introduction to mechanical vibrations and structural dynamics · pdf file1 introduction to...

1

Introduction to Mechanical Vibrations and Structural Dynamics

The one semester schedule :

1. Vibration - classification. Free undamped single DOF vibration, equation of motion,

solution, integrational constants, initial conditions.

2. Free damped single DOF vibration, equation of motion, solution, integrational constants,

initial conditions.

3. The bending stiffness, the spring assemblies.

4. Forced vibration, vibration under constant force.

5. Forced vibration, harmonically varying external force, particular solution, harmonic

response, amplitude and phase characteristic.

6. Forced vibration due to the centrifugal force.

7. The rotational vibration, free and forced, torsional stiffness.

8. The two DOF free vibration, the mode shape concept, the mode shape solution, the mode

shape orthogonality.

9. The two DOF free vibration - complete solution including integrational constants.

10. The modal coordinates, the modal transformation, the free (unconstraint system), the

symmetrical system.

11. The modal analysis of the n-DOF system, matrix formulation.

12. The harmonic response of the n-DOF system.

13. Vibration of rectilinear beams.

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1. Vibration - classification. Free undamped single DOF vibration, equation of motion,

solution, integrational constants, initial conditions.

The vibration classification.

free (natural) vibration forced vibration

undamped vibration damped vibration

linear vibration non-linear vibration

vibration of discrete mass (masses) vibration of continuously distributed mass

vibration with 1 degree of freedom (DOF) vibration with more DOF

Free undamped single DOF vibration

vibration

the mass the spring the spring

The spring stiffness :

k

0

F

undeformed spring

the spring free length

deformed spring

the spring lengthening

The main spring parameters :

0 - the spring free length, the length of undeformed spring (not lengthened nor pressed)

[units are m, mm, ... in square brackets the units will be mentioned in the whole text],

- the spring lengthening (elongation) [m, mm],

F - the lengthening force [N],

Fk - the spring stiffness [N/m, N/mm].

3

The example :

The stiffness of the spiral spring is :

3

4

Dn8

dGk

where : G - the elastic modulus (Young modulus) for shear deformation

[Pa N/m2, MPa N/mm

2],

d - the spring wire diameter [m, mm],

D - the spiral diameter [m, mm],

n - number of spring coils [-].

The spring potential (deformational) energy :

k

0

F

undeformed spring


deformed spring


2

21

0

2

21

00

P kxkdxxkdxFE

The action - reaction principle :

k

0

F

undeformed spring


deformed spring

FS


F - the external lengthening force, the action (black on figure),

FS - the spring force, the internal reaction force, caused by the spring deformation,

the spring resistance against deformation (red on figure).

kFFS

4

The equation of motion

x

k

k

m

m

0

vx ax

FS = k·x

the

equ

ilibri

um

positio

n

undeformed spring


deformed spring

xkFFam S

i

i

0xkxm

0xkam

where m is the mass, xa is the acceleration, k is stiffness, x is displacement.

0xx2

0

where : m

k0

The solution is :

00t tCx sin displacement [m]

000 tCxv cos 1st derivative, velocity [m/s]

00

2

0 tCxva sin 2nd

derivative, acceleration [m/s2]

placed into the equation of motion :

0xkxm

0tCktCm 0000

2

0 sinsin

0tCktm

kCm 0000 sinsin

0tCktkC 0000 sinsin

00

5

In the solution :

00t tCx sin

are : t - time, independent variable [s],

m

k0 - the natural circular frequency [s

-1],

2f 0

0 - the natural frequency, number of cycles per second [s-1

, Hz],

00

0

2

f

1T

- the period, the time of one cycle [s],

C, 0 - integrational constants.

The solution time curve :

t

x

C

0

0

2T

0

0t

00t tCx sin

The integrational constants physical meaning :

C - the amplitude, the displacement maximum value, [m, mm]

0 - the phase shift, the phase “angle”, [rad]

divided by circular frequency gives the shift of the sinus curve to the left

on the time axis (in seconds).

The integrational constants solution

The initial conditions - the status in the beginning of the vibration, in time t = 0

t = 0 ... x = x0 initial displacement,

v = v0 initial velocity.

