Balancing and Vibration Calculation for Crank Shaft and Weight Distribution

This article covers the calculation of balancing in a single plane, weight distribution of radius 50, 75 and 25 mm masses, and the outline of crank shaft. It also explains the calculation of forces along the axis, x, y, and z direction. The article provides solutions to problems related to balancing and vibration calculation for crank shaft and weight distribution.

Balancing and vibration

Vibration calculation
Contents
Solution -1.......................................................................................................................................2
Solution -2.......................................................................................................................................3
Solution -3.......................................................................................................................................5
Solution -4a......................................................................................................................................8
b)..................................................................................................................................................9
c)................................................................................................................................................11
Solution -5.....................................................................................................................................12
.......................................................................................................................................................12
a)................................................................................................................................................13
b)................................................................................................................................................13
c)................................................................................................................................................16
Solution -6.....................................................................................................................................17
a) Newton’s Second law of motion...........................................................................................17
b) using Lagrange’s equation,....................................................................................................19
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Vibration calculation
Solution -1
The given Problem describes the problem of balancing in a single plane, these are the following
data which is tabulated for ease of calculation given below
As given in problem
Condition Amplitude Phase
Vibration displacement Angle
mm Degree
Original Unbalance 0.165 15o CW
Trial Weight 0.225 35o
Weight = 50 at 45o CCW
The calculation of such kind of problem starts with defining vector form of the data as given in
question, we have assumed that, original unbalance vector = ⃗ V u
From, data, this can be defining as, the rectangular form of this vector can be given as
(V u )⃗  =0.165←15o = 0.165 cos (−15o ) +0.165 sin (−15o ) =0.16−0.0427 j ..(i)
In the same way trial weight can be given as⃗
V u +w=0.225< 35o=0.1843+ 0.13 j ……. (ii)
The trial weight can be represented as⃗
W w=50< 45o =¿ 35.36+35.36 j … (iii)
The further equation for measured vector can be given as follows⃗
V u=⃗ A A⃗ W w …………. (iv)⃗
V u +w=⃗ A A (⃗ W w+⃗ W w) …………. (v)
The resultant magnitude can be obtained from subtracting (iv) from (v)⃗
A A=⃗ V u+ w−⃗ V u⃗
W w
=¿ [ 0.184 3+ 0.130 3 j ] − [ 0.16 4−0.0427 1 j ]
35.36+35.36 j =(0.024 3+0.176 1 j)
35.36+35.36 j⃗
A A=(−0.024+0.176 j)(35.36−35.36)
( 35.36+35.36 ) 35.36−35.36 ¿ ¿=6.72−5.304 j
2500.659 =0.0024 +0.17271 j
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Vibration calculation
Since amplitude of resultant vector is calculated, on this basis we can calculated weight of the
balanced mass⃗
W u=⃗ V u+w⃗
AA
−⃗ W w Now putting the value
¿ [ 0.184+0.130 j ]
0.0024+0.17271 j −35.36+35.36 j=−0.6142−2.08123 j Ans
The placement balance mass will be just opposite of the result calculated,⃗
B=−⃗ W u = 6.148 +2.08123j
The balance vector in polar coordinate⃗
B=6.491<18.672oCCW.
As per above calculation, we must place 6.491 gm of weight at 18.672o CCW. Ans
Solution -2
The question illustrates the weight distribution of radius 50, 75 and 25 mm masses 1 kg, 3
kg, and 2 kg respectively,
To solve
the condition further we must decide a reference plane, and in this condition, the reference plane
is G, we must calculate all the distance from reference plane G, which is as follows.
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Vibration calculation
The different axial distance from the figure is given as
df =200 mm
d E=600 mm
d D=1200 mm
dC=2000mm
d B=2200 mm
d A =2600 mm
We can represent the weight vector as follows⃗
W C=1<90o⃗
W D =3<220o⃗
W E =2<330o
The above radius vector is positioned at different radius on axis, for solving the problem,
it is necessary that, we must, take one radius as reference radius. Which is given as follows. The
reference radius is taken as R for the given condition which is equal to 50 mm
The converted weight vector W c
' = rc
R x⃗ WC= 50
50∗1=1 W c
' =1< 90o, with reference to the
standard radius.
