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SOLUTION (15.1) Known: À pinion with 32 teeth and 8 diametral pitch meshes with a gear having 65 teeth. Find: Calculate the standard center distance. Schematic and Given Data: N, = 65 Assumptions: 1. The gears are spur gears. 2. The gears have teeth of standard involute profile. 3. The gears mesh along their pitch circles. Analysis: 1. From Eg. (15.3): P = N/d = Ny/d, = No/d, 2. P=8,Np= 32. Hence, d = 40in. 3. Nç=65. Hence, d = 8.12Sin. di+d, 4. The center distance, c = + = 6.0625 in. O] Comments: 1. Ifthe gears did not mesh at the theoretical pitch circles the measured (actual) center distance would not be equal to the sum of the theoretical pitch circle radii of the gears. 2. K should be evident that meshing gears must have the same diametral pitch. 151 SOLUTION (15.2) Known: A spur gear has a size 8 diametra! pitch. Find: Calculate the thickness of the spur gear tooth measured along the pitch circle. Schematic and Given Data: Tooth thickness t il Circular pitch Pitch circle Assumption: The gear has teeth of standard involute profile. Analysis: 1. From Eq. (15.5); pP=a P=8. Hence, p = 0.3927 in. 2. Tooth thickness, t= p/2 : t= 0.1963 in. n 15-2 SOLUTION (15.4) Known: A pinion with 20 teeth and 6 diametral pitch meshes with a gear having 55 teeth. Find: Calculate the standard center distance. Schematic and Given Data: Np=20 P=-6 N =55 Assumptions: 1. The gears are spur gears. 2. The gears have teeth of standard involute profile. 3. The gears mesh along their pitch circles. Analysis: 1. From Eq. (15.3): P=N/d=Nydo=Nodo 2. P=6, Np = 20. Hence, dp =333in. 3. Ng= 55. Hence, do=9.17in. +d, d, 4. The center distance, c = 5 P = 6.25 in. Comments: 1. Ifthe gears did not mesh at the theoretical pitch circles the measured (actual) center distance would not be equal to the sum of the theoretical pitch circle radii of the gears. 2. It should be evident that meshing gears must have the same diametral pitch. 15-4 SOLUTION (15.5) Known: A spur gear has à 6 diametral pitch. Find: Calculate the thickness of the spur gear tooth measured along the pitch circle. Schematic and Given Data: Tooth thickness t cd Circular pitch Pitch circle Assumption: The gear has teeth of standard involute profile. Analysis: 1. From Eq. (15.5); pP=a P=6. Hence, p = 0.5236 in. 2. Tooth thickness, t=p/2 : t = 0.262 in. = 15-5 SOLUTION (15.7D) Known: A web site address is given as http://www .grainger.com. Find: Search the web site and select a spur gear with 32 pitch, 14.5" pressure angle. and 20 teeth. List the manufacturer, description, and price of the gear. Analysis: The web site provides the following information: Mfg. Name: Boston Gear Description: Spur Gear-32 Pitch, Steel-14 1/2 Deg. Pressure Angle-20 Teeth Price: $11.28 SOLUTION (15.8) Known: À pinion of known pitch and number of teeth rotates at 2000 rpm and drives a gear at 1000 rpm. Find: Determine the number of teeth on the gear, theoretical center distance and circular pitch. Schematic and Given Data: 2000 rpm Assumptions: 1. The gears are spur gears. 2. The gears have teeth of standard involute profile. 3. The gears mesh along their pitch circles. Analysis: 1. A 2:1 speed ratio requires a 1:2 ratio in number of teeth. Hence, Ng =40 ER 2. dg= E =Sin. dp= 2 =2.5in. Theoretical center distance, c = da + do : c=3.75in. H 157 3. Circular pitch, p=L: p 4 in. = Comments: 1. Ifthe gear teeth were not of involute profile it would still be possible to have a constant speed ratio provided it is ensured that the pitch point is stationary. 2. Jfthe gears did not mesh along the pitch circles the speed ratio would not be constant and also the center distance would not be equal to the sum of the pitch circle radii of the gears. 3. Itshould be evident that meshing gears must have the same diametral pitch. SOLUTION (15.9) Known: A pair of spur gears provide a given speed ratio at a specified center distance and have a diametral pitch of 8. Find: Determine the numbers of teeth and the pitch diameters of the gears. Schematic and Given Data: P=8 dmg isd:l c=7.5in. Assumptien: The spur gears mesh at the pitch circles. Analysis: 1. Forc=75in,dp+dg=15in 2. For4:l speed ratio, dg = 4dp ; Hence Sdp = 15 in, dp=3in.; dg = 12in. E 3. ForP=8,Np=24;Ng =96 mn Comments: 1. The assumption to have spur gears meshing at the pitch circles ensures that the given center distance is the theoretical center distance and is equal to the sum of the pitch radii of the gears. 