Machine Design

SECTION 7 – SHAFT DESIGN

of 76

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 o20=φ ; 5.1=N 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

ksisy 55=

ksisu 85=

( ) ksiss un 5.42855.05.0 ===

φcosWA =

( )lbin

n

hpT −=== 6300

300

30000,63000,63

2

ADT =

( )2

126300

A=

lbA 1050=

φcosWA =

20cos1050 W=

lbW 1118=

Shear stress

( )33

63001616

dd

Tss

ππ==

3

800,100

dss mss

π==

0=ass


of 76

bending stress

From Table AT 2

4

FLM =

(a) Negligible R :

( )( )lbin

ALM −=== 4200

4

161050

4

( )333

400,13442003232

ddd

Ms

πππ===

0=ms

3

400,134

dssa

π==

SF

sKs

s

ss

af

m

y

ne +=

For profile keyway

0.2=fK

6.1=fsK

85.0=SF

( )( )( )( ) 33

661,100

85.0

400,1340.2

ddSF

sKs

af

e ===π

SF

sKs

s

ss

asfs

ms

ys

nses +=

294.1

1

55

5.42===

y

n

ys

ns

s

s

s

s

33

796,24800,100

294.1

1

dds

s

ss ms

ys

nses =

==

π

Octahedral-shear theory

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

3 500,42577.0

796,24

500,42

661,100

5.1

1

+

=

dd

ind 569.1=


of 76

use ind16

111=

(b) Considering both radial and tangential component.

( )( )lbin

WLM −=== 4472

4

161118

4

( )333

104,14344723232

ddd

Ms

πππ===

0=ms

3

104,143

dssa

π==

( )( )( )( ) 33

180,107

85.0

104,1430.2

ddSF

sKs

af

e ===π

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

3 500,42577.0

796,24

500,42

180,107

5.1

1

+

=

dd

ind 597.1=

use ind16

111=

(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


of 76

Solution:

For AISI 1141, cold-finished

ksisy 90=

ksisn 50=

8.1

1=

y

n

s

s

85.0=SF

n

hpT

000,63=

( )lbinT −== 000,21

600

200000,63max

( )lbinT −== 500,10

600

100000,63min

( ) ( ) lbinTTTm −=+=+= 750,15500,10000,212

1

2

1minmax

( ) ( ) lbinTTTa −=−=−= 250,5500,10000,212

1

2

1minmax

3

16

d

Tss

π=

( )33

000,252750,1516

ddsms

ππ==

( )33

000,24525016

ddsas

ππ==

SF

sKs

s

ss

asfs

ms

ys

nses +=

For profile keyway

0.2=fK

6.1=fsK

8.1

1==

y

n

ys

ns

s

s

s

s

( )( )333

894,94

85.0

000,846.1000,252

8.1

1

dddses =+

=

ππ

Bending stress, negligible radial load

lbinT −= 000,21 at 200 hp

For A:

TA =

2

20

( ) 000,2110 =A


of 76

lbA 2100= at 200 hp

For C:

TC =

2

10

( ) 000,215 =C

lbC 4200= at 200 hp

[ ]∑ = 0BM ( ) ( ) ( )152510 CDA =+

at 200 hp

( )( ) ( ) ( )( )15420025102100 =+ D

lbD 1680=

[ ]∑ = 0VF

DBCA +=+

at 200 hp

168042002100 +=+ B

lbB 4620=

At 200 hp: lbA 2100= , lbB 4620= , lbC 4200= , lbD 1680=

Shear Diagram

Maximum moment at B

( )( ) lbinM −== 000,21102100

( )333

000,672000,213232

ddd

Ms

πππ===

0=ms

3

000,672

dssa

π==

( )( )33

304,503

85.0

000,6720.20

ddSF

sKs

s

ss

af

m

y

ne =+=+=

π


of 76

3

894,94

dses =

Maximum Shear Theory

2

122

5.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

3 000,505.0

894,94

000,50

304,503

2

1

+

=

dd

ind 78.2=

use ind4

32=

475. 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 8.1=N and compute the shaft diameter, using only the tangential driving

loads for the first design.

Problem 475 – 477

Solution.

For AISI 1137, cold drawn

ksisy 93=

ksisu 103=

( ) ksiss un 5.511035.05.0 ===

806.1

1

93

5.51===

ys

ns

y

n

s

s

s

s

n

hpT

000,63=

( ) rpmrpmAin

Einn 6672000

.15

.5==


of 76

( )lbinT −== 9450

667

100000,63max

( )lbinT −== 945

667

10000,63min

( ) ( ) lbinTTTm −=+=+= 5.519794594502

1

2

1minmax

( ) ( ) lbinTTTa −=−=−= 5.425294594502

1

2

1minmax

3

16

d

Tss

π=

( )33

160,835.519716

ddsms

ππ==

( )33

040,685.425216

ddsas

ππ==

SF

sKs

s

ss

asfs

ms

ys

nses +=

For profile keyway

0.2=fK

6.1=fsK

85.0=SF

( )( )333

425,55

85.0

040,686.1160,83

806.1

1

dddses =+

=

ππ

Bending stress, using only tangential loads

For 100 hp:

lbinT −= 9450

TA =

2

15

( ) 94505.7 =A

lbA 1260=

For C:

TC =

2

10


of 76

( ) 94505 =C

lbC 1890=

[ ]∑ = 0BM CDA 14206 =+

( ) ( )1890142012606 =+ D

lbD 945=

[ ]∑ = 0VF

DBCA +=+

94518901260 +=+ B

lbB 2205=

Shear diagram

Maximum moment at B

( )( ) lbinM −== 756061260

( )333

920,24175603232

ddd

Ms

πππ===

0=ms

3

920,241

dssa

π==

( )( )33

189,181

85.0

920,2410.2

ddSF

sKs

s

ss

af

m

y

ne ==+=

π

3

425,55

dses =

Octahedral Shear Theory

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

3 500,51577.0

425,55

500,51

189,181

2

1

+

=

dd

ind 997.1=

use ind 2=


of 76

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

ksisy 93=

ksisu 103=

( ) ksiss un 5.511035.05.0 ===

806.1

1

93

5.51===

ys

ns

y

n

s

s

s

s

Pulley,

( )lbin

n

hpTA −=== 5188

425

35000,63000,63

For flat-belt

( ) ( )lb

D

TFFFFF

A

AA 692

30

51884222 1221 ==

=−=+=

Sprocket,

( )lbin

n

hpTB −=== 3706

425

25000,63000,63

For chain,

( )lb

D

TF

B

BB 741

10

370622===

Sheave,

( )lbin

n

hpTC −=== 1482

425

10000,63000,63


of 76

For V-belt,

( ) ( )lb

D

TFFFFF

C

CC 445

10

1482325.15.1 1221 ==

=−=+=

Consider shaft ABD.

35 hp

Shaft ABD

[ ]∑ = 0'DM

( ) ( ) BA FAF 4'48486 ++=++

( ) ( )7414'1269218 += A

lbA 791'=

[ ]∑ = 0VF

AFDF BA′+=′+

791741692 +=′+ D

lbD 840=′

Shear Diagram

Maximum M at A’.

