Structural Concrete - Hassoun

19.3 Loss of Prestress 745
4. Loss due to relaxation of steel: For low-relaxation strands, the loss is 2.5%.
Δf s = 0.025 × 148 = 3.7ksi
5. Slip in anchorage: For tensioning from one end only, assume a slippage of 0.15 in. The length of
the cable is 120 × 12 = 1440 in.
Δf s = ΔL
L × E s = 0.15 × 29, 000 = 3 ksi (Eq. 19.12)
1440
To allow for anchorage slip, set the tensioned force to 148 + 3 = 151 ksi on the pressure gauge
to leave a net stress of 148 ksi in the tendons.
6. Loss due to friction: The equation of parabolic profile is
e x = 4e
L (Lx − 2 x2 )
where e x is the eccentricity at a distance x measured from the support and e is eccentricity at
midspan:
d(e x )
= 4e (L − 2x)
dx L2 is the slope of the tendon at any point. At the support, x = 0 and the slope
d(e x )
= 4e
dx L = 4 × 20
120 × 12 = 0.056
The slope at midspan is 0; therefore, α px = 0.056. Using flexible metallic sheath, μ p = 0.5 and
K = 0.001. At midspan, x = 60 ft. Check if (μ p α px + Kl x ) ≤ 0.30:
μ p α px + Kl x = 0.5 × 0.056 + 0.001 × 60 = 0.0088 < 0.3
P px = P pj (1 + Kl px + μ p α px )
7. Total losses:
= P x (1 + 0.088) =1.088P x
= 1.088 × 148 = 161 K (force at jacking end)
Δf s = 161 − 148 = 13 ksi
Percent loss = 13 = 8.8% (Eq. 19.11)
148
Elastic shortening loss 5.2 ksi 3.5%
Shrinkage loss 5.8 ksi 3.9%
Creep of concrete loss 12.7 ksi 8.6%
Relaxation of steel loss 3.7 ksi 2.5%
Friction losses 13.0 ksi 8.8%
Total losses 40.4 ksi 27.3%
Effective prestress = 148 − 35.2 = 112.8ksi
Effective prestressing force(F) =(1 − 0.238)F i = 0.762F i
F = 0.762 × 1110 = 846 K
For F = η F i , η = 0.762.