Mechanical Engineering Design Chapter 16 Solutions

Chapter 16 16-1

Given: r = 300/2 = 150 mm, a = R = 125 mm, b = 40 mm, f = 0.28, F = 2.2 kN,  1 = 0,  2 = 120, and  a = 90. From which, sin a = sin90 = 1. Eq. (16-2): 0.28 pa (0.040)(0.150) 120 0 sin  (0.150  0.125 cos  ) d 1  2.993 10 4  pa N · m

Mf 

Eq. (16-3):

pa (0.040)(0.150)(0.125) 120 2 sin  d  9.478 104  pa N · m  0  1

MN 

c = 2(0.125 cos 30) = 0.2165 m Eq. (16-4):

F 

9.478 10 4  pa  2.993 10 4  pa 0.2165

 2.995 10 3  pa

p a = F/ [2.995(103)] = 2200/ [2.995(103)] = 734.5(103) Pa for cw rotation Eq. (16-7):

2200 

9.478 10 4  pa  2.993 10 4  pa 0.2165

p a = 381.9(103) Pa for ccw rotation A maximum pressure of 734.5 kPa occurs on the RH shoe for cw rotation. (b) RH shoe: Eq. (16-6): 0.28(734.5)103 (0.040)0.1502 (cos 0o  cos120o ) TR   277.6 N · m 1 LH shoe: 381.9 TL  277.6  144.4 N · m Ans. 734.5 T total = 277.6 + 144.4 = 422 N · m Ans.

Ans.

Ans.

Chapter 16, Page 1/27

(c)

RH shoe:

F x = 2200 sin 30° = 1100 N, F y = 2200 cos 30° = 1905 N

Eqs. (16-8):

1  A   sin 2   2  0o

Eqs. (16-9):

Rx 

120o

Ry 

 0.375,

734.5 103  0.040(0.150) 1 3 734.5 10  0.04(0.150)

R  [ 1007 

LH shoe:

2 / 3 rad

 1  B    sin 2  2 4  0

2

 1.264

[0.375  0.28(1.264)]  1100  1007 N

[1.264  0.28(0.375)]  1905  4128 N 1  41282 ]1/ 2  4249 N Ans.

F x = 1100 N, F y = 1905 N

Eqs. (16-10): Rx 

381.9 103  0.040(0.150)

[0.375  0.28(1.264)]  1100  570 N 1 381.9 103  0.040(0.150) [1.264  0.28(0.375)]  1905  751 N Ry  1 1/ 2 R   597 2  7512   959 N Ans.

______________________________________________________________________________ 16-2

Given: r = 300/2 = 150 mm, a = R = 125 mm, b = 40 mm, f = 0.28, F = 2.2 kN,  1 = 15,  2 = 105, and  a = 90. From which, sin a = sin90 = 1. Eq. (16-2): 0.28 pa (0.040)(0.150) 105 4 Mf  15 sin  (0.150  0.125 cos  ) d  2.177 10  pa 1

Chapter 16, Page 2/27

MN 

Eq. (16-3):

pa (0.040)(0.150)(0.125) 105 2 sin  d  7.765 10 4  pa  15  1

c = 2(0.125) cos 30° = 0.2165 m F 

Eq. (16-4):

7.765 10 4  pa  2.177 10 4  pa 0.2165

 2.58110 3  pa

p a = 2200/ [2.581(10 3)] = 852.4 (103) Pa = 852.4 kPa on RH shoe for cw rotation Ans.

RH shoe:

TR 

Eq. (16-6): LH shoe:

2200 

0.28(852.4)103 (0.040)(0.1502 )(cos15  cos105)  263 N · m 1 7.765 10 4  pa  2.177 10 4  pa

pa  479.110

3



0.2165 Pa  479.1 kPa on LH shoe for ccw rotation

Ans.

0.28(479.1)103(0.040)(0.150 2 )(cos15  cos105)  148 N · m 1  263  148  411 N · m Ans.

