Baixe shigley elementos de máquinas cap 13 e outras Exercícios em PDF para Engenharia Mecânica, somente na Docsity!
Chapter 13
d (^) P = 17 / 8 = 2 .125 in
d (^) G =
N 2
N 3
d (^) P =
(2.125) = 4 .375 in
N (^) G = Pd (^) G = 8(4.375) = 35 teeth Ans. C = (2. 125 + 4 .375)/ 2 = 3 .25 in Ans.
n (^) G = 1600(15/60) = 400 rev/min Ans. p = π m = 3 π mm Ans. C = [3(15 + 60)]/ 2 = 112 .5 mm Ans.
N (^) G = 20(2.80) = 56 teeth Ans. d (^) G = N (^) G m = 56(4) = 224 mm Ans. d (^) P = N (^) P m = 20(4) = 80 mm Ans. C = (224 + 80)/ 2 = 152 mm Ans.
13-4 Mesh : a = 1 / P = 1 / 3 = 0 .3333 in Ans.
b = 1. 25 / P = 1. 25 / 3 = 0 .4167 in Ans. c = b − a = 0 .0834 in Ans. p = π/ P = π/ 3 = 1 .047 in Ans. t = p / 2 = 1. 047 / 2 = 0 .523 in Ans.
Pinion Base-Circle : d 1 = N 1 / P = 21 / 3 = 7 in d 1 b = 7 cos 20° = 6 .578 in Ans.
Gear Base-Circle : d 2 = N 2 / P = 28 / 3 = 9 .333 in d 2 b = 9 .333 cos 20° = 8 .770 in Ans.
Base pitch : pb = pc cos φ = (π/3) cos 20° = 0 .984 in Ans. Contact Ratio : m (^) c = L (^) ab / pb = 1. 53 / 0. 984 = 1. 55 Ans. See the next page for a drawing of the gears and the arc lengths.
334 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
(a) A (^) O =
[(
) 2 ]^1 /^2
= 2 .910 in Ans.
(b) γ = tan−^1 (14/32) = 23 .63° Ans. = tan−^1 (32/14) = 66 .37° Ans. (c) d (^) P = 14 / 6 = 2 .333 in, d (^) G = 32 / 6 = 5 .333 in Ans.
(d) From Table 13-3, 0. 3 A (^) O = 0 .873 in and 10/ P = 10 / 6 = 1. 67
- 873 < 1. 67 ∴ F = 0 .873 in Ans.
13-
(a) pn = π/ 5 = 0 .6283 in pt = pn /cos ψ = 0. 6283 /cos 30° = 0 .7255 in p (^) x = pt /tan ψ = 0. 7255 /tan 30° = 1 .25 in
30
P G
213 "
513 "
A (^) O
10.5
Arc of approach 0.87 in Ans. Arc of recess 0.77 in Ans. Arc of action 1.64 in Ans. L (^) ab 1.53 in
10
O 2
O 1
14 (^) 12.6
P A B
336 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
13-8 From Ex. 13-1, a 16-tooth spur pinion meshes with a 40-tooth gear, m (^) G = 40 / 16 = 2. 5. Equations (13-10) through (13-13) apply. (a) The smallest pinion tooth count that will run with itself is found from Eq. (13-10)
N (^) P ≥ 4 k 6 sin^2 φ
1 + 3 sin^2 φ
6 sin^2 20°
1 + 3 sin^2 20°
≥ 12. 32 → 13 teeth Ans. (b) The smallest pinion that will mesh with a gear ratio of m (^) G = 2 .5, from Eq. (13-11) is
N (^) P ≥
[1 + 2(2.5)] sin^2 20°
- 52 + [1 + 2(2.5)] sin^2 20°
≥ 14. 64 → 15 pinion teeth Ans. (c) The smallest pinion that will mesh with a rack, from Eq. (13-12)
N (^) P ≥ 4 k 2 sin^2 φ
2 sin^2 20° ≥ 17. 097 → 18 teeth Ans. (d) The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-13) is
N (^) G ≤
N^2 P sin^2 φ − 4 k^2 4 k − 2 N (^) P sin^2 φ
≤ 132 sin^2 20° − 4(1) 2 4(1) − 2(13) sin^2 20° ≤ 16. 45 → 16 teeth Ans.