( )( ) .41526926 lbinM −==

( )333

864,13241523232

ddd

Ms

πππ===

0=ms

3

864,132

dssa

π==


of 76

SF

sKs

s

ss

af

m

y

ne +=

For sled-runner keyway (Table AT 13)

6.1=fK

6.1=fsK

85.0=SF

( )( )33

610,79

85.0

864,13260.10

ddSF

sKs

s

ss

af

m

y

ne =+=+=

π

at A’ lbinTT A −== 5188

( )333

008,8351881616

ddd

Tss

πππ===

sms ss =

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

33

630,14000,83

806.1

1

ddses =

=

π

Choose a design factor of 2.0

0.2=N

von Mises-Hencky theory of failure (Octahedral shear theory)

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

3 500,51577.0

630,14

500,51

610,79

2

1

+

=

dd

ind 48.1=

use ind2

11=

Consider shaft D-C

( )lbin

n

hpTC −=== 1482

425

10000,63000,63

For V-belt,

( ) ( )lb

D

TFFFFF

C

CC 445

10

1482325.15.1 1221 ==

=−=+=


of 76

[ ]∑ = 0'CM

CFD 38 =′′

( )44538 =′′D

lbD 167=′′

[ ]∑ = 0VF

CFDC +′′=′

445167 +=′C

lbC 612=′

Shear Diagram

( )( ) lbinM −== 13368167

( )333

752,4213363232

ddd

Ms

πππ===

0=ms , ssa =

( )( )33

616,25

85.0

752,4260.10

ddSF

sKs

s

ss

af

m

y

ne =+=+=

π

at C’, lbinTC −=1482

( )333

712,2314821616

ddd

Tss

πππ===

sms ss =

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

33

41800

712,23

806.1

1

ddses =+

=

π

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

3 500,51577.0

4180

500,51

616,25

2

1

+

=

dd


of 76

ind 011.1=

use ind 1=

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 44-

in. 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 cold-

drawn AISI 1137, use a design factor of 5.2=N with the octahedral shear

theory and account for the varying stresses. Determine the shaft diameters.

Problems 479-480

Solution:

Flat-Belt Drive (B)

( )lbin

n

hpTB −=== 6300

250

25000,63000,63

( ) ( )lb

D

TFFFFF

B

BB 573

44

63004222 2121 ==

=−=+=

Gear A, Doubled hp

( )lbin

n

hpTA −=

+== 600,12

250

2525000,63000,63

( )lb

D

TF

A

AA 3150

8

600,1222===

Loading:


of 76

Vertical:

lbFB BV 40545cos57345cos ===

[ ]∑ = 0DM

( ) VV EB 24815006 =+

( ) ( ) VE24405815006 =+

lbEV 510=

[ ]∑ = 0VF

VVV BDE +=+1500

4055101500 +=+ VD

lbDV 1605=

Shear Diagram


of 76

( )( ) lbinM

VD −== 900015006

( )( ) lbinMVB −== 816051016

( )( ) lbinMVA −== 25505105

Horizontal:

lbFB Bh 40545sin57345sin ===

[ ]∑ = 0DM

Ahh FEB 19248 =+

( ) ( )315019244058 =+ hE

lbEh 2359=

[ ]∑ = 0hF

Ahhh FEBD =++

31502359405 =++hD

lbDh 386=

Shear Diagram


of 76

lbinM

hD −= 0

( )( ) lbinMhB −== 30883868

( )( ) lbinMhA −== 795,1123595

( ) ( ) ( ) ( ) lbinMMMVh AAA −=+=+= 068,122550795,11

2222

( ) ( ) ( ) ( ) lbinMMMVh BBB −=+=+= 872581603088

2222

lbinM D −= 9000

Therefore

lbinM −= 068,12max

( )333

176,386068,123232

ddd

Ms

πππ===

Maximum moment subjected to alternating tension and compression

0=ms

3

176,386

dssa

π==

SF

sKs

s

ss

af

m

y

ne +=

For AISI 1137, cold-drawn,

ksisy 93=

ksisu 103=

( ) ksiss un 5.511035.05.0 ===

For sled-runner keyway (Table AT 13)

6.1=fK

6.1=fsK

85.0=SF

( )( )33

386,231

85.0

176,38660.10

ddse =+=

π


of 76

At A, 50 hp max. and 25 hp min.

50 hp

( )lbin

n

hpTA −=

+== 600,12

250

2525000,63000,63

( )lb

D

TF

A

AA 3150

8

600,1222===

( )333max

600,201600,121616

ddd

Tss

πππ===

25 hp

( )lbin

n

hpTA −=== 300,6

250

25000,63000,63

( )lb

D

TF

A

AA 1575

8

300,622===

( )333min

800,100300,61616

ddd

Tss

πππ===

( )33minmax

200,151800,100600,201

2

1

2

1

ddsss ssms

ππ=

+=+=

( )33minmax

400,50800,100600,201

2

1

2

1

ddsss ssas

ππ=

−=−=

SF

sKs

s

ss

asfs

ms

ys

nses +=

( )( )333

848,56

85.0

400,506.1200,151

806.1

1

dddses =+

=

ππ

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

3 500,51577.0

848,56

500,51

386,231

2

1

+

=

dd

ind 14.2=

say ind16

32=

THRUST LOADS


of 76

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 40<kLe .

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 2=N based on the maximum shear theory with varying stress.

Compute the shaft diameter.

Solution:

For cold-drawn monel shaft, Table AT 10

ksisy 75=

ksisn 42= at 108

at 2 x 105

ksisn 23.71102

1042

085.0

5

8

=

×≈

053.1

1

75

23.71===

ys

ns

y

n

s

s

s

s

( )lbin

n

hpT −=== 800,16

1500

400000,63000,63

( )333

800,268800,161616

DDD

Tss

πππ===

sms ss =

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

85.0=SF

assume 0.1== fsf KK

33

255,810

800,268

053.1

1

DDses =+

=

π

hpFvm η=

000,33

( )( )( )( ) fpmhrmiftknotmphknotsvm 3041min6015280152.130 ==

( ) ( )( )40070.0000,33

3041=

F

lbF 3040=

( )222

160,12304044

DDD

Fs

πππ===

ssm=

0=as


of 76

SF

sKs

s

ss

af

m

y

ne +=

22

36760

160,12

053.1

1

DDse =+

=

π

Maximum Shear Theory

2

122

5.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

2 230,715.0

255,81

230,71

3676

2

1

+

=

DD

2

12

3

2

2

2815.2

377.19

1

2

1

+

=

DD

By trial and error

ininD16

11166.1 ==

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 , lbG 580= , lbQ 926= ; which are the rectangular

components of the total reaction between the teeth; inDm 24= , inL 36= ,

ina 10= . Let the material be AISI C1045, cold drawn; 2=N . Considering

varying stresses and using the octahedral shear theory, determine the shaft

diameter.

Problems 482, 485, 486.

