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)

108 Reinforced concrete beams-the ultimate limit state
For the design strength 0.87fy to be reached, e~ = 0.87fyl £ 8 • Therefore
0.87fy = X - d' (0.0035)
£ 8 X
where Es = 200 kN/mm 2 , i.e.
d' = 1 - jy_
X 800
(4.6-10)
For fy = 460 N/mm 2 , this gives d' /x = 0.43. For the balanced condition of
xld = 0.5, we have
{[ = 2 x = 0.21
d' 1(d')
Therefore, as long as d' /d does not exceed 0.21 (i.e. d' lx does not exceed
0.43) the compression reinforcement can be assumed to reach the design
strength of 0.87fy·
Referring to the more general eqn (4.6-10), if d'/x exceeds 1 -
(fyl800), a reduced stress f~ should be used. From Fig. 4.6-4,
e~ = x - d' (0.0035)
X
Hence f~ = E8 e~ where Es = 200 kN/mm 2 , i.e.
f~ = 700 ( 1 - :') (4.6-11)
4.6(b) Designing from first principles
The I.Struct.E. Manual [14] gives a convenient procedure for practical
design, which will be explained later-after Example 4.6-4. In the mean
time, Examples 4.6-3 and 4.6-4 below will explain how designs can be
carried out from first principles.
Example 4.6-3
The design ultimate moment M for a rectangular beam of width b 250 mm
and effective depth d 700 mm is 300 kNm. If feu = 40 N/mm 2 and fy =
460 N/mm 2 , design the reinforcement. Work from first principles.
SOLUTION
Step(a)
Check concrete capacity Mu. From eqn (4.6-5),
Mu = 0.156fcubd 2
= (0.156)(40)(250)(700 2 ) Nmm
= 764.4 kNm
Since M < Mu, no compression steel is required.