Handout 7 (Undamped) Modal Analysis of MDOF Systems

ME617 - Handout 7 (Undamped) Modal Analysis of MDOF Systems

The governing equations of motion for a n-DOF linear mechanical system with viscous damping are:

MU+DU+KU()tt =F () (1)

where U,U,and U are the vectors of generalized displacement,

velocity and acceleration, respectively; and F()t is the vector of generalized (external forces) acting on the system. M,D,K represent the matrices of inertia, viscous damping and stiffness coefficients, respectively1.

The solution of Eq. (1) is uniquely determined once initial conditions are specified. That is,

att =→ 0UUUU(0) =oo , (0) = (2)

In most cases, i.e. conservative systems, the inertia and stiffness matrices are SYMMETRIC, i.e. MM= TT, KK= . The kinetic energy (T) and potential energy (V) in a conservative system are

11 TV==UMUTT, UKU (3) 22

1 The matrices are square with n-rows = n columns, while the vectors are n- rows.

MEEN 617 – HD#7 Undamped Modal Analysis of MDOF systems. L. San Andrés © 2008 1 In addition, since T > 0, then M is a positive definite matrix2. If V >0, then K is a positive definite matrix. V=0 denotes the existence of a rigid body mode, and makes K a semi-positive matrix.

In MDOF systems, a natural state implies a certain configuration of shape taken by the system during motion. Moreover a MDOF system does not possess only ONE natural state but a finite number of states known as natural modes of vibration. Depending on the initial conditions or external forcing excitation, the system can vibrate in any of these modes or a combination of them. To each mode corresponds a unique frequency knows as a natural frequency. There are as many natural frequencies as natural modes.

The modeling of a n-DOF mechanical system leads to a set of n- coupled 2nd order ODEs, Hence the motion in the direction of one DOF, say k, depends on or it is coupled to the motion in the other degrees of freedom, j=1,2…n.

In the analysis below, for a proper choice of generalized coordinates, known as principal or natural coordinates, the system of n-ODE describing the system motion is independent of each other, i.e. uncoupled. The natural coordinates are linear combinations of the (actual) physical coordinates, and conversely. Hence, the motion in physical coordinates can be construed or interpreted as the superposition or combination of the motions in each natural coordinate.

2 Positive definite means that the determinant of the matrix is greater than zero. More importantly, it also means that all the matrix eigenvalues will be positive. A semi-positive matrix has a zero determinant, with at least an eigenvalues equaling zero.

MU+KU()tt =F () (4) and att =→ 0UUUU(0) =oo , (0) = Presently, set the external force F=0, and let’s find the free vibrations response of the system.

MU+KU=0 (5)

The solution to the homogenous Eq. (5) is simply

U= φ cos(ω t −θ ) (6)

which denotes a periodic response with a typical frequency ω . From Eq. (6), 2 U= −φ ω cos(ωθt − ) (7)

Note that Eq. (6) is a simplification of the more general solution

st U= φ esiwith= ω and where i=− 1 (8)

Substitution of Eqs. (6) and (7) into the EOM (5) gives: MU+KU=0 → →−Mφ ωωθ2 cos(tt − )+Kφ cos( ωθ − )=0

2 →−⎣⎦⎡⎤M+Kωωθφ cos(t − )=0 and since cos(ω t −≠θ ) 0 for most times, then

Eq. (10) is usually referred as the standard eigenvalue problem (mathematical jargon):

Aφφ=λ (11) where Α=MK−12 andλ = ω

Eq.(9) is a set of n-homogenous algebraic equations. A nontrivial solution, φ ≠ 0 exists if and only if the determinant Δ of the system of equations is zero, i.e.

Δ= −M+Kω 2 =0 (12)

Eq. (12) is known as the characteristic equation of the system. It is a polynomial in ωλ2 = , i.e. 246 n Δ=0 =aa01 +ω + a 2ωω + a 3 + .... an ω n (13) i Δ=0 =aa0 +∑()i λ i=1 This polynomial or characteristic equations has n-roots, i.e. the set λ or ±ω since ω =± λ . {}k kn=1,2,.... {}k kn=1,2,....

The ω’s are known as the natural frequencies of the system. In the MATH jargon, the λ’s are known as the eigenvalues (of matrix A)

b) If M and K are positive definite, then

0<≤ω12ωωω ...... nn− 1 ≤ .

c) If K is semi-positive definite, then

0=≤ω12ωωω ...... nn− 1 ≤ , i.e. at least one natural frequency is zero, i.e. motion with infinite period. This is known as rigid body mode.

Note that each of the natural frequencies satisfies Eq. (9). Hence, associated to each natural frequency (or eigenvalues) there is a corresponding natural mode vector (eigenvector) such that

[−M+Kλiiin] φ()=0, = 1,... (14)

The n-elements of an eigenvector are real numbers (for undamped system), with all entries defined except for a constant. The eigenvectors are unique in the sense that the ratio between two elements is constant, i.e. ⎛⎞ϕ ()k j ==constant for any ji , 1,.... n ⎜⎟ϕ ⎝⎠()k i

The actual value of the elements in the vector is entirely arbitrary. Since Eq. (14) is homogenous, if φ is a solution, so it is αφ for any arbitrary constant α. Hence, one can say that the SHAPE of a natural mode is UNIQUE but not its amplitude.

For MDOF systems with a large number of degrees of freedom, n>>3, the eigenvalue problem, Eq. (11), is solved numerically.

MEEN 617 – HD#7 Undamped Modal Analysis of MDOF systems. L. San Andrés © 2008 5 Nowadays, PCs and mathematical computation software allow, with a single (simple) command, the evaluation of all (or some) eigenvalues and its corresponding eigenvectors in real time, even for systems with thousands of DOFs.

Long gone are the days when the graduate student or practicing engineer had to develop his/her own efficient computational routines to calculate eigenvalues. Handout # 9 discusses briefly some of the most popular numerical methods to solve the eigenvalue problem.

A this time, however, let’s assume the set of eigenpairs ω , φ is known. {}ii() in=1,2...

Properties of natural modes The natural modes (or eigenvectors) satisfy important orthogonality properties. Recall that each eigenpair ω ,φ satisfies the equation {}ii() in=1,2... ⎡⎤2 . (15) ⎣⎦−M+Kωiiinφ()=0, = 1,...

Consider two different modes, say mode-j and mode-k, each satisfying

22 ωωj Mφ()jj==K φ ()and kkkMφ ()K φ () (16)

T T Pre-multiply the equations above by φ()k and φ()j to obtain

Now, perform some matrix manipulations. The products φ TMφ and φ TKφ are scalars, i.e. not a matrix nor a vector. The transpose of a scalar is the number itself. Hence,

TT TTT ()φ()jkK φ ()=()K φφ () k( () j) TT = φ()kjK φ () TT = φ()kjK φ () since K=K and T TT T ()φ()jkMφφ ()= () kMφ () jsince M=M

for symmetric systems. Thus, Eqs. (17) are rewritten as

2 TT ω jjφ()Mφφ () k= () jK φ () k()a and (18) 2 TT ωkjφ()Mφφ () k= () jK φ () k()b

Subtract (b) from (a) above to obtain

22T ()ωωjk−=φ() jMφ () k0 (19)

if ω jk≠ ω , i.e. for TWO different natural frequencies; then it follows that

T T 2 for jk= φ()j Mφ ()jj= M and φ()j K φ ()jjjj==KMω (20)

where Kj and Mj are known as the j-modal stiffness and j-modal mass, respectively.

