Structural Concrete - Hassoun

744 Chapter 19 Introduction to Prestressed Concrete
Example 19.3
Calculate all losses of a 120-ft-span posttensioned beam that has an I-section with the following
details. Area of concrete section (A c ) = 760 in. 2 ; moment of inertia (I g ) = 1.64 × 10 5 in. 4 ; prestressing
force at transfer (F i ) = 1110 K; area of prestressing steel (A ps ) = 7.5 in. 2 ; f c ′ = 5ksi,E c = 5000 ksi and
E s = 29, 000 ksi; profile of tendon is parabolic; eccentricity at midspan = 20 in; and eccentricity at
ends = 0.
Solution
1. Loss due to elastic shortening:
Steel stress at transfer = F i
= 1110 = 148 ksi
A ps 7.5
Stress in concrete at end section = 1110
760
= 1.46 ksi
Stress in concrete at midspan = F i
+ F ie 2
− M D e
A c I I
Weight of beam = 760 × 150 = 790 lb∕ft
144
M D = 0.79 (120)2
8
= 1422 K ⋅ ft
Stress at midspan = 1110
760 + 1110(20)2 (1422 × 12)(20)
−
164, 000 164, 000
= 1.46 + 2.71 − 2.08 = 2.09 ksi
1.46 + 2.09
Average stress = = 1.78 ksi
2
Average strain = 1.78 = 1.78 = 0.356 × 10−3
E c 5000
Elastic loss is Δf s = ε c E s = 0.356 × 103 × 29, 000 = 10.3 ksi, assuming that the tendons are tensioned
two at a time. The first pair will have the greatest loss, whereas the last pair will have 0
loss. Therefore, average Δf s = 10.3/2 = 5.15 ksi.
Percent loss = 5.15
148 = 3.5%
2. Loss due to shrinkage of concrete:
Δf s (shrinkage) =0.0002E s = 0.0002 × 29, 000 = 5.8ksi
Percent loss = 5.8
148 = 3.9%
3. Loss due to creep of concrete: Assume C c = 1.5.
Elastic strain =
F i 1110
=
= 0.92 × 10−3
A c E c 760 × 5000
Δf s (creep) =C c (ε cr E s )
= 1.5(0.292 × 10 −3 × 29, 000) =12.7ksi
Percent loss = 12.7
148 = 8.6%