mechanical vibration

Eng5932 - Mechanical Vibrations

1 1.1

1

Introduction De…nitions Vibration – A motion that repeats itself after a time interval or oscillation (e.g. pendulum, plucked guitar string). Vibrating system – Consists of a mass, spring and possibly a damper.

– Mass (or inertia element for rotating systems) is a rigid body that stores kinetic energy – Spring stores potential energy – Damper dissipates energy

1.2

Spring Elements A spring element is generally assumed to have no mass and no damping. It will store potential energy due to stretch (or compression) or twist (for torsional springs). For a linear spring the spring force is proportional to the change in length of the spring element: F = kx (1) where, x is the measured deformation of the spring from its undeformed length, and k is the spring constant.


2

The work done deforming a spring from x1 to x2 is: Z x2 Z x2 k kx dx = (x22 F dx = U12 = 2 x1 x1

x21 )

(2)

and this work is stored as potential energy. The spring constant of a helical spring is: k=

Gd4 8nD3

(3)

wehere G is the shear modulus of the spring material, d is the diameter of the wire, D is the mean coil diameter, and n is the number of active turns of the coil. A rod can act as a spring (or consider an elevator cable when the winding drum suddenly stops). Consider a rod being stretched by a force F :

For linear elastic behaviour Hooke’s Law applies: =E Since

= F=A and

=

x=L, then: FL =E A x


3

or

AE x=k x L i.e. the linear spring constant for a rod (or cable in tension) is: F =

k=

1.3

AE L

(4)

Damping Elements Damping is the mechanism by which vibrational energy is converted into heat or sound. A damping element is assumed to have zero mass and elasticity. There are three standard models of damping: (1) viscous damping; (2) Coulomb or dry friction damping; and (3) hysteretic damping. Viscous Damping: – Fluid dynamic drag is used to dissipate energy (e.g. air drag on a pendulum, drag as a liquid is forced through an ori…ce (shock absorber)). – Viscous damping is the most common form of damping used in vibrating systems. – The damping force is proportional to the rate of change of length of the damping element or the relative velocity between the two ends of the damping element. F = cx_

(5)

– Note: Kinematics of a rigid body is used to give the position, velocity and acceleration of various components of a suspension system. The distance between the mounting points of the spring is used to give the spring force. The rate of change of the distance between the mounting points of the damper is used to give the damping force. If the spring and damper are coaxial, only one length is of interest. Coulomb Damping: – Coulomb damping is constant in magnitude but opposite in direction to the motion. – Due to the friction between rubbing surfaces (dry or with insu¢ cient lubrication). F = Hysteretic, Solid or Material Damping:

kN

(6)


4

– As a material is deformed, the work done on the material is stored as strain energy or heat. – Since energy is absorbed, a body subjected to material damping shpows a hysteresis loop on a stress-strain diagram.

1.3.1

A Simple Viscous Damper Consider two parallel plates separated by a distance h. The bottom plate is …xed and the top plate moves to the right at a constant speed v. The space between the plates is occupied by a ‡uid with dynamic viscosity .

Solution of the Navier-Stokes equation show that the velocity pro…le is linear, i.e. Couette ‡ow. The shear stress in the ‡uid is: =

@u v = @y h


5

The drag (i.e. frictional) force exerted by the ‡uid on the bottom plate is: F = A jy=0 =

A v h

where c = A=h is the damping constant and the damping force behaves linearly wrt v.


2

6

Free Vibration of a One Degree of Freedom System

2.1

Undamped Motion of a Single DOF System Consider a mass, m, resting on a ‡at frictionless surface and connected to a rigid support by a linear spring, k:

2.1.1

Equation of Motion We will need an equation governing the motion of the mass as it is moved from its equilibrium position. Such an equation of motion can be derived from Newton’s 2nd Law using the following steps: 1. Sketch the system and de…ne a suitable positive co-ordinate direction to de…ne the motion of the mass. 2. Determine the static equilibrium position of the system. It is common (and recommended) to de…ne the positive co-ordinate from the equilibrium position. 3. Draw a free body diagram (FBD) of the mass, assuming positive displacement, velocity and acceleration are given to it. Draw the forces acting on the mass. 4. Use Newton’s 2nd Law to derive an equation of motion.


7

+!

X

Fx = max

m• x=

kx

or m• x + kx = 0

(7)

Equation (7) is the equation governing the free vibration of the mass. What about a mass, m, suspended from a rigid support by a linear spring, k?

+# m• x=

X

Fx = max

kx + (W

k

st )

But, from statics: W =k

st

so the governing equation is: m• x + kx = 0 Note: The same governing equation arises when x is de…ned from the equilibrium position (this is the advantage of de…ning x measured from equilibrium).

2.2

Solution of the Governing Equation The governing equation for undamped free vibration of a 1 DOF mass-spring system is an homogeneous ordinary di¤erential equation (ODE): m• x + kx = 0

(8)


8

Assume a solution of the form: x(t) = Cest then: x(t) _ = Csest x •(t) = Cs2 est Substitution of the assumed solution into Eq. (8) gives: mCs2 est + kCest = 0 or ms2 + k = 0

(9)

which is called the characteristic equation. The solution of Eq. (9) is: s=

k m

1=2

=

where !n =

k m

i! n

(10)

1=2

(11)

is the natural frequency of the undamped system. It is the frequency of oscillation of the undamped system. The two values of s are the roots or eigenvalues of the characteristic equation. Since both values of s will satisfy the governing equation, the solution for x(t) is written as: x(t) = C1 ei!n t + C2 e

i! n t

(12)

where C1 and C2 are constants to be determined from initial conditions. It is di¢ cult to interpret the motion given by Eq. (12), so let’s make it look pretty. Euler’s formula states: e

i! n t

= cos ! n t

i sin ! n t

Substitution of Euler’s formula into Eq. (12) gives: x(t) = C1 [cos ! n t + i sin ! n t] + C2 [cos ! n t

i sin ! n t]

or x(t) = (C1 + C2 ) cos ! n t + i(C1

C2 ) sin ! n t


9

Since x(t) is a displacement, it must be real, therefore, (C1 + C2 ) must be real and (C1 C2 ) must be imaginary. For this to be true C1 and C2 must be complex conjugates: C1 = a + ib ; C2 = a ib so: x(t) = 2a cos ! n t

