1. Dynamic Analysis-A Finite –Element Approach
University of Illinois-Chicago
Chapter 10
Dynamic Analysis-A Finite –Element Approach
Principles of
Computer-Aided
Design and
Manufacturing
Second Edition 2004
ISBN
Author: Prof. Farid. Amirouche
University of Illinois-Chicago

2. 10.1 Introduction CHAPTER 10 10.1 Introduction
In the preceding chapters we examined the area of finite element associated with the first law of mechanics. The structure’s response to time variant loads needs to be captured by using differential equations that govern the motion. Applying Newton’s Second Law or Lagrange equations usually does this. Up to this point we have discussed only cases in which the load on the structure is static and therefore the response was also static and proportional to the structural stiffness and applied loads. It is apparent that the response of a structure will be different if we account for the inertia forces and the time variant effects due to the kinematics and forces applied to the system.
This chapter will be used to show how we can use finite element method to formulate and solve for dynamic problems. The finite element method was used successfully to determine element stiffness matrices in the study of trusses. The methodology is extended to derive the mass element matrices and their corresponding global mass matrix for the structure. In most cases the differential equations for a structure assuming small deformation take the following form.
(10.1)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

3. CHAPTER 10 10.1 Introduction If we assume a solution of the form
(10.2)
then the dynamic analysis of structures takes the following form.
(10.3)
where [K] is the global stiffness matrix, [M] is the global mass, 2 denotes the natural frequency, U the global displacement array and {F} the external excitation force vector.
One of the finite element’s most popular approaches in vibration is the computation of the natural frequencies of a mechanical system. Since both [K] and [M] can directly be calculated from finite element method provided that the geometric properties and the boundary conditions of the problem are known, we can evaluate [2] making use of numerical procedures readily available in most of the engineering numerical analysis books. In this chapter we will focus on the analysis of the eigenvalue problem and in the formulation of the mass matrix in particular and the corresponding stiffness by showing how both the stiffness and mass can be calculated and develop the necessary steps to compute [2].
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

4. 10.2 Element Stiffness and Mass Matrices
CHAPTER 10
10.2 Element Stiffness and Mass Matrices
10.2 Element Stiffness and Mass Matrices
Once more let us examine a beam under a compressive load, where the compressive force is not constant but rather a force that is harmonically exciting the beam as function of time. That is the external force of
(10.4)
Let F be applied at the end of the beam as shown in Figure 10.1
In previous chapters we developed the element stiffness matrix when a beam is discretized into N elements. Essentially each element acts as a spring with a stiffness value k equal to
Again observe the element shown in Figure 10.2 it is subjected to axial forces at each end (node), and the actual displacement of the element can be written as function of the node displacements u1 and u2. The stiffness matrix can be obtained from the energy function minimization on the basis of equilibrium. This is derived by first developing the potential and kinetic energy due to axial deformation under the load F.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

5. 10.2 Element Stiffness and Mass Matrices
CHAPTER 10
10.2 Element Stiffness and Mass Matrices
Figure 10.1 A beam excited longitudinally by a harmonic force F
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

6. 10.2 Element Stiffness and Mass Matrices
CHAPTER 10
10.2 Element Stiffness and Mass Matrices
Figure 10.2 A beam element with nodes 1 and 2 under compressive load.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

7. 10.2 Element Stiffness and Mass Matrices
CHAPTER 10
10.2 Element Stiffness and Mass Matrices
First we write the beam displacement in terms of u1 and u2 as follows.
(10.5)
Note how the displacement at x = 0 is u1 and at x = l it is u2. This is consistent with our assumed displacement function. Now we can compute the energy terms needed for our finite element formulation. The strain energy of the bar is given by the integral
(10.6)
The kinetic energy of the element can be computed from the integral
(10.7)
Where (x) is the density of the bar.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

8. 10.2 Element Stiffness and Mass Matrices
CHAPTER 10
10.2 Element Stiffness and Mass Matrices
Differentiating of u (x, t) with respect to x in equation (10.5) and substituting it into (10.6) yield and explicit expression for the potential energy of the form
(10.8)
The velocity equation is
(10.9)
Substituting equation (10.9) into (10.7) we obtain the kinetic energy equation in terms of
which can be written as
(10.10)
Where M is defined by
(10.11)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

