1. Mechanical Vibrations
Singiresu S.Rao
SI Edition
Chapter 6
Multidegree of Freedom Systems

2. Chapter Outline 6.1 Introduction
6.2 Modeling of Continuous Systems as Multidegree of Freedom Systems
6.3 Using Newton’s Second Law to derive Equations of Motion
6.4 Influence Coefficients
6.5 Potential and Kinetic Energy Expressions in Matrix Form

3. Chapter Outline 6.6 Generalized coordinates and Generalized forces
6.7 Using Lagrange’s Equations to derive Equations of Motion
6.8 Equations of Motion of Undamped systems in matrix form
6.9 Eigenvalue problem
6.10 Solution of the Eigenvalue problem

4. Chapter Outline 6.11 Expansion Theorem 6.12 Unrestrained systems
6.13 Free vibration of Undamped systems
6.14 Forced vibration of Undamped systems using Modal analysis
6.15 Forced vibration of Viciously damped systems
6.16 Self-excitation and Stability analysis

5. Introduction
For simplicity continuous systems are approximated as multidegree of freedom (MOF) systems.
Equations of motion of MOF systems can be obtained either from Newton’s 2nd law of motion or from Lagrange’s equations.
Analysis of MOF systems can be simplified using the orthogonality property of the mode shapes of the system’s natural frequencies.

6. Modeling of Continuous systems as MOF systems
Method 1: Lumped-mass system
Replace distributed mass of the system by finite number of lumped masses
Lumped masses are connected by massless elastic and damping members
E.g. Model the 3-storey building as a 3 lumped mass system

7. Modeling of Continuous systems as MOF systems
Method 2: Finite Element Method (FEM)
Replace geometry of the system by large number of small elements
Principles of compatibility and equilibrium are used to find an approx. solution to the original system
Details covered in Chapter 12

8. Using Newton’s 2nd Law to derive Equations of Motion
Step 1: Set up suitable coordinates to describe positions of the various masses in the system.
Step 2: Measure displacements of the masses from their static equilibrium positions
Step 3: Draw free body diagram and indicate forces acting on each mass
Step 4: Apply Newton’s 2nd Law to each mass

9. Example 6.1
Derive the equations of motion of the spring-mass damper system shown below.

10. Solution Free-body diagram is as shown:
Applying Newton’s 2nd Law gives:
Set i=1 with x0=0 and i=n with xn+1=0:

11. Solution
Notes:
The equations of motion can be expressed in matrix form:
[m] is the mass matrix
[c] is the damping matrix
[k] is the stiffness matrix

12. Example 6.2
Derive the equations of motion of the trailer-compound pendulum system as shown.

13. Solution Free-body diagram:
x(t) and θ(t) describes linear and angular displacement respectively.

14. Solution External forces on trailer: F(t), k1x,k2x,
Horizontal translation:
External forces on pendulum: M(t), mg
Rotational motion about hinge O:

15. Influence Coefficients
One set of influence coefficients is associated with each matrix involved in the equations of motion
Equation of Motion
Influence coefficient
Stiffness matrix
Stiffness influence coefficient
Mass matrix
Inertia influence coefficient
Inverse stiffness matrix
Flexibility influence coefficient
Inverse mass matrix
Inverse inertia coefficient

16. Stiffness Influence Coefficient, kij
kij is the force at point i due to unit displacement at point j and zero displacement at all other points.
Total force at point i,
Matrix form:

17. Stiffness Influence Coefficient, kij
Note:
kij = kji
kij for torsional systems is defined as the torque at point i due to unit angular displacement at point j and zero angular displacements at all other points.

18. Example 6.3
Find the stiffness influence coefficients of the system shown below.

19. Solution
Let x1, x2 and x3 be the displacements of m1, m2 and m3 respectively.
Set x1=1 and x2=x3=0.
Horizontal equilibrium of forces:
Mass m1: k1 = -k2 + k (E1)
Mass m2: k21 = -k (E2)
Mass m3: k31 = (E3)
Solving E1 to E3 gives
k11=k1+k2
k21 = -k2
k31 = 0

20. Solution Next set x2=1 and x1=x3=0. Horizontal equilibrium of forces:
Mass m1: k12 + k2 = 0 (E4)
Mass m2: k22 - k3 = k2 (E5)
Mass m3: k32 = - k (E6)
Solving E4 to E6 gives
k22 = k2+k3
k12 = -k2
k32 = -k3

