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)

116 Reinforced concrete beams-the ultimate limit state
Fig. 4.6-6
(25mm bars)
(a) design chart; and
(b) the BS 8110 simplified stress block
Here, feu = 40 N/mm 2 ; /y = 460 N/mm 2 •
SOLUTION
(a) Using design chart.
As 1 (two 25 mm) = 982 mm 2
As 2 (three 25 mm) = 1473 mm 2
As = As, + As2 = 2455 mm 2
A~ (two 25 mm) = 982 mm 2
_ As _ 2455 _ 1 490 ,{
(! - bd - 250 X -!(680 + 640) - . 0
n' - A~ - 982 - 0 59<X
.: - bd - 250 X -!(680 + 640) - . 0
From Fig. 4.5-2,
M = 5.3bd 2 = (5.3)(250)( 680 ~ 640 r(10- 6 ) = 577 kNm
d < 0.3
X
Comment
The d value used in the above calculations is taken as the mean of 640 mm
for As 1 and 680 mm for As2 and is hence only an approximation. The design
chart gives curves for xld = 0.3, 0.4 and 0.5 only; hence the precise xld
value cannot be determined.
(b) Using BS 8110 simplified stress block. With reference to the forces in
Fig. 4.6-7, let us assume at this stage that both the compression and
tension steels reach the full design strength of 0.87/y· Therefore, the
condition of equilibrium is (see Comment (a) below)
0.87/yAs = 0.405fcubx + 0.81/yA~