Concept Of Orthogonality.docx

CONCEPT OF ORTHOGONALITY The orthogonality property of the natural modes is most important in simplifying dynamics problems. The development of this property can be started by considering the inertial forces of the system. {Finertia } = [M]{ẍ } Recall the following equations and incorporating all natural modes, ẍ = ω2 Ceiωt {U} = {Ceiωt } The acceleration in the inertial force can be derived as {ẍ } = [U]{ω2 } Substituting this equation of acceleration to inertial force, {Finertia } = [M][U]{ω2 } The inertial forces can be separated to its different mode by getting the transpose of the equation {Finertia }T = {ω2 }[U]T [M]T To visualize this, we consider an actual case. For a 3–DOF system at one of its natural modes, say “r”, this would result to [Fr1

Fr2

Fr3 ] =

ω2r [u1r

ur2

m11 ur3 ] [m21 m31

m12 m22 m32

m13 m23 ] m33

From the above equation, a single inertia force can be represented by [Frj ] =

ω2r [u1r

u2r

m1j u3r ] {m2j } m3j

or in general, to make this applicable to any “n” degree of freedom system, this can be written as n

Frj =

ω2r ∑ uir mij i=1

Where: r = mode number i and j are arbitrary indices Virtual work can be introduced by multiplying the inertial forces associated with mode “r” by virtual displacement associated with mode “s”. n

ωrs = ∑ Frj ujs j=1

Where: ωrs = virtual work Substituting the equation of Frj to the virtual work equation, the result is n

n

ωrs = ∑ ω2r ∑ uir mij ujs j=1

n

i=1 n

ωrs = ω2r ∑ ∑ uir mij ujs j=1 i=1

The above equation constitute the equation of virtual work which is produced by the inertial force associated with mode “r” and the virtual displacement associated with mode “s.” Suppose we evaluate the virtual work produced by the inertial force associated with mode “s” and the virtual displacement associated with mode “r”. The virtual work equation is equal to n

ωsr =

n

ω2s ∑ ∑ uis mij ujr j=1 i=1

Note that “i" and “j” are arbitrary indices and so we can choose any other letter to present these indices, say “i" for “j” and “j” for “i". Hence, the equation can be changed to n

ωsr =

n

ω2s ∑ ∑ ujs mji uir i=1 j=1

Interchanging the order of the summation, n

ωsr =

n

ω2s ∑ ∑ uir mji ujs j=1 i=1

Note that [M] is symmetric, therefore mij = mji, hence n

ωsr =

n

ω2s ∑ ∑ uir mij ujs i=1 j=1

By Betti’s Law, the virtual work produced by the inertial force at mode “r” and the virtual displacement at mode “s” must be equal to the virtual work produced by the inertial force at mode “s” and the virtual displacements at mode “r”. ωrs = ωsr From the equations n

ωrs = and

n

ω2r ∑ ∑ uir mij ujs j=1 i=1

n

ωsr =

n

ω2s ∑ ∑ uir mij ujs i=1 j=1

Then, n

n

ω2r ∑ ∑ uir mij ujs j=1 i=1 n n

n

=

n

ω2s ∑ ∑ uir mij ujs i=1 j=1

(ω2r − ω2s ) ∑ ∑ uir mij ujs = 0 j=1 i=1

If r = s, then ω2r − ω2s = 0, therefore, n

n

∑ ∑ uir mij ujs = 0 if r ≠ s j=1 i=1

The preceding above equation is known as the orthogonality property of the natural modes. It should be observed that the orthogonality property can also be applied to the term involving stiffness matrix and damping matrix. Mathematically, these terms would have the same condition as that of the equation of orthogonality. n

n

∑ ∑ uir k ij ujs = 0 if r ≠ s j=1 i=1 n n

∑ ∑ uir cij ujs = 0 if r ≠ s j=1 i=1

However, this equation are valid provided that the mass matrix, stiffness matrix and damping matrix are symmetric. With the orthogonality property known, it would be very easy to comprehend how step 6 in the listed procedure of modal analysis reduce the simultaneous equations of MDOF system to simply a set of SDOF equations.

