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)

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98 Reinforced concrete beams-the ultimate limit state
where k 1 and k 2 values are given in Fig. 4.4-4.
Thus M 0 /bd 2 values may be computed for various values of the steel
ratio (!, up to the balanced value given by eqn ( 4.5-1 ). For the particular
case of feu = 40 N/mm 2 and /y = 460 N/mm 2 , we saw above that
e(balanced) = 0.0252. As an exercise, the reader should show that the
graph of Mulbd 2 against(! is the first portion of the bottom curve of the
beam design chart (Fig. 4.5-2); that is, up to the kink at Ajbd = 2.52%.
For (! exceeding the balanced value, the beam is over-reinforced, and
eqn (4.3-4) applies:
I.e.
Mu = kdcubx(d - k2x)
(4.5-3)
where k1 and k2 values are given in Fig. 4.4-4, and the neutral axis factor
xld is given by eqn (4.3-3):
~~su(~Y + o.o035(~) - o.oo35 = o (4.5-4)
in which Es is 200 kN/mm 2 , since(! now exceeds the balanced value.
As an exercise, the reader should choose some values of (! between
2.52% and 3.50%, and then use eqns (4.5-3) and (4.5-4) to complete the
verification of the design curve (the bottom one, fore' = 0) in Fig. 4.5-2.
For a doubly reinforced beam, i.e. a beam having both tension
"'e
13
12
11
10
9
E s
........ 7
z
-o
N 6
..0 5
-~ 4
3
2
0
0·5 1·0
-b-tl
·~·
f-- f 1.
'-- 1. d
f--
f--
1--- -
1/.
/
• As• J
/
0·5%
/ "'
1·5 2·0 2·5
feu 40 x/d = 0·3 .............
ty
460 x/d = 0·4 ------
3·0 3·5
4
L~
d'/d 0·15 x/d = 0·5 ---- ~~~1
/'
~
~ v
A
~ j..V -
~
I~ -
·0"
·0" -o
·5" ..0
·0" ..:-;;;
0·5%
1·5% 2·0% 2·5% 3·0% 3·5%
<(
II
-~
Fig. 4.5-2 Beam design chart-ultimate limit state (BS 8110)