Define a modal matrix Φ has as its columns each of the eigenvectors, i.e.

Φφ=[ 12 φ.. φn ] (21)

and the modal properties are written as

TT Φ MΦΦ==[M ]; K Φ [K ] (22)

where [M] and [K] are diagonal matrices containing the modal mass and stiffnesses, respectively.

The eigenvector set φ k=1,…n is linearly independent. Hence, any vector (v) in n-dimensional space can be described as a linear combination of the natural modes, i.e. n v ==∑a jjφΦ() a (23) j=1

⎡⎤a1 ⎢⎥a v =+φφaa ++=.. φ a[] φφφ .. ⎢⎥2 = Φa 11 2 2nn 1 2 n⎢⎥.. ⎢⎥ ⎣⎦an

The orthogonality property of the natural modes (eigenvectors) permits the simplification of the analysis for prediction of system response. Recall that the equations of motion for the undamped system are MU+KU()tt =F () (4) and att =→ 0UUUU(0) = 0 , (0) = 0

3 Consider the modal transformation U()tt= Φ q () (24) And with U()tt= Φ q (), then EOM (4) becomes:

MΦq+K Φq=F()t

which offers no advantage in the analysis. However, premultiply the equation above by ΦT to obtain

TTT ()Φ MΦ q+(Φ K Φ)q=Φ F()t (25) and using the properties of the natural modes, ΦTTMΦΦ==[M ]; K Φ [K ], then Eq. (25) becomes

T [MK]q+ [ ] q=Q=Φ F()t (26)

3 Eq. (24) sets the physical displacements U as a function of the modal coordinates q. This transformation merely uses the property of linear independence of the natural modes.

M11qKqQ += 11 1 M qKqQ += 22 22 2 (27) .....

M nnqKqQ += nn n

Or Mq += Kq Q withω =K j , (28) j jjjj nj M j jn=1,2...

The set of q’s are known as modal or natural coordinates (canonical or principal, too). The vector T Q =Φ F()t is known as the modal force vector.

Thus, the major advantage of the modal transformation (24) is that in modal space the EOMS are uncoupled. Each equation describes a mode as a SDOF system.

The unique solution of Eqs. (28) needs of initial conditions

specified in modal space, i.e. {qqoo, }. Using the modal transformation, Uoooo==ΦqU; Φq , it follows −−11 qoooo==Φ Uq; Φ U (28)

However, Eq. (28) requires of the inverse of modal matrix Φ, i.e. ΦΦ−1 = I . For systems with a large number of DOF, n>> 1, finding the matrix Φ−1 is computationally expensive.

MEEN 617 – HD#7 Undamped Modal Analysis of MDOF systems. L. San Andrés © 2008 10 A more efficient to determine the initial state {qqoo, }in modal coordinates follows. Start with the fundamental transformation, T Uoo=Φq , and premultiply this relationship by Φ M to obtain,

TT Φ MUoo=Φ MΦq since [M ] = ΦT MΦ , hence =[]M qo ,

−1 T qoo=[M ] Φ MU , (29a) −1 T q oo=[]M Φ MU or 11 qq==φTTMU , φ MU okookokk()() ()() (29b) MMkk

Eqs. (29) are much easier to calculate efficiently when n-DOF is large. Note that finding the inverse of the modal mass matrix [M]-1 is trivial, since this matrix is diagonal.

Comparing eqs. (28) and (29a) it follows that

−1 −1 T ΦΦ= [M ] M (30)

The solution of ODEs M jjqKqQ + jj= j with initial conditions qq, follows an identical procedure as in the {}oojj solution of the SDOF response. That is, each modal response adds the homogeneous solution and the particular solution. The particular solution clearly depends on the time form of the modal

Free response in modal coordinates Without modal forces, Q=0, the modal equations are

M jHjqKqQ +== jHj0 j (31a)

with solutions, for an elastic mode

qo qq=+cosωω tj sin t if ω ≠ 0 (31b) Hj ojj() nω () n j n j n j ; and for a rigid body mode

qqqt=+ if ω = 0 (31c) Hj ojj o n j j=1,2,….n

Forced response in modal coordinates

For step-loads, QSj, the modal equations are

M jjqKqQ + jj= Sj (32a)

and; for an elastic mode, ω ≠ 0, n j

qQoS qq=+cosωω tjj sin t +−⎡ 1 cos ω t⎤ (32a) jojj() nω () n jK ⎣ () n j⎦ njj

1 QS qq=+ qt + j t2 (32c) jojj o 2 M j j=1,2,….n For periodic loads,, the modal equations are

M qKqQ + =Ωcos( t ) (33a) jj jj Pj with solutions for an elastic mode, ω ≠ 0, and Ω ≠ ω n j n j

⎡⎤ QP 1 qC=++cosωω tS sin tj ⎢⎥ cos() Ω t jj() njj j () n ⎢⎥2 K j 1−Ωω ⎣⎢⎥()n j ⎦ (33b)

Note that if Ω=ω , a resonance appears that will lead to system n j destruction. For a rigid body mode, ω = 0, n j

QP qq=+ qt −j cos( Ω t ) (33c) jojj o 2 M jΩ

For arbitrary-loads Qj , the modal response is q 1 t qq=++cos(ω t )jo sin(ωωττ t ) Q sin⎡ ( t − )⎤ d jjoj nωω n jM ∫0 j()τ ⎣ n j⎦ njnjj (34) for an elastic mode, ω ≠ 0. n j

MEEN 617 – HD#7 Undamped Modal Analysis of MDOF systems. L. San Andrés © 2008 13 System Response in Physical Coordinates Once the response in modal coordinates is fully determined, the system response in physical coordinates follows using the modal transformation

U()tt= Φ q () = q ⎡⎤1()t ⎢⎥ ⎢⎥q2 U()tn==+++[]φφ 1 2.. φ φ 1qq 1 φ 2 2 .. φ nn q ⎢⎥.. ⎢⎥ ⎣⎢⎥qn ⎦

n U = φ q ()tjj∑ ()t (35) j=1

One important question follows: are all the modal responses important and need be accounted for to obtain the response in physical coordinates? If not, savings in computation time are evident. Hence, the physical response becomes

m U ≈<φ qmn, ()tjj∑ ()t (36) j=1

If m

MEEN 617 – HD#7 Undamped Modal Analysis of MDOF systems. L. San Andrés © 2008 14 Example: Consider the case of force excitation with frequency Ω≠ω and acting for very long times. The EOMs in physical n j space are MU+KU=FP cos(Ωt)

Let’s assume there is a little damping; hence, the steady state periodic response in modal coordinates is (see eq. (33b)):

⎡⎤ QP 1 qt≈Ωj ⎢⎥cos (37a) j ⎢⎥2 () K j 1−Ωω ⎣⎦⎢⎥()n j And thus,

n ⎛⎞⎡⎤ QP 1 UU=Ω==cos(tt )Φq ⎜⎟φ j ⎢⎥ cos() Ω P ∑⎜⎟j ⎢⎥2 j=1 ⎜⎟K j 1−Ωω ⎝⎠⎣⎦⎢⎥()n j (38) The physical response is also periodic with same frequency as the force excitation.