21. Solution Finally set x3=1 and x1=x2=0.
Horizontal equilibrium of forces:
Mass m1: k13 = (E7)
Mass m2: k23 + k3 = 0 (E8)
Mass m3: k33 = k (E9)
Solving E7 to E9 gives
k33 = k3
k13 = 0
k23 = -k3
Thus the stiffness matrix is:

22. Flexibility Influence Coefficient, aij
Have to solve n sets of linear equations to obtain all the kij’s in an n DOF system
Generating aij’s is simpler.
aij is defined as the deflection at point i due to unit load at point j,
xij = aijFj, where xij is the displacement at point i due to external force Fj
Matrix form:

23. Flexibility Influence Coefficient, aij
Note:
Stiffness and flexibility matrices are the inverse of each other.
aij = aji
aij for torsional systems is defined as the angular deflection of point i due to unit torque at point j.

24. Example 6.4
Find the flexibility influence coefficients of the system shown below.

25. Solution
Let x1, x2 and x3 be the displacements of m1, m2 and m3 respectively.
Set F1=1 and F2=F3=0.
Horizontal equilibrium of forces:
Mass m1: k1a11 = k2(a21 – a11) + 1 (E1)
Mass m2: k2(a21 – a11) = k3(a31 – a21) (E2)
Mass m3: k3(a31 – a21) = 0 (E3)
Solving E1 to E3 gives a11 = a21 = a31 = 1/k1

26. Solution Next set F2=1 and F1=F3=0. Horizontal equilibrium of forces:
Mass m1: k1a12 = k2(a22 – a12) (E4)
Mass m2: k2(a22 – a12) = k3(a32 – a22) +1 (E5)
Mass m3: k3(a32 – a22) = 0 (E6)
Solving E4 to E6 gives
a12 = 1/k1
a22 = a32 = 1/k1 + 1/k2

27. Solution Next set F3=1 and F1=F2=0. Horizontal equilibrium of forces:
Mass m1: k1a13 = k2(a23 – a13) (E7)
Mass m2: k2(a23 – a13) = k3(a33 – a23) (E8)
Mass m3: k3(a33 – a23) = 1 (E9)
Solving E7 to E9 gives
a13 = 1/k1
a23 = 1/k1 + 1/k2
a33 = 1/k1 + 1/k2 + 1/k3

28. Inertial Influence coefficients, mij
mij is defined as the impulse applied at point i to produce a unit velocity at point j and zero velocity at every other point.
Total impulse at point i,
Matrix form:
where

29. Example 6.5
Find the inertial influence coefficients of the system shown below.

30. Solution
Let x(t) and θ(t) be the linear and angular position of the trailer and pendulum.
Apply impulse m11 and m21 such that
Linear impulse – linear momentum equations:
m11 = (M+m)(1)
m21 = m(1)

31. Solution Next apply impulse m12 and m22 such that
Linear impulse – linear momentum equations:
m12 = m(1)
m22 =
Inertial matrix:

33. Potential and Kinetic Energy expressions in Matrix form
Elastic potential energy of the ith spring, Vi=0.5Fixi
Total potential energy
Since
Matrix form:

34. Potential and Kinetic Energy expressions in Matrix form
Kinetic energy of mass mi:
Total KE of system:
Matrix form

35. Generalized coordinates and Generalized forces
n independent coordinates are needed to describe the motion of a n DOF system.
E.g. Consider the triple pendulum as shown.

36. Generalized coordinates and Generalized forces
(xj,yj) are constrained by the following:
(xj,yj) are not independent, thus they cannot be called generalized coordinates.
If angular displacements θj are used to specify the locations of the masses mj, there will be no constraints on θj.
Thus they form a set of generalized coordinates and are denoted by qj= θj,j=1,2,3

37. Generalized coordinates and Generalized forces
When external forces act on the system, the new system configuration is obtained by changing qj by δqj, j=1,2,…,n
The corresponding generalized force Qj=Uj/δqj, where Uj is the work done in changing qj by δqj.
Qj will be a moment when qj is an angular displacement.