2b sin ! n t

Then the displacement x(t) can be written as: x(t) = A1 cos ! n t + A2 sin ! n t

(13)

where A1 and A2 are new constants evaluated from the initial conditions of the system. Two conditions are required (i.e. the same number as the order of the governing equation). Specifying the initial displacement and velocity at t = 0: x(t = 0) = xo x(t _ = 0) = x_ o and solving for A1 and A2 : A1 = xo x(t) _ =

A1 ! n sin ! n t + A2 ! n cos ! n t x_ o = ! n A2 A2 =

x_ o !n

So the solution to Eq. (8) is: x(t) = xo cos ! n t +

x_ o sin ! n t !n

(14)

i.e. harmonic motion. Equation (14) is the motion of any system whose governing equation can be written in the form of Eq. (8). Note: 1. If xo = 0: x(t) =

x_ o sin ! n t !n

2. If x_ o = 0: x(t) = xo cos ! n t


10

Equation (13) can be written in a more convenient form by de…ning: A1 = A cos

(15)

A2 = A sin

(16)

Square and add Eqs. (15) and (16) to …nd A: A=

A21

+

1=2 A22

x2o

=

x_ o !n

+

2

!1=2

Divide Eq. (16) by Eq. (15) to …nd : 1

= tan

A2 A1

= tan

1

x_ o xo ! n

Substitution of Eqs. (15) and (16) into Eq. (13) gives: x(t) = A cos cos ! n t + A sin sin ! n t Using the trigonometric identity: cos(x

y) = cos x cos y + sin x sin y

Then x(t) = A cos(! n t where the amplitude A and phase angle A =

x2o

= tan

)

(17)

!1=2

(18)

are: + 1

x_ o !n

2

x_ o xo ! n

(19)

This is a very convenient form of the solution for x(t) as it makes the motion quite obvious. The velocity and acceleration of mass m are: x(t) _ =

! n A sin(! n t

) = ! n A cos(! n t

+

)

(20)

x •(t) =

! 2n A cos(! n t

) = ! 2n A cos(! n t

+ )

(21)

i.e. x(t) _ leads x(t) by =2 (or 90o ) and x •(t) leads x(t) by

2

(or 180o ).


11


12

The motion x(t) can also be expressed as a sine wave by de…ning A1 and A2 in Eq. (13) as: A1 = A sin

o

A2 = A cos

o

Then: A = A21 + A22

o

= tan

1

1=2

A1 A2

=

x_ o !n

x2o + = tan

1

2

!1=2

xo ! n x_ o


13

Substitution of the de…nitions of A1 and A2 into Eq. (13) gives: x(t) = A sin

o cos ! n t

+ A cos

o sin ! n t

Using the trigonometric identity: sin(x + y) = sinx cos y + cos x sin y Then x(t) = A sin(! n t +

o)

with the same magnitude as Eq. (17), but a new phase angle, o

2.3

= tan

1

(22) o:

xo ! n x_ o

(23)

Free Vibration with Viscous Damping Consider a mass, m, resting on a ‡at frictionless surface and connected to a rigid support by a linear spring, k, and a viscous damper, c:

We will need an equation governing the motion of the mass as it is moved from its equilibrium position. Such an equation of motion can be derived using Newton’s 2nd Law. 2.3.1

Equation of Motion The FBD for a 1 DOF mass-spring-damper system is shown below:


14

To derive the equation of motion assume a positive displacement, velocity and acceleration of the mass m. Newton’s 2nd Law gives: +!

X

Fx = m• x=

Fs

Fd =

kx

cx_

where, for a linear spring Fs = kx, and a viscous damper Fd = cx. _ The governing equation can be rearranged to give: m• x + cx_ + kx = 0

(24)

This equation governs the motion of all 1 DOF systems in translation with a linear spring and viscous damper. 2.3.2

Solution of the Governing Equation The governing equation for viscously damped vibration of a 1 DOF system is an homogeneous ODE: m• x + cx_ + kx = 0 As for undamped motion, assume a solution of the form: x(t) = Cest Substitution of the assumed solution into the governing equation, Eq. (24), gives the following characteristic equation: mCs2 est + cCsest + kCest = 0 or ms2 + cs + k = 0

(25)


15

i.e. a quadratic equation in s. The roots of the characteristic equation are: r p k c c c2 4mk c 2 = s1;2 = 2m 2m 2m m

(26)

So the solution to Eq. (24) is: x(t) = C1 es1 t + C2 es2 t or x(t) = C1 e

c + 2m

q

2

c ( 2m )

k m

t

+ C2 e

c 2m

q

2

c ( 2m )

k m

t

(27)

where, C1 and C2 are constants obtained from initial conditions. This form of the solution is hard to interpret so other forms are used. De…ne: 1. Critical Damping, cc , as the damping constant c which makes the radical in Eq. (26) zero: cc 2 k =0 2m m or r p k cc = 2m = 2 km = 2m! n (28) m 2. Damping ratio, , as the ratio of the damping constant c to the critical damping constant cc : = c=cc (29) Using Eqs. (28) and (29):

c c cc = = !n 2m cc 2m

(30)

Using Eqs. (30) and the de…nition of ! n , Eq. (11), the roots s1 and s2 , Eq. (26), can be written in terms of and ! n : q 2 s1;2 = 1 !n (31) The solution of Eq. (24) can be rewritten in the following form: p2 p2 + 1 !n t 1)! n t x(t) = C1 e + C2 e( Note: if

(32)

= 0, the solution reduces to that for undamped free vibration, Eq. (12).