Recall that KM==ω 2 φT K φ and Q = φT F jnjj () j () j Pjj () P

However, nowadays the engineer in a hurry prefers to dump the

problem into a super computer; and for U=UP cos(Ωt ) , finds the solution −1 ⎡⎤2 U=KP ⎣ −Ω M⎦ FP (39) at a fixed excitation frequency Ω. Brute force substitutes beauty and elegance, time savings in lieu of understanding!

MEEN 617 – HD#7 Undamped Modal Analysis of MDOF systems. L. San Andrés © 2008 15 Example: Find natural frequencies and natural ORIGIN := 1 mode shapes of UNDAMPED system. Given EOMs for a 2DOF - undamped- system:

M 0 X 2⋅K −2K X ⎜⎛ 2 ⎟⎞ d2 ⎜⎛ 2 ⎟⎞ ⎜⎛ 2 2 ⎟⎞ ⎜⎛ 2 ⎟⎞ ⎛ 0 ⎞ ⋅ + ⋅ = ⎜ ⎟ (1) ⎜ ⎟ 2 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ K ⋅Z ⎝ 0 M1 ⎠ dt ⎝ X1 ⎠ ⎝ −2⋅K2 2⋅K2 + K1 ⎠ ⎝ X1 ⎠ ⎝ 1 ⎠

where M2=mo, M1=5 mo, K2=ko; K1=5 ko

m 0 X 2⋅k −2k X ⎜⎛ o ⎟⎞ d2 ⎜⎛ 2 ⎟⎞ ⎜⎛ o o ⎞⎟ ⎜⎛ 2 ⎟⎞ ⎛ 0 ⎞ ⋅ + ⋅ = ⎜ ⎟ ⎜ ⎟ 2 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ K ⋅Z ⎝ 0 5⋅mo ⎠ dt ⎝ X1 ⎠ ⎝ −2⋅ko 2⋅ko + 5⋅ko ⎠ ⎝ X1 ⎠ ⎝ 1 ⎠ (a) PROCEDURE TO FIND NATURAL FREQUENCIES AND NATURAL MODES: Assume the motions are periodic with frequency ω, ie

X2= a1⋅cos()ω⋅t X1= a2⋅cos()ω⋅t (2) Set the RHS of Eq. (1) equal to 0. Substitution of (2) into (1) gives

2 ⎛ 2⋅k − m ⋅ω −2k ⎞ a ⎜ o o o ⎟ ⎜⎛ 1 ⎟⎞ ⎛ 0 ⎞ ⋅ ⋅cos()ω⋅t = ⎜ ⎟ ⎜ 2 ⎟ ⎜ a ⎟ ⎝ 0 ⎠ ⎝ −2⋅ko 7⋅ko − 5⋅mo⋅ω ⎠ ⎝ 2 ⎠ cancel cos(ωt) since it is NOT zero for all times The homogeneous system of eqns

2 ⎛ 2⋅k − m ⋅ω −2k ⎞ a ⎜ o o o ⎟ ⎜⎛ 1 ⎞⎟ ⎛ 0 ⎞ ⋅ = ⎜ ⎟ ⎜ 2 ⎟ ⎜ a ⎟ ⎝ 0 ⎠ (3) ⎝ −2⋅ko 7⋅ko − 5⋅mo⋅ω ⎠ ⎝ 2 ⎠ has a non-trivial solution if the determinant of the system of equations equals zero, i.e. if

2 2 2 Δω()= ()7⋅ko − 5⋅mo⋅ω ⋅()2⋅ko − mo⋅ω − 4⋅ko = 0

2 Let λω= , and expanding the products in the determinant

2 2 2 2 0 = λ ⋅5mo − λ⋅()7⋅ko⋅mo + 10⋅ko⋅mo + 14⋅ko − 4⋅ko

Let ⎯ m ⎯2 ⎯ ⎛ o⎞ Leads to: ⎡ ⎤ λ = λ⋅⎜ ⎟ 0 = ⎣a⋅()λ + b⋅λ + c⎦ (4) ⎝ ko ⎠ with: a := 5 b := −17 c := 10

The roots (eigenvalues) of the characteristic equation are 2 0.5 2 0.5 −b − ()b − 4⋅a⋅c −b + ()b − 4⋅a⋅c 0.757 ⎛ ko ⎞ λ1 := λ2 := ⎛ ⎞ 2⋅a 2⋅a λ = ⎜ ⎟ ⎜ ⎟ ⎝ 2.643 ⎠ ⎝ mo⎠

and the natural frequencies are:

0.5 0.5 k 0.5 ω1 := ()λ1 ω2 := ()λ2 ⎛ 0.87 ⎞ ⎛ o ⎞ ω = ⎜ ⎟ ⎜ ⎟ ⎝ 1.626 ⎠ ⎝ mo⎠ Find the eigenvectors: The two equations in (3) are linearly dependent. Thus, one cannot solve for a1 and a2. Set φ1 := 1 arbitrarily; and from the first equation for ω 1 2 ()2⋅ko − mo⋅ω1 ()2⋅ko − 0.757⋅ko ()2− 0.757 φ2 := φ2 = = 2 2⋅ko 2⋅ko φ2 = 0.621

φ1 := φ ⎛ 1 ⎞ is the first eigenvector (natural mode) φ1 = ⎜ ⎟ ⎝ 0.621 ⎠

(b) Explanation: DOF1 (X2) and DOF2 (X1) move in phase, with X2>X1

for ω φ := 1 2 2 1 ()2⋅ko − mo⋅ω ()2⋅k − 2.643⋅k 2 o o ()2− 2.643 φ2 = = φ := 2⋅ko 2⋅ko 2 2 φ2 := φ ⎛ 1 ⎞ φ2 = ⎜ ⎟ is the 2nd eigenvector (natural mode) ⎝ −0.321 ⎠

(b) Explanation:DOF1 (X2) and DOF2 (X1) move 180 deg OUT of phase, with |X2|>|X1|

(c) find the numerical value for each natural frequency: 1000lb 5 lb 0.5 m := ko := 10 ⋅ Since ⎛ 0.87 ⎞ ⎛ ko ⎞ o g in ω := ⎜ ⎟⋅⎜ ⎟ ⎝ 1.626 ⎠ ⎝ mo⎠ 0 Note that mass must be ⎛ 170.947 ⎞ rad ω = ⎜ ⎟ expressed in physical units ⎝ 319.495 ⎠ sec consistent with the problem, i.e. ω 2 fn := ⎛ 27.207 ⎞ lb⋅ sec 2⋅π fn = ⎜ ⎟ Hz mo = 2.59 ⎝ 50.849 ⎠ in Perform same work using a calculator Use BUILT IN functions - ⎛ 1 0 ⎞ ⎛ 2 −2 ⎞ Not much learning M := ⎜ ⎟⋅mo K := ⎜ ⎟⋅ko ⎝ 0 5 ⎠ ⎝ −2 7 ⎠

