Dynamics v2.2

8/2/2019 Dynamics v2.2

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CE4006 - Structural Engineering Dr. Denis Kelliher

Dept. of Civil & Environmental Engineering, UCC 1

Introduction

Structures respond dynamically when the applied loading is time dependent [ ( )p t ], as

distinct from the static response due to constant loads (or at least loads which vary

slowly with respect to time). Typical dynamic loading of large scale civil and buildingstructures would include:

Earthquake Loading Wind Loading (particularly high rise buildings) Wave loading (on offshore structures) Blast loading Other dynamic sources (large vibrating mechanical plant supported within a

structure)

Hence, if any of the above loading conditions exists, then it may be necessary to

consider the response of structures to dynamic loading. Broadly speaking, there are two

primary structural dynamic responses: the response of the structure tofree vibration and

the response of the structure toforced vibration.

Free vibration

Free vibration occurs when the structure is initially displaced from equilibrium, and is

then let vibrate with no external forcing. Figure 1 shows a typical response for a simple

cantilever type structure. This concept is important since the response is dependent on

parameters such as the natural period (or frequency) of vibration, stiffness of the

structure and damping. Free vibration may be used to determine experimentally the

natural frequency and damping coefficient of real structures.

m

0t

v v

v

0

v

v0

0

Figure 1: (a) Vertical column with concentrated mass at top. (b) Structure initially

displaced and then released. (c) Typical displacement response at top of column over

time of a real system.

Forced vibration

As the name suggests, forced vibration is the response of a structure to an external time

varying force. A typical example is the response of a structure to earthquake

accelerations or ground movements.

This section is divided intowill introduce basic dynamic concepts and consider the free


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vibration of simple SDOF structural systems. It will then look at forced vibration of

SDOF sand the associated responses.

SDOF Dynamic Systems

Figure 2 shows an idealised mass-dashpot-spring (m-c-k) dynamic system, acted on by

a force p, which varies in the time domain

Mass m

Spring: k

Damper: cFrictionless

Rollers

Force: p(t)

v (-ve) v (+ve)

Figure 2: Idealised SDOF dynamic system

Using Newton's 2nd

Law of Motion, the governing basic equation of motion for a SDOF

system is the following 2nd

order equation:2

2

( ) ( )( ) ( )

d v t dv t m c kv t p t

dt dt (1)

Free vibration ( 0p )

Using the standard dot notation for differentiation with respect to time, t, we get0 mv cv kv (2)

Solution must have the general form stv Ae and substituting this into (2) to get2[ ] 0 stAe ms cs k (3)

Since 0stAe , then 2 0 ms cs k which results in the roots s1 and s2:2

1,22 2

c c ks

m m m

(4)

and the general solution to (2) becomes1 2s t s t v Ae Be (5)

Case 1 - Undamped vibration ( 0c )

For 0c , (4) results in 1,2s i k m and substituting these into (5) to get

( )i k m t i k m t

v t Ae Be (6a)

or (using Euler's equation):

( ) sin cosv t A k m t B k m t (6b)

It is easily observed from (6) that ( )v t is a periodic function of circular frequency

n k m where n is the natural frequency of vibration of the system with no

damping. This is simple harmonic motion (SHM). The roots of (4), for this undampedcase, may now be defined as 1,2 ns i . The natural frequency and period of natural


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vibration the of a system may be defined as

n

1 2and T

2

nn

n n

ff

(7)

This is the period at which the undamped system will vibrate, when undergoing free

vibration. Equation (6b) may now be written in terms ofn to get( ) sin cosn nv t A t B t (8a)

or ( ) sin( )nv t C t (8b)

where 2 2C A B , (the maximum amplitude of vibration) and 1tan ( )B A .ClearlyA andB depend on initial conditions.

tC

CC sin

nTv(t)

Figure 3: Response of a SDOF with no damping present

Case 2 - Damped vibration ( 0c )

The SHM system described above will continue to vibrate ad infinitum, with no changein the amplitude of vibration. However real systems will dissipate energy and this is

referred to as damping.

Damping is a property that is difficult to assess and model in a real system. However a

good approach is to consider the damping resistance as proportional to the velocity of

the system: viscous damping. It is related to the velocity of the system and is described

by a property called the damping coefficient, c. Obviously, little damping will result in

an oscillatory motion that will come to rest eventually where as much damping will

prevent any oscillation of the system. Therefore it is clear that there exists a value for c,

which is the critical point between an oscillation of the system and none. This is

referred to as the critical damping coefficient, cc. Considering equation (4), it is clear as

c is increased from 0 (no damping) to a large value, then the roots 1,2s will go fromimagiary complex root to real roots. Consequently equation (5) will have an oscillatory

component when the roots are imaginary and will be purely exponential when the roots

are real. The changeover from imaginary to real roots (i.e. from oscillatory to non-

oscillatory motion) occurs when the radical in (4) is zero, resulting in2

2

cc k

m m

(9)

Using the fact that 2n k m

2 2c nc m km (10)

Defining the damping ratio cc c and noting that 2 2c nc m c m , (4) may bewritten in the form


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1,2 21

n

si

(11)

For: 1 the roots are complex conjugate pair, resulting in an oscillatory response1 the roots are real, resulting in an non-oscillatory (aperiodic) response

Figure 4 shows the relationship between the roots 1,2s and the damping ratio .

sn

ns

Real

Imag

Figure 4: Relationship between roots 1,2s and damping ratio .

