Machine Design

September 1, 2017 | Author: Terenz Calangi | Category: Belt (Mechanical), Gear, Manufactured Goods, Classical Mechanics, Applied And Interdisciplinary Physics

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SECTION 7 – SHAFT DESIGN 471.

A short stub shaft, made of SAE 1035, as rolled, receivers 30 hp at 300 rpm via a 12-in. spur gear, the power being delivered to another shaft through a flexible coupling. The gear is keyed (profile keyway) midway between the bearings. The pressure angle of the gear teeth φ = 20 o ; N = 1.5 based on the octahedral shear stress theory with varying stresses. (a) Neglecting the radial component R of the tooth load W , determine the shaft diameter. (b) Considering both the tangential and the radial components, compute the shaft diameters. (c) Is the difference in the results of the parts (a) and (b) enough to change your choice of the shaft size?

Problem 471. Solution: For SAE 1035, as rolled s y = 55 ksi su = 85 ksi sn = 0.5su = 0.5(85) = 42.5 ksi A = W cos φ 63,000hp 63,000(30 ) T= = = 6300 in − lb n 300 AD T= 2 A(12 ) 6300 = 2 A = 1050 lb A = W cos φ 1050 = W cos 20 W = 1118 lb Shear stress 16T 16(6300 ) ss = = π d3 π d3 100,800 ss = sms = π d3 sas = 0

Page 1 of 76

SECTION 7 – SHAFT DESIGN bending stress

From Table AT 2 FL M= 4 (a) Negligible R : AL (1050 )(16 ) M= = = 4200 in − lb 4 4 32 M 32(4200 ) 134,400 s= = = π d3 π d3 π d3 sm = 0 134,400 sa = s = π d3 K f sa s se = n s m + sy SF For profile keyway K f = 2 .0 K fs = 1.6 SF = 0.85 K s (2.0)(134,400) = 100,661 se = f a = SF (0.85) π d 3 d3 K fs sas s ses = ns sms + s ys SF

(

)

sns sn 42.5 1 = = = s ys s y 55 1.294 sns  1  100,800  24,796  = sms =   3 s ys d3  1.294  π d  Octahedral-shear theory ses =

1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     2 1  100,661   24,796 =   +  3 1.5  42,500d   0.577 42,500d 3 d = 1.569 in

(

Page 2 of 76

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SECTION 7 – SHAFT DESIGN

use d = 1

11 in 16

(b) Considering both radial and tangential component. WL (1118)(16 ) M= = = 4472 in − lb 4 4 32 M 32(4472 ) 143,104 s= = = π d3 π d3 π d3 sm = 0 143,104 sa = s = π d3 K s (2.0)(143,104) = 107,180 se = f a = SF (0.85)(π d 3 ) d3 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     2 1  107,180   24,796 =   +  3 1.5  42,500d   0.577 42,500d 3 d = 1.597 in 11 use d = 1 in 16

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(c) The difference in the results of the parts (a) and (b) is not enough to change the choice of the shaft size. 472.

A cold-finished shaft, AISI 1141, is to transmit power that varies from 200 to 100 and back to 200 hp in each revolution at a speed of 600 rpm. The power is received by a 20-in. spur gear A and delivered by a 10-in. spur gear C. The tangential forces have each been converted into a force ( A and C shown) and a couple (not shown). The radial component R of the tooth load is to be ignored in the initial design. Let 2 and, considering varying stresses with the maximum shear theory, compute the shaft diameter.

Problems 472 – 474

Page 3 of 76

SECTION 7 – SHAFT DESIGN Solution: For AISI 1141, cold-finished s y = 90 ksi sn = 50 ksi sn 1 = s y 1.8 SF = 0.85 63,000hp T= n 63,000(200 ) Tmax = = 21,000 in − lb 600 63,000(100 ) Tmin = = 10,500 in − lb 600 1 1 Tm = (Tmax + Tmin ) = (21,000 + 10,500 ) = 15,750 in − lb 2 2 1 1 Ta = (Tmax − Tmin ) = (21,000 − 10,500 ) = 5,250 in − lb 2 2 16T ss = πd3 16(15,750 ) 252,000 sms = = π d3 π d3 16(5250 ) 24,000 = sas = π d3 π d3 K fs sas s ses = ns sms + s ys SF For profile keyway K f = 2 .0 K fs = 1.6

sns sn 1 = = s ys s y 1.8  1  252,000  (1.6)(84,000) 94,894  + ses =   = 3 0.85π d 3 d3  1.8  π d  Bending stress, negligible radial load T = 21,000 in − lb at 200 hp

For A:  20  A  = T  2  A(10 ) = 21,000

Page 4 of 76

SECTION 7 – SHAFT DESIGN A = 2100 lb at 200 hp For C:  10  C  = T 2 C (5) = 21,000 C = 4200 lb at 200 hp

[∑ M

]

= 0 A(10 ) + D(25) = C (15) at 200 hp (2100)(10) + D(25) = (4200)(15) D = 1680 lb

[∑ F

B

]

=0 A+C = B + D at 200 hp 2100 + 4200 = B + 1680 B = 4620 lb At 200 hp: A = 2100 lb , B = 4620 lb , C = 4200 lb , D = 1680 lb Shear Diagram V

Maximum moment at B M = (2100)(10 ) = 21,000 in − lb 32 M 32(21,000 ) 672,000 s= = = π d3 π d3 π d3 sm = 0 672,000 sa = s = π d3 K f sa (2.0)(672,000) = 503,304 s se = n s m + = 0+ sy SF 0.85π d 3 d3

Page 5 of 76

SECTION 7 – SHAFT DESIGN 94,894 d3 Maximum Shear Theory ses =

1

2 2 2 1  se   ses     =   +  N  sn   0.5sn     2 1  503,304   94,894 =   +  3 2  50,000d   0.5 50,000d 3 d = 2.78 in 3 use d = 2 in 4

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475.

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A shaft S, of cold-drawn AISI 1137, is to transmit power received from shaft W, which turns at 2000 rpm, through the 5-in. gear E and 15-in. gear A. The power is delivered by the 10-in. gear C to gear G, and it varies from 10 hp to 100 hp and back to 10 hp during each revolution of S. The design is to account for the varying stresses, with calculations based on the octahedral shear stress theory. Let N = 1.8 and compute the shaft diameter, using only the tangential driving loads for the first design.

Problem 475 – 477 Solution. For AISI 1137, cold drawn s y = 93 ksi su = 103 ksi sn = 0.5su = 0.5(103) = 51.5 ksi sn sns 51.5 1 = = = s y s ys 93 1.806 63,000hp T= n 5 in. E n= (2000 rpm) = 667 rpm 15 in. A

Page 6 of 76

SECTION 7 – SHAFT DESIGN 63,000(100 ) = 9450 in − lb 667 63,000(10 ) Tmin = = 945 in − lb 667 1 1 Tm = (Tmax + Tmin ) = (9450 + 945) = 5197.5 in − lb 2 2 1 1 Ta = (Tmax − Tmin ) = (9450 − 945) = 4252.5 in − lb 2 2 16T ss = πd3 16(5197.5) 83,160 sms = = π d3 π d3 16(4252.5) 68,040 = sas = π d3 π d3 K fs sas s ses = ns sms + s ys SF For profile keyway K f = 2 .0 Tmax =

K fs = 1.6 SF = 0.85  1  83,160  (1.6)(68,040) 55,425 + ses =  =  3  0.85π d 3 d3  1.806  π d  Bending stress, using only tangential loads

For 100 hp: T = 9450 in − lb  15  A  = T 2 A(7.5) = 9450 A = 1260 lb For C:  10  C  = T 2

Page 7 of 76

SECTION 7 – SHAFT DESIGN C (5) = 9450 C = 1890 lb

[∑ M

B

]

= 0 6 A + 20 D = 14C

6(1260) + 20 D = 14(1890) D = 945 lb

[∑ F

]

=0 A+C = B + D 1260 + 1890 = B + 945 B = 2205 lb Shear diagram V

Maximum moment at B M = (1260)(6) = 7560 in − lb 32 M 32(7560 ) 241,920 s= = = π d3 π d3 π d3 sm = 0 241,920 sa = s = π d3 K f sa (2.0 )(241,920 ) 181,189 s se = n s m + = = sy SF 0.85π d 3 d3 55,425 d3 Octahedral Shear Theory ses =

1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     2 1  181,189   55,425 =   +  3 2  51,500d   0.577 51,500d 3

(

d = 1.997 in use d = 2 in

Page 8 of 76

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SECTION 7 – SHAFT DESIGN 478.

A shaft made of AISI 1137, cold rolled, for a forage harvester is shown. Power is supplied to the shaft by a vertical flat belt on the pulley A. At B, the roller chain to the cutter exerts a force vertically upwards, and the V-belt to the blower at C exerts a force vertically upwards. At maximum operating conditions, the flat belt supplies 35 hp at 425 rpm, of which 25 hp is delivered to the cutter and 10 hp to the blower. The two sections of the shaft are joined by a flexible coupling at D and the various wheels are keyed (sled-runner keyway) to the shafts. Allowing for the varying stresses on the basis of the von Mises-Hencky theory of failure, decide upon the diameters of the shafts. Choose a design factor that would include an allowance for rough loading.

