6.8. Worked Examples with Engineering Applications

Example 1: Beam Under Uniform Load Given: $\frac{d^4y}{dx^4} = \frac{q_0}{EI}$ Where:

• EI = flexural rigidity of the beam
• q_0 = uniform load per unit length Solution: Rewriting: $\frac{d^4y}{dx^4} = 0$ Integrating four times: $\frac{d^3y}{dx^3} = \frac{q_0}{EI} x + C_1$ $\frac{d^2y}{dx^2} = \frac{q_0}{2EI} x^2 + C_1 x + C_2$ $\frac{dy}{dx} = \frac{q_0}{6EI} x^3 + C_1 x^2 + C_2 x + C_3$ $y(x) = \frac{q_0}{24EI} x^4 + \frac{C_1}{6} x^3 + \frac{C_2}{2} x^2 + C_3 x + C_4$ Boundary conditions (e.g., at fixed ends) are used to determine C_1 to C_4.

In this example, we have a beam that is under a uniform load. The governing differential equation is derived based on the physical principles of how beams respond to loads. We start with the fourth derivative of the deflection of the beam, which is directly proportional to the applied load and inversely proportional to the flexural rigidity (EI) of the beam. By rewriting the equation, we set it equal to zero and integrate four times to find the deflection equation. Each integration adds a constant of integration (C_1, C_2, C_3, C_4) because the solution to a differential equation includes these arbitrary constants. The specific values of these constants are determined by applying boundary conditions that correspond to the physical situation, such as points of fixed support on the beam.

Imagine a hammock stretched between two trees. The way the hammock bends downwards under the weight of a person is similar to how a beam bends under a uniform load. Just as the hammock sags more in the middle with additional weight, our calculations help us understand how much a beam will flex and bend based on how much load it carries.

Example 2: Vibration of a Damped System with Forcing Given: $m\frac{d^2y}{dt^2} + c\frac{dy}{dt} + ky = F_0 \cos(\omega t)$ This is a non-homogeneous second-order ODE representing a damped forced vibration.

• Solve the homogeneous equation: $mr^2 + cr + k = 0$
• Based on discriminant, determine y_h
• Guess PI using undetermined coefficients (if $\omega \neq \omega_0$) or resonance form (if $\omega = \omega_0$). This equation is foundational in seismic design, machine foundation modeling, and vibration control in structures.

In the second example, we look at a system where a mass is attached to a spring and is subjected to a damping force, simulating conditions such as a building swaying in an earthquake. The equation describes how this mass will move over time when subjected to an external oscillating force (like vibrations from a machine or an earthquake). The equation includes terms that represent the mass, damping, and stiffness of the system. We first solve the homogeneous part of the equation to understand the natural behavior of the system. This involves determining the roots of the corresponding characteristic equation. Depending on the relationship between the forcing frequency ($\omega$) and the system's natural frequency ($\omega_0$), we select different forms for the particular integral (PI) to capture the system's response accurately.

Think of a swing being pushed in rhythm. The swing has a natural frequency at which it would move if it were not being pushed. If someone pushes the swing at the right moments (matching the swing's natural frequency), the back-and-forth motion becomes greater — this is analogous to resonance. But if the pushes are out of sync, the swing moves normally, illustrating the concept of forced vibrations. Understanding how to model these dynamics helps engineers design safer buildings that can withstand vibrations.

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