2. Ifthe diametral pitch were chosen to be higher then the number of teeth would be more on both the pinion and gear (other parameters being kept the same). 15-8 SOLUTION (15.11) Known: A pair of spur gears of known velocity ratio, center distance, diametral pitch, and pressure angle are given. Find: Draw a full-size layout of the spur gears and label the following: (a) pitch circle, (b) base circle, (c) pressure angle, (d) addendum (for both the pinion and the gear), (e) dedendum (for the pinion only) Assumption: The gears mesh along their pitch circles. Schematic and Given Data and Analysis: Pinion axis =" (b) Base circles (a) Pitch circles Interference does existh Gear addendum should: be cut back this much. Gear axis For velocity ratio = 4. d, = 4d, Forc=10in,d,+d, =20im. Therefore, d+ 4d, = 20 in. andd,=4in.d = 16in. 1 Addendum ="; = =02in. Dedendum =LÉ. -0.25in. Note: This drawing is not drawn to scale. 15-10 SOLUTION (15.12) Known: For a pair of standard 20º full-depth spur gears the diametral pitch, velocity ratio, number of teeth on the pinion and its direction of rotation are given. Find: Draw a full-size layout of the spur gears in the region of tooth contact and show the following: (a) pitch circle, addendum circle, dedendum circle, and base circle of the gear, (b) interference, (c) path of contact, (d) angle of recess for the pinion and the gear. Assumption: The gears mesh along their pitch circles. Schematic and Given Data and Analysis: Gear addendum reduction needed circle , () Pitch circle DN Angle of recess Base circle Center of gear Dedendum ' circle —Np 2teeth em dp = P —4teehin. =6im.=dy de = (dpX velocity ratio) = 6in. (2) = 12 in. = de Addendum= 1/P= 1/4 in. Dedendum = 1.157/P=0.289in. Note: This drawing is not drawn to scale to eliminate interference +, Y | | 15-11 SOLUTION (15.14) Known: A gear has a known outside diameter, diametral pitch and pressure angle. Find: Determine the pitch diameter, the circular pitch, the addendum, the dedendum of the gear, and the number of gear teeth. Schematic and Given Data: Outside diameter of gear = 3.000 in. Circular pitch Addendum circle: / Dedendum Pitch Addendum circle dircle Dedendu; Assumptions: 1. The gear is a spur gear. 2. The gear has teeth of standard involute profile. Analysis: 1. Addendum, a = 1/P = 0.05 in. Pitch diameter, dp = [(outside diameter) - 2a] dp = 3.0 - 0)%0.05) = 2.9 in. = 2. Module, m = 1/P = 0.05 in. = 1.27 mm. From Eq (15.5), circular pitch, p = mm. Therefore, p = 3.99 mm. | 3. Dedendum, d = 1.25 a = 1.25(0.05) = 0.0625 in. Hence, d = 0.0625 in. [| From Eq (15.4), the number of gear teeth, N = dp/m Therefore, N = 58. E 15-13 SOLUTION (15.15) Known: A pair of mating gears of known pressure angle, numbers of teeth, and center distance is given. The pinion has stub teeth and the gear has full involute teeth. Find: Calculate the contact ratio and the diametral pitch. Schematic and Given Data: N=18 4=20deg. N, = 36 Assumptions: 1. The gears are spur gears. 2. The gears mesh along their pitch circles. Analysis: 1, The ratio of the number of teeth, Ng/Np = 36/18 = 2/1. Hence, dp/dg = 2/1. Center distance, c = (dp + dg)/2 Therefore, dp = 20/3. dg = 40/3 Diametral pitch, P= N/d = 2.7 [1 For full deep involute teeth, the gear addendum, a = 1/P = 0.37. The addendum circle radius of the gear rag = rg + à = 7.037 in. From the textbook, the addendum for a 20º stub system is 0.8/P. The addendum of the pinion, ap = 0.296 in. The addendum circle radius of the pinion, rap = Ig + ap = 3.63 in. 4. The base circle radius of the gear, rbg = Ipg COS 6 = (20/3) cos 20º = 626 in. The base circle radius of the pinion, rbp = (10/3) cos 20 = 3.33 cos 20º = 3.13 5. From Eq. (15.10), the base pitch, pb = (mdb/N), pp = 1.09 in. 6. From Eq. (15.9), contact ratio, o VRp- Ro + YHãg - The -esin O SE) CR Ph = 43.63? - 3.132 + 7.037? - 6.26? - 10 sin 20º 1.498 1.09 , Hence, CR = 1.5 nm 15-14 3. Tooth thickness, t = p/2 = 1.25 mm. E 4. Base circle diameter of the gear, dbg = dg cos 4 dbg = 2.625 cos 20º = 2.47 in. m Base circle diameter of the