Cable profiles
In a post-tensioned member, the tendon need not be straight. Its profile is chosen so that the prestress works against the load where the load is largest. A tendon draped in a parabola between the supports of a uniformly loaded beam, low at mid-span and near the centroid at the ends, mirrors the bending moment diagram, giving most help at mid-span and none where it is not needed.
A curved tendon pushes on the concrete along its length. A parabolic tendon with sag e over span L exerts an upward uniform load equal to 8 P e over L squared. This is the load-balancing idea: if the upward load from the tendon equals the downward permanent load, the beam carries that load with no bending at all, only uniform compression, and it neither sags nor hogs. Only the extra load is resisted by bending.
The tendon must stay within limits. If the prestress resultant falls outside the kern of the section, the opposite fibre goes into tension. The kern points for a rectangle are a sixth of the depth above and below the centroid. Combined with the moment from loads, this leads to a cable zone, an envelope inside which the tendon centroid must lie along the member for no tension, or for limited tension.
Continuous members add secondary, or parasitic, moments: the supports restrain the deformation the prestress tries to cause, adding moments that must be included in design. Profiles are also limited by cover, duct radius, anchorage positions and the need to avoid sharp curvature, which raises friction.
Choose the sag first by load balancing, then check the cable zone and cover limits.
Load balancing (parabolic profile): w_b = 8 P e / L²Rectangle kern distance = h / 6 each side of the centroidKern of general section: k_t = Z_b / A (above), k_b = Z_t / A (below)
Measuring sag from the soffit instead of from the line joining the tendon ends.
Problem. A simply supported beam of span 20 m has a parabolic tendon with effective force 1 200 kN and a sag of 0.4 m (centroid at the ends, 0.4 m below at mid-span). Find the balanced load. If the permanent load is 9.6 kN/m, what moment remains at mid-span?
- w_b = 8 × 1 200 × 0.4 / 20² = 3 840 / 400 = 9.6 kN/m upward.
- The permanent load is 9.6 kN/m downward, so the net load is zero.
- Net mid-span moment from permanent load = 0; the section carries only the uniform compression P/A.
- Any imposed load is then resisted by bending, giving a mid-span moment of q L² / 8 for an imposed uniform load q.
Answer. Balanced load 9.6 kN/m; under that permanent load the mid-span bending is zero
Quick quiz
1What upward load does a parabolic tendon apply?
8 P e / L² per unit length.
2Kern distance of a rectangle?
h / 6 above and below the centroid.
3What are secondary moments?
Moments in continuous members from support restraint of prestress deformations.
- 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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