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)

118 Reinforced concrete beams-the ultimate limit state
= (0.87)(460)(982)(640 - 65.48)(10- 6 ) = 225.6 kNm
0.87fyAs2( d2 - O.ix)
= (0.87)(460)(1473)(680 - 65.48)(10- 6 ) = 362.0 kNm
M = 597.6 kNm
Comments
(a) Note that the condition of equilibrium has to be worked out from
Fig. 4.6-7:
(4.6-15)
Equation ( 4.6-7) assumes that xl d = 0.5 and cannot therefore be
used here.
(b) If the strain distribution in Fig. 4.6-8 had shown that the design
strengths 0.87fy were not attained in some or all of the reinforcement,
then a trial and error procedure would have to be used (see Example
4.5-1).
In Examples 4.6-10 and 4.6-11 below, we shall derive the formulae, given
in BS 8110: Clause 3.4.4.4, for the zld factors in Table 4.6-1 above
(reproduced from Table 24 of the I.Struct.E. Manual [14]).
Example 4.6-10
A bending moment Misapplied to a rectangular beam section. If M ::5 Mu
of eqn (4.6-5), show that zld is given by the following BS 8110 formula:
~ = o.5 + ~( o.25 - 0~) (4.6-16)
Where K = Mlfcubd 2 ;
z = lever-arm distance measured from the tension reinforcement
to the centroid of the concrete compression block.
SOLUTION
ForM ::5 Mu of eqn (4.6-5) we have, from eqn (4.6-4),
K = (0.405x/d)(1 - 0.45xld)
From Fig. 4.6-1,
z = d- 0.45x
Eliminating x from these two equations, we have
(~Y - (~)
+ 1.u1K = o
from which
~ = 0.5 + ~( 0.25 - 0~9)