TL  Ttotal

Comparing this result with that of Prob. 16-1, a 2.6% reduction in torque is obtained by using 25% less braking material. ______________________________________________________________________________ 16-3

Given:  1 = 0°,  2 = 120°,  a = 90°, sin  a = 1, a = R = 3.5 in, b = 1.25 in, f = 0.30, F = 225 lbf, r = 11/2 = 5.5 in, counter-clockwise rotation. LH shoe: Eq. (16-2), with  1 = 0:  f pabr 2 f pabr  a  sin   r  a cos   d  Mf  r (1  cos  2 )  sin 2  2    sin  a 1 sin  a  2  0.30 pa (1.25)5.5  3.5 2  5.5(1  cos120o )  sin 120  1 2    14.31 pa lbf · in Eq. (16-3), with  1 = 0:  pabra 2 2 p bra   2 1  MN  sin  d  a  sin 2 2    sin  a 1 sin  a  2 4  

 pa (1.25)5.5(3.5) 120    1    sin 2(120)   1  2  180  4   30.41 pa lbf · in 

Chapter 16, Page 3/27

 180o   2  o c  2r cos    2(5.5) cos 30  9.526 in 2   30.41 pa  14.31 pa  1.690 pa F  225  9.526 pa  225 / 1.690  133.1 psi

Eq. (16-6):

f pabr 2 (cos 1  cos  2 ) 0.30(133.1)1.25(5.52 )  [1  (0.5)] sin  a 1  2265 lbf · in  2.265 kip · in Ans.

TL 

RH shoe: 30.41 pa  14.31 pa  4.694 pa 9.526 pa  225 / 4.694  47.93 psi 47.93 TR  2265  816 lbf ·in  0.816 kip·in 133.1 F  225 

T total = 2.27 + 0.82 = 3.09 kip  in Ans. ______________________________________________________________________________ 16-4

(a) Given:  1 = 10°,  2 = 75°,  a = 75°, p a = 106 Pa, f = 0.24, b = 0.075 m (shoe width), a = 0.150 m, r = 0.200 m, d = 0.050 m, c = 0.165 m. Some of the terms needed are evaluated here: 

2 2 2 2 1 2    A  r  sin  d  a  sin  cos  d  r   cos    a  sin    1  1 1 2 1

75

1   200   cos  10  150  sin 2    77.5 mm 2 10 75 /180 rad 2  1  2 B   sin  d    sin 2   0.528 1 2 4 10 /180 rad 75

C 

2



sin  cos  d  0.4514

1

Now converting to Pascals and meters, we have from Eq. (16-2), 0.24 106  (0.075)(0.200) f pabr Mf  A (0.0775)  289 N · m sin  a sin 75

Chapter 16, Page 4/27

From Eq. (16-3), MN

pabra 106 (0.075)(0.200)(0.150)  B  (0.528)  1230 N · m sin  a sin 75

Finally, using Eq. (16-4), we have F 

MN  M f c



1230  289  5.70 kN 165

Ans.

(b) Use Eq. (16-6) for the primary shoe. T  

fpabr 2 (cos 1  cos  2 ) sin  a

0.24 106  (0.075)(0.200)2 (cos 10  cos 75) sin 75

 541 N · m

For the secondary shoe, we must first find p a . Substituting 1230 289 pa and M f  6 pa into Eq. (16 - 7), 6 10 10 (1230 / 106 ) pa  (289 / 106 ) pa 5.70  , solving gives pa  619 103  Pa 165 MN 

Then T 

0.24 619 103   0.075  0.2002   cos 10  cos 75  sin 75

 335 N · m

so the braking capacity is T total = 2(541) + 2(335) = 1750 N · m

Ans.

(c) Primary shoes: pabr  C  f B   Fx sin  a 106 (0.075)0.200  [0.4514  0.24(0.528)](10 3 )  5.70  0.658 kN sin 75 pabr ( B  f C )  Fy Ry  sin  a 106 (0.075)0.200  [0.528  0.24(0.4514)] 10 3   0  9.88 kN sin 75 Rx 

Chapter 16, Page 5/27

Secondary shoes: Rx  

pabr (C  f B)  Fx sin  a

0.619 106  0.075(0.200)

sin 75  0.143 kN p br Ry  a ( B  f C )  Fy sin  a 

0.619 106  0.075(0.200)

 4.03 kN

sin 75

[0.4514  0.24(0.528)] 10 3   5.70

[0.528  0.24(0.4514)] 10 3   0

Note from figure that +y for secondary shoe is opposite to +y for primary shoe. Combining horizontal and vertical components, RH  0.658  0.143  0.801 kN RV  9.88  4.03  5.85 kN R  (0.801) 2  5.852  5.90 kN Ans. ______________________________________________________________________________ 16-5

Given: Face width b = 1.25 in, F = 90 lbf, f = 0.25. Preliminaries:  1 = 45°  tan1(6/8) = 8.13°,  2 = 98.13°,  a = 90°, a = (62 + 82)1/2 = 10 in Eq. (16-2): 

f pabr 2 0.25 pa (1.25)6 Mf  sin   r  a cos   d   sin  a 1 1

98.13



sin   6  10 cos   d

8.13

 3.728 pa lbf · in

Eq. (16-3): 

MN 

pabra 2 2 p (1.25)6(10) sin  d  a  sin  a 1 1

98.13



sin 2  d

8.13

 69.405 pa lbf · in

Eq. (16-4): Using Fc = M N  M f , we obtain 90(20)  (69.405  3.728) pa



pa  27.4 psi

Ans.

