A cantilever beam of length, 𝑳, and flexural rigidity, 𝑬𝑰, is subjected to an end moment, 𝑴, as shown in the figure. The deflection of the beam at 𝒙 = 𝑳 / 𝟐 is

The correct option is ML²/8EI.

Visual Analysis of the Given Figure:
The provided image displays a cantilever beam of length $L$ fixed at the left end ($x=0$). The coordinate axis $x$ starts from this fixed support and extends along the beam. At the free end on the right ($x=L$), a concentrated clockwise bending moment $M$ is applied, represented by a curved arrow.

Step-by-Step Derivation:

1. Bending Moment Equation:
Since there are no other external transverse loads or forces acting along the length of the beam, the internal bending moment $M\left(x\right)$ at any cross-section is uniform and equal to the applied end moment $M$:
$$M\left(x\right)=M$$

From the Euler-Bernoulli beam theory, the governing differential equation for the beam deflection is:
$$EI\frac{d^{2}y}{d{x}^2}=M$$
where:
• $y$ represents the deflection of the beam at distance $x$ from the fixed end,
• $EI$ represents the flexural rigidity of the beam.

2. Finding the Slope Equation:
Integrating the differential equation once with respect to $x$:
$$EI\frac{dy}{dx}=Mx+C_1$$
where $C_1$ is the first constant of integration.

We apply the boundary condition at the fixed end ($x=0$), where the slope is zero ($\frac{dy}{dx}=0$):
$$EI\left(0\right)=M\left(0\right)+C_1\Rightarrow C_1=0$$
So, the slope equation simplifies to:
$$EI\frac{dy}{dx}=Mx$$

3. Finding the Deflection Equation:
Integrating the slope equation with respect to $x$ yields:
$$EIy=\frac{M{x}^2}{2}+C_2$$
where $C_2$ is the second constant of integration.

We apply the boundary condition at the fixed end ($x=0$), where the deflection is zero ($y=0$):
$$EI\left(0\right)=\frac{M\left(0^2\right)}{2}+C_2\Rightarrow C_2=0$$
Thus, the deflection curve of the beam is described by:
$$y\left(x\right)=\frac{M{x}^2}{2EI}$$

4. Deflection at the Mid-Point of the Beam:
To find the deflection at the mid-span of the cantilever beam, we substitute $x=\frac{L}{2}$ into the deflection equation:
$$y\left(\frac{L}{2}\right)=\frac{M{\left(\frac{L}{2}\right)}^2}{2EI}$$
Simplifying the numerator and denominator:
$$y\left(\frac{L}{2}\right)=\frac{M\cdot \frac{L^2}{4}}{2EI}=\frac{M{L}^2}{8EI}$$

Thus, the deflection of the beam at $x=\frac{L}{2}$ is ML²/8EI.