Problem 478. Solution: For AISI 1137, cold rolled s y = 93 ksi su = 103 ksi sn = 0.5su = 0.5(103) = 51.5 ksi sn sns 51.5 1 = = = s y s ys 93 1.806 Pulley, 63,000hp 63,000(35) TA = = = 5188 in − lb n 425 For flat-belt  2T  4(5188) FA = F1 + F2 = 2(F2 − F1 ) = 2 A  = = 692 lb 30  DA  Sprocket, 63,000hp 63,000(25) TB = = = 3706 in − lb n 425 For chain, 2T 2(3706 ) FB = B = = 741 lb DB 10 Sheave, 63,000hp 63,000(10 ) TC = = = 1482 in − lb n 425

Page 9 of 76

SECTION 7 – SHAFT DESIGN For V-belt,  2T FC = F1 + F2 = 1.5(F2 − F1 ) = 1.5 C  DC Consider shaft ABD.

35 hp Shaft ABD

[∑ M

D'

=0

]

(6 + 8 + 4)FA = (8 + 4)A'+4FB 18(692) = 12 A'+4(741) A' = 791 lb

[∑ F

V

=0

]

FA + D′ = FB + A′ 692 + D′ = 741 + 791 D′ = 840 lb

Shear Diagram

Maximum M at A’. M = (6)(692) = 4152 in − lb. 32 M 32(4152 ) 132,864 s= = = π d3 π d3 π d3 sm = 0 sa = s =

132,864 π d3

Page 10 of 76

 3(1482 )  = = 445 lb 10 

SECTION 7 – SHAFT DESIGN K f sa sn sm + sy SF For sled-runner keyway (Table AT 13) K f = 1 .6 se =

K fs = 1.6 SF = 0.85 K f sa (1.60)(132,864) = 79,610 s se = n s m + = 0+ sy SF 0.85π d 3 d3

at A’ T = TA = 5188 in − lb 16T 16(5188) 83,008 ss = = = π d3 π d3 π d3 sms = ss sas = 0 K fs sas s ses = ns sms + s ys SF  1  83,000  14,630 = ses =   3  d3  1.806  π d  Choose a design factor of 2.0 N = 2 .0 von Mises-Hencky theory of failure (Octahedral shear theory) 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     2 1  79,610   14,630 =   +  3 2  51,500d   0.577 51,500d 3 d = 1.48 in 1 use d = 1 in 2

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Consider shaft D-C 63,000hp 63,000(10 ) TC = = = 1482 in − lb n 425 For V-belt,  2T  3(1482 ) FC = F1 + F2 = 1.5(F2 − F1 ) = 1.5 C  = = 445 lb 10  DC 

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SECTION 7 – SHAFT DESIGN

[∑ M

C'

=0

]

8 D′′ = 3FC 8 D′′ = 3(445) D′′ = 167 lb

[∑ F

V

=0

]

C ′ = D′′ + FC C ′ = 167 + 445 C ′ = 612 lb Shear Diagram

M = (167 )(8) = 1336 in − lb 32 M 32(1336 ) 42,752 s= = = π d3 π d3 π d3 sm = 0 , sa = s K f sa (1.60)(42,752) = 25,616 s se = n s m + = 0+ sy SF 0.85π d 3 d3 at C’, TC = 1482 in − lb 16T 16(1482 ) 23,712 ss = = = π d3 π d3 π d3 sms = ss sas = 0 K fs sas s ses = ns sms + s ys SF 4180  1  23,712  +0 = 3 ses =   3  d  1.806  π d  1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     2 1  25,616   4180 =   +  3 2  51,500d   0.577 51,500d 3

(

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SECTION 7 – SHAFT DESIGN d = 1.011 in use d = 1 in

479.

A shaft for a punch press is supported by bearings D and E (with L = 24 in.) and receives 25 hp while rotating at 250 rpm, from a flat-belt drive on a 44in. pulley at B, the belt being at 45o with the vertical. An 8-in. gear at A delivers the power horizontally to the right for punching operation. A 1500-lb flywheel at C has a radius of gyration of 18 in. During punching, the shaft slows and energy for punching comes from the loss of kinetic energy of the flywheel in addition to the 25 hp constantly received via the belt. A reasonable assumption for design purposes would be that the power to A doubles during punching, 25 hp from the belt, 25 hp from the flywheel. The phase relations are such that a particular point in the section where the maximum moment occurs is subjected to alternating tension and compression. Sled-runner keyways are used for A, B, and C; material is colddrawn AISI 1137, use a design factor of N = 2.5 with the octahedral shear theory and account for the varying stresses. Determine the shaft diameters.

Problems 479-480 Solution: Flat-Belt Drive (B) 63,000hp 63,000(25) TB = = = 6300 in − lb n 250  2T  4(6300) FB = F1 + F2 = 2(F1 − F2 ) = 2 B  = = 573 lb 44  DB  Gear A, Doubled hp 63,000hp 63,000(25 + 25) TA = = = 12,600 in − lb n 250 2T 2(12,600 ) FA = A = = 3150 lb DA 8 Loading:

Page 13 of 76

SECTION 7 – SHAFT DESIGN

Vertical:

BV = FB cos 45 = 573 cos 45 = 405 lb

[∑ M

D

=0

]

6(1500) + 8BV = 24 EV 6(1500) + 8(405) = 24 EV EV = 510 lb

[∑ F

V

=0

]

1500 + EV = DV + BV 1500 + 510 = DV + 405 DV = 1605 lb Shear Diagram

Page 14 of 76

SECTION 7 – SHAFT DESIGN

M DV = (6 )(1500 ) = 9000 in − lb M BV = (16 )(510 ) = 8160 in − lb M AV = (5)(510 ) = 2550 in − lb

Horizontal:

Bh = FB sin 45 = 573 sin 45 = 405 lb

[∑ M

D

=0

]

8 Bh + 24 Eh = 19 FA 8(405) + 24 Eh = 19(3150) Eh = 2359 lb

[∑ F

h

=0

]

Dh + Bh + Eh = FA Dh + 405 + 2359 = 3150 Dh = 386 lb Shear Diagram

Page 15 of 76

SECTION 7 – SHAFT DESIGN

M Dh = 0 in − lb M Bh = (8)(386 ) = 3088 in − lb M Ah = (5)(2359 ) = 11,795 in − lb MA = MB =

(M ) + (M ) (M ) + (M ) 2

Ah

2

Bh

2

=

(11,795)2 + (2550)2

2

=

(3088)2 + (8160)2

AV

BV

= 12,068 in − lb

= 8725 in − lb

M D = 9000 in − lb Therefore M max = 12,068 in − lb 32 M 32(12,068) 386,176 s= = = π d3 π d3 π d3 Maximum moment subjected to alternating tension and compression sm = 0 386,176 sa = s = π d3 K f sa sn sm + sy SF For AISI 1137, cold-drawn, s y = 93 ksi se =

su = 103 ksi sn = 0.5su = 0.5(103) = 51.5 ksi For sled-runner keyway (Table AT 13) K f = 1 .6 K fs = 1.6 SF = 0.85 (1.60)(386,176 ) = 231,386 se = 0 + 0.85π d 3 d3

Page 16 of 76

SECTION 7 – SHAFT DESIGN

At A, 50 hp max. and 25 hp min. 50 hp 63,000hp 63,000(25 + 25) TA = = = 12,600 in − lb n 250 2T 2(12,600 ) FA = A = = 3150 lb DA 8 16T 16(12,600 ) 201,600 ss max = = = π d3 π d3 π d3 25 hp 63,000hp 63,000(25) TA = = = 6,300 in − lb n 250 2T 2(6,300 ) FA = A = = 1575 lb DA 8 16T 16(6,300 ) 100,800 ss min = = = π d3 π d3 π d3 sms =

 1 ,800  151,200  = (ss max + ss min ) = 1  201,600 + 100 3 2 2 πd π d3 

sas =

 1 ,800  50,400  = (ss max − ss min ) = 1  201,600 − 100 3 2 2 πd π d3 

ses =

K fs sas sns sms + s ys SF

 1  151,200  (1.6)(50,400) 56,848  + ses =  =  3 0.85π d 3 d3  1.806  π d  1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     2 1  231,386   56,848 =   +  3 2  51,500d   0.577 51,500d 3 d = 2.14 in 3 say d = 2 in 16

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SECTION 7 – SHAFT DESIGN 481.

A cold-drawn monel propeller shaft for a launch is to transmit 400 hp at 1500 rpm without being subjected to a significant bending moment; and Le k < 40 . The efficiency of the propeller is 70 % at 30 knots (1.152 mph/knot). Consider that the number of repetitions of the maximum power at the given speed is 2x 105. Let N = 2 based on the maximum shear theory with varying stress. Compute the shaft diameter.

Solution: For cold-drawn monel shaft, Table AT 10 s y = 75 ksi sn = 42 ksi at 108 at 2 x 105 0.085

 108   sn ≈ 42 = 71.23 ksi 5  2 × 10   sn sns 71.23 1 = = = s y s ys 75 1.053 63,000hp 63,000(400 ) = = 16,800 in − lb n 1500 16T 16(16,800 ) 268,800 ss = = = π D3 π D3 π D3 s ms = ss

T=

s as = 0 K fs sas sns sms + s ys SF SF = 0.85 assume K f = K fs = 1.0 ses =

81,255  1  268,800   + 0 = ses =   3 D3  1.053  π D  Fvm = ηhp 33,000 vm = (30 knots )(1.152 mph knot )(5280 ft mi )(1 hr 60 min ) = 3041 fpm F (3041) = (0.70)(400) 33,000 F = 3040 lb 4F 4(3040 ) 12,160 s= = = 2 πD π D2 π D2 sm= s sa = 0

Page 18 of 76

SECTION 7 – SHAFT DESIGN

se =

K f sa sn sm + sy SF

3676  1  12,160  +0 = 2 se =   2  D  1.053  π D  Maximum Shear Theory 1

2 2 2 1  se   ses     =   +  N  sn   0.5sn     2 1  3676   81,255 =   +  2 2  71,230 D   0.5 71,230 D 3

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1

)

  

2

2  

1

2 2 1  1   2.2815   2 =   +   2  19.377 D 2   D 3   By trial and error 11 D = 1.66 in = 1 in 16

482.