Chapter 16, Page 6/27

Eq. (16-6):

fp br 2  cos 1  cos  2  0.25(27.4)1.25  6   cos8.13  cos 98.13  T  a  sin  a 1  348.7 lbf · in Ans. ______________________________________________________________________________ 2

16-6

For 3ˆ f :

f  f  3ˆ f  0.25  3(0.025)  0.325 From Prob. 16-5, with f = 0.25, M f = 3.728 p a . Thus, M f = (0.325/0.25) 3.728 p a = 4.846 p a . From Prob. 16-5, M N = 69.405 p a . Eq. (16-4): Using Fc = M N  M f , we obtain 90(20)  (69.405  4.846) pa



pa  27.88 psi

Ans.

From Prob. 16-5, p a = 27.4 psi and T = 348.7 lbf⋅in. Thus,  0.325   27.88  T    348.7  461.3 lbf ·in  0.25   27.4 

Ans.

Similarly, for 3ˆ f : f  f  3ˆ f  0.25  3(0.025)  0.175 M f  (0.175 / 0.25) 3.728 pa  2.610 pa

90(20) = (69.405  2.610) p a  p a = 26.95 psi  0.175  26.95  T    348.7  240.1 lbf · in Ans.  0.25  27.4  ______________________________________________________________________________ 16-7

Preliminaries:  2 = 180°  30°  tan1(3/12) = 136°,  1 = 20°  tan1(3/12) = 6°,  a = 90, sin a = 1, a = (32 + 122)1/2 = 12.37 in, r = 10 in, f = 0.30, b = 2 in, p a = 150 psi. Eq. (16-2):

Mf 

0.30(150)(2)(10) 136o 6 sin  (10  12.37 cos  ) d  12 800 lbf · in sin 90

Eq. (16-3):

MN 

150(2)(10)(12.37) 136 2 6 sin  d  53 300 lbf · in sin 90

LH shoe: c L = 12 + 12 + 4 = 28 in

Chapter 16, Page 7/27

Now note that M f is cw and M N is ccw. Thus,

Eq. (16-6):

FL 

53 300  12 800  1446 lbf 28

TL 

0.30(150)(2)(10) 2 (cos 6  cos136)  15 420 lbf · in sin 90

RH shoe: M N  53 300

pa  355.3 pa , 150

M f  12 800

pa  85.3 pa 150

On this shoe, both M N and M f are ccw. Also, c R = (24  2 tan 14°) cos 14° = 22.8 in Fact  FL sin14  361 lbf Ans. FR  FL / cos14  1491 lbf Thus,

1491 

Then,

TR 

355.3  85.3 pa  pa  77.2 psi 22.8

0.30(77.2)(2)(10) 2 (cos 6  cos136)  7940 lbf · in sin 90

T total = 15 420 + 7940 = 23 400 lbf · in Ans. ______________________________________________________________________________ 16-8 2

M f  2 ( fdN )(a cos   r ) 0

where dN  pbr d

2

 2 fpbr  (a cos   r ) d  0 0

From which 2

2

0

0

a cos  d  r  d

r 2 r (60)( / 180) a    1.209r sin  2 sin 60

Ans.

Chapter 16, Page 8/27

Eq. (16-15): a 

4r sin 60  1.170r 2(60)( / 180)  sin[2(60)]

Ans.

a differs with a ¢ by 100(1.170 1.209)/1.209 =  3.23 % Ans. ______________________________________________________________________________ 16-9

(a) Counter-clockwise rotation,  2 =  / 4 rad, r = 13.5/2 = 6.75 in Eq. (16-15): 4r sin  2 4(6.75) sin( / 4)   7.426 in a  2 2  sin 2 2 2 / 4  sin(2 / 4) e  2a  2(7.426)  14.85 in

Ans.

