ME 352 - Machine Design I
Name of Student:____________________________
Fall Semester 2016
Lab Section Number:_________________________
Homework No. 3 (30 points). Due at the beginning of lecture on Wednesday, September 14th.
Consider Problem 3.15, see Figure P3.15, page 160. For the given position of the input angle
of the inverted slider-crank mechanism, that is, 2  150o , perform a position analysis of the
mechanism using trigonometry (that is, the law of sines and the law of cosines).
Also, for the given position of the input angle determine the angular velocity of link 3 and
the angular velocity of link 4 using:
(i) The method of kinematic coefficients.
(ii) The method of instantaneous centers of velocity.
Compare the answers obtained from Part (i) with the answers obtained from Part (ii).
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Solution to Homework Set 3 (30 Points).
Position Analysis. The known link lengths and angles are R1  125 mm, R 2  75 mm, 1  0o ,
and 2  150o. The unknown position variables can be calculated from trigonometry.
Consider the triangle O2 AO4 . The distance from pin A (the pin connecting links 2 and 3) to
the the ground pin O 4 can be written from the law of cosines as
2
R 34
 R12  R 22  2 R1 R 2 cos 2
(1)
where the input angle 2  150o. Substituting the given numerical data into this equation gives
2
R 34
 (125 mm)2  (75 mm)2  2 125 mm  75 mm  cos 150
(2a)
Therefore, the distance from pin A to the ground pin O 4 is
R 34  193.62 mm
(2b)
R 34
R1

sin O 2 AO4 sin O4O 2 A
(3a)
Using the law of sines gives
Rearranging this equation and substituting in the numerical data gives
R sin O 4O2 A (125 mm) sin150

sin O2 AO4  1
R 34
193.62 mm
(3b)
Therefore, the angles are
O 2 AO4  18.83
and
3  34   (30  18.83)   11.17
(3c)
or
348.83
(3d)
Velocity Analysis.
(i) The method of kinematic coefficients.
A suitable choice of vectors for the inverted slider-crank mechanism are shown in Figure 1.
Figure 1. Vectors for the kinematic analysis of the mechanism.
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The vector loop equation (VLE) for the mechanism, see Figure 1, can be written as
I
?? 
R 2  R 34  R1  0
(4)
Since the angle 34  3 , see Eq. (3d), then the X and Y components of Equation (4) can be
written as
and
R2 cos  2  R34 cos 3  R1 cos 1  0
(5a)
R2 sin  2  R34 sin 3  R1 sin 1  0
(5b)
Differentiating Eqs. (5a) and (5b) with respect to the input position θ2 gives
and
 cos 3  0
 R2 sin  2  R34 sin 33  R34
(6a)
 sin 3  0
R2 cos  2  R34 cos 33  R34
(6b)
Recall that the angle 34  3 , therefore, the first-order kinematic coefficient of link 3 is
3  34
Then writing Eqs. (6) in matrix form gives
  R34 sin 3
 R cos 
3
 34
cos 3   3   R2 sin  2 

    R2 cos  2 
sin 3   R34
(7)
The determinant of the coefficient matrix can be written as
DET 
 R34 sin 3
cos 3
R34 cos 3
sin 3
  R34   193.62 mm
(8)
Using Cramer’s rule, the first-order kinematic coefficient of link 3 can be written as
R2 sin  2 cos 3
 R2 cos  2 sin 3
3 
DET
which can be written as
3 
R2 cos ( 2  3 )
 R34
(9a)
(9b)
Substituting Eqs. (2b) and (3d) and the known data into Eq. (9b), the first-order kinematic
coefficient of link 3 is
(75 mm) cos (150  11.17)
3 
  0.367 mm/mm
(9c)
 193.62 mm
Sign Convention: The positive sign indicates that link 3 is rotating in the same as the direction as
the input link 2, that is, counterclockwise.
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Similarly, the first-order kinematic coefficient of links 3 and 4 can be written as
 R34 sin 3
R2 sin  2
R cos 3  R2 cos  2
  34
R34
DET
which can be written as
 