Case 2a - Under-damped vibration ( or 1c

c c )

tC sin

C

e tn

Td

v(t)

Figure 5: Response of a SDOF with damping present which is less than critical

damping

From (11),2

1,2 ( 1 ) ns i and substitution into (5) gives2 2( 1 ) ( 1 )

( ) n ni t i t

v t Ae Be (12a)

or2 2

( ) sin( 1 ) cos( 1 )nt

n nv t e A t B t

(12b)

(12) is a periodic function of circular frequency2

1d n , which exponentially

decays. Equation 12 may be re-written as:


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( ) sin( )nt dv t Ce t

(12c)

This is the frequency at which the under-damped system will vibrate, during free

vibration. It is worth noting that for 0.2 that 0.98d n n (see Figure 6). For

structures, typically 20% .

Damping Ratio -

d

n

Figure 6: Typically for structures 20% so that d n

Case 2b - Critical damping ( or 1cc c )

From (11), there is one real root n

s and substitution into (5) gives

( )

ntv t Ae (13)

Therefore the response of the system is a non oscillatory as in Figure 7.

Aent

t

v(t)

Figure 7: Response of a SDOF with damping present which is less than critical

damping

Case 2c - Over-damped vibration ( or 1cc c )

From (11), 21,2 ( 1) ns and substitution into (5) gives

2 2( 1) ( 1)( ) n nt tv t Ae Be (14)


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This is an aperiodic function, which is similar to that in Figure 7.

Logarithmic Decrement

Referring to case 2a ( 1 ), Equation 12c and the response diagram in figure 8.

v(t)

t

P

QR S

dT

Figure 8: Logarithmic decrement

At the points P, Q, R, S, the amplitude of the points are , , , ,P Q R Sv v v v and since

sin( ) 1dt so that we may write:

exp2

QP R

Q R S

vv v ct

v v v m

(15a)

where t is the time interval between adjacent points. Since2

d

t

then

exp

QP R

Q R S d

vv v c

v v v m

(15b)

Defining logarithmic decrement as ln P Qv v then dc m . Substituting forand dc gives

22 1 (16)

For structures, typically 0.2 which means 21 1 and (16) may be written as:2 (17)

This is a useful equation for estimating the damping ratio of real structures. If the

structure can be given an initial disturbance and the resulting displacement response isrecorded, then equation (17) may be used to find a viscous damping ratio. An estimate

of the natural frequency may also be found.

Forced vibration ( ( )p p t )

In this section the response of a damped SDOF system to a cyclic load is considered.

Only the case where the applied cyclic load is a harmonic disturbing force is considered:

0( ) cos( )p t P t (18)so that the equation of motion becomes

2

02( ) ( ) ( ) cos( )d v t dv t m c kv t P t

dt dt (19)


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The solution of this equation consists of two parts:

The complementary solution i.e. the solution of the homogeneous case where the RHS

of equation (19) is equal to zero. This solution has already been considered in the

previous section and for 1 was periodic with exponential decay.The particular solution this entails the particular forcing function

0

cos( )P t .Ths response is then, simply, the sum of the particular (steady state response) and

complementary (starting transient response) solutions.

0

1

2

3

4

5

6

0.0 0.5 1.0 1.5 2.0 2.5 3.0

Freq. ratio:

DynamicMagnific

ationFactor)

max

dat

Figure 8: Response to Harmonic Forced Vibration

For the particular integral (steady state response) a solution of the form

cos( ) sin( )A t B t is used and after some algebra the solution of equation (19) maybe found to be:

0

1 22 2 2

cos( )( )

( )

P tv t

k m c

(20)

with2

tan( )

c

k m

as the pahse angle.

Equation (20) may be recast as ( ) cos( )v t V t where Vis amplitude of the steadystate vibrations. Define 0stV P k as the static displacement. Then using cc c and

2c nc k and defining

The frequency ratio - n

The gain factor - stG V V (or Dynamic Magnification Factor)

then2 2 2 1 2

1

[(1 ) (2 ) ]G

(21)

This relationship is plotted in Figure 9.

The value of at which G is a maximum is found from 2 21 2


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0 max occurs at 1

2 max occurs at 0.96

G

G

When 0 and 1 then RESONANCE occurs.

