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)

Step(b)
Find lever arm z. From Fig. 4.6-1(b),
Hence
concrete compression = 0.405fcuhx
lever arm z = d - 0.45x
Hence
z = 700 - 0.45x = 649 mm
Step(c)
Find A •. From Fig. 4.6-1(b),
Designing from first principles 109
M = 0.405fcuhx(d -
0.45x)
(300)(10 6 ) = (0.405)(40)(250)(x)(700 - 0.45x)
x 2 - 1560x + 165000 = 0
·M = 0.87/yAsz
x =114m
(or 1441 mm, which is inadmissible)
(where z = 649 mm from Step (b) above)
(300)(10 6 ) = (0.87)(460)A.(649)
As= 1155 mm 2
Provide four 20 mm bars (As = 1257 mm 2 )
Comment
Example 4.5-3 solves the same problem using BS 8110's design chart. See
also Example 4.6-5 which follows the I.Struct.E. Manual's procedure [14].
Example 4.6-4
Repeat Example 4.6-3 if M is 900 kNm. What is that xld ratio of the beam
section so designed?
SOLUTION
Step(a)
Check concrete capacity Mu·
Mu = 764.4 kNm (from Step (a) of Example 4.6-3)
Since M > Mu, compression steel is required.
Sup(b)
Find compression steel area A~. From eqn (4.6-6),
0.87/yA~(d - d') = M - Mu
(0.87)(460)A~(700 - 60 say) = (900 - 764.4)(10 6 )
A~= 530 mm 2