(b)

 = tan1(3/14.85) = 11.4°

M F x

R

 0  3F x  6.375P

 0  F x  R x

 F x  2.125P

 R x  F x  2.125P

F y  F x tan11.4o  0.428P  Fy   P  F y  R y R y  P  0.428P  1.428P

Left shoe lever.  M R  0  7.78S x  15.28F x 15.28 Sx  (2.125P)  4.174P 7.78 S y  f S x  0.30(4.174 P)  1.252P  Fy  0  R y  S y  F y

R y   F y  S y  0.428P  1.252 P  1.68P  Fx  0  R x  S x  F x R x  S x  F x  4.174P  2.125P  2.049P

Chapter 16, Page 9/27

(c) The direction of brake pulley rotation affects the sense of Sy, which has no effect on the brake shoe lever moment and hence, no effect on Sx or the brake torque.

The brake shoe levers carry identical bending moments but the left lever carries a tension while the right carries compression (column loading). The right lever is designed and used as a left lever, producing interchangeable levers (identical levers). But do not infer from these identical loadings. ______________________________________________________________________________ 16-10 r = 13.5/2 = 6.75 in,

b = 6 in,

 2 = 45° =  / 4 rad.

From Table 16-3 for a rigid, molded non-asbestos lining use a conservative estimate of p a = 100 psi, f = 0.33. Equation (16-16) gives the horizontal brake hinge pin reaction which corresponds to Sx in Prob. 16-9. Thus, p br 100(6)6.75 N  S x  a  2 2  sin 2 2   2  / 4   sin  2  45   2 2  5206 lbf





which, from Prob. 6-9 is 4.174 P. Therefore, 4.174 P = 5206

 P = 1250 lbf = 1.25 kip

Ans.

Applying Eq. (16-18) for two shoes, where from Prob. 16-9, a = 7.426 in T  2a f N  2(7.426)0.33(5206)  25 520 lbf · in  25.52 kip · in Ans. ______________________________________________________________________________

16-11 Given: D = 350 mm, b = 100 mm, p a = 620 kPa, f = 0.30,  = 270. Chapter 16, Page 10/27

Eq. (16-22): pabD 620(0.100)0.350   10.85 kN 2 2

P1 

Ans.

f   0.30(270)( / 180)  1.414

Eq. (16-19):

P 2 = P 1 exp( f  ) = 10.85 exp( 1.414) = 2.64 kN

Ans.

T  ( P1  P2 )( D / 2)  (10.85  2.64)(0.350 / 2)  1.437 kN · m Ans. ______________________________________________________________________________ 16-12 Given: D = 12 in,

Eq. (16-22):

f = 0.28, b = 3.25 in,  = 270°, P 1 = 1800 lbf.

pa 

2 P1 2(1800)   92.3 psi bD 3.25(12)

Ans.

f   0.28(270o )( / 180o )  1.319 P2  P1 exp( f  )  1800 exp(1.319)  481 lbf T  ( P1  P2 )( D / 2)  (1800  481)(12 / 2)  7910 lbf · in  7.91 kip · in Ans. ______________________________________________________________________________ 16-13

 M O = 0 = 100 P 2  325 F  P 2 = 325(300)/100 = 975 N Ans.  100    cos 1    51.32  160    270  51.32  218.7 f   0.30(218.7)  / 180   1.145 P1  P2 exp( f  )  975exp(1.145)  3064 N Ans. T   P1  P2  ( D / 2)  (3064  975)(200 / 2)  209 103  N · mm  209 N · m

Ans.

______________________________________________________________________________

Chapter 16, Page 11/27

16-14 (a) D = 16 in, b = 3 in n = 200 rev/min f = 0.20, p a = 70 psi

Eq. (16-22): P1 

Eq. (16-14):

pabD 70(3)(16)  1680 lbf  2 2 f   0.20(3 / 2)  0.942

P2  P1 exp( f  )  1680 exp(0.942)  655 lbf

D 16  (1680  655) 2 2  8200 lbf · in Ans. Tn 8200(200) H    26.0 hp 63 025 63 025 3P 3(1680)  504 lbf Ans. P  1  10 10 T  ( P1  P2 )

Ans.

(b) Force of belt on the drum:

R = (16802 + 6552)1/2 = 1803 lbf Force of shaft on the drum: 1680 and 655 lbf TP1  1680(8)  13 440 lbf · in TP2  655(8)  5240 lbf · in

Net torque on drum due to brake band: T  TP1  TP2  13 440  5240  8200 lbf · in

The radial load on the bearing pair is 1803 lbf. If the bearing is straddle mounted with the drum at center span, the bearing radial load is 1803/2 = 901 lbf.

Chapter 16, Page 12/27

(c) Eq. (16-21): 2P bD 2 P1 2(1680)    70 psi 3(16) 3(16)

p  p

  0

Ans.

