test4bsol.fall10

# test4bsol.fall10 - ME 560 Kinematics. Class Test 4. Friday,...

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ME 560 Kinematics. Class Test 4. Fall Semester 2010 Friday, November 19th Name of Student : ________________________________ OPEN BOOK AND OPEN NOTES Problem 1 (30 points). A spur gear with a diametral pitch of 2.5 tooth/in and 48 teeth is used to drive a spur pinion with 22 teeth. Both the gear and the pinion are cut on the 23° full-depth involute system. Determine the following: (i) the base pitch of the pinion gear; (ii) the arc of approach and the arc of recess; (iii) the angles of approach for the gear and the pinion; (iv) the angles of recess for the gear and the pinion; and (v) the contact ratio. Problem 2 (20 points). Consider the planetary, or epicyclic, gear train shown in the figure below. The input shaft is connected to the sun gear 2 which is rotating counterclockwise with a constant angular velocity 2 ω 65 rad/sec. = The output shaft is connected to the arm, link 3, and the ring gear 6 is locked. Part (i). Determine the angular velocity of the arm. Specify the magnitude and the direction. Part (ii). Determine the angular velocity of the idler gear 4. Specify the magnitude and the direction. Part (iii). Determine the angular velocity of the planet gear 5. Specify the magnitude and the direction. 3 6 2 4 24T 20T 32T 128T Figure. A Planetary Gear Train.

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2 Solution to Problem 1. (30 points). (i) 14 Points . The diametral pitch is specified as P = 2.5 teeth/inch (1) From Table 6.2, see page 258, the addendum for the gear and the pinion are 11 0.40 in 2.5 == = a P (2) Also, from Table 6.2, see page 258, the dedendum for the gear and the pinion are 1.25 1.25 0.50 in 2.5 === d P (3) From Eq. (6.4), see page 254, the circular pitch is 1.2566 in/tooth 2.5 = p P π (4) The base pitch, see Eq. (6.6), page 261, is b =c o s φ pp (5a) which can be written as b 1.2566 cos 23 1.1567 in/tooth o p (5b) (ii) Note that the gear is the driver and will be denoted as gear 2. The pinion is the driven gear and will be denoted as gear 3. This choice is in agreement with Figure 6.20, page 269. Therefore, the radius of the pitch circle of gear 2, see Eq. (6.1), page 252, is 2 2 48 9.6 in 22 x 2 . 5 = N R P (6a) The radius of the pitch circle of pinion 3, see Eq. (6.1), page 252, is 3 3 22 4.4 in x 2 . 5 = N R P (6b) The radius of the base circle of gear 2, see Eq. (6.15), page 272, is cos 9.6 cos23 8.8368 in = = o rR (7a) The radius of the base circle of pinion 3, see Eq. (6.15), page 272, is 33 cos 4.4 cos23 4.0502 in = = o (7b)
3 (ii) 6 Points . From Figure 6.20, page 269, and Eq. (6.10), see page 269, the arc of approach (from the driven gear; i.e., the pinion 3) is 22 33 3 CP = (R + a) (R cos R sin φ) − φ (8a) which can be written as CP (4.4 0.40) (4.4 cos 23 4.4 sin 23 0.8568 in =+− ) = oo (8b) Also, from Figure 6.19, page 268, and Eq. (6.11), see page 269, the arc of recess (from the driver gear; i.e., the gear 2) is 2 PD = (R + a) (R cos R sin φ (9a) which can be written as PD (9.6 0.40) (9.6 cos 23 9.6 sin 23 0.9298 in ) = (9b) Note that the arc of recess is greater than the arc of approach; i.e., PD CP. > Adding Eqs. (8b) and (9b) gives CD CP PD 0.8568 0.9298 1.7866 in =+= + = (10) Check: Equations (6.10) and (6.11) are only valid for the conditions where 2 CP R sin ≤φ and 3 PD R sin φ

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## test4bsol.fall10 - ME 560 Kinematics. Class Test 4. Friday,...

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