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ME 352 - Machine Design I Name of Student_____________________________ Summer Semester 2013 Lab Section Number__________________________ EXAM 1. OPEN BOOK AND CLOSED NOTES. Wednesday, June 26th, 2013 Use the blank paper provided for your solutions. Write on one side of the paper only. Where necessary, you can use the figures provided on the exam to show vectors and instant centers. Any work that cannot be followed is assumed to be in error. At the completion of your exam, please staple each problem separately and staple your crib sheet to the end of Problem 1. Problem 1 (25 Points). Part I. (15 Points). For the mechanism shown in Figure 1a: (i) Determine the mobility using the Kutzbach criterion. Clearly number each link and label the lower pairs and the higher pairs on the given figure. (ii) Define suitable vectors for a kinematic analysis. Label and show the direction of each vector clearly on the given figure. (iii) Write the vector loop equations that are required for a kinematic analysis of the mechanism. Clearly identify suitable input(s) for the mechanism. List: (a) the known quantities; (b) the unknown variables; and (c) any constraints. If you identified constraints in part (b) then write the constraint equations.
Figure 1a. A planar mechanism.
Part II. (10 Points). Using trigonometry, determine the mechanical advantage of the four-bar linkage in the position shown in Fig. 1b. The input link 2O A 6 in,= link AB 10 in,= and output link 4BO 4 in.=
Figure 1b. A four-bar linkage.
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ME 352 - Machine Design I Name of Student_____________________________ Summer Semester 2013 Lab Section Number__________________________ Problem 2 (25 Points). For the mechanism in the position shown in Figure 2, the angular velocity of the input gear 2 is 2 15rad / sω = counterclockwise. Gear 3 is rolling, without slip, on gear 2 at the point of contact E. The radius of gear 2 is 2 400 mm,ρ = the radius of gear 3 is 3 150 mm,ρ = and the distance from point C on link 3 to pin D connecting links 3 and 4 is CD 500 mm.= (i) Draw vectors clearly on the given figure that are suitable for a kinematic analysis of the mechanism. Then write the vector loop equation for the mechanism. (ii) Using your vector loop equation determine the first-order kinematic coefficients for the mechanism. (iii) Determine the angular velocities of links 3 and 4. Give the magnitudes and directions of the vectors.
Figure 2. A planar mechanism.
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ME 352 - Machine Design I Name of Student_____________________________ Summer Semester 2013 Lab Section Number__________________________ Problem 3 (25 Points). For the mechanism in the position shown in Figure 3, the input link 2 is moving downward with a constant velocity 2V 90 mm / s.= (i) List the primary instant centers and the secondary instant centers for the mechanism. (ii) Using the given Kennedy circle, show the location of all the instant centers on the given figure. Using the location of the instant centers, determine: (iii) The first-order kinematic coefficients of links 3, 4, and 5. (iv) The magnitudes and the directions of the angular velocities of links 3, 4, and 5. (v) The magnitude and direction of the velocity of point D which is fixed in link 4.
Figure 3. A planar mechanism. Scale: full size, that is, 1 mm = 1 mm.
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ME 352 - Machine Design I Name of Student_____________________________ Summer Semester 2013 Lab Section Number__________________________ Problem 4 (25 Points). For the mechanism in the position shown in Figure 4, the angular velocity of the input link 2 is 2 50 rad / sω = counterclockwise. The link lengths are 2O A 500 mm,= AB 400 mm,= and the radius of the wheel which is rolling without slipping on link 1 is 4 140 mm.ρ = (i) Write a vector loop equation that is suitable for a kinematic analysis of this mechanism. Draw your vectors clearly on the given figure. (ii) Using your vector loop equation determine the first-order kinematic coefficients for the mechanism. (iii) Determine the angular velocities of links 3 and 4. Give the magnitude and direction of each vector. (iv) Determine the velocity of point B fixed in link 4. Give the magnitude and direction of this vector.
Figure 4. A planar mechanism.
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Solution to Problem 1. Part I. (i) 4 points. The links of the mechanism and the joints are as shown in Figure 1(a).
Figure 1(a). The links and joints of the mechanism.
The Kutzbach mobility criterion can be written as
1 2M 3(n 1) 2 j 1j= − − − (1)
The number of links, the number of lower pairs (or 1j joints), and the number of higher pairs (or 2j joints), respectively, are
1 2n 7, j 8, and j 1= = = (2) Substituting Equation (2) into Equation (1), the mobility of the mechanism is
M 3(7 1) 2(8) 1 1= − − − = (3) This is the correct answer for this mechanism, that is, for a single input there is a unique output. (ii) 4 Points. Suitable vectors for a kinematic analysis of the mechanism are shown in Figure 1(b).