Examples of Free Undamped Vibrations - SDOF

Note in the examples that increasing mass means decreasing frequency.

Example 1

m

mk

x

EI

Neglect mass of column

v

8L m 6 260 10 EI Nm 1,000m kg

Circular Frequency: nk

m

For a cantilever:6

3 3

3 3(60 10 )351, 562.5 /

8

EIk N m

L

351,562.518.75 /

1,000 n rad s

Natural Frequency:1

2.98 0.3352

n

n n

n

f Hz T sec/cyclef


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Example 2

m

2EI

6m

10m

EI

mk

x

A

B C

DNeglect mass of columns

Rigid Beamv

6 2

60 10 EI Nm 2,000m kg

Stiffnesses (sway condition):3 3

12 12

6 18

ABEI EI EI

k

L

3 3

12 12(2 ) 3

10 125 CD

EI EI EIk

L

6 61 360 10 4.77 10 /

18 125

AB CDk k k N m

64.77 1048.85 /

2,000

n rad s

17.76 0.129

2

nn n

n

f Hz T sec/cyclef

Example 3

L

M = AL

m

k

x

Flexural Stiffness = EIIgnore Mass of BEam

For a SS beam:3

48

EIk

L

2

3

48 1

n

k EI

m L AL

Define 24

1

EI

A L 2 2

48n


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Multi-degree of freedom dynamic systems

In this section we will look at multi-DOF dynamic systems, but to avoid complexity, the

DOF is considered unidirectional. As an introduction, the 2-DOF system shown below

is considered.

Example 4 (A discrete 2-DOF system)

Consider the two masses, springs and dashpots connected in series as shown below.

c

k

Mass m

Frictionless Rollers

1

1

1

Mass m 2

2

1 c 2 c 3

p (t)

k k

1p (t)

2

v v

2 3

The dynamic response is completely defined if, over time t, 1v and 2v are known.

Hence the solution variable 12

( )

( )

v tt

v t

v . Applying equilibium to the 2 DOF

dynamic system above, we can write the following equations:

1 1 1 1 2 1 2 2 1 2 1 2 2

2 2 2 2 3 2 2 1 2 3 2 2 1

( ) ( ) ( )

( ) ( ) ( )

m v p t c c v c v k k v k v

m v p t c c v c v k k v k v

Rearranging and writing in matrix form:

1 2 2 1 2 21 1 1 1 1

2 2 3 2 2 32 2 2 2 2

0 ( )

0 ( )

c c c k k k m v v v p t

c c c k k k m v v v p t

or in more compact form: Mv + C Kv f v

where

is the mass matrix

is the damping matrix

is the stiffness matrix

is the time dependent force matrix

M

C

K

f

A few points to note:

M is a diagonal matrix with all off-diagonal terms being zero each mass is lumped at a

DOF and does not influence the force at any other DOF. In other words the mass

matrix is uncoupled. This is known as a lumped mass system, i.e. the

K is the structural stiffness matrix i.e. that which is used for the static analysis.

C is equivalent in form to the stiffness matrix.


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This is a set of two coupled differential equations and the solution is outside the scope

of this course. Suffice to say there are analytical closed form solutions available for

special cases and numerical approaches would be employed for more general

applications.

Discrete n-DOF system

The system of equations governing the dynamic response of a discrete system is easily

extended form the special 2-DOF system described above.

Mass mi

Mass mI+1

c

p (t)

k

i

i+1

I+1

v vi

p (t)

Consider two individual masses,

im & 1im , and the associated DOF, ( )iv t & 1( )iv t .

The masses are connected by a spring with stiffness, k, and a dashpot with damping, c.

Both masses are subject to forces ( )i

p t & 1( )ip t . Writing a force equation for each

mass:

1 1

1 1 1 1 1

( )

( )

i i i i i i i

i i i i i i i

m v p t cv cv kv kv

m v p t cv cv kv kv

1 1 1 1 1

0 ( )

0 ( )

i i i i i

i i i i i

m v v v p t c c k k

m v v v p t c c k k

or e e e e M v + C v K v f (A)

This equation may be written for each pair of masses, connected by a spring and/or a

viscous damper (dashpot). The similarity between these sets of equations and the

element stiffness matrices developed in an earlier section of this course should be noted.

In fact the subsystem above is essentially an unidirectional linear dynamic element.