9. 10.3 Axial Deformation CHAPTER 10 10.3 Axial Deformation
One Element equation of motion
The equations of vibration can be obtained from the preceding expressions for the kinetic energy T( t) and strain energy V (t) by using the variational or Lagrangian approach.
The equation of motion can be calculated from energy in the structure from
(10.13)
Where ui is the coordinate of the system, which is assumed to have n degrees of freedom, and where
is the external force applied at coordinates ui.
Here the energies T and V are the total Kinetic energy and strain energy respectively in the structure.
Analysis of the boundary conditions indicates that the time response at the clamped end must be zero. Hence the total kinetic energy becomes
(10.14)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

10. CHAPTER 10 10.3 Axial Deformation (10.15)
Substitution of these two energy expressions into the Lagrange equation
(10.16)
Which constitutes a simple finite element model of the cantilever bar using only one element.
This finite element model of the bar can now be solved, given a set of initial conditions, for the nodal displacement u2(t). The solution to the equation is
(10.17)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

11. CHAPTER 10 10.3 Axial Deformation
Three Elements equation of motion
The total strain energy is the sum of the strain energy associated with each element , that is,
(10.18) = =
The vector of derivatives of this total strain energy
indicated in the Lagrangian equation is found to be
(10.19)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

12. CHAPTER 10 10.3 Axial Deformation
In a similar fashion, total kinetic energy can be expressed by adding the contribution of the
kinetic energy from all 3 elements where we write
(10.20)
The first term in the Lagrangian equation, which is responsible for the mass matrix, then becomes
(10.21)
Combining all the Lagrangian terms the equations of motion take the following form
(10.22)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

13. CHAPTER 10 10.3 Axial Deformation
Where u(t)=[u2 u3 u4]T is the vector of nodal displacements. Here the matrix
(10.23)
is the global mass matrix and the corresponding K
(10.24)
is the global stiffness matrix.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

14. CHAPTER 10 10.3 Axial Deformation
The finite element method allows us to derive both the mass and stiffness of a structure by means of the energy method. It is obvious that the size of these matrices can expand rapidly if we choose a large number of elements. When solving a linear problem we are mostly interested in the vibration characteristics of the problem and not necessarily its time response. Hence we reduce the problem to a so-called an eigen value problem by first assuming a solution to equations of the form given by (10.22) that is given by
(10.25)
We have essentially reduced the problem to a free-vibration where the right hand side of the governing equations is set to zero. Substitution of equation (10.25) into (10.22) yields
(10.26)
We can solve for w2 by setting the following determinant equal to zero.
(10.27)
Equation (10.26) represents the generalized eigen-value problem.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

15. 10.4 Bending or Transverse Deformation
CHAPTER 10
10.4 Bending or Transverse Deformation
10.4 Bending or Transverse Deformation of a Beam
Let us consider a beam subject to transverse loading, which causes the beam to bend. The deflection of the neutral axis at any location x is represented by the variable v, for small deflections we have the flexure formula which defines the stress, the moment and the vertical displacement measured from the neutral axis as
(10.28) Or
(10.29)
Where I is the second moment of area. From standard strength of materials the above equation
can be expressed further as
(10.30)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

16. Figure 10.3 Simply supported beam
CHAPTER 10
10.4 Bending or Transverse Deformation
Figure 10.3 Simply supported beam
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

17. 10.4 Bending or Transverse Deformation
CHAPTER 10
10.4 Bending or Transverse Deformation
The total potential energy of the beam equals the strain energy less the potential energy due to the external loads. Let us represent the energy equation by
(10.31)
The strain energy contribution is obtained from the following integral where we can assume that the energy for a differential element will also have the same equation except the limits on the integral will be those associated with the element. If we ignore the contribution of the energy due to the external forces, the energy equation reduces to
(10.31)
The displacement u at each node is expressed in terms of vertical and longitudinal components as shown in Figure 10.3, hence the deflection can be expressed as fourth order polynomial
(10.32)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

