Numerical Solutions for Mechanical Vibration

Numerical
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Mechanical Vibration

Numerical
Contents
Solution -1.......................................................................................................................................3
Solution -2.......................................................................................................................................4
Solution -3.......................................................................................................................................6
Solution -4.......................................................................................................................................9
a)..................................................................................................................................................9
b)................................................................................................................................................12
c)................................................................................................................................................13
Solution -5.....................................................................................................................................15
.......................................................................................................................................................15
a)................................................................................................................................................15
b)................................................................................................................................................16
c)................................................................................................................................................19
Solution -6.....................................................................................................................................20
a) Newton’s Second law of motion...........................................................................................20
b) Lagrange’s equation,.............................................................................................................22
References......................................................................................................................................25
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Solution -1
As given in question,
It is a case of measurement of single plane balancing, with following data
Unbalance amplitude = 0.165 mm at phase angle degree = 15o
Trial weight = 50 gm at phase angle degree 45o CCW and amplitude 0.225 mm.
First, we will express the above condition in the form of vector, and assumption was taken as
counter clockwise will be positive direction
Suppose the vector of original unbalance = ⃗ V u ⃗
V u=0.165←15o = = 0.165*Cos(-15o) +0.165*Sin(-15o) = 0.16 - 0.0427j
Measured vector of trial weight, ⃗ V u +w⃗
V u +w=0.225 <35o = 0.18431+0.130j
Now trial weight vector⃗
W w=50< 45o = 35.36 + 35.36j
Now we know that, measured vector can be expressed,⃗
V u=⃗ A A⃗ W w ............. (a)⃗
V u +w=⃗ A A (⃗ W w+⃗ W w ) ............. (b)
By subtracting equation (a) from (b), we will get ⃗
A A=⃗ V u+ w−⃗ V u⃗
W w
¿ [ 0.184+0.130 j ]− [ 0.16−0.0427 j ]
35.36+35.36 j = (0.024+ 0.176 j)
35.36+35.36 j =
(−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+0.17271j
Since vector ⃗ A A is obtained, now we can calculate ⃗ W u, with the following relation
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W u=⃗ V u+w⃗
AA
−⃗ W w
Putting the value in above equation,
¿ [ 0.184+0.130 j ]
0.0024+0.17271 j−35.36 +35.36 j
= -6.15-2.0812j
This is the required unbalance vector is calculated, to make it stable we have to place the mass in
opposite side,
i.e. ⃗ B❑=−⃗ W u
or⃗ B❑=6.15+ 2.0812 j
The balance vector in polar coordinate⃗
B❑=6.49< 18.67oCCW.
As per above calculation, we must place 6.49 gm of weight at 18.67o CCW.
Solution -2
As given in question, the weight 1 kg, 3 kg and 2 kg, are in the radius of 50, 75 and 25 mm, in
the planes C, D, and E, the support bearing is at the place B, and F
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In this condition we must determine plane A and G. Taking G as the reference plane, the axial
position of each weight and unbalance weight is given below
lf =200 mm
lE =600 mm
lD =1200 mm
lC=2000 mm
lB =2200 mm
lA =2600 mm
The weight vector for each position C, D, E can be given as ⃗
W C=1< 90o⃗
W D=3<220o⃗
W E =2<330o
As can see that, the weight vector is positioned at different radius position, we must convert
reference radius position.
W c
' = r c
R x⃗ W C= 50
50∗1=1Then new weight vector W c
' =1< 90o, at reference position of 50
mm.
W D
' =r D
R x⃗ W D= 75
50∗3=1Then new weight vector W c
' =4.5<220o, at reference position of
50 mm.
W E
' = r E
R x⃗ W E= 25
50∗2=1Then new weight vector W c
' =1< 330o, at reference position of
50 mm.
Now I must convert this unbalance weight into rectangular form of vector
The weight vector W AC
' =1< 90o, then rectangular form of vector will be 0+ 0.77j at radius
position of 50 mm.
The weight vector W AD
' =4.5< 220o, then rectangular form of vector will be -1.59 - 1.34j at radius
position of 50 mm.
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The weight vector W AE
' =1<330o, then rectangular form of vector will be 0.2 -0.12j at radius
position of 50 mm.
The total unbalance weight in rectangular form = -1.39-0.69j.
Total unbalance vector = ⃗ W A =⃗ W AC +⃗ W AE +⃗ W AD
The required weight at position A in rectangular form = -1.39-0.69j or 1.552<206.39o Ans
Similarly, we must calculate unbalance weight in plane G
W GC
' =lA −lC
lA
x⃗ W 'C= 2600−2000
2600 ∗1=0.2301Then new weight vector W c
' =0.2301<90o,
or (0, 0.23j) at reference position of 50 mm.
W GD
' =lA −lD
lA
x⃗ W 'D = 2600−1200
2600 ∗4.5=2.423Then new weight vector W c
' =2.423<220o,
or (-1.86, -1.56j) at reference position of 50 mm.
W ¿
' = lA−lE
lA
x⃗ W 'E= 2600−600
2600 ∗1=0.231Then new weight vector W c
' =0.77<330o, or
(0.67, -.39j) at reference position of 50 mm.
Total unbalance vector = ⃗ W G=⃗ W GC +⃗ W ¿+⃗ W GD⃗
W G=¿-1.19-1.72j or 2.09<23532 at radius 50 mm
Therefore, required weight at A = 1.552 kg and at G = 2.09 kg at a radius of 50 mm Ans
Solution -3
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