A shaft receives 300 hp while rotating at 600 rpm, through a pair of bevel gears, and it delivers this power via a flexible coupling at the other end. The shaft is designed with the average forces ( at the midpoint of the bevel-gear face); the tangential driving force is F , G = 580 lb , Q = 926 lb ; which are the rectangular components of the total reaction between the teeth; Dm = 24 in , L = 36 in , a = 10 in . Let the material be AISI C1045, cold drawn; N = 2 . Considering varying stresses and using the octahedral shear theory, determine the shaft diameter.

Problems 482, 485, 486. Solution: For AISI C1045, cold drawn s y = 85 ksi su = 100 ksi

Page 19 of 76

SECTION 7 – SHAFT DESIGN sn = 0.5su = 0.5(100) = 50 ksi SF = 0.85 sn sns 50 1 = = = s y s ys 85 1.7 63,000hp 63,000(300 ) T= = = 31,500 in − lb 600 600 16T 16(31,500 ) 504,000 ss = = = π D3 π D3 π D3 s ms = ss s as = 0 K fs sas s ses = ns sms + s ys SF 94,370  1  504,000   + 0 = ses =   3 D3  1.7  π D  D  F m  = T  2   24  F   = 31,500  2  F = 2625 lb Vertical:

D   24  Q m  = 926  = 11,112 in − lb  2   2  G = 580 lb

[∑ M

B

]

=0 

− Av (36 ) +

QDm − G (10 ) = 0 2

QDm = G (10 ) + Av (36 ) 2 11,112 = 580(10) + Av (36)

Av = 148 lb

Page 20 of 76

SECTION 7 – SHAFT DESIGN

[∑ F =0] v

Av + Bv = 580 lb 148 + Bv = 580 lb Bv = 432 lb Shear Diagram

Moment Diagram

M Cv = 11,112 in − lb M Bv = 5328 in − lb

Horizontal:

[∑ M

B

=0

]

Ah (36) = (2625)(10) Ah = 729 lb

[∑ F

h

=0

]

Bh = Ah + F Bh = 725 + 2625

Page 21 of 76

SECTION 7 – SHAFT DESIGN Bh = 3354 lb Shear Diagram

M Ch = 0 M Bh = (36 )(729 ) = 26,244 in − lb

Maximum M M = MB =

(M ) + (M ) 2

Bh

(26,244)2 + (5328)2 = 26,780 in − lb 32(26,780 ) 4(926 ) 856,960 3,704 = + = + 2

BV

=

32 M 4Q + 3 π D π D2 π D3 π D2 π D3 4Q 32 M 3704 856,960 smin = − = − π D 2 π D3 π D 2 π D3 1 sm = (smax + smin ) 2 1  856,960 3704 3704 856,960  3704 = sm =  + + − π D2 π D2 π D 3  π D 2 2  π D3 1 sa = (smax − smin ) 2 856,960 sa = π D3 K f sa s se = n s m + sy SF smax =

assume K f = 1.0 at B  1  3704   1.0  856,960  964 320,916 +  = 2 + se =    2   3 D3  1.7  π D   0.85  π D  D Octahedral Shear Theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn    

Page 22 of 76

π D2

SECTION 7 – SHAFT DESIGN 1 2

 694 320,916   1 2 2 2 2 +       2 3   1 94,370 6.42   3.27   1 D  +    =  =  D + 3  + 3   3  2 2  50,000 D   D     0.577(50,000)D    72 D      By trial and error, use 1 D = 2 in 2 2

483.

The worm shown is to deliver 65.5 hp steadily at 1750 rpm. It will be integral with the shaft if the shaft size needed permits, and its pitch diameter 3 in. The 12in. pulley receives the power from a horizontal belt in which the tight tension F1 = 2.5F2 . The forces (in kips) on the worm are as shown, with the axial force taken by bearing B. The strength reduction factor for the thread roots may be taken as K f = 1.5 , shear or bending. The shaft is machined from AISI 1045, as rolled. (a) For N = 2.2 (Soderberg criterion) by the octahedral-shear theory, compute the required minimum diameter at the root of the worm thread (a first approximation). (b) What should be the diameter of the shaft 2.5 in. to the left of the centerline of the worm? (c) Select a shaft size D and check it at the pulley A.

Problem 483. Solution: For AISI 1045, as rolled s y = 59 ksi su = 96 ksi sn = 0.5su = 48 ksi sn sns 48 1 = = = s y s ys 59 1.229 63,000hp 63,000(65.5) T= = = 2358 in − lb 1750 1750 (F1 − F2 ) 12  = T 2 (2.5F2 − F2 )(6) = 2358 F2 = 262 lb F1 = 2.5F2 = 655 lb

Page 23 of 76

SECTION 7 – SHAFT DESIGN FA = F1 + F2 = 655 + 262 = 917 lb Horizontal

[∑ M

B

=0

]

(917 )(6) + (1570)(6.5) = 13Eh Eh = 1208 lb

[∑ F

h

=0

]

917 + Eh = Bh + 1570 917 + 1208 = Bh + 1570 Bh = 555 lb Shear Diagram

M Ah = 0 M Bh = (917 )(6 ) = 5502 in − lb M Ch = (1208)(6.5) = 7852 in − lb

Vertical:

3 M ′ = (2540 )  = 3810 in − lb 2 ∑ M E =0

[

]

M ′ + (1170)(6.5) = 13Bv 3810 + (1170)(6.5) = 13Bv Bv = 878 lb

Page 24 of 76

SECTION 7 – SHAFT DESIGN

[∑ F =0] v

Ev + Bv = 1170 Ev + 878 = 1170 Ev = 292 lb

Shear Diagram

Moment Diagram

M Av = 0 M Bv = 0 M Cv = 5707 in − lb

M= MA = MB = MC =

(M h )2 + (M v )2

(0)2 + (0)2 = 0 (5502)2 + (0)2 = 5502 in − lb (7852)2 + (5707)2 = 9707 in − lb

(a) Minimum diameter at the root of the warm thread. K f = K fs = 1.5 M = M C = 9707 in − lb F = 2540 lb smax =

32 M 4F 32(9707 ) 4(2540 ) 310,624 10,160 + = + = + 3 2 π Dr π Dr π Dr3 π Dr2 π Dr3 π Dr2

Page 25 of 76

SECTION 7 – SHAFT DESIGN

smin = −

310,624 10,160 + π Dr3 π Dr2

1 (smax + smin ) 2 10,160 sm = π Dr2 1 sa = (smax − smin ) 2 310,624 sa = π Dr3 K f sa s se = n s m + sy SF sm =

 1  10,160   1.5  310,624  2632 174,485 se =  +   2   3  = D2 + D3  1.229  π Dr   0.85  π Dr  r r 16T 16(2358) 12,000 ss = = = π Dr3 π Dr3 Dr3 sms = ss sas = 0 K fs sas s ses = ns sms + s ys SF

9764  1  12,000  ses =  +0= 3  3  Dr  1.229  Dr  N = 2.2 , Octahedral shear theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

 2632 174,485  2 2 1 + 2 2 2 2   2  3    1  Dr Dr   9764 1 3.635   1      +    = +  = + 2    0.577(48,000)Dr3   2.2  48,000 Dr3   2.84 Dr3    18.24 Dr       

By trial and error Dr = 2.023 in 1 say Dr = 2 in 16 (b) D – shaft diameter 2.5 in. to the left of the center line of worm

Page 26 of 76

SECTION 7 – SHAFT DESIGN 3 in 16 Figure AF 12 3 r 16 = ≈ 0.1 d  3 2.023 − 2 2   16  D 2.023 = = 1 .2 d  3 2.023 − 2 2   16  K f = K t = 1.65 r=

K fs = K ts = 1.34 at 2.5 in to the shaft M h = (917 )(6) + (362)(6.5 − 2.5) = 6950 in − lb

M v = (878)(6.5 − 2.5) = 3512 in − lb M=

(6950)2 + (3512)2

= 7787 in − lb

10,160 π D2 32 M 32(7787 ) 249,184 sa = = = π D3 π D3 π D3 K f sa s se = n s m + sy SF sm =

 1  10,160   1.65  249,184  2632 153,970 +  = se =  +   2   3 D2 D3  1.229  π D   0.85  π D  9764 ses = D3 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

 2632 153,970  2 2 1 2 2 2 2 +       2 3   1 9764 1 3.21   1   D  +    =  =  D + 3  +   3  2 2.2  48,000 18.24 D D   2.84 D 3     0.577(48,000)D         

By trial and error D = 1.9432 in 15 say D = 1 in 16

Page 27 of 76

SECTION 7 – SHAFT DESIGN

15 in = 1.9375 in 16 At the pulley A, or 3 in. right of centerline M h = (917 )(3) = 2751 in − lb

(c) Selecting D = 1

Mv = 0 M = 2751 in − lb For sled runner keyway K f = 1 .6 K fs = 1.6

sm = 0 32 M 32(2751) sa = = = 3853 psi 3 πD π (1.9375)3 K f sa s se = n s m + sy SF  1 .6  se = 0 +  (3853) = 7253 psi  0.85  9764 ses = = 1343 psi (1.9375)3 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

2 2 2   1  7253   1343   =   + N  48,000   0.577(48,000)   N = 6.30 > 2.2 , therefore o.k.

484.