NORMALIZATION In this section, the process of reducing the equation of motion of MDOF system to a set of SDOF equations will be presented. This is accomplished by introducing a new coordinate, Q, to represent the original relative global coordinate, x, of the system so that by using of orthogonality property uncoupling of equations could be achieved. This process is called normalization and enumerated as step 5 and 6 of the listed procedure for modal superposition. As an illustration, consider an undamped free vibrating 2-DOF system. The general equation of motion of MDOF system for undamped free vibrating system is [M]{ẍ } + [K]{x} = {0}

For a 2-DOF system, m11 [m 21

m12 ẍ 1 k11 m22 ] {ẍ 2 } + [k 21

k12 x1 0 ]{ } = { } k 22 x2 0

From the modal matrix (step 5), {x} = [U]{Q} The relative global coordinate, x, in terms of the normalized coordinate, Q, can be introduced and is equal to x1 U {x } = [ 11 U21 2

U12 q1 ]{ } U22 q2

The second derivative of x can be computed as ẍ U { 1 } = [ 11 U21 ẍ 2

U12 q̈ 1 ]{ } U22 q̈ 2

Substitute these equations to the first equation: m11 [m 21

m12 U11 m22 ] [U21

U12 q̈ 1 k ] { } + [ 11 U22 q̈ 2 k 21

k12 U11 ][ k 22 U21

U12 q1 0 ]{ } = { } U22 q2 0

Premultiply this equation by [U]T as stated in step 6 of the listed procedure for modal superposition. [

U11 U12

U21 m11 ][ U22 m21

m12 U11 m22 ] [U21

U12 q̈ 1 U ] { } + [ 11 U22 q̈ 2 U12

U21 k11 ][ U22 k 21

k12 U11 ][ k 22 U21

U12 q1 0 ]{ } = { } U22 q2 0

Proceeding with the multiplication of matrices and making use of the indicial notation, this can be simplified as [Mrs ]{q̈ s } + [K rs ]{qs } = {0} Where: n

n

Mrs = ∑ ∑ uir mij ujs ; r = 1,2; s = 1,2 j=1 i=1 n n

K rs = ∑ ∑ uir k ij ujs ; r = 1,2; s = 1,2 j=1 i=1

Recalling the orthogonality property, Mrs = 0 if r ≠ s K rs = 0 if r ≠ s with the orthogonality in mind, we can write all the elements of the matrices as M [ 11 0

0 q̈ K ] { 1 } + [ 11 M22 q̈ 2 0

0 q1 0 ]{ } = { } K 22 q2 0

It can be noticed that in this equation, the uncoupling has been accomplished, thus, we can write this simply as M11 q̈ 1 + K11 q1 = 0 M22 q̈ 2 + K 22 q2 = 0 In the above equation, normalization has been completed. The previously simultaneous equations has been reduced to a set of independent equation in which each equation can be treated as a SDOF equation.

FORCED VIBRATION Forced vibration of lumped mass MDOF system can be induced directly to a lumped mass or indirectly in the form of ground motion. Let us consider the case where the forcing functions are acting directly to the masses. The equation of motion for damped forced vibration is given by [M]{ẍ } + [C]{ẋ } + [K]{x} = {F(t)} We can change the coordinate from x to q by using the relation {x} = [U]{Q̈} [M][U]{Q̈} + [C][U]{Q̇} + [K][U]{Q} = {F(t)} Premultiply by [U]T : [U]T [M][U]{Q̈} + [U]T [C][U]{Q̇} + [U]T [K][U]{Q} = [U]T {F(t)} By the orthogonality property, the above equation can be reduced to independent equations n

M1 q̈ 1 + C1 q̇ 1 + K1 q1 = ∑ uji Fj (t) j=1 n

Mrr q̈ r + Crr q̇ r + K rr qr = ∑ ujr Fj (t) j=1

Crr K rr q̈ r + q̇ r + q = Mrr Mrr r

n

∑ j=1

ujr Fj (t)

