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Chapter_13_Solutions

# Chapter_13_Solutions - Chapter 13 d P 17 8 2.125 in N 1120...

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Chapter 13 13-1 17 / 8 2.125 in P d   2 3 1120 2.125 4.375 in 544 GP N dd N   8 4.375 35 teeth . GG NP d A n s n s n s s 2.125 4.375 / 2 3.25 in . CA  ______________________________________________________________________________ 13-2 1600 15 / 60 400 rev/min . G nA 3 mm . pm A n 3 15 60 2 112.5 mm .   n s n s ______________________________________________________________________________ 13-3 16 4 64 teeth . G NA   64 6 384 mm . dN m A n s   16 6 96 mm . PP m A n s n s s n s s 384 96 / 2 240 mm . ______________________________________________________________________________ 13-4 Mesh: 1/ 1/ 3 0.3333 in . aP A n 1.25 / 1.25 / 3 0.4167 in . bP A 0.0834 in . cba A n s  / / 3 1.047 in . pP A n / 2 1.047 / 2 0.523 in . tp A n s Pinion Base-Circle: 11 /2 1 / 3 7 i P n 1 7cos20 6.578 in . b dA n s Gear Base-Circle: 22 / 28 / 3 9.333 in P 2 9.333cos20 8.770 in . b n s Base pitch:   cos / 3 cos20 0.984 in . bc p pA  n s Contact Ratio: / 1.53/ 0.984 1.55 . ca b b mLp A n s See the following figure for a drawing of the gears and the arc lengths. Chapter 13, Page 1/35

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______________________________________________________________________________ 13-5 (a) 1/2 22 0 14 / 6 32 / 6 2.910 in . A A n s        (b)  1 tan 14 / 32 23.63 . Ans  1 tan 32 /14 66.37 . Ans  (c) Ans. 14 / 6 2.333 in P d 32 / 6 5.333 in . G dA n s Chapter 13, Page 2/35
(d) From Table 13-3, 0.3 A 0 = 0.3(2.910) = 0.873 in and 10/ P = 10/6 = 1.67 0.873 < 1.67 0.873 in . FA n s ______________________________________________________________________________ 13-6 (a) / / 4 0.7854 in nn pP  / cos 0.7854 / cos30 0.9069 in tn pp / tan 0.9069 / tan30 1.571 in xt (b) Eq. (13-7): cos 0.7854cos25 0.7380 in . nb n n p pA n s (c) cos 4cos30 3.464 teeth/in  11 tan tan / cos tan (tan 25 / cos30 ) 28.3 . Ans    (d) Table 13-4: 1/ 4 0.250 in . aA n s n s 1.25 / 4 0.3125 in . bA 20 5.774 in . P dA n s 36 10.39 in . G n s ______________________________________________________________________________ 13-7 19 teeth, 57 teeth, 20 , 2.5 mm PGn n NN m (a) 2.5 7.854 mm . p mA  n s 7.854 9.069 mm . cos cos30 n t p p Ans 9.069 15.71 mm . tan tan30 t x p p Ans (b) 2.5 2.887 mm . cos cos30 n t m n s Chapter 13, Page 3/35

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1 tan 20 tan 22.80 . cos30 t Ans     (c) 2.5mm . n am A n s  1.25 1.25 2.5 3.125 mm . n bm A n s 19 2.887 =54.85 mm . Pt t N dN m P A n s   57 2.887 164.6 mm . G dA n s ______________________________________________________________________________ 13-8 (a) Using Eq. (13-11) with k = 1, = 20º, and m = 2,       22 2 2 2 2 2 12 s i n i n 21 2 2 1 2 2 sin 20 14.16 teeth 122s i n 2 0 P k Nm m m m    Round up for the minimum integer number of teeth. N P = 15 teeth Ans. (b) Repeating (a) with m = 3, N P = 14.98 teeth. Rounding up, N P = 15 teeth. Ans. (c) Repeating (a) with m = 4, N P = 15.44 teeth. Rounding up, N P = 16 teeth. Ans. (d) Repeating (a) with m = 5, N P = 15.74 teeth. Rounding up, N P = 16 teeth. Ans. Alternatively, a useful table can be generated to determine the largest gear that can mesh with a specified pinion, and thus also the maximum gear ratio with a specified pinion. The Max N G column was generated using Eq. (13-12) with k = 1, = 20º, and rounding up to the next integer. Min N P Max N G Max m = Max N G / Min N P 13 16 1.23 14 26 1.86 15 45 3.00 16 101 6.31 17 1309 77.00 18 unlimited unlimited With this table, we can readily see that gear ratios up to 3 can be obtained with a minimum N P of 15 teeth, and gear ratios up to 6.31 can be obtained with a minimum N P of 16 teeth. This is consistent with the results previously obtained.
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Chapter_13_Solutions - Chapter 13 d P 17 8 2.125 in N 1120...

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