exercises with solutionsportal.uc3m.es/portal/page/portal/congresos_jornadas/siomms_2016… · =...

3rd Students International Olympiad on MECHANISM AND MACHINE SCIENCE 20-21 of October, 2016, Madrid, Spain SIOMMS 2016

1

EXERCISES WITH SOLUTIONS

PROBLEM B-1 (8 points)

Using Freudenstein’s equation and considering the independent variables 2 and d,

determine the dimensions of the bars L2 and L3, and the height h of the mechanism shown

in the figure. The precision points are:

2 = 0º d = 120 mm.

2 = 90º d = 98 mm.

2 = 150º d = 57 mm.

N.B. - When formulating the closed vector equation, use the nomenclature shown. Also, observe the directions and senses of the vectors, and the definitions of the angles, all as shown.

SOLUTION:

The equation of closure will be: L L d h2 3

L cos L cos d

L sin L sin h

2 2 3 3

2 2 3 3

L cos L cos d

L sin L sin h

3 3 2 2

3 3 2 2

L cos L cos d

L sin L sin h

22 23 3 2 2

22 23 3 2 2

L L d h d L cos h L sin2 2 2 23 2 2 2 2 22 2

L L d h d L cos h L sin2 2 2 22 3 2 2 2 22 2 0

Slider position

Height


2

L d cos h L sin h L L d2 2 2 22 2 2 2 2 32 2

K dcos K sin K d21 2 2 2 3

K L

K h L

K h L L

1 2

2 2

2 2 23 2 3

2

2

2 = 0º d = 120 mm.

K cos º K sin º K 21 2 3120 0 0 120

2 = 90º d = 97,98 mm.

K cos º K sin º K 21 2 398 90 90 98

2 = 150º d = 57.01 mm.

K cos º K sin º K 21 2 357 150 150 57

K K

K K

' K K K

1 3

2 3

1 2 3

120 14400

9604

49 36 3249

KK

K K

31

2 3

14400

120

9604

K

' ' K K

' K ' ' K K

' K '

33 3

3 3 3

3

1440049 36 0 5 9604 3249

120

0 411 5923 2 0 5 4802 3249

0 911 4370 2

K ' 3 4797 15

K 'K '

3

1

14400 14400 4797 1580 02

120 120

K ' ' 2 9604 4797 15 4806 85


3

K L

K h L

K h L L

1 2

2 2

2 2 23 2 3

2

2 K ,

L ' mm. 12

80 0240 01

2 2

K 'h ' mm.

L ,

2

2

4806 8560 07

2 2 40 01

L h L K L h L K 2 2 2 2 23 2 3 3 2 3

L h L K ' ' ' ' , mm. 2 2 2 23 2 3 60 07 40 01 4797 15 10006 35 100 03


4


A machine operates in a steady-state condition. Wittenbauer’s diagram (the kinetic energy

versus the reduced moment of inertia) for this state is given in the figure.

After adding a flywheel, the steady-state condition of the machine changes to a new state

where the minimum and maximum angles are min = 49.5º and max = 52º. The axes scales

used are: kE = 90 N·m/mm and kJ = 0.2 kg·m2/mm.

Determine the average rotational speed (in r.pm.) and the coefficient of machine speed

fluctuation (regulation).

SOLUTION:

Angles min = 49.5º and max = 52º provide the maximum and minimum angular speeds of

the machine with flywheel:

max max

902 tg( ) 2 tg(52º ) 33,94 rad / s

0,2 E

J

k

k

min min

902 tg( ) 2 tg(49,5º ) 32,46 rad / s

0,2 E

J

k

k

And the average angular speed running is:

max minav

52º 49,5º50,75º

2 2

av av

902 tg( ) 2 tg(50,75º ) 33,19 rad / s

0,2 E

J

k

k

max minav

33,94 32,46or 33,20 rad / s

2 2

E

Jred

maxmin

kE

kJ


5

av

6033,19 316,94 r.p.m.

2

Hence, the coefficient of the mechanism speed fluctuation is found as:

max min

av

33,94 32,460,045

33,19


6


For the gear box shown, we know the number of teeth for the following gears: Z1 = 20,

Z2 = 34 and Z3 = 18. Shafts 0, B and Z are coaxial.

The gear sets have modules m12 = 4 mm and m34 = 3 mm.

a) Determine number of teeth for gear 4 (Z4).

b) Determine the ratio Z

B

, i.e., from shaft Z to shaft B, when 0 = 0 .

c) Determine the ratio Z

0

, i.e., from shaft Z to shaft 0, when O

B2

.

N.B. - All gears are cut with a standard rack (α = 20º , ha = m , c = 0.25·m), with no

displacement.