Solution:

For AISI C1045, cold drawn

ksisy 85=

ksisu 100=


of 76

( ) ksiss un 501005.05.0 ===

85.0=SF

7.1

1

85

50===

ys

ns

y

n

s

s

s

s

( )lbin

hpT −=== 500,31

600

300000,63

600

000,63

( )333

000,504500,311616

DDD

Tss

πππ===

sms ss =

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

33

370,940

000,504

7.1

1

DDses =+

=

π

TD

F m =

2

500,312

24=

F

lbF 2625=

Vertical:

lbinD

Q m −=

=

112,11

2

24926

2

lbG 580=

[ ]0∑ =BM �

( ) ( ) 0102

36 =−+− GQD

A mv

( ) ( )36102

vm AG

QD+=

( ) ( )3610580112,11 vA+=

lbAv 148=


of 76

[ ]0∑ =vF

lbBA vv 580=+

lbBv 580148 =+

lbBv 432=

Shear Diagram

Moment Diagram

lbinM

vC −= 112,11

lbinMvB −= 5328

Horizontal:

[ ]0∑ =BM

( ) ( )( )10262536 =hA

lbAh 729=

[ ]0∑ =hF

FAB hh +=

2625725 +=hB


of 76

lbBh 3354=

Shear Diagram

0=

hCM

( )( ) lbinMhB −== 244,2672936

Maximum M

( ) ( ) ( ) ( ) lbinMMMMVh BBB −=+=+== 780,265328244,26

2222

( ) ( )232323max

704,3960,8569264780,2632432

DDDDD

Q

D

Ms

ππππππ+=+=+=

3232min

960,8563704324

DDD

M

D

Qs

ππππ−=−=

( )minmax2

1sssm +=

23223

3704960,85637043704960,856

2

1

DDDDDsm

πππππ=

−++=

( )minmax2

1sssa −=

3

960,856

Dsa

π=

SF

sKs

s

ss

af

m

y

ne +=

assume 0.1=fK at B

3232

916,320964960,856

85.0

0.13704

7.1

1

DDDDse +=

+

=

ππ

Octahedral Shear Theory

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N


of 76

( )

2

12

3

2

32

2

1

2

3

2

32 27.342.6

72

1

000,50577.0

370,94

000,50

916,320694

2

1

+

+=

+

+

=DDDD

DD

By trial and error, use

inD2

12=

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 12-

in. pulley receives the power from a horizontal belt in which the tight tension

21 5.2 FF = . 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 5.1=fK , shear or bending. The shaft is machined from AISI 1045, as

rolled. (a) For 2.2=N (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

ksisy 59=

ksisu 96=

ksiss un 485.0 ==

229.1

1

59

48===

ys

ns

y

n

s

s

s

s

( )lbin

hpT −=== 2358

1750

5.65000,63

1750

000,63

( ) TFF =

−

2

1221

( )( ) 235865.2 22 =− FF

lbF 2622 =

lbFF 6555.2 21 ==


of 76

lbFFFA 91726265521 =+=+=

Horizontal

[ ]0∑ =BM

( )( ) ( )( ) hE135.615706917 =+

lbEh 1208=

[ ]0∑ =hF

1570917 +=+ hh BE

15701208917 +=+ hB

lbBh 555=

Shear Diagram

0=

hAM

( )( ) lbinMhB −== 55026917

( )( ) lbinMhC −== 78525.61208

Vertical:

( ) lbinM −=

=′ 3810

2

32540

[ ]0∑ =EM

( )( ) vBM 135.61170 =+′

( )( ) vB135.611703810 =+

lbBv 878=


of 76

[ ]0∑ =vF

1170=+ vv BE

1170878 =+vE

lbEv 292=

Shear Diagram

Moment Diagram

0=

vAM

0=vBM

lbinMvC −= 5707

( ) ( )22

vh MMM +=

( ) ( ) 00022

=+=AM

( ) ( ) lbinM B −=+= 55020550222

( ) ( ) lbinMC −=+= 97075707785222

(a) Minimum diameter at the root of the warm thread.

5.1== fsf KK

lbinMM C −== 9707

lbF 2540=

( ) ( )232323max

160,10624,31025404970732432

rrrrrr DDDDD

F

D

Ms

ππππππ+=+=+=


of 76

23min

160,10624,310

rr DDs

ππ+−=

( )minmax2

1sssm +=

2

160,10

r

mD

sπ

=

( )minmax2

1sssa −=

3

624,310

r

aD

sπ

=

SF

sKs

s

ss

af

m

y

ne +=

3232

485,1742632624,310

85.0

5.1160,10

229.1

1

rrrr

eDDDD

s +=

+

=

ππ

( )333

000,1223581616

rrr

sDDD

Ts ===

ππ

sms ss =

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

33

97640

000,12

229.1

1

rr

esDD

s =+

=

2.2=N , Octahedral shear theory

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

32

2

1

2

3

2

32

84.2

1635.3

24.18

1

000,48577.0

9764

000,48

485,1742632

2.2

1

+

+=

+

+

=rrrr

rr

DDDD

DD

By trial and error

inDr 023.2=

say inDr16

12=

(b) D – shaft diameter 2.5 in. to the left of the center line of worm


of 76

inr16

3=

Figure AF 12

1.0

16

322023.2

16

3

≈

−

=d

r

2.1

16

322023.2

023.2=

−

=d

D

65.1== tf KK

34.1== tsfs KK

at 2.5 in to the shaft

( )( ) ( )( ) lbinM h −=−+= 69505.25.63626917

( )( ) lbinM v −=−= 35125.25.6878

( ) ( ) lbinM −=+= 77873512695022

2

160,10

Dsm

π=

( )333

184,24977873232

DDD

Msa

πππ===

SF

sKs

s

ss

af

m

y

ne +=

3232

970,1532632184,249

85.0

65.1160,10

229.1

1

DDDDse +=

+

=

ππ

3

9764

Dses =

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

32

2

1

2

3

2

32

84.2

121.3

24.18

1

000,48577.0

9764

000,48

970,1532632

2.2

1

+

+=

+

+

=DDDD

DD

By trial and error

inD 9432.1=

say inD16

151=


of 76

(c) Selecting ininD 9375.116

151 ==

At the pulley A, or 3 in. right of centerline

( )( ) lbinM h −== 27513917

0=vM

lbinM −= 2751

For sled runner keyway

6.1=fK

6.1=fsK

0=ms

( )( )

psiD

Msa 3853

9375.1

2751323233

===ππ

SF

sKs

s

ss

af

m

y

ne +=

( ) psise 7253385385.0

6.10 =

+=

( )psises 1343

9375.1

97643

==

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

000,48577.0

1343

000,48

72531

+

=

N

2.230.6 >=N , 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 BF 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 2=N ; 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).


of 76

Problem 484.

