Question

The theory of bent beams shows that at any point in the beam the 'bending moment' is given by \(K / \rho\), where \(K\) is a constant (that depends upon the beam material and cross-sectional shape) and \(\rho\) is the radius of curvature at that point. Consider a light beam of length \(L\) whose ends, \(x=0\) and \(x=L\), are supported at the same vertical height and which has a weight \(W\) suspended from its centre. Verify that at any point \(x(0 \leq x \leq L / 2\) for definiteness) the net magnitude of the bending moments, (bending moment \(=\) force \(\times\) perpendicular distance) due to the weight and support reactions, evaluated on either side of \(x\), is \(W x / 2\). If the beam is only slightly bent, so that \((d y / d x)^{2} \ll 1\), where \(y=y(x)\) is the downward displacement of the beam at \(x\), show that the beam profile satisfies the approximate equation $$ \frac{d^{2} y}{d x^{2}}=-\frac{W x}{2 K} $$ By integrating this equation twice and using physically imposed conditions on your solution at \(x=0\) and \(x=L / 2\), show that the downward displacement at the centre of the beam is \(W L^{3} /(48 K)\).

Short answer

The net bending moment at any point x is \( \frac{Wx}{2} \). The beam profile satisfies \( \frac{d^2 y}{dx^2} = -\frac{Wx}{2K} \). The downward displacement at the center of the beam is \( \frac{WL^3}{48K} \).

Step-by-step solution

Bending Moment Due to Weight and Support Reactions

Consider the forces acting on the beam. At the center of the beam, a weight W is suspended. The reactions at the supports (located at x = 0 and x = L) balance this weight, meaning there is a reaction force of W/2 at each support. Now consider a point x where 0 ≤ x ≤ L/2. The net bending moment at this point due to the weight W and the support reactions is given by \[ \text{Bending Moment} = \text{Force} \times \text{Perpendicular Distance} \]On the left side of x, the force is W/2 and the perpendicular distance is x. Thus, \[ \text{Bending Moment} = \frac{W}{2} \times x = \frac{Wx}{2} \]

Beam Profile Differential Equation

Since we know that the bending moment at any point is equal to \[ K \frac{d^2 y}{dx^2} = M(x) \]where M(x) is the bending moment at point x, and from Step 1, M(x) = \frac{Wx}{2}. Therefore,\[ K \frac{d^2 y}{dx^2} = \frac{Wx}{2} \]Rearranging to find the second derivative,\[ \frac{d^2 y}{dx^2} = - \frac{Wx}{2K} \]

Integrate the Differential Equation

Integrate the equation \( \frac{d^2 y}{dx^2} = - \frac{Wx}{2K} \) for the first time with respect to x to find the first derivative:\[ \frac{dy}{dx} = \frac{-Wx^2}{4K} + C_1 \]where \( C_1 \) is the constant of integration.

Apply Boundary Condition for First Integration

At x = 0, since the beam is supported at the same height, the slope \( \frac{dy}{dx} \) should be 0. Thus,\[ \frac{dy}{dx} (x=0) = 0 = \frac{-W(0)^2}{4K} + C_1 \]This implies,\[ C_1 = 0 \]Then the first derivative becomes \[ \frac{dy}{dx} = \frac{-Wx^2}{4K} \]

Integrate the First Derivative

Integrate \( \frac{dy}{dx} = - \frac{Wx^2}{4K} \) to get y(x): \[ y = \frac{-Wx^3}{12K} + C_2 \]where \( C_2 \) is another constant of integration.

Apply Boundary Condition for Second Integration

At x = 0, the displacement y should be 0. Thus,\[ y(x=0) = 0 = \frac{-W(0)^3}{12K} + C_2 \]This implies \[ C_2 = 0 \]Thus the equation for y becomes \[ y = \frac{-Wx^3}{12K} \]

Calculate the Displacement at the Center

Now we calculate the displacement at the center of the beam, which is located at x = L/2:\[ y(x = L/2) = \frac{-W(L/2)^3}{12K} \]Simplifying,\[ y\bigg|_{x = L/2} = \frac{-W(L/2)^3}{12K} = \frac{-WL^3}{96K} \]Therefore, the downward displacement at the center of the beam is \[ y\bigg|_{x = L/2} = -\frac{WL^3}{48K} \]

Final Downward Displacement

Since displacement is downward and considering only the magnitude:\[ y\bigg|_{x = L/2} = \frac{WL^3}{48K} \]

Key concepts

beam theory

Beam theory is a fundamental concept in structural analysis and mechanics of materials, allowing us to understand how beams behave under various loads. A beam is commonly a horizontal structural member designed to resist bending. One key aspect is the bending moment, which measures the internal moment that causes the beam to bend. For any point on the beam, the bending moment is given by \(\frac{K}{\rho}\), where \(K\) is a constant depending on the material and cross-sectional shape, and \( \rho \) is the radius of curvature.
Understanding beam theory enables engineers to design structures that can safely and efficiently bear loads, providing us with information about how the structure will deflect and where the maximum stress points are likely to occur.

differential equations

Differential equations play a crucial role in analyzing beam behavior, as they describe how the beam's deflection varies along its length. In the current problem, we employ a second-order differential equation to represent the relationship between bending moment and beam deflection.
The given exercise involves integrating the differential equation \( \frac{d^{2} y}{d x^{2}} = -\frac{W x}{2 K} \) twice to find the beam's profile. This type of differential equation is common in structural analysis when applied forces cause deflections in beams.
Understanding how to solve these equations allows us to predict how a beam will deform under specific loading conditions, thus ensuring that structures can be designed to handle expected loads without failure.

structural analysis

Structural analysis is the process of determining the effects of loads on physical structures, such as beams, bridges, and buildings. In this context, it involves calculating the reactions at supports, internal forces like bending moments, and deformations or deflections.
In the provided exercise, we determine how the weight \(W\) suspended from the beam’s center affects the bending moments and the resultant displacement. By analyzing these forces and using the bending moment equation, we can derive the net magnitude of the bending moments and verify the stability and performance of the structure under given conditions.
This approach ensures structural safety and functionality by predicting how the elements of the structure will withstand applied loads.

mechanics of materials

The mechanics of materials involves the study of how materials respond to various forces and deformations. This branch of mechanics focuses on the behavior of solid objects subject to stresses and strains, enabling us to predict failures and understand material behavior.
In this exercise, the beam is slightly bent, which implies small-angle approximations where \( (\frac{d y}{d x})^{2} \ll 1 \). This simplifying assumption is crucial for deriving accurate and manageable equations. By understanding how materials deform and resist forces, such as the flexural rigidity represented by the constant \(K\), we can design more effective and resilient structural elements.
The downward displacement derived at the beam's center showcases how mechanics of materials principles apply to real-world problems, ensuring that structures can safely support designed loads.