Hence the assembly procedure for the damping and stiffness matrices is essentially thesame. Therefore the final set of equations in matrix form would be the same as equation

(A) above:

Mv + Cv Kv f (B)


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Example 5 (Structural Analogy)

M

EI

EI

EI

M

M

EI

EI

EI

3 Storey Sway Frame (Rigid Floors) and

it's equivalent dynamic modelC

C

C

h

h

h

x3

x2

x1

c

k

M

1

c

p (t)

k

1

p (t)3

p (t)2

p (t)1

M

2

c

p (t)

k

1 M

3

Element 1 Element 2 Element 3v v v

1

1 2 3 2

3

0 00 0

0 0

m vm m m m m v

m v

Mv

01

3

1

01 124

1 1

Elemv SupportEI

vh

K x

12

3

2

1 124

1 1

ElemvEI

vh

K x 23

3

3

1 124

1 1

ElemvEI

vh

K x

1

23

3

2 1 0241 2 1

0 1 1

vEIv

hv

Kx

01

1

01 1

1 1

Elemv Support

Cv

C x

12

2

1 1

1 1

Elemv

Cv

C x

23

3

1 1

1 1

Elemv

Cv

C x

1

2

3

2 1 0

1 2 1

0 1 1

v

C v

v

Cx

Therefore the final set of three differential equations is:

1 1 1 1

2 2 2 23

3 3 3 3

0 0 2 1 0 2 1 0 ( )24

0 0 1 2 1 1 2 1 ( )

0 0 0 1 1 0 1 1 ( )

m v v v p t EI

m v C v v p t h

m v v v p t


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Natural (Free) vibration frequencies and modes

Each SDOF system has a natural frequency, which depends on the stiffness of the

spring and the mass. This quantity is associated with the free vibration of the system,

i.e. when there is no damping and no external force. It will become apparent that in a

MDOF system, there will be n natural frequencies (one per degree of freedom) and anassociated mode shape with each natural frequency. To derive these for a MDOF

system, damping is ignored and no external forcing is applied. Therefore equation (B)

reduces to:

Mv + Kv 0 (C)

For a harmonic response (free vibration), solutions of these equations must be of the

form:

( ) ( )it q t iv a

where 1 2T

n ia a aia is a set of nodal displacements describing the

deflected shape of the system during vibration anddoes not vary with time.

( ) sin( )i i i iq t A t is the simple harmonic function describing the

variation of the displacements with time where iA and

i are constants depending on initial conditions.

Hence 2 ( )i iq t iv a Substituting these into Equation (C) and simplifying gives:

2 ( ) 0i i

q t i iKa Ma

Since ( ) 0iq t (trivial solution), then the natural frequency i and the mode shape ia must satisfy the following equation:

2

ii iKa Ma or2 0i iK M a

This is a matrix eigenvalue problem, since for a non-trivial solution the determinant of

the matrix must be zero, i.e.2

0 K M .

This is an n degree polynomial equation known as the characteristic or frequency

equation. It has n real roots, i since M and K are both symmetric and positive

definite.

Once a natural frequency is known then an associated eigen vector ia can be chosen

which represents the mode shape of the vibration. This is not an absolute amplitude of

vibration only the relative values of the nodal displacements.

The solution of the eigen value problem is not trivial, but there are well known and

robust algorithms available to solve it.

Example 4 revisited.Letting C 0 and f 0 for free vibration and substituting for M and K into equation


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(B) results in:

1 2 2 1 1 12

2 2 3 2 2 2

0

0

k k k a m a

k k k a m a

Expanding2 0 K M gives

2

1 2 1 2

2

2 2 3 2

0k k m k

k k k m

4 2

1 2 2 1 2 1 2 3 1 2 1 3 2 3[ ( ) ( )] 0m m m k k m k k k k k k k k

The solution of this equation gives two values for 2i ,

2

2 1 2 1 2 3 2 1 2 1 2 3 1 2 1 2 1 3 2 32

1 2

( ) ( ) [ ( ) ( )] 4 ( )

2i

m k k m k k m k k m k k m m k k k k k k

m m

which corresponds to two natural frequencies for the system, 1 2 .For example if 1 2m m m and 1 2 3k k k k

2 2 2 2 22

2 2

4 (4 ) 12 4 4 2

2 2

mk mk m k mk m k k k

m m m

Hence 2 21 23

andk k

m m

Example 5

2,000kg

6m

5m

EI5m

1,200kg

EI

EI EI

x

k

1

k

x2

Element 1 Element 2

1 ,

2 0 0 k

g

2 ,

0 0 0 k

g

k

1 2

3

For the above structure, calculate the two natural frequencies and sketch the mode

shapes. Take6 2

60 10 EI Nm .