Solution:

For stainless 410, cold-worked

ksisy 85=

ksisn 53=

85.0=SF

Belt drive

( )lbin

n

hpTB −=== 5000

315

25000,63000,63

( ) ( )lb

D

TFFFFF

B

BB 1250

16

50004222 2121 ==

=−=+=

Propeller

( )lbin

n

hpTP −=

−== 000,55

315

25300000,63000,63

Thrust

( )000,33hpFvm η=

( ) ( )( )( )000,332530060.01500 −=F

lbF 3630=

Vertical loading

[ ]0∑ =EM

( )( ) C60125020 =

lbC 417=

[ ]0∑ =vF

BFCA =+

1250417 =+A


of 76

lbA 833=

Shear Diagram

( )( ) lbinM B −== 660,1683320

Maximum T at B

lbinTTT PB −=+= 000,60

(a) Shaft size assuming no buckling

lbinM −= 660,16

lbF 3630=

( )222

520,14363044

DDD

Fsm

πππ===

( )333

120,533660,163232

DDD

Msa

πππ===

For sled-runner keyway

6.1=fK

6.1=fsK

604.1

1

85

53===

ys

ns

y

n

s

s

s

s

SF

sKs

s

ss

af

m

y

ne +=

3232

430,3192882120,533

85.0

6.1520,14

604.1

1

DDDDse +=

+

=

ππ

( )333

000,960000,601616

DDDss mss

πππ====

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

32

510,1900

000,960

604.1

1

DDses =+

=

π


of 76

2

122

1

+

=

ns

es

n

e

s

s

s

s

N

2=N , Octahedral Shear Theory, nns ss 577.0=

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

12

3

2

32

2

1

2

3

2

32 230.6027.6

39.18

1

000,53577.0

510,190

000,53

430,3192882

2

1

+

+=

+

+

=DDDD

DD

By trial and error

inD 6.2=

say ininD 625.28

52 ==

(b) Equivalent Column Stress

απ 2

4

D

Fs =

inLe 82106012 =++=

( ) inDk 65625.0625.24

1

4

1===

12012565625.0

82>==

k

Le

Use Euler’s equation

( )( )

486.41030

1258532

2

2

2

=×

=

=ππ

αE

k

Ls e

y

( )( )

( ) psiD

Fs 3000486.4

625.2

36304422

===π

απ

Since 1>α , it is different enough to call for another shaft size.

Solving for N by maximum shear theory.

( ) ( )psi

DDse 078,18

625.2

430,319

625.2

2882430,31928823232

=+=+=

( )psises 533,10

625.2

510,1903

==

Equation (8.4)


of 76

( ) psis

ss 880,132

078,18533,10

2

2

12

22

12

2 =

+=

+=τ

( )91.1

880,13

000,535.05.0===

τns

N

Equation (8.11) nns ss 5.0=

( )

2

1222

122

000,535.0

533,10

000,53

078,181

+

=

+

=

ns

s

n s

s

s

s

N

91.1=N

CHECK PROBLEMS

485. A 3-in. rotating shaft somewhat as shown (482) carries a bevel gear whose mean

diameter is inDm 10= and which is keyed (profile) to the left end. Acting on the

gear are a radial force lbG 8.1570= , a driving force lbQ 6.3141= . The thrust

force is taken by the right-hand bearing. Let ina 5= and inL 15= ; 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

ksisy 48=

ksisu 80=

ksiss un 405.0 ==

2.1

1

48

40===

ys

ns

y

n

s

s

s

s

( )( )lbin

FDT m −=== 416,31

2

102.6283

2

( )( )

psiD

Tss 5926

3

416,31161633

===ππ

sms ss =

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

( ) psises 4940059262.1

1=+

=

Vertical


of 76

( )( )

lbinQDm −== 708,15

2

106.3141

2

[ ]0∑ =EM

Vm AG

QD155

2+=

( ) VA158.15705708,15 +=

lbAV 6.523=

[ ]0∑ =vF

GBA VV =+

8.15706.523 =+ VB

lbBV 2.1047=

Shear Diagram

Moment Diagram

lbinMVC −= 708,15

lbinMVB −= 7854


of 76

Horizontal

[ ]0∑ =BM

( )2.6283515 =hA

lbAh 4.2094=

[ ]0∑ =hF

FAB hh +=

2.62834.2094 +=hB

lbBh 6.8377=

Shear Diagram

0=

hCM

( )( ) lbinMhB −== 416,314.209415

Maximum Moment

( ) ( ) lbinMMM BvBh−=+=+= 383,327854416,31

2222

Since thrust force is taken by the right-hand bearing

0=mss

( )( )

psiD

Msas 217,12

3

383,32323233

===ππ

SF

sKs

s

ss

af

m

y

ne +=

Assume 0.1=fK at the bearing B

( ) psise 373,14217,1285.0

0.10 =

+=


of 76

Maximum shear theory nns ss 5.0=

2

122

5.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

000,405.0

4940

000,40

373,141

+

=

N

3.2=N

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-shear-

stress theory considering varying stresses.

Problem 487, 488

Solution:

For AISI 1117, as rolled

ksisy 3.44=


of 76

ksisu 6.70=

ksiss un 3.355.0 ==

255.1

1

3.44

3.35===

ys

ns

y

n

s

s

s

s

85.0=SF

( )lbinT −== 400,10

315

52000,63

Pulley C

( ) ( )lb

D

TFFFFF

C

C 231118

400,104222 1221 ==

=−=+=

Pulley D

( ) ( )lb

D

TFFFFF

D

D 166425

400,104222 1221 ==

=−=+=

Horizontal

[ ]0∑ =AM

( ) hB59231115 =

lbBh 588=

[ ]0∑ =hF

2311=+ hh BA

2311588 =+hA

lbAh 1723=

Shear Diagram

( )( ) lbinM

hC −== 845,25151723


of 76

( )( ) ( )( ) lbinMhD −=−= 557,1026588151723

Vertical

[ ]0∑ =BM

( ) vA59166418 =

lbAv 508=

[ ]0∑ =vF

1664=+ vv BA

1664508 =+ vB

lbBv 1156=

Shear Diagram

( )( ) lbinMvC −== 762015508

( )( ) lbinMvD −== 808,20181156

( ) ( ) lbinMMMvh CCC −=+=+= 945,267620845,25

2222

( ) ( ) lbinMMMvh DDD −=+=+= 333,23808,20557,10

2222

Maximum M at C

lbinMM C −== 945,26

0=ms

3

32

D

Msa

π=

ininD 4375.216

72 ==


of 76

( )( )

psisa 952,184375.2

945,26323

==π

assume 0.1== fsf KK

SF

sKs

s

ss

af

m

y

ne +=

( ) ( )( )psise 300,22

85.0

952,180.10

255.1

1=+

=

( )( )

psiD

Tss 3658

4375.2

400,10161633

===ππ

psiss sms 3658==

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

( ) psises 291503658255.1

1=+

=

Octahedral shear theory nns ss 577.0=

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

300,35577.0

2915

300,35

300,221

+

=

N

544.1=N

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

pressure angle of

o

2

114=φ (

v

h

v

h

B

B

A

A==φtan ). Both gears are keyed (profile) to

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?


of 76

Problem 489, 490

Solution:

