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)

268 Eccentrically loaded columns and slender columns
Then it can be designed either for (N, M~) or for (N, M~) depending on
which of the following two conditions is valid:
(a) For Mxlh' ~ Mylb':
h'
M~ = Mx + {Jb,MY
(7.3-3)
(b) For Mxlh' < Mylb':
b'
M~ =My+ {Jh,Mx
(7.3-4)
where h' = the effective section dimension in a direction perpendicular to
the major axis x-x, as shown in Fig. 7.3-3;
b' = the effective section dimension perpendicular to the minor
axis y-y, as shown in Fig. 7.3-3;
fJ = a coefficient to be obtained from Table 7.3-1;
Mx (My) should not be taken as less than Nemin about the x-x (y-y) axis.
Table 7.3-1 Values of {J for eqns (7.3-3) and (7.3-4) (BS 8110: Clause 3.8.4.5)
N
fcubh
0
0.1 0.2
0.3 0.4 0.5 2:::0.6
1.00
0.88 0.77
0.65 0.53 0.42 0.30
Examples 7.3-1 and 7.3-2 below illustrate the design procedures for
uniaxial and biaxial bending. Section 7.3(b), which follows these examples,
explains the behaviour of columns under biaxial bending.
Example 7.3-1 (Uniaxial bending)
Design the longitudinal reinforcement for a 500 by 300 mm column section
if:
(a) N = 2300 kN and Mx = 300 kNm,
(b) N = 2300 kN and My = 120 kNm,
where Mx is the bending moment about the major axis and My is the
bending moment about the minor axis. Given: feu = 40 N/mm2 and [y =
460 N/mm2.
SOLUTION
(a) Nand Mx.
N _ (2300)(10 3 ) _ 2
bh - (300)(500) - 15.33 N/mm
Mx (300)(10 6 ) 2
bh2 = (300)(5002) = 4.00 N/mm
From the column design chart in Fig. 7.3-1,
Asc = 2.3%