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ENMEC4060 Vibration and Machine Dynamics Assessment

An unbalanced flywheel shows an amplitude of 0.165 and a phase angle of 15° clockwise from the phase mark. When a trial weight of magnitude 50 is added at an angular position 45° counter-clockwise from the phase mark, the amplitude and the phase angle become 0.225 and 35° counterclockwise, respectively. Find the magnitude and angular position of the balancing weight required. Assume that the weights are added at the same radius.

Answer:

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
p>

 

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 =

From, data, this can be defining as, the rectangular form of this vector can be given as

In the same way trial weight can be given as

The trial weight can be represented as

The further equation for measured vector can be given as follows

The resultant magnitude can be obtained from subtracting (iv) from (v)

Since amplitude of resultant vector is calculated, on this basis we can calculated weight of the balanced mass

The placement balance mass will be just opposite of the result calculated,

= 6.148 +2.08123j

The balance vector in polar coordinate CCW.

As per above calculation, we must place 6.491 gm of weight at 18.672o CCW. Ans

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.

We can represent the weight vector as follows

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 , with reference to the standard radius.

, with reference to the standard radius R = 50 mm.

, 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  is 0+0.772j with reference to the standard radius R = 50 mm.

The rectangular form of weight is -1.59-1.34j with reference to the standard radius R = 50 mm.

The rectangular form of weight is 0.2-0.12j with reference to the standard radius R = 50 mm.

The sum of unbalance vector =

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 with reference to the standard radius R = 50 mm, the new weight vector , or (0, 0.23j)

The rectangular form of weight with reference to the standard radius R = 50 mm, the new weight vector , or (-1.862, -1.561j).

The rectangular form of weight with reference to the standard radius R = 50 mm, the new weight vector , or (-.67, -0.39j).

The sum of unbalance vector =

If we add the above rectangular vector, we will get sum of unbalanced rectangular vector.

= -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

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
 
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
 
The component of forces along x direction is given as

In the same way, the primary and secondary forces along y direction is given
 
For simplifying the equation, we must simply the variable first.
 
Which is as follows
 
The given equation,
 
As the angle given in question, replacing these values in equation
        
Similarly, for  by putting the value of α and t = 0
    
Now equation 
    
As the angle given in question, replacing these values in equation

Simplifying with trigonometrical theories (b)
    
Same calculation for y and z axis
     
We must calculate the moment in z axis
 
Similarly, moment along y axis,

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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