ME 352 - Machine Design I
Name of Student:___________________________
Fall Semester 2016
Lab Section Number:_______________________
Homework No. 7 (30 points). Due at the beginning of lecture on Friday, October 21st.
Solve Problem 14.15, page 679. For this offset slider-crank mechanism, solve the dynamic force
analysis problem by the method of inspection; that is, solve one equation for one unknown variable, or
in the worst case scenario solve two equations for two unknown variables.
Note that the standard procedure for the force analysis is to first perform a kinematic analysis of the
mechanism, that is, determine: (i) the first and second-order kinematic coefficients of the mechanism;
and (ii) the first and second-order kinematic coefficients of the centers of mass of links 3 and 4.
Also, note the given assumptions: (i) the crank is balanced, that is, the center of mass of link 2 is
coincident with the crankshaft at O2; (ii) there is no friction in the mechanism; (iii) the external force FB
is acting on the slider (link 4) at point B; and (iv) gravity is acting into the paper, that is, in the negative
Z-direction.
If you would like to check your answers for this homework assignment then you could write a
computer program in Matlab which will use matrix inversion to determine the unknown variables.
Solution to Homework Set 7.
(i) Kinematic analysis of the offset slider-crank mechanism. The vectors for the mechanism are as
shown in Figure 1.
Figure 1. The vectors for the offset slider-crank mechanism.
The vector loop equation (VLE) for the slider-crank mechanism, see Figure 1, can be written as
 I  ? ? 
R2  R3  R4  R1  0
(1)
The X and Y components of Equation (1) are
and
R2 cos  2  R3 cos 3  R4 cos  4  R1 cos 1  0
(2a)
R2 sin  2  R3 sin 3  R4 sin  4  R1 sin 1  0
(2b)
Substituting the numerical data into Equation (2b) gives
0.1 sin 120  0.38 sin 3  R4 sin 0  0.06 sin ( 90)  0
which gives
or
sin 3   0.3858 radians
3   22.69
or
3   337.31
(3a)
(3b)
(3c)
Substituting Equation (3b) and the numerical data into Equation (2a) gives
0.1 cos 120  0.38 cos ( 22.69)  R4 cos 0  0.06 cos ( 90)  0
(4a)
Therefore, the displacement of the slider is
R4  x  0.30 m
2
(4b)
Differentiating Equation (2a) and (2b) with respect to the input position variable  2 gives
and
 R2 sin  2  R3 sin 3 3  R4 cos  4  0
(5a)
R2 cos  2  R3 cos 3 3  R4 sin  4  0
(5b)
Writing Equations (5a) and (5b) in matrix form gives
  R3 sin θ3
 R cos θ
3
 3
 cos θ 4   θ3   R2 sin θ 2 

 sin θ 4   R 4    R2 cos θ 2 
(6)
The determinant of the coefficient matrix in Equation (6) can be written as
DET 
 R3 sin θ3
 cos θ 4
R3 cos θ3
 sin θ 4
 R3 sin θ3 sin θ 4  R3 cos θ3 cos θ 4  R3 cos (θ3  θ 4 )
(7)
Using Cramer’s rule, the first-order kinematic coefficient for link 3 can be written as
R2 sin  2
 cos  4
 R2 cos  2  sin  4
 R2 sin  2 sin  4  R2 cos  2 cos  4  R2 cos ( 2   4 )


3 
DET
DET
R3 cos (3   4 )
(8a)
which can be written as
 3 
 0.1 cos (120  0)
  0.1426 rad/rad
0.38 cos ( 22.69  0)
(8b)
The positive sign indicates that link 3 is rotating in the clockwise direction for a clockwise rotation of
the input link 2. The first-order kinematic coefficient for link 4 can be written as
R4 
 R3 sin 3
R3 cos 3
R2 sin  2
 R2 cos  2
DET

R2 R3 cos  2 sin 3  R2 R3 sin  2 cos 3 R2 R3 sin (3   2 )

DET
R3 cos (3   4 )
