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)

190 Reinforced concrete beams-the serviceability limit states
k "d h (3)(82)(0.001657) 0 34
crac WI t = [ J = . mm
1 + 2 750 - 303.6
82 - 40
(d) Side face: 250 mm below neutral axis. From Fig. 5.6-1(d), the reader
should verify that
acr = 129 mm
a' = 250 + x (where x is 303.6 mm as in (a))
= 553.6 mm
From Example 5.2-1(b),
E1 (designated E250 then) = 0.00098
Using eqn (5.6-2),
(450)(750 - 303.6)(553.6 - 303.6)
Em = 0.00098 - 3(200)(103)(3769)(690 - 303.6)
= 0.00092
From eqn (5.6-1),
Example 5.6-2
crack width =
(3)(129)(0.00092) = 0 _ 25 mm
[ 129 - 40 J
1 + 2 750 - 303.6
If calculations show that BS 8110's crack width limit is exceeded at a
particular point, what remedial actions may be taken?
SOLUTION
Possible remedial actions include:
(a) The area As of the tension reinforcement may be provided in the
form of a larger number of smaller bars. This has the effect of
reducing the distance acr.
(b) A larger As than that required for the ultimate limit state may be
provided. This has the effect of increasing Ic and x (Figs 5.2-3 and
5.2-4) and hence reducing the strain Em·
(c) Or, an additional bar may be placed directly adjacent to the point
where the crack width is excessive. From eqn (5.6-4) the crack width
is now 3cminEm; thus if (say) Cmin =50 mm and Em= (5/yfB)IEs (see
eqn 5.3-1(b)), the crack width will be reduced to
(3)(50)(5/8)(460) - 0 22
(200)(103) - . mm
Finally it should be pointed out that the methods in Sections 5.4 and 5.6
are intended for flexural cracks in ordinary beams. They do not apply to
shear cracks (Chapter 6) which fortunately are rarely a problem under
service loading. Nor do these methods apply to deep beams where service