A large and very high-speed turbine is to operate at an angular velocity of 18 000 rev/min and will have a rotor with a principal mass moment of inertia of 225 N.m.s2 . It has been suggested that because this turbine will be installed at the North Pole with its axis horizontal, perhaps the rotation of the earth will cause gyroscopic loads on its bearings. Estimate the size of these additional loads due to gyroscopic effect.

 

 

  

 

  

Question 1:                                                                                                                            (9 marks)

It is required to carryout dynamic force analysis of the four bar mechanism as shown in the figure.

 

Assume the following data:

rad

𝜔2 = 20

 

𝛼2 = 160

s

rad s2

𝑂𝐴 = 250 mm

𝑂𝐺2 = 110 mm

𝐴𝐵 = 300 mm

𝐺3 = 150 mm

𝐵𝐶 = 300 mm

𝐶𝐺4 = 140 mm

𝐶  = 550 mm

∠𝐴𝑂𝐶  = 60⁰

 

Link	Mass (kg)	Mass Moment if Inetia (kgm2)
2	20.7	0.01872
3	9.66	0.01105
4	23.47	0.0277

 

a)       Draw the acceleration diagram                                                                                                          (2 marks)

b)       Calculate the angular accelerations of all links                                                                                (4 marks)

c)        Calculate the inertia forces                                                                                                                   (3 marks)

 

Solution:-

 

 

 

 

 

From velocity and acceleration analysis,

V_A=250mm x 20 = 0.25mx20 =5m/s

V_B=4m/s, V_BA=4.75m/s

 

𝛼^rA = 0.25x20² =100m/s2

𝛼^tA = 0.25x160=40m/s2

 

A^r_B=VB²/CB=4²/0.3=53.33m/s2

A^r_BA=V_BA²/BA=4.75²/0.3= 75.2m/s2

Og2=AG2=48m/s2

Og3=AG3=120m/s2

 

𝛼3=ABA/AB=19/0.3=63.13 rad/s2

 

𝛼4=Ag/CB=129/0.3=430 rad/s2

 

Inertia Forces (Accelerating Forces)

Fg2=m2xAg2=20.7x48=993.6N (in direction of Og2)

Fg3=m3xAg3=9.66x120=1159.2N (in the direction of Og3)

Fg4=m4xAg4=23.47x65.4=1534.94N (in the direction of Og4)

 

h₂=(IG₂x 𝛼₂)/F₂=(0.01872x160)/993.6=3.01x10^-3m

h₃=(IG₃x 𝛼₃)/F₃=(0.01105x63.3)/1159.2=6.03x10^-4m

h₄=(IG₄x 𝛼₄)/F₄=(0.0277x430)/1534.94=7.76x10^-3m

 

The inertia forces F_i2, F_i3, F_i4 have magnitudes equal and direction opposite to the respective accelerating forces and will be tangents to the circles of radius h2, h3 and h4 from G2, G3 and G4 so as to oppose 𝛼2,  𝛼3 and  𝛼4.

 

F_i2 =993.6N

F_i3 =1159.2N

F_i4=1534.94N

 

Question 2:                                                                                                                                                (11 marks)

a)       For the following cam follower mechanism, define: prime circle, pitch circle and trace point.

(3 marks)

 

 

Prime circle: It is the smallest circle that can be drawn from the centre of the cam and tangent to the pitch curve. For a roller follower, the prime circle is larger than the base circle by the radius of the roller.

 

Pitch circle: It is a circle drawn from the centre of the cam through the pitch points.

 

Trace point: It is a reference point on the follower and is used to generate the pitch curve. In a roller follower, the centre of the roller represents the trace point.

 

 

 

 

 

 

 

b)       The reciprocating radial roller follower of a plate cam is to rise 40 mm with simple harmonic motion in 180o of cam rotation and return with simple harmonic motion in the remaining 180o. If the roller radius is 7.5 mm and the prime-circle radius is 40 mm, construct the:

i.        displacement diagram                                                                                      (2 marks)

ii.        the pitch curve                                                                                                  (1 marks)

iii.        the cam profile for clockwise cam rotation.                                                   (3 marks)

If you are using the analytical method, you will need to use Excel or MATLAB and provide the screenshot of the drawing. You can access Excel and MATLAB through the EIT remote lab.

If you are using the graphical method, take a picture of your working and insert it as a screenshot.

 

 

 

 

c)        Identify the type of cam follower mechanism for the following arrangements:   (2 marks)

 

 

 

Answer:- Type of Cam Follower:-

 

a.      Knife Edge follower

b.      Flat Faced Follower

c.      Roller Follower

d.      Spherical Follower

 

Question 3:                                                                                                                                                (12 marks)

a)       Explain how planetary gear train is used for getting different gear ratios in an automobile.

 

Planetary gear sets are used and combined in a complex manner so that transmissions with seven or eight speeds forward plus reverse are possible. Shifts are made by engaging or releasing one or more internal clutches that drive a gear set member, or by engaging or releasing other clutches or bands that hold a gear set member stationary. An automatic transmission might have as many as seven of these power control units (clutches or bands). One-way clutches are also used that self-release and overrun when the next gear is engaged. The control units can operate without the interruption of the power flow.

 

PLANETARY GEAR SET OPERATION

 

Planetary gear sets are so arranged that power enters through one of the members and leaves through one of the other members while the third member is held stationary in reaction. Power flow through a planetary gear set is controlled by clutches, bands, and one-way clutches. One or more clutches will control the power coming to a planetary member and one or more reaction members can hold a gear set member stationary. The third planetary member will be the output.

(a) If the planet carrier is held with the sun gear rotating, the planet gears simply rotate in the carrier and act as idler gears between the sun and ring gears.