(9a)
which can be written as
R4 
(0.1) (0.38) sin ( 22.69  120)
  0.0657 m/rad
0.38 cos ( 22.69  0)
(9b)
The negative sign indicates that link 4 is moving to the right, that is, the velocity of link 4 is to the right,
for a clockwise rotation of the input link 2. Differentiating Equations (5a) and (5b) with respect to the
input position variable  2 gives
 R2 cos  2  R3 cos  3 32 R3 sin  3 3  R4 cos  4  0
(10a)
 R2 sin  2  R3 sin  3 32  R3 cos  3 3  R4 sin  4  0
(10b)
and
3
Writing Equations (10a) and (10b) in matrix form gives
  R3 sin θ3  cos θ 4   θ3   R2 cos  2  R3 cos 3 32

   
2
 R3 cos θ3  sin θ 4   R 4   R2 sin  2  R3 sin 3 3 
Using Cramer’s rule, the second-order kinematic coefficient for link 3 is
R2 cos  2  R3 cos 3 32  cos θ 4
R sin  2  R3 sin 3 32  sin θ 4
R sin ( 2   4 )  R3 sin (3   4 ) 32
θ3  2
 2
DET
R3 cos (3   4 )
(11)
(12a)
which can be written as
θ3 
(0.1) sin (120  0)  (0.38) sin ( 22.69  0)( 0.1426)2
  0.2385 rad/rad 2
0.38cos ( 22.69  0)
(12b)
and the second-order kinematic coefficient for link 4 is
R 4 
 R3 sin θ3
R3 cos θ3
R2 cos  2  R3 cos 3 32
R2 sin  2  R3 sin 3 32  R2 R3 cos ( 2  3 )  R32 32

DET
R3 cos (3   4 )
(13a)
which can be written as
R 4 
 (0.1)(0.38) cos (120  22.69)  (0.38) 2 (0.1426) 2
  0.07783 m/rad 2
0.38cos ( 22.69  0)
(13b)
The angular acceleration of the connecting rod (link 3) can be written as
 3   3 2 2   3  2
(14a)
Therefore, the angular acceleration of the connecting rod is
 3  ( 0.2385)(  18) 2  (  0.1426)( 0)   77.274 rad/s 2
(14b)
The positive sign indicates that the angular acceleration of link 3 is counterclockwise.
The acceleration of the piston (link 4) can be written as
  R   2  R  
R
4
4
2
4
2
(15a)
Therefore, the acceleration of the piston is
  (  0.0778)( 18) 2  ( 0.0657)( 0)   25.207 m/ s 2
R
4
(15b)
The positive sign indicates that link 4 is accelerating to the right.
The vector equation for the mass center of link 3 (i.e., point G3) can be written as
??
 I C
RG3  R2  R33
4
(16a)
where R2  0.1 m,  2  120, R33  0.26 m, and the constraint
33  3    3  22  337.31  22  359.31
(16b)
which indicates that, in this position, the vector R33 is 0.69 below the fixed X-axis (see Figure 3).
The X and Y components of Equation (16a) are
and
X G3  R2 cos  2  R33 cos 33   0.2100 m
(17a)
YG3  R2 sin  2  R33 sin 33   0.0835 m
(17b)
Differentiating Equations (17a) and (17b) with respect to the input position  2 , the first-order kinematic
coefficients of the mass center G3 are
and
X G 3   R2 sin  2  R33 sin 33  3 =  0.0862 m/rad
(18a)
YG3  R2 cos  2  R33 cos 33 3 =  0.0130 m/rad
(18b)
Then differentiating Equations (18a) and (18b) with respect to the input position  2 , the second-order
kinematic coefficients of the mass center G3 are
X G3   R2 cos  2  R33 cos  33  32  R33 sin  33  3 =  0.0455 m/rad 2
(19a)
YG3   R2 sin  2  RG 3 sin  33  32  R33 cos  33  3 =  0.0245 m/rad 2
(19b)
and
The X and Y components of the acceleration of the mass center G3 are
A G 3 X  X G3 22  X G 3  2  ( 0.0455)( 18) 2  ( 0.0862)( 0)   14.742 m/s 2
(20a)
A G 3Y  YG3 22  YG3  2  ( 0.0245)( 18) 2  ( 0.0130)( 0)   7.938 m/s 2
(20b)
and
Summary: The geometry, the first-order kinematic coefficients, and the second-order kinematic
coefficients of the mechanism, for the given input position, are presented in the following table.