Figure 1(b). Vectors for the mechanism.
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(iii) 7 Points. (a) The input link is chosen to be link 2 (see Figure 1a) and the input variable is the angle 2θ . Note that link 4 would also be suitable as an input link since it pinned to the ground link. The slider
(link 7) would also be suitable as an input link. Other choices for the input link are not really practical. (b) The six unknown variables are the four angles 3θ , 4θ , 5θ , and 6θ and the two distances 36R and 7R . (c) There are four constraints, the four constraint equations are
33 3θ θ φ= − (5a)
36 33 (180 )θ θ α= + °− (5b)
66 6θ θ β= − (5c) and
333 3θ θ γ= + (5d) Since there are 6 unknown variables then three independent vector loop equations are required. The three vector loop equations can be written as
Loop 1: 1 2 3 4+ + oR R R R
√√ √Ι √ √?+ =？ (4a)
Loop 2: 31 2
2 33 36 6 66 7 11 1
? oCC CR R R R R R R R
√? ?√Ι √ ?√ √√ √√+ + + + − + + = (4b)
Loop 3: 1 2 4
2 33 36 6 5 333 4 1
? oC C CR R R R R R R R
? ? √√Ι √ √? √? √√+ + + + + + + = (4c) Part II. 10 points. The mechanical advantage of the four-bar linkage, see Chapters 1 and 3, is defined as
4 4
2 2
sin sinsin sin
R BOMAR AO
ψ ψφ φ
= = (1)
where the angles ψ and φ are as shown in Figure 1(c).
Figure 1(c). The four-bar linkage.
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From the triangle 2AO B , the angles can be written as
2sin AOAB
ψ = (2a)
and 2sin BO
ABφ = (2b)
Sustututing Equations (2) into Equation (1), the mechanical advantage of the four-bar linkage can be written as
24
4
2 22
AOBO BOABMA BO BOAOAB
= = (3a)
Therefore, the mechanical advantage of the four-bar linkage can be written as
4
2
BOMABO
= (3b)
Notice that the triangle 2AO B is a right-angled triangle with the angle 2 90AO B∠ = ° . Therefore, the length of 2O B is
2 2 2 22 2 10 6 8 inchesBO AB O A= − = − = (4)
Substituting Equation (4) and the known length of the output link, that is, 4BO 4 in,= into Equation (3), the mechanical advantage of the four-bar linkage is
4 0.58
MA = = (5)
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Solution to Problem 2. (i) 7 Points. A suitable choice of vectors for the mechanism are as shown on Figure 2.
Figure 2. A planar mechanism. The vector loop equation (VLE) for the mechanism can be written as
√ C √ ? √ ? √ √ 7 3 4 1R R R R 0− − + = (1)
where link 7 is the arm. The input of this mechanism is the angular position of gear 2. (ii) 12 points. The rolling contact constraint between gear 2 and link 3 can be written as
3 72
3 2 7
θ θρρ θ θ
Δ −Δ± =
Δ −Δ (2)
The correct sign is positive because there is internal contact between the two gears.
Differentiating Equation (2) with respect to the input position 2θ gives
3 72
3 2 7
θ θρρ θ θ
′ ′−+ =
′ ′− (3a)
Rearranging Equation (3a) and substituting the kinematic coefficient of the input link 2 1θ′ = into the resulting equation, the first-order kinematic coefficient for the arm (that is, link 7) can be written as
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3 3 27
3 2
ρ θ ρθρ ρ′ −′ =−
(3b)
Substituting the given dimensions into Equation (3b), the first-order kinematic coefficient for the arm can be written as
37
150 400150 400θθ′ −′ =−
(3c)
The X and Y components of the VLE, see Equation (1), are
7 7 3 3 4 4 1 1R cos R cos R cos R cos 0θ − θ − θ + θ = (4a) and
7 7 3 3 4 4 1 1R sin R sin R sin R sin 0θ − θ − θ + θ = (4b)
Differentiating Equations (4) with respect to the input position 2θ gives
7 7 7 3 3 3 4 4 4R sin R sin R sin 0′ ′ ′− θ θ + θ θ + θ θ = (5a) and
7 7 7 3 3 3 4 4 4R cos R cos R cos 0′ ′ ′+ θ θ − θ θ − θ θ = (5b) Then substituting Equation (3b) into Equations (5) gives
3 3 27 7 3 3 3 4 4 4
3 2R sin R sin R sin 0
⎛ ⎞′ρ θ −ρ ′ ′− θ + θ θ + θ θ =⎜ ⎟ρ −ρ⎝ ⎠ (6a)
and
3 3 27 7 3 3 3 4 4 4
