Losses of prestress
The force in a tendon is never the force applied by the jack. Some is lost immediately and some over the years, and design must use the force that actually remains. Typical total losses are of the order of 15 to 25 percent of the initial prestress, which is why tendons are stressed to a high fraction of their strength in the first place.
Immediate losses occur at or soon after transfer. Elastic shortening: when the concrete is compressed, it shortens, and bonded steel shortens with it, losing stress equal to the modular ratio times the concrete stress at the tendon level. Friction: in post-tensioning, the tendon rubs against the duct as it is pulled, through curvature (characterised by a coefficient of friction) and through unintended wobble of the duct, so the force falls along the tendon from the jack end. Anchorage slip: the wedges draw in by a few millimetres when the jack releases, a loss concentrated near the anchor.
Time-dependent losses build up over months and years. Creep of concrete under sustained compression shortens the member further. Shrinkage of concrete as it dries does the same. Relaxation of the steel is a loss of stress at constant length, much smaller for low-relaxation strand. These losses interact, and codes give methods or allowances for each.
Accurate losses matter both ways. Underestimate them and the member cracks in service; overestimate them and the member may be over-compressed at transfer or camber upwards more than expected. IS 1343 and IRC 112 give the methods used in India.
Stress long tendons from both ends to halve the friction loss at mid-length.
Elastic shortening (pre-tensioned): Δf_p = m f_c, m = E_p / E_cFriction: P_x = P₀ e^−(μ α + k x)Shrinkage loss Δf = ε_sh E_p; creep loss Δf = φ m f_c (φ = creep coefficient)
Applying the elastic-shortening formula for pre-tensioned members to the last tendon stressed in a post-tensioned member.
Problem. A post-tensioned tendon is jacked to P₀ = 1 500 kN. At a section 20 m from the jack the cumulative angle change is 0.15 rad. Take μ = 0.20 and wobble k = 0.002 per m. Find the force there. Separately, a pre-tensioned member has concrete stress 12 MPa at steel level with m = 6; find the elastic shortening loss.
- Exponent: μ α + k x = 0.20 × 0.15 + 0.002 × 20 = 0.03 + 0.04 = 0.07.
- P_x = 1 500 × e^−0.07 = 1 500 × 0.9324 = 1 398.6 kN, a friction loss of 101.4 kN or 6.8 percent.
- Elastic shortening: Δf_p = 6 × 12 = 72 MPa.
- For strand initially at 1 395 MPa, that is about 5.2 percent.
Answer. About 1 399 kN at 20 m (6.8 percent friction loss); elastic shortening loss 72 MPa
Quick quiz
1Which losses happen immediately?
Elastic shortening, friction and anchorage slip.
2Which steel property causes a loss at constant length?
Relaxation.
3Why is friction loss larger far from the jack?
Friction accumulates with curvature and length along the duct.
- IS 1343 (India): Design and construction of prestressed concrete.
- IRC:112 (India): Code of practice for concrete road bridges.
- EN 1992 (Eurocode 2) (Europe): Design of concrete structures.
Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- N. Krishna Raju, Prestressed Concrete, McGraw Hill Education (India), latest edition.
- T. Y. Lin, N. H. Burns, Design of Prestressed Concrete Structures, Wiley, 3rd edition.
- E. G. Nawy, Prestressed Concrete: A Fundamental Approach, Pearson, latest edition.
Free and official sources
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