18. 10.4 Bending or Transverse Deformation
CHAPTER 10
10.4 Bending or Transverse Deformation
Which defines a cubic spline. The coefficients can be found by making use of boundary conditions at both ends. The equation above becomes
(10.33)
Note how we have substituted u for v in the above equation. We can write equation (10.33) in a matrix form where we can isolate the shape function and the displacement vector and express u as
(10.34)
Where the shape functions are
(10.35)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

19. 10.4 Bending or Transverse Deformation
CHAPTER 10
10.4 Bending or Transverse Deformation
By substituting equation (10.34) into the strain energy equation we obtain
(10.36)
By taking the second derivatives of the shape functions we get
(10.37)
By substitution of the above equation into and recalling that the displacement at the equilibrium position occurs such that the value of the system total energy is minimum, we get
(10.38)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

20. 10.4 Bending or Transverse Deformation
CHAPTER 10
10.4 Bending or Transverse Deformation
(10.39)
Hence, the stiffness matrix for a beam element due to deflection (bending) is
(10.40)
Once again we have denoted by [ke] the element stiffness matrix to be consistent with our formulation of the finite element method. Now we can tackle the formulation of the mass matrix.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

21. 10.4 Bending or Transverse Deformation
CHAPTER 10
10.4 Bending or Transverse Deformation
Consider the uniform bar shown in Figure The element has a velocity in the x direction. If we discretize the bar into N elements, we evaluate the kinetic energy of the bar by integrating over the total length the bar. To obtain the kinetic energy of the element we first define the kinetic energy equation as follows
(10.41)
(10.42)
Where
The displacement field can be expressed as a function of basis function or shape functions such that
(10.43)
Where q (t) defines the generalized coordinates.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

22. 10.4 Bending or Transverse Deformation
CHAPTER 10
10.4 Bending or Transverse Deformation
Substituting the above equation into the kinetic energy of the bar yields
(10.44)
which leads to
(10.45)
The mass matrix Mij is then defined by the following expression
(10.46)
or simply
(10.47)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

23. 10.4 Bending or Transverse Deformation
CHAPTER 10
10.4 Bending or Transverse Deformation
Ni & Nj are the shape functions used in determining the element stiffness k(e) matrix, while i & j are interpolating functions with the following property: when  is equal to N, Mij is referred as the consistent mass matrix. For the sake of simplicity the shape functions N are used for the evaluation of the mass matrix.
For a uniform beam subject to bending the mass matrix is found to be
(10.48)
m is the total mass of the beam or element and is defined by
= Mass/unit volume
A: cross section area
L: length of the beam
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

24. 10.4 Bending or Transverse Deformation
CHAPTER 10
10.4 Bending or Transverse Deformation
Example 10.1
Find the coefficients M11 & M12 of the mass matrix using the shape function defined in the evaluation of the element stiffness matrix given by equation (10.35).
Solution:
Using equation (10.47) and substituting both subscripts i and j by 1 we obtain:
Factorizing the constants  and A we get
Note that
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

25. 10.4 Bending or Transverse Deformation
CHAPTER 10
10.4 Bending or Transverse Deformation
The above integral becomes = =
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

26. 10.4 Bending or Transverse Deformation
CHAPTER 10
10.4 Bending or Transverse Deformation
In a similar fashion we derive the element M44
N4N4
M44 =
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

27. 10.5 Bernoulli Beam CHAPTER 10 10.5 Bernoulli Beam
Consider a beam element whose nodal deformations are described by the displacements and a rotation at each end as shown in Figure We notice that the node displacements are along the longitudinal and normal axis. Essentially we have two translations at each node plus one rotation. A beam with both axial and bending deformation is referred as a Bernoulli Beam, which bears the name of one of the greatest scientists in Beam Theory.
Each joint depicts a longitutidinal and lateral deformation and a rotation. Hence the element is subject to axial and bending deformation. The transverse shear is neglected. We write the total strain energy as a combination of both the longitudinal deformation as well the bending deformation. Therefore, the energy due to the strain deformation can calculated from
(10.49)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

28. CHAPTER 10 10.5 Bernoulli Beam
Combining the results obtained in equation (10.12) and equation (10.40), we obtain the elastic stiffness matrix of the beam as
(10.50)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

