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)

Principles of column interaction diagrams 257
Curve I: Concrete only
Curve ][:Concrete + As2
Curve m:Concrete +As2 +A's1
[
a;
l
0·3
= t~h 0·2
H
0·1
0·10 0·12 0·14
Fig. 7.1-7 Column interaction diagram (A; 11bh = 1%, Adbh = 1 %,feu=
40N/mm 2 ,fy = 460N/mm 2 , d2/h = 0.15, d'/h = 0.15)
the reinforcement. Using a column interaction diagram, determine:
(a) the magnitude of the load, the neutral axis depth x, the depth de of
the stress block and the stress fs2 in the steel at incipient failure;
(b) the additional percentage of steel required at the compression face at
(c)
d'/h = 0.15 to increase the ultimate load to 900 kN;
with this additional steel, the failure load if the eccentricity were
increased to 200 mm; and
(d) the answer for the magnitude of the load in part (a) above if feu had
been 30 N/mm 2 and /y 410 N/mm 2 •
SOLUTION
(a)
Referring to Fig. 7.1-8, curve II is relevant. The line e/h = 0.3 (i.e.
150 mm/500 mm) intersects curve II at 0 1• By measurement, a(Q 1)
= 0.23. Therefore
N = 0.23/eubh = (0.23)(40)(150)(500)(10- 3 ) = 690 kN
Draw vector Q0QJ/C2B to intersect curve I at 0 0 •
~ at 0 0 = 0.68 by interpolation
x = (0.68)(500) = 340 mm