Dynamics and Stability – Exercises Module 7: Hamilton’s Principle and Ritz Method

Course week(s) 7. Hamilton’s Principle and Ritz Mehtod

Corresponding to each module, as described in the lectures section, the teachers handed out one or more sample problems. These are worked out examples that illustrate how to work with a certain method.

Next to the given sample problems, several recommended exercises are given. These exercises can be found in the textbook of this course (Analytical Mechanics, by Josef S. Tőrők). More on this textbook can be found in the readings section.

Sample problems:

Strain waves bar This sample problem illustrates how to apply the Ritz method to find approximate solution for variational problems. The problem is about the analysis of strain waves in a bar.

Example problem 1 In this problem, a cantilever beam is illustrated and it is shown how to formulate the equations of motion, initial conditions and boundary condition by making use of the Lagrangian.

Example problem 2 This problem also deals with a cantilever beam, this time subjected to a distributed load (which is not uniform, thus must be represented by an integral function). It is shown how to calculate the equations of motion, initial conditions and boundary condition by making use of Hamilton’s Principle.

Recommended exercises:

Solve the exercises in the provided order:

3.23 (Notice that κ is the elastic constant of the foundation rather that the curvature, and that the subscript “eff” in that formula is meaningless. Do not forget to include the own elastic potential energy of the beam with uniform rigidity EI)
3.24 (Consider uniform rigidity EI and a spring constant k. The equilibrium equation is nothing new; concentrate on the boundary condition)
3.37
3.40 (That is Ritz method with degrees-of-freedom A, B and C and shape functions 1, t and t².)
3.34(a) (Solve this problem using Ritz method with the shape functions 1, x and x². Consider two cases separately: (i) The natural boundary condition found in Week 6 for this problem is imposed as constraint for x=1. (ii) No constraint is imposed for x=1 and a two variable minimisation problem is solved. Why are the results different? Why is the solution in the second case not fulfilling the natural boundary condition? Why are both solutions acceptable?)

Answers exercises This file contains answers and hints corresponding to the following recommended exercises of this module: 3.23, 3.24, 3.37, 3.40, 3.34(a).

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