38. Using Lagrange’s Equations to derive Equations of Motion
= qj/ t is the generalized velocity
Qj(n) is the non-conservative generalized force corresponding to qj.
Qj(n) can be computed as follows:
Fxk, Fyk and Fzk are the external forces acting on the kth mass in the x, y and z directions
xk ,yk and zk are the displacements of the kth mass in the x, y and z directions

39. Equations of Motion of Undamped systems in Matrix form
Lagrange’s Equations:
Fi is the non-conservative generalized force corresponding to ith generalized coordinate xi
Kinetic energy:
Potential energy:
where

40. Equations of Motion of Undamped systems in Matrix form
Differentiating T wrt
Differentiating wrt time:
T is a function of only
* wrt : with respect to

41. Equations of Motion of Undamped systems in Matrix form
Differentiating V wrt xi:
Substitute into Lagrange’s equations, we obtained the desired equations of motion:
where

42. Equations of Motion of Undamped systems in Matrix form
Note:
For a conservative system, Fi =0
Therefore equations of motion become
If xi are the same as the actual displacements, [m] is a diagonal matrix

43. Eigenvalue Problem
To solve , assume a solution of the form xi(t)=XiT(t), i=1,2,…,n
where Xi is constant and T is a function of time
Substituting xi(t)=XiT(t) into
we obtain
Rewrite as

44. Eigenvalue Problem Left side of equation is independent of i
Right side of equation is independent of t
Therefore both sides equal to a constant, which we named ω2.

45. Eigenvalue Problem Left hand side: Solution gives: T(t)=C1cos(ωt+Φ)
C1 is the amplitude
Φ is the phase angle
i.e. all coordinates can perform a harmonic motion with freqency ω and phase angle Φ

46. Eigenvalue Problem Right hand side: or (Eigenvalue problem)
For a non-trivial solution, determinant Δ of the coefficient matrix must be zero.
i.e. Δ= |kij-ω2mij| = |[k]- ω2[m]| =0 (Characteristic equation)
ω2 is the eigenvalue

48. Solution of the Eigenvalue Problem
Multiplying by [k]-1:
[I] is identity matrix
[D]=[k]-1[m] is dynamical matrix.
For a non-trivial solution of
characteristic determinant Δ=|λ[I]-[D]|=0
Use numerical methods to solve if DOF of system is large

49. Example 6.7
Find the natural frequencies and mode shapes of the system shown below for k1=k2=k3=k and m1=m2=m3=m

50. Solution Dynamical matrix [D]=[k]-1[m] ≡[a][m] Flexibility matrix
Mass matrix
Thus

51. Solution
Frequency equation:
Δ=|λ[I]-[D]|=
Dividing throughout by λ,

52. Solution
Once the natural freq are known, the eigenvectors can be calculated using

53. Solution 1st mode: Substitute λ1=5.0489 into (E1):
3 unknowns X1(1),X2(1), X3(1) in 3 equations
Can express any 2 unknowns in terms of the remaining one.

54. Solution X2(1) + X3(1) = 4.0489 X1(1) 3.0489 X2(1) – 2X3(1) = X1(1)
Solving the above, we get
X2(1)=1.8019X1(1) and X3(1)=2.2470X1(1)
Thus first mode shape
where X1(1) can be chosen arbitrarily.

55. Solution 2nd mode: Substitute λ2=0.6430 into (E1):
3 unknowns X1(2),X2(2), X3(2) in 3 equations
Can express any 2 unknowns in terms of the remaining one.

56. Solution –X2(2) – X3(2) = 0.3570X1(2) -1.3570X2(2) – 2X3(2) = X1(2)
Solving the above, we get
X2(2)=0.4450X1(2) and X3(2)= X1(2)
Thus 2nd mode shape
where X1(2) can be chosen arbitrarily.

57. Solution 3rd mode: Substitute λ3=0.3078 into (E1):
3 unknowns X1(3),X2(3), X3(3) in 3 equations
Can express any 2 unknowns in terms of the remaining one.

58. Solution -X2(3) - X3(3) = 0.6922X1(3) -1.6922X2(3) – 2X3(3) = X1(3)
Solving the above, we get
X2(3)= X1(3) and X3(3)=0.5544X1(3)
Thus 3rd mode shape
where X1(3) can be chosen arbitrarily.