It should be obvious that due to the 2 1 term in the radical di¤erent types of solution will arise if < 1, = 1 and > 1. These three cases correspond to underdamped, critically damped, and overdamped cases, respectively. The solutions for these di¤erent cases will now be considered separately.


2.3.3

16

Case 1: Underdamped Motion ( < 1, c < cc ) p p 2 2 1=i 1 , therefore, the roots s1 and s2 can If < 1, then 2 1 < 0, and be written as: q 2 s1;2 = i 1 !n =

!n

i! d

(33)

where ! d is the damped frequency of oscillation: q 2 !d = 1 !n

(34)

The solution to Eq. (24) can be written in the following form: ! n +i! d )t

x(t) = C1 e(

+ C2 e(

! n i! d )t

or x(t) = e

!n t

C1 ei!d t + C2 e

i! d t

(35)

Compare Eq. (35) with Eq. (12) for undamped motion. The solution will be harmonic with p an exponentially decaying amplitude. The frequency of oscillation will be ! d = 2 1 ! n , i.e. less than ! n for undamped motion. Due to the similarities of Eqs. (12) and (35) the same techniques may be used to rewrite Eq. (35) in more easily interpreted forms. x(t) = e

!n t

[(C1 + C2 ) cos ! d t + i(C1

C2 ) sin ! d t]

or x(t) = e

!n t

C10 cos ! d t + C20 sin ! d t

Then: x(t) = Xe

!n t

sin (! d t +

x(t) = Xe

!n t

cos (! d t

o)

(36)

)

(37)

or

Where the …nal two forms of the solution clearly show that the motion is harmonic with frequency ! d and exponential decay in the amplitude. The constants C10 and C20 can be derived from the initial conditions: x(t = 0) = xo x(t _ = 0) = x_ o


17

and solving for C10 and C20 : x(t) _ =

!ne

!n t

C10 = xo

C10 cos ! d t + C20 sin ! d t + e x_ o = C20 =

(38) !n t

! d C10 sin ! d t + ! d C20 cos ! d t

! n C10 + ! d C20 x_ o + ! n xo !d

and o are de…ned as follows: q X = (C10 )2 + (C20 )2

(39)

The amplitude X and phase angles

o

= tan

= tan

1

(C10 =C20 ) 1 (C20 =C10 )

(40) (41) (42)

The amplitude X is the maximum amplitude that would exist at t = 0, for zero phase angle. A typical underdamped system response is shown below.


18

The motion is harmonic, with exponential decay of amplitude and damped frequency of oscillation, ! d , which is always less than ! n since < 1. Vehicle suspension systems are examples of underdamped systems. Case 2: Critically Damped Systems ( = 1, c = cc )

2.3.4 If

= 1 the roots s1 and s2 are not distinct: cc = 2m

s1 = s2 =

!n =

To determine the motion take the limit of Eq. (36) as !n t

lim x(t) = lim e !1

!1 !n t

= e

!n

(43)

! 0.

C10 cos ! d t + C20 sin ! d t

(C10 + C20 ! d t

Remember: !d =

q 1

2

!n

lim (cos ) = 1 !0

lim (sin ) = !0

So the solution can be written as: x(t) = (C10 + C20 ! d t)e

!n t

Using Eqs. (38) and (39), the motion of a critically damped system is given by: x(t) = (xo + (x_ o + ! n xo ) t) e

!n t

(44)


19

The motion is not harmonic, it is aperiodic and decreases to zero as t increases due to the e !n t term. Critical damping, cc , is the lowest value of c that will not produce oscillations. A typical example of critically damped motion is shown below with an underdamped case of the same system for comparison.

2.3.5

Case 3: Overdamped System ( > 1, c > cc ) For this case, the roots s1 and s2 are both real, distinct and less than 1. q 2 s1 = + 1 !n q 2 s2 = 1 !n and s2 < s1 .

(45) (46)


20

The solution is: x(t) = C1 e(

+

p

2

1)! n t

+ C2 e(

p

2

1)! n t

(47)

For initial conditions: x(t = 0) = xo x(t _ = 0) = x_ o the constants are: C1 = C2 =

xo ! n ( + 2! n xo ! n ( 2! n

p

p

p

p

2

1) + x_ o

2

1 2

2

1) 1

x_ o

(48)

(49)

The motion is aperiodic. It has a similar form to the critically damped solution, however, it overshoots the maximum amplitude of the critically damped system and takes longer to return to equilibrium.


21

A good example of an overdamped system is the closer on a screen door which takes forever and an age to close. 2.3.6

Summary

Underdamped motion: – is harmonic with exponential decay of amplitude.

p – The frequency of oscillation of underdamped motion (! d = 1 less than the frequency of oscillation of undamped motion (! n ).

2

! n ) is always

– Underdamped motion will not reach the same peak amplitude as undamped motion, unless the phase angle is zero. – The underdamped case is the only solution that results in oscillatory motion. – Underdamped systems return to equilibrium quickly, but continue to oscillate. In the context of a suspension system the wheels would be in position to respond to another bump more quickly than a critically damped system. Critically damped motion:


22

– is aperiodic with exponential decay to equilibrium position. – Critically damped systems are quickest to return to the equilibrium postion. – A critically damped system would transmit too much force to car passengers. Vehicle suspensions are underdamped. Overdamped systems: – take a long time to return to the equilibrium position. This is not practical for suspension systems. These systems are useful for controlled, slow return to equilibrium without oscillation. Screen door closers are overdamped. 2.3.7

Logarithmic Decrement Consider the motion of an underdamped system.