A propeller shaft as shown is to receive 300 hp at 315 rpm from the right through a flexible coupling. A 16-in. pulley is used to drive an auxiliary, taking 25 hp. The belt pull FB is vertically upward. The remainder of the power is delivered to a propeller that is expected to convert 60% of it into work driving the boat, at which time the boat speed is 1500 fpm. The thrust is to be taken by the right-hand bearing. Let N = 2 ; material cold-worked stainless 410. Use the octahedral shear theory with varying stresses. (a) Determine the shaft size needed assuming no buckling. (b) Compute the equivalent column stress. Is this different enough to call for another shaft size? Compute N by the maximum shear stress theory, from both equations (8.4) and (8.11).

Page 28 of 76

SECTION 7 – SHAFT DESIGN

Problem 484. Solution: For stainless 410, cold-worked s y = 85 ksi sn = 53 ksi SF = 0.85 Belt drive 63,000hp 63,000(25) TB = = = 5000 in − lb n 315  2T  4(5000) FB = F1 + F2 = 2(F1 − F2 ) = 2 B  = = 1250 lb 16  DB  Propeller 63,000hp 63,000(300 − 25) TP = = = 55,000 in − lb n 315 Thrust Fvm = ηhp(33,000) F (1500) = (0.60)(300 − 25)(33,000) F = 3630 lb Vertical loading

[∑ M

E

=0

]

(20)(1250) = 60C C = 417 lb

[∑ F =0] v

A + C = FB A + 417 = 1250

Page 29 of 76

SECTION 7 – SHAFT DESIGN A = 833 lb

Shear Diagram

M B = (20)(833) = 16,660 in − lb Maximum T at B T = TB + TP = 60,000 in − lb (a) Shaft size assuming no buckling M = 16,660 in − lb F = 3630 lb 4F 4(3630 ) 14,520 = = sm = 2 πD π D2 π D2 32 M 32(16,660 ) 533,120 sa = = = π D3 π D3 π D3 For sled-runner keyway K f = 1 .6 K fs = 1.6

sn sns 53 1 = = = s y s ys 85 1.604 se =

K f sa sn sm + sy SF

 1  14,520   1.6  533,120  2882 319,430 + = se =  +   2   3  D2 D3  1.604  π D   0.85  π D  16 16(60,000 ) 960,000 ss = sms = = = 3 πD π D3 π D3 sas = 0 K fs sas s ses = ns sms + s ys SF

190,510  1  960,000   + 0 = ses =   2 D3  1.604  π D 

Page 30 of 76

SECTION 7 – SHAFT DESIGN 1 2

1  se   ses   =   +    N  sn   sns     N = 2 , Octahedral Shear Theory, sns = 0.577 sn 2

2

1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1 2

 2882 319,430   1 2 2 2 2 +       2 3   1 190,510 1 6.027   6.230   D  +    =  =  D +  +   3  2 2  53,000 18.39 D D 3   D 3     0.577(53,000 )D          By trial and error D = 2.6 in 5 say D = 2 in = 2.625 in 8 2

(b) Equivalent Column Stress 4F s= α π D2 Le = 12 + 60 + 10 = 82 in 1 1 k = D = (2.625) = 0.65625 in 4 4 Le 82 = = 125 > 120 k 0.65625 Use Euler’s equation 2

L  sy  e  2 85(125) k   α= = 2 = 4.486 π 2E π 30 ×103 4F 4(3630 ) s= (4.486) = 3000 psi α= 2 πD π (2.625)2 Since α > 1 , it is different enough to call for another shaft size.

(

)

Solving for N by maximum shear theory. 2882 319,430 2882 319,430 se = + = + = 18,078 psi 2 2 3 D D (2.625) (2.625)3 190,510 ses = = 10,533 psi (2.625)3 Equation (8.4)

Page 31 of 76

SECTION 7 – SHAFT DESIGN 1

1

2  2  s 2  2   18,078   2 2 τ =  ss +    = (10,533) +    = 13,880 psi  2    2     0.5sn 0.5(53,000) N= = = 1.91 τ 13,880 Equation (8.11) sns = 0.5sn 1

1

2 2 2  18,078  2  10,533  2  2 1  s   ss     =   +    =   +  ( ) N  sn   sns   53 , 000 0 . 5 53 , 000          N = 1.91

CHECK PROBLEMS 485.

A 3-in. rotating shaft somewhat as shown (482) carries a bevel gear whose mean diameter is Dm = 10 in and which is keyed (profile) to the left end. Acting on the gear are a radial force G = 1570.8 lb , a driving force Q = 3141.6 lb . The thrust force is taken by the right-hand bearing. Let a = 5 in and L = 15 in ; material, AISI C1040, annealed. Base calculations on the maximum shearing stress theory with variable stress. Compute the indicated design factor N . With the use of a sketch, indicate the exact point of which maximum normal stress occurs.

Solution: For AISI C1040, annealed, Figure AF 1 s y = 48 ksi su = 80 ksi sn = 0.5su = 40 ksi sn sns 40 1 = = = s y s ys 48 1.2 FDm (6283.2 )(10 ) T= = = 31,416 in − lb 2 2 16T 16(31,416 ) ss = = = 5926 psi π D3 π (3)3 sms = ss sas = 0 K fs sas s ses = ns sms + s ys SF  1  ses =  (5926 ) + 0 = 4940 psi  1 .2  Vertical

Page 32 of 76

SECTION 7 – SHAFT DESIGN

QDm (3141.6 )(10 ) = = 15,708 in − lb 2 2 ∑ M E =0

[

]

QDm = 5G + 15 AV 2 15,708 = 5(1570.8) + 15 AV

AV = 523.6 lb

[∑ F =0] v

AV + BV = G 523.6 + BV = 1570.8 BV = 1047.2 lb Shear Diagram

Moment Diagram

M CV = 15,708 in − lb M BV = 7854 in − lb

Page 33 of 76

SECTION 7 – SHAFT DESIGN Horizontal

[∑ M

B

=0

]

15 Ah = 5(6283.2) Ah = 2094.4 lb

[∑ F

h

=0

]

Bh = Ah + F Bh = 2094.4 + 6283.2 Bh = 8377.6 lb Shear Diagram

M Ch = 0 M Bh = (15)(2094.4 ) = 31,416 in − lb

Maximum Moment 2 M = M B2h + M Bv =

(31,416)2 + (7854)2

= 32,383 in − lb

Since thrust force is taken by the right-hand bearing sms = 0 32 M 32(32,383) sas = = = 12,217 psi π D3 π (3)3 K f sa s se = n s m + sy SF Assume K f = 1.0 at the bearing B  1 .0  se = 0 +  (12,217 ) = 14,373 psi  0.85 

Page 34 of 76

SECTION 7 – SHAFT DESIGN Maximum shear theory sns = 0.5sn 1

2 2 2 1  se   ses     =   +  N  sn   0.5sn     1

2 2 2 1  14,373   4940     =   + N  40,000   0.5(40,000)   N = 2 .3

Location of maximum normal stress

487.

A 2 7/16-in. countershaft in a machine shop transmits 52 hp at 315 rpm. It is made of AISI 1117, as rolled, and supported upon bearing A and B, 59-in. apart. Pulley C receives the power via a horizontal belt, and pulley D delivers it vertically downward, as shown. Calculate N based on the octahedral-shearstress theory considering varying stresses.

Problem 487, 488 Solution: For AISI 1117, as rolled s y = 44.3 ksi

Page 35 of 76

SECTION 7 – SHAFT DESIGN su = 70.6 ksi sn = 0.5su = 35.3 ksi sn sns 35.3 1 = = = s y s ys 44.3 1.255 SF = 0.85 63,000(52 ) T= = 10,400 in − lb 315 Pulley C  2T  4(10,400)  = FC = F1 + F2 = 2(F2 − F1 ) = 2 = 2311 lb D 18  C Pulley D  2T  4(10,400)  = FD = F1 + F2 = 2(F2 − F1 ) = 2 = 1664 lb D 25  D Horizontal

[∑ M

A

=0

]

15(2311) = 59 Bh Bh = 588 lb

[∑ F

h

=0

]

Ah + Bh = 2311 Ah + 588 = 2311 Ah = 1723 lb Shear Diagram

M Ch = (1723)(15) = 25,845 in − lb

Page 36 of 76

SECTION 7 – SHAFT DESIGN M Dh = (1723)(15) − (588)(26 ) = 10,557 in − lb

Vertical

[∑ M

B

=0

]

18(1664) = 59 Av Av = 508 lb

[∑ F =0] v

Av + Bv = 1664 508 + Bv = 1664 Bv = 1156 lb Shear Diagram

M Cv = (508)(15) = 7620 in − lb M Dv = (1156 )(18) = 20,808 in − lb M C = M C2h + M C2v =

(25,845)2 + (7620 )2

M D = M D2 h + M D2 v =

(10,557 )2 + (20,808)2

Maximum M at C M = M C = 26,945 in − lb sm = 0 32 M sa = π D3 7 D = 2 in = 2.4375 in 16 Page 37 of 76

= 26,945 in − lb = 23,333 in − lb

SECTION 7 – SHAFT DESIGN 32(26,945) = 18,952 psi π (2.4375)3 assume K f = K fs = 1.0 sa =

se =

K f sa sn sm + sy SF

(1.0)(18,952) = 22,300 psi  1  se =  (0 ) + 0.85  1.255  16T 16(10,400 ) ss = = = 3658 psi π D 3 π (2.4375)3 sms = s s = 3658 psi sas = 0 ses =

K fs sas sns sms + s ys SF

 1  ses =  (3658) + 0 = 2915 psi  1.255  Octahedral shear theory sns = 0.577 sn 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

2 2 2   1  22,300   2915   =   + N  35,300   0.577(35,300)   N = 1.544

489.