59. Solution
When X1(1) = X1(2) = X1(3) =1, the mode shapes are as follows:

60. Orthogonality of Normal modes
Recall:
ωi and modal vector satisfy it such that
Eq.1
Another freq ωj and its will also satisfy it such that Eq.2
Pre-multiplying Eq.1 and Eq.2 by
respectively,

61. Orthogonality of Normal modes
Subtracting one from the other,
ωi2 ≠ ωj2 thus
Similarly we can obtain
This shows that are orthogonal wrt both [k] and [m]

62. Orthogonality of Normal modes
When i=j,
Generalized mass coefficient of the ith mode:
Generalized stiffness coefficient of the ith mode:
If we normalize
Kii reduces to

63. Example 6.8
Orthonormalize the eigenvectors of E.g. 6.7 with respect to the mass matrix.
Solution
For i=1, or

64. Solution
For i=2, or
For i=3,

65. Repeated Eigenvalues
Modal shapes are not unique when characteristic equation has repeated roots
Let λ1=λ2=λ and λ3 be a different eigenvalue.
Recall:

66. Repeated Eigenvalues Multiply Eq.4 by constant p and add to Eq.5:
This shows that the new mode shape also satisfies Eq.3
Hence the mode shape corresponding to λ is not unique.

67. Example 6.9
Determine the eigenvalues and eigenvectors of a vibrating system for which
Solution:
Eigenvalue equation

68. Solution Characteristic eq: |[k] – λ[m]|= λ2(λ-4)=0 λ1=0, λ2=0 , λ3=4
Using λ3=4, (E1) gives
-3X1(3) – 2X2(3) + X3(3)=0
-2X1(3) – 4X2(3) – 2X3(3)=0
X1(3) – 2X2(3) – 3X3(3)=0
If X1(3) =1,

69. Solution
λ1=λ2=0 indicates the system is degenerate. Using λ1=0 we obtain
X1(1) – 2X2(1) + X3(1)=0
-2X1(1) + 4X2(1) – 2X3(1)=0
All these eq are of the form X1(1)=2X2(1) – X3(1)
Hence

70. Solution
If X2(1) =1 and X3(1) =1,
If X2(1) =1 and X3(1) =-1,

71. Expansion Theorem
The eigenvectors are linearly independent and form a basis in the n-dimensional space.
Pre-multiplying by , the value of the constants ci can be determined.
If are normalized,
This is known as the expansion theorem.

72. Unrestrained Systems
Systems that are not attached to any stationary frame. E.g 2 moving railway cars
Equation of motion for free vibration:
For rigid body translation: ω=0, q(t)=α+βt where α and β are constants.
Let be the modal vector corresponding to rigid body mode.
Eigenvalue problem:

73. Unrestrained Systems When ω=0,
If system undergoes rigid body translation,
Therefore determinant of [k] must be 0, i.e. [k] is singular.
Thus potential energy
V is positive if
[k] will then be a positive semidefinite matrix.
When ω=0, we can have at most 6 rigid body modes, 3 for translation and 3 for rotation.

74. Example 6.10
3 freight cars are coupled by 2 springs as shown below. Find the natural frequencies and mode shapes of the system for m1=m2=m3=m and k1=k2=k.

75. Solution
Kinetic energy of the system:

76. Solution Potential energy of the system:
Elongation of spring k1=(x2-x1)
Elongation of spring k2=(x3-x2)
V=0 if x1= x2=x3=c (rigid body motion)

77. Solution Eigenvalue problem Set determinant of coefficient matrix = 0:
Expanding: m3ω6 – 4m2kω4 + 3mk2ω2=0
By setting
As m≠0,

78. Solution
For ω1=0, gives
If we fix X1(1)=1, we can solve for X2(1) and X3(1): X2(1)= X3(1)=1
Thus 1st mode

79. Solution For ω2=(k/m)1/2, gives
If we fix X1(2)=1, we can solve for X2(2) and X3(2): X2(2)=0, X3(2)= -X1(2)= -1
Thus 2nd mode

80. Solution For ω3=(3k/m)1/2, gives
If we fix X1(3)=1, we can solve for X2(3) and X3(3): X2(3)= -2X1(3) =-2, X3(3)= -0.5X2(3)= 1
Thus 3rd mode

81. Free Vibration of Undamped Systems
Equation of motion for free vibration of undamped system:
General solution:
If initial displacements and velocities are
Then

82. Example 6.11
Find the free vibration response of the spring-mass system shown below corresponding to the initial conditions x1(0)=x10, x2(0)=x3(0)=0. Assume that ki=k and mi=m for i=1,2,3.