2
3
degrees
degrees
a  R1
m
120
337.31
0.06
x  R4
m
0.30

C
 3
R4
 3
R4
degrees degrees rad/rad m/rad rad/rad 2 m/rad 2
32  0.1426  0.0657  0.2385  0.0778
22
The first-order and second-order kinematic coefficients of the mechanism.
(ii) The dynamic force analysis of the offset slider-crank mechanism. Recall that gravity acts into the
paper (i.e., the negative Z-direction) and the effects of friction can be neglected.
The free-body diagram for link 2 is shown in Figure 2. The sum of the external forces in the Xdirection acting on link 2 can be written as
F
X
 m2 AG 2 X
EXT
5
(1a)
Since the mass center G2 is coincident with the ground pivot O2 then Equation (1a) can be written as
F12 X  F32 X  0
(1b)
The sum of the external forces in the Y-direction acting on link 2 can be written as
F
y
 m2 AG 2Y
(2a)
EXT
or as
F12Y  F32Y  0
(2b)
Figure 2. The free-body diagram for link 2.
The sum of the external moments acting on link 2 about the fixed pivot O2 (which is also the center of
mass of link 2) can be written as
M
 I O2  2
(3a)
( R2 X F32Y  R2Y F32 X )  T12  I O2  2
(3b)
O2
EXT
The scalar form of this equation is
or
( R2 cos  2 F32Y  R2 sin  2 F32 X )  T12  I O2  2
(3c)
Substituting  2  0 into Equation (3c) gives
R2 cos  2 F32Y  R2 sin  2 F32 X  T12  0
(3d)
Note that the torque T12 is the torque acting from the ground link 1 on the crankshaft (or the input link 2)
and is assumed to be in the counterclockwise direction. Depending on the application of this mechanism,
this torque is commonly referred to as either the driving torque (a pump or a compressor) or the reaction
torque (an engine).
There are three equations, see Equations (1b), (2b), and (3d), and five unknown variables for the free
body diagram of link 2. The five unknown variables are the four components of the internal reaction
forces F12 X , F12Y , F32 X and F32Y and the driving torque (or reaction torque) T12 .
6
The free-body diagram for link 3 is shown in Figure 3. The sum of the external forces in the Xdirection acting on link 3 can be written as
F
X
 m3 AG 3 X
(4a)
EXT
or as
F23 X  F43 X  FCX  m3 AG 3 X
(4b)
The sum of the external forces in the Y-direction acting on link 3 can be written as
F
Y
 m3 AG 3Y
(5a)
EXT
or as
F23Y  F43Y  m3 AG 3Y
(5b)
Figure 3. The free-body diagram for link 3.
The sum of the external moments acting about the mass center of link 3 can be written as
M
G3
 I G3  3
(6a)
EXT
Also, the sum of the external moments acting about point A on link 3 can be written as
M
A
 I A  3  m3 RG 3/ A  AA
(6b)
M
A
 I G3  3  m3 RG 3/ A  AG 3
(6c)
EXT
or as
EXT
Note that Equations (6a) and (6b) are not as convenient to use as Equation (6c). Therefore, Equation (6c)
will be used here and will be written as the scalar equation
( R3 cos 3 F43Y  R3 sin 3 F43 X )  RAC sin  AC ( FCX )  I G3 3  m3  R33 cos G 3 AG 3Y  R33 sin G 3 AG 3 X 
7
(6d)
Equations (4b), (5b), and (6d) contain two new unknown variables; namely, the internal reaction
forces F43X and F43Y. Therefore, there are a total of six equations, Equations (1b), (2b), (3d), (4b), (5b)
and (6d), and seven unknown variables, F12 X , F12Y , F32 X , F32Y , T12 , F43X and F43Y.