2P2 2(655)   27.3 psi Ans. 3(16) 3(16) ______________________________________________________________________________ p

  270



16-15 Given:  = 270°, b = 2.125 in, f = 0.20, T =150 lbf · ft, D = 8.25 in, c 2 = 2.25 in (see figure). Notice that the pivoting rocker is not located on the vertical centerline of the drum. (a) To have the band tighten for ccw rotation, it is necessary to have c 1 < c 2 . When friction is fully developed,

P1 / P2  exp( f  )  exp[0.2(3 / 2)]  2.566 If friction is not fully developed, P 1 /P 2 ≤ exp( f  ) To help visualize what is going on let’s add a force W parallel to P 1 , at a lever arm of c 3 . Now sum moments about the rocker pivot.

M

 0  c3W  c1P1  c2 P2

From which

c2 P2  c1P1 c3 The device is self locking for ccw rotation if W is no longer needed, that is, W ≤ 0. It follows from the equation above W 

P1 c  2 P2 c1

When friction is fully developed 2.566  2.25 / c1 2.25 c1   0.877 in 2.566

When P 1 /P 2 is less than 2.566, friction is not fully developed. Suppose P 1 /P 2 = 2.25,

Chapter 16, Page 13/27

then c1 

2.25  1 in 2.25

We don’t want to be at the point of slip, and we need the band to tighten. c2  c1  c2 P1 / P2

When the developed friction is very small, P 1 /P 2 → 1 and c 1 → c 2

Ans.

(b) Rocker has c 1 = 1 in

P1 c 2.25  2   2.25 P2 c1 1 ln( P1 / P2 ) ln 2.25 f    0.172  3 / 2 Friction is not fully developed, no slip. T  ( P1  P2 )

P D D  P2  1  1 2  P2 2

Solve for P 2 2T 2(150)(12)   349 lbf [( P1 / P2 )  1]D (2.25  1)(8.25) P1  2.25P2  2.25(349)  785 lbf 2P 2(785) p  1   89.6 psi Ans. 2.125(8.25) bD

P2 

(c) The torque ratio is 150(12)/100 or 18-fold. 349 P2   19.4 lbf 18 P1  2.25P2  2.25(19.4)  43.6 lbf 89.6 p   4.98 psi Ans. 18 Comment: As the torque opposed by the locked brake increases, P 2 and P 1 increase (although ratio is still 2.25), then p follows. The brake can self-destruct. Protection could be provided by a shear key. ______________________________________________________________________________ 16-16 Given: OD = 250 mm, ID = 175 mm, f = 0.30, F = 4 kN.

Chapter 16, Page 14/27

(a) From Eq. (16-23), 2  4000  2F pa    0.194 N/mm 2  194 kPa Ans.  d ( D  d )  (175)(250  175) Eq. (16-25): Ff 4000(0.30) T  (D  d )  (250  175)10  3  127.5 N · m Ans. 4 4 (b) From Eq. (16-26),

4F 4(4000)   0.159 N/mm 2  159 kPa 2 2 2 2  ( D  d )  (250  175 )

pa  Eq. (16-27): T 



f pa ( D3  d 3 ) 



(0.30)159 103  2503  1753 10 3 

Ans.

3

12 12  128 N · m Ans. ______________________________________________________________________________

16-17 Given: OD = 6.5 in, ID = 4 in, f = 0.24, p a = 120 psi. (a) Eq. (16-23):  pa d  (120)(4) F  (D  d )  (6.5  4)  1885 lbf Ans. 2 2 Eq. (16-24) with N sliding planes:  fpa d 2  (0.24)(120)(4) T  (D  d 2 ) N  (6.52  42 )(6) 8 8  7125 lbf · in Ans. (b)

T 

 (0.24)(120d ) 8

(6.52  d 2 )(6)

d, in T, lbf · in 2 5191 3 6769 4 7125 Ans. 5 5853 6 2545 (c) The torque-diameter curve exhibits a stationary point maximum in the range of diameter d. The clutch has nearly optimal proportions. ______________________________________________________________________________ 16-18 (a) Eq. (16-24) with N sliding planes:

Chapter 16, Page 15/27

 f pa d ( D 2  d 2 ) N

T 

8



 f pa N 8

D d  d  2

3

Differentiating with respect to d and equating to zero gives dT  f pa N 2  D  3d 2   0  8 dd D d*  Ans. 3  f pa N 3 f pa N d 2T  6 d  d 2 8 4 dd

which is negative for all positive d. We have a stationary point maximum. (b)

d* 

6.5  3.75 in 3

Eq. (16-24): T* 



Ans.