pinion, dbp = dp cos é dbp = 0.53125 cos 20º = 0.49921 in. m 5. From Eg. (15.10), base pitch, p, = (mdb)/N). Therefore, p, = 0.092 in. o] 6. Addendum circle radius of the gear, tag = 1.34 in. Addendum circle radius of the pinion, rap = 0.296 in. From Eq. (15.9); contact ratio, CR= vp” Hp + VH Tg -esin Q Po - 0.296- 0.249] + 41,34" 1235] - 1,5781810 20º | 595 0.092 . = 7. Arc of action = (CR)p = (1.525X0.098) = 0.15 in. O] Comment: Although an assumption was made that interference (contact below the base circle) would not take place, a calculation needs to be performed to determine whether there will be interference when standard full-depth teeth are used. Following SAMPLE PROBLEM 15.1, we first determine the base circle radii of pinion and gear. From Eq. 15.11, rop = (0.53125/2) cos 20º = 0.2496 in. and reg = (2.625/2) cos 20º = 12333 in. Substitution in Eg. 15.8 gives rap(max) = 0.59467 for the pinion and rag(max) = 1.34624 for the gear. The limiting outer gear radius is equivalent to an addendum of Tag(max) - Lg = 0.03374 in., whereas a standard full-depth tooth has an addendum of 1/P =.03125 in. So, the use of standard teeth should not cause interference. SOLUTION (15.17) Known: For a pair of mating gears, the diametral pitch, center distance and the number of teeth are given. The center distance is increased by 0.125 in. Find: Determine the contact ratio and the pressure angle. Schematic and Given Data: Np=17 P=32 c=1.703L in. 15-16 Assumptions: . The gears are physically the same spur gears as in Problem 15.16. 2. The base and outside diameters of the gear and pinion remain the same as in Problem (15.16). 3. The gears mesh along their actual (not theoretical) pitch circles. 4. The gears have teeth of standard involute profile. Analysis: 1. Ifthe center distance is increase by 0.125 in., the pinion and gear will no longer mesh -- see the following diagram. 2. Ifthe center distance is increased by 0.0125 in., the gears will mesh along their actual (not theoretical) pitch circles. +O; To 1.5781 4 03125 Loss $ Addendum . Pitch in Addendum |” | circle circle à O, 3. From the analysis of Problem (15.16) and the given data, center distance, c = 1.5781 + 0.0125 = 1.5906 in. 4. NyNg = dp/de = 17/84. Hence, dg = 4.9412dp. c= (dp + de)/2 = 1.5906. Solving for dg and dp, dz = 2.64575 in., dp = 0.53545 in. 5. dbg= 24667 = de cos 4 =2.64575 cos d. Solving for 4, 6 = 21.20º. Therefore, pressure angle, 6 = 21.20º. |] 15-17 Assumption: The spur gears mesh along their pitch circles. Analysis: l. From Eg. (15.4); dp = m Np = 8(30) = 240 mm de =m Ng = 6(60) = 360 mm addendum = a = m = 8 mm dedendum = 1.25m = 10 mm Td T(240 cos(0.35)) From Eq. (15.10); p, = 4 PRN 30 pp = 23.6 mm contact ratio = length of line of action/p, from the drawing, CR = 38/23.6 = 1.610. n Comments: 1. 2. Increasing the contact ratio of the gear pair for the same numbers of teeth can be achieved by increasing the diametral pitch for the gears. If the gears have a 6 mm module, then the contact ratio CR can be calculated as follows. From Eg. (15.4); dp = m Np = 6(30) = 180 mm de =m Ng = 6(60) = 360 mm addendum = a = m = 6 mm dedendum = 1.25m = 7.5 mm Tdpp x(180 cos(0.35)) From Eq. (15.10); py = —5— Eq. ( ); Ph N, 30 Pp = 177 mm contact ratio = length of line of action/p, from the drawing, CR = 38/17.7 = 2.15. SOLUTION (15.19) Known: A pair of standard spur gears of known pressure angle, center distance and velocity ratio is given. Number of teeth on pinion is specified. Find: (a) (Db) tc) (d) Determine diametral, circular and base pitches. Sketch drawings showing geometric details. Determine if interference results from choice of standard tooth proportions. Determine length of path of contact from drawing and compute contact ratio. 15-19 Schematic and Given Data: Base Circle Pitch circle Addendum Dedendum Path of contact =lin. Gear axis Note: This drawing is not drawn to scale. Assumption: The spur gears mesh along their pitch circles. Ana lysis: c=10=rp+rg and rg=4rp; hence, rp = 2 in. and rg = 8 in. P = Np/dp = 20/4 , hence, P=5 p=71/5 = 0.6280 in. Pp = P cos 6 = 0.628 cos 20º = 0.590 in. Addendum = À = 1 in. =0.2in Ps oo Dedendum = 1.25(1/5) = 0.25 in. Tap(max) = 4.0 in., Tag(max) = 8.3 in. Hence, no interference. Path of contact = 1.0 in. 15-20 