The free-body diagram for link 4 is shown in Figure 4. The sum of the external forces in the Xdirection acting on link 4 can be written as
F
X
 m4 AG 4 X
(7a)
EXT
or as
F34 X  FBX  m4 AG 4 X
(7b)
The sum of the external forces in the Y-direction acting on link 4 can be written as
F
 m4 AG 4Y
Y
(8a)
EXT
Since link 4 can only accelerate in the X-direction; i.e., AG 4Y  0, then Equation (8a) can be written as
F34Y  F14Y  0
(8b)
Figure 4. The free-body diagram for link 4.
The sum of the external moments acting about the mass center of link 4 can be written as
M
G4
 I G4  4
(9a)
EXT
or as
( R14 X F14Y  R14Y F14 X )  I G4  4
(9b)
Since link 4 can only translate; i.e., the angular acceleration  4  0, and the internal reaction force
F14 X  0 (since there is no friction) then Equation (9b) can be written as
R14 X F14Y  0
(9c)
Equations (7b), (8b), and (9c) contain two new unknown variables; namely, the internal reaction force
F14Y and the distance R14X. Therefore, there are a total of nine equations, Equations (1)-(9), and a total of
nine unknown variables.
8
Summary: The nine equations from the dynamic force analysis are
F12 X  F32 X  0
(1b)
F12Y  F32Y  0
(2b)
R2 cos  2 F32Y  R2 sin  2 F32 X  T12  0
(3d)
F23 X  F43 X  FCX  m3 AG 3 X
(4b)
F23Y  F43Y  m3 AG 3Y
(5b)
( R3 cos 3 F43Y  R3 sin 3 F43 X )  RAC sin  AC ( FCX )  I G3 3  m3  R33 cos G 3 AG 3Y  R33 sin G 3 AG 3 X 
F34 X  FBX  m4 AG 4 X
(6d)
(7b)
F34Y  F14Y  0
(8b)
R14 X F14Y  0
(9c)
and
The nine unknown variables are
F12 X , F12Y , F32 X , F32Y , F43 X , F43Y , F14Y , R14 X and T12
(iv) Solving the dynamic force analysis problem by the Method of Inspection.
Recall that the given data and the results obtained from the kinematic analysis are
m 2  m 4  2.5 kg, m3 = 7.4 kg,
IG 2  0.005 N-m-s 2 , IG3  0.0136 N-m-s 2
 2   0 rad/s 2 ,  3 =  77.27 rad/s 2 ,  4  0 rad/s 2 , A G2X  0 m/s 2 ,
A G3X   14.74 m/s 2 , A G3Y   7.94 m/s 2 ,
R33  0.26 m,
 33  3  22  359.31,
A G4X  A 4   25.21 m/s 2 ,
RAC  0.4 m,
and
A G2Y  0 m/s 2
A G4Y  0 m/s 2
 AC  3  32  9.31
Rearranging Equation (7b), the internal reaction force acting from link 3 on link 4 in the X-direction
can be written as
F34 X  m4 AG 4 X  FBX
(10a)
Substituting the known data into this equation, the internal reaction force acting from link 3 on link 4 in
the X-direction is
F34 X  2.5  ( 25.21)  2000   2063.03 N
(10b)
Therefore, the internal reaction force acting from link 3 on link 4 in the X-direction is
F43 X   2063.03 N
(10c)
Rearranging Equation (4b), the internal reaction force acting from link 2 on link 3 in the X-direction
can be written as
F23 X  m3 AG 3 X  F43 X  FCX
(11a)
9
Substituting the known data and Equation (10c) into Equation (11a), the internal reaction force acting
from link 2 on link 3 in the X-direction is
F23 X  7.4  ( 14.74)  ( 2063.03)  1000   3172.11 N
(11b)
Therefore, the internal reaction force acting from link 3 on link 2 in the X-direction is
F32 X   3172.11 N
(11c)
Substituting Equation (11c) into Equation (1b), the internal reaction force between link 1 and link 2
in the X-direction is
F12 X   F32 X   3172.11 N
(12a)
Therefore, the internal reaction force acting from link 2 on link 1 in the X-direction is
