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)

114 Reinforced concrete beams-the ultimate limit state
values of b and d, if feu = 40 N/mm 2 and /y = 460 N/mm 2 • What is the
neutral axis depth factor xld for such a section?
SOLUTION
The design formulae based on BS 8110's simplified stress block are derived
on the assumption that, for any doubly reinforced beam, xld = 0.5. Hence
the steel areas As and A~ must obey the definite relation given in the
I.Struct.E. Manual [14], namely: As = Mu/(0.87/yz) + A~. Equation
(4.6-7) expresses this relation in the equivalent form
0.87/y(} = 0.2fcu + 0.87/y(}'
Thus we can either specify that(} = 3% or that e' = 1.5%, but not both.
Suppose we choose to specify (} = 3%; then
(0.87)(460)(0.03) = o.2(40) + (0.87)(460)e'
whence e' = 1.0%. From eqn (4.6-6),
A' M- Mu
s = 0.87fy(d - d')
Dividing by bd 2 and rearranging
(A~) ( d') M Mu
0·87/y bd 1 - d = bd2 - bd2
(0.87)(460)(0.01)(1 - 0.15 say) = b~2 - ~~
Substituting in M = (1600)(10 6 ) and Mu = 0.156fcubd 2 , we obtain
bd 2 = (165.9)(10 6 )
Ford = 2.5b, we have b = 298 mm, d = 746 mm say 300 mm by 750 mm
(xld = 0.5).
Comments
See also Example 4.5-5, in which the above problem is solved using
BS 8110's design chart.
Examples 4.6-3 to 4.6-7 are design examples. Examples 4.6-8 and 4.6-9
below deal with the analysis of given beam sections. The reader should
note that the formulae based on BS 8110's simplified stress block should
not be used unless the following conditions are met:
(a) For a singly reinforced beam section, x/d ~ 0.5.
(b) For a doubly reinforced beam section, xld = 0.5.
Example 4.6-8
Determine the ultimate moment of resistance M and the xl d ratio of the
beam section in Fig. 4.6-5, using:
(a) design chart; and
(b) the BS 8110 simplified stress block.