358 ME Homework Solution 4
Problem 1 A 2.5-diametral pitch, 17-tooth pinion is to drive a 50-tooth gear. The spur gears are cut on the 20o-full-depth-involute system. a) Calculate the following quantities: velocity ratio 2/3, addendum a, dedendum b, clearance c, hole depth ht, working depth hk, pitch diameters d2 and d3, center distance C, circular pitch p, arc tooth thickness at the pitch circle tp (assuming zero backlash), base circle radii rb2 and rb3, base pitch pb, length of path of approach ua, length of path of recess ur, length of path of contact u, and contact ratio mc. b) Check for undercutting. c) Using MATLAB, calculate and plot the backlash that is introduced by increasing the center distance C by an amount 0 C 0.05C . Solution a)
2 N 3 50 = = = 2.9412 3 N 2 17 a= b= 1.000 1.000 = = 0.4in P 2.5 1 in 1.250 1.250 = = 0.5in P 2.5 1 in
c = b a = 0.5in 0.4in = 0.1in h t = a + b = 0.4in + 0.5in = 0.9in h k = 2a = 2 0.4in = 0.8in
d2 =
N2 17 = = 6.8in P 2.5 1 in
d3 =
N3 50 = = 20in P 2.5 1 in
C = r2 + r3 =
6.8in 20in + = 13.4in 2 2 p= = = 1.2566in P 2.5 1 in tp = p 0.7854in = = 0.6283in 2 2 = d 2 cos = 6.8in cos 20o = 6.3899in
d b2
d b 3 = d 3 cos = 20in cos 20o = 18.7939in
p b = p cos = 1.2566in cos 20o = 1.1809in
20in 20in 18.7939in sin 20o u a = (r3 + a ) rb 3 r3 sin = + 0.4in 2 2 2 6.8in u a = 1.0360in < 1.1629in = sin 20o = r2 sin 2
2 2
2
2
6.8in 6.8in 6.3899in u r = (r2 + a ) rb 2 r2 sin = sin 20o + 0.4in 2 2 2 20in u r = 0.8944in < 3.4202in = sin 20o = r3 sin 2 u = u a + u r = 1.0360in + 0.8944in = 1.9304in
2 2
2
2
mc = b)
u 1.9304in = = 1.6347 p b 1.1809in
6.8in 6.3899in 2 2 2 o = rb 2 + C 2 sin 2 r2 a 2 = 0.4in < 2.1868in = + (13.4in ) sin 20 2 2 20in 18.7939in 2 2 2 o a 3 = 0.4in < 0.4550in = = rb 3 + C 2 sin 2 r3 + (13.4in ) sin 20 2 2 both satisfied inequalities -> no undercutting
2
2
Problem 2
For a gear train formed by seven spur gears, the numbers of teeth are as shown in the figure. Compute the speed ratio e of the gear train. Determine the speed 8 and direction of rotation for gear 8 if gear 2 is driven at 1200 rpm (CCW).
Solution
e=
8 = e2 = 0.05682 1200rpm = 68.18rpm (CCW )
Problem 3
N 2 N 4 N 5 N 7 N 2 N 4 N 7 18 15 16 5 = = = = 0.05682 N 3 N 5 N 6 N 8 N 3 N 6 N 8 44 36 48 88
In the planetary gear train shown in the figure, the internal gear 7 turns at 60 rpm (CCW). Using the tabular method, determine the speed and direction of rotation of arm 3.
Solution
Step Gear 2 Arm 3 1 1 1 2 3 1 0 0 1
Gear 4 1 20 1 ( 1) = 2 40 3 2
Gears 5 / 6 Gear 7 1 1 40 1 10 36 10 20 = = 9 154 9 77 18 2 1 57 9 77
3 1 77 = = = 1.351 7 57 57 77 3 = 1.351(60rpm ) = 81.05rpm (CCW )
Problem 4
In the planetary gear train shown in the figure, shaft B is stationary and shaft C is driven at 380 rpm (CCW). Using the tabular method, determine the speed and direction of rotation of shaft A.
Solution
Step 1 2 3 Gear 2 Gear 3 / Arm 4 Gear 5 24 4 1 1 = 18 3 0 4 3 0 1 1 0 Gears 6 / 7 1 18 18 ( 1) = 42 42 60 42 Gear 8 1 3 20 18 = 14 40 42 11 14
4 2 56 = 3 = = 1.697 11 8 33 14 2 = 1.697(380rpm ) = 644.8rpm (CW )
Problem 5
In the Lvai typa-L planetary gear train shown in the figure, the teeth numbers are as follows: N2=16, N4=19, N5=17, N6=24 and N7=95. Internal gear 7 is fixed and gear 2 is driven at 100 rpm (CW). Using the tabular method, determine the speed and direction of rotation of arm 3.
Solution
Step 1
Gear 2 Arm 3 Gear 4 1 1 1 1615 19 85 17 95 85 = 2 0 = 384 16 24 19 24 24 1231 109 3 1 24 384 3 384 1 = = = 0.3119 1231 2 1231 384 3 = 4 = 0.3119( 100rpm ) = 31.19rpm (CCW )
Gears 5 / 6 Gear 7 1 1 95 95 ( 1) = 1 24 24 71 0 24