The ratio of displacements at two times separated by one period of damped oscillation is: x1 Xe !n t1 cos(! d t1 ) = ! t n 2 cos(! t x2 Xe ) d 2 But t2 = t1 +

d

where

d

= 2 =! d .

Since the angle in the cosine terms would be di¤erent by 360o (2 radians or one full oscillation) the cosines would be equivalent. e !n t1 x1 = x2 e !n (t1 +

d)

De…ne the logarithmic decrement as the natural log of the displacement ratios: = ln

x1 = !n x2

d

= !n

2 p !n 1

2

=p

2 1

2

=

c 2 ! d 2m

(50)


23

Note: in the log domain the amplitude decay is constant. In the …gure above the dashed line is Xe !n t . The damping ratio, , can be de…ned as a function of : =q

(51) (2 )2 +

2

How can logarithmic decrement be used? 1. Given a system with a known ! n and both amplitude ratio and d are speci…ed ! , i.e. the damping ratio and thus the damping constant. Suspension design. 2. Given a system with known ! n and analysis.

! , i.e. the amplitude ratio. Suspension

Experimental measurement of time and displacement can be used to determine the damping constant. It is not necessary to use only one period of oscillation, however, as any numnber of periods may be used: x1 xm+1

=

x1 x2 x3 x2 x3 x4

xm xm+1

where m is the number of cycles. Note: m may be a partial cycle. Since

x1 =e x2

then

x1 xm+1

or ln and =

!n

= em

x1 xm+1 1 ln m

d

!n

d

=m x1 xm+1

(52)


3

24

Forced Vibration - Harmonic Excitation Forced vibration is caused by an external applied force or displacement. Harmonic excitation implies the applied force is of the form F (t) = Fo ei(!t+ ) of F (t) = Fo cos(!t + ) or F (t) = Fo sin(!t + ). The phase angle, , is often chosen to be zero as only the steady state response of the system is of interest.

3.1

Equation of Motion Consider a 1 DOF spring-mass-damper system subjected to an external appled force.

The governing equation is: m• x + cx_ + kx = F (t)

(53)

Equation (53) is a nonhomogeneous ODE that has solution x(t) that is a sum of an homogeneous solution xh (t) and a particular solution xp (t). The homogeneous solution satis…es: m• x + cx_ + kx = 0 i.e. the equation for free vibration. The solution xh (t) will, therefore, decay to zero for any under, over or critically damped system. The solution to Eq. (53) will, after some transient during which the homogenous solution decays to zero, reduce to the particular solution, xp (t).


25

Harmonic response will be considered separately for undamped and damped motion.

3.2

Undamped Motion Consider Eq. (53) for the undamped case when F (t) = Fo cos(!t): m• x + kx = Fo cos !t

(54)

xh (t) = C1 cos ! n t + C2 sin ! n t

(55)

The homogeneous solution is:

where ! n =

p k=m.

Since F (t) is harmonic, the particular solution is also harmonic. Propose: xp (t) = X cos !t

(56)

Substitue Eq. (56) into Eq. (54): m! 2 X cos !t + kX cos !t = Fo cos !t then X=

k

Fo = m! 2

st

1

! !n

2

(57)

where st

is the static de‡ection due to force Fo .

= Fo =k

(58)


26

The total solution is: x(t) = xh (t) + xp (t) = C1 cos ! n t + C2 sin ! n t +

Fo cos !t k m! 2

Using the initial conditions: x(t = 0) = xo , x(t _ = 0) = x_ o : xo = C1 +

Fo k m! 2

C1 = xo

Fo k m! 2

So:

x(t) _ =

C1 ! n sin ! n t + C2 ! n cos ! n t

(59)

k

Fo ! sin !t m! 2

then x_ o = C2 ! n or C2 =

x_ o !n

(60)

And the total solution for the motion is: x(t) =

xo

Fo k m! 2

cos ! n t +

x_ o sin ! n t + !n

Fo k m! 2

cos !t

(61)

Interesting stu¤ can be seen by writing Eq.(57) in a non-dimensional form: X st

1

= 1

! !n

2

(62)

which is callled the amplitude ratio (or amplitude factor or magni…cation factor). It is the ratio of the dynamic amplitude to the static amplitude of the motion. A plot of X=

st

versus !=! n yields:


27

As shown by the …gure there are three distinct behaviours. Case 1: 0 < !=! n < 1 – The frequency of the input force is less than the natural frequency of the system and the denominator in Eq. (62) is positive. – The steady state motion is given by Eq. (61) and the harmonic response of the system, xp (t), is in phase with the forcing function.


28

Case 2: 1 < !=! n < 1 – The frequency of the input force is greater than the natural frequency of the system and the denominator in Eq. (62) is negative, therefore, the amplitiude ratio is negative. The motion of the system, xp (t), Eq. (56), will be out of phase with the external force.

– As !=! n ! 1, X ! 0, i.e. zero response to a high frequency input. In other words there is insu¢ cient time for the system to respond to the input. Later this concept will be used to show why you should always drive faster. Case 3: !=! n = 1 ! Resonance – The frequency of the input force equals the natural frequency of the system.


29

– X = 1, i.e. resonance and if energy is continuously fed into the system it will fail. Let’s prove it. – Equation(61) can be rewritten as: x(t) = xo cos ! n t +

Fo x_ o sin ! n t + (cos !t !n k m! 2

or x(t) = xo cos ! n t +

x_ o sin ! n t + !n

0

B cos !t

st @

cos ! n t) 1

cos ! n t C A 2

! !n

1

(63)

– Consider the response at ! = ! n . The last term in Eq. (63) is unde…ned when ! = ! n , so use l’Hôpital’s Rule: 0 1 1 0 B cos !t lim @ !!! n 1

d B d! (cos !t B lim !!! n @ d d! 1 # " t sin !t = lim 2!

cos ! n t C A = 2

! !n

!!! n

=

cos ! n t) C C A 2

! !n

! 2n

!nt sin ! n t 2

– So, the system response at resonance is: x(t) = xo cos ! n t +

x_ o sin ! n t + !n

st ! n t

2

which is harmonic with linear increase in amplitude.

sin ! n t

(64)


3.2.1

30

Total Response Similar to free vibration, x(t), can be written as follows: x(t) = A cos(! n t

st

)+ 1

! !n

2

cos !t

(65)

Note: Eq. (3.17) in the text is incorrect. The negative sign in front of the second term should be a positive, and then Eq. (3.170 is redundant.

