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)

110 Reinforced concrete beams-the ultimate limit state
Step(c)
Find tension steel area As. From eqn (4.6-7),
0.87/yAs = 0.2fcubd + 0.87/yA~
(0.87)( 460)As = (0.2)( 40)(250)(700) + (0.87)( 460)(530)
As= 4030 mm 2
Provide two 20 mm top bars (A~ = 628 mm 2 )
Provide five 32 mm bottom bars (As = 4021 mm 2 )
Step(d)
The x/d ratio. As explained in the paragraph preceding eqn (4.6-6), the
x/d ratio is 0.5.
Comments
(a)
Example 4.5-4 solves the same problem using BS 8110's design
chart. See also Example 4.6-6 which uses the I.Struct.E. Manual's
procedure [14].
(b) Examples 4.6-3 and 4.6-4 deal with design. For the use of BS 8110's
simplified stress block in analysis, see Examples 4.6-8 and 4.6-9.
4.6(c) Design procedure for rectangular beams
(BS 8110/I.Struct.E. Manual)
The design procedure given below is that of the I.Struct. E. Manual [14].
Comments have been added to explain the derivation of the formulae and
the technical background. As stated at the beginning of this section, the
same procedure can be used for up to 10% moment redistribution. For
moment redistribution exceeding 10%, see Section 4. 7. For flanged beams,
see Section 4.8. Consider the beam section in Fig. 4.6-1(a). Suppose the
design bending moment is M. Proceed as follows.
Stepl
Calculate Mu for concrete.
Mu = K'fcubd2
where K' = 0.156.
Comments
See eqn (4.6-5).
Step2
If the design moment M::::; Mu of Step 1: the tension reinforcement As is
given by
A = M (4.6-12)
s (0.87/y)z
where the lever arm z is obtained from Table 4.6-1.