13-9 From Ex. 13-2, a 20° pressure angle, 30° helix angle, pt = 6 teeth/in pinion with 18 full depth teeth, and φ t = 21 .88°. (a) The smallest tooth count that will mesh with a like gear, from Eq. (13-21), is
N (^) P ≥ 4 k cos ψ 6 sin^2 φ t
1 + 3 sin^2 φ t
4(1) cos 30° 6 sin^2 21 .88°
1 + 3 sin^2 21 .88°
≥ 9. 11 → 10 teeth Ans. (b) The smallest pinion-tooth count that will run with a rack, from Eq. (13-23), is
N (^) P ≥ 4 k cos ψ 2 sin^2 φ t
≥ 4(1) cos 30◦ 2 sin^2 21 .88° ≥ 12. 47 → 13 teeth Ans.
Chapter 13 337
(c) The largest gear tooth possible, from Eq. (13-24) is
N (^) G ≤
N^2 P sin^2 φ t − 4 k^2 cos^2 ψ 4 k cos ψ − 2 N (^) P sin^2 φ t
≤ 102 sin^2 21 .88° − 4(1^2 ) cos^2 30° 4(1) cos 30° − 2(10) sin^2 21 .88° ≤ 15. 86 → 15 teeth Ans.
13-10 Pressure Angle: (^) φ t = tan−^1
tan 20° cos 30°
Program Eq. (13-24) on a computer using a spreadsheet or code and increment N (^) P. The first value of N (^) P that can be doubled is N (^) P = 10 teeth, where N (^) G ≤ 26 .01 teeth. So N (^) G = 20 teeth will work. Higher tooth counts will work also, for example 11:22, 12:24, etc. Use 10:20 Ans.
13-11 Refer to Prob. 13-10 solution. The first value of N (^) P that can be multiplied by 6 is N (^) P = 11 teeth where N (^) G ≤ 93 .6 teeth. So N (^) G = 66 teeth. Use 11:66 Ans.
13-12 Begin with the more general relation, Eq. (13-24), for full depth teeth.
N (^) G =
N^2 P sin^2 φ t − 4 cos^2 ψ 4 cos ψ − 2 N (^) P sin^2 φ t Set the denominator to zero 4 cos ψ − 2 N (^) P sin^2 φ t = 0 From which
sin φ t =
2 cos ψ N (^) P
φ t = sin−^1
2 cos ψ N (^) P For N (^) P = 9 teeth and cos ψ = 1
φ t = sin−^1
= 28 .126° Ans.
(a) pn = π m (^) n = 3 π mm Ans. pt = 3 π/cos 25° = 10 .4 mm Ans. p (^) x = 10. 4 /tan 25° = 22 .3 mm Ans.
18 T 32 T
25 , n 20 , m 3 mm
Chapter 13 339
e =
n (^) a =
(1200) = 11 .84 rev/min cw Ans.
(a) n (^) c =
(540) = 162 rev/min cw about x. Ans.
(b) d (^) P = 12 /(8 cos 23°) = 1 .630 in d (^) G = 40 /(8 cos 23°) = 5 .432 in d (^) P + d (^) G 2 = 3 .531 in Ans.
(c) d =
= 8 in at the large end of the teeth. Ans.
13-18 (a) The planet gears act as keys and the wheel speeds are the same as that of the ring gear. Thus n (^) A = n 3 = 1200(17/54) = 377 .8 rev/min Ans.
(b) n (^) F = n 5 = 0, n (^) L = n 6 , e = − 1
− 1 = n 6 − 377. 8 0 − 377. 8
- 8 = n 6 − 377. 8 n 6 = 755 .6 rev/min Ans. Alternatively, the velocity of the center of gear 4 is v 4 c ∝ N 6 n 3. The velocity of the left edge of gear 4 is zero since the left wheel is resting on the ground. Thus, the ve- locity of the right edge of gear 4 is 2v 4 c ∝ 2 N 6 n 3. This velocity, divided by the radius of gear 6 ∝ N 6 , is angular velocity of gear 6–the speed of wheel 6.
∴ (^) n 6 = 2 N^6 n^3 N 6 = 2 n 3 = 2(377.8) = 755 .6 rev/min Ans.
(c) The wheel spins freely on icy surfaces, leaving no traction for the other wheel. The car is stalled. Ans.
13-19 (a) The motive power is divided equally among four wheels instead of two.
(b) Locking the center differential causes 50 percent of the power to be applied to the rear wheels and 50 percent to the front wheels. If one of the rear wheels, rests on a slippery surface such as ice, the other rear wheel has no traction. But the front wheels still provide traction, and so you have two-wheel drive. However, if the rear differential is locked, you have 3-wheel drive because the rear-wheel power is now distributed 50-50.
