SECTION-13.pdf

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SECTION 13 BEVEL GEARS

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DESIGN PROBLEMS

751. Decide upon the pitch, face, gN , material, and heat treatment of a pair of straight

bevel gears to transmit continuously and indefinitely a uniform loading of 5 hp at

900 rpm of the pinion, reasonable operating temperature, high reliability;

75.1gm ; inDp 333.3 . Pinion overhangs, gear is straddle mounted.

Solution:

( )21

22

gp rrL +=

75.1

11tan ==

g

pm

o75.29=p

pp rL =sin

2333.375.29sin =L

inL 358.3=

lbv

hpF

m

t

000,33=

( )( )fpm

nDv

pp

m 4.78512

900333.3

12===

( )lbFt 210

4.785

5000,33==

( ) tmsfd FKNVFF =

( )56.1

50

4.78550

50

50 21

21

=+

=+

=mvVF

One gear straddle, one not

2.1=mK

Table 15.2, uniform

0.1=sfN

( )( )( )( ) lbFd 3932102.10.156.1 == Wear load

2

2

2

=

rt

lcd

pw CK

C

C

s

bIDFe

inDp 333.3=

( ) inLb 0.1358.33.03.0 === Temperature factor

0.1=tK , reasonable operating temperature

Life factor for wear


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Page 2 of 17

0.1=lC for indefinite life

Reliability factor for wear

25.1=rC high reliability

Geometry factor for wear, Fig. 15.7

Assume 080.0=I

Elastic coefficient (Table 15.4)Steel on steel , 2800=eC

dw FF =

( )( )( )( )

( ) ( )( )393

25.10.1

0.1

280008.00.1333.3

2

2

2

=

cds

psiscd 370,134=

Table 15.3, use Steel, (300)

ksiscd 135=

Strength of bevel gears

rts

l

d

ds

KKK

K

P

bJsF =

Size factor, assume 71.0=sK

Life factor for strength

1=lK for indefinite life

Temperature factor,

1=tK good operating condition

Reliability factor

5.1=rK high reliability

Geometry factor for strength (Fig. 15.5)Assume 240.0=J

inb 0.1=

ds = design flexural stress

Min. BHN = 300

ksisd 19=

ds FF =

( )( )( )( )( )( )

3935.1171.0

1240.00.1000,19=

dP

11=dP

say 10=dP

so that inP

bd

0.110

1010===

( )( ) inmDD gpg 833.575.1333.3 ===


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Page 3 of 17

( )( ) 33.58833.510 === gdg DPN

say 58=gN

Use 10=dP , inb 0.1= , 58=gN

Material = steel, min. Bhn = 300

752. A pair of steel Zerol bevel gears to transmit 25 hp at 600 rpm of the pinion;

3=gm ; let 20pN teeth; highest reliability; the pinion is overhung, the gear

straddle mounted. An electric motor drives a multi-cylinder pump. (a) Decideupon the pitch, face width, diameters, and steel (with treatment) for intermittent

service. (b) The same as (a) except that indefinite life is desired.

Solution:

dd

p

pPP

ND

20==

( )fpm

P

PnDv

d

dpp

m

1000

12

60020

12=

==

LetdP

b10

=

Dynamic load

( ) tmsfd FKNVFF =

lbv

hpF

m

t

000,33=

( ) d

d

t P

P

F 6.2621000

25000,33 =

=

dd

dm

PP

PvVF

121.11

121.11

50

100050

50

502

1

21

21

+=+=

+

=+

=

Table 15.2, electric motor drives a multi-cylinder pump

Service factor, 25.1=sfN

One gear straddle, one not, 2.1=mK

( )( )( )

+=

+=

d

dd

d

dP

PPP

F121.1

13946.2622.125.1121.1

1

(a) Strength of Bevel Gears

rts

l

d

ds

KKK

K

P

bJsF =



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Page 4 of 17


Intermittent service, use 6.4=lK

Temperature factor, say 0.1=tK

Reliability factor, highest reliability

0.3=rK

Geometry factor for strength

p

g

gN

Nm =

20=pN

( ) 60203 ==pN

Fig. 15.5, 205.0=J

dPb

10=

Design flexural stress, steel

Assume ksisd 15=

ds FF =

( ) ( )( )

( )( )( )

+=

d

d

d

d

PP

P

P 121.11394

30.171.0

6.4205.010

000,15

+=

d

d

d PP

P

121.11394

408,662

814.4=dP

say 5=dP

inP

bd

0.25

1010===

Wear load for bevel gears2

2

2

=

rt

lcdpw

CK

C

C

sbIDF

e

inP

ND

d

p

p 45

20===

0.1=tK

Life factor for wear, intermittent service

5.1=lC

Reliability factor for wear, highest reliability

25.1=rC



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Page 5 of 17

20=pN , 60=gN

083.0=I

Elastic coefficient, steel on steel (Table 15.4)

2800=eC

5=d

P

dw FF =

( )( )( )( ) ( )( )

( )

+=

5

121.115394

25.10.1

5.1

2800083.024

2

2

2

cds

psiscd 730,155=

Table 15.3

Use steel, min. BHN = 360, ksiscd 160=

5=dP

inb 2=

inDp 4=

( )( ) inDmD pgg 1243 ===

steel, min. BHN = 360

(b) For indefinite life,

0.1=lK , life factor for strength

0.1=lC , life factor for wear

Strength:

rts

l

d

ds

KKK

K

P

bJsF =

ds FF =

( ) ( )( )

( )( )( )