3.2.2

Beat Phenomena Beating occurs when ! is close to, but not equal to, ! n . The amplitude will build up and diminish in a regular pattern.


3.3

31

Damped Motion Consider a 1 DOF system with a linear spring and viscous damper subjected to a harmonic applied force. The governing equation is: m• x + cx_ + kx = Fo cos !t

(66)

Assume a particular solution of form: xp (t) = X cos(!t

)

(67)

with derivatives: x_ p (t) =

!X sin(!t

)

x •p (t) =

! 2 X cos(!t

)

Substitution of the assumed solution into Eq. (66) gives: X (k

m! 2 ) cos(!t

)

c! sin(!t

) = Fo cos(!t)

(68)

The trigonometric relations: cos(!t

) = cos(!t) cos( ) + sin(!t) sin( )

sin(!t

) = sin(!t) cos( )

cos(!t) sin( )

can be used in Eq. (68) to give: X (k

m! 2 )(cos(!t) cos( ) + sin(!t) sin( ))

c!(sin(!t) cos( )

cos(!t) sin( )) = Fo cos !t

or X cos !t((k

m! 2 ) cos + c! sin ) + sin !t((k

m! 2 ) sin

c! cos ) = Fo cos !t (69)


32

Equating LHS and RHS coe¢ cients of cos !t and sin !t in Eq. (69): X (k

m! 2 ) cos + c! sin

= Fo

(70)

X (k

m! 2 ) sin

=0

(71)

c! cos

Equations (70) and (71) can be solved for X and . To solve for X, square each equation and then add the resultants to eliminate : X 2 (k

m! 2 )2 cos2

m! 2 ) cos sin + (c!)2 sin2

+ 2c!(k

+ (k

m! 2 )2 sin2

or X 2 (k

m! 2 )2 + (c!)2 = Fo2

So: X=

Fo [(k

m! 2 )2

+ (c!)2 ]1=2

(72)

and from Eq. (71): = tan

1

k

c! m! 2

(73)

A typical plot of a forcing function and steady-state system response, xp (t), is shown below. Here, the system motion lags the forcing function.

Using the de…nitions of undamped natural frequency, ! n , damping ratio, , static de‡ection, st , and frequency ratio, r: p ! n = k=m =

c c = ! c = 2 m! n cc 2m! n st

=

Fo k

2c!(k

m! 2 )


33

r=

! !n

(74)

the amplitude ratio (or magni…cation factor, M ), X= (72): X =

h

=

h

Fo =k (1

m! 2 2 k )

(1

( !!n )2 )2

st ,

st

i1=2

+ (2

! 2 !n )

2 + ( c! k )

can be obtained from Eq.

i1=2

So: M=

X st

2

=4 1

! !n

2

!2

! + 2 !n

2

3 5

1=2

=p (1

Similarly, , Eq. (73) can be written in terms of r and : 2 3 ! 2 r 6 2 !n 7 1 = tan 1 4 2 5 = tan 1 r2 ! 1 !n

1 r2 )2 + (2 r)2

(75)

(76)

Plots of M = X= st and phase angle, , versus frequency ratio are shown below for various values of .

Magni…cation factor comments:


34

1. Any amount of damping reduces M at all frequency ratios. Damping slows system response and leads to the phase angle, see Eq. (76). 2. As

", M # at any r, i.e. more damping ! less motion.

3. As r ", M #, i.e. the system does not respond to high frequency inputs.

4. Damping produces a signi…cant reduction in M near r = 1, i.e. damping is very important near resonance as it controls the motion. 5. For what value of r would M be a maximum, and what is Mmax ? dM = 0 = (1 dr

0=

q 1

r= then (1

2

2

2 2 ))2 + (2

(1

= (4 2 (1 1 p = 2 1

r2 )( 2r) + (2 )2 (2r)

2(1

r2 ) + (2 )2

2(1

– So:

Mmax =

1 2

3=2

r2 )2 + (2 r)2

2

))

1=2

(77) q

1=2

1

2 2 )2

2

(78)

– If is known, Eq. (77) gives r and Eq. (78) gives Mmax and Xmax = Mmax st gives an estimate of the maximum displacement. – Measurement of Xmax will give an estimate of (if Fo and k are known). Phase angle comments: 1. For

> 0, 0 < r < 1, then 0o <

< 90o , i.e. the response lags the input.

2. For > 0 r > 1 then 90o < < 180o , i.e. response is out of phase with input (similar to undamped system). 3.3.1

Total Response (Damped System) The complete expression for the response of the damped 1 DOF system to a harmonic input is: x(t) = Xo e !n t cos(! d t ) (79) o ) + X cos(!t Note: the …rst term in Eq. (79) is the homogeneous solution and it will decay to zero, and the solution will reduce to the steady-state solution, i.e. the second term which is the particular solution.


Xo and

o

35

are obtained from initial conditions: x(t = 0) = xo and x(t _ = 0) = x_ o xo = Xo cos( = Xo cos(

x_ = Xo (

!ne

!n t

cos(! d t

o)

x_ o = Xo (

! n cos(

= Xo (

! n cos(

o) o)

!de o)

o)

+ X cos(

)

+ X cos( ) !n t

sin(! d t

! d sin(

+ ! d sin(

o ))

o ))

(80) o ))

X! sin(!t

X! sin(

+ X! sin( )

)

) (81)

It is easier to solve for Xo and o with numbers from a particular problem, rather than derive a messy general solution.