A shaft for a general-purpose gear-reduction unit supports two gears as shown. The 5.75-in. gear B receives 7 hp at 250 rpm. The 2.25-in. gear A delivers the power, with the forces on the shaft acting as shown; the gear teeth have a o 1 A B pressure angle of φ = 14 ( tan φ = h = h ). Both gears are keyed (profile) to 2 Av Bv the shaft of AISI 1141, cold rolled. (a) If the fillet radius is 1/8 in. at bearing D, where the diameter is 1 3/8 in., compute N based on the octahedral-shear-stress theory (Soderberg line). The shaft diameter at A is 1 11/16 in. What is N here?

Page 38 of 76

SECTION 7 – SHAFT DESIGN

Problem 489, 490 Solution: For AISI 1141, cold rolled s y = 90 ksi sn = 50 ksi sn sns 50 1 = = = s y s ys 90 1.8 SF = 0.85 63,000(7 ) T= = 1764 in − lb 250 16T sms = πD 3 sas = 0 Gear B:  5.75  Bv   = T = 1764 in − lb  2  Bv = 614 lb Bh = Bv tan φ = 614 tan 14.5 = 159 lb Gear A:  2.25  Av   = T = 1764 in − lb  2  Av = 1568 lb Ah = Av tan φ = 1568 tan 14.5 = 406 lb Vertical

Page 39 of 76

SECTION 7 – SHAFT DESIGN

[∑ M

D

=0

]

8Cv = 4(1568) − 3(614) Cv = 554 lb

[∑ F =0] v

Cv + Dv = Av + Bv 554 + Dv = 1568 + 614 Dv = 1628 lb Shear Diagram

M Av = (554 )(4 ) = 2216 in − lb M Dv = (614 )(3) = 1842 in − lb

Horizontal

[∑ M

C

=0

]

4(406) + 8Dh = 11(159) Dh = 16 lb

[∑ F

h

=0

]

Ch + Bh = Ah + Dh Ch + 159 = 406 + 16 Ch = 263 lb Shear Diagram

Page 40 of 76

SECTION 7 – SHAFT DESIGN M Ah = (263)(4 ) = 1052 in − lb M Dh = (159 )(3) = 477 in − lb M A = M A2h + M A2v =

(1052)2 + (2216)2

= 2453 in − lb

M D = M D2 h + M D2 v =

(477 )2 + (1842)2

= 1903 in − lb

(a) At bearing D 1 r = in 8 3 d = 1 in 8 r 0.125 = ≈ 0.10 d 1.375 D 1.375 + 0.25 = ≈ 1 .2 d 1.375 K t ≈ K f = 1 .6 K ts ≈ K fs = 1.34

M = MD sm = 0 32 M 32(1903) sa = = = 7456 psi π d 3 π (1.375)3 K f sa s se = n s m + sy SF se = 0 + sms

(1.6)(7456) = 14,035 psi

0.85 16T 16(1764 ) = = = 3456 psi 3 π D π (1.375)3

sas = 0 ses =

K fs sas sns sms + s ys SF

 1  ses =  (3456 ) + 0 = 1920 psi  1 .8  Octahedral shear theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn    

Page 41 of 76

SECTION 7 – SHAFT DESIGN 1

2 2 2   1  14,035   3456   =   +  N  50,000   0.577(50,000)   N = 3.28

(b) At A For profile keyway K f = 2.0 , K fs = 1.6 11 in = 1.6875 in 16 M = M A = 2453 in − lb sm = 0 32 M 32(2453) sa = = = 5200 psi 3 πd π (1.6875)3 K f sa s se = n s m + sy SF

d =1

se = 0 + sms

(2.0)(5200) = 12,235 psi

0.85 16T 16(1764 ) = = = 1870 psi 3 π D π (1.6875)3

sas = 0 ses =

K fs sas sns sms + s ys SF

 1  ses =  (1870 ) + 0 = 1040 psi  1 .8  Octahedral shear theory

1  se   ses   =   +  N  sn   0.577 sn   2

2

  

1 2

1

2 2 2   1  12,235   1040   =   + N  50,000   0.577(50,000)   N = 4.043

The high-speed shaft of a worm-gear speed reducer, made of carburized AISI 8620, SOQT 450 F, is subjected to a torque of 21,400 in-lb. Applied to the right

Page 42 of 76

SECTION 7 – SHAFT DESIGN end with no bending. The force on the worm has three components: a horizontal force opposing rotation of W = 6180 lb , a vertical radial force S = 1940 lb , and a rightward thrust of F = 6580 lb . The shaft has the following dimensions: a = 6 , 7 9 3 9 b = 4 , c = 10 , d = 4 , e = 2 , f = 13 , g = 11.646 , h = 10.370 , 8 16 4 16 13 D2 = 4 , D3 = 4 , D4 = 3.3469 , D5 = 3.253 , r1 = 0.098 , D1 = 3.740 , 16 3 1 r2 = r3 = , r4 = 0.098 , r5 = , all in inches. The pitch diameter of the worm, 4 16 6.923 in., is the effective diameter for the point of application of the forces. The root diameter, 5.701 in. is used for stress calculations. The left-hand bearing takes the thrust load. Calculate N based on the octahedral-shear-stress theory with varying stresses. (Data courtesy of Cleveland Worm and Gear Company.)

Problem 491 Solution: Table AT 11n For AISI 8620, SOQT 450 F s y = 120 ksi su = 167 ksi sn = 0.5su = 83.5 ksi sn sns 83.5 1 = = = s y s ys 120 1.437 SF = 0.85 T = 21,400 in − lb Vertical

Page 43 of 76

SECTION 7 – SHAFT DESIGN  6.923   6.923  M ′ = F  = 6580  = 22,777 in − lb  2   2  ∑ M A =0

[

]

22,777 + (11.646)(1940) = (11.646 + 10.370)Gv Gv = 2061 lb

[∑ F =0] v

S + Av = Gv 1940 + Av = 2061 Av = 121 lb Shear Diagram

Moment Diagram

M Av = 0 M Bv = −(121)(1.2035) = −146 in − lb M Cv = −(121)(1.2035 + 4.875) = −736 in − lb M Dv = −(121)(1.2035 + 4.875 + 5.5675) = −1409 in − lb at left side M Dv = −1409 + M ′ = −1409 + 22,777 = 21,368 in − lb at right side M Ev = 21,368 − (2061)(4.4325) = 12,233 in − lb M Fv = 12,233 − (2061)(4.5625) = 2830 in − lb M Gv = 2830 − (2061)(1.375) = 0 Horizontal

Page 44 of 76

SECTION 7 – SHAFT DESIGN

[∑ M

A

=0

]

(11.646)(6180) = (11.646 + 10.370)Gh Gh = 3269 lb

[∑ F

h

=0

]

Ah + Gv = W Ah + 3269 = 6180 Ah = 2911 lb Shear Diagram

Moment Diagram

M Ah = 0 M Bh = (2911)(1.2035) = 3500 in − lb M Ch = (2911)(1.2035 + 4.875) = 17,695 in − lb M Dh = 33,900 in − lb M Eh = 33,900 − (3269 )(4.4325) = 19,410 in − lb M Fh = 19,410 − (3269 )(4.5625) = 4495 in − lb M Fh = 4495 − (3269 )(1.375) = 0

Combined M = M h2 + M v2 MA = MB = MC = MD = MD = ME =

(0)2 + (0)2 = 0 in − lb (3500)2 + (146)2 = 3503 in − lb (17,695)2 + (736)2 = 17,710 in − lb (33,900)2 + (1409)2 = 33,930 in − lb (left) (33,900)2 + (21,368)2 = 40,073 in − lb (right) (19,410)2 + (12,233)2 = 22,944 in − lb

Page 45 of 76

SECTION 7 – SHAFT DESIGN

MF = MG =

(2830)2 + (4495)2 = 5312 in − lb (0)2 + (0)2 = 0 in − lb

Bending stresses (Maximum) At A, s A = 0 32 M B 32(3503) At B, s B = = = 682 psi π D13 π (3.740)3 32 M C 32(17,710 ) = = 1618 psi At C, sC = π D23 π (4.8125)3 32 M D 32(40,073) At D, s D = = = 2203 psi π Dr3 π (5.701)3 32 M E 32(22,944 ) At E, s E = = = 3652 psi π D33 π (4)3 32 M F 32(5312 ) At F, s F = = = 1443 psi 3 π D4 π (3.3469 )3 At G, sG = 0 Shear Stresses: 16T 16(21,400 ) ssA = ssB = = = 2083 psi π D13 π (3.740)3 16T 16(21,400 ) ssC = = = 978 psi π D23 π (4.8125)3 16T 16(21,400 ) ssD = = = 588 psi π Dr3 π (5.701)3 16T 16(21,400 ) ssE = = = 1703 psi π D33 π (4)3 16T 16(21,400 ) ssF = ssG = = = 2907 psi π D43 π (3.3469)3 Tensile stresses: F = 6580 lb 4F 4(6580 ) s′A = s′B = = = 599 psi 2 π D1 π (3.740)2 4F 4(6580 ) sC′ = = = 362 psi 2 π D2 π (4.8125)2 4F 4(6580 ) s′D = = = 258 psi 2 π Dr π (5.701)2 4F 4(6580 ) s′E = = = 524 psi 2 π D3 π (4 )2

Page 46 of 76

SECTION 7 – SHAFT DESIGN

s′E = s′F =

4F 4(6580 ) = = 748 psi 2 π D4 π (3.3469 )2

r1 0.098 = = 0.03 D1 3.740 D2 4.8125 = = 1 .3 D1 3.740 Figure AF 12 K f ≈ K t = 2 .3

At B:

