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 procedure for rectangular beams 113
(3a} Compression reinforcement is required. From eqn (4.6-13),
A' M- Mu
s = 0.87/y(d - d')
where d' = 60 mm, say,
(900 - 764.4)(10 6 ) 2
= (0.87}(460)(700-60) = 530 mm
(3b} Check d'lx. From Table 4.6-1,
~ = 0.5 (see Comment (c) under Table 4.6-1}
X = (0.5)(700} = 350
d' 60 ( /y)
x = 350 = 0·17 < 1 - 800
Therefore the compression steel reaches the design strength of
0.87/y and the A~ calculated in Step (3a) above is acceptable.
(3c} As = O.~fyz + A~ (see eqn 4.6-14}
(where z = 0.775d from Table 4.6-1}
(764.4)(10 6 )
= (0.87)(460)(0.775)(700) + 530
= 4051 mm 2
Provide two 20 mm top bars ( 628 mm 2 )
Provide five 32 mm bottom bars ( 4021 mm 2 )
Comments
Example 4.5-4 solves the same problem using BS 8110's design chart,
which gives A~ = 981 mm 2 and As = 3955 mm 2 , so that A~ + As =
4936 mm 2 which exceeds the A~ + A. obtained using the simplified stress
block here. The higher total steel area in Example 4.5-4 arises for two
reasons:
(a) The design chart in Fig. 4.5-2 assumes d'/d = 0.15. In the present
example, d' /d = 60/700 = 0.09.
(b) In Example 4.5-4, the compression steel ratio(/ is taken as 0.5%,
which is the lowest value shown in the design chart (Fig. 4.5-2). If(/
had been taken as 0.3%, for example, a smaller total for A~ + As
would have resulted (though in using Fig. 4.5-2 a curve for e' =
0.3% would have to be added by interpolation).
Example 4.6-7
The effective depth d of a beam section is to be 2~ times its width b. The
design ultimate moment is 1600 kNm. If e (= A.fbd) is not to exceed 3%
and e'(= A~lbd) is not to exceed 1.5%, determine the minimum required