unisa mom3602 theory of machines iii mock exam 2015 _preparation practice problems

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MOM3602 Theory of Machines III Mock Exam 2015 (100 Marks in 3 Hours)

Preparation/Practice Problems: First try to solve without using your textbooks/notes and then go to your textbooks/notes if stuck

QUESTION 1 [20 Marks] Consider the mass-spring-damper system as illustrated below in Figure 1.

Figure 1.

The system is known to be under-damped with � � 1 so that the displacement of the system is of the form

�� cos ��1 � ��

where is a constant amplitude, � is the damping factor, �� is the angular natural frequency, � is the time, and � is a constant phase lag. Experimental measurements are conducted on the system as illustrated in Figure 2.

Figure 2.

The displacement at time � � 0.25s is measured as ��0.25� � �6.7716 "10�#m and the displacement at time

� � 0.50s is measured as ��0.50� � �0.3834 "10�#m. Determine the following:

• Using the specified experimental data determine the expression for the displacement by finding the values of

the parameters and � and write down the expression ��

• Use the determined expression for the displacement to calculate the velocity of the system at time � � 0.65s

UNISA MOM3602 Theory of Machines III Mock (Preparation/Practice) Exam 2015

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QUESTION 2 [10 Marks] Consider the mechanical system illustrated in Figure 3.

Figure 3.

The system is composed of three rods which are all rigidly mounted together and allowed to rotate by the same angle ( in a counter-clockwise direction.

The rod of length )* has a mass of +* and an additional point mass of mass ,* which is a distance of *#)* from the

point of rotation, and a spring of spring constant -* is located a distance of �#)* from the point of rotation.

The rod of length )� has a mass of +� and an additional point mass of mass ,� which is a distance of �# )� from the

point of rotation, and a damper of damping constant .� is located a distance of *# )� from the point of rotation.


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The rod of length )# has a mass of +# and an additional point mass of mass ,# which is a distance of )# from the

point of rotation, and a damper of effective damping constant ./00 is located a distance of �# )# from the point of rotation,

and a spring of effective spring constant -/00 which is located a distance of *# )# from the point of rotation. The effective

spring constant -/00 is made up of two springs of equal spring constants -# which are connected in parallel, and the

effective damping constant ./00 is made up of two dampers of equal damping constants .# which are connected in

series. If the systems equation of motion using the rotational form of Newton’s second law of form

1�2/345�6�� 7�/849:65/�� d�(d��

where 1�2/345�6�� is the resultant counter-clockwise moment and 7�/849:65/�� is the equivalent rotation moment of

inertia may be expressed in the form

,/8(< + ./8(> + -/8( � 0

then determine the equivalent mass ,/8, equivalent damping constant ./8, and equivalent spring constant -/8.

EXTRA HINTS:

• Springs in series *?@A = *

?B + *?C +⋯

• Springs in parallel -/8 = -* + -� +⋯

• Dampers in series *

E@FF = *EB + *

EC +⋯

• Dampers in parallel ./8 = .* + .� +⋯

QUESTION 3 [10 Marks] The needle indicator of an electronic instrument is connected to a torsional viscous damper and a torsional spring. If the rotary inertia of the needle indicator about its pivot point is 25 kg.m2 and the spring constant of the torsional spring is 100 N.m/rad, determine the damping constant of the torsional damper if the instrument is to be critically damped. QUESTION 4 [15 MARKS] In a Hartnell governor as illustrated in Figure 4, the length of the ball arm is 190 mm, that of the sleeve arm is 140 mm, and the mass of each ball is 2.7 kg. The distance of the pivot of each bell-crank lever from the axis of rotation is 170 mm, and the speed, when the ball arm is vertical, is 300 rev/min. The speed is to increase 0.6% for a lift of 12 mm of the sleeve.

Figure 4.


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Determine the following:

a) Neglecting the dead load on the sleeve, find the necessary stiffness of the spring and the required initial compression

b) What spring stiffness and initial compression would be required if the speed is to remain the same for the changed position of the sleeve (i.e. the governor is to be isochronous)?

QUESTION 5 [30 Marks] The turning moment diagram for an engine is given by

1 � G2100 sin�(� + 900 sin�2(� ,when0 ≤ ( ≤ P375 sin�(� , whenP ≤ ( ≤ 2P

where 1 is in units of N.m and the angle is in radians and is illustrated below in Figure 5.

Figure 5.

The resisting torque is constant and the speed is 850 rev/min. The total moment of inertia of the rotating parts of the engine and the driven member is 270 kg.m2. Using any suitable combination of analytical, numerical or graphical techniques determine:

a) The power b) The fluctuation in speed c) The maximum instantaneous angular acceleration of the engine d) The value of the crank angle corresponding to the state where the maximum instantaneous angular

acceleration occurs


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QUESTION 6 [15 Marks] Consider a cam with curved flanks and a flat-ended follower as illustrated below in Figure 6.

(a) Follower on Flank Surface AB (b) Follower on Nose Surface BC

Figure 6.