Mrr

Note: Damping can be better estimated if modal damping ratio would be used instead of damping coefficient. That is Crr = 2hr ωr Mrr and K rr = ω2 Mrr Then,

n

∑ 2

q̈ r + 2hr ωr q̇ r + ω qr =

j=1

ujr Fj (t)

Mrr

Where: n

n

Mrr = ∑ ∑ uir mij ujr j=1 i=1

A closed form solution of the above equation can be arrived at if the forcing function is sinusoidal (sine or cosine functions). Assuming a sine function for the external force F(t), then the solution is Fj (t) = Foj sin(pj t) Where: Fj (t) = sinusoidal forcing function acting at the jth lumped mass Foj = amplitude of the jth sinusoidal forcing function pj = frequency of the jth sinusoidal forcing function in rad/sec Since forcing function is sinusoidal, then the solution is composed of particular solution, qpr and complimentary solution, qcr, such that qr = qpr + qcr The complimentary solution is derived in the same manner as derived in the previous discussions. qcr = e−hr wr t [Ar cos(ωdr t) + Br sin(ωdr t)] Where: ωdr = ωr √1 − h2r The complimentary solution arises from the initial conditions of the structure. In the absence of these initial conditions, the complimentary solution does not necessarily become zero since the coefficients Ar and Br are evaluated from the total response and not from the complimentary response alone. Since this type of response diminishes when there is damping it commonly termed as the transient response. The derivation of the particular solution is done in the same manner as derived in the previous discussions.

n

ujr Foj sin(pj t − ϕj )

∑

2

j=1 (

qpr =

2

√(ω2r − p2j ) + (2ωr hr pj ) n

)

n

∑

∑

i=1

j=1

uir mij ujr

Where ϕ is called the phase angle which is expressed mathematically as ϕj = tan−1 ( ϕj = tan−1 (

2ωr hr pj ω2r − p2j 2ωr hr pj ω2r − p2j

) when ωr > pj + π) when ωr ≤ pj

The behavior of the particular solution is mainly dependent on the forcing function. This means that since the forcing function is sinusoidal, we will expect the particular response to be sinusoidal also. Due to this behavior, the particular solution is commonly called the steady state response. The actual response is the sum of the complimentary and the particular solution. This is to have an exact solution. From the complimentary solution, the coefficients Ar and Br have to be evaluated. These coefficients are usually computed from the prescribed initial displacements and initial velocities. If the load function is a sine function, then the values of the coefficients are evaluated by the following: n

Foj sin(−ϕj )

∑ Ar = qor −

2

√(ω2r − p2j ) + (2ωr hr pj ) ] j=1 [ Mrr n

ujr Foj pj cos(−ϕj )