1 2 21

1 2 2 3 12

0 0

0 0

i i

i i

v k k k vm

v k k k vm


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66

1 3 3

12 24(60 10 )2 11.52 10

5

EIk N m

h

66

2 3 3

12 12(60 10 )5.76 10

5

EIk N m

h

66

2 3 3

12 12(60 10 ) 3.33 106

EIk N mh

1,200 0

0 2,000kg

M

617.28 5.76

105.76 9.09

N m

K

Characteristic equation

2 6 217.28 5.76 1,200 0

0 10 05.76 9.09 0 2,000

K M

6 2 6

6 6 2

17.28 10 1,200 5.76 100

5.76 10 9.09 10 2000

6 4 9 2 122.4 10 45.47 10 123.89 10 0

9 9 2 6 12

2

6

45.47 10 (45.47 10 ) 4(2.4 10 )(123.89 10 )

2(2.4 10 )

9472.9 6173.8

1

2

57rad/sec

125

Mode Shapes (Eigen vectors)

2 2 2

1 3299.1 / rad s

12 6

2

13.32 5.76 00 10

5.76 2.49 0

a

a

K M a

Let 1 1.0a and then 2 2.3125a , Normalising gives1

2

0.43

1.0

a

a

2 2 2

1 15646.7 / rad s

12 6

2

1.50 5.76 00 10

5.76 22.20 0

a

a

K M a

Let 1 1.0a and then 2 0.260a , giving1

2

1.0

0.260

a

a


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2,000kg

1,200kg

1.01.0

0.430.43

2,000kg

1,200kg

1.0

0.26 0.26

Mode 1 - mode shape Mode 2 - mode shape

1

1

1

5757

9.072

10.11

9.07

rad/sec

f Hz

T s

2

2

2

125125

19.92

10.05

19.9

rad/sec

f Hz

T s


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Continuous Systems: Flexure of Beams

Here we will consider only the free undamped case: concentrate on natural frequencies

of vibration (SHM). Only slender beams are considered for which shear defoemation

and rotary inertia are negligible. Also axial force effects are neglected and vibration is

considered about a principal plane.

z

v(x,t)

y

x

Figure 9: Simple continuous beam undergoing free vibration (a) Deflected shape at

time, t. (b) Axis definition.

Figure 9 shows a simple continuous beam which is vibrating and for which the

deformed shape at any instant is ( , )v x t measured from the static equilibrium position

(gravity effects are also neglected). From simple beam theory2 2

2 2( , ) ( , )EI v x t q x t

x x

(22)

where ( , )q x t is the distributed loading per unit length. Here2

2

( , )

( , )

v x t

q x t A t

(23)where A is the cross-sectional area so that for constant EI

4

40EI v Av

x

(24)

For free vibrations ( , )v x t is a harmonic function of time so that

( , ) ( )sin( )v x t V x t (25)Substituting Equation 25 into 24 results in

44

4

( )( ) 0

d V xV x

dx (26)

where2

4 AEI

(27)

This 4th

order equation has the general solution of the form

1 2 3 4( ) sin cos sinh coshV x A x A x A x A x (28)For specific beam configurations the end conditions may be used to determine the 4

constantsi

A using:

1 2 3 4

2

1 2 3 4

3

1 2 3 4

cos sin cosh sinh

sin cos sinh cosh

cos sin cosh sinh

V A x A x A x A x

V A x A x A x A x

V A x A x A x A x

(29)


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Example (i): Simply supported beam

L

udm = A Flexural Stiffness = EI

Bending moment and deflection are zero at both ends

i.e.(0) 0 ( ) 0

(0) 0 ( ) 0

V V L

EIV EIV L

where

2

2

d VV

dx

For end 0x 2 4 2 42 4

(0) 0 00

(0) 0 0

V A AA A

V A A

For end x L 1 3

1 3

( ) 0 0 sin sinh

( ) 0 0 sin sinh

V L A L A L

V L A L A L

The only nontrival solution is 3 0A and sin 0L leading to:1,2,3,L m m

Hence 42 2

m m where2

4

1

EI

A L

Mode 2

Mode 3

Mode 4

Mode 5

Mode 11 2

19.87

EI

A L

2 2

1

39.48

EI

A L

3 2

188.83

EI

A L

4 2

1157.91

EI

A L

5 2

1246.74

EI

A L

The mode shapes are therefore defined as sin 1, 2, 3,m mm

V A x mL

so that

( , ) sin sin( )m m m mm

v x t A x t L

(30)

and1

( , ) sin sin( )m m mm

mv x t A x t

L

(31)

where &m mA are determined by the initial conditions of the system


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Example (ii): Cantilever beam

L


Bending moment and shear are zero at free end. At the encastre end the deflection and

slope are zero

i.e.(0) 0 ( ) 0

(0) 0 ( ) 0

V EIV L

V EIV L

For end 0x 3 1

4 2

0 ; 0

0 ; 0

A Ax V

A Ax V

Hence 1 2( ) (sin sinh ) (cos cosh )V x A x x A x x (32)

For end x L we get:

1

2

; 0 sin sinh cos cosh 0

; 0 cos cosh sin sinh 0

Ax L V L L L L

Ax L V L L L L

(33)

The only non-trivial solution to these equations is when the determinant is zero, hence

cos cosh 1 0L L (34)

There is no closed form solution to this equation, but it may be solved numerically.