29. CHAPTER 10 10.5 Bernoulli Beam
In a similar fashion we obtain the mass element matrix for the beam using equation (10.11) and (10.48) as follows
(10.51)
So we have established a procedure by which a mass and a stiffness element can be computed. In a similar way we will explore how the load vector is computed so we can complete the governing dynamical equations. The load vector represents the contribution of the external forces and moments applied to the beam.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

30. 10.6 Force Vector CHAPTER 10 10.6 Force Vector (10.52) (10.53)
When the load is distributed over the beam we need to find the generalized force vector that represents the contribution of the distributed at the degrees of freedom used to denote the beam deformation. In this case the load will affect the longitudinal direction that is normal to the beam axis and will ultimately affect the bending moments at the nodes. The contribution along the axial direction along the beam neutral axis will be zero. If we let the load F (x, t) be w the generalized force vector can be expressed as
(10.52)
For the case of a beam subjected to bending at the end points we have a displacement and a rotation, resulting in both a reaction force and a moment.
(10.53)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

31. CHAPTER 10 10.6 Force Vector Example 10.2b)
Figure 10.4 Two beams: a) simply supported at the right and fixed at the left end, and
b) fixed at both ends.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

32. CHAPTER 10
10.6 Force Vector
Two distinct uniform beams with different boundary conditions are being evaluated. They are both of length L, have a cross section area A and a modulus of elasticity E. The boundary conditions are such that the first beam is simply supported with a pin and roller at one end and fixed at the other, while the second beam has both ends fixed. a) Derive the mass and stiffness matrix for a 2-element model. For both cases and explain how the boundary conditions alter the global matrix K and M.
Solution:
Figure 10.5 Bernoulli Beam
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

33. CHAPTER 10
10.6 Force Vector
Let the beam be discretized as shown above in Figure 10.5 where the displacement and rotation at each node are represented by Uix, Uiy and i. (i=1, number of nodes).
Let {U} denote the global displacement array, and then we can determine the size of the array by simply examining that at each node we have 3 degrees of freedom, hence the global displacement vector {U} should be of size 9.
The global K matrix is defined by summing the stiffness of elements.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

34. CHAPTER 10 10.6 Force Vector where
This element stiffness for a beam undergoing 2 translations and one rotation at each joint is given by equation (10.50).
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

35. CHAPTER 10 10.7 Boundary Conditions 10.7 Boundary Conditions
From the beam end supports we deduce the following criteria for
The case where the beam is fixed at one end and simply supported at the other
b) The case with fixed-fixed conditions at both ends of the beam
U1x = U1y = 0
U3x = U 3y = 0
Furthermore let the axial deformation U2x be negligible, than the global stiffness
matrix reduces for case (a) to
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

36. CHAPTER 10 10.7 Boundary Conditions
Following the outlined procedure above, we develop the global mass matrix by first noting that
for a uniform beam the element mass is
We obtain the global matrix by adding the element mass matrix for each element
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

37. CHAPTER 10 10.7 Boundary Conditions
Applying the boundary conditions for case (a) the mass matrix reduces to
[M] =
Combining [M] &[K], and following the assumed solution outlined in equation (10.25) we can express the characteristic equation for the eigen value problem as the determinant of the following equation:
The eigen values of which can be computed by determining the determinant of the matrix and setting it equal to zero. This will result into the natural frequencies of the assumed beam with the boundary conditions outlined in case a. In a similar fashion we can build the mass and stiffness matrices for case b. We notice that for case b) the boundary conditions result into 5 displacements to be zero, hence reducing the size of the global matrices by 5. We should then expect 5 eigen values for case a) and 4 eigen values for case b).
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

38. Figure 10.6 A planar beam structure in space
CHAPTER 10
10.8 Planar Beam Structure
10.8 Planar Beam Structure
Figure 10.6 A planar beam structure in space
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

39. CHAPTER 10 10.8 Planar Beam Structure
Figure10. 7 A simple beam with an orientation  in a 2D space.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