The converted weight vector ,W D
' =r D
R x⃗ W D= 75
50∗3=1W c
' =4.5<220o, with reference to
the standard radius R = 50 mm.
The converted weight vector ,W E
' =r D
R x⃗ W E= 25
50∗2=1W c
' =4.5<330o, with reference to
the standard radius R = 50 mm.
The further calculation is possible only after converting the rectangular form of weight
vector with reference to the standard radius R = 50 mm.
The rectangular form of weight W AC
' =1< 90o is 0+0.772j with reference to the standard
radius R = 50 mm.
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Vibration calculation
The rectangular form of weight W AD
' =4.5<220ois -1.59-1.34j with reference to the
standard radius R = 50 mm.
The rectangular form of weight W AE
' =1< 330ois 0.2-0.12j with reference to the standard
radius R = 50 mm.
The sum of unbalance vector = ⃗ W A =⃗ W AC+⃗ W AE +⃗ W AD
If we add the above rectangular vector, we will get sum of unbalanced rectangular vector.
= -1.39-0.69j or 1.552 <206.39o Ans
Similarly, for plane G the unbalance vector is given as.
The rectangular form of weight W GC
' =lA −lC
lA
x⃗ W ' C= 2600−2000
2600 ∗1=0.2302with
reference to the standard radius R = 50 mm, the new weight vector W c
' =0.2301<90o, or
(0, 0.23j)
The rectangular form of weight W G D
' =lA −lD
lA
x⃗ W 'C= 2600−1200
2600 ∗4.5=2.423with
reference to the standard radius R = 50 mm, the new weight vector W c
' =2.423<220o, or
(-1.862, -1.561j).
The rectangular form of weight W ¿
' = lA−lE
lA
x⃗ W ' C= 2600−600
2600 ∗4.5=0.231with
reference to the standard radius R = 50 mm, the new weight vector W c
' =0.772< 33 0o, or
(-.67, -0.39j).
The sum of unbalance vector = ⃗ W G=⃗ W GC+⃗ W ¿+⃗ W GD
If we add the above rectangular vector, we will get sum of unbalanced rectangular vector.⃗
W G= -1.19-1.72j or 2.09 <235.32o with reference to the standard radius R = 50 mm
The calculated weight is A = 1.552 kg, and at G = 2.09 kg, with reference to the standard
radius R = 50 mm Ans
Solution -3
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Vibration calculation
The figure represents the outline of crank shaft, which is axial distance of a and radius and
angular position of radius is at 120o each. The angular position of each arm is given as follows
α 1=a6=0o , a2=a5 =120o , a3=a4=240o, The crank length is r, connecting rod = l and
reciprocating mass = m. rotating mass = mC Taking cylinder 1 as the reference plane.
The distance of different cylinder is given as follow. The cylinder 1 is taken as reference plane.
The distance of cylinder 1 = d1 = a
The distance of cylinder 2 = d2 = 2a
The distance of cylinder 2 = d3 = 3a
The distance of cylinder 2 = d4 = 4a
The distance of cylinder 2 = d5 = 5a
The angular velocity = ω. The forces along the axis, x, y, and z direction can be calculated as
( Fx )i= ( m+ mc ) r ω2 cos ( ωt+ αi ) +m r 2 ω2
l cos (2 ωt+ 2αi) ……(a)
The component of forces along x direction is given as
( Fx )total= ( m+ mc ) r ω2
∑
i=1
N
cos ( ωt+ αi ) + m r2 ω2
l ∑
i=1
N
cos (2 ωt +2 αi )… … (b)
In the same way, the primary and secondary forces along y direction is given as
( F y )total=−(mc)r ω2
∑
i=1
N
sin ( ωt +αi ) …………(c)
For simplifying the equation, we must simply the variable first.
