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 formulae and procedure-BS 8110 simplified stress block 125
(b) The formulae in Step (3b) were derived in Example 4.6-2; i.e. eqns
(4.6-10) and (4.6-11).
(c) To derive the formula in Step (3c), equate the forces in Fig. 4.7-1:
0.87/yAs = 0.405fcubx + 0.87/yA~ (4.7-12)
Noting from Fig. 4.7-1(b) that Mu = (0.405fcubx)z we have
0.87/yAs = ~u + 0.87/yA~
A Mu A'
s = 0.87/yz + s
(4.7-13)
Of course, if the compression steel A~ does not reach 0.87/y then As
should be calculated from:
0.87/yAs = ~u + f~A~ (4.7-14)
where f~
is as given in Step (3b) above.
Example 4. 7-3
Repeat Example 4.6-5, assuming 15% moment redistribution.
SOLUTION
Step]
Mu = K'fcubd 2 (where K' is 0.144 from Table 4.7-1)
= (0.144)(40)(250)(700 2 ) Nmm
= 705.6 kNm
Step2
M(= 300 kNm) < Mu of Step 1
Calculate
.....-f-o--8-1-fy-As-,
0·45fcu
I" I
0·9X
t
z
_j
(X= 1pb -0·4)d)
(a)
(b)
Fig. 4.7-1