K fs ≈ K ts = 1.7 K f sa sn sm + sy SF sm = s′B = 599 psi

se =

sa = s B = 682 psi

(2.3)(682 ) = 2262 psi  1  se =  (599 ) + 0.85  1.437  K fs sas s ses = ns sms + s ys SF sms = ssB = 2083 psi sas = 0  1  ses =  (2083) + 0 = 1450 psi  1.437  Octahedral shear theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

2 2 2   1  2262   1450   =   + N  83,500   0.577(83,500)   N = 24.7

r2 0.75 = = 0.16 D2 4.8125 Dr 5.701 = = 1 .2 D2 4.8125 Figure AF 12 K f ≈ K t = 1 .5

At C:

K fs ≈ K ts = 1.2

Page 47 of 76

SECTION 7 – SHAFT DESIGN

se =

K f sa sn sm + sy SF

sm = 362 psi sa = 1618 psi

(1.5)(1618) = 3107 psi  1  se =  (362 ) + 0.85  1.437  K fs sas s ses = ns sms + s ys SF sms = ssC = 978 psi sas = 0  1  ses =  (978) + 0 = 681 psi  1.437  Octahedral shear theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

2 2 2   1  3107   681   =   + N  83,500   0.577(83,500)   N = 25.1

At D: Assume K f = 1.5 as in Prob. 483 se =

K f sa sn sm + sy SF

sm = 258 psi sa = 2203 psi

(1.5)(2203) = 4067 psi  1  se =  (258) + 0.85  1.437  K fs sas s ses = ns sms + s ys SF sms = s sD = 588 psi sas = 0  1  ses =  (588) + 0 = 409 psi  1.437  Octahedral shear theory

Page 48 of 76

SECTION 7 – SHAFT DESIGN

1  se   ses   =   +  N  sn   0.577 sn   2

2

  

1 2

1

2 2 2   1  4067   409   =   + N  83,500   0.577(83,500)   N = 20.2

r3 0.75 = = 0.19 D3 4 Dr 5.701 = = 1.43 D3 4 Figure AF 12 K f ≈ K t = 1.45

At E:

K fs ≈ K ts = 1.25 K f sa sn sm + sy SF sm = s′E = 524 psi se =

sa = s E = 3652 psi

(1.45)(3652) = 6595 psi  1  se =  (524 ) + 0.85  1.437  K fs sas s ses = ns sms + s ys SF sms = s sE = 1703 psi sas = 0  1  ses =  (1703) + 0 = 1185 psi  1.437  Octahedral shear theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

2 2 2   1  6595   1185   =   + N  83,500   0.577(83,500)  

N = 12

At F:

r4 0.098 = = 0.03 D4 3.3469

Page 49 of 76

SECTION 7 – SHAFT DESIGN D3 4 = = 1 .2 D4 3.3469 Figure AF 12 K f ≈ K t = 2 .3 K fs ≈ K ts = 1.7 K f sa sn sm + sy SF sm = s′F = 748 psi

se =

sa = s F = 1443 psi

(2.3)(1443) = 4425 psi  1  se =  (748) + 0.85  1.437  K fs sas s ses = ns sms + s ys SF sms = s sF = 2907 psi sas = 0  1  ses =  (2907 ) + 0 = 2023 psi  1.437  Octahedral shear theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

2 2 2   1  4425   2023   =   + N  83,500   0.577(83,500)  

N = 14.8

Then N = 12 at r3 = 492.

3 in (E) 4

The slow-speed shaft of a speed reducer shown, made of AISI 4140, OQT 1200 F, transmits 100 hp at a speed of 388 rpm. It receives power through a 13.6 in. gear B. The force on this gear has three components: a horizontal tangential driving force Ft = 2390 lb , a vertical radial force S = 870 lb , and a thrust force Q = 598 lb taken by the right-hand bearing. The power is delivered to a belt at F that exerts a downward vertical force of 1620 lb.; sled runner keyways. Use the octahedral shear theory with the Soderberg line and compute N at sections C and D. (Data courtesy of Twin Disc Clutch Company.)

Page 50 of 76

SECTION 7 – SHAFT DESIGN

Problem 492, 493 Solution: For AISI 4140, OQT 1200 F s y = 83 ksi su = 112 ksi sn = 0.5su = 56 ksi sn sns 56 1 = = = s y s ys 83 1.482 SF = 0.85 63,000(100 ) T= = 16,237 in − lb 388 Vertical

 13.6   13.6  M ′ = Q  = (598)  = 4066.4 in − lb  2   2  ∑ M A =0

[

]

Page 51 of 76

SECTION 7 – SHAFT DESIGN 5 5 3 7 11 13 3  3  3 1 + 1 (870 ) + 1 + 1 + 3 + 1 + 1 + + 2 (1620 ) + 4066.4 8 32 32 16 4  16 8   16 8 5 3 7   3 = 1 + 1 + 3 + 1 Gv 8 32   16 8 Gv = 3573 lb

[∑ F =0] v

Av + S + F = Gv Av + 870 + 1620 = 3573 Av = 1083 lb Shear Diagram

Moment Diagram

M Av = 0  3 M Pv = −(1083)1  = −1286 in − lb  16   5 M Bv = −1286 + (− 1083)1  = −3046 in − lb at the left  8 M Bv = −3046 + 4066.4 = 1021 in − lb at the right  3 M Cv = 1021 − (1953) 3  = −5570 in − lb  8  7  M Gv = −5570 − (1953)1  = −7950 in − lb  32 

Page 52 of 76

SECTION 7 – SHAFT DESIGN  11  M Dv = −7950 + (1620 )1  = −5773 in − lb  32   13  M Ev = −5773 + (1620 )  = −4457 in − lb  16   3 M Fv = −4457 + (1620 ) 2  = 0 in − lb  4

Horizontal

[∑ M

A

=0

]

19   13   13  2 (2390 ) +  2 + 4 Gh 32   16   16 Gh = 908 lb

[∑ F

h

=0

]

Ah + Gh = Ft Ah + 908 = 2390 Ah = 1482 lb Shear Diagram

M Ah = 0  3 M Ph = (1482 )1  = 1760 in − lb  16   5 M Bh = 1760 + (1482 )1  = 4168 in − lb  8

Page 53 of 76

SECTION 7 – SHAFT DESIGN  3 M Ch = 4168 − (908) 3  = 1104 in − lb  8  7  M Ch = 1104 − (908)1  = 0 in − lb  32  M Dh = 0 in − lb M Eh = 0 in − lb M Fh = 0 in − lb

Combined M = M h2 + M v2 M A = 0 in − lb MP = MB = MC = MD = ME = MF =

(1760)2 + (1286)2 = 2180 in − lb (4168)2 + (3046)2 = 5163 in − lb (1104)2 + (5570)2 = 5678 in − lb (0)2 + (5773)2 = 5773 in − lb (0)2 + (4457 )2 = 4457 in − lb (0)2 + (0)2 = 0 in − lb

1 in = 0.125 in 8 d = 2.750 in D = 2.953 in r 0.125 = = 0.05 d 2.750 D 2.953 = = 1.10 d 2.750 Figure AF 12 K f 1 ≈ K t = 1 .9

at C: r =

K fs1 ≈ K ts = 1.3 For sled runner keyway K f 2 = 1 .6 K fs 2 = 1.6 K f = 0.8 K f 1 K f 2 = 0.8(1.9 )(1.6 ) = 2.4 K fs = 0.8 K fs1 K fs 2 = 0.8(1.3)(1.6 ) = 1.7

Page 54 of 76

SECTION 7 – SHAFT DESIGN

se =

K f sa sn sm + sy SF

4Q 4(598) = = 101 psi 2 πd π (2.750)2 32 M C 32(5678) sa = = = 2781 psi π d3 π (2.750)3 (2.4)(2781) = 7920 psi  1  se =  (101) + 0.85  1.482  K fs sas s ses = ns sms + s ys SF sm =

16T 16(16,237 ) = = 3976 psi π d 3 π (2.750)3 sas = 0

sms =

 1  ses =  (3976 ) + 0 = 2683 psi  1.482  Octahedral shear theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

2 2 2   1  7920   2683   =   + N  56,000   0.577(56,000)   N =6

1 in = 0.0625 in 16 d = 2.953 in 3 D = 3 in = 3.375 in 8 r 0.0625 = = 0.02 d 2.953 D 3.375 = = 1.14 d 2.953 Figure AF 12 K f ≈ K t = 2 .4

at D: r =

K fs ≈ K ts = 1.6 se =

K f sa sn sm + sy SF

Page 55 of 76

SECTION 7 – SHAFT DESIGN 4Q 4(598) = = 87.3 psi 2 πd π (2.953)2 32 M C 32(5773) sa = = = 2284 psi π d3 π (2.953)3 (2.4)(2284) = 6508 psi  1  se =  (87.3) + 0.85  1.482  K fs sas s ses = ns sms + s ys SF sm =

16T 16(16,237 ) = = 3211 psi π d 3 π (2.953)3 sas = 0 sms =

 1  ses =  (3211) + 0 = 2167 psi  1.482  Octahedral shear theory 1

2 2 2 1  se   ses     =   +  N  sn   0.577 sn     1

2 2 2   1  6508   2167   =   + N  56,000   0.577(56,000)   N = 7 .5

TRANSVERSE DEFLECTIONS 494.

The forces on the 2-in. steel shaft shown are A = 2 kips , C = 4 kips . Determine the maximum deflection and the shaft’s slope at D.