When the follower is in contact with the flank surface AB (i.e. “follower on flank”) the displacement of the follower is specified as

� � QR* � QR, forwhenthefollowerisonflanksurfaceAB

When the follower is in contact with the flank surface BC (i.e. “follower on nose”) the displacement of the follower is specified as

� = QR� − QR, forwhenthefollowerisonflanksurfaceAB

Determine:

a) A derivation for the expression of the follower displacement in terms of \ , ] , and ( showing all of your

geometrical and mathematical calculations for when the follower is on the flank surface AB

b) A derivation for the expression of the follower displacement in terms of ^, �, _, and ] showing all of your

geometrical and mathematical calculations for when the follower is on the flank surface BC

c) Considering only the case for a “follower on flank” assume that the expression for the follower displacement on

surface AB is � = �\ − ]��1 − cos (� and use differentiation to calculate the formulae for the velocity

` = aba� and then use differentiation again to work out the acceleration c = a:

a�

d) Using your derived formulae for � and ` and c sketch on a single graph the curves for the displacement,

velocity, and velocity on a vertical axis and indicating on the horizontal axis the regions where the “follower on

flank” AB is and where the “follower on nose” BC is (just show the general shape of the �, `, and c curves)


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LIST OF FORMULAE

TRIGNOMETRIC IDENTITIES

sin��(� + cos��(� = 1, sin�R + d� = sin�R�cos�d� + cos�R�sin�d� cos�R + d� = cos�R�cos�d� − sin�R�sin�d�, tan�(� = efg�h�

ije�h�, 1 + tan��(� = sec��(� 1 + cot��(� = cosec��(�, sin �k� − (� = cos�(�, cos �k� − (� = sin�(�, tan �k� − (� = cot�(�

sin�−(� = −sin�(�, cos�−(� = +cos�(�, tan�−(� = −tan�(�, sin�2(� = 2sin�(�cos�(�, cos�2(� = cos��(� − sin��(�

tan�l ± `� = nog�4�±nog�:�*∓nog�4� nog�:�, Pradian = 180°, 1radian = �*rsk �

° , 1° = � k*rs� radian

FREE AND DAMPED VIBRATIONS

, aCba�C + . aba� + -� = 0, �� = te3�, ,u� + .u + - = 0

v�� + w� + . = 0, � = �x±√xC�z6E�6

u* = − E�{ +|� E

�{�� − ?

{, u� = − E�{ −|� E

�{�� − ?

{, �� = t*e3B� + t�e3C� .E = 2,|?

{ = 2√-, = 2,��, � = EE}, General roots u*,� = ~−� ± �� − 1��

General solution �� = t*e��C�*� �� + t�e��C�*� �� , �� = ��1 − �� , �� = 2Pc ,

c = *�

Under-damped if � < 1 then u* = ~−� + ��1 − �� and u� = ~−� − �1 − �� where � =√−1and �� = t*e3B� + t�e3C� = �� cos~�1 − �� − ��

Critically-damped if � = 1 then �� = �t* + t��e� ��

Over-damped if � > 1 then �� = t*3B� + t�3C� = t*��C�*� �� + t��C�*� ��


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Alternative Formulae Notation

Logarithmic decrement

� � ln bBbC � ��1� � �k�

�*��C ��k �

∙ E�{, 1� � �k

� , � � �� cos~�1 − �� − ��

�� = �* + 1�, � = ��k�C��C, Amplitude �* at time �* and amplitude �{�* at time �{�* = �* +,1�

has bB

b��B = bBbC

bCb�

b�b�⋯ b�

b��B, b�b��B = e� �� ,

bBb��B = ~� ��{ = e{� ��, � = *

{ ln � bBb��B�

KINETICS/DYNAMICS

Moment of inertia for a thin rod about the centre-of-mass is 72�� = **�+)�, Angle ( = 3

� where u is the arc-

length and ] is the radius

Parallel axis theorem 7|| = 7�� ++��, Moment of inertia from radius of gyration - is 7 = +-�

Resisting spring force �3�29�� = −-�, Resisting damping force ��6{�9�� = −. aba� Newton’s second law for translations �2/345�6�� = , aCb

a�C , Newton’s second law for rotations 12/345�6�� =73�3�/{ aCh

a�C

Torque 1 = � × ]

CAMS

Curved Flanks & Flat-Ended Follower

Follower in contact with flank: � = �\ − ]��1 − cos (�, ` = ��\ − ]� sin (, c = ��\ − ]� cos (


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\ � �C�2C��C�� ije��2�� ije�� , sin� = � efg�

��2

Follower in contact with nose: � = �^ cos� + _� − ], ` = �^ sin�, c = −��^ cos�

Straight Flanks & Roller-Ended Follower

Roller in contact with flank: � = �] + _s� sec ( −�] + _s�, ` = ��] + _s� sec ( tan (,

c = ��] + _s��2sec#( − sec (�, tan � = � efg��2� , cos � = ��2

�

Roller in contact with nose:

� = ^ cos� +��¡� − sin��¢ − �] + _s�, ` = −�^ £sin� + efg��¤��C�efgC¤¥

c = −��^ ¦cos� + efg�¤��Cije��¤��C�efgC¤��C §, ¡ = 2�2�

� , c = −��^ �1 + *��, Lift ¨ = ^ + _ − ]

Normal force between roller-ended follower and flank of tangent cam

� cos ( = ,c + ©�� + ª�, Reaction torque «� = � ×�] +_s� tan (

Circular cam & flat-ended follower:

� = ^�1 − cos (�, ` = �^ sin (, c = ��^ cos (


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Moments of inertia

7 � ,-�

7¬ � ,)�12

7¬ = ,_�2

7� = 7¬ +,ℎ�

Equivalent moment of inertia at machine 7® = 7{6E¯9�/ + � °�±²±³°�´}µ¶�@�

� 7{��2

Crank Effort Diagrams

Energy fluctuation

Centrifugal stress · = \`� = \��_�

0̧ = 12 7��*� − �� 7 = ¸¹-3��

º3 = �* − �� = 0̧7��

� = «7

»�sin�,(��d(x

6= £− 1

, cos�,(�¥6x

»�cos�,(��d(x

6= £1, sin�,(�¥6

x

sin�2(� = 2 sin ( cos (

cos 2( = cos�( −sin�(

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unisa mom3602 theory of machines iii mock exam 2015 _preparation practice problems

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