SOLUTION:

a) Since the gear train is REVERTED and given that the second gear set is internal, the

following equation must be apply:

1 2

1 2 4 3 1 2 4 3

m ma r r r r a (Z Z ) (Z Z )

2 2

It can be seen that:

4 4 4

4 3 3a (20 34) (Z 18) 108 (Z 18) Z 90

2 2 2

b) The apparent transmission ratio will be:

3 1 31A A

2 4 2 4

Z Z ZZ 20 18 2

Z Z Z Z 34 90 17

The transmission ratio between the last axle and the arm when the first axle is fixed

(0 = 0), will be:


7

Z B Z Z ZA A

O B B B B

2 191 1 1

17 17

c) Now the epicycloidal gear train is used as a differential, additionally it is known that

OB

2

, so

Z B AA B Z O

O B A A

1

1 1

O A A AZ O Z O O

A A A A A

11 1 1

2 1 1 1 2 1 2 (1 )

Z A Z

O O

21

1 1517

2 2 34


8


Given that: OA = BC = BD = CD = 250mm; AB = OC = 350mm; = 30º; M1 = M2 = M3 = M4 = 0;

M5 = 0.4 kg. Additionally, vB = 100 mm/s; aB = 10 mm/s2, both with sense towards point O.

1. Draw the mechanism for the specified position (Note, the above drawing is not to scale).

2. Determine the angular velocity of link 1.

3. Determine the velocity and acceleration of point D.

4. Determine the torque that must be acting on link 1, as the only external applied force.

Consider the mechanism as a planar mechanism on a horizontal plane, and that all the joints

are frictionless.

O

C

A

D

B

1

2

3

4

5

T ?

vB, aB


9

SOLUTION:

SPECIFIED POSITION:

VELOCITY IMAGE:

�⃗�𝐴 = �⃗�𝐵 + �⃗�𝐴𝐵; �⃗�𝐴 ⊥ 𝑂𝐴, �⃗�𝐴𝐵 ⊥ 𝐴𝐵

�⃗�𝐶 = �⃗�𝐵 + �⃗�𝐶𝐵; �⃗�𝐶 ⊥ 𝑂𝐶, �⃗�𝐶𝐵 ⊥ 𝐶𝐵

Velocity of point D can be obtained from that of point C

by making the image velocity of link 4 (which is the

equilateral triangle bcd), or by using the instantaneous

center of rotation of link 4, I4 (note that it is equidistant

from points C and D).

𝑣𝐴 = 𝑣𝐶𝐵 = 80 𝑚𝑚

𝑠; 𝜔1 =

𝑣𝐴

𝑂𝐴= 0.32

𝑟𝑎𝑑

𝑠 𝐶𝐶𝑊;

𝑣𝐷 = 𝑣𝐶 = 𝑣𝐴𝐵 = 50 𝑚𝑚

𝑠.

O

C

A

D

B

a

c

b

d

o

�⃗�𝐴

�⃗�𝐷

I4


10

ACCELERATION IMAGE:

�⃗�𝐶𝑛 + �⃗�𝐶

𝑡 = �⃗�𝐵 + �⃗�𝐶𝐵𝑛 + �⃗�𝐶𝐵

𝑡 ;

𝑎𝐶𝑛 =

𝑣𝐶2

𝑂𝐶= 7.14

𝑚𝑚

𝑠2( 𝐶 → 𝑂); �⃗�𝐶

𝑡 ⊥ 𝑂𝐶; �⃗�𝐶𝐵𝑛 =

𝑣𝐶𝐵2

𝐶𝐵= 25.6

𝑚𝑚

𝑠2( 𝐶 → 𝐵); �⃗�𝐶𝐵

𝑡 ⊥ 𝐶𝐵.

From this equation, acceleration of point C is graphically obtained (point c’ in the image

acceleration)

Acceleration of point D can be obtained from that of point C by making the image

acceleration of link 4 (which is the equilateral triangle b’c’d’). We get: �⃗�𝐷 = 30.9𝑚𝑚

𝑠2 .

TORQUE:

Principle of virtual power states:

�⃗⃗� · �⃗⃗⃗�1 − 𝑀4�⃗�𝐵 · �⃗�𝐵 = 0 ⇒ �⃗⃗� · �⃗⃗⃗�1 = 𝑀4𝑎𝐵𝑣𝐵 ⇒ 𝑇 = 1.25 𝑁𝑚𝑚, 𝐶𝐶𝑊

o

'

c'

b

'

d

'

�⃗�𝐶𝑛

�⃗�𝐶𝐵𝑛

�⃗�𝐵

�⃗�𝐷


11


The gear train in the figure has a transmission ratio of = 3/13 (between the input [Gear 1] and the output [Gear 3]) and comprises of three gears.

All the gears have been manufactured with a normalized rack with a reference pressure

angle of = 20º and a module m = 10.

Gears 1 and 2 have Z1 = 12 teeth and Z2 = 35 teeth, respectively.

