Question

deal with column buckling problems. For each of the following boundary conditions find the smallest eigenvalue (the buckling load) of \(y^{\prime \prime}+\lambda y^{\prime \prime}=0,\) and also find the corresponding eigenfunction (the shape of the buckled column). $$ \begin{array}{ll}{\text { (a) } y(0)=y^{\prime \prime}(0)=0,} & {y(L)=y^{\prime \prime}(L)=0} \\ {\text { (b) } y(0)=y^{\prime \prime}(0)=0,} & {y(L)=y^{\prime}(L)=0} \\ {\text { (c) } y(0)=y(0)=0,} & {y(L)=y^{\prime}(L)=0}\end{array} $$

Short answer

Question: Determine the smallest eigenvalue (buckling load) and corresponding eigenfunction (shape of the buckled column) for the given boundary conditions of a second-order differential equation. Case (a): Boundary Conditions: $y(0) = y''(0) = 0$ and $y(L) = y''(L) = 0$ Answer: In this case, the eigenfunction is $y(x) = 0$, and there is no eigenvalue since the column does not buckle under these conditions. Case (b): Boundary Conditions: $y(0) = y''(0) = 0$ and $y(L) = y'(L) = 0$ Answer: In this case, the eigenfunction is $y(x) = C_2$, and there is no eigenvalue since the column does not buckle under these conditions. Case (c): Boundary Conditions: $y(0) = y(0) = 0$ and $y(L) = y'(L) = 0$ Answer: In this case, the eigenfunction is $y(x) = 0$, and there is no eigenvalue since the column does not buckle under these conditions.

Step-by-step solution

General Solution for the Differential Equation

To find the eigenvalues, we first have to find the general solution for the given differential equation: $$y^{\prime \prime} + \lambda y^{\prime \prime} = 0$$ We can rewrite this as: $$y^{\prime \prime}(1 + \lambda) = 0$$ Now, we'll integrate it to get the general solution: $$ y^\prime = C_1$$ and integrate once more: $$ y = C_1x + C_2$$ Now, let's solve for each given boundary condition.

Case (a)

For the first boundary condition, we have $$ y(0) = y^{\prime \prime}(0) = 0 \\ y(L) = y^{\prime \prime}(L) = 0 $$ Applying the boundary conditions to our general solution, we get $$ 0 = C_1 \cdot 0 + C_2 \\ 0 = C_1 \cdot L + C_2 $$ Solving these linear equations, we find that \(C_1 = 0\) and \(C_2 = 0\). Thus, the eigenfunction for this case is \(y(x) = 0\), and there is no eigenvalue since the column does not buckle under these conditions.

Case (b)

For the second boundary condition, we have $$ y(0) = y^{\prime \prime}(0) = 0 \\ y(L) = y^{\prime} (L) = 0 $$ Applying the boundary conditions to our general solution, we get $$ 0 = C_1 \cdot 0 + C_2 \\ 0 = C_1 \cdot L $$ Solving these linear equations, we find that \(C_1 = 0\) and \(C_2\) is arbitrary; thus, the eigenfunction for this case is \(y(x) = C_2\). However, there is no eigenvalue for this case since the column does not buckle under these conditions.

Case (c)

For the third boundary condition, we have $$ y(0) = y(0) = 0 \\ y(L) = y^{\prime}(L) = 0 $$ Applying the boundary conditions to our general solution, we get $$ 0 = C_1 \cdot 0 + C_2 \\ 0 = C_1 \cdot L $$ Solving these linear equations, we find that \(C_1 = 0\) and \(C_2 = 0\). Thus, the eigenfunction for this case is \(y(x) = 0\), and there is no eigenvalue since the column does not buckle under these conditions. In summary, for all three cases, the eigenfunction is \(y(x) = 0\) (meaning the column remains unbuckled), and there is no eigenvalue since the buckling load does not exist for these conditions.

Key concepts

Eigenvalues

In the study of column buckling, eigenvalues play a crucial role in understanding the load at which a column will buckle. The process involves determining specific values of \( \lambda \), which are called eigenvalues, that satisfy the differential equation under certain boundary conditions. A column remains stable until it reaches a load equivalent to its smallest eigenvalue. Beyond this point, it can start to buckle, changing its shape.

• The smallest eigenvalue is especially significant, as it indicates the lowest load at which buckling occurs.
• This critical value is determined by applying the given boundary conditions to the differential equation.
• In our example, however, no eigenvalue manifests for the scenarios provided, signifying that under these conditions the column does not experience buckling.

Understanding eigenvalues helps engineers design structures that can withstand certain loads without deforming, ensuring safety and reliability.

Boundary Conditions

Boundary conditions are essential for solving differential equations, especially in the context of physical systems like column buckling. They define how the solution behaves at the boundaries of the domain, in this case, the ends of the column.

• Boundary conditions are set based on physical constraints or symmetries in the problem.
• For example, a fixed end might impose that both displacement and slope are zero at that point.

In our exercise:

• Case (a) specifies that both ends of the column have zero deflection and zero bending moment. Thus, \( y(0) = y(L) = 0 \) and \( y''(0) = y''(L) = 0 \).
• Case (b) modifies the condition at \(x = L\) to a fixed slope, \( y'(L) = 0 \), while upholding \( y(L) = 0 \).
• Case (c) maintains zero displacement at both ends and zero derivative at \(x = L\).

In all these instances, the boundary conditions were pivotal in interpreting the outcomes of the differential equation solutions.

Differential Equations

Differential equations are fundamental in describing the behavior of systems such as column buckling. They help simulate how physical quantities, like displacement, vary over a structure. The differential equation in this exercise is:\[ y'' + \lambda y'' = 0 \]

• This form suggests that solutions involve constants and are dependent on the boundary conditions applied.
• The simplicity of the equation reflects specific conditions of symmetric loading and constraints.
• These conditions separated the exercise from more complex equations involving higher-order derivatives or non-linear terms.

In practice, engineers meet challenges when differential equations become less straightforward. However, understanding simple cases like this builds a foundational understanding of more complex interactions.

Eigenfunctions

Eigenfunctions represent the shape a structure takes when subject to its corresponding eigenvalue. For column buckling, they illustrate how the column deforms as it is loaded at this specific value. Finding eigenfunctions involves solving the differential equation and applying boundary conditions, allowing us to visualize potential deflection patterns.

• In the exercise, no eigenfunctions arose for any of the specified cases, suggesting no buckling at the given loads.
• For a typical non-zero eigenvalue scenario, eigenfunctions would represent a non-trivial shape showcasing the buckled form of the column.
• Zero eigenfunctions, as in this exercise, indicate stability and no change in shape under the given loads and constraints.

In engineering, visualizing these functions aids in pre-emptive planning, ensuring structures maintain integrity under expected loads.