Mode 1 2 3 4 5

( )mL 1.875 4.694 7.855 10.996 14.137

Using Equation 33, it is possible to rewrite Equation 32 as:

1

sin sinh( ) sin sinh (cos cosh )

cos cosh

L LV x A x x x x

L L

The term in brackets represents the mode shape of vibration and substituting the roots ofEquation 34 (i.e. values of ( )mL , where m is the mode of vibration) will give the mode

shape corresponding to the particular frequency. As before we may write

2 4 2

m m


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Mode 2

Mode 3

Mode 4

Mode 5

Mode 1

3.52 1

22.03 1

61.70 1

120.90

1

199.86 1

Example (iii): Sway beam

L


Consider the sway beam above, with flexural stiffness, EI, and UDM, A. At theencastre edn, the deflection and slore are zero and at the sway (roller) end the slope and

shear are zero:

i.e.(0) 0 ( ) 0

(0) 0 ( ) 0

V EIV L

V EIV L

For end 0x 24

130;00;0

AAAA

VxVx

Substituting forA3 andA4, and applying two more boundary conditions

1 2( ) (sin sinh ) (cos cosh )V x A x x A x x (35)For end x L

1

2



Ax L V L L L L

Ax L V L L L L

(36)

The only non-trivial solution to these equations (36) is when the determinant is zero,

hence

cos sinh sin cosh 0L L L L (37)


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There is no closed form solution to this equation, but it may be solved numerically.

Mode 1 2 3 4 5

( )mL 2.365 5.498 8.639 11.781 14.923

Using (37), it is possible to rewrite (35) as

1

cos coshcos cosh (sin sinh )

sin sinh

L LV A x x x x

L L

The term in brackets represents the mode shape of vibration and substituting the roots of

(37) (i.e. values of ( )mL , where m is the mode of vibration) will give the mode shape

corresponding to the particular frequency. As before we may write2 4 2

m m Expressions for the first five natural frequencies of free vibration are given in the table.

The corresponding mode shapes are also shown.

Mode 2

Mode 3

Mode 4

Mode 5

Mode 1

1 2

15.59

EI

A L

22

130.23

EI

A L

3 2

174.64

EI

A L

4 2

1138.79

EI

A L

5 2

1222.68

EI

A L


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Approximate methods for Continuous Systems

As seen from the previous section, it is difficult to solve the governing differential

equations except for the simplest of cases. Therefore other methods, which allow the

calculation of natural frequencies of continuous systems are desirable. Energy methods

are one approach to this problem and Rayleighs method will be considered here.

Rayleighs Method

For a conservative system, the total energy is constant. Thus for free un-damped

vibration where the energy is partially kinetic and partially strain energy, it holds thatconstantE T U

where: E, T & U denote the Total Energy, Kinetic Energy and Strain Energy of the

system respectively.

Mi-1 MiKj-1 Kj

xx

x

i-1

i

j-1

xj

Flexural Stiffness = EI(x)Distributed Mass = A(x)

Figure 10: Typical beam system with concentrated masses and supporting springs.

Now consider a beam system with concentrated masses iM at locations ix and with

discrete support springs jK at locations jx . For such a beam undergoing free harmonic

vibration, the displacement may be written as (from equation 25):( , ) ( )sin( )v x t V x t (38)where ( )V x is the mode shape. It follows that the velocity is

( , ) ( )cos( )v x t V x t (39)Clearly: when the displacement is a maximum, then the velocity is zero;

when the displacement is zero, then the velocity is a maximum.

Therefore max max0 0E T U T U

So that max maxT U (40)For the beam system the strain energy is:

2 21 1

[ ]2 2Beam Springs j jU U U EI v dx K v 2 2

max

1 1[ ( )] [ ( )]

2 2j jU EI V x dx K V x (41)

The corresponding kinetic energy is:

2 21 1

2 2Beam Masses i iT T T Av dx M v

2 22 2

max [ ( )] [ ( )]2 2

i iT A V x dx M V x

(42)Hence from Equation 40 it follows that:


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2 2

2

2 2

[ ( )] [ ( )]

[ ( )] [ ( )]

j j

i i

EI V x dx K V x

A V x dx M V x

(Rayleighs Quotient) (43)

where ( )V x is the mode shape and is the natural frequency for this mode. It is

noteworthy that the for any kinematically admissable function ( )V x , Rayleighs methodgives an upperbound estimate of the frequency.

Example (vii)

Simply supported beam of span L with central concentrated mass M AL . Negect

the mass of the beam and take ( ) sinV x x L (kinematically admissable)

Hence

2

( ) sinx

V xL L

4

2

4 40

2 2 2 2

2 4

sin 1 48.7

2 2sin

2

L

UPPER BOUND

xEI dx EIL L

A LAL

Compare this solution with the exact solution from Example (iii): 2 248

EXACT

Example (viii)

As Example (vii) but include the mass of the beam.4

2

402 2 2

2

2

0

sin

(1 2 )sin sin

2

L

L

x

EI dxL L

xA dx AL

L

Example (ix)

As example (viii), except the concentrated mass 0M AL i.e. 0 . Therefore2 4 2

This is the exact solution since the assumed mode shape ( ) sinV x x L is the truemode shape for this case.

