The correct answer is $\boxed{\frac{{{\text{WL}}}}{{4{\text{P}}}}}$.
The maximum dip $h$ of a pre-stressed rectangular beam which carries two concentrated loads $W$ at $\frac{{\text{L}}}{3}$ 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, can be calculated using the following formula:
$$h = \frac{{3{\text{WL}}}}{{4{\text{P}}}}$$
where:
- $h$ is the maximum dip of the beam,
- $W$ is the concentrated load,
- $L$ is the length of the beam, and
- $P$ is the tension in the tendon.
The formula can be derived by considering the equilibrium of the beam. The beam is subjected to two concentrated loads $W$ at $\frac{{\text{L}}}{3}$ from either end, and a tensile force $P$ in the tendon. The beam is also supported by two reactions at the ends. The reactions at the ends must be equal and opposite in order to maintain equilibrium.
The bending moment at any point along the beam can be calculated using the following formula:
$$M = \frac{{W{\text{L}}}}{{6}} – \frac{{P{\text{x}}}}{{2}}$$
where:
- $M$ is the bending moment,
- $W$ is the concentrated load,
- $L$ is the length of the beam,
- $x$ is the distance from the left end of the beam, and
- $P$ is the tension in the tendon.
The maximum bending moment occurs at the center of the beam, where $x = \frac{{L}}{2}$. The maximum bending moment is then:
$$M_{\max} = \frac{{W{\text{L}}}}{{12}} – \frac{{P{\text{L}}}}{{4}}$$
The maximum deflection of the beam can be calculated using the following formula:
$$\delta = \frac{{M_{\max}{\text{L}}^{3}}}{{3EI}}$$
where:
- $\delta$ is the maximum deflection,
- $M_{\max}$ is the maximum bending moment,
- $L$ is the length of the beam,
- $E$ is the Young’s modulus of the beam, and
- $I$ is the moment of inertia of the beam.
The moment of inertia of a rectangular beam can be calculated using the following formula:
$$I = \frac{{\text{bh}^{3}}}{{12}}$$
where:
- $I$ is the moment of inertia,
- $b$ is the width of the beam,
- $h$ is the height of the beam.
Substituting the formula for $M_{\max}$ into the formula for $\delta$, we get:
$$\delta = \frac{{\frac{{W{\text{L}}}}{{12}} – \frac{{P{\text{L}}}}{{4}}{\text{L}}^{3}}}{{3EI}}$$
$$\delta = \frac{{WL^{4}}}{{144EI}} – \frac{{PL^{3}}}{{36EI}}$$
$$\delta = \frac{{WL^{4}}}{{144EI}} – \frac{{2PL^{3}}}{{36EI}}$$
$$\delta = \frac{{WL^{4}}}{{144EI}} – \frac{{WL^{3}}}{{18EI}}$$
$$\delta = \frac{{WL^{3}}}{{18EI}}$$
The maximum dip $h$ is equal to the maximum deflection $\delta$ divided by two. Therefore, the maximum dip $h$ is:
$$h = \frac{{\delta}}{2} = \frac{{\frac{{WL^{3}}}{{18EI}}}}{2} = \frac{{WL^{3}}}{{36EI}}$$
$$h = \frac{{3{\text{WL}}}}{{4{\text{P}}}}$$