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)

Derivation of design formulae 105
concrete compression = (0.45fcu)(0.9x)b
Equating this to the steel tension:
0.405fcubx = 0.87/yAs
= 0.405fcubx
(4.6-1)
:!: = 2 15 [y As
d · feu bd
(4.6-2)
The moment M corresponding to the forces in Fig. 4.6-l(b) is simply the
concrete compression or the steel tension times the lever arm z, where
z = d- 0.45x
Using the concrete compression, say,
M = (0.405fcubx)(d - 0.45x)
(4.6-3)
= (0.405x/d)(l - 0.45xld)fcubd 2
= Kfcubd 2
(4.6-4)
As expected, M increases with xld and hence with A, (see eqn 4.6-2). In
design, BS 8110 limits xld to not exceeding 0.5. When xld = 0.5, the forces
are as shown in Fig. 4.6-l(c); the moment Mu, which corresponds to these
forces, represents the maximum moment capacity of a singly reinforced
beam. From Fig. 4.6-l(c):
Mu = (0.2fcubd)z
= O.l56fcubd 2 (since z = 0. 775d)
= K'fcubd 2 (4.6-5)
where K' = 0.156. Of course, the same result of O.l56fcubd 2 can be
obtained by writing xld = 0.5 in eqn (4.6-4).
Where the applied bending moment M exceeds Mu of eqn (4.6-5), the
excess (M - Mu) is to be resisted by using an area A~ of compression
reinforcement (Fig. 4.6-2(a)) such that the neutral axis depth remains at
the maximum permitted value of 0.5d (i.e. depth of stress block = 0.9x =
t 0·87o/4~~
0·45d
~
Fig. 4.6-2
(a) (b)