The deformation of a beam due to flexural stresses (bending), is calculated as:

$$\frac{\,d^2W}{\,dx^2}=\frac{\,d\theta}{\,dx}=\frac{M_y(x)}{E \cdot I_y}$$

where: \(\begin{cases} W(x) & \text{beam displacement} \\ \theta(x) & \text{beam slope} \\ E \cdot I_y & \text{flexural rigidity}\end{cases}\)

So, we need to get an expression for the bending momentum as a function of the \(x\) coordinate. For a punctual force, the momentum follows a linear evolution, if \(x=0\) is located in the free extreme: \(M(x) = F \cdot x\)

Now we are ready to introduce this into the previous equation and integrate:

$$\frac{\,d^2W}{\,dx^2}=\frac{M_y(x)}{E \cdot I_y}=\frac{F \cdot x}{E \cdot I_y}$$

then:

$$\,d^2W=\frac{F \cdot x}{E \cdot I_y}\,dx^2 \rightarrow \iint \,d^2W = \frac{F}{E \cdot I_y} \cdot \iint x\,dx\,dx$$

Integrating:

$$\int\,dW = \frac{F}{E \cdot I_y} \cdot \int (\frac{x^2}{2}+C_1)\,dx$$

and integrating again:

$$W = \frac{F}{E \cdot I_y} \cdot (\frac{x^3}{6}+C_1\cdot x + C_2)$$

The last part is to get the value of the integration constants by applying boundary conditions:

1. In the fixed extreme the deflection is zero: \(W(x=L)=0\)
2. In the fixed extreme the slope will be zero: \(\frac{\,dW(x=L)}{\,dx}=0\)

From (2), we get that: \(0 = \frac{L^2}{2} + C_1 \rightarrow C_1 = -\frac{L^2}{2}\)

While from (1), we get: \(0 = \frac{L^3}{6}-\frac{L^3}{2} + C_2 \rightarrow C_2 = \frac{L^3}{3}\)

so, introducing them in the original equation, we get that the deformation of the beam follows the function:

$$W = \frac{F}{E \cdot I_y} \cdot (\frac{x^3}{6}-\frac{L^2}{2}\cdot x + \frac{L^3}{3}) = \frac{F}{6 \cdot E \cdot I_y} \cdot (x^3 - 3L^2 \cdot x + 2L^3)$$

Finally, the maximum deformation will happen in the free extreme, so:

$$W(x=0)=\frac{F}{6 \cdot E \cdot I_y} \cdot (2L^3)$$

Obtaining:

$$W_{max}=\frac{F \cdot L^3}{3 \cdot E \cdot I_y}$$