In the solution :

00t tCx sin

000 tCxv cos

6

use the initial conditions :

0000t x0Cx sin

00000t v0Cv cos

simply :

00 xC sin

000 vC cos

the solution of the two equations with two unknowns :

2

0

2

02

0

vxC

0

00

0v

x arctan

The amplitude and phase shift are determined by the initial conditions.

The alternative expression of the displacement solution :

tBtAxv

tBtAtCx

0000t

0000t

cossin

sincossin

where :

0

0

CB

CA

cos

sin

are integrational constants (in opposite to C and 0, A and B have no direct physical meaning)

Integrational constants solution from initial conditions :

0000000t

0000t

vB0B0Av

xA0B0Ax

cossin

sincos

The integrational constants are then :

0

0

0

vB

xA

and :

2

0

2

02

0

22 vxBAC

0

00

0v

x

B

A arctanarctan

7

The note to the numerical calculation of the arctg function :

The function arctg has always 2 roots.

For example : arctg 0.5 = 26.6°, arctg 0,5 = 206,6°. Because both tg 26.6° = tg 206,6° = 0.5

A C

B

0

I quadrant II quadrant

III quadrant IV quadrant

B

A0 arctan

The common calculator always returns the root in the interval 0 -90°, 90°.

But if B<0, the proper root is shifted 180º or rad.

B

A0 arctan

B < 0 B > 0

A > 0 0 90°, 180° (II. quadrant)

41532

1.arctan

while calculator returns -26.6° !

0 0, 90° (I. quadrant)

6262

1.arctan

A < 0 0 -180°, -90° (III. quadrant)

6206

2

1.arctan

or

4153

2

1.arctan

while calculator returns 26.6° !

0 -90°, 0 (IV. quadrant)

6262

1.arctan

8

2. Free damped single DOF vibration

The damping is a physical phenomenon, which causes the decrease of vibration until it

completely disappears.

The causes of damping :

- the environmental resistance against motion (in liquid, in air),

- the internal friction inside material structure - the material damping,

- technical device - damper.

the symbol of damping

b

Fb F

The force resistance against motion - the damping force :

vbFb

here : b - coefficient of damping [N·s·m-1

],

v - velocity [m/s].

The equation of motion :

x

k m

vx ax

b

Fs

Fb

vbxkFFFam bS

i

i

0xkxbxm

0xkvbam

0xx2x2

0

where : m2

b

is the decay constant

the solution is :

0

t

t teCx sin

9

where : 22

0 - natural circular frequency of damped vibration [s-1

],

2f - the natural frequency (the num. of cycles per sec.) [Hz],

2

f

1T - the period (the time of one cycle) [s],

C, 0 - integrational constants.

The solution time curve :

0 1·T 2·T 3·T 4·T 5·T 6·T 7·T

C

t

x

0

t

t teCx sin

teC

x(t

)

x(t

+T

)

T

The integrational constants solution

The initial conditions - the status in the beginning of the vibration, in time t = 0

t = 0 ... x = x0 initial displacement,

v = v0 initial velocity.

used in the solution, (alternative expression) :

tBtAeteCx t

0

t sincossin

tBAtABexv t sincos

AB0BA0ABev

A0B1A10B0Aex

0

0

0

0

sincos

sincos

0xA

000 xvAv

B

2

2

002

0

22 xvxBAC

00

0

0xv

x

B

A

arctanarctan

10

Further :

0

is the damping ratio,

222

0Tt

t

1

22T

x

x

ln is the logarithmic decrement.

22 4

The exponential function

teCx

determines the time of the vibration disappearing :

0

C

t [s]

x [m]

t

tmax eCx

1time constant

xmax = 37 % C

the tangential line

Let us define so called time constant :

1

The function is then :

t

t eCeCx

The numerical values of the function can by calculated as follows :

t = C370eCeCeCx 1

tmax_

, xmax = 37% C

t = 2· C140eCeCeCx 2

22

2tmax_

, xmax = 14% C


33

3tmax_

, xmax = 5% C


44

4tmax_

, xmax = 2% C


55

5tmax_

, xmax = 0,7% C

That can be seen that in the time t = 5· the rest of the vibration amplitude is less than 1% of

the initial value C. This time can be understood as the vibration disappearing time.

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3. The bending stiffness, the bending vibration, the spring assemblies

The bending stiffness

Suppose the mass particle on the free end of the flexible beam (of the negligible mass).