− 1 Let ZM:= ⋅K

λ := sort() eigenvals() Z ⎛ 2.921× 104 ⎞ 1 λ = ⎜ ⎟ 5 2 ⎝ 1.021× 10 ⎠ sec .5 ω1 := ()λ1 ⎛ 170.914 ⎞ rad ω ⎛ 27.202 ⎞ .5 ω = ⎜ ⎟ = ⎜ ⎟ Hz ω2 := ()λ2 ⎝ 319.495 ⎠ sec 2⋅π ⎝ 50.849 ⎠

natural modes: φ1 := eigenvec() Z ,λ1 0.849 ⎛ ⎞ ()φ1 φ1 = ⎜ ⎟ 2 ⎝ 0.528 ⎠ = 0.622 φ1 ()1 φ2 := eigenvec() Z ,λ2 ⎛ 0.952 ⎞ φ2 = ⎜ ⎟ ⎝ −0.306 ⎠ φ2 ()2 = −0.322 φ2 ()1 which are the same ratios as for the vectors found earlier 1000lb 5 lb ft Example: Undamped Modal Analysis m := k := 10 ⋅ g= 32.174 ORIGIN:= 1 o g o in 2 sec Equations of motion: natural frequencies, modal matrix (eigenvectors)

m 0 2 X 2k⋅ −2k X ⎛ 170.95 ⎞ rad ⎛ 1 1 ⎞ ⎜⎛ o ⎟⎞ d ⎜⎛ 2 ⎟⎞ ⎜⎛ o o ⎟⎞ ⎜⎛ 2 ⎟⎞ ⎛ 0 ⎞ ω := ⎜ ⎟⋅ Φ := ⎜ ⎟ ⋅ + ⋅ = ⎜ ⎟ n ⎜ ⎟ 2⎜ ⎟ ⎜ ⎟ ⎜ ⎟ k ⋅Z ⎝ 319.5 ⎠ sec ⎝ 0.621 −0.321 ⎠ ⎝ 0 5m⋅ o ⎠ dt ⎝ X1 ⎠ ⎝ −2⋅ko 7k⋅ o ⎠ ⎝ X1 ⎠ ⎝ o ⎠

given: Zo := 0.01⋅ in provides a Fo := ko⋅Zo constant force Define matrices: m 0 2k⋅ −2k ⎜⎛ o ⎟⎞ ⎜⎛ o o ⎟⎞ ⎛ 0lb⋅ ⎞ at t=0s, Initial conditions: M := K := F := ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ F system is at REST ⎝ 0 5m⋅ o ⎠ ⎝ −2⋅ko 7k⋅ o ⎠ ⎝ o ⎠ T T (a) FIND modal masses and stiffnesses MM := Φ ⋅ΦM⋅ KM := Φ ⋅ΦK⋅

− 3 2 ⎛ 7.584 8.534× 10 ⎞ lb⋅ sec non-diagonal elements are very small= non zero b/c of M = ⎜ ⎟ M ⎜ − 3 ⎟ in roundoff in numerical calculator ⎝ 8.534× 10 3.925 ⎠ 2 modal masses and stiffnesses: Mode 1 Mm := MM Km := ⎛ωn ⎞ ⋅MM 1 11, 1 ⎝ 1⎠ 11, M := M Mode 2 m2 M22, 2 Km := ⎛ωn ⎞ ⋅MM 2 ⎝ 2⎠ 22, 2 ⎛ 5 ⎞ ⎛ 7.584 ⎞ lb⋅ sec ⎜ 2.216× 10 ⎟ lb Mm = ⎜ ⎟ Km = ⎝ 3.925 ⎠ in ⎜ 5 ⎟ in ⎝ 4.006× 10 ⎠ (b) Find initial moddal displacements and velocities and modal force vector (Q) At time t=0s, the system is at REST at its static equilibrium position, hence the initial conditions are null displacements and null velocities. Of course, the same applies to modal space, i.e. null initial displacements and velocities ⎛ 0 ⎞ X1 ⎛ 0 ⎞ ft for generality, define: Xo := ⎜ ⎟⋅ft Vo := ⎜ ⎟⋅ Calculate inverse of A matrix − 1 ⎝ 0 ⎠ X2 ⎝ 0 ⎠ sec Φinv := Φ and in modal coordinates qo := Φinv⋅Xo qo_dot := Φinv⋅Vo velocity (disp & velocities) ⎛ 0 ⎞ ⎛ 0 ⎞ ft No need for actual calculation q = ⎜ ⎟ ft q = ⎜ ⎟ o ⎝ 0 ⎠ o_dot ⎝ 0 ⎠ sec - a knowledge statement suffices Define modal force T Q := Φ ⋅F ⎛ 621 ⎞ Q = ⎜ ⎟ lb Both natural modes will be excited ⎝ −321 ⎠ (c) Modal EOMs and modal responses ⎛ 2 ⎞ ⎜ d ⎟ The EOMs in modal space are uncoupled M q + K ⋅q = Q m ⎜ 2 i⎟ m i i i1= , 2 and equal to i⎝ dt ⎠ i Using the cheat sheet, and since the Initial conditions Q Q 1 2 are null, the response in modal coordinates are δ := δ := m1 K m2 K m1 m2 ⎛ 170.95 ⎞ rad q1()t := δm ⋅⎛1− cos⎛ωn ⋅t⎞⎞ q2()t := δm ⋅⎛1− cos⎛ωn ⋅t⎞⎞ ωn = ⎜ ⎟ 1 ⎝ ⎝ 1 ⎠⎠ 2 ⎝ ⎝ 2 ⎠⎠ where: ⎝ 319.5 ⎠ sec − 3 ⎛ 2.802× 10 ⎞ are the "static" deflections in modal space. δ2 << δ1, thus first where δ = ⎜ ⎟ in m ⎜ − 4 ⎟ modal response is MORE important ⎝ −8.013 × 10 ⎠ (d) The response in physical coordinates, X1 and X2, equals (from transformation x=Aq) with ⎛ 1 1 ⎞ X1()t := q1()t + q2()t Φ = ⎜ ⎟ X1()t = δm ⋅δ⎛1− cos⎛ωn ⋅t⎞⎞ + m ⋅⎛1− cos⎛ωn ⋅t⎞⎞ ⎝ 0.621 −0.321 ⎠ 1 ⎝ ⎝ 1 ⎠⎠ 2 ⎝ ⎝ 2 ⎠⎠

X2(t ):= 0.621⋅ q1()t − 0.321⋅ q2()t for graph below: 1 ⎛ 2⋅π ⎞ X2()t = δm ⋅0.621⋅δ⎛1− cos⎛ωn ⋅t⎞⎞ + m ⋅()−0.321 ⋅⎛1− cos⎛ωn ⋅t⎞⎞ Tlarge := 10⋅⎜ ⎟ 1 ⎝ ⎝ 1 ⎠⎠ 2 ⎝ ⎝ 2 ⎠⎠ ⎜ ωn ⎟ ⎝ 1 ⎠ − 4 − 3 δ ⋅()−0.321 = 2.572× 10 in δ ⋅0.621 = 1.74× 10 in m2 m1

Explanation: Since q1 and q2 are non-zero, then physical motion, X1 &X2, shows excitation of the TWO fundamental modes of vibration - BUT response for second mode is much less

GRAPHs not needed for exam: Note that there is no damping or 0.006 attenuation of motions. 0.005 0.004 Not too complicated physical response. It 0.003 shows dominance of first mode (lowest natural freq or largest period) 0.002 0.001 2⋅π 0 = 0.037 sec displacements (inch) displacements ω 0.001 n1 0.002 2⋅π 0 0.046 0.092 0.14 0.18 0.23 0.28 0.32 0.37 = 0.02 sec ω time (sec) n2 X1 X2 terminal value Terminal condition: If damping is present and since the applied force is a constant, the system will achieve a new steady state condition.

