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)

Flanged beams 131
(b) If Kr of eqn (4.8-5) does not exceed K' (= 0.156) of eqn (4.6-5), we
know that the moment capacity of the web component is not
exceeded, i.e. compression steel is not required.
(c) To derive eqn (4.8-6), consider the actual As in Fig. 4.8-4(a) to be
made up of Asr of the flange component (Fig. 4.8-4(b)) and Asw of
the web component (Fig. 4.8-4(c)). By considering Fig. 4.8-4(b),
or
Mur = 0.87/yAsr(d - 0.5hr)
Mur
A s r = 0.87,y '" ( d - 0.5hr )
By considering Fig. 4.8-4(c),
or
M -
Mur = 0.87/yAswZ
M- Mur
Asw = 0.87/y:z
(4.8-6(a))
(4.8-6(b))
where z is from Table 4.6-1. Equation ( 4.8-6) follows from As =
Asf + Asw• i.e.
eqn (4.8-6) = eqn (4.8-6(a)) + eqn (4.8-6(b))
(d) In the unlikely event that Kr > 0.156, redesign the section or consult
Example 4.8-1 for design of compression steel.
I.Struct.E. Manual's design procedure (Case II: 0-30% moment
redistribution)*
The procedure below is that of the I.Struct. E. Manual [14]. Comments
have been added to explain the derivation of the formulae and the
technical background. The design steps are identical to those of Case I
above, except as mentioned below.
Stepl
As Step 1 of Case I, except that the xld and zld ratios are now to be
obtained from Table 4.7-2.
Comment
Comment (c) below Table 4. 7-2 explains how to use the table for various
% moment redistribution.
Step2
As Step 2 of Case I, except that in calculating As from eqn (4.8-3), the
lever arm z is now obtained from Table 4.7-2. (Note the similarities
between eqns 4.8-3 and 4.7-9.)
Step3
As Step 3 of Case I.
* This procedure (Case II) may be omitted on first reading. Readers may move on to Section
4.9 and then read Section 4.7 before returning to this page.