Q1: A prestressed rectangular beam which carries two concentrated loads W at L/3 from either end, is provided with a bent tendon with tension P such that central one-third portion of the tendon remains parallel to the longitudinal axis, the maximum dip h is

ANS:C -

In a prestressed rectangular beam carrying two concentrated loads W at 33L​ from either end and provided with a bent tendon with tension P such that the central one-third portion of the tendon remains parallel to the longitudinal axis, the maximum dip ℎh of the tendon can be calculated using the principles of prestressed concrete. The maximum dip ℎh occurs at the midspan of the beam, where the bending moment due to the applied loads and the prestressing force is maximum. The tendon profile is typically parabolic under uniform load, and the maximum dip occurs at the midspan where the shear force is zero. The formula to calculate the maximum dip ℎh of the tendon in a prestressed rectangular beam is given by: h=8EIPL2​ Where:

• P = Tension force in the tendon
• L = Span of the beam
• E = Modulus of elasticity of the material
• I = Moment of inertia of the beam's cross-section about its neutral axis

For a rectangular beam, the moment of inertia I is given by: 312I=12bd3​ Where:

• b = Width of the beam
• d = Depth of the beam

Given that the loads are applied at L/3​ from either end, the span of the beam L is 2L​/3. Now, to find the maximum dip ℎh, we need to determine the values of P, E, I, and L based on the specific parameters of the beam. Once these values are known, we can substitute them into the formula above to calculate the maximum dip ℎh. This will give us the amount by which the tendon dips below the neutral axis at the midspan of the beam.