Which is as follows
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Vibration calculation
∑
i=1
N
cos (ωt +α i ), ∑
i=1
N
cos (2 ωt+2 αi ), ∑
i=1
N
sin ( ωt+αi )
The given equation, ∑
i=1
N
cos ( ωt +α i ) =∑
i=1
N
cos ( 0 t+ αi ) =∑
i=1
N
cos ( αi )
⟹∑
i=1
N
cos ( αi )=cos ( α1 ) +cos ( α2 ) +cos ( α3 ) + cos ( α 4 ) +cos ( α5 ) + cos ( α6 )
As the angle given in question
a1=a6 =0o , a2=a5 =120o , a3=a4=240o, replacing these values in equation
⟹∑
i=1
N
cos ( αi )=cos ( 0o ) +cos ( 120o ) + cos ( 240o )+cos ( 240o )+ cos ( 120o ) + cos ( 0o )=1−0.5−0.5−0.5−0.5+1=0
Similarly, for ∑
i=1
N
cos (2 ωt+2 αi ) by putting the value of α and t = 0
∑
i=1
N
cos (2 ωt+ 2 αi )=cos ( 2∗0o ) +cos ( 2∗12 0o ) + cos ( 2∗24 0o ) +cos ( 2∗24 0o ) +cos ( 2∗12 0o ) + cos ( 2∗0o )=1−0.5−0.5
Now equation ∑
i=1
N
sin ( ωt+ αi ) =∑
i=1
N
sin ( 0 t +αi )=∑
i=1
N
sin ( αi )
∑
i=1
N
sin ( αi ) =sin ( αi ) +sin ( α 2 )+ sin ( α 3 )+ sin (α4 ) +sin ( α5 ) +sin ( α6 )
∑
i=1
N
sin ( αi ) =sin ( αi ) +sin ( α 2 )+ sin ( α 3 )+ sin (α4 ) +sin ( α5 ) +sin ( α6 )
As the angle given in question
a1=a6 =0o , a2=a5 =120o , a3=a4=240o, replacing these values in equation
∑
i=1
N
sin ( αi ) =sin ( 0o ) + sin ( 120o ) +sin ( 240o ) +sin ( 240o ) +sin ( 120o ) +sin ( 0o )
∑
i=1
N
sin ( αi ) =0+ √3
2 − √3
2 − √3
2 + √ 3
2 +0 = 0
Simplifying with trigonometrical theories (b)
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Vibration calculation
( Fx ) total= ( m+mc ) r ω2
∑
i=1
N
cos ( ωt+ αi ) + m r2 ω2
l ∑
i=1
N
cos (2 ωt +2 αi )
( Fx )total= ( m+ mc ) r ω2 xo+m r 2 ω2
l x 0=0
Same calculation for y and z axis
( Fx )total=0∧ ( F y ) total=0
We must calculate the moment in z axis
M z = ( m+mc ) r ω2
∑
i =1
N
cos ( ωt +αi ) +m r2 ω2
l ∑
i=1
N
cos (2ωt +2 αi) x li ……(d)
Similarly, moment along y axis,
M x=∑
i =0
N
( Fx )i li From equation (iii)
M x=− (mc )r ω2
∑
i=1
N
sin ( ωt+αi )∗di
¿ ,− ( mc ) r ω2
∑
i=1
N
di∗sin ( α i ) ………. (e)
By putting the trigonometrical value as above, Putting the value of a1=a6 =0o , a2=a5 =120o ,
a3=a4=240o, d1=0, l2=a,d2=a, d3=2a, 4=3 a, l5=4 a, l6=5 a.
∑
i=1
N
cos (ωt +α i )∗di=0∗cos ( 0o ) +a∗cos ( 120o ) +2 a∗cos ( 24 0o ) +3 a∗cos ( 240o ) + 4 a∗cos ( 120o ) +5 a∗cos (0o )
¿ 0−0.5 a−a−1.5 a−2 a+5 a=0
The moment of for ∑
i=1
N
cos (2 αi )∗di is zero, and value for ∑
i=1
N
sin ( 2α i )∗di = 0
Therefore, from equation (d),
M z =0
M x=0
As calculation above show that, the moment in x, and z direction is in balanced condition,
Therefore, the all crank shaft in is fully balanced.
