Introuction To numerical methods - Philadelphia University

Philadelphia University

Engineering Faculty Mechanical Engineering Department

Eng. Laith Batarseh

Mechanical Vibrations Forced Vibration Systems

Harmonically Excited Vibration

Physical system

Equation of motion:

This equation is none – homogenous ODE

The solution of this equation involves the solution of the homogenous part (discussed in the Ch.2) and a particular solution

tFkxxcxm ...

Harmonically Excited Vibration

Homogenous part of mathematical model represents a

free vibration damped system:

As seen form the previous discussion for this type of

vibration, the amplitude will eventually dies out.

For steady state study of forced vibration, the particular

solution is the only solution remains in the seen.

0...

kxxcxm

Un-damped Forced Vibration System

Un-damped system:

Because the main cause of forced vibration in rotating

machinery is the unbalance, the most convenient and simple

form of forcing vibration F(t) = Fo cos(ωt). So, the governing

equation become:

The homogenous solution ( Free vibration):

x(t) = C1cos(ωnt) + C2sin(ωnt)

to find the particular solution, assume: xp(t) = Xcos(ωt)…Eq.2

where: X is constant that donates the maximum amplitude of

xp(t).

tFkxxm ..

1.cos..

EqtFkxxm o


Substitution of Eq.2 into Eq.1 to find X:

where: δst = Fo/k denotes the deflection of mass (m) under the

force Fo and it is called static deflection.

The general solution now become:

using the I.Cs:

22

1

n

sto

mk

FX

tmk

FtCtCtx o

nn

cossincos221

n

oo

ooo

xCxtx

mk

FxCxtx

.

.0

.

0

2

21


Substitution C1 and C2 in the general solution:

The ratio between maximum amplitude and the static

deflection (X/δst) can be expressed as:

As you can see, there are three possible cases:

When 0<r<1

When r>1

When r = 1. this case is called resonance and the amplitude of

vibration approaches the infinity.

tmk

Ft

xt

mk

Fxtx o

n

n

o

no

o

cossin

.

cos22

2

2

1111 rX

nst

Frequency ratio Magnification factor

Resonance

All cases:

At resonance

Particular solution


Total response:

A and ϕ can be determined as in the previous analysis of free

vibration:

1 cos1

cos

1 cos1

cos

2

2

rfortr

tAtx

rfortr

tAtx

stn

stn

22

2

2

2

2

1

.

n

ooo

x

mk

FxCCA

2

1

1

21

.

tantan

mkFx

x

C

C

oo

no

Homogenous solution


Example 3.1: Plate Supporting a Pump:

A reciprocating pump, weighing 68 kg, is mounted at the middle of a steel

plate of thickness 1 cm, width 50 cm, and length 250 cm. clamped along

two edges as shown in Fig. During operation of the pump, the plate is

subjected to a harmonic force, F(t) = 220 cos(62.832t) N. if E=200 Gpa,

Find the amplitude of vibration of the plate.


Example 3.1: solution

The plate can be modeled as fixed – fixed beam has the

following stiffness:

The maximum amplitude (X) is found as:

mNx

xxkSo

mxxxbhIBut

l

EIk

/ 82.400,10210250

10667.4110200192 ,

10667.41101105012

1

12

1

192

32

99

493223

3

mm

mk

FX o 32487.1

832.626882.400,102

2202

-ve means that the

response is out of

phase with

excitation

Damped Forced Vibration System

Damped system:

To find the particular solution, assume: xp(t) = Xcos(ωt – ϕ).

Substitute the assumed solution into the governing equation

and rearrange the terms:

Use the trigonometric relations

Therefore:

Solve for X and ϕ:

tFkxxcxm o cos...

tFtctmkX o cossincos2

sincoscossinsin

sinsincoscoscos

ttt

ttt

0cossin

sincos

2

2

cmkX

FcmkX o

222 cmk

FX o

2

1tan

mk

c


Divide this equation by k and make the

following substations:

You will eventually get:

222 cmk

FX o

m

kn

cc

c

k

Fost

n

r

2

1

222 1

2tan and

21

1

r

r

rr

X

st

Also called

amplitude ratio (M)


Graphical representation for X and ϕ.


Notes on the graphical representation for X.

For ζ = 0 , the system is reduced to become un-damped.

for any amount of (ζ) ; ζ > 0 , the amplitude of vibration

decreases (i.e. reduction in the magnification factor M). This is

correct for any value of r.

