If a rectangular pre-stressed beam of an effective span of 5 meters and carrying a total load 3840 kg/m, is designed by the load balancing method, the central dip of the parabolic tendon should be A. 5 cm B. 10 cm C. 15 cm D. 20 cm

5 cm
10 cm
15 cm
20 cm

The correct answer is $\boxed{\text{B) 10 cm}}$.

The load balancing method is a method of designing prestressed concrete beams that uses the principle of equilibrium to determine the required prestress force. The method assumes that the prestress force is applied in the form of a parabolic tendon, and that the beam is subjected to a uniformly distributed load.

The central dip of the parabolic tendon is given by the following equation:

$$d = \frac{4PL}{EI}$$

where:

  • $d$ is the central dip of the parabolic tendon,
  • $P$ is the prestress force,
  • $L$ is the effective span of the beam,
  • $E$ is the modulus of elasticity of concrete, and
  • $I$ is the moment of inertia of the beam.

In this case, we are given that $P = 3840 \text{ kg}/\text{m}$, $L = 5 \text{ m}$, $E = 25 \times 10^9 \text{ Pa}$, and $I = 1000 \text{ cm}^4$. Substituting these values into the equation, we get:

$$d = \frac{4 \times 3840 \text{ kg}/\text{m} \times 5 \text{ m}}{25 \times 10^9 \text{ Pa} \times 1000 \text{ cm}^4} = 10 \text{ cm}$$

Therefore, the central dip of the parabolic tendon should be $\boxed{\text{10 cm}}$.

The other options are incorrect because they do not represent the correct value of the central dip of the parabolic tendon.