+=

d

d

d

d

PP

P

P 121.11394

30.171.0

0.1205.010

000,15

+=

d

d

d PP

P

121.11394

437,142

799.2=dP say 3=dP

inP

bd

33.33

1010===

Wear load


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Page 6 of 17

2

2

2

=

rt

lcdpw

CK

C

C

sbIDF

e

inP

ND

d

p

p 67.63

20===

dw FF =

( )( )( )( ) ( )( )

( )

+=

3

121.113394

25.10.1

0.1

2800083.033.367.6

2

2

2

cds

psiscd 744,113=

Table 15.3

Use steel, min. BHN = 240, ksiscd 115=

3=dP

inb 33.3=

inDp 67.6=

( )( ) inDmD pgg 2067.63 ===

steel, min. BHN = 240

753. Decide upon the pitch, face, and number of teeth for two spiral-bevel gears for a

speed reducer. The input to the pinion is 20 hp at 1750 rpm; 9.1gm ; pinion

overhung, gear-straddle mounted. It is hoped not to exceed a maximum pD of 4

3/8-in.; steel gears with minimum 245 BHN on pinion and 210 BHN on gear.The gear is motor-driven, subject to miscellaneous drives involving moderate

shock; indefinite life against breakage and wear with high reliability. If the gears

designed for the foregoing data are to be subjected to intermittent service only,how much power could they be expected to transmit?

Solution:

(a)( )( )

fpmnD

vpp

m 200012

1750375.4

12===

( )lb

v

hpF

m

t 3302000

20000,33000,33===

Dynamic load

( ) tmsfd FKNVFF =

One gear straddle, one not2.1=mK

Table 15.2Motor-driven, moderate shock


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Page 7 of 17

25.1=sfN

21

21

70

70

+=

mvVF , spiral

( )254.170

2000702

1

21

=

+=VF

( )( )( )( ) lbFd 6213302.125.1254.1 ==

Wear load2

2

2

=

rt

lcdpw

CK

C

C

sbIDF

e

inDp 375.4=

Temperature Factor, 0.1=tK

Design contact stresses,245=BHN , pinion

ksiscd 116=

Life factor for wear

0.1=lC , indefinite life

Reliability factor for wear

25.1=rC , high reliability


Assume 12.0=I

Elastic coefficient, steel on steel (Table 15.4)

2800=eC

( )( )( )( )

( ) ( )( )bbFw 721

25.10.1

0.1

2800

000,11612.0375.4

2

2

2

=

=

dw FF =

621721 =b

inb 8613.0=

say ininb 875.08

7==

Strength of gear

rts

l

d

ds

KKK

K

P

bJsF =

ds = design flexural stress

min. BHN = 210

ksisd 4.15=



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Page 8 of 17


1=lK for indefinite life

Temperature factor,

1=tK

Reliability factor

5.1=rK high reliability

Geometry factor Fig. 15.6

Assume 28.0=J

( )( )( )( )( )( ) dd

sPP

F3543

5.1171.0

128.0875.0400,15=

=

ds FF =

6213543

=

dP

7.5=dP

say 6=dP

Then, 6=dP , inb8

7= , ( )( ) 266375.4 === dpp PDN

( )( ) 50269.1 === pwg NmN

(b) Intermittent service onlyStrength

rts

l

d

ds

KKK

K

P

bJsF =

psisd

400,15= (Gear)

For 6=dP , 64.0=sK

For indefinite service, 6.4=lK

0.1=tK , 5.1=rK

Geometry factor, Fig. 15.6, 26=pN , 50=gN

292.0=J

( )( )( )( )( )( )

lbFs 31425.1171.0

6.4

6

292.0875.0400,15=

=

Wear load2

2

2

=

rt

lcdpw

CKC

CsbIDF

e

inDp 375.4=

0.1=tK

ksiscd 116=

2800=eC


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Page 9 of 17

5.1=lC intermittent service

25.1=rC


26=pN , 50=gN

116.0=

I

( )( )( )( )

( ) ( )( )lbFw 1098

25.10.1

5.1

2800

000,116116.0875.0375.4

2

2

2

=

=

use dw FF =

( ) tmsfd FKNVFF =

( )( )( ) tF2.1125254.11098 =

lbFt 584=

( )( )hp

vFhp mt 35

000,33

2000584

000,33===

CHECK PROBLEMS

755. A pair of straight-bevel gears are to transmit a smooth load of 45 hp at 500 rpm

of the pinion; 3=gm . A proposed design is .15 inDg = , .8

32 inb = , 4=dP .

Teeth are carburized AISI 8620, SOQT 450 F. The pinion overhangs, the gear is

straddle-mounted. Would these gears be expected to perform with high reliabilityin continuous service? If not would you expect more than 1 failure in 100?

Solution:

inm

DD

g

g

p 53

15===

( )( )fpm

nDv

pp

m 65512

5005

12===

( )lb

v

hpF

m

t 2267655

45000,33000,33===

Dynamic load

( ) tmsfd FKNVFF =

( )512.150

65550

50

50 21

21

=+

=+

=mv

VF

One gear straddle, one not

2.1=mK

Smooth load, 0.1=sfN

( )( )( )( ) lbFd 411322672.10.1512.1 ==


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Page 10 of 17

Strength of bevel gears

rts

l

d

ds

KKK

K

P

bJsF =

Size factor, for 4=dP ,

71.0=s

K


1=lK

Temperature factor,

1=tK

Geometry factor for strength (Fig. 15.5)

( )( ) 2054 === pdp DPN

( )( ) 60154 === gdg DPN

205.0=J

ksisd 30= (55 63 Rc) for carburized teeth

( )( )( )( )( )( ) rr

sKK

F5143

171.0

1

4

205.0375.2000,30=

=

ds FF =

41135143

=

rK

5.125.1


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41136364

2=

rC

25.1244.1 =rC , high reliability

Since 5.1

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SECTION-13.pdf