The pinion [Gear 1] is manufactured with a positive rack displacement of 1.8 mm, and Gear 2 is manufactured without rack displacement.

Knowing that the working distance between centres (O1O2) is = 238.17 mm, complete the following:

For the gear-pair 1-2, specify the working pressure angle, the working (radial) clearance, and the contact ratio

If Gear 3 is manufactured with a negative rack displacement of 1.1 mm, and the mounting between Gears 2 and 3 is implemented without interference and without

backlash, calculate the working pressure angle (’23) and the working distance between centres for the gear-pair 2-3 (O2O3).

Justify numerically if undercutting or interference exists on Gear 3.

SOLUTION:

Since the gear-set (Z 1-Z 2) has module m = 10 mm, a pinion Z 1 = 12 and a gear Z 2 = 35, their reference pitch radii will be:

1 1

1 1 1

m Z 10 12r r r 60 mm.

2 2


12

1 2

2 2 2

m Z 10 35r r r 175 mm.

2 2

The mounting is a non-standard inter-axial distance [V-mount / montaje en V]

( 1 2 12 12 1 2O O 238,17 a ' a r r 60 175 235 ), with a working pressure angle of ’

that can be determined as:

cosa ' a

cos '

1 1a 235' cos cos cos cos20º ' 22.00º

a' 238,17

The working pitch radii are then:

1 1 1

cos cos20ºr ' r 60 r ' 60,81mm.

cos ' cos22º

2 2 2

cos cos20ºr ' r 175 r ' 177,36 mm.

cos ' cos22º

Since the gears are cut using a standard normalized rack, the addendum radii and the dedendum radii can be calculated as:

a fr r (1 x) m r r (1 c x) m siendo c 0,25

a1r 60 1 0,18 10 71,8 mm.

a2r 175 1 0 10 185 mm.

f1r 60 1 0,25 0,18 10 49,3 mm.

f 2r 175 1 0,25 10 162,5 mm.

The base radii are:

b1 1r r cos 60 cos20º 56,38 mm.

b2 2r r cos 175 cos20º 164,45 mm.

The (radial) clearance can be calculated from taking the working distance between centres, and subtracting both the addendum radius of one gear and the dedendum radius of the other:

a1 f 2 a2 f1Working clearance (c') a ' (r r ) a ' (r r )

Working clearance (c') 238,17 (71,8 162,5) 238,17 (185 49,3) 3,87 mm.

The gearing length is equal to:

2 2 2 2

a1 b1 a2 b2 1 2g r r r r r ' r ' sen '

2 2 2 2g 71,8 56,38 185 164,45 60,81 177,36 sen22º 39,98 mm.


13

The contact ratio is equal to the gearing length divided by the circular base pitch (pb):

b

g g 39,981,35

p m cos 10 cos20º

Knowing the transmission ratio and the number of teeth of the pinion (Z1), the number of teeth on gear (Z3) can be determined:

1 2 1 1

3

2 3 3

Z Z Z Z 12Z 52 dientes

Z Z Z 3 /13

The reference distance between centre will be:

23 2 3 2 3

m 10a r r (Z Z ) (35 52) 435mm.

2 2

With the rack displacement known, the displacement coefficient is:

3

desplazamiento en talla 1,1x 0,11

m 10

With the displacement coefficients now known, and specifying gear mounting without interference and without backlash, the working pressure angle can be calculated:

2 3

2 3

x xEv ' 2 tg Ev

Z Z

0 0,11Ev ' 2 tg20º Ev20º 0,013984

35 52 ' 19,6º

Finally, the working distance between centres a’23 will be:

23 23

cos cos20ºa ' a 435 433,91mm.

cos ' cos19,6º

The condition that must be satisfied in order to test for undercutting is:

2

3 b3 3r (1 x ) m r cos r cos

Calculating the reference pitch radius:

3 3

3 3 3

m Z 10 52r r r 260 mm.

2 2

It can be verified that undercutting DOES NOT occur during manufacture:

2260 (1 0,11) 10 260 cos 20º 248,9 229,59


14

GENERAL REFERENCES

Curso de la Teoría de Mecanismos y Máquinas, G.G. Baránov, 2nd edition, 1985, Editorial

MIR, Moscu.

Cinemática y Dinámica de Máquinas, A de Lamadrid & A del Corral, 7th Edition, 1992,

Madrid.

Theory of Machines and Mechanisms, J. Uicker, G. Pennock, J. Shigley, Oxford University

Press; 4th Edition.

Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines,

R.L. Norton, 3rd edition, McGraw-Hill Mechanical Engineering.

Kinematic Synthesis of Linkages. R.S. Hartenburg, J. Denavit, McGraw-Hill, USA, 1964.

Mecanismos: fundamentos cinemáticos para el diseño y optimización de maquinaria,

López-Cajún C.S., Ceccarelli M., Ciudad de México: Trillas, 2nd Edition, 2008.

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