 (0.24)(120) 6.5 / 3  2 2 6.5  6.5 / 3  (6)  7173 lbf · in 8





 

(c) The table indicates a maximum within the range: 3 ≤ d ≤ 5 in d (d) Consider:  0.80 0.45  D Multiply through by D, 0.45D  d  0.80D 0.45(6.5)  d  0.80(6.5) 2.925  d  5.2 in *

1 d  0.577    d * /D  3 D which lies within the common range of clutches. Yes. Ans. ______________________________________________________________________________

16-19 Given: d = 11 in,

l = 2.25 in, T = 1800 lbf · in,

D = 12 in,

f = 0.28.

 0.5 

  tan 1    12.53  2.25 

Chapter 16, Page 16/27

Uniform wear Eq. (16-45):

 f pa d 2 D  d2  8sin   (0.28) pa (11) 2 1800  12  112   128.2 pa  8sin12.53 1800  14.04 psi Ans. pa  128.2 T 

Eq. (16-44):

 pa d

F 

2

(D  d ) 

 (14.04)11 2

(12  11)  243 lbf

Ans.

Uniform pressure Eq. (16-48):

 f pa D3  d 3   12 sin   (0.28) pa 1800  123  113   134.1 pa  12sin12.53 1800 pa  Ans.  13.42 psi 134.1 T 

Eq. (16-47):

 pa

 (13.42)

122  112   242 lbf Ans. 4 4 ______________________________________________________________________________ F 

(D 2  d 2 ) 

16-20 Uniform wear

Eq. (16-34): Eq. (16-33): Thus,

1 ( 2  1) f pa ri  ro2  ri 2  2 F = ( 2   1 ) p a r i (r o  r i ) T 

(1 / 2)( 2  1) f pa ri  ro2  ri 2  T  f ( 2  1) pa ri (ro  ri )( D) f FD r  ri D / 2  d / 2 1 d  o   1   O.K . D 2D 2D 4

Ans.

Uniform pressure Eq. (16-38):

T 

1 ( 2  1) f pa  ro3  ri3  3

Chapter 16, Page 17/27

F 

Eq. (16-37):

1 ( 2  1) pa  ro2  ri 2  2

Thus,

(1 / 3)( 2  1) f pa  ro3  ri3  T 2  ( D / 2)3  (d / 2)3      f FD (1 / 2) f ( 2  1) pa  ro2  ri 2  D 3   ( D / 2) 2  (d / 2)2 D   2( D / 2)3 1  (d / D)3  1  1  (d / D) 3      O.K . Ans. 3( D / 2)2 1  (d / D) 2  D 3 1  (d / D) 2  ______________________________________________________________________________ 16-21

  2 n / 60  2 500 / 60  52.4 rad/s T 

H





2(103 )  38.2 N· m 52.4

Key: T 38.2   3.18 kN r 12 Average shear stress in key is 3.18(103 )    13.2 MPa Ans. 6(40) Average bearing stress is F 3.18(103 ) b     26.5 MPa Ab 3(40) Let one jaw carry the entire load. F 

Ans.

1  26 45      17.75 mm 2 2 2 38.2 T F    2.15 kN rav 17.75

rav 

The bearing and shear stress estimates are

b 

2.15 103 

 22.6 MPa Ans. 10(22.5  13) 2.15(103 )  0.869 MPa Ans.   10 0.25 (17.75) 2 

______________________________________________________________________________

Chapter 16, Page 18/27

16-22

1  2 n / 60  2 (1600) / 60  167.6 rad/s 2  0 From Eq. (16-51), I1I 2 Tt1 2800(8)    133.7 lbf · in · s 2 I1  I 2 1  2 167.6  0

Eq. (16-52): E 

I1 I 2 133.7 2 (167.6  0) 2  1.877 106  lbf  in 1  2   2  I1  I 2  2

H = E / 9336 = 1.877(106) / 9336 = 201 Btu

In Btu, Eq. (16-53): Eq. (16-54):

H 201   41.9F Ans. C pW 0.12(40) ______________________________________________________________________________ T 

16-23

n1  n2 260  240   250 rev/min 2 2 C s = ( 2   1 ) /  = (n 2  n 1 ) / n = (260  240) / 250 = 0.08 n

Eq. (16-62):

Ans.

 = 2 (250) / 60 = 26.18 rad/s From Eq. (16-64): I 

6.75 103  E2  E1   123.1 N · m · s 2 2 2 Cs 0.08(26.18)

I 

m 2  do  di2  8



m

8I 8(123.1)   233.9 kg 2 d  di 1.52  1.42

Table A-5, cast iron unit weight = 70.6 kN/m3 Volume:

2 o

  = 70.6(103) / 9.81 = 7197 kg / m3.