Example (x)

L


Assume

2 3

( ) 3 2x x

V xL L

(kinetmatically admissable)2

1( ) 6 12

xV x

L L

2

2

max 2 30

1 1 6[ ( )] 6 12

2 2

LEI x EIU EI V x dx dx

L L L


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22 32 2

2 2

max0

13[ ( )] 3 2

2 2 70

L x xT A V x dx A dx AL

L L

For a conservative system max maxT U which leads to:

2 2 23 413 6 420 1 32.370 13

EI EIALL A L

5.68 or4

0.905 0.9052

EIf f

AL

Note:

Compare above with Example (vi) [analytical solution]

1 5.59ANALYTICAL

or 1 0.877ANALYTICAL

f (3.5 % error)

This indicates that the assumed mode shape is a good approximation to the actual mode

shape corresponding to the first mode of free vibration.

Dunkerlys Equation for Multi Degree of Freedom (MDOF) Systems

A MDOF system, of degree n, will have associated with it n natural fequencies, each of

which will have a corersponding mode shape. Dunkerleys equation is:

2 2 2 2 2 2

1 2 11 22

1 1 1 1 1 1

n nn (44)

or for a contimuous system

2 2 2 2

1 2 11 22

1 1 1 1

(45)

where: i is the thi natural frequency of the system and,

ii ii iik M is the natural frequency of the system with only iiM present.

Neglecting modes 2 onwards, then a lower boundto the first natural frequency is given

by2 2 2 2

1 11 22 33

1 1 1 1

(46)

This lower bound method complements Rayleighs method, which is upper bound.

Example (xi)

L

Flexural Stiffness = EIDistributed Mass = A

M = AL

M = AL

A = 0

Subsystem (i)

42 2

112

- example (vii)

A

Subsystem (ii)

2 2 4

22 - example (ix)


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Combining using Dunkerleys equation:

2 2 2 4 2 4 2 4 2

1 11 22

1 1 1 2 1 (2 1)

42 2

1 (2 1)

Note:

The combined solution here using Dunkereleys equation is exactly the same as that

from example (viii) using Rayleighs method.

Example (xii)

h

h

2M

AEI

AEI

hAEI

2M

2M

AEI

AEI

AEI

A

A

A

M

M

M M

M

M

A

A

AV

V

A

B

VC VC

3 Storey SwayFrame (Rigid

Floors)

Dynamic Model with approximate

mode shape(1st Nat. Freq.)

(i) (ii) (iii) (iv)

Sub-systems

This problem consists of a 3 storey frame with rigid floors. The assumed mode shape

for a single storey is to be taken as2 3 2

2

6( ) 3 2 ( )

6( ) 1 2

x x x xV x V x

h h h h h

xV xh h

A piecewise combination of the above mode shape will give the assumed shape for 2

and 3 storeys. Take the storey mass as a factor of the column self-weight, i.e.

M AL and define:

2

4

1

EI

A h


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Subsystem Mode Shapes

Subsystem (i)

2 2

0

( )

( )

A

h

A

V x

V x

V

V

Subsystem (ii)

2 2 2 2

0 0

( ) ( ) ( ) ( )

( ) 2 ( )

B A

h h

B A

h V x V h V x

V x V x

V V

V V

Subsystem (iii) & (iv)

2 2 2 2

0 0

( ) ( ) ( ) 2 ( ) ( )

( ) 3 ( )

C A B

h h

C B

h h V x V h V x

V x V x

V V V

V V

Apply Rayleigh Method to each Subsystem

Subsystem (i)

2 2

max 30 0

1 1 6dx ( ) dx

2 2

h h

A

EIU EI EI V x

hV [see Example (x)]

2 2 2

2sub(i) sub(i) sub(i)2

max ( ) ( )(1)2 2 2

AT m h Ah AhV

2

sub(i) 2 2

max max sub(i)3

6 12

2

EIU T Ah

h

Subsystem (ii)

2 2

max 30 0

1 1 12dx 2 ( ) dx

2 2

h h

B

EIU EI EI V x

hV

2 2

2sub(ii) sub(ii) 2 2

max sub(ii)( ) ( )(1 1) 22 2

BT m h Ah AhV

2 2 2

max max sub(ii) sub(ii)3

12 62

EIU T Ah

h

Subsystem (iii)

2 2

max 30 0

1 1 18dx 3 ( ) dx

2 2

h h

C

EIU EI EI V x

h

V

2 2

2sub(iii) sub(iii) 2 2

max sub(iii)

9( ) ( )(2 1)

2 2 2

CT m h Ah AhV

2 2 2

max max sub(iii) sub(iii)3

18 9 4

2

EIU T Ah

h

Subsystem (vi)

