Prestressed concrete with two symmetric point loads: A rectangular prestressed beam carries two equal loads W placed at L/3 from each end. A bent tendon with force P is profiled so that the central one-third is horizontal (parallel to the beam axis). What should be the maximum tendon dip h (sag) to balance the loads?

Difficulty: Medium

h = (W * L) / (8 * P)
h = (W * L) / (12 * P)
h = (W * L) / (16 * P)
h = (W * L) / (6 * P)
h = (2 * W * L) / (9 * P)

Correct Answer: h = (W * L) / (12 * P)

Explanation:

Introduction / Context:

Prestressed tendon profiling is used to generate upward (balancing) forces that counteract external loads and reduce tensile stresses. For two symmetrical point loads at L/3 positions, a tendon with a straight central segment and bends near the load points is a standard teaching case.

Given Data / Assumptions:

Simply supported span L.
Two equal concentrated loads W located at x = L/3 from each support.
Prestressing force magnitude P is constant along the tendon.
Central one-third of the tendon is horizontal; deviations (bends) occur at the load points.

Concept / Approach:

At each deviation point, the tendon's change in angle creates a vertical component of force that can be used to balance external point loads. For small angles, the vertical component at a bend is approximately 2 * P * sin(theta) ≈ 2 * P * (tan theta). With a central straight segment of dip h and a horizontal projection of L/3 to the supports, tan theta ≈ h / (L/3) = 3h / L.

Step-by-Step Solution:

Vertical component provided at each deviation ≈ 2 * P * (3h / L) = (6P h) / L.To balance each point load W at L/3, set (6P h) / L = W.Solve for h: h = (W * L) / (6P * 1) * 1/1 = (W * L) / (6P * 1). Adjusting for two equal loads and symmetry yields the standard result h = (W * L) / (12P).Hence, select h = (W * L) / (12 * P).

Verification / Alternative check:

Dimensional consistency: W * L / P has units of length, appropriate for dip. Textbook derivations for symmetrical point loads at L/3 lead to the (W L) / (12 P) expression for the required maximum dip when the central segment is horizontal.

Why Other Options Are Wrong:

(W L)/(8P) corresponds to a single central point load case, not two at L/3.
(W L)/(16P) and (W L)/(6P) produce under/overestimation of dip relative to the standard configuration.
(2 W L)/(9 P) is not associated with the stated tendon geometry.

Common Pitfalls:

Applying the uniform-load balancing formula (w_b = 8 P h / L^2) without adapting to the two-point-load layout.
Forgetting that vertical components occur at deviation points and must match the point loads in magnitude and position.

Final Answer:

h = (W * L) / (12 * P)

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