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Vibration calculation
Solution -4a
The given figure represents torsional system, consists of two discs mounted on the shaft. The
general equation of motion for the such system is given as
J1 ¨θ1+ ( k t1 +kt 2 ) θ1−k t 2 θ2=0
J2 ¨θ2−k t 2 θ1( kt 2 + kt 3 )θ2 =0
If we put the value of kt3 = 0, kt1 = kt2 = kt, J1 = J0, J2 = 2J0):
J0 ¨θ1 +2 k1 θ1−kt θ2=0
2 J 0 ¨θ2−kt θ1+ kt θ2=0………. (I)
The above equation can be given in harmonic solution
θt ( t ) =φi Cost (ωt +ϕ ) where, I = 1,2,3… …….(ii)
The frequency is
2 ω4 jo
2−5 ω2 jo ki+ ko
2 =0 ……………………………(iii)
Then natural frequency can be given by
ω1= √ kt
4 jo
(5− √17 ¿)¿and ω2= √ kt
4 jo
(5− √ 17¿)¿
The amplitude ratio can be given by
r1= φ2
( 1 )
φ1
( 1 ) =2−¿ ¿
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Vibration calculation
r2= φ2
( 2 )
φ1
( 2 ) =2−¿ ¿
b)
As the figure describes, the equation of motion for given pendulum can be given as
m1 ¨x1 + ( k1 +k2 ) x1− pSinθ=0 ……. (a)
Taking moment around the pendulum bob
m1 l2 ¨θ−m2 glSinθ=0 ………….. (b)
Assuming small angles θ, the Sinθ ≅ θ
Now P=m2 gCosθ≅ m2 g
To obtain the equation of motion we will put these values in equation (a) and (b)
m1 ¨x1 + ( k1 +k2 ) x1−m2 gθ=0
m1 l2 ¨θ−m2 glθ =0 Ans
If the system works as harmonic motion, then equation can be written as
x1 ( t )= X1 cos ( ωt+ ϕ ) …………. (c)
x2 ( t ) = X2 cos ( ωt+ϕ )
The equation of motion can also be written as
{−m1 ω2+ ( k1 +k2 ) + m2 g
l } X1− m2 g
l X 2………. (d)
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Vibration calculation
−m2 g
l X1+ {−m2 ω2 + m2 g
l }X2 =0 ………. (e)
The coefficients of the equation give following determinant
[ −m1 ω2 ( k1 +k2 ) + m2 g
l
−m2 g
l −m1 ω2 + m2 g
l ] =0 … … … ..( f )
m1 m2 ω4− m1 m2 g
l ω2 + m2 g
l ( k1 + k2+ m2 g
l )=0
The roots of the equation provide the natural frequency of vibration of the system
ω1
2 , ω2
2=
(−m1 m2 g
l )± √ (−m1 m2 g
l )2
−4 m1 m2
m2 g
l (k1 +k2 + m2 g
l )
2m1 m2
Ans.
c)
The equation given as per figure, we should take moment about mass
m1 l1
2 ¨θ1=−w1 ( l1 sin θ1 ) +QSin θ2 ( lCos θ1 ) −QCos θ2 (l1 sin θ1)
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Vibration calculation
Suppose Q = W2
Then,
m2 l2
2 ¨θ2 +m2 l2 ( l1 ¨θ1 )=−w2 ( l2 sin θ2 ) …. (c)
We know that, θ1= x1
l1
and θ2= x2−x1
l2
Putting the value in equation (i) and (ii)
m1 l1
2 ¨x1 + [ w1 +w2 ( l1 +l2
l2 ) ] x1− w2 l1
l2
x2=0 (d)
Then the equation of the motion is
m2 l2 ¨x2 +w2 x1 +w2 x2=0
Putting the value of m1 = m2 = m, l1 = l2 = l, w1 = w2 = mg the equation (iii) and (iv) gives
equation of motion.