83. Solution
(From Example 6.7)

84. Solution Applying the initial conditions:
A1cosΦ1+ A2cosΦ2+ A3cosΦ3 = x10
1.8019A1cosΦ A2cosΦ2–1.2468A3cosΦ3 =0
2.247A1cosΦ1–0.802A2cosΦ A3cosΦ3 =0

85. Solution Solving the above eqs gives: Thus solution of system:
A1=0.1076x10, A2=0.5431x10, A3=0.3493x10,
Φ1=Φ2=Φ3=0
Thus solution of system:

86. Forced Vibration of Undamped System using Modal Analysis
Forced vibration occur when external forces act on a system.
Equations of motion is a set of 2nd order ODE:
Eq.6
Modal analysis is a more convenient method to solve this when the DOF is large.
First we must solve the eigenvalue problem and find the natural freq. and the corresponding normal modes.

87. Forced Vibration of Undamped System using Modal Analysis
Eigenvalue problem:
According to expansion theorem, solution of Eq.6:

88. Forced Vibration of Undamped System using Modal Analysis
Since [X] is not a function of time,
Rewrite Eq.6 as
Pre-multiplying by [X]T,
Normalizing the normal modes:
[X]T[m][X]=[I], [X]T[k][X]=[ ω2 ]
Define generalized forces
Therefore
Solution of this 2nd order ODE:

89. Forced Vibration of Undamped System using Modal Analysis

90. Example 6.12
Using modal analysis, find the vibration response of a 2 DOF system with equations of motion:
Assume the following data:m1=10, m2=1, k1=30, k2=5, k3=0 and

91. Solution
Orthogonalizing normal modes wrt mass matrix:

92. Solution E1 can be expressed as Solution
Using the initial conditions, we can find

93. Solution

94. Forced Vibration of Viscously Damped Systems
Modal analysis only applies to undamped sys
Viscously damped system is opposed by a force proportional to velocity but in the opposite direction
We shall consider the equations of motion of viscously damped systems using Lagrange’s equations

95. Forced Vibration of Viscously Damped Systems
Rayleigh’s dissipation function:
[c] is the damping matrix
Lagrange’s equations:

96. Forced Vibration of Viscously Damped Systems
Equations of motion of damped system:
Special case: Proportional damping
[c]=α[m]+β[k] where α and β are constants.
Equation of motion becomes:

97. Forced Vibration of Viscously Damped Systems
Pre-multiplying by [X]T:
By normalizing , we obtain
By writing α+ωi2β=2ζiωi where ζi is the modal damping ratio for the ith normal mode,

98. Forced Vibration of Viciously Damped Systems
Each of the n equations is uncoupled from all of the others.
Hence solution when ζi<1 is

99. Example 6.13
Derive the equations of motion of the system as shown:

100. Solution
Use Lagrange’s equations with Rayleigh’s dissipation function.
Kinetic energy:
Potential energy:
Rayleigh’s dissipation function:
Lagrange’s equations:

101. Solution
Substituting (E.1) to (E.3) into (E.4), differential equations of motion:

102. Self-Excitation and Stability Analysis
Friction leads to negative damping, causing system instability (self-excited vibration).
Equations of motion for the system shown below is as follows:

103. Self-Excitation and Stability Analysis
We assume solution of the form
s is a complex number to be determined
Cj is the amplitude of xj
Real part of s determines the damping.
Imaginary part gives the natural frequency.

104. Self-Excitation and Stability Analysis
For free vibration
For nontrivial solution of Cj, determinant of coefficients of Cj is set to zero:
Expanding,
Let the roots be sj=bj+iωj, j=1,2,…,m

105. Self-Excitation and Stability Analysis
If bj are all -ve, system is stable.
If one or more bj are +ve, system is unstable.
If bj=0, system is at borderline between stability and instability

106. Self-Excitation and Stability Analysis
Solution to this is lengthy.
Simplified procedure: Routh-Hurwitz stability criterion
Define the following mth order determinant Tm:

107. Self-Excitation and Stability Analysis
Define the following subdeterminants:
Replace all ai with i>m or i<0 by zeros.
A necessary and sufficient condition for stability is that all coefficients a0, a1,…,am and all determinants T1,T2,…,Tm must be +ve (Routh-Hurwitz stability criterion)