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)

132 Reinforced concrete beams-the ultimate limit state
Step4
Kc is calculated as in Step 4 of Case I, using eqn (4.8-5).
If Kc:::; K', obtained from Table 4.7-1, then obtain the lever arm z
from Table 4.7-2 and calculate As from eqn (4.8-6).
Comments
(a) The equations in this step are derived in the comments below Step 4
of Case I.
(b) The essential differences between Step 4 here and Step 4 of Case I
are:
(1) Kc is now compared with K' of Table 4.7-1 (and not with K' =
0.156 of eqn 4.6-5);
(2) in using eqn (4.8-6), z is now obtained from Table 4.7-2
(and not from Table 4.6-1).
(c) If Kt > K' of Table 4.7-1, redesign the section or consult Example
4.8-1 for the design of the compression steel.
Example 4.8-1
With reference to Step 4 of the design procedure above, if Kc > K',
explain how to design the compression steel.
SOLUTION
The I.Struct.E. Manual's advice [14] on this point is: 'consult BS 8110 for
design of compression steel'. With reference to Fig. 4.8-4(c), the moment
capacity of the web component is
Muw = K'fcubwd 2 (4.8-7)
where K' is obtained from Table 4.7-1. The design moment M is partly
resisted by Mur (see eqn 4.8-4) and partly by Muw (eqn 4.8-7). The
difference M - Mur - Muw will be resisted by compression steel:
0.87/yA~(d - d') = M - Mur - Muw (4.8-8)
Equation ( 4.8-8) gives the required area A~ of the compression steel. The
tension steel area As is then determined from the equilibrium of forces:
0.87/yAs = flange compression + web compression
+ steel compression
0.87/yAs = 0.45/cu(b - bw)hc + 0.45fcubw(0.9x) + 0.87/yA~
(4.8-9)
where, in the second term on the right-hand side, the neutral axis depth xis
obtained from Table 4.7-2.
Comment
Compare eqn (4.8-8) with eqn (4.7-10); compare eqn (4.8-9) with eqn
(4.7-12).