V = m /  = 233.9 / 7197 = 0.0325 m3 V   t  d o2  d i2  / 4   t 1.52  1.4 2  / 4  0.2278t

Equating the expressions for volume and solving for t, 0.0325 t   0.143 m  143 mm 0.2278

Ans.

Chapter 16, Page 19/27

______________________________________________________________________________ 16-24 (a) The useful work performed in one revolution of the crank shaft is U = 320 (103) 200 (103) 0.15 = 9.6 (103) J Accounting for friction, the total work done in one revolution is U = 9.6(103) / (1  0.20) = 12.0(103) J Since 15% of the crank shaft stroke accounts for 7.5% of a crank shaft revolution, the energy fluctuation is E 2  E 1 = 9.6(103)  12.0(103)(0.075) = 8.70(103) J (b) For the flywheel,

Since Eq. (16-64):

Ans.

n  6(90)  540 rev/min 2 n 2 (540)   56.5 rad/s   60 60

C s = 0.10 E E 8.70(103 ) I  2 21   27.25 N · m · s 2 2 Cs 0.10(56.5)

Assuming all the mass is concentrated at the effective diameter, d, md 2 I  mr  4 4I 4(27.25)  75.7 kg Ans. m 2  d 1.22 ______________________________________________________________________________ 2

16-25 Use Ex. 16-6 and Table 16-6 data for one cylinder of a 3-cylinder engine. Cs  0.30 n  2400 rev/min or 251 rad/s 3(3368) Tm   804 lbf · in Ans. 4 E2  E1  3(3531)  10 590 in · lbf E E 10 590  0.560 in · lbf · s 2 Ans. I  2 21  2 0.30(251 ) Cs ______________________________________________________________________________

Chapter 16, Page 20/27

16-26 (a) (1)

(T2 )1   F21rP  

T2 T rP  2 rG n

Ans.

Equivalent energy

(2)

(1 / 2) I 222  (1 / 2)( I 2 )1 12  I 22 ( I 2 )1  2 I 2  22 1 n 2

(3)

2

Ans.

2

I G  rG   mG   rG   rG             n4 I P  rP   mP   rP   rP 

From (2)

( I 2 )1 

(b) I e  I M  I P  n 2 I P 

IL n2

IG n4 I P   n2I P n2 n2

Ans.

Ans.

______________________________________________________________________________ 16-27 (a) Reflect I L , I G2 to the center shaft

Chapter 16, Page 21/27

Reflect the center shaft to the motor shaft

I e  I M  I P  n2I P 

I P m2 I  2 I P  2L 2 2 n n mn

(b) For R = constant = nm, I e  I M

(c) For R = 10,

Ans.

IP R2I P I  I P  n I P  2  4  L2 n n R 2

Ans.

I e 2(1) 4(102 )(1)  0  0  2n(1)  3  00 n n5 n n6  n2  200 = 0

From which n*  2.430 Ans. 10 m*   4.115 2.430

Ans.

Notice that n*and m* are independent of I L . ______________________________________________________________________________ 16-28 From Prob. 16-27, IP R2I P I   L2 2 4 n n R 1 100(1) 100  10  1  n 2 (1)  2   2 n n4 10 100 1  12  n 2  2  4 n n

Ie  I M  I P  n2I P 

Chapter 16, Page 22/27

Optimizing the partitioning of a double reduction lowered the gear-train inertia to 20.9/112 = 0.187, or to 19% of that of a single reduction. This includes the two additional gears. ______________________________________________________________________________ 16-29 Figure 16-29 applies, t2  10 s, t1  0.5 s t2 - t1 10  0.5   19 t1 0.5

The load torque, as seen by the motor shaft (Rule 1, Prob. 16-26), is TL 

1300(12)  1560 lbf · in 10

The rated motor torque T r is Tr 

63 025(3)  168.07 lbf · in 1125

For Eqs. (16-65):

2 (1125)  117.81 rad/s 60 2 (1200)  125.66 rad/s s  60 168.07 Tr a    21.41 lbf  in  s/rad 125.66  117.81  s  r 168.07(125.66) Trs b   2690.4 lbf · in s  r 125.66  117.81

r 

Chapter 16, Page 23/27

The linear portion of the squirrel-cage motor characteristic can now be expressed as Eq. (16-68):

