ENMEC4060: Vibration and Machine Dynamics Assignment Solutions

This document provides a detailed solution to a mechanical engineering assignment on vibration and machine dynamics (ENMEC4060). The solution covers several problems, including calculating the balancing weight and angular position for an unbalanced flywheel, determining weight distributions on a shaft, analyzing the balancing of a crankshaft, and examining a torsional system. The solutions involve vector calculations, the application of Newton's second law and Lagrange's equation, and trigonometric principles to solve for forces, moments, and system equilibrium. Each problem is presented with step-by-step calculations, diagrams, and explanations, making it a valuable resource for students studying vibration and machine dynamics.

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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