For AISI 1141, cold rolled

ksisy 90=

ksisn 50=

8.1

1

90

50===

ys

ns

y

n

s

s

s

s

85.0=SF

( )lbinT −== 1764

250

7000,63

3

16

D

Tsms

π=

0=ass

Gear B:

lbinTBv −==

1764

2

75.5

lbBv 614=

lbBB vh 1595.14tan614tan === φ

Gear A:

lbinTAv −==

1764

2

25.2

lbAv 1568=

lbAA vh 4065.14tan1568tan === φ

Vertical


of 76

[ ]0∑ =DM

( ) ( )6143156848 −=vC

lbCv 554=

[ ]0∑ =vF

vvvv BADC +=+

6141568554 +=+ vD

lbDv 1628=

Shear Diagram

( )( ) lbinMvA −== 22164554

( )( ) lbinMvD −== 18423614

Horizontal

[ ]0∑ =CM

( ) ( )1591184064 =+ hD

lbDh 16=

[ ]0∑ =hF

hhhh DABC +=+

16406159 +=+hC

lbCh 263=

Shear Diagram


of 76

( )( ) lbinMhA −== 10524263

( )( ) lbinMhD −== 4773159

( ) ( ) lbinMMMvh AAA −=+=+= 245322161052

2222

( ) ( ) lbinMMMvh DDD −=+=+= 19031842477

2222

(a) At bearing D

inr8

1=

ind8

31=

10.0375.1

125.0≈=

d

r

2.1375.1

25.0375.1≈

+=

d

D

6.1=≈ ft KK

34.1=≈ fsts KK

DMM =

0=ms

( )( )

psid

Msa 7456

375.1

1903323233

===ππ

SF

sKs

s

ss

af

m

y

ne +=

( )( )psise 035,14

85.0

74566.10 =+=

( )( )

psiD

Tsms 3456

375.1

1764161633

===ππ

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

( ) psises 1920034568.1

1=+

=

Octahedral shear theory

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N


of 76

( )

2

122

000,50577.0

3456

000,50

035,141

+

=

N

28.3=N

(b) At A

For profile keyway

0.2=fK , 6.1=fsK

inind 6875.116

111 ==

lbinMM A −== 2453

0=ms

( )( )

psid

Msa 5200

6875.1

2453323233

===ππ

SF

sKs

s

ss

af

m

y

ne +=

( )( )psise 235,12

85.0

52000.20 =+=

( )( )

psiD

Tsms 1870

6875.1

1764161633

===ππ

0=ass

SF

sKs

s

ss

asfs

ms

ys

nses +=

( ) psises 1040018708.1

1=+

=


2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

000,50577.0

1040

000,50

235,121

+

=

N

043.4=N

THRUST LOADS

491. 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


of 76

end with no bending. The force on the worm has three components: a horizontal

force opposing rotation of lbW 6180= , a vertical radial force lbS 1940= , and a

rightward thrust of lbF 6580= . The shaft has the following dimensions: 6=a ,

8

74=b , 10=c ,

16

94=d ,

4

32=e ,

16

913=f , 646.11=g , 370.10=h ,

740.31 =D , 16

1342 =D , 43 =D , 3469.34 =D , 253.35 =D , 098.01 =r ,

4

332 == rr , 098.04 =r ,

16

15 =r , all in inches. The pitch diameter of the worm,

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

ksisy 120=

ksisu 167=

ksiss un 5.835.0 ==

437.1

1

120

5.83===

ys

ns

y

n

s

s

s

s

85.0=SF

lbinT −= 400,21

Vertical


of 76

lbinFM −=

=

=′ 777,22

2

923.66580

2

923.6

[ ]0∑ =AM

( )( ) ( ) vG370.10646.111940646.11777,22 +=+

lbGv 2061=

[ ]0∑ =vF

vv GAS =+

20611940 =+ vA

lbAv 121=

Shear Diagram

Moment Diagram

0=

vAM

( )( ) lbinMvB −−=−= 1462035.1121

( )( ) lbinMvC −−=+−= 736875.42035.1121

( )( ) lbinMvD −−=++−= 14095675.5875.42035.1121 at left side

lbinMMvD −=+−=′+−= 368,21777,2214091409 at right side

( )( ) lbinMvE −=−= 233,124325.42061368,21

( )( ) lbinMvF −=−= 28305625.42061233,12

( )( ) 0375.120612830 =−=vGM

Horizontal


of 76

[ ]0∑ =AM

( )( ) ( ) hG370.10646.116180646.11 +=

lbGh 3269=

[ ]0∑ =hF

WGA vh =+

61803269 =+hA

lbAh 2911=

Shear Diagram

Moment Diagram

0=

hAM

( )( ) lbinMhB −== 35002035.12911

( )( ) lbinMhC −=+= 695,17875.42035.12911

lbinMhD −= 900,33

( )( ) lbinMhE −=−= 410,194325.43269900,33

( )( ) lbinMhF −=−= 44955625.43269410,19

( )( ) 0375.132694495 =−=hFM

Combined

22

vh MMM +=

( ) ( ) lbinM A −=+= 00022

( ) ( ) lbinM B −=+= 3503146350022

( ) ( ) lbinMC −=+= 710,17736695,1722

( ) ( ) lbinM D −=+= 930,331409900,3322

(left)

( ) ( ) lbinM D −=+= 073,40368,21900,3322

(right)

( ) ( ) lbinM E −=+= 944,22233,12410,1922


of 76

( ) ( ) lbinM F −=+= 53124495283022

( ) ( ) lbinMG −=+= 00022

Bending stresses (Maximum)

At A, 0=As

At B, ( )

( )psi

D

Ms B

B 682740.3

3503323233

1

===ππ

At C, ( )

( )psi

D

Ms C

C 16188125.4

710,17323233

2

===ππ

At D, ( )( )

psiD

Ms

r

DD 2203

701.5

073,40323233

===ππ

At E, ( )

( )psi

D

Ms E

E 36524

944,22323233

3

===ππ

At F, ( )

( )psi

D

Ms F

F 14433469.3

5312323233

4

===ππ

At G, 0=Gs

Shear Stresses:

( )( )

psiD

Tss sBsA 2083

740.3

400,21161633

1

====ππ

( )( )

psiD

TssC 978

8125.4

400,21161633

2

===ππ

( )( )

psiD

Ts

r

sD 588701.5

400,21161633

===ππ

( )( )

psiD

TssE 1703

4

400,21161633

3

===ππ

( )( )

psiD

Tss sGsF 2907

3469.3

400,21161633

4

====ππ

Tensile stresses: lbF 6580=

( )( )

psiD

Fss BA 599

740.3

65804422

1

===′=′ππ

( )( )

psiD

FsC 362

8125.4

65804422

2

===′ππ

( )( )

psiD

Fs

r

D 258701.5

65804422

===′ππ

( )( )

psiD

FsE 524

4

65804422

3

===′ππ


of 76

( )( )