T M = 21.41 + 2690.4 lbf · in 19

 1560  168.07  T2  168.07    1560  T2 

One root is 168.07 which is for infinite time. The root for 10 s is desired. Use a successive substitution method T2 0.00 19.30 24.40 26.00 26.50

New T 2 19.30 24.40 26.00 26.50 26.67

Continue until convergence to T 2 = 26.771 lbf ⋅ in Eq. (16-69): a  t2  t1  21.41(10  0.5)   110.72 lbf · in · s 2 ln T2 / Tr  ln(26.771 / 168.07) T b   a T  b 26.771  2690.4 max  2   124.41 rad/s Ans. a 21.41 min  117.81 rad/s Ans. 124.41  117.81    121.11 rad/s 2 124.41  117.81 max  min Cs    0.0545 Ans. (max  min ) / 2 (124.41  117.81) / 2 1 1 E1  I r2  (110.72)(117.81) 2  768 352 in · lbf 2 2 1 1 E2  I 22  (110.72)(124.41) 2  856 854 in · lbf 2 2 E  E2  E1  856 854  768 352  88 502 in · lbf I 

Eq. (16-64): E  Cs I  2  0.0545(110.72)(121.11) 2  88 508 in · lbf, close enough Ans.

Chapter 16, Page 24/27

During the punch 63 025H n TL (60 / 2 ) 1560(121.11)(60 / 2 ) H   28.6 hp  63 025 63 025 T 

The gear train has to be sized for 28.6 hp under shock conditions since the flywheel is on the motor shaft. From Table A-18,







m 2 W 2 d o  di2  d o  di2 8 8g 8gI 8(386)(110.72) W  2  2 d o  di d o2  di2 I 



If a mean diameter of the flywheel rim of 30 in is acceptable, try a rim thickness of 4 in di  30  (4 / 2)  28 in d o  30  (4 / 2)  32 in 8(386)(110.72)  189.1 lbf W  322  282 Rim volume V is given by V 

l

d 4

2 o

 di2  

l 4

(322  282 )  188.5l

where l is the rim width as shown in Table A-18. The specific weight of cast iron is  = 0.260 lbf / in3, therefore the volume of cast iron is V 

W





189.1  727.3 in 3 0.260

Equating the volumes, 188.5 l  727.3 727.3 l   3.86 in wide 188.5

Proportions can be varied. ______________________________________________________________________________ 16-30 Prob. 16-29 solution has I for the motor shaft flywheel as

Chapter 16, Page 25/27

I = 110.72 lbf · in · s2 A flywheel located on the crank shaft needs an inertia of 102 I (Prob. 16-26, rule 2) I = 102(110.72) = 11 072 lbf · in · s2 A 100-fold inertia increase. On the other hand, the gear train has to transmit 3 hp under shock conditions. Stating the problem is most of the solution. Satisfy yourself that on the crankshaft: TL  1300(12)  15 600 lbf · in Tr  10(168.07)  1680.7 lbf · in r  117.81 / 10  11.781 rad/s s  125.66 / 10  12.566 rad/s a  21.41(100)  2141 lbf · in · s/rad b  2690.35(10)  26903.5 lbf · in TM  2141c  26 903.5 lbf · in 19

 15 600  1680.5  T2  1680.6    15 600  T2 

The root is 10(26.67) = 266.7 lbf · in

  121.11 / 10  12.111 rad/s Cs  0.0549 (same) max  121.11 / 10  12.111 rad/s Ans. min  117.81 / 10  11.781 rad/s Ans.

E 1 , E 2 , E and peak power are the same. From Table A-18 6 8gI 8(386)(11 072) 34.19 10  W  2   d o  di2 d o2  di2 d o2  di2

Scaling will affect d o and d i , but the gear ratio changed I. Scale up the flywheel in the Prob. 16-29 solution by a factor of 2.5. Thickness becomes 4(2.5) = 10 in. d  30(2.5)  75 in do  75  (10 / 2)  80 in di  75  (10 / 2)  70 in

Chapter 16, Page 26/27

W 

34.19 106 

 3026 lbf 802  702 W 3026 V    11 638 in 3 0.260  V 



l (802  702 )  1178 l 4 11 638 l   9.88 in 1178

Proportions can be varied. The weight has increased 3026/189.1 or about 16-fold while the moment of inertia I increased 100-fold. The gear train transmits a steady 3 hp. But the motor armature has its inertia magnified 100-fold, and during the punch there are deceleration stresses in the train. With no motor armature information, we cannot comment. ______________________________________________________________________________ 16-31 This can be the basis for a class discussion.

Chapter 16, Page 27/27