MEE 341 Problems Final Exam, Spring 2018

Chapter 9: solve problems: 9-1 -> 9-22 9-25 -> 9-28 9-34 9-36 9-38 9-39 9-42 9-43 9-53 9-54 9-55 9-57 9-58 9-63 9-64 9-65 (Figure 9-35 mentioned in these 3 exercises is the Levai catalogued 12 possible types of basic planetary gearsets available in lecture 06 slides on page 05) DESIGN OF MACHINERY SOLUTION MANUAL 9-1-1 PROBLEM 9-1 Statement: A 22-tooth gear has AGMA standard full-depth involute teeth with diametral pitch of 4. Calculate the pitch diameter, circular pitch, addendum, dedendum, tooth thickness, and clearance. Given: Tooth number N  22 Diametral pitch pd 4 in  1   Solution: See Table 9-1 and Mathcad file P0901. 1. Calculate the pitch diameter using equation 9.4c and the circular pitch using equation 9.4d. Pitch diameter d N pd  d 5.5000in  Circular pitch pc  pd  pc 0.7854 in  2. Use the equations in Table 9-1 to calculate the addendum, dedendum, tooth thickness and clearance. Addendum a 1.0000 pd  a 0.2500in  Dedendum b 1.2500 pd  b 0.3125in  Tooth thickness t 0.5 pc   t 0.3927 in  Clearance c 0.2500 p d  c 0.0625in  Note: The circular tooth thickness is exactly half of the circular pitch, so the equation used above is more accurate than the one in Table 9-1. Also, all gear dimensions should be displayed to four decimal places since manufacturing tolerances for gear teeth profiles are usually expressed in ten-thousandths of an inch.. DESIGN OF MACHINERY SOLUTION MANUAL 9-5-1 PROBLEM 9-5 Statement: A spur gearset must have pitch diameters of 4.5 and 12 in. What is the largest standard tooth size, in terms of diametral pitch, that can be used without having any interference or undercutting and what are the number of teeth on each gear that result from using this diametral pitch? Assume that both gears are cut with a hob. a. For a 20-deg pressure angle b. For a 25-deg pressure angle Given: Pitch diameters: d1 4.5 in   d2 12 in   Solution: See Table 9-4 and Mathcad file P0905. 1. To avoid undercutting, use the minimum tooth numbers given in Table 9-4b. a. Pressure angle of 20 deg. Nmin  21 pdmin Nmin d 1  pdmin 4.667 in  1  From Table 9-2, the smallest standard diametral pitch (largest tooth size) that can be used is 5. But, since the pinion pitch diameter is not an integer, using 5 would result in a noninteger number of teeth. Therefore, we must go to the next larger (even) pitch (smaller tooth size) of pd 6 in  1   . The resulting tooth numbers are: N1 pd d1   N1  27 N2 pd d2   N2 72  b. Pressure angle of 25 deg Nmin  14 pdmin Nmin d 1  pdmin 3.111 in  1  From Table 9-2, the smallest standard diametral pitch (largest tooth size) that can be used is 4. pd 4 in  1   . The resulting tooth numbers are: N1 pd d1   N1  18 N2 pd d2   N2  48 DESIGN OF MACHINERY SOLUTION MANUAL 9-6-1 PROBLEM 9-6 Statement: Design a simple, spur gear train for a ratio of -9:1 and a diametral pitch of 8. Specify pitch diameters and numbers of teeth. Calculate the contact ratio. Given: Gear ratio mG  9 Diametral pitch pd 8 in  1   Assumptions: The pinion is not cut by a hob and can, therefore, have fewer than 21 teeth (see Table 9-4b) for a 20-deg pressure angle. Design Choice:Pressure angle  20 deg   Solution: See Mathcad file P0906. 