psiD

Fss FE 748

3469.3

65804422

4

===′=′ππ

At B: 03.0740.3

098.0

1

1 ==D

r

3.1740.3

8125.4

1

2 ==D

D

Figure AF 12

3.2=≈ tf KK

7.1=≈ tsfs KK

SF

sKs

s

ss

af

m

y

ne +=

psiss Bm 599=′=

psiss Ba 682==

( ) ( )( )psise 2262

85.0

6823.2599

437.1

1=+

=

SF

sKs

s

ss

asfs

ms

ys

nses +=

psiss sBms 2083==

0=ass

( ) psises 145002083437.1

1=+

=


2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

500,83577.0

1450

500,83

22621

+

=

N

7.24=N

At C: 16.08125.4

75.0

2

2 ==D

r

2.18125.4

701.5

2

==D

Dr

Figure AF 12

5.1=≈ tf KK

2.1=≈ tsfs KK


of 76

SF

sKs

s

ss

af

m

y

ne +=

psism 362=

psisa 1618=

( ) ( )( )psise 3107

85.0

16185.1362

437.1

1=+

=

SF

sKs

s

ss

asfs

ms

ys

nses +=

psiss sCms 978==

0=ass

( ) psises 6810978437.1

1=+

=


2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

500,83577.0

681

500,83

31071

+

=

N

1.25=N

At D:

Assume 5.1=fK as in Prob. 483

SF

sKs

s

ss

af

m

y

ne +=

psism 258=

psisa 2203=

( ) ( )( )psise 4067

85.0

22035.1258

437.1

1=+

=

SF

sKs

s

ss

asfs

ms

ys

nses +=

psiss sDms 588==

0=ass

( ) psises 4090588437.1

1=+

=



of 76

2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

500,83577.0

409

500,83

40671

+

=

N

2.20=N

At E: 19.04

75.0

3

3 ==D

r

43.14

701.5

3

==D

Dr

Figure AF 12

45.1=≈ tf KK

25.1=≈ tsfs KK

SF

sKs

s

ss

af

m

y

ne +=

psiss Em 524=′=

psiss Ea 3652==

( ) ( )( )psise 6595

85.0

365245.1524

437.1

1=+

=

SF

sKs

s

ss

asfs

ms

ys

nses +=

psiss sEms 1703==

0=ass

( ) psises 118501703437.1

1=+

=


2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

500,83577.0

1185

500,83

65951

+

=

N

12=N

At F: 03.03469.3

098.0

4

4 ==D

r


of 76

2.13469.3

4

4

3 ==D

D

Figure AF 12

3.2=≈ tf KK

7.1=≈ tsfs KK

SF

sKs

s

ss

af

m

y

ne +=

psiss Fm 748=′=

psiss Fa 1443==

( ) ( )( )psise 4425

85.0

14433.2748

437.1

1=+

=

SF

sKs

s

ss

asfs

ms

ys

nses +=

psiss sFms 2907==

0=ass

( ) psises 202302907437.1

1=+

=


2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

500,83577.0

2023

500,83

44251

+

=

N

8.14=N

Then 12=N at inr4

33 = (E)

492. 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 lbFt 2390= , a vertical radial force lbS 870= , and a thrust force

lbQ 598= 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.)


of 76

Problem 492, 493

Solution:

For AISI 4140, OQT 1200 F

ksisy 83=

ksisu 112=

ksiss un 565.0 ==

482.1

1

83

56===

ys

ns

y

n

s

s

s

s

85.0=SF

( )lbinT −== 237,16

388

100000,63

Vertical

( ) lbinQM −=

=

=′ 4.4066

2

6.13598

2

6.13

[ ]0∑ =AM


of 76

( ) ( )

vG

+++=

+

+++++++

+

32

71

8

33

8

51

16

31

4.406616204

32

16

13

32

111

32

71

8

33

8

51

16

31870

8

51

16

31

lbGv 3573=

[ ]0∑ =vF

vv GFSA =++

35731620870 =++vA

lbAv 1083=

Shear Diagram

Moment Diagram

0=

vAM

( ) lbinMvP −−=

−= 1286

16

311083

( ) lbinMvB −−=

−+−= 3046

8

5110831286 at the left

lbinMvB −=+−= 10214.40663046 at the right

( ) lbinMvC −−=

−= 5570

8

3319531021

( ) lbinMvG −−=

−−= 7950

32

7119535570


of 76

( ) lbinMvD −−=

+−= 5773

32

11116207950

( ) lbinMvE −−=

+−= 4457

16

1316205773

( ) lbinMvF −=

+−= 0

4

3216204457

Horizontal

[ ]0∑ =AM

( ) hG

++

32

194

16

1322390

16

132

lbGh 908=

[ ]0∑ =hF

thh FGA =+

2390908 =+hA

lbAh 1482=

Shear Diagram

0=

hAM

( ) lbinMhP −=

= 1760

16

311482

( ) lbinMhB −=

+= 4168

8

5114821760


of 76

( ) lbinMhC −=

−= 1104

8

339084168

( ) lbinMhC −=

−= 0

32

719081104

lbinMhD −= 0

lbinMhE −= 0

lbinMhF −= 0

Combined

22

vh MMM +=

lbinM A −= 0

( ) ( ) lbinM P −=+= 21801286176022

( ) ( ) lbinM B −=+= 51633046416822

( ) ( ) lbinMC −=+= 56785570110422

( ) ( ) lbinM D −=+= 57735773022

( ) ( ) lbinM E −=+= 44574457022

( ) ( ) lbinM F −=+= 00022

at C: ininr 125.08

1==

ind 750.2=

inD 953.2=

05.0750.2

125.0==

d

r

10.1750.2

953.2==

d

D

Figure AF 12

9.11 =≈ tf KK

3.11 =≈ tsfs KK

For sled runner keyway

6.12 =fK

6.12 =fsK

( )( ) 4.26.19.18.08.0 21 === fff KKK

( )( ) 7.16.13.18.08.0 21 === fsfsfs KKK


of 76

SF

sKs

s

ss

af

m

y

ne +=

( )( )

psid

Qsm 101

750.2

5984422

===ππ

( )( )

psid

Ms C

a 2781750.2

5678323233

===ππ

( ) ( )( )psise 7920

85.0

27814.2101

482.1

1=+

=

SF

sKs

s

ss

asfs

ms

ys

nses +=

( )( )

psid

Tsms 3976

750.2

237,16161633

===ππ

0=ass

( ) psises 268303976482.1

1=+

=


2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

000,56577.0

2683

000,56

79201

+

=

N

6=N

at D: ininr 0625.016

1==

ind 953.2=

ininD 375.38

33 ==

02.0953.2

0625.0==

d

r

14.1953.2

375.3==

d

D

Figure AF 12

4.2=≈ tf KK

6.1=≈ tsfs KK

SF

sKs

s

ss

af

m

y

ne +=


of 76

( )( )

psid

Qsm 3.87

953.2

5984422

===ππ

( )( )

psid

Ms C

a 2284953.2

5773323233

===ππ

( ) ( )( )psise 6508

85.0

22844.23.87

482.1

1=+

=

SF

sKs

s

ss

asfs

ms

ys

nses +=

( )( )

psid

Tsms 3211

953.2

237,16161633

===ππ

0=ass

( ) psises 216703211482.1

1=+

=


2

122

577.0

1

+

=

n

es

n

e

s

s

s

s

N

( )

2

122

000,56577.0

2167

000,56

65081

+

=

N

5.7=N

TRANSVERSE DEFLECTIONS

494. The forces on the 2-in. steel shaft shown are kipsA 2= , kipsC 4= . Determine

the maximum deflection and the shaft’s slope at D.

