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)

StepS
Calculation of short-term and long-term deflections (BS 8110) 185
Use the curvature-area theorems. From Example 5.5-1(c) and (b), the
required long-term deflection is
Example 5.5-4
a = 9 ~6 / 2 (6.87 X 10- 6 ) [Step 3]
+ ~F(l.04 X 10- 6 ) [Step 4]
= 21.2 mm
The structural member in Example 5.5-3 is a simply supported beam. If it
had been the interior span of a continuous beam, explain how the solution
to Example 5.5-3 would have to be modified.
SOLUTION
Calculate the total long-term curvatures 1/r" 1/r2 , and 1/1r 3 at the left
support, the right support and the midspan, respectively. Then use the
method of superposition illustrated in Example 5.5-2 and eqn (5.5-2).
Example 5.5-5
BS 8110: Part 2: Clause 3.6 states that, in deflection calculations, the
shrinkage curvature may be taken as
1 fesaeSs
res=--~where
the symbols are as defined earlier under eqn (5.5-6). Derive the
equation.
SOLUTION
Consider again the typical beam section in Fig. 5.5-4 and the arguments
that lead to eqn (5.5-5):
1 ez - ft
shrinkage curvature - = d
res
(5.5-7)
where t: 2 is the concrete strain at the top level and t: 1 is that at the level of
the tension steel, as shown in Fig. 5.5-5. Let
Is = compressive stress in the steel due to the concrete shrinkage;
fc1 = concrete tensile stress at the level of the tension reinforcement,
due to the concrete shrinkage; and
fez = concrete tensile stress at the top fibres of the beam section,
due to the concrete shrinkage.
From the geometry of Fig. 5.5-5,
(5.5-8)
(5.5-9)