ml ¨x1 +3 mg x1−mg x2=0 ……. (e)
ml ¨x2−3 mg x1 +mg x2=0 …… (f)
The above system should be in harmonic motion
x ( t ) =xCosωt Putting the value in equation (e)
−𝜔2 1𝑚𝑙𝑋 −ω2 ml X1+ 3 mg X1 −mg X2=0 ……(f)
−ω2 ml X2−mg X1 +mg X2=0
After such calculation, the equation for natural frequency can be given as
ω4 m2 l2−4 m2 lg ω2 +2 m2 g2=0
Then root of the equation,
ω1
2 , ω2
2= ( 2∓ √2 ) g
l
The ratio of the amplitude
X1
X2
= mg
−ωml+3 mg =¿
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Vibration calculation
Then mode vector of the system
First mode=⃗ x (1 ) ( t )= { X1
(1 ) mg
−X1
( 1 )−ωml +3 mg }
Second mode =⃗ x ( 2 ) ( t ) = { X1
( 2 ) mg
− X1
( 2 ) +ωml−3 mg }
Solution -5
The figure in in question,
represents as coupled pendulum with mas m1 , m2 and Spring constant K
The oscillation of m1 , m2, represented by displacement x1, and x2 with angle θ1 and θ2
The displace of the system x1=dSin θ1 ………. (a)
And x2=dSinθ2 …………. (b) is the form of system
a)
The equation of motion can be represented as
−mglSin ( θ1 )−k ( x1−x2 ) d=J ¨θ1 or,
J ¨θ1 +mglSin ¿ ……… (c)
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Vibration calculation
The MOI for the system is given as J and acceleration as ¨θ1
By putting the value of equation (a), (b) and value of J
ml2 ¨θ1 +mglSin θ1+k ( dsinθ1 −dsinθ2 ) d=0
ml2 ¨θ1 + ( mglSin θ1+ k d2 ) θ1−k d2 θ2=0
Now the equation for pendulum 1 can be given as
ml2 ¨θ1 + ( mgl +k d2 ) θ1−k d2 θ2=0 ……(d)
The equation for Q is shown below
ml2 ¨θ2 + ( mgl +k d2 ) θ2−k d2 θ1=0……(e)
The equation (d) and (e) can be expressed in the matrix equation is
[ml2 0
0 m l2 ] [ ¨θ1
¨θ2 ]+ [(mgl+k d2) −k d2
−k d2 (mgl+k d2) ]=0
The equation in matrix form
[θ1
θ2 ]= [ X1 cos(wt +ϕ )
X2 cos(wt +ϕ ) ]
X1 and X2 is amplitude, ω is frequency and φ is phase angle
b)
We can derive the represent the frequency as follows
[ −ml2 ω2 +(mgl+k d2 ) −k d2
−k d2 −ml2 ω2+(mgl +k d2) ] =0
Further solving the equation
(−ml2 ω2 + (mgl+ k d2 ) )2
− ( −k d2 )2
=0
⟹ m2 l4 ω4−2 m l2 ( mgl+k d2 ) ω2 + ( mgl +k d2 )
2
−k 2 d4 =0
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Vibration calculation
The Y represents ω2, putting the value in equation
m2 l4 ω4 −2 ml2 ( mgl+k d2 ) X2 + ( m2 g2 l2 +2 mglk d2 ) =0
The roots of the above equation can be written as
Y 1,2=− (−2 ml2 ( mgl+ k d2 ) ) ± √ (−2 ml2 ( mgl+ k d2 ) )
2
−4 m2 l4 ¿ ¿ ¿
The roots of the equation is
Y 1= 2 ml2 ( mgl+k d2 ) −2 ml2 k d2
2 m2 l4
¿ 2m2 l3 g
2 m2 l4 = g
l
Another value of Y is
Y 2= 2 ml2 ( mgl+k d2 ) +2 ml2 k d2
2 m2 l4
Y 2= 2 m2 l2 gl+4 ml2 k d2
2m2 l4
Y 2= g
l + 2 k d2
ml2
Put the value of Y
ω1
2= g
l , ω1= √ g
l (if the value of ω = 2πf)
f 1= 1
2 π ( √ g
l )
In such condition the natural frequency of the system = 1
2 π ( √ g
l )
Substituting the value of Y in second root
We get natural frequency ¿ 1