Problems 494-496

Solution:


of 76

[ ]0=BM

( ) ( )15425102 =+ D

kipsD 6.1=

[ ]0=vF

DBCA +=+

6.142 +=+ B

kipsB 4.4=

Shear Diagram

Moment Diagram

4

64

DE

M

EI

M

π=

A B C D

( )kipinM − 0 -20 16 0

44 10DEI

M

0 -135.8 108.6 0


of 76

Scale ininss 10=

EI

M, Scale inper

Ds

EIM 4

410200 −×=

Slope θ , Scale inradDs

42.0=θ

y deflection, Scale ininDsy

40.2=

Deflection:

At A: inD

yA 4

625.0=


of 76

At C: inD

yC 4

375.0=

Slope:

At A: radD

4

075.0=θ

At B: radD

4

0125.0=θ

At D: radD

4

05625.0=θ

Maximum deflection:

( )inyy A 04.0

2

625.04

===

Shaft’s slope at D

( )rad0035.0

2

05625.04

==θ

495. The forces on the steel shaft shown are kipsA 2= , kipsC 4= . Determine the

constant shaft diameter that corresponds to a maximum deflection of 0.006 in. at

section C.

Solution:

(see Problem 494)

006.0375.0

4==

DyC

inD 812.2=

say inD8

72=

496. The forces on the steel shaft shown are kipsA 2= , kipsC 4= . Determine a

constant shaft diameter that would limit the maximum deflection at section A to

0.003 in.

Solution:

(see Problem 494)

003.0625.0

4==

DyA

inD 80.3=

say inD8

73=

497. A steel shaft is loaded as shown and supported in bearings at 1R and 2R .

Determine (a) the slopes at the bearings and (b) the maximum deflection.


of 76

Problem 497

Solution:

[ ]∑ = 01RM

( ) ( ) 28

72

4

12

8

71

4

12

8

72100

8

11

8

73000 R

+++=

++−

+

lbR 4442 −=

[ ]∑ = 0F

3000210021 =++ RR

300021004441 =+−R

lbR 13441 =

Loading

Shear Diagram

Moment Diagram


of 76

lbinM A −= 0

( ) lbinM B −=

= 1176

8

71344

( ) lbinMC −=

+= 2688

8

11

8

71134

( ) lbinM D −=

−= 825

8

1116562688

( )( ) lbinM E −−=−= 83111656825

( )( ) lbinM F −−=+−= 3871444831

( ) lbinMG −=

+−= 0

8

7444387

A B1 B2 C D1 D2 E F1 F2 G

( )kipsinM − 0 1.18 1.18 2.69 0.83 0.83 -0.83 -0.39 -0.39 0

( )inD 1 ½ 1 ½ 2 2 2 1 ¾ 1 ¾ 1 ¾ 1 ½ 1 ½

( )( )410EI

M 0 1.58 0.50 1.14 0.35 0.60 -0.60 -0.28 -0.52 0

Scale ininss 2=


of 76

EI

M, Scale inper

Ds

EIM 4

4102 −×=


44104 −×=θ


44108 −×=

(a) Slopes at the bearings

at 1R , ( ) radA

44 105.1104375.0 −− ×=×=θ

at 2R , radG 0=θ

(b) Maximum deflection

at C, ( ) inyC

44 105.11081875.0 −− ×=×=


of 76

498. (a) Determine the diameter of the steel shaft shown if the maximum deflection is

to be 0.01 in.; kipsC 5.1= , kipsA 58.1= , inL 24= . (b) What is the slope of the

shaft at bearing D? See 479.

Problems 498, 505, 506.

Solution:

Vertical

[ ]∑ = 0DM

( ) ( ) vE24424.085.16 =+

kipEv 516.0=

[ ]∑ = 0vF

vv ED +=+ 5.1424.0

516.05.1424.0 +=+vD

kipDv 592.1=

Shear Diagram

0=CM ; ( ) kipsinM D −−=−= 95.16

( ) kipsinM B −−=+−= 264.8092.089


of 76

( ) kipsinM A −−=+−= 588.2516.011264.8

( ) 0516.05588.2 =+−=EM

C D B A E

( )kipsinM − 0 -9 -8.264 -2.588 0

( ) 44 10×DEI

M 0 -61.1 -56.1 -17.6 0

Scale ininss 8=

EI

M, Scale inper

Ds

EIM 4

410120 −×=

Slope θ , Scale inradDs4096.0=θ


of 76


4768.0=

Deflections.

inD

yvC 4

384.0=

inD

yvB 4

288.0=

inD

yvA 4

168.0=

Slope

radDvD 4

057.0=θ

Horizontal

[ ]∑ = 0DM

( ) ( )58.11924424.08 =+ hE

kipEh 1095.1=

[ ]∑ = 0hF

58.1424.0 =++ hh ED

58.1424.01095.1 =++hD


of 76

kipDh 0465.0=

Shear Diagram

Moments

0=CM

0=DM

( ) kipinM B −−=−= 372.00465.08

( ) kipsinM A −−=−+−= 5475.54705.011372.0

( ) 01095.155475.5 =+−=EM

C D B A E

( )kipsinM − 0 0 -0.372 -5.5475 0

( ) 44 10×DEI

M 0 0 -2.53 -37.7 0

Scale ininss 8=

EI

M, Scale inper

Ds

EIM 4

4104 −×=


4032.0=θ


of 76


4256.0=

Deflections.

inD

yhC 4

064.0=

inD

yhB 4

072.0=

inD

yhA 4

096.0=

Slope

radDhD 4

012.0=θ

Resultant deflection:

( )2

122

vh yyy +=

( ) ( )[ ]44

2

122

390.0384.0064.0

DDyC =

+=

( ) ( )[ ]44

2

122

297.0288.0072.0

DDyB =

+=

( ) ( )[ ]44

2

122

194.0168.0096.0

DDyA =

+=

Slope:


of 76

( )2

122

vh θθθ +=

( ) ( )[ ]rad

DDD 44

2

122

05823.0057.0012.0=

+=θ

(a) Diameter D.

Maximum deflection = inD

yC 01.0390.0

4==

inD 50.2=

(b) slope of the shaft at bearing D

( )rad

DD 0015.0

5.2

05823.005823.044

===θ

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

( )( )

( ) ( )in

EI

WLy 052634.0

64

178.010303

5.685.0

3 4

6

33

=

×

==π

( )rpm

y

g

Wy

Wygn oo

c 818052634.0

386303030 2

1

2

12

1

2=

=

=

=

∑∑

πππ

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:


of 76

Table AT 2

Deflection of B.