F21 X   F12 X   3172.11 N
(12b)
Rearranging Equation (6d), the internal reaction force acting from link 4 on link 3 in the Y-direction
can be written as
R3 cos 3 F43Y  I G3 3  m3  R33 cos  G 3 AG 3Y  R33 sin  G 3 AG 3 X   R3 sin 3 F43 X  RAC sin  AC ( FCX ) (13a)
Substituting the known data and Equation (10c) into Equation (13a), the internal reaction force acting
from link 4 on link 3 in the Y-direction can be written as
(0.38cos 337.3) F43Y  0.0136 ( 77.27)  7.4[ 0.26 cos 359.3o ( 7.94)  0.26sin 359.3o ( 14.74) ]
 0.38sin 337.3(  2063.03)  0.4sin 9.3(  1000)
(13b)
or
0.35 F43Y   1.05  15.28  0.05  302.53  64.64 Nm   223.71 Nm
(13c)
Therefore, the internal reaction force acting from link 4 on to link 3 in the Y-direction is
F43Y   639.17 N
(13d)
This implies that the internal reaction force acting from link 3 on link 4 in the Y-direction is
F34Y   639.17 N
(14)
Rearranging Equation (5b), the internal reaction force acting from link 2 on link 3 in the Y-direction
can be written as
F23Y  m3 AG 3Y  F43Y
(15a)
Substituting the value for m3 AG 3Y and Equation (13d) into this equation, the internal reaction force
acting from link 2 on link 3 in the Y-direction is
F23Y  7.4  ( 7.94)  639.17   697.93 N
10
(15b)
Therefore, the internal reaction acting from link 3 on link 2 in the Y-direction is
F32Y   697.93 N
(16)
Substituting Equation (14) into Equation (8b), the internal reaction force acting from link 1 on link 4
in the Y-direction is
F14Y   F34Y   639.17 N
(17)
Note that Equation (9c) implies that either: (i) the distance R14 X  0; or (ii) the Y-component of the
internal reaction force F14Y  0 . Since (ii) is not possible, that is, the internal reaction force F14Y cannot
be zero (i.e., there must be contact between link 1 and link 4, either a single point of contact or two
points of contact, and also the mechanism has a mobility of one) then the conclusion is that the distance
R14 X  0
(18)
Substituting Equation (16) into Equation (2b), the internal reaction force acting from link 1 on link 2
in the Y-direction is
F12Y   F32Y   697.93 N
(19a)
and the internal reaction force acting from link 2 on link 1 in the Y-direction is
F21Y   F12Y   697.93N
(19b)
Finally, rearranging Equation (3d), the reaction torque can be written as
T12   R2 cos  2 F32Y  R2 sin  2 F32 X
(20a)
Substituting Equations (11c) and (16) into this equation, the reaction torque is
T12   0.1 cos 120  ( 697.93)  0.1 sin 120  ( 3172.11)
(20b)
T12   34.90  274.71 Nm
(20c)
which can be written as
Therefore, the reaction torque is
T12   239.21 N-m
(21a)
The negative sign indicates that the assumption that the reaction torque (for the specified position of the
slider-crank mechanism) is acting counterclockwise is not correct. The reaction torque must be acting
clockwise, that is
T12   239.21 k N-m
(21b)
The crank torque, from the crankshaft (or the input link 2) acting on the ground link 1, is
T21   T12   239.21 N-m
(22a)
The positive sign indicates that the crank torque (for the specified position of the slider-crank
mechanism) is acting in the counterclockwise direction, that is
T21   239.21 k N-m
11
(22b)