10.6.2. Application: Beam Deflection with One Fixed End

A cantilever beam of length L, fixed at x = 0, subjected to a load q(x).

A cantilever beam is a type of beam that is fixed at one end and free at the other. When subjected to a load q(x), the beam experiences deflection due to this load. The length of the beam is represented as L, and the fixed end is at the point x = 0. Understanding how loads affect the deflection of the beam is crucial in structural engineering to ensure safety and stability.

You can think of a diving board as a cantilever beam. The end where the diver stands (the free end) can bend down when someone jumps on it, while the other end (the fixed end) is secured to the platform.

The Euler-Bernoulli beam equation is: $\frac{d^4y}{dx^4} = \frac{q(x)}{EI}$

The Euler-Bernoulli beam equation describes the relationship between the load applied to the beam and the resulting deflection. Here, $y$ is the deflection of the beam, $q(x)$ represents the load per unit length, and $E$ is the modulus of elasticity of the material, while $I$ is the moment of inertia of the beam's cross-section. The fourth derivative of deflection with respect to x indicates how the curvature of the beam changes along its length, which is critical for understanding deflection behavior.

Imagine the deflection as a bow and arrow. The stiffer the bowstring (like having a high EI), the less it bends under the force you apply. The load q(x) can be thought of as the strength of your pull on the string.

With boundary conditions (fixed end at x=0): y(0)=0, y′(0)=0.

Boundary conditions are constraints that describe the behavior of the beam at its ends. The condition $y(0) = 0$ indicates that there is no deflection at the fixed end, meaning it must stay at that point. The second condition $y'(0) = 0$ means that the slope (or angle) of the beam is also zero at the fixed end, implying that it does not tilt or rotate at this point. These conditions are crucial to accurately solving for the deflection across the length of the beam.

Think of a streetlight pole anchored into the ground. The point where it’s anchored (boundary condition) cannot move at all (y(0)=0) and it stands straight up (y′(0)=0) without tilting.

We apply the Fourier Cosine Transform to both sides: Let Y(s)=F {y(x)}, then: $F \left\{ \frac{d^4y}{dx^4} \right\} = s^4 Y(s)$.

Applying the Fourier Cosine Transform converts the differential equation in the spatial domain into an algebraic equation in the frequency domain. By denoting $Y(s)$ as the cosine transform of the deflection function $y(x)$, we reframe the problem to be easier to solve. The transform of the fourth derivative leads to a multiplication by $s^4$, changing the equation's structure and facilitating the solution.

Consider getting a math problem simplified into a form that's easier to solve, like turning a complex subtraction problem into addition by breaking it down step by step.

Thus, $s^4 Y(s) = \frac{F \{ q(x) \}}{EI}$ ⇒ $Y(s) = \frac{F \{ q(x) \}}{EI s^4}$. Taking the inverse transform yields the deflection y(x).

After applying the Fourier Cosine Transform, we found an expression for $Y(s)$ that relates the transformed deflection to the transformed load. The inverse transformation restores our original function, allowing us to find the deflection $y(x)$ of the beam in the spatial domain. This inverse process is essential in engineering for determining how structures respond to applied loads.

It's like decoding a message that was encoded in a secret language. Taking the inverse transform is like translating it back to plain language so you can understand how the beam is responding to loads.

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