4

36

Harmonic Motion of the Base Consider a 1 DOF viscously damped system subjected to a harmonic motion of the base.

The motion of the base is de…ned by: y(t) = Y sin !t

(82)

The spring force and damping force will be functions of both x and y: Fs = k(x

y)

Fd = c(x_

y) _

the governing equation for the motion is: m• x + c(x_

y) _ + k(x

y) = 0

(83)

Note: there is no forcing function. Since y(t) = Y sin !t, the governing equation can be written as: m• x + cx_ + kx = cy_ + ky = c!Y cos !t + kY sin !t = A sin(!t

)

(84)

where p k 2 + (c!)2 c! = tan 1 k

A = Y

(85) (86)


37

Comparison of Eq. (84) with Eq. (53) shows that they are of the same form (i.e. a damped 1 DOF system with a harmonic forcing function). The forcing motion is not transmitted directly to the mass, however, as it is transmitted through the spring and damper (e.g. a suspension system). The particular solution is then: p Y k 2 + (c!)2 xp (t) = sin(!t ) (87) 1 ((k m! 2 )2 + (c!)2 )1=2 where 1

= tan

c! k m! 2

1

(88)

i.e. similar to Eqs. (??, (??) and (??). Using the trigonometric identities for angle addition, Eq. (87) can be written as: xp (t) = X sin(!t where X = Y = tan

(k

k 2 + (c!)2 m! 2 )2 + (c!)2

1

k(k

1=2

mc! 3 m! 2 ) + (c!)2

=

)

(89)

1 + (2 r)2 (1 r2 )2 + (2 r)2

= tan

1

1=2

2 r3 1 + (4 2 1)r2

(90) (91)

The ratio of the amplitudes of the response to base input, X=Y , is called the displacement transmissibility, Td . The variation of Td and .

with frequency ratio are shown below for various values of


38

Td has a maximum value at: rm

1 = 2

q

1=2

1+8

2

1

The force, F , acting on the base is

F = c(x_

y) _ + k(x

y) =

m• x

(92)

Using Eq. (89) for x: F = m! 2 X sin(!t

) = FT sin(!t

)

where FT = m! 2 X is the amplitude of the force transmitted to the base. Using Eq. (90): X FT = r2 (93) kY Y where FT =kY is the force transmissibility. The behaviour of force transmissibilty with frequency ratio is plotted below for various values of damping ratio.


De…ne z = x written as:

39

y as the motion of the mass relative to the base, then Eq. (84) can be m• z + cz_ + kz =

m• y = m! 2 Y sin !t

(94)

Comparison of Eq. (??) with Eq. (84) gives the solution: z(t) = Z sin(!t where Z=p 1

m! 2 Y (k

m! 2 )2 + (c!)2

= tan

1

k

c! m! 2

=Yp

1)

(95) r2

(1

= tan

r2 )2 + (2 r)2

1

1

2 r r2

The variation of Z=Y with frequency ratio for several values of

(96) (97)

is shown below.


40


5

41

System Modelling

5.1

Combinations of Springs In practical systems several linear springs may be used in combination (e.g. four springs used in a car). An e¤ective spring constant can be developed for spring combinations in parallel and series.

5.1.1

Springs in Parallel Consider the case where two linear springs support a platform.

~ is applied to the platform such that it does not rotate, the springs will If a force W obtain the same static de‡ection, st . Using the FBD of the platform when it is in static equilibrium: X +" Fy = 0 = F1 + F2 W

But

F1 = k1

st

F2 = k2

st

therefore W = k1

st

+ k2

st

We want an equivalent spring keq that will support force W and attain the same static de‡ection, st . W = keq st


42

So keq = k1 + k2 For any combination of parallel springs: keq =

n X

ki

(98)

i=1

5.1.2

Springs in Series Consider two linear springs k1 and k2 attached in series:

~ is applied to spring 2 the springs will de‡ect a di¤erent amount, unless When load W k1 = k2 . Consider an FBD for each spring:

Each spring will carry the same load, W : W = k1

1

= k2

2

= keq

st

= keq (

1

+

2)

De…ne an equivalent linear spring keq that would carry load W , but de‡ect 1 + 2.

st

=


43

Note: 1

=

2

=

keq W = k1 k1 keq W = k2 k2

st

st

Then: st

or

=

1

+

=

keq k1

2 st

+

keq k2

st

1 1 1 = + keq k1 k2

For any combination of springs in series: n

X 1 1 = keq ki

(99)

i=1

5.1.3

General Method for Combinations of Springs Since springs are used to store potential energy it is possible to equate the potential energy stored due to the deformation of a new equivalent spring to that stored in the original spring system. e.g. springs in parallel

1 k1 2

1 1 + k2 2st = keq 2 2 ! keq = k1 + k2

2 st

2 st


44

e.g. spings in series:

1 k1 2

2 1

1 + k2 2

2 2

1 = keq 2

2 st

But

1

=

2

=

keq W = k1 k1 keq W = k2 k2

Then 1 keq 2

2 st

1 = k1 2 !

5.2

keq k1

2 st

st

st

1 + k2 2

keq k2

2 st

1 1 1 = + keq k1 k2

Combinations of Damping Elements To obtain equivalent damping elements use the method of equivalence of force between the original combination and the equivalent damper.

5.2.1

Viscous Dampers in Parallel Assume the velocity at the end of each damper is equivalent.