40. CHAPTER 10 10.8 Planar Beam Structure
Planar structures are common in engineering or their analysis is vital to design, longevity and safety. In the previous section we derived the element stiffness or mass of a beam but we assumed the beam neutral axis to be horizontal coinciding with the x axis of the coordinates X, Y. Obviously if a beam is positioned in space and we have several of them whose inclinations with respect to the global frame XY are different, then their mass and stiffness element matrix will contribute differently to the global stiffness and mass matrix. What follows is derivation of element [M(e)] or [K(e)] when its subjected to such orientation in space.
We have previously defined the element stiffness or mass for a uniform beam when the beam is in a longitudinal position. Let {u1x, u1y, 1,u2x, u2y, 2} represent the displacement at each node with respect to a local coordinate system of the beam with x along the neutral axis of the beam. Then we can transform the local coordinate to a global ones along XY by using a transformation matrix R such that
(10.54)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

41. CHAPTER 10 10.8 Planar Beam Structure Or simply (10.55)
The new element mass and stiffness matrices become
(10.56)
(10.57)
Note that the element mass and stiffness matrices on the right hand side of equation (10.56) and (10.57) are those associated of zero orientation or simply the matrices developed for the element coinciding with the horizontal (x-axis).
This orientation is justified for the measurement of the orientation angle commencing from the horizontal line to the neutral axis of the beam element (CCW).
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

42. 10.9 Eigen Value Problem CHAPTER 10 10.9 Eigen Value Problem
Once the dynamical equations are formulated we need to solve for the eigen values and eigen vectors of the system. What follows are some common methods used in the solution of such problems. The actual solution procedure used to solve the eigen problem is discussed and some considerations are outlined for further study.
A very common eigen problem is encountered in vibration analysis. Examining equation (10.26) we can generalize the solution of the problem to the following form
(10.58)
The solution for the n eigen values and corresponding eigen values can be written as
(10.59)
where the column of  represent the eigen vectors whereas the eigen values are stored in the matrix , which is a diagonal matrix listing the eigen values.
The eigen values are the roots of the characteristic polynomial,
(10.60)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

43. CHAPTER 10 10.9 Eigen Value Problem
This is easily shown in the solution of free vibration problems where the equations of motion
are represented in a matrix as
(10.61)
Assuming a solution of the form
(10.62)
we get
(10.63)
or simply
(10.64)
Where
and X is analogous to  in equation (10.58)
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

44. CHAPTER 10 10.9 Eigen Value Problem Example 10.3
Consider a linear system composed of two masses and springs interconnected to form an oscillatory system used to characterize the dynamics of a train-car model. We are interested in the solution of this problem from the point of view in learning about the frequency response and the system modes of vibration.
Figure 10.8 A two-degree of freedom mass-spring system
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

45. CHAPTER 10 10.9 Eigen Value Problem
The equations of motion can easily be derived using Newton’s Law where we denote by x1 and x2 the displacements of mass 1 and 2 respectively.
(10.65)
Where
and
Find:
a) the eigen values
b) the corresponding eigen vectors.
The physical parameters have the following values: m1=9 kg, m2=1 kg, k1=72 N/m and k2 =4.5 N/m.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

46. CHAPTER 10 10.9 Eigen Value Problem
Solution: For these values the characteristic equation
So that 12=9 and 22=4. There are two roots, and each corresponds to two values of the constant  in the assumed form of the solution:
Now we need to solve for the eigen vectors corresponding to each eigen values.
We proceed by noticing that the equations of motion reduce to
Where u1 denoted the vector associated with the first eigen value 1. Let such a vector u1 be expressed in terms of its components we get
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

47. CHAPTER 10 10.9 Eigen Value Problem And for 2 the vector u2 satisfies
Similarly, the two equations and two unknowns are found from the simple relation we have derived from the assumed solution such as
These expressions can be solved for the vectors u1 and u2 where we can see after substitution of the assumed values the above equations become:
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

48. CHAPTER 10 10.9 Eigen Value Problem
The solution for the first mode shape vector u1 is as follows:
We find the second in a similar fashion to be
Hence,
and
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

49. CHAPTER 10 10.9 Eigen Value Problem Example 10.4
Calculate the displacements of the two masses – two springs system in the previous problem
for the given initial conditions :
Solution:
The solution of the free vibration problem at hand can be expressed in a matrix form as follows:
At t=0 this yields
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

50. CHAPTER 10 10.9 Eigen Value Problem
Differentiating the above equations and evaluating them at t=0 yields
so that
Adding the above two equations yields
the velocity equation reduces to
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