340 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
13-20 Let gear 2 be first, then n (^) F = n 2 = 0. Let gear 6 be last, then n (^) L = n 6 = −12 rev/min.
e =
, e = n (^) L − n (^) A n (^) F − n (^) A
(0 − n (^) A )
= − 12 − n (^) A
n (^) A =
= − 17 .49 rev/min (negative indicates cw) Ans.
Alternatively, since N ∝ r , let v = N n (crazy units).
v = N 6 n 6 N 6 = 20 + 30 − 16 = 34 teeth v A N 4
v N 4 − N 5 ⇒ v A = N 4 N 6 n 6 N 4 − N 5
n (^) A = v A N 2 + N 4
N 4 N 6 n 6 ( N 2 + N 4 )( N 4 − N 5 )
=
= 17 .49 rev/min cw Ans.
13-21 Let gear 2 be first, then n (^) F = n 2 = 180 rev/min. Let gear 6 be last, then n (^) L = n 6 = 0.
e =
, e = n (^) L − n (^) A n (^) F − n (^) A
(180 − n (^) A )
= (0 − n (^) A )
n (^) A =
180 = − 82 .29 rev/min
The negative sign indicates opposite n 2 ∴ n (^) A = 82 .29 rev/min cw Ans.
Alternatively, since N ∝ r , let v = N n (crazy units).
v A N 5
v N 4 − N 5
N 2 n 2 N 4 − N 5
v A = N 5 N 2 n 2 N 4 − N 5
n (^) A = v A N 2 + N 4
N 5 N 2 n 2 ( N 2 + N 4 )( N 4 − N 5 )
=
= 82 .29 rev/min cw Ans.
5 4
v 0
v N 2 n 2
N 2
vA
2
4 5
v
v 0 vA
2
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13-25 n 2 = n (^) b = n (^) F , n (^) A = n (^) a , n (^) L = n 5 = 0
e = −
n (^) L − n (^) A n (^) F − n (^) A −
( n (^) F − n (^) A ) = 0 − n (^) A
With shaft b as input −
n (^) F +
n (^) A +
n (^) A = 0
n (^) A n (^) F
n (^) a n (^) b
n (^) a =
n (^) b , in the same direction as shaft b , the input. Ans.
Alternatively, v A N 4
n 2 N 2 N 3 + N 4
v A = n 2 N 2 N 4 N 3 + N 4 n (^) a = n (^) A = v A N 2 + N 3
n 2 N 2 N 4 ( N 2 + N 3 )( N 3 + N 4 )
= 18(21)( n (^) b ) (18 + 72)(72 + 21)
n (^) b in the same direction as b Ans.
13-26 n (^) F = n 2 = n (^) a , n (^) L = n 6 = 0
e = −
, e = n (^) L − n (^) A n (^) F − n (^) A
0 − n (^) b n (^) a − n (^) b
−
0 − n (^) b n (^) a − n (^) b
n (^) b n (^) a
Ans.
Yes, both shafts rotate in the same direction. Ans. Alternatively, v A N 5
n 2 N 2 N 3 + N 5
N 2
N 3 + N 5
n (^) a , v A =
N 2 N 5
N 3 + N 5
n (^) a
n (^) A = n (^) b = v A N 2 + N 3
N 2 N 5
( N 2 + N 3 )( N 3 + N 5 )
n (^) a
n (^) b n (^) a
Ans.
n (^) b rotates ccw ∴^ Yes Ans.
13-27 n 2 = n (^) F = 0, n (^) L = n 5 = n (^) b , n (^) A = n (^) a
e = +
35
v 0 vA n 2 N 2 2
3
4
2
v 0 vA
n 2 N 2
Chapter 13 343
25 36
n (^) b − n (^) a 0 − n (^) a n (^) b n (^) a
Ans.
Same sense, therefore shaft b rotates in the same direction as a. Ans. Alternatively,
v 5 N 3 − N 4
( N 2 + N 3 ) n (^) a N 3
v 5 = ( N 2 + N 3 )( N 3 − N 4 ) n N 3
n (^) b = v 5 N 5
( N 2 + N 3 )( N 3 − N 4 ) n (^) a N 3 N 5 n (^) b n (^) a
same sense Ans.
