Enmec4060 Vibration And Machine Dynamics Assessment Answers

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:

Equations of motion about P and Q

Assuming θ1 and θ2 are small, the moment equilibrium equations for the two moves about P and Q.

 At point P, the equation of motion will be;

j + mglsin θ1+ k(x1-x2) d =0

Taking account of horizontal displacement, we have

 X1= dsin θ1

X2=d sin θ2

Note θ2 is the displacement for pendulum 2

Substituting j with ml², x1 with dsin θ1 and x2 with d sin θ2, the simplified equation will be:

Ml²+ mgl sin θ1+ k (dsin θ1- dsin θ2) d=0 simplified as

Ml²+ (mgl +kd²) θ1 – kd² θ2 =0 for the 1^st pendulum

At point Q, the equation of motion will be;

j + mglsin θ2+ k(x2- x1) d =0

Note the acceleration for pendulum 2 will be

Following the same process as point P, the simplified equation is shown below

Ml²+ (mgl +kd²) θ2 – kd² θ1 =0 for the 2^nd pendulum eqn 2

Natural frequencies and modal shapes

Assuming motion to be S.H.M with

θi (t) = θi cos (ωt- ?) with i= 1,2,3   eqn 3

Where θ1 and θ2 are the amplitudes for pendulum 1 and 2 respectively, ? as the phase angle and ω as the natural frequency.

With equations 1, 2 and 3 the equation of motion can be expressed in form of a matrix as shown below

-ω² ml^{2+    . =}           eqn 4

Taking into account non trivial solutions, the frequency equation will be

= 0

With det (-ω² m + k) =0 similarly expressed as

-ω⁴ –ω² [ + =0

Substituting ω⁴ with Y² and ω² with Y

The above equation can be rewritten as

M²l⁴x²- 2ml²(mgl +kd²) x² + (m²g²l²+ 2mglkd²) =0

The roots of the above are

X1=  and x2=

Substituting Y1 for ω₁²

ω₁₌ (g/l) ^1/2

Also ω₁ can stand for 2

Therefore, the 1^st natural frequency can be written as

(g/l) ^1/2

Following the same steps, the 2^nd natural frequency will be

() ^1/2

Applying into our matrix equation the amplitude vector, the following equations will be generated.

-ml² ω²- kd²x2 + (mgl+ kd²) x1 =0

-ml² ω²+ (mgl+ kd²) x2- kd²x1=0

Applying our natural frequencies and the amplitudes, we get the first and second amplitude ratios respectively as follows.

Amplitude ratio one=

Amplitude ratio two=

The generated amplitude vectors will be

λ1=θ₁¹

λ2=θ₁²

Thus, the two nodal motions will be

λ1=θ₁¹ cos (ω₁t + ?₁)

λ2=θ₁² cos (ω₂t + ?₂)

Linear vibration response

Using linear superstition of natural modes, the free vibration response of this response is given by

= c₁ + c

By choosing c1=c2=1 with no loss in generality, the above nodal equations leads to

 θ₁¹cos (ω₁t + ?₁) + θ₁²cos (ω₂t + ?₂) eqn a

 θ₁¹cos (ω₁t + ?₁) - θ₁²cos (ω₂t + ?₂) eqn b

θ₁¹_, ?1, θ₁²and   ?₂ are to be determined from the initial conditions with

θ1 (0) =a               (0) =0

θ2 (0) =0               (0) =0

These above conditions are given in the problem

a= θ₁¹cos ?₁ + θ₁²cos ?₂

0= θ₁¹cos ?₁ - θ₁²cos ?₂ eqn c

0= -ω₁θ₁¹sin ?₁ – ω₂ θ₁²sin ?₂

0= -ω₁θ₁¹sin ?₁ + ω₂ θ₁²sin ?₂

Equation c can thus be solved for θ₁¹_, ?1, θ₁²and   ?_{2 to} obtain the responses as follows

 a cos * cos  

 a sin * sin   

By using the newton’s second law of motion, the equation for each mass can be expressed as follows:

m_i =

m_i is the mass for each body

 is the acceleration for each mass

 is the force acting on each mass

The motion equation for each mass is as follows

m_i  = -ku +k(u-u) – c+ c₁(+ f₁(t)

The displacement for each mass is ui

Stiffness of each column is ki

Damping of each floor is ci

Velocity of each mass is

The equation for mass 1 will be

M₁  – c₂ + (k+ku – k2u2 = f₁ (t) eqn 1

For mass 2, the equation will be

M₂  (K2 + k3) u2 – k2u1- k3u3 + (c2+c3) – c2 - c3 = f₂ (t) eqn 2

For mass 3, the equation will be

M₃  (K3 + k4) u3 – k3u2- k4u4 + (c3+c4) – c3 - c4 = f₃ (t) eqn 3

For mass 4, the equation will be

M₄  k4u4- k4u3 + c4 – c4 = f₄ (t) eqn 4

Equations 1 to 4 can be expressed in a matrix form as follows 

(m)  + (c)  + (k)  =

Stiffness matrix is m

Damping matrix is c

Mass matrix is m

Using langrage’s equation of motion

The equation for each mass is written as follows

  +  = Qj^uj 

Potential energy for the system is V

Rayleigh dissipation function is R

Kinetic energy is T given as

T=

T=  + ++

Rayleigh dissipation function R is expressed as

R=

R=  + +

Potential energy for this system is as shown below

V=

V=  + +

The generated force corresponding to the generated coordinate Qj^uj is

Qj^uj =

Qj^uj = + ++

To determine the langrage’s equation, we will have to determine the variable   

Determine variable 

Determine  

Determine the variable  

= k₁u₁-k₂ (u₂-u₁)

Determine the variable Q₁^a

Q₁^a = + ++

=

=

Rebuilding our langrage’s equation

  +  = Q₁ ^(a) with substitution of the following variables

 for

0 for

  for

+ = Q₁ ^(a)

+ = Q₁ ^(a)

Also substitute

K1u1-k2 (u₂-u₁) for

F1 (t) for Q₁ ^(a)

+ F₁(t)

+ F₁(t)

Thus the motion equation for mass 1 is

+ F₁(t)

Repeating the above steps, the equations for subsequent masses are

Motion equation for mass 2 will be

- F₂(t)

Motion equation for mass 3 will be

- F₃(t)

Motion equation for mass 4 will be

+ F₁(t)

Equations 1 to 4 can be expressed in a matrix form as follows

  +   +   =    

(m)  + (c)  + (k)  =

Stiffness matrix is m

Damping matrix is c

Mass matrix is m