(b) If the sun or ring is held, the planet gears will walk around that stationary gear; they rotate on their shafts as the carrier rotates.

(c) If two parts are driven and no parts are held, the planet gears are stationary on their shafts, and the whole assembly rotates as a unit.

b)      Determine gear ratios for all the speeds for the following truck transmission gear box.

(6 marks)

 

 

Solution:-

 

T2=17T

T3=43T

T4=36T

T5=27T

T6=17T

T7=24T

T8=33T

T9=43T

T10=18T

T11=22T

 

Speed 1 = 2-3-6-9

Gear Ratio 1 = (T3/T2)x(T6/T9) = (43/17)x(17/43) = 1

 

Speed 2 = 2-3-5-8

Gear Ratio 2 = (T3/T2)x(T5/T8) = (43/17)x(27/33) = 2.06

Speed 3 = 2-3-4-7

 

Gear Ratio 1 = (T3/T2)x(T4/T7) = (43/17)x(36/24) = 3.79

 

Reverse = 2-3-6-10-11-9

 

Gear Ratio R = (T3/T2)x(T6/T10)x(T11/T9) = (43/17)x(17/18)x(22/43) = 1.22

 

Question 4:                                                                                                                            (8 marks)

Determine the bearing reactions at A and B for the system given below. If it rotates at 350 rev/min. Determine the magnitude and the angular orientation of the balancing mass if it is located at a radius of 50 mm.

 

 

 

Solution:-

 

w=360rev/min, R=50mm

w= (350x2π)/60 = 36.65 rad/ sec

 

F₁=m₁xR₁xw² = 2x0.025x36.65²= 67.16N

F₂=m₂xR₂xw² = 1.5x0.035x36.65²= 70.51N

F₁=m₃xR₃xw² = 3x0.040x36.65²= 161.18N

 

F1 =67.16 sin90 = 67.16j N

F2 = 70.51 (ang -165) = -70.51 sin75i – 70.51cosj = - 68.10i-18.24j

F3 = 161.18N (ang.-75) = 161.18cos75i-161.18sin75j = 41.71i-155.68j

 

∑F = F1+F2+F3

∑F = -26.39i – 106.76j N

 

∑F = 109.95N (ang.-103.9ᵒ)

 

Since all rotating masses are in a single plane the correction mass must be in that plane.

Fc = -∑F = 109.65N (ang.76.1)

 

Fc = Mc x Rc x W²

109.65 = Mc x 0.050 x 36.65²

Mc=1.637 1kg

Θc = 76.1ᵒ

 

Question 5:                                                                                                                                              (10 marks)

A large and very high-speed turbine is to operate at an angular velocity of 18 000 rev/min and will have a rotor with a principal mass moment of inertia of 225 N.m.s2 . It has been suggested that because this turbine will be installed at the North Pole with its axis horizontal, perhaps the rotation of the earth will cause gyroscopic loads on its bearings. Estimate the size of these additional loads due to gyroscopic effect.        (5 marks)

Given :-

I = 225 Nms2 or kgm2

w = 18000rev/min

now, angular speed of precession:

The angular velocity of precession is normal to the earths orbital plane and its magnitude is the angular speed 50.2”/year,nearly.

Therefore, angular speed of precession = 360 ᵒ/ period of revolution in years

=360/25800 = 0.01395ᵒ/year

Importantly, the inclination of the polar axis to the normal to the ecliptic remains the same 23.4ᵒ.

i.e Wp = 0.01395ᵒ/year

=0.01395/(365x24x60x60) x (π/180)

Wp=7.720 x 10^-12 rad / sec

w= 18000/60 x (π/180) = 5.2359 rad/sec.

Gyroscopic Couple acting on the turbine

C=I x w x wp

=225x5.235x7.720x10^-12

=9.993x10^-9 Nm

Therefore, the value of couple is too small.

It might not have any effect. Hence, the additional load does not matter on the system.

 

a)     The diameter of the driver pulley is 25 cm and that for the driven pulley is 55 cm for an open belt drive. The driver pulley rotates at 345 rpm. The angle of contact for the driver pulley is

    rad. Calculate:

i.        the centre distance between two pulleys                                                       (1 mark)

ii.        the angle of contact for driven pulley,                                                            (1 mark)

iii.        length of the belt,                                                                                            (1 mark)

iv.        angular speed of the follower pulley                                                              (1 mark)

v.        The velocity of the belt.                                                                                   (1 mark)

 

Solution: - For Open belt drive

 

Driver Pulley d2= 25 cm N2 = 345 rpm

Driven Pulley d1 = 55 cm

Angle of contact = 3.054 rad

 

Centre distance between two pulleys(x)

 

Θ = 3.054 rad = 3.054 x (180/π) = 174.98ᵒ

 

Θ = 180 - 2𝛼

 174.98ᵒ = 180 - 2𝛼

𝛼 = 2.509ᵒ

 

Sin 𝛼 = d2-d1 / 2x

Sin2.509 = 55-25 / 2x

X = 342.65 cm

 

Length of belt = L = 2x + [π(D₂ + D₁)]/2 + [(D₂ + D₁)²]/4x

 

L = 733.08 cm

 

Velocity of belt, V = (π x d2 x N2) / 60

=( π x 0.25 x 345 ) / 60

V = 4.516 m / sec.

 

As we know, N2/N1 = d1/ d2

345/N1 = 55/25

N1=156.81 rpm

 

Angular speed of follower pulley,

w= (2 x π x N1) / 60

   = (2 x π x 156.81) / 60

W = 16.42 rad/sec

 

REFERENCES

1.      A textbook of mechanics by R S Khurmi S Chand and Company LTD.

2.      A textbook of theory of machine by R S khurmi .

3.      THEORY OF MACHINE by Joseph Shigley.

4.      Some engineering notes (like made easy and ace etc).