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)

Braced frame analysis 413
36 kN/m and the characteristic imposed load qk is 45 kN/m; use the subframe
of Fig. 11.4-1 (g).
SOLUTION
As previously mentioned, the first difficulty encountered is the assessment
of the relative stiffness values of the various members, which are assumed
to be rigidly connected. Some guesses have to be made in order to make a
start and the ones indicated in the figure are based on the assumption
that the first- and second-floor loadings will be the same and greater than
that on the roof, and that the second-floor columns are of smaller section
than those of the other two floors.
In all the following examples the moments and shears calculated are
those obtained direct from elastic analyses, i.e. prior to the redistribution
of moments. The actual process and implications of redistribution have
already been covered in Section 4.9. The discerning student will have
noted that the horizontal dimensions are as those in Example 4. 9-1. Thus
that example can be taken here as a demonstration of the sub-frame shown
in Fig. 11.4-1(g).
Strictly speaking, the relative span dimensions and also the relative
magnitudes of the dead (gk = 36 kN/m) and imposed loads (qk = 45 kN/m)
preclude (see BS 8110: Clause 3.4.3) the use of the coefficients given in
Table 11.4-1, but their quick calculation is helpful in giving a comparison
with those values shown in Fig. 4.9-7(b). It is important to remember that
in those circumstances where the use of the coefficients is allowed and they
are used, no redistribution of the moments so obtained is permitted. A
more extensive list of similar coefficients for two-, three-, four- and fivespan
continuous beams and also incorporating point load effects is given in
Reference 6.
The maximum distributed load
= 1.4gk + 1.6qk = (1.4 X 36) + (1.6 X 45) = 122 kN/m
(to 3 significant figures)
F = 122 x span
(1) Near the middle of the end span (sagging)
0.09Fl = 0.09(122 x 8)(8) = 702 kNm; (730 kNm; -3.8%)
Note: The bracketed figures are the equivalent values from Fig.
4.9-7(b) and the corresponding percentage difference.
(2) At the first interior support (hogging)
O.llFl = 0.11(122 x 10)(10) = 1342 kNm (1097 kNm; +22%)
Note: The greater of the two spans meeting at the support is used.
(3) At the middle of the interior span (sagging)
0.07 Fl = 0.07(122 X 10)(10) = 854 kNm (769 kNm; + 11%)
It can be seen, notwithstanding their strictly speaking inapplicability
here, that the coefficients of Table 11.4-1 do give a rapid and useful
estimate of the bending moments. The student is advised to compare their