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)

BS 8110 design charts-their construction and use 101
(where k 2 = 0.444 from Fig. 4.4-4)
= (0.03)(400)[1 - (0.444)(0.5)] + (0.01)(400)[(0.444)(0.5)
- (0.15)]
= 9.62 (and xld = 0.5 as found above)
This agrees with the Mufbd 2 value given by the BS 8110 chart (Fig. 4.5-2).
Example 4.5-2
The beam design chart in Fig. 4.5-2 shows the neutral axis depth factors
x/ d for various steel ratios Q and e'. For a given combination of feu, /y, and
d' /d, explain how such x/d ratios may be determined.
SOLUTION
For a general beam section, such as that in Fig. 4.2-1, the trial and error
procedure in Example 4.5-1 above is appropriate.
In the particular case of a singly reinforced beam (e' = 0), a more direct
(but not necessarily quicker) approach may be used.
Step I
Compare the actual steel ratio with the balanced ratio from eqn (4.5-1):
(b I d) kdcu 0.0035 (4 5 9)
Q a ance = 0.87/y 0.0035 + Ey · -
where k 1 is determined from Fig. 4.4-4 and the strain Ey at which the
design strength 0.87/y is first reached is given by Fig. 4.5-1.
If the actual e value does not exceed the balanced value, the steel
stress reaches 0.87/y under the design ultimate moment. The neutral axis
depth factor is therefore given by eqn (4.2-6) withf 5 replaced by 0.87/y:
i.e.
Step2
0.87/y = kt~c;b X
-=--e
X 0.87/y
d kdcu
(4.5-10)
If the actual Q value exceeds the balanced value in Step 1, the beam is
over-reinforced. The xl d ratio should then be calculated from eqn
(4.5-4):
kdcu(!) 2 + 0.0035(!) - 0.0035 = 0
eEs d d
where Es is 200 kN/mm 2 , since e now exceeds the balanced value.
Example 4.5-3
The design ultimate moment for a beam of width 250 mm and effective