For the case of r = 0, the magnification factor equal 1.

The amplitude of the forced vibration approaches zero when the

frequency ration approaches the infinity (i.e. M→0 when r → ∞)


Notes on the graphical representation for ϕ.

For ζ = 0 , the phase angle is zero for 0<r<1 and 180o for r>1.

For any amount of (ζ) ; ζ > 0 and 0<r<1 , 0o<ϕ<90o.

For ζ > 0 and r>1 , 90o<ϕ<180o.

For (ζ) ; ζ > 0 and r=1 , ϕ= 90o.

For (ζ) ; ζ > 0 and r>>1 , ϕ approaches 180o.


Total response

The total response is x(t) = xh(t) + xp(t)

For under-damped system, the general solution is given as:

where:

Assume the I.Cs: and substitute it in

the general solution:

tXtXetx d

tn coscos nd 21

oo xtxxtx.

0.

and 0

1.

sinsincos.

coscosEq

XXXx

XXx

oodoono

ooo


Total response

Solve the Eq.1 to find Xo and ϕo:

cos

sincos.

tan

sincos.1

cos

2

2

2

Xx

XXxx

XXxxXxX

od

oonoon

o

oonoon

d

oo


Example 3.2:

Find the total response of a single-degree-of-freedom system with m

= 10 kg, c = 20 N-s/m, k=4000 N/m, xo = 0.01m and = 0 under

the following conditions:

a. An external force F(t) = Focos(ωt) acts on the system with Fo = 100 N

and ω = 10 rad/sec

b. Free vibration condition : F(t) = 0

Solution

a. From the given data

ox.

sradm

kn /20

10

4000

05.0

20102

20

2

nm

c

sradd

d

nd

/975.19

2005.01

1

2

2

5.020

10

n

r


Example 3.2:

Solution

Using I.Cs to find Xo and ϕo

mk

Fost 025.0

4000

100

m

rr

X st 3326.0

21222

o

r

r814.3

1

2tan

2

1

o

od

oonoon

o

oonoon

d

oo

Xx

XXxx

XXxxXxX

587.5cos

sincos.

tan

0233.0sincos.1

cos

2

2

2


Example 3.2:

Solution

b. For free vibration:

Substitute the values of X and ϕ into the general solution :

o

no

ono

n

nooono

x

xx

xxxxX

866.21

.

tan

010012.01

.2

.

2

1

2

222

tXetx n

tn 21cos

tetx t 975.19cos010012.0

Base excitation

Example 3.2:

Solution

b. For free vibration:

Substitute the values of X and ϕ into the general solution :

o

no

ono

n

nooono

x

xx

xxxxX

866.21

.

tan

010012.01

.2

.

2

1

2

222

tXetx n

tn 21cos

tetx t 975.19cos010012.0

Forced Vibration System under Base Excitation

Physical system:

Mathematical model: 0....

yxkyxcxm

The forcing function for the base excitation


Substitute the forcing function into the math. Model:

The particular solution: xp(t) = Xcos(ωt – ϕ).

k

c

ckYA

Where

tAtYctkYkxxcxm

1

22

tan

:

sincossin...

This is harmonic

excitation force

22

31

22

31

222

2

222

22

141

2tantan

21

21

r

r

cmkk

mc

rr

r

cmk

ck

Y

X

Displacement

transmissibility


Graphical representation for X and ϕ.

Forced Vibration System due to Unbalance

Assume the following general case:

The centrifugal force equal meω2

The vertical component (meω2

sin(ωt) is the effective one because

the direction of motion is vertical.

the vertical component can be

moved to the center of rotation due

to vector definition.

The equation of motion can derived

as:

The derived equation of monition is

a harmonic excited vibrating

system which discussed before.

tmekxxcxm sin...

2

Forced Vibration System General Periodic force

In some cases, the external force F(t) is periodic with period:

τ=2π/ω.

Periodic force can expanded in Fourier series as:

Where:

The order of the expansion is (j)

As the number of terms increases, as the accuracy of

expanding increases.

11

sincos2 j

j

j

jo tjbtja

atF

0

0

.sin2

.cos2

dttjtFb

dttjtFa

j

j

//upload.wikimedia.org/wikipedia/commons/7/72/Fourier_transform_time_and_frequency_domains_(small).gif

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Introuction To numerical methods - Philadelphia University