Problems 494-496 Solution:

Page 56 of 76

SECTION 7 – SHAFT DESIGN

[M B = 0] 2(10) + 25D = 4(15) D = 1.6 kips [Fv = 0] A+C = B + D 2 + 4 = B + 1 .6 B = 4.4 kips

Shear Diagram

Moment Diagram

M 64 M = EI Eπ D 4

M (in − kip ) M  4 4   D 10  EI 

Page 57 of 76

A 0

B -20

C 16

D 0

0

-135.8

108.6

0

SECTION 7 – SHAFT DESIGN Scale ss = 10 in in

M 200 × 10−4 , Scale sM = per in EI EI D4

Slope θ , Scale sθ = 0.2 D 4 rad in

y deflection, Scale s y = 2.0 D 4 in in

Deflection: At A: y A =

0.625 in D4

Page 58 of 76

SECTION 7 – SHAFT DESIGN

At C: yC =

0.375 in D4

Slope: 0.075 rad D4 0.0125 At B: θ = rad D4 0.05625 At D: θ = rad D4

At A: θ =

Maximum deflection: 0.625 y = yA = = 0.04 in (2 )4 Shaft’s slope at D 0.05625 θ= = 0.0035 rad (2)4 495.

The forces on the steel shaft shown are A = 2 kips , C = 4 kips . Determine the constant shaft diameter that corresponds to a maximum deflection of 0.006 in. at section C.

Solution: (see Problem 494) 0.375 yC = = 0.006 D4 D = 2.812 in 7 say D = 2 in 8 496. The forces on the steel shaft shown are A = 2 kips , C = 4 kips . Determine a constant shaft diameter that would limit the maximum deflection at section A to 0.003 in. Solution: (see Problem 494) 0.625 yA = = 0.003 D4 D = 3.80 in 7 say D = 3 in 8 497.

A steel shaft is loaded as shown and supported in bearings at R1 and R2 . Determine (a) the slopes at the bearings and (b) the maximum deflection.

Page 59 of 76

SECTION 7 – SHAFT DESIGN

Problem 497 Solution: ∑ M R1 = 0

[

]

(3000 ) 7 + 1 1  − (2100) 7 + 2 1 + 1 =  7 + 2 1 + 2 + 7  R2 8 8 R2 = −444 lb

[∑ F = 0]

R1 + R2 + 2100 = 3000 R1 − 444 + 2100 = 3000 R1 = 1344 lb Loading

Shear Diagram

Moment Diagram

Page 60 of 76

8

4

 8

4

8

SECTION 7 – SHAFT DESIGN

M A = 0 in − lb 7 M B = (1344 )  = 1176 in − lb 8 7 1 M C = (1134 ) + 1  = 2688 in − lb 8 8  1 M D = 2688 − (1656 )1  = 825 in − lb  8 M E = 825 − (1656)(1) = −831 in − lb M F = −831 + (444)(1) = −387 in − lb 7 M G = −387 + (444 )  = 0 in − lb 8

M (in − kips ) D(in ) M 10 4 EI

(

)( )

A 0 1½

B1 1.18 1½

B2 1.18 2

C 2.69 2

D1 0.83 2

D2 0.83 1¾

E -0.83 1¾

F1 -0.39 1¾

F2 -0.39 1½

G 0 1½

0

1.58

0.50

1.14

0.35

0.60

-0.60

-0.28

-0.52

0

Scale ss = 2 in in

Page 61 of 76

SECTION 7 – SHAFT DESIGN M 2 × 10−4 , Scale sM = per in EI EI D4

Slope θ , Scale sθ = 4 × 10 −4 D 4 rad in

y deflection, Scale s y = 8 × 10 −4 D 4 in in

(a) Slopes at the bearings

(

)

at R1 , θ A = 0.375 4 × 10 −4 = 1.5 × 10 −4 rad at R2 , θ G = 0 rad (b) Maximum deflection at C, yC = 0.1875(8 × 10 −4 ) = 1.5 × 10 −4 in

Page 62 of 76

SECTION 7 – SHAFT DESIGN 498.

(a) Determine the diameter of the steel shaft shown if the maximum deflection is to be 0.01 in.; C = 1.5 kips , A = 1.58 kips , L = 24 in . (b) What is the slope of the shaft at bearing D? See 479.

Problems 498, 505, 506. Solution: Vertical

[∑ M

D

=0

]

6(1.5) + 8(0.424) = 24 Ev Ev = 0.516 kip

[∑ F

v

=0

]

Dv + 0.424 = 1.5 + Ev Dv + 0.424 = 1.5 + 0.516 Dv = 1.592 kip Shear Diagram

M C = 0 ; M D = −6(1.5) = −9 in − kips M B = −9 + 8(0.092) = −8.264 in − kips

Page 63 of 76

SECTION 7 – SHAFT DESIGN M A = −8.264 + 11(0.516) = −2.588 in − kips M E = −2.588 + 5(0.516) = 0 M (in − kips ) M D 4 ×10 4 EI

(

)

C 0 0

D -9 -61.1

Scale ss = 8 in in

M 120 × 10 −4 , Scale sM = per in EI EI D4

Slope θ , Scale sθ = 0.096 D 4 rad in

Page 64 of 76

B -8.264 -56.1

A -2.588 -17.6

E 0 0

SECTION 7 – SHAFT DESIGN y deflection, Scale s y = 0.768 D 4 in in

Deflections. 0.384 yCv = in D4 0.288 y Bv = in D4 0.168 y Av = in D4 Slope 0.057 θ Dv = rad D4 Horizontal

[∑ M

D

=0

]

8(0.424 ) + 24 Eh = 19(1.58) Eh = 1.1095 kip

[∑ F

h

=0

]

Dh + Eh + 0.424 = 1.58 Dh + 1.1095 + 0.424 = 1.58

Page 65 of 76

SECTION 7 – SHAFT DESIGN Dh = 0.0465 kip Shear Diagram

Moments MC = 0 MD = 0 M B = 8(− 0.0465) = −0.372 in − kip M A = −0.372 + 11(− 0.4705) = −5.5475 in − kips M E = −5.5475 + 5(1.1095) = 0 M (in − kips ) M D 4 ×10 4 EI

(

)

C 0 0

D 0 0

Scale ss = 8 in in

M 4 × 10−4 , Scale sM = per in EI EI D4

Slope θ , Scale sθ = 0.032 D 4 rad in

Page 66 of 76

B -0.372 -2.53

A -5.5475 -37.7

E 0 0

SECTION 7 – SHAFT DESIGN

y deflection, Scale s y = 0.256 D 4 in in

Deflections. 0.064 yC h = in D4 0.072 y Bh = in D4 0.096 y Ah = in D4 Slope 0.012 θ Dh = rad D4 Resultant deflection:

(

y = yh2 + yv2

1 2

)

1 2 2

[(0.064) + (0.384) ] 2

yC =

D

4

[(0.072) + (0.288) ] D

4

[(0.096) + (0.168) ]

Slope: Page 67 of 76

=

0.297 D4

=

0.194 D4

1 2 2

2

yA =

0.390 D4

1 2 2

2

yB =

=

D

4

SECTION 7 – SHAFT DESIGN

θ = (θ + θ 2 h

1 2 2 v

)

1 2 2

[(0.012) + (0.057) ] = 2

θD

D

4

=

(a) Diameter D. Maximum deflection = yC =

0.390 = 0.01 in D4

D = 2.50 in

(b) slope of the shaft at bearing D 0.05823 0.05823 θD = = = 0.0015 rad D4 (2.5)4

CRITICAL SPEED 499.

A small, high-speed turbine has a single disk, weighing 0.85 lb., mounted at the midpoint of a 0.178-in. shaft, whose length between bearings is 6 ½ in. What is the critical speed if the shaft is considered as simply supported? Solution:

Table AT 2 3 WL3 ( 0.85)(6.5) y= = = 0.052634 in 4 3EI 6  π (0.178 )  3 30 × 10   64  

(

)

1

1

1

30  g o (∑ Wy ) 2 30  g o  2 30  386  2 nc = = = 818 rpm   = π  ∑ Wy 2  π  y  π  0.052634 

500.

The bearings on a 1 ½-in. shaft are 30 in. apart. On the shaft are three 300-lb disks, symmetrically placed 7.5 in. apart. What is the critical speed of the shaft?

Solution:

Page 68 of 76

SECTION 7 – SHAFT DESIGN

Table AT 2 Deflection of B. y B = y B1 + y B2 + y B3 y B1 =

(300)(22.5)(7.5)([ 30)2 − (22.5)2 − (7.5)2 ] = 0.01273 in 4 6  π (1.5 )  6(30 × 10 ) (30)

   64  (300)(15)(7.5)(30)2 − (15)2 − (7.5)2 = 0.01556 in y B2 =  π (1.5)4  6 30 × 106   (30)  64  2 2 2 ( 300)(7.5)(7.5)(30) − (7.5) − (7.5) y B3 = = 0.00990 in 4 6  π (1.5 )  6 30 × 10   (30) 64   y B = 0.01273 + 0.00990 + 0.01556 = 0.03819 in

[

(

]

)

[

(

]

)

Deflection of C. yC = yC1 + yC2 + yC3 yC1

2 2 2 ( 300)(7.5)(30 − 15)( [ 30) − (7.5) − (30 − 15) ] = = 0.01556 in 4 6  π (1.5 )  6(30 × 10 ) (30)

   64  (300)(15)(30 − 15)(30)2 − (15)2 − (30 − 15)2 = 0.02264 in yC2 = 4 6  π (1.5 )  6 30 ×10   (30)  64  (300)(7.5)(15)(30)2 − (7.5)2 − (15)2 = 0.01556 in yC3 =  π (1.5)4  6 30 × 106   (30)  64  yC = 0.01556 + 0.02264 + 0.01556 = 0.05376 in Deflection of D. y D = y D1 + y D2 + y D3

[

(

)

[

(

Page 69 of 76

]

]

)

SECTION 7 – SHAFT DESIGN

y D1 =

(300)(7.5)(30 − 22.5)([ 30)2 − (7.5)2 − (30 − 22.5)2 ] = 0.00990 in 4 6  π (1.5 )  6(30 × 10 ) (30)

   64  (300)(15)(30 − 22.5)(30)2 − (15)2 − (30 − 22.5)2 = 0.01556 in y D2 = 4 6  π (1.5 )  6 30 ×10   (30)  64  (300)(7.5)(22.5)(30)2 − (7.5)2 − (22.5)2 = 0.01273 in y D3 =  π (1.5)4  6 30 × 106   (30 )  64  y D = 0.00990 + 0.01556 + 0.01273 = 0.03819 in

[

(

]

)

[

(

]

)

1 2

1

30  g o (∑ Wy ) 30  g o ( y B + yC + y D )  2 nc = =   π  ∑ Wy 2  π  y B2 + yC2 + y D2  1

30  386(0.03819 + 0.05376 + 0.03819)  2 nc =   = 888 rpm π  (0.03819)2 + (0.05376)2 + (0.03819)2 

501.