3 2R cos R cos R cos 0
⎛ ⎞′ρ θ −ρ ′ ′θ − θ θ − θ θ =⎜ ⎟ρ −ρ⎝ ⎠ (6b)
Then writing Equations (6) in matrix form gives
3 7 27 3 3 4 4 7 73
3 2 3 2
3 7 247 3 3 4 4 7 73 2 3 2
R sin R sin R sin R sin
R cos R cos R cos R cos
⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ρ ρ− θ + θ + θ − θ′θ⎢ ⎥ ⎢ ⎥⎡ ⎤⎜ ⎟ ⎜ ⎟ρ −ρ ρ −ρ⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎢ ⎥ =⎢ ⎥ ⎢ ⎥⎢ ⎥⎛ ⎞ ⎛ ⎞ρ ρ⎢ ⎥ ⎢ ⎥⎢ ⎥′θ+ θ − θ − θ + θ⎣ ⎦⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ρ −ρ ρ −ρ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦
(7)
Substituting the known data into Equation (7) gives
3
4
+ 353.55 mm 348.46 mm 0 mm
203.55 mm 603.55 mm 400 mm
′θ+ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥′⎢ ⎥ ⎢ ⎥+ − −θ⎣ ⎦ ⎣ ⎦⎣ ⎦
(8a)
The determinant of the coefficient matrix in Equation (8a) is
2DET ( 353.55 mm)( 603.55 mm) ( 348.46 mm)( 203.55 mm) 284314 mm= − − + + = − (8b)
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Using Cramer’s rule, the first-order kinematic coefficient for link 3 from Equation (8a) is
23 2
139384 mm 0.4902 rad / rad284314 mm
+′θ = = −−
(9)
The negative sign indicates that link 3 is rotating clockwise as link 2 rotates counterclockwise. Also, the first-order kinematic coefficient for link 4 is
24 2
141420 mm 0.4974 rad / rad284314 mm
−′θ = = +−
(10)
The positive sign indicates that link 4 rotates counterclockwise as gear 2 rotates counterclockwise.
Substituting Equation (9) into Equation (3c), the first-order kinematic coefficient for the arm is
7150( 0.4902) 400 1.89 rad/rad
150 400θ − −′ = = +
− (11)
The positive sign indicates that the arm is rotating counterclockwise as gear 2 rotates counterclockwise. (iii) 6 Points. The angular velocity of link 3 can be written as
3 3 2′ω = θ ω (12a) Substituting Equation (9) and the angular velocity of the input link 2 into Equation (12a) gives
( )3 0.4902 rad ( 15rad / s) 7.35 rad / sω = − + = − (12b) The negative sign indicates that link 3 is rotating clockwise.
The angular velocity of link 4 can be written as
4 4 2′ω = θ ω (13a) Substituting Equation (10) and the angular velocity of the input link 2 into Equation (13a) gives
4 ( 0.4974 rad) ( 15 rad / s) 7.46 rad / sω = + + = + (14b) The positive sign indicates that link 4 is rotating counterclockwise.
Note that the angular velocity of the arm (link 7) can be written as
7 7 2′ω = θ ω (15a) Substituting Equation (11) and the angular velocity of the input link 2 into Equation (15a) gives
( )7 1.89 rad ( 15 rad / s) 28.35 rad / sω = + + = + (15b) The positive sign indicates that the arm (link 7) is rotating counterclockwise.
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Solution to Problem 3. (i) 1 Point. The number of links in the mechanism is four, therefore, the total number of instant centers for this mechanism is six; i.e.,
( 1) 5 x 4 102 2
n nN −= = = (1)
There are 6 primary instant centers; namely, I12, I15, I45, I23, I14, and I34. Therefore, there are 4 secondary instant centers; namely, I24, I25, I13, and I35. Note: The point of intersection of the line through I15I45 and the line through the contact point and perpendicular to the sliding direction is the instant center I14.
The procedure to locate the 4 secondary instant centers is: (i) The point of intersection of the line through I23I34 and the line through I12I14 is the instant center I24. (ii) The point of intersection of the line through I15I12 and the line through I45I24 is the instant center I25. (iii) The point of intersection of the line through I14I34 and the line through I12I23 is the instant center I13. (iv) The point of intersection of the line through I45I34 and the line through I25I23 is the instant center I35. (ii) 10 Points. The locations of the 10 instant centers for this mechanism are shown on Figure 3.