13-28 (a) ω = 2 π n / 60
H = T ω = 2 π T n / 60 ( T in N · m, H in W)
So T =
60 H (10^3 )
2 π n = 9550 H / n ( H in kW, n in rev/min)
Ta =
= 398 N · m
r 2 = m N 2 2
= 42 .5 mm So F 32 t = Ta r 2
= 9 .36 kN
F 3 b = − Fb 3 = 2(9.36) = 18 .73 kN in the positive x -direction. Ans. See the figure in part (b) on the following page.
2
a T (^) a 2 398 N • m
F t 32
3
4
v 5 v 0
( N 2 N 3 ) n (^) a
Chapter 13 345
Input torque: T 2 =
63 025 H
n
T 2 =
= 1576 lbf · in
For 100 percent gear efficiency
T arm =
= 7878 lbf · in Gear 2
W t^ =
= 788 lbf
F (^) 32 r = 788 tan 20° = 287 lbf
Gear 4
F (^) A 4 = 2 W t^ = 2(788) = 1576 lbf
Gear 5
Arm
T out = 1576(9) − 1576(4) = 7880 lbf · in Ans.
13-30 Given: P = 2 teeth/in, n (^) P = 1800 rev/min cw, N 2 = 18 T , N 3 = 32 T , N 4 = 18 T , N 5 = 48 T. Pitch Diameters: d 2 = 18 / 2 = 9 in; d 3 = 32 / 2 = 16 in; d 4 = 18 / 2 = 9 in; d 5 = 48 / 2 = 24 in.
4" 5"
1576 lbf
1576 lbf
T out
(^5) W t^ 788 lbf
F r^ 287 lbf
2 W t^ 1576 lbf
W t
F r
4
n 4 F (^) A 4
W t W t
F r^ F r
2
n 2^ T^2 ^ 1576 lbf • in
F ta 2
W t
F ra 2 F r 42
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Gear 2
Ta 2 = 63 025(200)/ 1800 = 7003 lbf · in W t^ = 7003 / 4. 5 = 1556 lbf W r^ = 1556 tan 20° = 566 lbf
Gears 3 and 4
W t^ (4.5) = 1556(8), W t^ = 2766 lbf W r^ = 2766 tan 20◦^ = 1007 lbf
Ans.
13-31 Given: P = 5 teeth/in, N 2 = 18 T , N 3 = 45 T , φ n = 20°, H = 32 hp, n 2 = 1800 rev/min.
Gear 2 T in =
= 1120 lbf · in
d (^) P =
= 3 .600 in
d (^) G =
= 9 .000 in
W (^) 32 t =
= 622 lbf
W (^) 32 r = 622 tan 20° = 226 lbf
F (^) at 2 = W (^) 32 t = 622 lbf, F (^) ar 2 = W (^) 32 r = 226 lbf
Fa 2 = (622^2 + 2262 ) 1 /^2 = 662 lbf
Each bearing on shaft a has the same radial load of R (^) A = R (^) B = 662 / 2 = 331 lbf.
2
a
T in
W t 32
W r 32
F ra 2
F ta 2
b
3
4
y
x
W r^ 566 lbf
W t^ 1556 lbf
W t^ 2766 lbf
W r^ 1007 lbf
2
a
W t^ 1556 lbf
W r^ 566 lbf
T (^) a 2 7003 lbf • in
348 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
The compressive loads at A and D are absorbed by the base plate, not the bolts. For W (^) 32 t , the tensions in C and D are ∑ M (^) AB = 0 1681(4. 875 + 15 .25) − 2 F (15.25) = 0 F = 1109 lbf
If W (^) 32 t reverses, 15.25 in changes to 13.25 in, 4.815 in changes to 2.875 in, and the forces change direction. For A and B ,
1681(2.875) − 2 F 1 (13.25) = 0 ⇒ F 1 = 182 .4 lbf For W (^) 32 r
M = 612(4. 875 + 11. 25 /2) = 6426 lbf · in
a =
(14/2) 2 + (11. 25 /2) 2 = 8 .98 in
F 2 =
= 179 lbf
At C and D , the shear forces are:
FS 1 =
[153 + 179(5. 625 / 8 .98)]^2 + [179(7/ 8 .98)]^2
= 300 lbf At A and B , the shear forces are:
FS 2 =
[153 − 179(5. 625 / 8 .98)]^2 + [179(7/ 8 .98)]^2
= 145 lbf
C
a D
153 lbf
153 lbf F^2
F 2^ F 2
F 2
612 4 ^ 153 lbf
14 612 lbf
153 lbf
B
C
