is not less
than 1 hour
Nominal cover = 20 mm
Fire resistance OK
Step 2 Loading-per metre width of slab
Self-weight = (0.180 m) (24 kN/m 3 ) (5.5 m) = 23.8
Patitions and finishes = (1.5) (5.5) = 8.3
Characteristic dead load Gk = 32.1 kN/m width
Characteristic imposed load Qk = (3) (5.5) = 16.5 kN/m width
Design load F = 1.4 Gk + 1.6Qk
Step 3 Ultimate moments
From Table 11.4-2,
= 44.9 + 26.4
= 71.3 kN/m width
Gk = 32.1 kN/m Qk = 16.5 kN/m
F = 71.3 kN/m
Mat supports = 0.063Fl = (0.063) (71.3) (5.5) = 24.7 kNm/m
M at midspan = 0.063Fl = 24.7 kNm/m
Step 4 Main reinforcement
Effective depth d = 180 - 20 - 1 bar ifJ = 154 mm, say
M (24.7) (10 6 )
Supports: fcubdz = ( 40) (1000) (1542) = 0.026
From Table 4.6-1,
zld = 0.94 xld = 0.13
(Note: As explained at the end of Section 4.5,