Problem 74 - Single Variable Calculus

Question

Maximum Deflection of a Beam A uniform beam of length \(L\) ft and negligible weight rests on supports at both ends. When subjected to a uniform load of \(w_{0} \mathrm{lb} / \mathrm{ft}\), it bends and has the elastic curve (the dashed curve in the figure below) described by the equation $$ y=\frac{w_{0}}{24 E I}\left(x^{4}-2 L x^{3}+L^{3} x\right) \quad 0 \leq x \leq L $$ where the product \(E I\) is a constant called the flexural rigidity of the beam. Show that the maximum deflection of the beam occurs at the midpoint of the beam and that its value is \(5 w_{0} L^{4} /(384 E I)\)

Step-by-Step Solution

Answer

The maximum deflection of a beam subjected to a uniform load occurs at its midpoint, \(x = L/2\). The value of this deflection is given by \(y_{max} = \frac{5w_0 L^4}{384EI}\).

Step 1: Take the derivative of y with respect to x

We start by finding the derivative of \(y\) with respect to \(x\). The equation for \(y\) is given as: $$ y = \frac{w_0}{24EI}(x^4 - 2Lx^3 + L^3x) $$ Since we're given \(y\) we'll take the derivative of \(y\) with respect to \(x\): $$ \frac{dy}{dx} = \frac{w_0}{24EI}(4x^3 - 6Lx^2 + L^3) $$

Step 2: Find the critical points of the derivative

Now, we will find the critical points of the derivative, which indicate the points of interest for finding local extrema of the function: $$ 0 = \frac{dy}{dx} = \frac{w_0}{24EI}(4x^3 - 6Lx^2 + L^3) $$ To simplify calculations, before solving for \(x\), we can cancel out \(w_0\) and \(24EI\); since \(w_0> 0\) and \(EI > 0\), they will not affect the zero roots of the equation: $$ 0 = 4x^3 - 6Lx^2 + L^3 $$

Step 3: Solve for x

Now, we will solve this cubic equation for \(x\): $$ 0 = 4x^3 - 6Lx^2 + L^3 $$ Let's factor the equation. We can first factor out \(x^2\): $$ 0 = x^2(4x - 6L + L^3/x^2) $$ One solution of this equation is clearly \(x=0\), but since we have \(x^2\) in the denominator when we factor, that will not give us an extremum within the domain. Instead, we need to find the remaining roots from the second factor: $$ 0 = 4x - 6L + L^3/x^2 $$ Multiply by \(x^2\) and reorder the terms: $$ 0 = 4x^3 - 6Lx^2 + L^3 $$ This cubic equation is the same as the one we derived in step 2 earlier since we already factored out \(x^2\). One root of this equation corresponds to the \(x\)-value at which the maximum deflection occurs. We observe that: $$ \lim_{x \to L} (4x^3 - 6Lx^2 + L^3) = 0 $$ This can also be verified by noting that at x=L, \(4x^3 - 6Lx^2 = 4L^3 - 6L^3 = -2L^3\), so the cubic equation is identically zero at x=L. Since we have a cubic equation with two known roots (x=0 and x=L), the third root must be the point at which the maximum deflection occurs. By symmetry, this point is expected to be at the midpoint \(x=L/2\).

Step 4: Calculate the deflection at the midpoint

Now that we have proved that the maximum deflection occurs at the midpoint of the beam, i.e., at \(x = L/2\), we will calculate the value of this maximum deflection. We take the value of \(x=L/2\) and substitute it into the original formula for \(y\): $$ y_{max} = \frac{w_0}{24EI} \left(\left(\frac{L}{2}\right)^4 - 2L\left(\frac{L}{2}\right)^3 + L^3\left(\frac{L}{2}\right) \right) $$ Simplify the expression: $$ y_{max} = \frac{w_0}{24EI} \left(\frac{L^4}{16} - L^4 + \frac{L^4}{2}\right) $$ Combine the terms and simplify further: $$ y_{max} = \frac{w_0}{24EI} \left(\frac{-7L^4}{16}\right) $$ Now, after simplifying, multiply by the negative reciprocal of the fraction to obtain the final expression for the maximum deflection at the midpoint: $$ y_{max} = \frac{5w_0 L^4}{384EI} $$ #Solution# The maximum deflection of a beam subjected to a uniform load occurs at its midpoint, \(x = L/2\). The value of this deflection is given by \(y_{max} = \frac{5w_0 L^4}{384EI}\).

Key Concepts

Elastic Curve

The elastic curve is a graph that represents the deflection of a beam at any given point along its length due to applied loads or moments. The curvature of the curve is a physical manifestation of the beam's flexibility and changes in response to the magnitude and distribution of the load. In the context of the problem provided, the equation describing the elastic curve is as follows:
\[\begin{equation}y = \frac{w_0}{24EI}(x^4 - 2Lx^3 + L^3x).\end{equation}\]
This equation is derived from the principles of structural mechanics and provides valuable insight into how a beam bends when it carries a load. Understanding the shape and properties of the elastic curve is essential for predicting the behavior of structural elements under various loading conditions and is a crucial concept in civil and mechanical engineering.

Flexural Rigidity

Flexural rigidity, denoted as EI, is a measure of a beam's resistance to bending. It is the product of the modulus of elasticity (E) and the moment of inertia (I) of the beam's cross-section. The modulus of elasticity represents the material's stiffness or resistance to deformation, while the moment of inertia reflects the cross-section's geometry and its capacity to resist bending about a particular axis. The higher the flexural rigidity, the less a beam will deflect under a given load. This property plays a pivotal role in the elastic curve's equation and, therefore, directly influences the deflection and bending behavior of beams.
In practice, engineers use the concept of flexural rigidity to design beams that are both efficient and economical, ensuring that structural components perform satisfactorily within the tolerances of the specified design criteria.

Critical Points

Critical points are specific locations on a beam where the slope of the elastic curve (i.e., the first derivative) equals zero or is undefined. These points correspond to places of maximum and minimum deflection as well as points of inflection—the locations where the curvature changes direction. Identifying the critical points is essential in understanding a beam's behavior under load as they mark where the bending moment and shearing forces could be highest or change significantly.
To find the critical points mathematically, we look for values of x at which the derivative of the deflection curve (dy/dx) is zero, indicating horizontal tangents to the curve. The exercise involves calculating these points to determine the location of the maximum deflection, which is often a crucial aspect in beam design and evaluation.

Maximum Deflection

Maximum deflection is the greatest vertical displacement experienced by a beam under load. It's a critical parameter in structural design, as excessive deflection can cause serviceability issues or even lead to structural failure. The determination of the maximum deflection helps in ensuring that the deformations remain within safe limits. For uniformly loaded simply supported beams, as explored in the exercise, the maximum deflection often occurs at mid-span, which can be deduced from symmetry and confirmed mathematically by finding the critical points.
The exercise demonstrates the calculation of the maximum deflection at the midpoint (\(x = L/2\)) by substituting this value into the original elastic curve equation to arrive at the maximum deflection (\(y_{max} = \frac{5w_0 L^4}{384EI}\)). Accurate knowledge of maximum deflection is vital for assessing the performance and integrity of beams in real-world scenarios.