MATH 312 Section 5.2: Linear Models: BVPs

Transcription

1 MATH 312 Section 5.2: Linear Models: BVPs Prof. Jonathan Duncan Walla Walla University Spring Quarter, 2008

2 Outline 1 Deflection of a Beam 2 Eigenvalues and Eigenfunctions 3 Conclusion

3 Understanding the Set-up One useful application of linear BVPs is to compute the deflection or bending of a beam under stress. Beam Deflection To model the deflection of a bean, we look for a function representing the curve of the axis of symmetry (also called the deflection curve) of the beam. To develop our DE, we appeal to the theory of elasticity which deals with the bending moment, M(x) at a point x along the beam.

4 Elasticity Theory Elasticity Theory starts with the following assumption. Elasticity Theory The bending moment at a point M(x) is related to the load at that point, w(x), by the equation: d 2 M dx 2 = w(x) Bending Moment Additionally, the bending moment is proportional to the curvature κ of the elastic curve. M(x) = EI κ Where E is Young s modulus of elasticity for the material of the beam, and I is the moment of inertia of the cross section of the beam. EI is called the flexural rigidity of the beam.

5 Elasticity Theory, Continued... Putting this together with come Calculus helps develop the theory. Curvature Recall from calculus that the curvature of a function y is Note: y κ = (1 + (y ) 2 ) 2 3 When the amount of deflection is y(x) is small, this curvature becomes κ = y. Differential Equation Putting all of this together yields our differential equation d 2 M dx 2 = EI d 2 dx 2 y = EI d 4 y dx 4 EI d 4 y dx 4 = w(x)

6 Boundary Conditions: Embedded Depending on how the ends of the beam are supported, we may have several different boundary conditions. Embedded at Both Ends When both ends of the beam are embedded have the following boundary conditions. No displacement from axis: y(0) = 0, y(l) = 0 No deflection from axis: y (0) = 0, y (L) = 0

7 Boundary Conditions: Cantilevered Another possibility is to have only one end of the beam embedded and the other free. Embedded at One End (Cantilever) When only one end of the beam is embedded, we have the following boundary conditions. Embedded as before: y(0) = 0, y (0) = 0 Bending moment and Shear Force: y (L) = 0, y (L) = 0

8 Boundary Conditions: Simple Support One other type of support system is that of a beam simply supported at both ends. Simply Supported Beam When both ends are simply supported as shown, we have the following boundary conditions. No displacement from Axis: y(0) = 0, y(l) = 0 Bending Moment is Zero: y (0) = 0, y (L) = 0

9 Boundary Conditions: Summary In any beam problem, the boundary conditions will appear at each end of the beam and depend on the support type at that endpoint. Boundary Conditions The following boundary conditions apply at an beam end based on the support type at that end. Conditions y = 0, y = 0 y = 0, y = 0 y = 0, y = 0 Support Type Embedded End Free End Simply Supported End

10 An Example We now apply this theory to an example. Example Find the curve of deflection for a simply supported beam of length L with w(x) = w 0 a constant load. Graph the deflection curve if w 0 = 24EI and L = 1. EI d 4 y dx 4 = w 0 d 4 y dx 4 = w 0 Ei d 4 y dx 4 = 0 y c = C 1 + C 2 x + C 3 x 2 + C 4 x 3 d 4 y dx 4 = w 0 y p = Ax 4 y p = w 0 EI 24EI x 4 y(x) = w 0 ( x 4 2Lx 3 + L 3 x ) 24EI

11 Example, Continued Continuing the example... Let w 0 = 24EI and L = 1 to get y(x) = x 4 2x 3 + x Note that the image above is inverted because y(x) measures deflection in the downward direction.

12 A Common Boundary Value Problem Many applied problems require the solution of a two-point BVP to a 2nd order linear differential equation of a particular form. The Function of Interest In many instances we wish to solve a BVP involving y + λy = 0 While the trivial solution y = 0 always works in this setting, we wish to find non-trivial solutions to the BVP. Eigenvalues The values of λ which will produce non-trivial solutions are called the eigenvalues of the equation. Eigenfunctions Non-trivial solutions to this differential equation are called eigenfunctions.

13 An Example Let s see what can happen with an example boundary value problem of this form. Example Solve the boundary value problem y + λy = 0 subject to y( π) = 0 and y(π) = 0. Case I: λ = 0 Case II: λ < 0 y = C 1 x + C 2 only solution is: y = 0 y = C 1 e λx + C 2 e λx only solution is: y = 0 Case III: λ > 0 y = C 1 cos λx + C 2 sin ( nx λx y = cos 2 ) ( nx ), y = sin 2

14 Important Concepts Things to Remember from Section Setting-up BVPs for beam displacement 2 Identifying boundary conditions for beam displacements 3 Identifying eigenvalues and eigenfunctions for BVPs of the form y + λy = 0.