15.25" 4.875 1681 lbf
F
F
D
F 1
F 1
A
Chapter 13 349
The shear forces are independent of the rotational sense. The bolt tensions and the shear forces for cw rotation are,
Tension (lbf) Shear (lbf) A 0 145 B 0 145 C 1109 300 D 1109 300
For ccw rotation,
Tension (lbf) Shear (lbf) A 182 145 B 182 145 C 0 300 D 0 300
13-33 T in = 63 025 H / n = 63 025(2.5)/ 240 = 656 .5 lbf · in
W t^ = T / r = 656. 5 / 2 = 328 .3 lbf γ = tan−^1 (2/4) = 26 .565° = tan−^1 (4/2) = 63 .435° a = 2 + (1.5 cos 26.565°)/ 2 = 2 .67 in W r^ = 328 .3 tan 20° cos 26.565° = 106 .9 lbf W a^ = 328 .3 tan 20° sin 26.565° = 53 .4 lbf W = 106. 9 i − 53. 4 j + 328. 3 k lbf R AG = − 2 i + 5. 17 j , R AB = 2. 5 j ∑ M 4 = R AG × W + R AB × F B + T = 0 Solving gives R AB × F B = 2. 5 F (^) Bz i − 2. 5 F (^) Bx k R AG × W = 1697 i + 656. 6 j − 445. 9 k
So (1697 i + 656. 6 j − 445. 9 k ) +
- 5 F (^) Bz i − 2. 5 F (^) Bx k + T j
F (^) Bz = − 1697 / 2. 5 = − 678 .8 lbf T = − 656 .6 lbf · in F (^) Bx = − 445. 9 / 2. 5 = − 178 .4 lbf
y 2
(^2 ) B A
W r G^ W t
W a
T in Not to scale
z x
a
F (^) Ay
F (^) Az
F (^) Bz
F (^) Ax
F (^) Bx
Chapter 13 351
13-35 Sketch gear 2 pictorially.
Pt = Pn cos ψ = 4 cos 30° = 3 .464 teeth/in
φ t = tan−^1
tan φ n cos ψ = tan−^1
tan 20° cos 30°
Sketch gear 3 pictorially,
d (^) P =
= 5 .196 in Pinion (Gear 2) W r^ = W t^ tan φ t = 800 tan 22.80° = 336 lbf W a^ = W t^ tan ψ = 800 tan 30° = 462 lbf W = − 336 i − 462 j + 800 k lbf Ans. W = [(−336) 2 + (−462) 2 + 8002 ]^1 /^2 = 983 lbf Ans. Gear 3 W = 336 i + 462 j − 800 k lbf Ans. W = 983 lbf Ans.
d (^) G =
= 9 .238 in
TG = W t^ r = 800(9.238) = 7390 lbf · in
13-36 From Prob. 13-35 solution,
Notice that the idler shaft reaction contains a couple tending to turn the shaft end-over- end. Also the idler teeth are bent both ways. Idlers are more severely loaded than other gears, belying their name. Thus be cautious.
800
336
462
4
800 800
336 336
(^3462)
462
800
2 336
462
W a
T (^) G
W r
W t
x
3
z y
W a
W r
T
W t
x
z y
2
352 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Gear 3 : Pt = Pn cos ψ = 7 cos 30° = 6 .062 teeth/in
tan φ t = tan 20° cos 30°
= 0 .4203, φ t = 22 .8°
d 3 =
= 8 .908 in
W t^ = 500 lbf W a^ = 500 tan 30° = 288 .7 lbf W r^ = 500 tan 22.8° = 210 .2 lbf W 3 = 210. 2 i + 288. 7 j − 500 k lbf Ans.
Gear 4 : d 4 =
= 2 .309 in
W t^ = 500
= 1929 lbf
W a^ = 1929 tan 30° = 1114 lbf W r^ = 1929 tan 22.8° = 811 lbf W 4 = − 811 i + 1114 j − 1929 k lbf Ans.
Pt = 6 cos 30° = 5 .196 teeth/in
d 3 =
= 8 .083 in
φ t = 22 .8°
d 2 =
= 3 .079 in
T 2 =
= 916 lbf · in
W t^ =
T
r
= 595 lbf
W a^ = 595 tan 30° = 344 lbf W r^ = 595 tan 22.8° = 250 lbf W = 344 i + 250 j + 595 k lbf R DC = 6 i , R DG = 3 i − 4. 04 j
T 3
C
B A
D
T 2
y 3
2
x
z
y
x W r W a W t
W t
W r
W a
r 4
r 3