45

Since the ends of each damper will have the same velocities: F1 = c1 (x_ 2

x_ 1 )

F2 = c2 (x_ 2

x_ 1 )

Feq = ceq (x_ 2

x_ 1 )

But the equivalent damper will exert the force of both original dampers: Feq = F1 + F2 So: ceq = c1 + c2 For any combination of dampers in parallel: n X ceq = ci

(100)

i=1

5.2.2

Viscous Dampers in Series Consider three viscous dampers in series to be replaced by and equivalent viscous damper ceq that will have the same velocites at its end points.


46

Each damper will carry the same force: F1 = F2 = F3 = Feq where F1 = c1 (x_ 2

x_ 1 )

F2 = c2 (x_ 3

x_ 2 )

F3 = c3 (x_ 4

x_ 3 )

Feq = ceq (x_ 4

x_ 1 )

From kinematics: (x_ 4 Therefore

Then

x_ 1 ) = (x_ 2

x_ 1 ) + (x_ 3

x_ 2 ) + (x_ 4

x_ 3 )

Feq F1 F2 F3 = + + ceq c1 c2 c3 1 1 1 1 = + + ceq c1 c2 c3

For any combination of viscous dampers in series: n

X1 1 = ceq ci i=1

(101)


6 6.1

47

Two Degree of Freedom Systems Governing Equations Consider the two degree of freedom (2 DOF) system shown below:

Each mass can move in one dimension, therefore the system has two degrees of freedom. This system is the most general case where a forcing function is applied to each mass. Use the FBD for each mass to derive the governing equations:

To de…ne directions of forces assume relative magnitudes for the motion of each mass, e.g. x2 > x1 and x_ 2 > x_ 1 . m1 x •1 = c2 (x_ 2 m2 x •2 = or

c2 (x_ 2

x_ 1 ) + k2 (x2 x_ 1 )

k2 (x2

x1 ) x1 )

k1 x1 k3 x2

c1 x_ 1 + F1 (t) c3 x_ 2 + F2 (t)


48

m1 x •1 + (c1 + c2 )x_ 1 m2 x •2

c2 x_ 2 + (k1 + k2 )x1

c2 x_ 1 + (c2 + c3 )x_ 2

k2 x2 = F1 (t)

k2 x1 + (k2 + k3 )x2 = F2 (t)

(102)

or m1 0 0 m2

x •1 x •2

+

(c1 + c2 ) c2 c2 (c2 + c3 )

x_ 1 x_ 2

+

(k1 + k2 ) k2 k2 (k2 + k3 )

x1 x2

F1 (t) F2 (t)

=

(103)

Note: The governing equation for any 2 DOF system can be written in the following general form: m11 m12 m21 m22

x •1 x •2

+

c11 c12 c21 c22

x_ 1 x_ 2

+

k11 k12 k21 k22

x1 x2

=

F1 (t) F2 (t)

(104)

Here: m11 m12 m21 m22

=

m1 0 0 m2

c11 c12 c21 c22

=

(c1 + c2 ) c2 c2 (c2 + c3 )

k11 k12 k21 k22

=

(k1 + k2 ) k2 k2 (k2 + k3 )

For di¤erent systems, the entries in the matrices on the RHS of the equations would change.

6.2

Free Vibration of an Undamped 2 DOF System Simplify the system above for the case of undamped free vibration:


49

From the general form of the governing equation for a 2 DOF system Eq. (104) the equations of motion for this system are: m11 m12 m21 m22

x •1 x •2

+

k11 k12 k21 k22

x1 x2

=

0 0

(105)

For this system, m12 = m21 = 0. This will have signi…cance later. So, the governing equations are: m11 x •1 + k11 x1 + k12 x2 = 0 m22 x •2 + k21 x1 + k22 x2 = 0

(106)

Can m1 and m2 oscillate at the same frequency and phase angle, but with di¤erent amplitude? x1 (t) = X1 cos(!t + ) x2 (t) = X2 cos(!t + )

(107)

The same phase angle would imply each mass passes through its equilibrium position at the same time. From Eq. (107):

x •1 =

! 2 X1 cos(!t + )

x •2 =

! 2 X2 cos(!t + )

(108)

Substitution of Eqs. (107) and (108) into Eq. (105): m11 ! 2 + k11 X1 + k12 X2 cos(!t + ) = 0 m22 ! 2 + k22 X2 + k21 X1 cos(!t + ) = 0 or

( m11 ! 2 + k11 ) k12 k21 ( m22 ! 2 + k22 )

X1 X2

=

0 0

(109)

(110)

For a nontrivial solution: ( m11 ! 2 + k11 ) k12 k21 ( m22 ! 2 + k22 )

=0

(111)

So: (m11 m22 )! 4

(m11 k22 + m22 k11 )! 2 + (k11 k22

k12 k21 ) = 0

(112)


50

This equation is quadratic in ! 2 . For convenience, rewrite the equation in the following form: a! 4 + b! 2 + d = 0 (113) where: a = m11 m22 b =

(m11 k22 + m22 k11 )

d = k11 k22

k12 k21

(114)

Solving for the two roots gives: !1 =

!2 =

p

b2 2a p b + b2 2a

b

4ad

4ad

!1=2 !1=2

(115)

(116)

These are the two natural frequencies of the system! Substitution of Eqs. (115) and (116) into Eq. (110) allows determination of amplitude ratios at the two natural frequencies. ( m11 ! 21 + k11 )X1 + k12 X2 = 0 ( m22 ! 21 + k22 )X2 + k21 X1 = 0 and ( m11 ! 22 + k11 )X1 + k12 X2 = 0 ( m22 ! 22 + k22 )X2 + k21 X1 = 0 Then the amplitude ratios: X2 X1

(1)

r1 =

X2 X1

(2)

r2 =

=

( m11 ! 21 + k11 ) = k12

k21 m22 ! 21 + k22

(117)

=

( m11 ! 22 + k11 ) = k12

k21 m22 ! 22 + k22

(118)

! 1 , ! 2 , r1 and r2 de…ne the natural modes of the system. The bracketed superscripts indicate the …rst and second natural modes of the system.