∑

2

√(ω2r − p2j ) + (2ωr hr pj ) ] j=1 [ Mrr ωdr

q̇ or + ωr hr Ar − Br = Where: n

qor =

n

∑

∑ j=1 n

∑ j=1

i=1

uir mij xoj

n

∑ i=1

uir mij ujr

VIBRATION DUE TO GROUND MOTION Forced vibration of structures may be induced by ground motion. Ground motion are expressed as ground acceleration. In the previous cases of vibration, the reference of all motion have been always fixed, this time we have a case in which the ground or base of the structure is moving. (Insert figure here) In the figure, xg denotes ground movement while x1 and x2 denotes relative motion of the structure. Relative motion pertains to all motions reckoned from the original position of the structure but with the base held fixed. Absolute movement on the other hand, pertains to all movements reckoned from the imaginary reference considering the base movements. Investigation of the forces that are developed due to ground motion or base movements reveals that the inertial force that will be developed in each lumped mass is equal to its mass multiplied by the absolute acceleration of the mass. Absolute acceleration means the ground acceleration plus the relative acceleration developed in each mass on the course of vibration of the system or simply the acceleration reckoned from the reference axis. Mathematically speaking, absolute acceleration is ẍ + ẍ g . It should be note that the resisting forces due to the stiffness and damping of the structure are internally developed, that is, these forces arises only from the relative movements. Hence, the simultaneous equations of motion for forces acting on an MDOF system subjected to ground motion or base movement is [M]{ẍ + ẍ g } + [C]{ẋ } + [K]{x} = {0} Transferring the term of the inertial force [M]{ẍ g } to the right hand side of the equation: [M]{ẍ } + [C]{ẋ } + [K]{x} = −[M]{ẍ g } It can be observed that the elements of the column vector {ẍ g } are identical, that is, each element is actually the ground acceleration. Therefore, it would be simpler to write this column as vector {I}ẍ g. [M]{ẍ } + [C]{ẋ } + [K]{x} = −[M]{I}ẍ g Changing the coordinates from x to q: [M][U]{Q̈} + [C][U]{Q̇} + [K][U]{Q} = −[M]{I}ẍ g Premultiply by [U]T : [U]T [M][U]{Q̈} + [U]T [C][U]{Q̇} + [U]T [K][U]{Q} = −[U]T [M]{I}ẍ g The above equation can be further simplified by assigning the unit vector equal to the product of the modal matrix and participating vector {β}, that is {I} = [U]{β}. [U]T [M][U]{Q̈} + [U]T [C][U]{Q̇} + [U]T [K][U]{Q} = −[U]T [M][U]{β}ẍ g Uncoupling the above equation by making use of the orthogonality property: Mrr q̈ r + Crr q̇ r + K rr q r = −Mrr βr ẍ g

Divide the equation by Mrr: q̈ r +

Crr K rr q̇ r + q = −βr ẍ g Mrr Mrr r

But: Crr = 2ωr hr Mrr and K rr = ω2r Mrr Then, q̈ r + 2ωr hr q̇ r + ω2r qr = −βr ẍ g Considering that the ground excitation is sinusoidal, the solution of the above equation is a closed form solution. Let: ẍ g = xog sin(pt) Where: xog = amplitude of ground acceleration p = frequency of the ground excitation q̈ r + 2ωr hr q̇ r + ω2r qr = −βr xog sin(pt) The solution of the preceding equation is composed of the complimentary part and the particular part. The complimentary solution is qcr = e−ωr hr t [Ar cos(ωd t) + Br sin(ωd t)] Where: ωd = ωr √1 − h2r The normalized particular solution of a damped MDOF system subjected to ground or base excitation is: qpr = Where:

−βr √(ω2r − p2 )2 + (2ωr hr p)2

xog sin(pt − ϕ)

2ωr hr p ) when ωr > p ω2r − p2 2ωr hr pj ϕ = tan−1 ( 2 + π) when ωr ≤ p ωr − p2 ϕ = tan−1 (

The coefficients Ar and Br can be computed by substituting the normalized initial displacement and initial velocity into the normalized total response the result of which is: βr xog sin(−ϕ) − p2 )2 + (2ωr hr p)2 βr xog cos(−ϕ) q̇ or + ωr hr Ar + √(ω2r − p2 )2 + (2ωr hr p)2 Br = ωd

Ar = qor +

(ω2r

The participating factor {β} can be computed as follows: {I} = [U]{β} {β} = [U]−1 {I} An alternative way of computing for the participating factor can be developed as follows: {I} = [U]{β} Premultiply by [U]T [M]: [U]T [M]{I} = [U]T [M][U]{β} Making use of the orthogonality property, the right hand side of the equation can be uncoupled while the corresponding elements of the left hand side can be computed by performing the matrix operation. The result of this is presented below: n

n

n

n

∑ ∑ uir mij ujr βr = ∑ ∑ uir mij j=1 i=1

j=1 i=1 n

n

∑

∑

i=1

j=1

βr =

n

∑ j=1

uir mij

n

∑ i=1

uir mij ujr

So, having all the equations needed, the response in the normalized coordinate system can be evaluated. The response in the “x” coordinate can be determined by converting in “q” coordinate to “x” coordinate. This is done as follows: {X} = [U]{Q} In indicial notation this is expressed as n

xj = ∑ ujr qr r=1

Take note that “x” is the relative displacement. If the absolute displacement is needed then it is necessary to evaluate the ground displacement and add it to the relative displacement. The ground displacement can be evaluated by twice integrating the given ground acceleration.