Equivalent Load of a Parabolic Tendon

Mediumtendonequivalent load

A simple-span parabolic tendon has zero eccentricity at both ends and downward eccentricity emid = 0.20 m at midspan. The span is L = 20 m and the horizontal prestress component is constant at H = 1,000 kN. Using the small-slope model, what uniform load does the tendon exert on the concrete?

A2.0 kN/m upward
B4.0 kN/m upward
C4.0 kN/m downward
D80 kN/m upward

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Answer: B. 4.0 kN/m upward

For a parabola e(x) = 4·emid·x(L−x)/L², curvature magnitude is 8·emid/L², so wb = 8·H·emid/L². wb = 8(1,000)(0.20)/20² = 1,600/400 = 4.0 kN/m. A tendon sagging below the centroid pushes the concrete upward. Check: total 4(20) = 80 kN. 2.0 kN/m uses 4 instead of 8. 4.0 kN/m downward gets the direction wrong. 80 kN/m forgets to divide by L² (divides by L only).

w_{b}

= \frac{8 H e _{mi d}}{L ^{2}}

= \frac{8 ( 1000 ) ( 0.20 )}{2 0 ^{2}}

= 4.0

kN/m upward

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Equivalent Load of a Parabolic Tendon

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