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

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

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

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

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.

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

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

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

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

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:

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

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

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

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

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

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:

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.

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

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 + k11 (E1) Mass m2: k21 = -k2 (E2) Mass m3: k31 = 0 (E3) Solving E1 to E3 gives k11=k1+k2 k21 = -k2 k31 = 0

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 = - k3 (E6) Solving E4 to E6 gives k22 = k2+k3 k12 = -k2 k32 = -k3

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

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:

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.

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

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

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

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

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

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

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)

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

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

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

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.

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

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.

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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.

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.

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.

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)=-0.8020X1(2) Thus 2nd mode shape where X1(2) can be chosen arbitrarily.

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.

Solution -X2(3) - X3(3) = 0.6922X1(3) -1.6922X2(3) – 2X3(3) = X1(3) Solving the above, we get X2(3)=-1.2468X1(3) and X3(3)=0.5544X1(3) Thus 3rd mode shape where X1(3) can be chosen arbitrarily.

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

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,

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]

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

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

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

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

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.

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

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,

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

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

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.

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:

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.

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.

Solution Kinetic energy of the system:

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)

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

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

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

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

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

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.

Solution (From Example 6.7)

Solution Applying the initial conditions: A1cosΦ1+ A2cosΦ2+ A3cosΦ3 = x10 1.8019A1cosΦ1+0.445A2cosΦ2–1.2468A3cosΦ3 =0 2.247A1cosΦ1–0.802A2cosΦ2+0.5544A3cosΦ3 =0

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:

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.

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

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:

Forced Vibration of Undamped System using Modal Analysis

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

Solution Orthogonalizing normal modes wrt mass matrix:

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

Solution

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

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

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:

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,

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

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

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

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

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:

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.

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

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

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

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)