The particle can vibrate.

kb

m

y

m

kb

m

Let us express so called bending stiffness of the fixed beam.

y

E·J

F

The bending of the fixed beam under the force can be expressed as :

JE3

Fy

3

where F - the bending force [N],

- the beam length [m],

E - the beam material Young modulus [Pa N/m2],

J - the beam cross section quadratic moment of inertia [m4].

The bending stiffness can be expressed as :

3b

JE3

y

Fk

in terms [N/m]

The beam cross section quadratic moment of inertia can be calculated as :

d

full circle

b

t

rectangle 4

641 dJ 3

121 tbJ

12

All cases of vibration can be subsequently solved as the mass-spring system.

Of course the beam-support system can be different :

/2 /2

m E·J

kb

m

3b

JE48k

b a

m E·J

kb

m

22bba

JE3k

b a

m E·J

kb

m

2bb

JE3k

13

The stiffness of the hydraulic circuit

Suppose the hydraulic cylinder, exposed to the force.

F

A

y

V0

V

the hydraulic cylinder

the hydraulic liquid

the piston

the force

F - the force, pressing the hydraulic liquid [N],

V0 - the initial volume of the liquid [m3],

V - the volume change (decrease), the compressed volume [m3],

K - the module of the liquid volume compressibility [Pa] (approx. K = 1÷2 GPa),

p - the hydraulic pressure [Pa],

A - the piston cross section area [m2],

y - the piston displacement [m].

The pressure - relative compression formula (analogous to the Hook’s law for solid materials)

0V

VKp

here 0V

V represents the relative volume change (decrease).

Pressure can be expressed as :

A

Fp

The compressed volume can be expressed as :

yAV

14

Then :

0V

VKp

is :

0V

yAK

A

F

and finally :

0

2

hydV

AK

y

Fk

can be understood as hydraulic stiffness in terms [N/m].

The following solution is equal to the mass - spring system.

S

y

khyd V0

V

m m

15

The hydraulic circuit can contain more than only the hydraulic cylinder.

S

y

V0

V

m

In this case V0 is not only the volume under the cylinder piston, but volume of the hydraulic

liquid in the whole circuit.

16

The spring assemblies

The two springs can be combined in two ways.

The parallel assembly :

The two springs with stiffnesses k1 and k2 are combined side by side as shown on figure.

k1

F

k2

k1

k2

Fs1

Fs2

The external force F drags both springs through common base. There are two important keys :

1. the deformation of both springs is the same,

2. the spring forces Fs1 and Fs2 in both springs (red on figure) are summarized and in

equilibrium with the external force F (black on figure).

22s

11s

kF

kF

Fkk

Fkk

FFF

21

21

2s1s

The sum of stiffneses (k1+k2) can be interpreted as so called total stiffness :

21total kkk

and the system behaves as with one spring

k1

k2

ktotal

m

m

real system

substitutional system

17

The serial assembly :

The two springs with stiffneses k1 and k2 are combined in one line as shown on figure.

total = F/ktotal

1 = Fs1/k1

k1

F

k2

Fs1 Fs2

1 = Fs1/k1

Fs2 A B

The external force F drags both springs in point B. There are two important keys :

1. the total deformation total is the sum of both partial deformations of both springs,

21total

2. the spring forces Fs1 and Fs2 (red on figure) are in equilibrium in point A and are equal,

the spring force Fs2 and the external force F (black on figure) are in equilibrium in point B

and are equal.

FFF 2s1s

The partial deformations can be expressed as :

1

1s

1k

F

2

2s

2k

F

likewise :

total

totalk

F

2

2s

1

1s

total

21total

k

F

k

F

k

F

In the nominator F = Fs1 = Fs2 can be canceled and finally :

21total k

1

k

1

k

1

21

total

k

1

k

1

1k

21

21total

kk

kkk

k1 k2

B

ktotal

m

m

real system


18

The classification of the spring assembly can be confusing due to the visual form.

k1

m

k2

The assembly sketch on fig. looks similar to the assembly sketch of the serial assembly.

But only “looks similar”.

But ...

1

F

Fs1 Fs2

2

x

1. The body displacement x represents deformation of both springs 1 and 2, then :

x21

2. The sum of spring forces must be in the equilibrium with external force :

FFF 2s1s

That is clear, that the example represents the parallel assembly and total stiffness is :

21total kkk

k1

m

k2

ktotal

m

real system


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