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
d f =200 mm
d E=600 mm
d D=1200 mm
dC=2000 mm
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
' = r c
R x⃗ W C= 50
50∗1=1W c
' =1<90o, with reference to the
standard radius.
The converted weight vector , W D
' = r D
R x⃗ W D =75
50 ∗3=1 W c
' =4.5<220o, with reference to
the standard radius R = 50 mm.
The converted weight vector , W E
' = rD
R x⃗ W E =25
50 ∗2=1 W 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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