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)

282 Eccentrically loaded columns and slender columns
r
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. .
SOOmm!!.
·+-·
lC
l__ I
y
Cross section
(a) (b)
Fig. 7.5-1 Slender column (/e/400 = 16.25 > 15)
SOLUTION
For bending about a minor axis, Step 2 of Section 7.5 is relevant. First
calculate Mi and Madd· From eqn (7.4-8),
Mi = OAM1 + 0.6M2
= (0.4)(100) + (0.6)(120)
= 112 kNm
From eqn (7.4-6),
Madd = Nf3aKh
From Table 7.5-1, for lelb' = 6500/400 = 16.25, f3a = 0.13. The optional
reduction factor K is taken as unity (see Comment (c) at the end of the
solution). h is the depth in the plane of bending, i.e. 400 mm. Hence
Madd = (2500)(0.13)(1)(0.4) = 130 kNm
From eqn (7.5-1),
Mt = Mi + Madd
From eqn (7.5-2),
= 112 + 130 = 242 kNm
Mt = M2 = 120 kNm
From eqn (7.5-3),
1 130
Mt = M1 + zMadd = 100 + 2 = 165 kNm
From eqn (7.5-4),