2 2

max 30 0

1 1 18dx 3 ( ) dx

2 2

h h

C

EIU EI EI V x

hV ( as subsystem (iii))


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22 2 2sub(iv)

max

22 2 2sub(iv)

0

dx dx dx2

2 ( ) 1 ( ) ( ) dx2

C B A

h

T A A A

AV x V x V x

V V V

Expand and arrange to get:2

sub(iv) 2

max0

2

sub(iv) 2

sub(iv)

3 ( ) 6 ( ) 5 dx2

13 1 3193 6 5

2 35 2 70

hA

T V x V x

Ah Ah

2 2 2

max max sub(iv) sub(iv)3

18 319 1260

70 319

EIU T Ah

h

Apply Dunkerleys Equation to combine subsystems

2 2 2 2 2

1 sub(i) sub(ii) sub(iii) sub(iv)

1 1 1 1 1

2 2 2

1

1 1 1 1 1 319 1 319

12 6 4 1260 2 1260

2 2

1

1

(0.5 .2531746)

Comparison with Exact

This approach is compared with a numerical model based on a lumped mass system,

using the values given. The "exact" value, shown on the table, is computed using

DRAIN-2DX software (developed at the University of California, Berkeley).

152 152 23kg UC 11 4 -5 4 -3 22.0 10 / 1.25 10 2.92 10

23 / 3.6 65.428xx

E N m I m A m

A kg m h m r mm

2 2647.15 s

n fn Tn Tn Exact cycles / s /s s s 0 50.56 8.047 0.124 0.136

10 11.10 1.766 0.566 0.524

20 7.94 1.264 0.791 0.729

30 6.51 1.037 0.965 0.888

40 5.65 0.900 1.112 1.022

50 5.06 0.806 1.241 1.141

60 4.63 0.736 1.359 1.248

70 4.28 0.682 1.466 1.347

80 4.01 0.638 1.567 1.43990 3.78 0.602 1.662 1.526


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100 3.59 0.571 1.751 1.608

First four normal mode shapes for case with =50.

Problem Excercises

Using Rayleighs method combined with Dunkerleys equation, estimate the first natural

frequency of the two frames below. For both problems take m Ah and express nT as a function of. If you wish, calculate nT for a range of values of and plot them

against the Exact values, given on the attached plots.

00.20.40.60.8

11.21.41.61.82

0 10 20 30 40 50 60 70 80 90 100

Tn

(secs)

Natural Period of Vibration - Example (ii)

Approx Exact

0.0

3.6

7.2

10.8

Mode 1

0.0

3.6

7.2

10.8

Mode 2

0.0

3.6

7.2

10.8

Mode 3

0.0

3.6

7.2

10.8

Mode 4


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Problem (i)

2M

2M

2M

h

h

AEI

AEI

hAEI

AEI

AEI

AEI

A

A

A

M

M

M

3 Storey Sway

Frame

Dynamic Model

with approximatemode shape(1st Mode of

Vibration)

Structural Model

x

KinematicallyAdmissable

Mode

Shape

V(x)

2 3

( ) 3 23 3

x xV x

h h

Problem (ii)

2M

2M

2M

h

h

AEI

AEI

hAEI

AEI

AEI

AEI

A

A

A

M

M

M

3 Storey SwayFrame

Dynamic Model with approximate

mode shape(1st Mode of

Vibration)

Structural Model

x

KinematicallyAdmissableMode

Shape

V(x)

( ) 1 cos6

xV x

h


30/32



Problem Data and Results11 4 -5 4 -3 22.0 10 / 1.25 10 2.92 10

23 / 3.6 65.428xx

E N m I m A m

A kg m h m r mm

Tn values (secs)

Prob. (i) Prob. (ii)

Exact Approx Exact Approx0 0.3974 0.6323

10 1.5522 2.7437

20 2.1589 3.8287

30 2.6292 4.6679

40 3.0273 5.3778

50 3.3788 6.0043

60 3.6970 6.5714

70 3.9900 7.0932

80 4.2628 7.5793

90 4.5192 8.0359

100 4.7618 8.4680


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Natural Period of Vibration - Problem (i)

0.0

1.0

2.0

3.0

4.0

5.0

0 10 20 30 40 50 60 70 80 90 100

Tn(

secs)

0

3.6

7.2

10.8

Mode 1

0

3.6

7.2

10.8

Mode 2

0

3.6

7.2

10.8

Mode 3

0

3.6

7.2

10.8

Mode 4


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0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.08.0

9.0

0 10 20 30 40 50 60 70 80 90 100

Tn

(secs)

Natural Period of Vibration - Problem (ii)

0

3.6

7.2

10.8

Mode 1

0

3.6

7.2

10.8

Mode 2

0

3.6

7.2

10.8

Mode 3

0

3.6

7.2

10.8

Mode 4

Dynamics v2.2

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