2 π ( √ g
l + 2k d2
ml2 )
He matric form of
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Vibration calculation
[−ml2 ω2 +(mgl+ k d2 ) −k d2
−k d2 −ml2 ω2+(mgl +k d2) ] [ X1
X2 ]=0
Further simplification gives the equation as follows
(−ml2 ω2 + ( mgl+ k d2 ) ) X1−k d2 X2 =0 ………… (f)
−k d2 X1 + (−ml2 ω2+ ( mgl+ k d2 ) ) X2=0 ………… (g)
The further simplification of equation (d), we get
X1
X2
= k d2
−ml2 ω2 +( mgl+k d2 )
The X1 and X2 is related with ω1, it can be denoted as X1
(1)∧X 2
(1)and same way for ω2 is
X2
(1)∧X 2
(2)
If we apply it for first natural frequency
X1
(1 )
X2
(1 ) = k d2
−m l2 ω2+(mgl+k d2)
1
r1
= k d2
−ml2 ω2+(mgl +k d2) (r1 is denoted as amplitude ratio)
In the same manner second, natural frequency will be
1
r2
= k d2
−ml2 ω2 +(mgl +k d2 ) where r2 is the second amplitude ratio
The amplitude vector from above equation
Xi =X1
(1 )
X2
(1 ) (Substituting r1X1 for X2
Xi = X1
(1)
r1 X1
(1) = X1
(1)
[ 1
r1 ]
Now inducing r1
Xi =X1
(1 )
[ 1
k d2
−m l2 ω2+(mgl+ k d2) ]
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Vibration calculation
We can also derive the second amplitude vector as above
X2 =X1
(2 )
[ 1
k d2
−m l2 ω2+(mgl+k d2) ]
Now from the above calculation, motion vector can be given as follows
X1 =X1
(1 )
[ 1
k d2
−m l2 ω2+ ( mgl+k d2 ) ]cos (ω1 t +ϕ1)
X2 =X1
( 2 )
[ 1
k d2
−m l2 ω2+ ( mgl+ k d2 ) ] cos (ω2 t +ϕ2) Ans
c)
Th representation of equation is as follows⃗
θ ( t )=c1⃗ θ (1 ) ( t )+c2⃗ θ ( 2 ) ( t )
In which c1 and c2 are constants
If we consider the c1 and c2 as one, the solution of the vibration system can be given as ⃗
θ1 ( t ) =X1
(1 )cos (ω1 t+ ϕ1)+X 1
(2) cos( ω2 t+ϕ2 )
where ϕ1 and ϕ2 are phase angles of the equations
Further we can write the equation
˙
θ1 ( t )=−ω1 . X1
(1 ) sin ( ω1 t +ϕ1 )−ω2 . X1
(2 ) sin (ω2 t +ϕ2 )
For t = 0
˙θ1 ( 0 ) = X1
( 1 ) cos ( ϕ1 ) + X1
( 2 ) sin ( ϕ2 )
˙
θ2 ( 0 ) = X1
( 1 ) cos ( ϕ1 ) −X1
( 2 ) cos ( ϕ2 )
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Vibration calculation
˙˙θ1 ( 0 ) =−ω1 X1
( 1 ) sin ( ϕ1 ) −ω2 X1
( 2 ) sin ( ϕ2 )
˙˙θ1 ( 0 ) =−ω1 X1
( 1 ) sin ( ϕ1 ) + ω2 X1
( 2 ) sin ( ϕ2 )
Substituting the value of θ1 ( 0 ) = 0, θ2 ( 0 ) = 0 and ˙θ2 ( 0 ) = 0
˙
a=X1
( 1 ) cos ( ϕ1 ) + X1
( 2 ) sin ( ϕ2 )
˙
0=X1
( 1 ) cos ( ϕ1 ) − X1
( 2 ) sin ( ϕ2 )
Now we must differentiate the equation
˙0=−ω1 X1
( 1 ) sin ( ϕ1 )−ω2 X1
(2 ) sin ( ϕ2 )
˙
0=−ω1 X1
( 1 ) sin ( ϕ1 ) +ω2 X1
( 2 ) sin ( ϕ2 )
Further solving the equation, we obtain the following relations
θ1 ( t )=a . cos ( ω2−ω1
2 )tCos ( ω2+ ω1
2 )t
θ2 ( t )=a . sin ( ω2−ω1
2 )tSin ( ω2+ω1
2 )t Ans
Solution -6
a) Newton’s Second law of motion
As the figure given in question,
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Vibration calculation
For using Newton’s second law
∑
j
F y=mi ¨xi,
