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)

280 Eccentrically loaded columns and slender columns
designed for Nand Mt. using the same procedure as in Step 2. Of course,
M 1 and M 2 are now the initial end moments about the major axis.
Case 2: If either Condition (1) or Condition (2) is not satisfied, then the
column is designed as biaxially bent, with zero initial moment about the
minor axis-see procedure in Step 4 below.
Step 4 Biaxial bending
(a)
Calculate M 1y· The total moment M 1y about the minor axis is calculated
as the greatest value given by eqns (7.5-1) to (7.5-4), exactly
as in Step 2.
(b) Calculate M 1x. The total moment M 1x about the major axis is
calculated as the greatest value given by eqns (7.5-1) to (7.5-4) as in
Step 2, except that:
(1) M 1 and M 2 are now the initial end moments about the major
axis.
(2) Madd is calculated, as usual, from eqn (7.4-6): Madd = NPaKh.
However, in obtaining Pa either from eqn (7.4-5) or Table
7.5-1, b' is in this particular case to be taken as h, i.e. the
dimension in the plane of bending.
(c)
Design for N, M 1x, M 1y· The column is then designed for the ultimate
axial load N plus the two total moments M 1x and M 1y using eqns
(7.3-3) and (7.3-4) in Section 7.3(a).
Step 5 The reduction factor K
The additional moment Madd in eqns (7.5-1) and (7.5-3) is given by eqn
(7.4-6):
Madd = NPaKh (K :5 1)
where K is an optional reduction factor which can be read off Fig. 7.3-1
(and similar design charts in BS 8110: Part 3) and is defined by
(7.5-5)
where N = the ultimate axial load;
Nbal = the axial load corresponding to the balanced condition of
maximum compressive strain in the concrete of 0.0035
occurring simultaneously with a maximum tensile strain in
the reinforcement equal to the design yield strain, i.e.
0.87fy1E 8 • For /y = 460 N/mm 2 , the design yield strain is
0.002.
Nuz = the capacity of the column section under 'pure' axial load as
given by eqn (7.5-6):
(7.5-6)
Comments on Step 5
(a) The load Nbal is that corresponding to the kinks in the column interaction
diagrams (e.g. Fig. 7.3-1). Of course, the kinks in Fig. 7.3-1
correspond to the point H in Fig. 7.1-6.
(b) Figure 7.3-1 shows, as one would expect from eqn (7.5-5), that