In the limit as t approaches very, very large values 2 X X d ⎜⎛ 1 ⎟⎞ ⎛ 0 ⎞ ⎜⎛ 1end⎟⎞ = ⎜ ⎟ ; hence ===> K⋅ = F 2⎜ ⎟ ⎜ ⎟ dt ⎝ X2 ⎠ ⎝ 0 ⎠ ⎝ X2end⎠ And equations of motion reduce to: 2k⋅ −2k X ⎜⎛ o o ⎟⎞ ⎜⎛ 1end⎞⎟ ⎛ 0 ⎞ ⋅ = ⎜ ⎟ 3 ⎜ ⎟ ⎜ ⎟ F F = 110× lb ⎝ −2⋅ko 7k⋅ o ⎠ ⎝ X2end⎠ ⎝ o ⎠ o

And solving this system of equations using Cramer's rule 2 2 Δ := 14⋅ ko − 4k⋅ o determinant of system of eqns. Fo⋅2⋅ko 2k⋅ o⋅Fo X1end := X2end := Δ Δ − 3 − 3 X1end = 210× in X2end = 210× in

Note that the graph of undamped periodic motions Z(t) and X(t) shows oscillatory motions abut these terminal or end values. recall ⎛ − 3 ⎞ Zo = 0.01 in OR − 1 210× K ⋅F = ⎜ ⎟ in ⎜ − 3 ⎟ Zo ⎝ 210× ⎠ = 5 X1end 0.006 COMPARE actual 0.005 response with a 0.004 response 0.003 neglecting q2. Indeed mode 2 does 0.002 not afffect the physical 0.001 0 response, except for (inch) displacements motion X2 sligthly 0.001 0.002 0 0.12 0.25 0.37 time (sec) X1 X2 terminal

Normalization of eigenvectors (natural modes)

Recall that the components of an eigenvector φ j are ARBITRARY but for a multiplicative constant. If one of the elements of the eigenvector is assigned a certain value, then this vector becomes unique, since then n-1 remaining elements are automatically adjusted to keep constant the ratio between any two elements in the vector.

In practice, the eigenvectors are normalized. The resulting vectors are called NORMAL MODES.

Some typical NORMS are

L1 norm: q ==1maxq (39a) ()jj( k )

22 2 L2 norm: q ==1qq + + .... + q (39b) ()jjjj12 n

Or making the mass modal matrix equal to the identity matrix, [M]=I, i.e.

T φ()jjjMφ ()==M 1 (39c) hence T 22 φ()j K φ ()jjjjnj==KMω =ω (39d)

This normalization has obvious advantages since it will reduce the number of operations when conducting the modal analysis. However, the physical significance of the modal equations is lost. Note that the modal Eqs. (26) become:

Your lecturer recommends this normalization procedure be conducted only for systems with large number of degrees of freedom, n>>>1.

Note that the normalization process is a mere convenience, devoid of any physical significance.

Rayleigh’s Energy Method The method is a procedure to determine an approximate value (from above) for the fundamental natural frequency of a MDOF system. At times, the full solution of the eigenvalue problem is of NO particular interest and an estimate of the system lowest natural frequency suffices.

Recall that the pairs ω ,φ satisfyKφ =Mω 2 φ {}ii() in=1,2... ()ii () i

with properties ΦTTM ΦΦ==[M ]; K Φ [K ] i.e. with modal stiffness and masses calculated from:

TT 2 Ki KMii==φ()Kφφ () i;,and ii ()Mφ () iω i = (41) M i

1 T 2 K 2 φ()iiKφ () That is, ω ==i (42) i M 1 T i 2 φ()iiMφ ()

Above, the numerator relates to the potential or strain energy of the system for the i-mode, and the denominator to the kinetic energy for the same mode.

1 uKuT R()u = 2 1 T (43) 2 uMu

R()u is a scalar whose value depends not only on the matrices M & K, but also on the choice of the vector u.

Clearly, if the arbitrary vector u coincides with (or is a multiple of) one of the natural mode vectors, then Rayleigh’s quotient will deliver the exact natural frequency for that particular mode. It can also be shown that the quotient has a stationary value, i.e. a minimum, in the neighborhood of the system natural modes (eigenvectors). To show this, since u is an arbitrary vector and the natural modes are a set of linearly independent vectors, then one can represent n u==∑φΦjjc c (44) j=1 T Where c ={}cc12.. cn is the vector of coefficients in the expansion. Substitution of the expression above into Rayleigh’s quotient gives

T 1 TT 2 ()ΦcK(Φc) c (Φ KΦ)c R()u ==T 1 TT 2 ()ΦcM ()Φc c ()Φ MΦ c

ccT [K ] R()u = (45) ccT []M

Assume the modes have been normalized with respect to the mass matrix, i.e. n 22 T 2 c ω cc⎡⎤ω ∑ ini ⎣⎦n i=1 R()u ==T n (46a) cIc 2 ∑ci i=1

Next, consider that the arbitrary vector u (which at this time can be regarded as an assumed mode vector) differs very little from the

natural mode (eigenvector) φ()r . This means that in the expansion

of vector u, the coefficients ccir<<=; for i 1,2,... n and ir ≠ Or

cciiri==≠ς ;ς <<1 for i 1,2,... nir and

Then, Rayleigh’s quotient is expressed as

n cc22ω + 2ςω 22 rnri r∑ i n iir=≠1, R()u = n 22 2 ccrr+ ∑ ς i iir=≠1,

nn2 222 ςω ωςω++1 ini ninri∑∑ω ( nr ) R()u ==iir=≠1,ω 2 iir =≠ 1, (46b) nnnr 22 11++∑∑ςςii iir=≠1, iir =≠ 1, 2 The quantities {}ς i are small, of second order, hence R(u) differs from the natural frequency by a small quantity of second order.

The most important property of Rayleigh’s quotient is that it shows a minimum value in the neighborhood of the fundamental mode, i.e. when r=1. n 2 ςω 1 ini + ∑ ω ( n1 ) ω 22i=2 ni R()u ==ωωn n ,sinceω > 1 (47) 1 ()n1 2 1+ ∑ς i i=2 Then each term in the numerator is greater than the corresponding one in the denominator. Hence, it follows that

22 R()u =≥ω ω (48) n1 i.e., Rayleigh’s quotient provides an upper bound to the first (lowest) natural frequency of the undamped MDOF system.

Clearly, the equality holds above if one selects u=cc1(1)1φ ;0≠ .

Closure Rayleigh’s energy method is generally used when one is interested in a quick (but particularly accurate) estimate of the fundamental natural frequency of a continuous system, and for which a solution to the whole eigenvalue problem cannot be readily obtained. The method is based on the fact that the natural frequencies have stationary values in the neighborhood of the natural modes. In addition, Rayleigh’s quotient provides an upper bound to the first (lowest) natural frequency. The engineering value of this approximation can hardly be overstated. Rayleigh’s energy method is the basis for the numerical computing of eigenvectors and eigenvalues as will be seen later.

