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)

Slender columns (BS 8110) 285
Asc = 1.4% of bh (= 2800 mm 2 )
Provide 4 size 32 bars (3216 mm 2 )
Example 7.5-3
Design the longitudinal reinforcement for the braced slender column in
Fig. 7.5-1, for biaxial bending, if N = 2500 kN, M 1x = 200 kNm, M2x =
250 kNm, M 1y = 100 kNm, M2y = 120 kNm, feu= 40 N/mm 2 and /y =
460 N/mm 2 •
SOLUTION
For biaxial bending, Step 4 of Section 7.5 is relevant.
Step4(a)
Calculate Mty·
Step4(b)
Calculate Mtx·
M 1 y = 242 kNm (as M 1 in Example 7.5-1)
Mix = 230 kNm (as Mi in Example 7.5-2)
From eqn (7 .4-6),
Madd = NfJaKh
where h = 500 mm and K is for simplicity being taken as unity (see
Comments at the end of the solution). To obtain fJa in Table 7.5-1, we
note that, in evaluating lei b' for biaxial bending, b' is to be taken as the
depth in the plane of bending under consideration, i.e. in this case, b' =
h = 500 mm. Hence lelb' = 65001500 = 13, so that fJa = 0.085 from
Table 7 .5-1. Hence
Madd = (2500)(0.085)(1)(500)(10-3) = 106 kNm
From eqn (7.5-1)
Mtx = Mix + Madd
= 230 + 106 = 336 kNm
By inspection, eqns (7.5-2) to (7.5-4) are not critical. Hence M 1 x is
taken as 336 kNm.
Step4(c)
We shall first check which of eqns (7.3-3) and (7.3-4) is applicable.
With the symbols h' and b' as defined under those equations, we have
Mtx = (336)(10 6 ) = (791)(103)
h' (0.85, say)(500)
Mty = (242)(10 6 ) = (7l2)(103)
b' (0.85, say)(400)
Hence M 1xlh' > M 1yfb' and eqn (7.3-3) is applicable:
h'
M;x = Mtx + fJ b' Mty