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)

Short-term and long-term deflections 373
SOLUTION
Step 1 Loss of prestress
From Example 9.4-5,
loss of prestress ols = 424 N/mm 2 , within middle third of span
I . Is - ols 1290 - 424
prestress oss ratio a = Is = 1290
Comments on Step 1
= 0.67, within middle third of span
The 424 N/mm 2 loss of prestress is in a sense fictitious, because the effect of
the applied load has been ignored. The applied-load bending moment is
greatest at midspan and decreases to zeo at the support. For practical
design purposes, it can safely be said that no serious consequences will
result from neglecting the applied-load stresses this way.
Step 2 Long-term curvature due to prestress
From eqn (9.8-4),
r (prestress, long term) = E) a + - 2-cp
1 Pe ( 1 + a )
Within middle third of span:
1
- (prestress, long term)
r
= 451.5 X 10 3 X 100 (o 67 + 1 + 0.67 X 2)
34 X 103 X 4.5 X 108 . 2
= 6.905 x 10- 6 mm- 1 (hogging)
At the supports, es = 0; therefore
.! (prestress) = 0
r
Of course, if es had not been zero at the supports, it would have been
necessary also to calculate the loss ratio a for the support sections in
Step 1.
Step 3 Long-term deflection due to prestress
It is sufficiently accurate to assume that the curvature distribution is
similar to the tendon profile. Using the curvature-area theorem and
referring to Fig. 9.8-1(a).
midspan deflection = ~[ 1 -1(7) 2 ]~
= (10 x8 103 f [1 - Kn2] X 6.905 X 10-6
= 73.5 mm (upwards)