51. CHAPTER 10 10.9 Eigen Value Problem in the above two equation gives
Substituting
in the above two equation gives
And
which has a solution of
A1= -9/10 mm, A2 = 9/10 mm.
Thus
Above is a classic example of a two degrees-of-freedom vibration problem where students can evaluate the modes shapes and frequencies and deduce the transient solution resulting from the solution of the free vibration problem. We should also note how the mode shapes and frequencies contribute to the dynamic solution of lumped mass systems where the magnitude of vibration is dependent both on the initial conditions as well as the mode shapes or eigen vectors.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

52. 10.10 Modal Analysis CHAPTER 10 10.10 Modal Analysis
Computing the eigen values and eigen vectors of a mechanical system can become strenuous and cumbersome if the number of degrees of freedom increases. When a system possesses a large number of degrees of freedom the determinant of a matrix which results into the characteristic of the equations can lead to polynomial to the power 2*N. Keep in mind that the roots of such a polynomial need to be found and the solution of N equations and N unknowns for each eigen value needs to be resolved. What follows is a modal analysis approach in extracting the eigen values and eigen vectors once the generalized mass and stiffness matrix are found.
Let the general equation of a free vibration system be given by
(10.66)
And let the initial conditions imposed on the system be such that
The steps involved in calculating the eigen values and eigen vectors require that we make the mass matrix an identity matrix. This is simply obtained by pre multiplying the equations of motion above by the inverse matrix of M.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

53. CHAPTER 10 10.10 Modal Analysis
Let us assume that our mass matrix is an identity matrix then it is easy to calculate as a first step
1) Calculate the M-1/2.
Then we proceed by computing the normalized stiffness matrix following
2) Compute the
mass normalized stiffness matrix
3) Calculate the symmetric eigen value problem for
to get 2i and vi.
4) Normalize vi and form the matrix of eigen vectors
5) Calculate the matrix of mode shapes
and
6) Calculate the modal initial conditions:
7) Substitute the components of , into equations of solution of modal equations to get the solutions in the modal coordinates
8) Obtain the final solution by where S is the transformation from
the modal coordinate to the generalized coordinates of the system.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

54. CHAPTER 10 10.10 Modal Analysis Example 10.5
Calculate the solution of the n-degree of freedom system shown for n = 3 by modal analysis. Use the values m1=m2=m3=1 kg and k1=k2=k3=1 N/m, and the initial condition x1(0)=1 m with all other initial displacements and velocities zero.
Solution The mass and stiffness matrices for n=3 for the values given become
Where
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

55. Figure 10.9 A Mass-spring system of N Degrees of Freedom.
CHAPTER 10
10.10 Modal Analysis
Figure A Mass-spring system of N Degrees of Freedom.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

56. CHAPTER 10 10.10 Modal Analysis
Following the steps outlined in the modal analysis procedure we get 1. 2. 3.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

57. CHAPTER 10 10.10 Modal Analysis The roots of this cubic equation are
Thus the system’s natural frequencies are
To calculate the first eigenvector, substitute 1= into
. This yields
and solve for the vector v1=
Multiplying out this last expression yields three equations, only two of which are independent:
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

58. CHAPTER 10 10.10 Modal Analysis
Solving the first and third equations yields
The second equation is dependent and does not yield any new information.
Substituting these values into the vector v1 yields
4. Normalizing the vector yields
Solving for v31 and substituting back into the expression for v1 yields the normalized
version of the eigenvector v1 as
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

59. CHAPTER 10 10.10 Modal Analysis
Similarly v2 and v3 can be calculated and normalized to be
The matrix P is then given by
5. The matrix or
And
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

60. CHAPTER 10 10.10 Modal Analysis
6. The initial conditions in modal coordinates becomes
And
7. The modal solutions are each of the form given by
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago

61. CHAPTER 10 10.10 Modal Analysis
8.The solution in physical coordinates is next calculated from
x=S r (t)=
This systematic approach is far more convenient than the procedure outlined in the previous section.
One needs to keep in mind that there are a number of other techniques that can be used in the solution of eigen value problems.
Students are urged to look up these methods in other textbooks in finite element analysis and vibration.
Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN
Author: Prof. Farid. Amirouche, University of Illinois-Chicago