The different letter represented in equation, such as m which represents mass, acceleration is a
force acting on body, Fy, in this condition motion of the equation is given by
m1 ¨x1=− ( k1
2 + k1
2 ) x1+ ( k2
2 + k2
2 ) ( x2−x1 ) −c1 ˙x1+c2 ¿
m1 ¨x1=−k1 x1 +k2 ( x2−x1 ) −c1 ˙x1 +c2 ¿
The is representing displacement, the is as usual stiffness of the spring, ˙x1 is representing
velocity
Arranging the above formula
m1 ¨x1 + ( c1 +c2 ) ˙x1−c2 ˙x2+ ( k1 +k2 ) x1−k2 x2=F1 (t ) ……. (a)
The second story equation is given as
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Vibration calculation
m1 ¨x2=− ( k2
2 + k 2
2 ) ( x2−x1 ) +( k3
2 + k3
2 ) ( x2−x1 ) +c2 ¿
Or,
m1 ¨x2=−k2 ( x2−x1 ) + k3 ( x3−x2 ) +c2 ¿
It can be rearranged as
m2 ¨x2 + x2 ( k2+ k3 ) −k2 x1−k3 x3 + ( c2 +c3 ) ˙x2 −c2 ˙x1−c3 ˙x3 =F2 (t ) …(b)
In the similar way motion of equation of 3rd mass and 4th mass can be determined as follows
m3 ¨x3 + x3 ( k3 +k 4 )−k3 x2−k4 x4 + ( c2 +c3 ) ˙x3 −c3 ˙x2−c4 ˙x4 =F3 (t ) …(b)
m4 ¨x4 +k 4 x 4−k 3 x2−k 4 x3 +c4 ˙x4 −c4 ˙x3=F4 (t) …(iv)
All the above equation can be arranged in matrix form as follows
|m1 0 0 0
0 m2 0 0
0 0 m3 0
0 0 0 m4
|| ¨x1
¨x2
¨x3
¨x4
|+
|c1 +c2 c2 0 0
−c2 c2 +c3 −c3 0
0 −c3 c3 + c4 −c4
0 0 −c4 c4
|| ¨x1
¨x2
¨x3
¨x4
|+
|k 1+ k2 k 2 0 0
−k2 k2 + k3 −k3 0
0 −k 3 k3 + k4 −k4
0 0 −k 4 k4
||x1
x2
x3
x4
|=
|F1 (t)
F2 (t)
F3 (t)
F4 (t )
|
The other format for equation is ⃗
F= [ m ]⃗ ¨x + [ c ]⃗ ˙x + [ k ]⃗ x
Where m, c, k is the mass, damping and stiffness matrix respectively.
b) using Lagrange’s equation,
The representation of motion by LaGrange’s’ equation is given as
d
dt ( ∂ T
∂ ˙x j )− ∂T
∂ ˙x j
+ ∂T
∂ ˙x j
+ ∂ V
∂ ˙x j
=Q j
(n ) …… … …..(e )
The different symbol used in the above equation is T for kinetic energy, R for Rayleigh’s
dissipation function, V for potential energy, and force as Q j
( n )
The determination of kinetic energy is as follows
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Vibration calculation
T =∑
i
1
2 mi ˙xi
2
The equation of kinetic energy for the system is given as
T = 1
2 m1 ˙x1
2 +1
2 m2 ˙x2
2+ 1
2 m3 ˙x3
2 + 1
2 m4 ˙x4
2
Rayleigh’s dissipation function can be given as.
R=∑
i
1
2 ci ˙xi
2
then equation of Rayleigh’s dissipation using masses
R=1
2 c1 ˙x1
2+ 1
2 c2 ( ˙˙x2− ˙x1 )2+ 1
2 c3 ( ˙x3− ˙x2 )2+ 1
2 c4 ( ˙x4− ˙x3 )2
The determination of Rayleigh’s dissipation is as follows
V =∑
i
1
2 k ˙xi
2
then equation of Rayleigh’s dissipation using masses
V = 1
2 k1 ˙x1
2 +1
2 k2 ( ˙˙x2 − ˙x1 )2 + 1
2 k3 ( ˙x3− ˙x2 )2 +1
2 k4 ( ˙x4 − ˙x3 ) 2
The total system of the building is can be given as
Q j
( n ) =Fi ( t ) ( ∂ x1
∂ ˙x j ) +F2 ( t ) ( ∂ x2
∂ ˙x j ) + F3 ( t ) ( ∂ x3
∂ ˙x j ) +F4 ( t ) ( ∂ x4
∂ ˙x j )
Now we can evaluation Lagrange’s equation with respect to mass m1. For this we must
determine the variable d
dt ( ∂ T
∂ ˙x j ).