R34
R2 R34 sin (3   2 )
 R34
(10a)
(10b)
Substituting Eqs. (2b) and (3d) and the known data into Eq. (10b), the first-order kinematic
coefficient of links 3 and 4 is
   (75 mm) sin ( 11.17  150)   24.21 mm 2 / mm
R34
(10c)
Sign Convention: The positive sign indicates that the length of the vector R34 is increasing with
a counterclockwise rotation of the input link 2 (or the length of the vector R34 is decreasing with
a clockwise rotation of the input link 2).
The angular velocity of link 3 can be written, from the chain rule, as
3  3 2
(11a)
Substituting Eq. (9c) and the input angular velocity 2  60 rad/s counterclockwise into Eq.
(11a), the angular velocity of link 3 is
3  ( 0.367 mm/mm)  ( 60 rad/sec)   22 rad/sec
(11b)
Sign Convention: The positive sign indicates that the direction of the angular velocity of link 3 is
counterclockwise (that is, the same direction as the angular velocity of the input link 2).
The angular velocity of link 4 must be the same as the angular velocity of link 3, that is
4   22 rad/sec
(11c)
Aside: The relative velocity between links 3 and 4 can be written as
 2
V34  R34  R34
(12a)
Substituting Eq. (10b) and the input angular velocity 2  60 rad/s counterclockwise into Eq.
(12a), the relative velocity between links 3 and 4 is
V34  ( 24.21 mm 2 /mm)  ( 60 rad/sec)   1452.6 mm/sec
(12b)
Sign Convention: The positive sign indicates that the velocity of point A (fixed in link 3) relative
to the fixed pin O 4 is directed along link 3 away from pin O 4 for a counterclockwise rotation of
the input link 2.
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(ii) The method of instantaneous centers of velocity.
The number of instant centers in a planar mechanism can be obtained from the equation
N
n (n  1)
2
(13a)
Since the number of links is n  4 then the number of instant centers is
N
4x3
6
2
(13b)
There are four primary instant centers and two secondary instant centers. The four primary
instant centers are I12, I23, I34 and I41 (as indicated by the solid lines on the Kennedy circle in
Figure 2a) and the two secondary instant centers are I13 and I24 (as indicated by the dashed lines
on the Kennedy circle).
Figure 2a. The Kennedy Circle.
The locations of the instant centers are shown on Figure 2b (Scale: 1 mm on figure is 3 mm of
mechanism).
Figure 2b. Inversion of the Slider-Crank Mechanism.
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The instant center of links 2 and 3 is coincident with pin A. Therefore, the velocity of point
A fixed in link 3 is equal to the velocity of point A fixed in link 2; i.e.,
(14a)
VA3 = VA2
Equation (14a) can be written in terms of the instant centers as
( I13 I 23 ) 3  ( I12 I 23 ) 2
(14b)
Rearranging this equation, the first-order kinematic coefficient of link 3 can be written as
3 
From
the
scaled
drawing,
see
3
I I
 12 23
2
I13 I 23
Figure
2b,
(15a)
the
distances
I12 I 23  75 mm
and
I13 I 23  204.57 mm. Therefore, the first-order kinematic coefficient of link 3 is
3  
75 mm
  0.367 mm/mm
204.57 mm
(15b)
Sign Convention: The sign is positive because the relative instant center I 23 lies outside the two
absolute instant centers I12 and I13 (that is, the relative instant center does not lie between the
two absolute instant centers).
The angular velocity of link 3 is the same as the angular velocity of link 4 and can be written
from Eq. (11a) as
3  4  3 2
(16a)
Substituting the angular velocity ω2 = 60 rad/sec counterclockwise and Eq. (15b) into Eq. (16a),
the angular velocity of links 3 and 4 is
3  4  ( 0.367 mm/mm)  ( 60 rad/sec)   22 rad/sec
(16b)
Sign Convention: The positive sign indicates that the directions of the angular velocity of links 3
and 4 are the same as the direction of the angular velocity of the input link 2 which is
counterclockwise.
Aside. The relative velocity between point A fixed in link 3 and the fixed point O 4 (which is
along link 3) can be written as
V34  ( I13 I14 ) 3  ( I13 I14 ) 32
(17a)
Rearranging this equation, the first-order kinematic coefficient of links 3 and 4 can be written as
 
R34
V3/4
2
 ( I13 I14 ) 3
(17b)
From the scaled drawing of the mechanism, the distance
I13 I14  66.035 mm
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(18)
Substituting Eqs. (15b) and (18) into Eq. (17b), the first-order kinematic coefficient of links 3
and 4 is
  (  66.035 mm )  ( 0.367 mm/mm)   24.21 mm
(19)
R34
Sign Convention: The sign is positive because the distance between pin A and ground pin O 4 is
increasing for a positive change in the input position.
Alternative approach. Substituting Eq. (15a) into Eq. (17b), the first-order kinematic coefficient
of links 3 and 4 can also be written as
 
R34
( I13 I14 ) ( I12 I 23 )
( I13 I 23 )
(20)
Substituting the measurements I12 I 23  75 mm, I13 I 23  204.57 mm, and I13 I14  66.035 mm
into Eq. (20), the first-order kinematic coefficient of links 3 and 4 is
 
R34
that is
(66.035 mm) x (75 mm)
204.57 mm
   24.21 mm
R34
(21a)
(21b)
Sign Convention: The sign is positive because the distance between pin A and ground pin O 4 is
increasing for a positive change in the input position.
CHECK: Comparing the answers for the first-order kinematic coefficients obtained by the two
methods, shows that the answers are in very good agreement.
The accuracy will depend on the accuracy of the scaled drawing. Note that the accuracy will
generally improve as the drawing is made larger.
Aside. The relative velocity between links 3 and 4 can be written as
 2
V34  R34  R34
(22)
Substituting the angular velocity ω2 = 60 rad/sec counterclockwise and Eq. (21b) into Eq. (22) ,
the relative velocity between links 3 and 4 is
V34  ( 24.21 mm 2 /mm) x ( 60 rad/sec)   1452.6 mm/sec
(23)
Sign Convention: The positive sign indictes that the length of the vector R34 is increasing with a
counterclockwise rotation of the input link 2, see Eq. (12b).
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