51

Use of the amplitude ratios allows for a general solution for the motion of the system to be written as: " # " # (1) (1) x (t) X cos(! t + ) 1 1 1 1 ~x(1) (t) = = = …rst mode (119) (1) (1) x2 (t) r1 X1 cos(! 1 t + 1 ) " # " # (2) (2) x (t) X cos(! t + ) 2 2 1 1 ~x(2) (t) = = = second mode (120) (2) (2) x2 (t) r2 X1 cos(! 2 t + 2 ) Using superposition, the motion of each mass can be written as: (1)

x1 (t) = X1 cos(! 1 t + (1) r1 X1 cos(! 1 t

x2 (t) =

(1)

1)

+

(1) X1

(2) X1

=

=

1 r2

r1 1

r2

r1

1

= tan

1

2

= tan

1

" "

1)

(2)

where the general solution for X1 , X1 , fr2 x1 (0) f r1 x1 (0)

1

(2)

2)

(2) r2 X1 cos(! 2 t

+

+ X1 cos(! 2 t + +

and

2

2)

(121)

is given by:

f r2 x_ 1 (0) + x_ 2 (0)g2 x2 (0)g + ! 21 2

fr1 x_ 1 (0) + x_ 2 (0)g2 x2 (0)g + ! 22 2

r2 x_ 1 (0) + x_ 2 (0) ! 1 [r2 x1 (0) x2 (0)] r1 x_ 1 (0) + x_ 2 (0) ! 2 [ r1 x1 (0) x2 (0)]

#1=2 #1=2

(122)

See the derivation for Eq. (5.18) in the text. The i-th mode can be excited by setting initial conditions as follows: (i)

x1 (t = 0) = X1 x_ 1 (t = 0) = 0 (i) x2 (t = 0) = ri X1 x_ 2 (t = 0) = 0

(123)


52

So what’s that about? Consider the following simple system:

– From Eq. (103) the governing equation is: m 0 0 m

x •1 x •2

+

2k k

k 2k

x1 x2

m11 m12 m21 m22

=

m 0 0 m

k11 k12 k21 k22

=

2k k

0 0

=

(124)

– So, for reference:

k 2k

– From Eq. (114): a = m2 b =

(2km + 2km) =

d = 4k

2

2

k = 3k

4km

2

– From Eqs. (115) and (116): !1 =

!2 =

4km

4km +

p

p

(4km)2 2m2

4(m2 3k 2 )

(4km)2 2m2

4(m2 3k 2 )

!1=2 !1=2

=

=

r r

k m

(125)

3k m

(126)

– From Eqs. (117) and (118) X2 X1

(1)

r1 =

X2 X1

(2)

r2 =

= =

m

k m

+ 2k

k m

3k m

k

+ 2k

= =

k m

k m

m

3k m

+ 2k

k + 2k

=1 =

(127) 1

(128)


53

– The bracketed superscripts indicate the 1st and 2nd natural modes. 1st mode - amplitudes of both masses are equivalent. 2nd mode - amplitudes of both masses are the same, but the motion is out of phase (due to the negative sign). – The modes are often shown on a mode diagram.

– As shown in the mode diagram for the second mode, the point in the middle of the second spring remains …xed for all time. This point is called a node. – The general solution for the motion of the system is (using Eq. (121)). ! ! r r k 3k (1) (2) x1 (t) = X1 cos t + 1 + X1 cos t+ 2 m m ! ! r r k 3k (2) (1) X1 cos (129) t+ 1 t+ 2 x2 (t) = X1 cos m m (1)

(2)

where 1 , 2 , X1 and X1 are dependent on the intial conditions: x1 (0), x2 (0), x_ 1 (0) and x_ 2 (0). See Eq. (122) or Eq. (5.18) in the text. – Let’s use some numbers. Consider the case where m = 100 kg, k = 10 kN= m. Then: r k !1 = = 10 rad= s m r 3k !2 = = 17:32 rad= s m – Consider the following three initial states: 1. First mode: x1 (0) = 0:02 m x2 (0) = r1 x1 (0) = 0:02 m x_ 1 (0) = x_ 2 (0) = 0


54

From Eq. (122) or Eq. (5.18) in the text: =0 1 (1) X1 = ( 1 (0:02) 0:02) = 0:02 m 1 1 1 (2) X1 = ( 1 (0:02) + 0:02) = 0 m 1 1 Then from Eq. (121) or Eq. (129): 1

=

2

x1 (t) = 0:02 cos(10t) m x2 (t) = 0:02 cos(10t) m 2. Second mode: x1 (0) = 0:02 m x2 (0) = r2 x1 (0) =

0:02 m

x_ 1 (0) = x_ 2 (0) = 0 From Eq. (122) or Eq. 5.18 in the text: =0 1 (1) X1 = ( 1 (0:02) + 0:02) = 0 m 1 1 1 (2) ( 1 (0:02) 0:02) = 0:02 m X1 = 1 1 Then from Eq. (121) or Eq. (129): 1

=

2

x1 (t) = 0:02 cos(17:32t) m x2 (t) =

0:02 cos(17:32t) m

3. Arbitrary initial conditions: x1 (0) = 0:02 m x2 (0) =

0:01 m

x_ 1 (0) = x_ 2 (0) = 0 From Eq. (122) or Eq. (5.18) in the text: =0 1 (1) X1 = ( 1 (0:02) 1 1 1 (2) X1 = ( 1 (0:02) 1 1 Then from Eq. (121) or Eq. (129): 1

=

2

( 0:01)) = 0:005 m 0:01) = 0:015 m

x1 (t) = 0:005 cos(10t) + 0:015 cos(17:32t) m x2 (t) = 0:005 cos(10t)

0:015 cos(17:32t) m

mechanical vibration

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