Figure 3. The locations of the instant centers.
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(iii) 6 Points. The velocity of point A fixed in link 3 is equal to the velocity of point A fixed in link 2, that is
A A 23 2V V V= = (1a)
which can be written as 13 23 3 2( )I I Rω = (1b)
Rearranging this equation, the first-order kinematic coefficient of link 3 can be written as
33
2 13 23
1R I Iωθ ′ = = (1c)
The distance 13 23I I is measured as 13 23 33.3 mm.I I = Therefore, the first-order kinematic coefficient of link 3 is
31 0.030 rad/mm
33.3 mmθ ′ = = (2a)
Note that the correct sign of this kinematic coefficient is positive because link 3 is rotating counterclockwise about the instant center I13 as the length of the input vector is increasing, that is
3 0.030 rad/mmθ ′ = + (2b)
The velocity of link 2 can be written as
14 24 4 2( )I I Rω = (3a) Rearranging this equation gives the first-order kinematic coefficient of link 4, that is
414 24
1I I
θ ′ = (3b)
The distance 14 24I I is measured as 14 24 10.7 mm.I I = Therefore, the first-order kinematic coefficient of link 4 is
41 0.0933 rad / mm
10.7 mmθ ′ = = (4a)
Note that the correct sign is negative because link 4 is rotating clockwise about the instant center I14 while the length of the input vector is increasing, that is
4 0.0933 rad/mmθ′ = − (4b)
The velocity of link 2 can be written as
15 25 5 2( )I I Rω = (5a) Therefore, the first-order kinematic coefficient of link 5 can be written as
515 25
1I I
θ ′ = (5b)
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The distance 15 25I I is measured as 15 25 16.7 mm.I I = Therefore, the first-order kinematic coefficient is
51 0.060 rad / mm
16.7 mmθ ′ = = (6a)
Note that the correct sign is negative because link 5 is rotating clockwise about the instant center I15 while the length of the input vector is increasing, that is
5 0.060 rad/mmθ′ = − (6b) (iv) 6 Points. The angular velocity of link 3 can be written as
3 3 2Rω θ′= (7)
The input vector 2R is defined here as downward, therefore, the vector is becoming longer for the given
input velocity 2ˆV 90 j mm / s.= − Therefore, the time rate of change of the input vector is
2 90 mm/sR = + (8b) Substituting Equations (2b) and (8b) into Equation (7) gives
3 3 2 ( 0.030)( 90) 2.7 rad secRω θ′= = + + = + (9) The positive sign implies that the direction of the angular velocity of link 3 is counterclockwise.
The angular velocity of link 4 can be written as
4 4 2Rω θ′= (10a) Substituting Equations (4b) and (8b) into Equation (10a) gives
4 ( 0.0933)( 90) 8.40 rad secω = − + = − (10b) The negative sign implies that the direction of the angular velocity of link 4 is clockwise.
The angular velocity of link 5 can be written as
5 5 2Rω θ′= (11a) Substituting Equations (6b) and (8b) into Equation (11a) gives
5 ( 0.06)( 90) 5.4 rad secω = − + = − (11b) The negative sign implies that the direction of the angular velocity of link 5 is clockwise. (v) 2 Points. The velocity of point D (where D is fixed in link 4) can be written as
14 4 14 4 2( ) ( )DV I D I D Rω θ ′= = (12a)
The distance 14I D is measured as 14 95 mm.I D = Substituting this measurement and Equation (10b) into Equation (12a), the velocity of point D is
(95 mm)( 8.40 rad/s) 798 mm/sDV = − = − (12b) The negative sign indicates that the velocity of point D is directed to the left.
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Solution to Problem 4. (i) 8 Points. A set of vectors for a kinematic analysis of the mechanism are shown in Fig. 4(a).