Example for estimation of first natural frequency d2 M U  KU = 0 M identity( N ) mass 2 dt

2 1 0 0 0   U1 U2 U3 U4 U5  1 2 1 0 0    K stiff 0 1 2 1 0  K K K K K K  0 0 1 2 1    M M M M M  0 0 0 1 2  Estimate for first natural frequency - Use Rayleigh-Ritz method:

T ASSUME u () .5 .75 1 .75 .5 1

T u Ku calculate Rayleigh's quotient 11 Ru() T u Mu 11

n_ap  Ru() n_ap  16.903 Estimate

Actual   16.369 n1 rad/s

 n_ap % difference = 1  100  3.262  n   1 

1.5

0.5

0 0 0123456 M1 u (approx) Estimate for second natural frequency - Use Rayleigh-Ritz method:

T ASSUME u.5.50.5 () .5 1

T u Ku calculate Rayleigh's quotient 11 Ru() T u Mu 11

n_ap  Ru() n_ap  31.623 Estimate

Actual   31.623 n2 rad/s   n_ap  14 % difference = 1  100  1.11 10  n   2 

1.5

0 0

1 0123456 M2 u (approx) 3 ORIGIN 1 N5 number of DOFS mass 1 stiff 10

Example for estimation of first natural frequency d2 M U  KU = 0 M identity( N ) mass 2 dt

 2 1 0 0 0  U1 U2 U3 U4 U5  1 2 1 0 0    K stiff 0 1 2 1 0  K K K K K  0 0 1 2 1  M   M M M M  0 0 0 1 1  Estimate for first natural frequency - Use Rayleigh-Ritz method:

T ASSUME u () .1.3.5.71 1

T u Ku calculate Rayleigh's quotient 11 Ru() T u Mu 11

  Ru() n_ap n_ap  10.935 Estimate

Actual   9.001 n1 rad/s

 n_ap % difference = 1  100  21.485  n   1 

1.5

0.5

0 0 0123456 M1 u (approx) Mode Acceleration Method Recall that the response in physical coordinates is

m U ≈<φ qmn, ()tjj∑ ()t (36) j=1 where m

This method, however, fails to give an accurate solution even when a static load is applied (See Structural Dynamics, by R. Craig, J. Wiley Pubs, NY, 1981.).

The difficulty is overcome by using the procedure detailed below. Recall that the system motion is governed by the set of equations

MU+KU()tt =F () (4)

And, if there are no rigid body modes, i.e. all natural frequencies are greater than zero, then

−1 U=KF()tt( ()− MU) (51) where K −1 is a flexibility matrix. From Eq. (36), m U ≈<φ qmn , ∑ ()j j()t (52) j=1

Hence, Eq, (51) can be written as

m UKFKM≈−−−11φ q ()tt () ∑ ()j j()t (53) j=1

Using the fundamental identity,

211 − Kφ()ii=⇒=ω Mφφ () i2 () iKMφ () i ωi Write Eq. (53) as m ⎛⎞φ()j UKF≈−−1 ⎜⎟q ()tt () ∑⎜⎟2 j()t (54) j=1 ⎝⎠ω j

−1 Note that UKFSt= () (55)

is the displacement response vector due to a “pseudo-static” force F(t), i.e. without the system inertia accounted for. Hence write Eq. (54), as

m ⎛⎞φ()j UU≈−⎜⎟q ()tst () ∑⎜⎟2 j()t ; m

The second term above can be thought as the “inertia induced response.”

Example: Consider the case of force excitation with frequency Ω≠ω and acting for very long times. The EOMs in physical n j space are:

⎡⎤ QP 1 qt≈Ωj ⎢⎥cos (37a) j ⎢⎥2 () K j 1−Ωω ⎢⎥⎣ ()n j ⎦ Recall that, using the mode displacement method, the response in physical coordinates is:

m ⎛⎞⎡⎤ QP 1 U ≈Ω⎜⎟φ j ⎢⎥cos()t ∑⎜⎟j ⎢⎥2 (38) j=1 ⎜⎟K j 1−Ωω ⎝⎠⎣⎦⎢⎥()n j

From each of the modal responses, ⎡⎤ QP 1 j 2 ⎢⎥ qtj ≈−Ω() 2 cos() Ω ; K j ⎢⎥1−Ωω ⎣ ()j ⎦

2 ⎡⎤ −q QP ⎛⎞Ω 1 j ≈Ωj ⎢⎥cos t (57) 22⎜⎟ 2 () ωωjjjK ⎢⎥1−Ωω ⎝⎠⎣⎦()j

2 Since K jjj=ω M ; then using the mode acceleration method, the response is

⎧⎫m 2 ⎡⎤ ⎪⎪QP ⎛⎞Ω 1 UU≈+φ j ⎢⎥cos Ωt ⎨⎬SP ∑ j ⎜⎟2 2 ()(58 j=1 K jjω ⎢⎥1−Ωω ⎩⎭⎪⎪⎝⎠⎣⎦()j )

MEEN 617 – HD#7 Undamped Modal Analysis of MDOF systems. L. San Andrés © 2008 23 −1 where the pseudo-static response is UKFSP= P . Now, in the limit, as the excitation frequency decreases, i.e., as Ω → 0, the second term in Eq. (58) above disappears, and hence the physical response becomes: −1 UU==SP KF P (59)

which is the exact response, regardless of the number of modes chosen. Hence, the mode acceleration method is more accurate than the mode displacement method. Known disadvantages include more operations.

Finding the flexibility matrix is, in actuality, desirable. In particular, if derived from measurements, the flexibility matrix is easier to determine than the stiffness matrix.

MEEN 617 – HD#7 Undamped Modal Analysis of MDOF systems. L. San Andrés © 2008 24 STEP FORCED RESPONSE of Undamped 2-DOF ORIGIN := 1 mechanical system Dr. Luis San Andres (c) MEEN 363, 617 February 2008

The undamped equations of motion are: d2 (1) M ⋅ X + KX⋅ = Fo dt2

where M,K are matrices of inertia and stiffness coefficients, and X, V=dX/dt, d2X/dt2 are the vectors of physical displacement, velocity and acceleration, respectively. The FORCED undamped response to the initial conditions, at t=0, Xo,Vo=dX/dt, follows:

======The equations of motion are:

⎛ M11 M12 ⎞ d2 ⎛ x1 ⎞ ⎛ K11 K12 ⎞ ⎛ x1 ⎞ ⎛ F1o ⎞ ⋅ + ⋅ = (2) ⎜ ⎟ 2⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ M21 M22 ⎠ dt ⎝ x2 ⎠ ⎝ K21 K22 ⎠ ⎝ x2 ⎠ ⎝ F2o ⎠

1. Set elements of inertia and stiffness matrices DATA FOR problem

6 6 ⎛ 100 0 ⎞ ⎛ 210⋅ −1⋅ 10 ⎞ N M := ⎜ ⎟ ⋅ kg K := ⎜ ⎟ ⋅ 6 6 m ⎝ 0 50 ⎠ ⎝ −1⋅ 10 210⋅ ⎠ n := 2# of DOF

Note M and K are symmetric matrices ⎛ 0 ⎞ ⎛ 0.0 ⎞ m initial conditions Xo := ⎜ ⎟ ⋅ m Vo := ⎜ ⎟ ⋅ ⎝ 0 ⎠ ⎝ 0 ⎠ sec