d
dt ( ∂ T
∂ ˙x j )= d
dt ( ∂ ( 1
2 m1 ˙x1
2+ 1
2 m2 ˙x2
2+ 1
2 m3 ˙x3
2 + 1
2 m4 ˙x4
2
)
∂ ˙x j
)= d
dt ( ∂ ( 1
2 mi ˙xi
2
)
∂ ˙x j
)
The equation of kinetic energy ¿ d
dt ( m1 ¨x1 )=m1 ¨x1
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Vibration calculation
∂ T
∂ ˙x j
=
∂ ( 1
2 m1 ˙x1
2+ 1
2 m2 ˙x2
2 + 1
2 m3 ˙x3
2 + 1
2 m4 ˙x4
2
)
∂ ˙x j
=0
Similarly, we must calculate the value of R = ∂ R
∂ ˙xi
∂ R
∂ ˙xi
=
∂ ( 1
2 c1 ˙x1
2 + 1
2 c2 ( ˙˙x2− ˙x1 )2 + 1
2 c3 ( ˙x3 − ˙x2 )2 + 1
2 c4 ( ˙x4 − ˙x3 )2
)
∂ ˙x j
=¿
∂ R
∂ ˙xi
=c1 ¨x1−c2( ¨x2− ¨x1 )
Similarly, for V and Q can be determined as follows
∂ R
∂ ˙xi
=k 2( ¨x2− ¨x1 )−k 3( ¨x3 − ¨x2 )
And Q2
( n )=F2 ( t )
Similarly, we must determine the LaGrange’s equating for function =3. In this condition
equation of motion
m1 ¨x1 + ( c1 +c2 ) ˙x1−c2 ˙x2+ ( k1 +k2 ) x1−k2 x2=F1 (t ) ……. (e)
The variable d
dt ( ∂ T
∂ ˙x j ) can be given as ¿ m2 ¨x2
And variable ∂ T
∂ ˙x2
=0
∂ R
∂ ˙x2
=c2 ( ˙x2− ˙x1 ) −c3 ( ˙x3 − ˙x2 )
∂V
∂ ˙x2
=k 2 ( ˙x2− ˙x1 )−c3 ( ˙x3− ˙x2 )
The variable Q2
( n )=F2 ( t )
The equation of second mass
m2 ¨x2 + ( k 1+k2 ) x2−k2 x1−k3 x3 + ( c2 +c3 ) ˙x2 −c2 ˙x1−c3 ˙x3 =F2 (t) …. (f)
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Vibration calculation
Similarly, the equation for third mass
m3 ¨x3 + ( k3+k4 ) x3−k3 x2−k4 x4 + ( c3 +c4 ) ˙x3−c3 ˙x2−c4 ˙x4=F3 (t) …(g)
And for determining and rearranging the equation for third mass.
m4 ¨x4 +k 4 x 4−k 4 x3 +c4 ˙x4 −c4 ˙x3=F4 (t) …………(h)
We can arrange the equation (f) to equation (h) in matrix form as follows
The matrix form of above equation is as follows
|m1 0 0 0
0 m2 0 0
0 0 m3 0
0 0 0 m4
|| ¨x1
¨x2
¨x3
¨x4
|+
|c1 +c2 c2 0 0
−c2 c2 +c3 −c3 0
0 −c3 c3 + c4 −c4
0 0 −c4 c4
|| ¨x1
¨x2
¨x3
¨x4
|+
|k 1+ k2 k 2 0 0
−k2 k2 + k3 −k3 0
0 −k 3 k3 + k4 −k4
0 0 −k 4 k4
||x1
x2
x3
x4
|=
|F1 (t)
F2 (t)
F3 (t)
F4 (t )
|
The equation can also be represented as⃗
F= [ m ]⃗ ¨x + [ c ]⃗ ˙x + [ k ]⃗ x
Where m, c, k is the mass, damping and stiffness matrix respectively.
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References
Callister, W., & Rethwisch, D. (2010). Materials Science and Engineering : An Introduction (8th
ed.). New York: John Wiley & Sons.
Haym Benaroya, M. L. (2017). Mechanical Vibration: Analysis, Uncertainties, and Control (1st
ed.). Boca Raton: CRC.
Inman, D. J. (2014). Engineering Vibration. New Jersey: Pearson Education.
Palm, W. J. (2010). Mechanical vibration (2nd ed.). New York: Wiley.
Rao, S. (2013). Mechanical Vibration (6th ed.). New Jersey: pearson.
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