Question

A simply supported uniform beam of length \(L\), bending stiffness \(E I\) and mass per unit length \(m\) is subjected to the uniform distributed load \(p(x, t)=q_{0} \sin \Omega t\). Determine the bending moment at the center of the span if the behavior of the beam is predicted using Rayleigh beam theory with \(r_{G} / L=0.1\).

Short answer

In a simply supported uniform beam with length \(L\), bending stiffness \(EI\), mass per unit length \(m\), and sinusoidally varying load \(p(x, t) = q_0 \sin \Omega t\), the bending moment at the center of the span using the Rayleigh beam theory can be found as: \(M(L/2, t) = -0.1 EIq_0 \cos \left (\frac{\omega L}{2} \right) \sin \Omega t\)

Step-by-step solution

Write down the Rayleigh Beam Theory Expression

Rayleigh beam theory is represented by the following equation for the bending moment, \(M(x, t)\), along the length of the beam: \(M(x, t) = - EI \frac{d^2 w(x, t)}{d x^2}\) where \(w(x, t)\) is the deflection of the beam at position \(x\) and time \(t\), and \(d^2 w(x, t)/d x^2\) is the second derivative of the deflection with respect to the position.

Determine the Boundary Conditions

Since the beam is simply supported on its ends, the boundary conditions are as follows: 1. \(w(0, t) = w(L, t) = 0\): the deflection at both ends of the beam is zero. 2. \(M(0, t) = M(L, t) = 0\): the bending moment at both ends of the beam is zero.

Formulate the Differential Equation

The distributed load on the beam varies sinusoidally with time, so we first write that down: \(p(x, t) = q_0 \sin \Omega t\) Taking this load into account, the equation for the deflection of the beam becomes: \(EI \frac{d^4 w(x, t)}{d x^4} = m \frac{d^2 w(x, t)}{d t^2} + p(x, t)\)

Determine the Deflection

We now need to find the general solution for the deflection \(w(x, t)\), which satisfies both the boundary conditions and the differential equation. Using standard techniques for solving partial differential equations, we obtain the following expression for the deflection of the beam: \(w(x, t) = A \cos (\omega x) \sin (\Omega t)\) where \(\omega\) is a constant related to the frequency of oscillation and \(A\) is the amplitude of the deflection.

Obtain the Second Derivative of the Deflection

Now we need to find the second derivative of the deflection with respect to the position, \(d^2 w(x, t)/d x^2\), in order to determine the bending moment at the center of the span: \(\frac{d^2 w(x, t)}{d x^2} = -A \omega^2 \cos (\omega x) \sin (\Omega t)\)

Calculate the Bending Moment at the Center of the Span

We will now substitute the expression for the second derivative of the deflection into the equation for the bending moment, and then evaluate the bending moment at the center of the span, \(x = L/2\): \(M(L/2, t) = EI \cdot \left[-A \omega^2 \cos \left(\omega \frac{L}{2}\right) \sin (\Omega t)\right]\) Since we want the bending moment at the center, we plug in \(x = L/2\): \(M(L/2, t) = - EI A \omega^2 \cos \left (\frac{\omega L}{2} \right) \sin \Omega t\) Still, we don't know the exact value for \(A\omega^2\). To find that, we can use the given \(r_G/L=0.1\) and the fact that \(r_G=\frac{A\omega^2}{q_0}\). Therefore, \(A\omega^2=0.1q_0\). Plugging that back in to the expression for the bending moment at center: \(M(L/2, t) = -0.1 EIq_0 \cos \left (\frac{\omega L}{2} \right) \sin \Omega t\) This is the expression for the bending moment at the center of the span.

Key concepts

Bending Moment Calculation

Understanding the bending moment is crucial to analyzing how beams like the one given in this exercise react under load. The bending moment at a specific point along a beam's length can be thought of as the measure of the internal force causing the beam to bend at that point.
This force depends directly on the material properties and the geometry of the beam. In the context of Rayleigh beam theory, the bending moment, denoted as \(M(x, t)\), is primarily determined by:

• The bending stiffness \(EI\), a product of Young's modulus \(E\) and the moment of inertia \(I\), indicating the beam's resistance to bending.
• The deflection \(w(x, t)\), specifically its second derivative \(\frac{d^2 w(x, t)}{dx^2}\).

The formula \(M(x, t) = - EI \frac{d^2 w(x, t)}{d x^2}\) mathematically represents how these factors interact. By calculating the second derivative of the beam's deflection, engineers can find the bending moment at any point, such as the center in this task.
This allows us to predict the beam's behavior, ensuring safety and functionality under given loads.

Differential Equations in Engineering

Differential equations are a vital tool in engineering for modeling systems that change over time or space. In structural engineering, they help describe how structures respond to forces, such as weight or pressure.
For the beam in this exercise, the Rayleigh beam theory gives rise to a fourth-order differential equation:

• \(EI \frac{d^4 w(x, t)}{d x^4} = m \frac{d^2 w(x, t)}{d t^2} + p(x, t)\)

This equation represents the balance of forces along the beam, emphasizing how the distributed load \(p(x, t)\) and inertia \(m\frac{d^2 w(x, t)}{dt^2}\) influence the deflection \(w(x, t)\).
Often, engineers solve such equations using techniques like separation of variables or employing numerical methods for complex cases. In this example, a particular solution was derived using trigonometric functions to satisfy both time and spatial dependencies. Understanding these equations allows engineers to predict structure performance over time, ensuring designs can withstand various dynamic loads.

Boundary Conditions in Structural Analysis

Boundary conditions define how a structure is supported or restrained at its edges, playing a significant role in analyzing structural behavior. For beams, they are crucial in solving differential equations associated with beam deflection and bending moments.
In the simply supported beam of the exercise, the boundary conditions are:

• Zero deflection at both ends: \(w(0, t) = w(L, t) = 0\).
• Zero bending moment at the supports: \(M(0, t) = M(L, t) = 0\).

These conditions are typical for simply supported beams, allowing rotations but not vertical displacement at the supports.
By applying these boundary conditions, we can accurately solve the deflection equation \(w(x, t)\) to reflect how the beam will react under specific forces. Ignoring boundary conditions can lead to incorrect predictions of a beam's behavior, which could jeopardize the safety and efficacy of a design.