Applied force vector: ⎛ 10000 ⎞ Fo := ⎜ ⎟ ⋅ N ⎝ −5000 ⎠

2. Find eigenvalues (undamped natural frequencies) and eigenvectors Set determinant of system of eqns = 0 ⎡ 2 2 2 2 ⎤ Δ = ⎣ ()K11 − M11⋅ ω ⋅ ()K22 − M22⋅ ω − ()K12 − M12⋅ ω ⋅ ()K21 − M21⋅ ω ⎦ =(2a)0 4 2 2 2 (2b) Δ = a⋅ ω + b⋅ ω + c = ()a⋅ λ + b⋅ λ + c = 0with λω=

where the aM:= 11, ⋅ M22, − M12, ⋅ M21, coefficients (2c) are: bK:= 12, ⋅ M21, − K11, ⋅ M22, − K22, ⋅ M11, + K21, ⋅ M12,

cK:= 11, ⋅ K22, − K12, ⋅ K21, The roots of equation (2b) are: ⎡ 2 .5 ⎤ ⎡ 2 .5 ⎤ ⎣ −b − ()b − 4⋅ a⋅ c ⎦ ⎣ −b + ()b − 4⋅ a⋅ c ⎦ (3) λ1 := λ2 := 2⋅ a 2⋅ a also known as eigenvalues. The natural frequencies follow as:

j := 1 .. n .5 ⎛ 112.6 ⎞ rad ω := ()λ ω = j j ω ⎜ ⎟ sec (4) f := ⎝ 217.53 ⎠ 2⋅ π ⎛ 17.92 ⎞ f = ⎜ ⎟ Hz ⎝ 34.62 ⎠ Note that: Δω()1 = Δω()2 = 0 For each eigenvalue, the eigenvectors (natural modes) are j := 1 .. n ⎡ 1 ⎤ Set arbitrarily first element of vector = 1 ⎢ ⎥ aj := ⎢ K11, − M11, ⋅ λj ⎥ ⎢ ⎥ ⎣ −()K12, − M12, ⋅ λj ⎦ ⎛ 1 ⎞ ⎛ 1 ⎞ a1 = ⎜ ⎟ a2 = ⎜ ⎟ (5) ⎝ 0.73 ⎠ ⎝ −2.73 ⎠ MODAL matrix 〈〉j A := aj A is the matrix of eigenvectors (undamped ⎛ 1 1 ⎞ modal matrix): each column corresponds to an A = ⎜ ⎟ 0.73 2.73 eigenvector ⎝ − ⎠

Plot the mode shapes:

Aj , 1 0.37 Aj , 2

2.73 j DOF mode 1 mode 2 3. Modal transformation of physical equations to (natural) modal coordinates (6) Using transformation: XAq= ⋅ EOMs (1) become uncoupled in modal space:

d2 (7) Mm ⋅ q + Km ⋅ q = Qm dt2

T (8) with modal force vector: Qm = A ⋅ Fo

and initial conditions (modal displacement=q and modal velocity dq/dt=s)

− 1 T − 1 T qo = Mm ⋅ ()A ⋅ M ⋅ Xo so = Mm ⋅ ()A ⋅ M ⋅ Vo (9)

The modal responses are of the form: k=1....n

so Qm k k (10a) qk = qo ⋅ cos ()ωk⋅ t + ⋅ sin()ωk⋅ t + ⋅ ()1 − cos ()ωk⋅ t ω ≠ 0 k ωk Km k kk, for an elastic mode OR Qm 1 k 2 for ω = 0 qk = qo + so ⋅ t + ⋅ ⋅ t k (10b) k k 2 Mm for a rigid body mode kk,

And, the response in the physical coordinates is given Xt()= Aqt⋅ () (5) by the superposition of the modal responses, i.e. === CHECK ======Verify the orthogonality properties of the natural mode shapes

− 14 T ⎛ 126.79 −2.24 × 10 ⎞ Mm := A ⋅ M ⋅ A Mm = ⎜ ⎟ kg ⎝ −1.58 × 10− 14 473.21 ⎠ T K := A ⋅ K ⋅ A m ⎛ 1.61× 106 3.18× 10− 10 ⎞ N Km = ⎜ ⎟ − 10 7 m ⎝ 3.51× 10 2.24× 10 ⎠

⎛ 112.6 ⎞ -1 ω = ⎜ ⎟ s ⎝ 217.53 ⎠

======4. Find Modal and Physical Response for given initial condition and Constant Force vector

Recall the vectors of initial conditions ⎛ 0 ⎞ ⎛ 0 ⎞ m Xo = ⎜ ⎟ m Vo = ⎜ ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ s

and Constant forces: DATA FOR problem being analyzed: ⎛ 110× 4 ⎞ N Fo = ⎜ ⎟ m 3 m ⎝ −5 × 10 ⎠ 4.a Find initial conditions in modal coordinates (displacement = q, velocity = s)

Set inverse of modal mass matrix − 1 T Ainv := Mm ⋅ ()A ⋅ M

qo := Ainv⋅ Xo so := Ainv⋅ Vo

⎛ 0 ⎞ ⎛ 0 ⎞ -1 qo = ⎜ ⎟ m so = ⎜ ⎟ ms ⎝ 0 ⎠ ⎝ 0 ⎠ 4.b Find Modal forces:

T 3 Qm := A ⋅ Fo ⎛ 6.34× 10 ⎞ Qm = ⎜ ⎟ N ⎝ 2.37× 104 ⎠

4.c Build Modal responses:

so Qm 1 1 q1()t := qo ⋅ cos ()ω1⋅ t + ⋅ sin()ω1⋅ t + ⋅ ()1 − cos ()ω1⋅ t 1 ω1 Km 11,

so Qm 2 2 q2()t := qo ⋅ cos ()ω2⋅ t + ⋅ sin()ω2⋅ t + ⋅ ()1 − cos ()ω2⋅ t 2 ω2 Km 22,

for plots: 4.d Build Physical responses: Xt():= a ⋅ q ()t + a ⋅ q ()t 6 1 1 2 2 Tplot := f1 4.e Graphs of Modal and Physical responses:

Response in modal coordinates 0.01 0.005

0 0 0.056 0.11 0.17 0.22 0.28 0.33 time (s) q1 q2

Response in physical coordinates 0.01

0.005

0.01 0 0.056 0.11 0.17 0.22 0.28 0.33 time (s) x1 x2

5. Interpret response: analyze results, provide recommendations

S-S displacement − 3 − 1 ⎛ 510× ⎞ K ⋅ Fo = ⎜ ⎟ m Recall natural frequencies & periods ⎝ 0 ⎠

⎛ 17.92 ⎞ 1 ⎛ 0.056 ⎞ f = ⎜ ⎟ Hz = ⎜ ⎟ s 34.62 f 0.029 ⎛ 1 1 ⎞ ⎝ ⎠ ⎝ ⎠ A = ⎜ ⎟ ⎝ 0.73 −2.73 ⎠ ⎛ 112.6 ⎞ -1 ω = ⎜ ⎟ s ⎝ 217.53 ⎠ STEP FORCED RESPONSE of Undamped 2-DOF ORIGIN := 1 mechanical system Dr. Luis San Andres (c) MEEN 363, 617 February 2008