Figure 4(a). Suitable vectors for the mechanism. The vector loop equation (VLE) can be written as
√ I √ ? √ √ ? √ √ √ 2 3 7 9 1R R R R R 0+ + + + = (1)
where the vector 7R is the arm and the vector 9R follows the path of pin B. (ii) 10 Points. The X and Y components of the VLE, see Equation (1), are
2 2 3 3 7 7 9 9 1 1R cos R cos R cos R cos R cos 0θ + θ + θ + θ + θ = (2a) and
2 2 3 3 7 7 9 9 1 1R sin R sin R sin R sin R sin 0θ + θ + θ + θ + θ = (2b) The angles for the given position of the mechanism are
2 120 ,θ = ° 3 210 ,θ = ° 9 270 ,θ = ° and 7 180θ = ° (3)
Differentiating Equations (2) with respect to the input position 2θ gives
2 2 3 3 3 9 9R sin R sin R cos 0′ ′− θ − θ θ + θ = (4a) and
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2 2 3 3 3 9 9R cos R cos R sin 0′ ′+ θ + θ θ + θ = (4b) Writing Equations (4) in matrix form gives
3 3 9 3 2 2
3 3 9 9 2 2
R sin cos R sinR cos sin R R cos
′− θ + θ θ + θ⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥′+ θ + θ − θ⎣ ⎦⎣ ⎦ ⎣ ⎦
(5)
Substituting the known data into Equation (5) gives
3
9
400sin(210 ) mm cos(270 ) rad 500sin(120 ) mmR400cos(210 ) mm sin(270 ) rad 500cos(120 ) mm′θ− ° ° + °⎡ ⎤⎡ ⎤ ⎡ ⎤
=⎢ ⎥⎢ ⎥ ⎢ ⎥′+ ° ° − °⎣ ⎦ ⎣ ⎦⎣ ⎦ (6a)
which simplifies to
3
9
200 mm 0 rad 433.01 mmR346.41 mm 1 rad 250 mm′θ+ +⎡ ⎤⎡ ⎤ ⎡ ⎤
=⎢ ⎥⎢ ⎥ ⎢ ⎥′− − +⎣ ⎦ ⎣ ⎦⎣ ⎦ (6b)
The determinant of the coefficient matrix in Equation (6b) is
DET ( 200 mm)( 1 rad) 200 mm= + − = − (7a) Using Cramer’s rule, the first-order kinematic coefficient for the angle of link 3 is
3433.01 mm 2.165 rad / rad
200 mm−′θ = = +−
(8)
The positive sign indicates that link 3 is rotating counterclockwise for a counterclockwise rotation of the input link 2.
From Cramer’s rule, the first-order kinematic coefficient for point B is
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200 250 433.01 346.41 mmR 999.995 mm / rad200 mm
+ × + ×′ = = −−
(9a)
Therefore, the first-order kinematic coefficient for point B is
9R 1.0 m / rad′ = − (9b)
The negative sign indicates that the length of the vector 9R is becoming shorter for a counterclockwise rotation of the input link 2, that is point B is moving downward.
The rolling contact constraint between link 4 and the ground link 1 can be written as
9 4 4 7( )R ρ θ θ± Δ = Δ −Δ (10) The correct sign is positive because as the wheel 4 rotates counterclockwise the magnitude of the vector
9R becomes longer. Also, the change in the angular position of the vector 7R is zero, that is, 7 0,θΔ = therefore, Equation (10) can be written as
9 4 4R ρ θ+ Δ = Δ (11)
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Differentiating Equation (11) with respect to the input position 2θ gives
9 4 4R ρ θ′ ′+ = (12a) Then rearranging this equation, the first-order kinematic coefficient for link 4 can be written as
94
4
Rθρ′
′ = + (12b)
Substituting Equation (9b) and the radius of link 4 into Equation (12b), the first-order kinematic coefficient for link 4 is
4( 999.995 mm) 7.143 rad/rad
140 mmθ −′ = + = − (13)
The negative sign indicates that link 4 is rotating clockwise as the input link 2 rotates counterclockwise. (iii) 4 Points. The angular velocity of link 3 can be written as
3 3 2′ω = θ ω (14a) Substituting Equation (8) and the given angular velocity of the input link 2 into Equation (14a), the angular velocity of link 3 is
3 ( 2.165 rad / rad) ( 50 rad / s) 108.25 rad / sω = + = + (14b) The positive sign indicates that link 3 is rotating counterclockwise as the input link 2 rotates counterclockwise.
The angular velocity of link 4 can be written as
4 4 2′ω = θ ω (15a) Substituting Equation (13b) and the given angular velocity of the input link 2 into Equation (15a), the angular velocity of link 3 is
4 ( 7.143 rad / rad) (50 rad / s) 357.14 rad / sω = − = − (15b) The negative sign indicates that link 4 is rotating clockwise as the input link 2 rotates counterclockwise. (iv) 3 Points. The velocity of point B fixed in link 4 can be written as
B 9 2V R′= ω (16a) Substituting Equation (9b) and the given angular velocity of the input link 2 into Equation (16a), the velocity of point B is
BV ( 1.0 m / rad)(50 rad / s) 50 m / s= − = − (16b)
The negative sign indicates that the velocity of point B is directed downward from point B as the input link 2 rotates counterclockwise, that is
BV 50 j m / s= − (17)