1. From inspection of Table 9-5a, we see that 17 teeth is the least number that the pinion can have for a gear ratio of 9. therefore, let the number of teeth on the pinion be Np  17 and Ng mG Np   Ng 153  2. Using equation 9.4c, calculate the pitch diameters of the pinion and gear. dp Np pd  dp  2.1250 in dg Ng pd  dg 19.1250 in  3. Calculate the contact ratio using equations 9.2 and 9.6b and those from Table 9-1. rp 0.5 dp   rp  1.0625in rg 0.5 dg   rg 9.5625in  ap 1 pd  ap  0.1250 in ag 1 pd  ag 0.1250 in  Center distance C rp  rg  C 10.6250 in  Z rp  ap  2 rp cos      2  rg  ag  2 rg cos      2   C sin      Z 0.6287in  Contact ratio mp p d Z   cos     mp  1.704 DESIGN OF MACHINERY SOLUTION MANUAL 9-10-1 PROBLEM 9-10 Statement: Design a compound, spur gear train for a ratio of -70:1 and diametral pitch of 10. Specify pitch diameters and numbers of teeth. Sketch the train to scale. Given: Gear ratio: mG  70 Diametral pitch: p d 10 in  1   Solution: See Mathcad file P0910. 1. Since the ratio is negative, we want to have an odd number of stages or an even number with an idler. Let the number of stages be Possible number of stages j 2 3   4  Ideal, theoretical stage ratios r j( ) mG 1 j  then j 2 3 4  r j( ) 8.367 4.121 2.893  2. Two stages would result in a stage ratio less than 10 but will require an idler, so we will use three stages. The average ratio for three stages is about 21:5. Using a pressure angle of 20 deg, let the stage ratios be Stage 1 ratio r1 21 5  Stage 2 ratio r2 20 5  Stage 3 ratio r3 25 6  and let the driver gears have tooth numbers of Tooth number index i 2 3   7  N2  20 N4  20 N6 18  then the driven gears will have tooth numbers of N3 r1 N2   N5 r2 N4   N7 r3 N6   N3  84 N5  80 N7 75  The pitch diameters are: di Ni pd  Tooth numbers: di in 2.0000 8.4000 2.0000 8.0000 1.8000 7.5000  Ni 20 84 20 80 18 75  i 2 3 4 5 6 7  i 2 3 4 5 6 7  Checking the overall gear ratio: N3 N5  N7  N2 N4  N6   70.000 DESIGN OF MACHINERY SOLUTION MANUAL 9-19-1 PROBLEM 9-19 Statement: Design a compound, reverted, spur gear transmission that will give two shiftable ratios of +3:1 forward and -4.5:1 reverse with diametral pitch of 6. Specify pitch diameters and numbers of teeth. Sketch the train to scale. Given: Gear ratio mG  3 Diametral pitch p d 6 in   1  Solution: See Mathcad file P0919. 1. Since the forward ratio is positive, we want to have an even number of stages. Let the number of stages be 2. 2. Using a pressure angle of 25 deg, let the stage ratios be Stage 1 ratio r1  2 Stage 2 ratio r2 1.5  3. Following the procedure of Example 9-3, Tooth number index i 2 3   5  N2  N3 N4  N5 = = K and, r1 N3 N2  r2 N5 N4  Solving independently for N2 and N4, N2 K r1 1   N4 K r2 1   where Kmin r1  1   r2  1     Kmin 7.500  By iteration, find a multiple of Kmin that will result in a minimum, integer number of teeth on N2 and N4. K 6 Kmin   N2 K r1 1   N4 K r2 1   K  45.000 N2  15 N4 18  These are acceptable tooth numbers for gears with a 25-deg pressure angle that are cut by a hob. 4. The driven gears for the forward train will have tooth numbers of N3 r1 N2   N3  30 N5 r2 N4   N5 27  The pitch diameters are: di Ni pd  di in 2.5000 5.0000 3.0000 4.5000  Ni 15 30 18 27  i 2 3 4 5  Checking the overall gear ratio: N3 N2     N5 N4      3.000  5. The reverse train will also have two stages and use the first forward stage and an idler gear to get the change in direction. Let the stage ratios be Stage 1 ratio r1  2 Stage 2 ratio r2 2.25  6. Let the number of teeth on the reverse stage driver gear be N6  12 then the number of teeth on the driven gear will be Driven reverse gear N7 r2 N6   N7  27 DESIGN OF MACHINERY SOLUTION MANUAL 9-26a-1 PROBLEM 9-26a Statement: Figure P9-2 shows a compound planetary gear train (not to scale). Table P9-2 gives data for gear numbers of teeth and input velocities. For the data in row a, find 2. Given: Tooth numbers: N2  50 N3  25 N4  45 N5  30 N6 40  Input velocities:   20 arm 50  Solution: See Figure P9-2 and Mathcad file P0926a. 1. The formula method will be used in this solution. