Structural Concrete - Hassoun

19.7 Deflection 765
called camber. The self-weight of the beam and any external gravity loads acting on the beam will
cause a downward deflection. The net deflection will be the algebraic sum of both deflections.
In computing deflections, it is important to consider both the short-term, or immediate, deflection
and the long-term deflection. To ensure that the structure remains serviceable, the maximum
short- and long-term deflections at all critical stages of loading must not exceed the limiting values
specified by the ACI Code (see Section 6.3 in this text).
The deflection of a prestressed concrete member may be calculated by standard deflection
equations or by the conventional methods given in books on structural analysis. For example, the
midspan deflection of a simply supported beam subjected to a uniform gravity load w is equal
to (5wL 4 /384EI). The modulus of elasticity of concrete is E c = 33ω 1.5√ f c ′ = 57, 000 √ f c
′ for
normal-weight concrete.
The moment of inertia of the concrete section I is calculated based on the properties of the
gross section for an uncracked beam. This case is appropriate when the maximum tensile stress
in the concrete extreme fibers does not exceed the modulus of rupture of concrete, f r = 7.5 √ f c
′
(class U beams). When the maximum tensile stress based on the properties of the gross section
exceeds 7.5 √ f c, ′ the effective moment of inertia, I e , based on the cracked and uncracked sections
must be used as explained in Chapter 6 (class T and C beams). Typical midspan deflections for
simply supported beams due to gravity loads and prestressing forces are shown in Table 19.3.
Example 19.6
For the beam of Example 19.4, calculate the camber at transfer and then calculate the final anticipated
immediate deflection at service load.
Solution
1. Deflection at transfer:
a. Calculate the downward deflection due to dead load at transfer, self-weight in this case. For a
simply supported beam subjected to a uniform load,
Δ D (dead load) = 5wL 4
384EI
From Example 19.4, w D = 388 lb/ft, L = 48 ft, E ci = 3600 ksi, and I = 66, 862 in. 4
5(0.388∕12)(48 × 12)4
Δ D = = 0.192 in. (downward)
384(3600)(66, 862)
b. Calculate the camber due to the prestressing force. For a simply supported beam harped at
one-third points with the eccentricity e 1 = 14.5 in. at the middle third and e 2 = 0 at the ends,
Δ p = 23(F i e 1 )L2
216E ci I
(Table 19.3)
23(365.9 × 14.5)(48 × 12)2
= =−0.779 in. (upward)
216(3600)(66, 862)
c. Final camber at transfer is −0.779 + 0.192 =−0.587 in. (upward).
2. Deflection at service load: The total uniform service load is W T = 0.388 + 0.9 + 1.1 = 2.388 K/ft,
and E c = 4000 ksi. The downward deflection due to W T is
Δ w = 5W T L4 5(2.388∕12)(48 × 12)4
= =+1.067 in. (downward)
384E c I 384(4000)(66, 862)