The undamped equations of motion are: d2 (1) M ⋅ X + KX⋅ = Fo dt2

======The equations of motion are: WITH RIGID BODY MODE ⎛ M11 M12 ⎞ d2 ⎛ x1 ⎞ ⎛ K11 K12 ⎞ ⎛ x1 ⎞ ⎛ F1o ⎞ ⋅ + ⋅ = (2) ⎜ ⎟ 2⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ M21 M22 ⎠ dt ⎝ x2 ⎠ ⎝ K21 K22 ⎠ ⎝ x2 ⎠ ⎝ F2o ⎠

1. Set elements of inertia and stiffness matrices DATA FOR problem

6 6 ⎛ 100 0 ⎞ ⎛ 110⋅ −1⋅ 10 ⎞ N M := ⎜ ⎟ ⋅ kg K := ⎜ ⎟ ⋅ 6 6 m ⎝ 0 50 ⎠ ⎝ −1⋅ 10 110⋅ ⎠ n := 2# of DOF

Note M and K are symmetric matrices ⎛ 0 ⎞ ⎛ 0.0 ⎞ m initial conditions Xo := ⎜ ⎟ ⋅ m Vo := ⎜ ⎟ ⋅ ⎝ 0 ⎠ ⎝ 0 ⎠ sec

Applied force vector: ⎛ 1000 ⎞ Fo := ⎜ ⎟ ⋅ N ⎝ −980 ⎠

where the aM:= 11, ⋅ M22, − M12, ⋅ M21, coefficients (2c) are: bK:= 12, ⋅ M21, − K11, ⋅ M22, − K22, ⋅ M11, + K21, ⋅ M12,

cK:= 11, ⋅ K22, − K12, ⋅ K21, , , , , The roots of equation (2b) are: ⎡ 2 .5 ⎤ ⎡ 2 .5 ⎤ ⎣ −b − ()b − 4⋅ a⋅ c ⎦ ⎣ −b + ()b − 4⋅ a⋅ c ⎦ (3) λ1 := λ2 := 2⋅ a 2⋅ a also known as eigenvalues. The natural frequencies follow as:

j := 1 .. n .5 ⎛ 0 ⎞ rad ωj := ()λj ω = ω ⎜ ⎟ sec (4) f := ⎝ 173.21 ⎠ 2⋅ π ⎛ 0 ⎞ f = ⎜ ⎟ Hz ⎝ 27.57 ⎠ Note that: Δω()1 = Δω()2 = 0 For each eigenvalue, the eigenvectors (natural modes) are j := 1 .. n ⎡ 1 ⎤ Set arbitrarily first element of vector = 1 ⎢ ⎥ aj := ⎢ K11, − M11, ⋅ λj ⎥ ⎢ ⎥ ⎣ −()K12, − M12, ⋅ λj ⎦ ⎛ 1 ⎞ ⎛ 1 ⎞ a1 = ⎜ ⎟ a2 = ⎜ ⎟ (5) ⎝ 1 ⎠ ⎝ −2 ⎠ MODAL matrix 〈〉j A := aj A is the matrix of eigenvectors (undamped ⎛ 1 1 ⎞ A = ⎜ ⎟ modal matrix): each column corresponds to an ⎝ 1 −2 ⎠ eigenvector

Plot the mode shapes:

Aj , 1 0 Aj , 2

2 j DOF mode 1 mode 2

3. Modal transformation of physical equations to (natural) modal coordinates

Using transformation: X A (6) Using transformation: X = A ⋅ q (6) EOMs (1) become uncoupled in modal space:

d2 (7) Mm ⋅ q + Km ⋅ q = Qm dt2

T (8) with modal force vector: Qm = A ⋅ Fo

and initial conditions (modal displacement=q and modal velocity dq/dt=s)

− 1 T − 1 T qo = Mm ⋅ ()A ⋅ M ⋅ Xo so = Mm ⋅ ()A ⋅ M ⋅ Vo (9)

The modal responses are of the form: k=1....n

so Qm k k (10a) qk = qo ⋅ cos ()ωk⋅ t + ⋅ sin()ωk⋅ t + ⋅ ()1 − cos ()ωk⋅ t ω ≠ 0 k ω Km k k kk, for an elastic mode OR Qm 1 k 2 for ω = 0 qk = qo + so ⋅ t + ⋅ ⋅ t k (10b) k k 2 Mm for a rigid body mode kk,

T ⎛ 150 0 ⎞ Mm := A ⋅ M ⋅ A Mm = ⎜ ⎟ kg ⎝ 0 300 ⎠ T Km := A ⋅ K ⋅ A ⎛ 0 0 ⎞ N ⎛ 0 ⎞ -1 Km = ⎜ ⎟ ω = ⎜ ⎟ s ⎝ 0 910× 6 ⎠ m ⎝ 173.21 ⎠

======

4. Find Modal and Physical Response for given initial condition and Constant Force vector

Recall the vectors of initial conditions ⎛ 0 ⎞ ⎛ 0 ⎞ m Xo = ⎜ ⎟ m Vo = ⎜ ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ s

d Cttf and Constant forces: DATA FOR problem being analyzed: ⎛ 110× 3 ⎞ N Fo = ⎜ ⎟ m ⎝ −980 ⎠ m 4.a Find initial conditions in modal coordinates (displacement = q, velocity = s)

Set inverse of modal mass matrix − 1 T Ainv := Mm ⋅ ()A ⋅ M ⎛ 0.67 0.33 ⎞ Ainv = ⎜ ⎟ ⎝ 0.33 −0.33 ⎠ qo := Ainv⋅ Xo so := Ainv⋅ Vo

⎛ 0 ⎞ ⎛ 0 ⎞ -1 qo = ⎜ ⎟ m so = ⎜ ⎟ ms ⎝ 0 ⎠ ⎝ 0 ⎠ 4.b Find Modal forces: T Qm := A ⋅ Fo ⎛ 20 ⎞ Qm = ⎜ ⎟ N ⎝ 2.96× 103 ⎠

4.c Build Modal responses:

Qm 2 1 t response for rigid body mode q1()t := qo + so ⋅ t + ⋅ 1 1 Mm 2 11,

so Qm 2 2 q2()t := qo ⋅ cos ()ω2⋅ t + ⋅ sin()ω2⋅ t + ⋅ ()1 − cos ()ω2⋅ t 2 ω2 Km 22,

for plots: 4.d Build Physical responses: 6 Xt():= a1⋅ q1()t + a2⋅ q2()t Tplot := f2 4.e Graphs of Modal and Physical responses:

Response in modal coordinates 0.004

0.003

0.002

0.001

0 0 0.022 0.044 0.065 0.087 0.11 0.13 0.15 0.17 0.2 0.22 time (s) time (s) q1 q2

Response in physical coordinates

0.002

0.002 0 0.022 0.044 0.065 0.087 0.11 0.13 0.15 0.17 0.2 0.22 time (s) x1 x2

5. Interpret response: analyze results, provide recommendations

S-S displacement - NONE

Recall natural frequencies & periods

⎛ 0 ⎞ ⎛ 1 1 ⎞ ⎛ 0 ⎞ -1 f = ⎜ ⎟ Hz A = ⎜ ⎟ ω = ⎜ ⎟ s ⎝ 27.57 ⎠ ⎝ 1 −2 ⎠ ⎝ 173.21 ⎠