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)

370 Prestressed concrete simple beams
Calculate the short-term deflections. (Assume unit weight of concrete =
23.6 kN/m3.)
SOLUTION
(a) Deflection due to prestressing. The curvature diagram is that shown in
Fig. 9.8-1(a); in this example
r = Eel = 34 X 103 X 4.5 X 108 = . X mm oggmg
1 Pes 350 X 1290 X 100 2 95 10-6 -1 (h · )
From Fig. 9.8-1(a), the midspan deflection is
a= ~[ 1 -1(7YH
= (10 x8 103)2 [1 - 1(!)2]2.95 x 10-6
= 31 mm (upwards)
(b) Deflection due to non-permanent load. At midspan, the curvature is
1 M
r Eel
_ .! X 1.5 (N/mm) X (10 X 103)2
- 8 34 X 103 X 4.5 X 108
= 1.23 X 10-6 mm- 1 (sagging)
The curvature distribution is parabolic, and Fig. 5.5-1(c) applies.
The midspan deflection is then as given in Example 5.5-1(c), namely:
[2 1
a=--
9.6 r
= (10 ;.6103)2 X 1.23 X 10-6
= 12.8 mm (downwards)
(c) Deflection due to permanent load
. 5 X 10 4
Self-wetght = 106 X 23.6 kN/m = 1.18 kN/m
Permanent load = 1.18 + 1.5 = 2.68 kN/m
From the result of (b) above, the midspan deflection is
a = 2 i~ 8 x 12.8 = 22.8 mm (downwards)
(d) Short-term deflections. The short-term deflection when the nonpermanent
load is acting is
a = -31 mm + 12.8 mm + 22.8 mm
= 5 mm (say)