The correct answer is $\frac{{{\text{W}}{{\text{L}}^3}}}{{3{\text{E}}I}}$.
The maximum deflection due to a load $W$ at the free end of a cantilever of length $L$ and having flexural rigidity $EI$ is given by the following equation:
$$\delta = \frac{{{\text{W}}{{\text{L}}^3}}}{{3{\text{E}}I}}$$
where:
- $\delta$ is the maximum deflection,
- $W$ is the load,
- $L$ is the length of the cantilever,
- $E$ is the Young’s modulus, and
- $I$ is the moment of inertia.
The equation can be derived using the following steps:
- Assume that the cantilever is a uniform beam with a constant cross-section.
- Apply the principle of superposition to find the deflection of the beam due to the load $W$.
- The deflection of the beam due to the load $W$ is given by the following equation:
$$\delta = \frac{{M}}{{EI}}$$
where:
- $M$ is the bending moment at the free end of the beam,
- $E$ is the Young’s modulus, and
$I$ is the moment of inertia.
The bending moment at the free end of the beam is given by the following equation:
$$M = WL$$
where:
- $W$ is the load, and
$L$ is the length of the cantilever.
Substituting equation (2) into equation (3) gives the following equation:
$$\delta = \frac{{WL}}{{EI}}$$
- The maximum deflection occurs at the free end of the beam, where the bending moment is maximum.
- The maximum deflection is given by the following equation:
$$\delta = \frac{{{\text{W}}{{\text{L}}^3}}}{{3{\text{E}}I}}$$