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be 3 and last be 6. Then, using equation 9.13c, write the relationship among the first, last, and arm. Larm Farm  arm   arm  = R = 2. Calculate R using equation 9.14 and inspection of Figure P9-2. R N3 N5  N4 N6   R 0.4167  R is positive in this case because gears 3 and 6 rotate in the same direction. 3. Solve the right-hand equation in step 1 for 3.   arm  R  arm   118.000  4. Solve for 2 using equation 9.7.  N3 N2     59.000  2 will be in the opposite direction as 6. DESIGN OF MACHINERY SOLUTION MANUAL 9-38-1 PROBLEM 9-38 Statement: Figure P9-6b shows a differential. Gear A is driven at 10 rpm CCW and gear B is driven CW at 24 rpm. The tooth numbers are indicated in the figure. Find the speed of gear D. Units: rpm 2   rad min  1   Given: Tooth numbers: NA'  18 NB'  18 NC  18 ND 30  Input speeds: A 10 rpm   B 24 rpm  Solution: See Figure P9-6b and Mathcad file P0938. 1. Determine the speed of the sun gear using the formula method for analyzing an epicyclic train. To start, choose a first and last gear that mesh with gears that have planetary motion. Let the first gear be A' and last be B'. Then, using equation 9.13c, write the relationship among the first, last, and arm. Larm Farm B' arm  A' arm  = R = Calculate R using equation 9.14 and inspection of Figure P9-6b. R NA' NB'      R 1.00000  Solve the right-hand equation above for D with A = 0. A'  A B' B  arm R A'  B'  R 1   arm 7.000 rpm  Gear D is attached to the arm shaft so, D  arm D 7.00 rpm  DESIGN OF MACHINERY SOLUTION MANUAL 9-57-1 PROBLEM 9-57 Statement: A pinion with a 3.000-in pitch diameter is to mesh with a rack. What is the largest standard tooth size, in terms of diametral pitch, that can be used without having any interference? a. For a 20-deg pressure angle b. For a 25-deg pressure angle Given: Pitch diameter: d 3.000 in   Solution: See Table 9-4 and Mathcad file P0957. 1. Assume thatthe pinion is generated by means other than being cut by a hob. a. Pressure angle of 20 deg. Nmin  18 pdmin Nmin d  pdmin 6.000 in  1  From Table 9-2, the smallest standard diametral pitch (largest tooth size) that can be used is 6. pd 6 in  1   . The resulting number of teeth on the pinion is: N pd d   N 18  b. Pressure angle of 25 deg Nmin  12 pdmin Nmin d  pdmin 4.000 in  1  From Table 9-2, the smallest standard diametral pitch (largest tooth size) that can be used is 4. pd 4 in  1   . The resulting number of teeth on the pinion is: N pd d   N  12 DESIGN OF MACHINERY SOLUTION MANUAL 9-64-1 PROBLEM 9-64 Statement: Figure P9-35h shows (schematically) a compound epicyclic train. The tooth numbers are 80, 20, 25, and 85 for gears 2, 3, 4, and 5, respectively. Gear 2 is driven at 200 rpm CCW and gear 5 is fixed to ground. Determine the speed and direction of the arm. What is the efficiency of this train if the basic gearsets have E0 = 0.98? Units: rpm 2   rad min  1   Given: Tooth numbers: N2  80 N3  20 N4  25 N5 85  Input speeds: 2 200 rpm   5 0 rpm   Basic gearset efficiency: E0 0.98  Solution: See Figure P9-35h and Mathcad file P0964. 1. Determine the speed of the arm using the formula method for analyzing an epicyclic train. To start, choose first and last gears that mesh with gears that have planetary motion. Let the first gear be 2 and last be 5. Then, using equation 9.13c, write the relationship among the first, last, and arm. Larm Farm 5 arm  2 arm  = R = Calculate R using equation 9.14 and inspection of Figure P9-7b. R N2 N3     N4 N5       R 1.17647  Solve the right-hand equation above for arm with  = 0. arm R R 1  2  arm 1333rpm  2. Find the basic ratio  for the train using equation 9.15.   R  1.176  3. The combination of  > 1, shaft 1 fixed, and input to gear 2 corresponds to Case 3 in Table 9-12 giving an efficiency of   E0  1  E0   1       0.884