Camber from a Straight Tendon
Mediumcamberdeflection
A simply supported prismatic beam spans 12 m with constant EI = 120,000 kN·m². A straight tendon carries constant P = 600 kN at e = 0.10 m below the centroid along the full span. Ignore self-weight, losses and shear deformation. The prestress produces a constant hogging moment Pe. What is the elastic midspan camber?
- A9.00 mm upward
- B7.50 mm upward
- C4.50 mm upward
- D18.0 mm upward
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Answer: A. 9.00 mm upward
Constant moment M0 = Pe = 600(0.10) = 60 kN·m. Constant curvature M0/EI gives a parabolic deflected shape with midspan displacement M0L²/(8EI) = 60(12²)/(8·120,000) = 8,640/960,000 = 0.009 m = 9.00 mm, upward (hogging). 7.50 mm uses the coefficient 5/48, carried over from the UDL-style formula, which is not valid for constant moment. 4.50 mm uses 1/16 and 18.0 mm uses 1/4 in place of 1/8.
m mm
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