321 BBBB yyyy ++=

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyB 01273.0

3064

5.110306

5.75.22305.75.223004

6

222

1=

×

−−=

π

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyB 01556.0

3064

5.110306

5.715305.7153004

6

222

2=

×

−−=

π

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyB 00990.0

3064

5.110306

5.75.7305.75.73004

6

222

3=

×

−−=

π

inyB 03819.001556.000990.001273.0 =++=

Deflection of C.

321 CCCC yyyy ++=

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyC 01556.0

3064

5.110306

15305.73015305.73004

6

222

1=

×

−−−−=

π

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyC 02264.0

3064

5.110306

153015301530153004

6

222

2=

×

−−−−=

π

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyC 01556.0

3064

5.110306

155.730155.73004

6

222

3=

×

−−=

π

inyC 05376.001556.002264.001556.0 =++=

Deflection of D.

321 DDDD yyyy ++=


of 76

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyD 00990.0

3064

5.110306

5.22305.7305.22305.73004

6

222

1=

×

−−−−=

π

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyD 01556.0

3064

5.110306

5.223015305.2230153004

6

222

2=

×

−−−−=

π

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyD 01273.0

3064

5.110306

5.225.7305.225.73004

6

222

3=

×

−−=

π

inyD 03819.001273.001556.000990.0 =++=

( ) ( ) 2

1

222

2

1

2

3030

++

++=

=

∑∑

DCB

DCBoo

cyyy

yyyg

Wy

Wygn

ππ

( )( ) ( ) ( )

rpmnc 88803819.005376.003819.0

03819.005376.003819.038630 2

1

222=

++

++=

π

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:

lbW 501 =

lbW 503 =

( ) ( ) ( ) lbW 1608034

284.02

2 =

=

π weight of shaft

inlbw 280

1602 ==

Deflection of B.

321 BBBB yyyy ++=


of 76

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyB 002844.0

8064

3510306

2258802250504

6

222

1=

×

−−=

π

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyB 002296.0

8064

310306

2222802222504

6

222

3=

×

−−=

π

( )( ) ( ) ( )( ) ( )[ ]( ) ( )

inyB 006843.0

64

310306

2222802802224

6

323

2=

×

−−=

π

inyB 011983.0002296.0006843.0002844.0 =++=

Deflection of C.

321 CCCC yyyy ++=

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyC 003317.0

8064

3510306

40802280408022504

6

222

1=

×

−−−−=

π

( )( )( ) ( ) ( ) ( )[ ]( ) ( ) ( )

inyC 003317.0

8064

310306

4022804022504

6

222

3=

×

−−=

π

( )( ) ( ) ( )( ) ( )[ ]( ) ( )

inyC 008942.0

64

310306

4040802804024

6

323

2=

×

−−=

π

inyC 015576.0003317.0008942.0003317.0 =++=

By symmetry

inyy BD 011983.0==

(a) Deflection curve


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(b) Critical speed

( ) 2

1

2

30

=

∑∑Wy

Wygn

o

cπ

( )( ) ( )( ) ( )( ) 69046.3011983.050015576.0160011983.050 =++=∑Wy

( )( ) ( )( ) ( )( ) 053177.0011983.050015576.0160011983.0502222

=++=∑Wy

( )rpmnc 1563

053177.0

69046.338630 2

1

=

=

π

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

ksisy 53=

ksisu 92=

ksisy 9.153.0 =

ksisu 56.1618.0 =

use ksisyd 9.153.0 ==τ

kipsinM −= 43

kipsinT −= 27

( )( ) ( ) 2

12

22

4

3

8

1

1

16

+++

−=

BFDMKTK

BD ms

d

α

πτ

Reduce to

( )( ) ( )[ ]2

122

4

3

1

16MKTK

BD ms

d

+−

=πτ

For mild shock load, rotating shafts

75.1=mK

25.1=sK

0=B

( )( )( )[ ] ( )( )[ ]{ }2

1223 000,4375.1000,2725.1

900,15

16+=

πD

inD 98.2=

say inD 3=


of 76

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

ksisy 130=

ksisu 151=

ksisy 393.0 =

ksisu 18.2718.0 =

use ksisud 18.2718.0 ==τ

( )lbinT −== 4492

1150

8263000

lbinM −= 7500

lbF 000,15=

(a) ( )

( ) ( ) 2

12

22

4

3

8

1

1

16

+++

−=

BFDMKTK

BD ms

d

α

πτ

For mild shock load

75.1=mK

25.1=sK

0=B

1=α

( )( )( )[ ] ( )( )

( )( ) 2

12

23

8

000,151750075.1449225.1

27180

16

++=

DD

π

[ ]{ }2

123 875.1125.1353.311874.0 DD ++=

inD 4668.1=

say inD 5.1=

(b) ( )( )

( )( )

ksipsiD

F

D

Ms 124.31124,31

5.1

000,154

5.1

7500324322323

==+=+=ππππ


of 76

( )( )

ksipsiD

Tss 7785.65.6778

5.1

4492161633

====ππ

+

=

22

1

ys

s

y s

s

s

s

N

Maximum shear theory

yys ss 5.0=

( )

+

=

22

1305.0

7785.6

130

124.311

N

83.3=N

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 o20=φ . 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

ksisy 55=

ksisu 85=

ksisy 5.163.0 =

ksisu 3.1518.0 =



of 76

Data are the same as 471.

From Problem 471.

(a) kipsinlbinM −=−= 2.44200

kipsinlbinT −=−= 3.66300

( )( ) ( ) 2

12

22

4

3

8

1

1

16

+++

−=

BFDMKTK

BD ms

d

α

πτ

Reduce to

( )( ) ( )[ ]2

122

4

3

1

16MKTK

BD ms

d

+−

=πτ

For mild shock load, rotating shafts

75.1=mK

25.1=sK

0=B

( )( )( )[ ] ( )( )[ ]{ }2

1223 2.475.13.625.1

3.15

16+=

πD

inD 5306.1=

say inD16

91=

(b) kipsinlbinM −=−= 472.44472

kipsinlbinT −=−= 3.66300

( )( )( )[ ] ( )( )[ ]{ }2

1223 472.475.13.625.1

3.15

16+=

πD

inD 5461.1=

say inD16

91=

(c) Not enough to change the shaft size.

505. Two bearings D and E, a distance inD 24= . 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:


of 76

Data and figure is the same as in Problem 479. Also figure is the same as in Problem 498.

For AISI 1137, cold-finished

ksisy 93=

ksisu 103=

ksisy 9.273.0 =

ksisu 54.1818.0 =


From Problem 479

kipsinlbinMM B −=−== 343.14343,14

kipsinlbinTT A −=−== 6.12600,12

For heavy shock load

5.2=mK

75.1=sK

0=B

( )( ) ( ) 2

12

22

4

3

8

1

1

16

+++

−=

BFDMKTK

BD ms

d

α

πτ

( )( ) ( )[ ]2

122

4

3

1

16MKTK

BD ms

d

+−

=πτ

( )( )( )[ ] ( )( )[ ]{ }2

1223 343.145.26.1275.1

54.18

16+=

πD

inD 2613.2=

say inD16

52=

- end -

Machine Design

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