A fan for an air-conditioning unit has two 50-lb. rotors mounted on a 3-in. steel shaft, each being 22 in. from an end of the shaft which is 80 in. long and simply supported at the ends. Determine (a) the deflection curve of the shaft considering its weight as well as the weight of the rotors, (b) its critical speed.

Solution:

W1 = 50 lb W3 = 50 lb π  2 W2 = (0.284 ) (3) (80 ) = 160 lb weight of shaft 4 160 w2 = = 2 lb in 80 Deflection of B. y B = y B1 + y B2 + y B3

Page 70 of 76

SECTION 7 – SHAFT DESIGN

y B1 =

(50)(50)(22)([ 80)2 − (58)2 − (22)2 ] = 0.002844 in 4 6  π (35 )  6(30 × 10 ) (80)

   64  (50)(22)(22)(80)2 − (22)2 − (22)2 = 0.002296 in y B3 = 4 6  π (3)  6 30 × 10   (80)  64  (2)(22)(80)3 − 2(80)(22)2 − (22)3 = 0.006843 in y B2 =  π (3)4  6 30 × 106    64  y B = 0.002844 + 0.006843 + 0.002296 = 0.011983 in

[

(

]

)

[

]

(

)

Deflection of C. yC = yC1 + yC2 + yC3 yC1

2 2 2 ( 50)(22)(80 − 40)( [ 80) − (22) − (80 − 40) ] = = 0.003317 in 4 6  π (35)  6(30 ×10 ) (80)

   64  2 2 2 ( 50)(22 )(40)(80) − (22) − (40) yC3 = = 0.003317 in 4 6  π (3)  6 30 × 10   (80)  64  (2)(40)(80)3 − 2(80)(40)2 − (40)3 = 0.008942 in yC2 = 4 6  π (3)  6 30 × 10    64  yC = 0.003317 + 0.008942 + 0.003317 = 0.015576 in

[

(

]

)

[

]

(

By symmetry y D = y B = 0.011983 in (a) Deflection curve

Page 71 of 76

)

SECTION 7 – SHAFT DESIGN (b) Critical speed 1

30  g o (∑ Wy ) 2 nc =   π  ∑ Wy 2 

∑Wy = (50)(0.011983) + (160)(0.015576) + (50)(0.011983) = 3.69046 ∑Wy = (50)(0.011983) + (160)(0.015576) + (50)(0.011983) = 0.053177 2

2

2

2

1

30  386(3.69046)  2 nc = = 1563 rpm π  0.053177  ASME CODE 502.

A cold-rolled transmission shaft, made of annealed AISI C1050, is to transmit a torque of 27 in-kips with a maximum bending moment of 43 in-kips. What should be the diameter according to the Code for a mild shock load?

Solution: For AISI C1050, annealed s y = 53 ksi su = 92 ksi 0.3s y = 15.9 ksi 0.18su = 16.56 ksi use τ d = 0.3s y = 15.9 ksi M = 43 in − kips T = 27 in − kips 1

2 2   16 α FD 1 + B 2   2 3   (K sT ) +  K m M + D = 8 πτ d 1 − B 4      Reduce to 1 16 2 2 2 3 D = (K sT ) + ( K m M ) πτ d 1 − B 4 For mild shock load, rotating shafts K m = 1.75 K s = 1.25 B=0 1 16 2 2 2 3 D = [(1.25)(27,000)] + [(1.75)(43,000)] π (15,900) D = 2.98 in say D = 3 in

(

)

(

)[

{

Page 72 of 76

(

)

]

}

SECTION 7 – SHAFT DESIGN

503.

A machinery shaft is to transmit 82 hp at a speed of 1150 rpm with mild shock. The shaft is subjected to a maximum bending moment of 7500 in-lb. and an axial thrust load of 15,000 lb. The material is AISI 3150, OQT 1000 F. (a) What should be the diameter when designed according to the Code? (b) Determine the corresponding conventional factor of safety (static-approach and maximum-shear theory).

Solution: For AISI 3150, OQT 1000 F s y = 130 ksi su = 151 ksi 0.3s y = 39 ksi 0.18su = 27.18 ksi use τ d = 0.18su = 27.18 ksi 63000(82 ) T= = 4492 in − lb 1150 M = 7500 in − lb F = 15,000 lb   16 α FD 1 + B 2 2 3 (K sT ) +  K m M + (a) D = πτ d 1 − B 4  8  

(

(

)

1

)   

2

2  

For mild shock load K m = 1.75 K s = 1.25 B=0 α =1 1

2  ( 16 1)(15,000)D   2  2 3 D = [(1.25)(4492)] + (1.75)(7500) +   π (27180)  8  

{

1 2 2

D 3 = 0.1874 31.53 + [13.125 + 1.875D ] D = 1.4668 in say D = 1.5 in

(b) s =

}

32 M 4F 32(7500 ) 4(15,000 ) + = + = 31,124 psi = 31.124 ksi 3 2 πD πD π (1.5)3 π (1.5)2

Page 73 of 76

SECTION 7 – SHAFT DESIGN

ss =

16T 16(4492 ) = = 6778.5 psi = 6.7785 ksi π D 3 π (1.5)3

2 2 1  s   ss   = + N  s y   s ys     Maximum shear theory s ys = 0.5s y 2 2 1  31.124   6.7785     =   + N  130   0.5(130)   N = 3.83

504.

short stub shaft, made of SAE 1035, as rolled, receives 30 hp at 300 rpm via a 12-in. spur gear, the power being delivered to another shaft through a flexible coupling. The gear is keyed midway between the bearings and its pressure angle φ = 20o . See the figure for 471. (a) Neglecting the radial component of the tooth load, determine the shaft diameter for a mild shock load. (b) Considering both tangential and radial components, compute the shaft diameter. (c) Is the difference in the foregoing results enough to change your choice of the shaft size?

Solution:

Figure for 471.

For SAE 1035, as rolled s y = 55 ksi su = 85 ksi 0.3s y = 16.5 ksi 0.18su = 15.3 ksi use τ d = 0.18su = 15.3 ksi

Page 74 of 76

SECTION 7 – SHAFT DESIGN

Data are the same as 471. From Problem 471. (a) M = 4200 in − lb = 4.2 in − kips T = 6300 in − lb = 6.3 in − kips   16 α FD 1 + B 2 2   ( ) D3 = K T + K M + s  m πτ d 1 − B 4  8   Reduce to 1 16 2 2 2 D3 = K T + K M ( ) ( ) s m πτ d 1 − B 4 For mild shock load, rotating shafts K m = 1.75 K s = 1.25 B=0 1 16 D3 = [(1.25)(6.3)]2 + [(1.75)(4.2)]2 2 π (15.3) D = 1.5306 in 9 say D = 1 in 16

(

)

(

)[

(

1

)   

2

2  

]

{

}

(b) M = 4472 in − lb = 4.472 in − kips T = 6300 in − lb = 6.3 in − kips 16 D = [(1.25)(6.3)]2 + [(1.75)(4.472)]2 π (15.3) D = 1.5461 in 9 say D = 1 in 16

{

3

1 2

}

(c) Not enough to change the shaft size. 505.

Two bearings D and E, a distance D = 24 in . Apart, support a shaft for a punch press on which are an 8-in. gear A, a 44-in. pulley B, and a flywheel C, as indicated (498). Weight of flywheel is 1500 lb.; pulley B receives the power at an angle of 45o to the right of the vertical; gear A delivers it horizontally to the right. The maximum power is 25 hp at 250 rpm is delivered, with heavy shock. For cold-finish AISI 1137, find the diameter by the ASME Code.

Solution:

Page 75 of 76

SECTION 7 – SHAFT DESIGN

Data and figure is the same as in Problem 479. Also figure is the same as in Problem 498. For AISI 1137, cold-finished s y = 93 ksi su = 103 ksi 0.3s y = 27.9 ksi 0.18su = 18.54 ksi use τ d = 0.18s u = 18.54 ksi From Problem 479 M = M B = 14,343 in − lb = 14.343 in − kips T = TA = 12,600 in − lb = 12.6 in − kips For heavy shock load K m = 2.5 K s = 1.75 B=0   16 α FD 1 + B 2 (K sT ) +  K m M + 4 πτ d 1 − B  8   1 16 2 2 2 ( ) ( ) D3 = K T + K M s m πτ d 1 − B 4 D3 =

(

)

(

)[

(

]

16 [(1.75)(12.6)]2 + [(2.5)(14.343)]2 π (18.54) D = 2.2613 in 5 say D = 2 in 16 D3 =

{

2

1 2

}

- end -

Page 76 of 76

)   2

  

1 2