Those are a few chapters of solution "Design of machine Element" by Virgil M. & Roy M. Faires

1. SECTION 12 – HELICAL GEARS
Page 1 of 14
DESIGN PROBLEMS
701. For continuous duty in a speed reducer, two helical gears are to be rated at 7.4 hp
at a pinion speed of 1750 rpm; 75.2≈wm ; the helix angle 15o
; 20o
F.D. teeth in
the normal plane; let 21=pN teeth, and keep pDb 2< . Determine the pitch, face,
gN , and the material and heat treatment. Use through-hardened teeth with a
maximum of 250 BHM (teeth may be cut after heat treatment).
Solution:
o
15=ψ
o
n 20=φ
12
pp
m
nD
v
π
=
dd
p
p
PP
N
D
21
==
rpmnp 1750=
( )
d
d
m
P
P
v
9621
12
1750
21
=






=
π
( )( )
d
d
m
t P
P
v
hp
F 38.25
9621
4.7000,33000,33
=






==
pDb 2≤
dd PP
b
4221
2 =





=
( )
( )
lb
CbFv
CbFv
FF
tm
tm
td
2
1
2
2
cos05.0
coscos05.0
ψ
ψψ
++
+
+=
Table AT 25
Assume 1660=C
o
15=ψ
lb
P
P
P
P
P
P
PF
d
d
d
d
d
d
dd
2
1
2
2
15cos
42
166038.25
9621
05.0
15cos15cos
42
166038.25
9621
05.0
38.25












++

















+





+=

2. SECTION 12 – HELICAL GEARS
Page 2 of 14
lb
P
P
P
P
P
P
PF
d
d
d
d
d
d
dd
2
1
65050
38.25
481
65050
38.25
465
38.25






++






+
+=
For continuous service: dw FF ≥
ψ2
cos
gp
w
QKbD
F =
( ) 467.1
175.2
75.22
1
2
=
+
=
+
=
g
g
m
m
Q
Table At 26, Bhn = 250
Sum of BHN = 500, o
n 20=φ
131=gK
( )( )
22
670,181
15cos
131467.12142
ddd
w
PPP
F =











=
dw FF ≥
By trial and error method
dP dF wF
7 3967 3708
6 4758 5046
use 6=dP
in
P
D
d
p 5.3
6
2121
===
in
P
b
d
7
6
4242
===
fpm
P
v
d
m 1604
6
96219621
===
Fig. AF 19, permissible error = 0.0018 in
Fig. AF 20
Use carefully cut gears, 6=dP
Error = 0.001 in is o.k.
For material
Strength
df
s
PK
sbY
F
ψcos
=

3. SECTION 12 – HELICAL GEARS
Page 3 of 14
23
15cos
21
cos 33
===
ψ
p
ep
N
N
Table AT 24, Load near middle
23=epN , FDn
o
20=φ
565.0=Y
assume 0.2=fK
dsfs FNF =
assume 0.2=sfN
( )( )
( )( )
( )( )24758
62
15cos565.07
=
s
psis 892,29=
use
3
u
n
s
s =
( ) psisu 676,89892,293 ==
Use C1050, OQT 1100 F,
ksisu 122= , 250248 <=BHN
Ans.
6=dP
inb 7=
( )( ) 582175.2 === pwg NmN
Material. C1050, OQT 1100 F
703. A pair of helical gears, subjected to heavy shock loading, is to transmit 50 hp at
1750 rpm of the pinion.; 25.4=gm ; o
15=ψ ; minimum .
4
3
4 inDp = ; continuous
service, 24 hr/day; 20o
F.D. teeth in the normal plane, carefully cut; through-
hardened to a maximum BHN = 350. Decide upon the pitch, face width, material
and its treatment.
Solution:
( )( ) fpmvm 2176
12
175075.4
==
π
( )( )
( )
lb
v
hp
F
m
t 758
2176
50000,33000,33
===
Dynamic load:
( )
( )
lb
CbFv
CbFv
FF
tm
tm
td
2
1
2
2
cos05.0
coscos05.0
ψ
ψψ
++
+
+=
Fig. AF 19, fpmvm 2176=
Permissible error = 0.0014 in

4. SECTION 12 – HELICAL GEARS
Page 4 of 14
Use carefully cut gears, ine 001.0= , 5=dP as standard
Table AT 25,
Steel and steel, 20o
FD
1660=C
( )( )
( ) ( )
lb
b
b
Fd
2
1
2
2
15cos1660758217605.0
15cos15cos1660758217605.0
758
++
+
+=
( )
( )
lb
b
b
Fd
2
1
8.15487588.108
8.15487581.105
758
++
+
+=
Wear load:
ψ2
cos
gp
w
QKbD
F =
( ) 619.1
125.4
25.42
1
2
=
+
=
+
=
g
g
m
m
Q
Table At 26, 20o
FD,
Sum of BHN =2(350)=700
270=gK
( )( )( ) b
b
Fw 2225
15cos
270619.175.4
2
==
dw FF ≥ , .69.4
tan
2
2min in
P
Pb
d
a ===
ψ
π
By trial and error method
b dF wF
5 5203 11125
6 5811 13350
use inb 5=
Material:
Strength:
dfdnf
s
PK
sbY
PK
sbY
F
ψcos
==
ψ3
cos
p
ep
N
N =
( )( ) 22375.45 === pdp DPN
25
15cos
22
3
==epN
Table AT 24, Load near middle

5. SECTION 12 – HELICAL GEARS
Page 5 of 14
25=epN , FDn
o
20=φ
580.0=Y
assume 7.1=fK
( )( )
( )( )
s
s
Fs 32955.0
57.1
15cos580.05
==
dsfs FNF =
for 24 hr/day service, heavy shock loading
75.1=sfN
( )( )520375.132955.0 =s
psis 629,27=
use
3
u
n
s
s =
( ) psisu 887,82629,273 ==
Table AT 9
Use 4150, OQT 1200 F,
ksisu 159= , 350331<=BHN
Ans.
5=dP
inb 5=
Material. 4150, OQT 1200 F
705. Design the teeth for two herringbone gears for a single-reduction speed reducer
with 80.3=wm . The capacity is 36 hp at 3000 rpm of the pinion; o
30=ψ ; F.D.
teeth with o
20=nφ . Since space is at a premium, the initial design is for 15=pN
teeth and carburized teeth of AISI 8620; preferably pDb 2< .
Solution:
dd
p
p
PP
N
D
15
==
pDb 2≈
d
p
P
Db
30
2 ==
12
pp
m
nD
v
π
=
( )
d
d
m
P
P
v
781,11
12
3000
15
=






=
π

6. SECTION 12 – HELICAL GEARS
Page 6 of 14
( )( )
d
d
m
t P
P
v
hp
F 101
781,11
36000,33000,33
=






==
Dynamic load
( )
( )
lb
CbFv
CbFv
FF
tm
tm
td
2
1
2
2
cos05.0
coscos05.0
ψ
ψψ
++
+
+=
o
n 20=φ
o
30=ψ
Assume 1660=C , Table AT 25, 20o
FD
lb
P
P
P
P
P
P
PF
d
d
d
d
d
d
dd
2
1
2
2
30cos
30
1660101
781,11
05.0
30cos30cos
30
1660101
781,11
05.0
101












++

















+





+=
lb
P
P
P
P
P
P
PF
d
d
d
d
d
d
dd
2
1
350,37
101
589
350,37
101
510
101






++






+
+=
Wear load
ψ2
cos
gp
w
QKbD
F =
( ) 583.1
180.3
80.32
1
2
=
+
=
+
=
g
g
m
m
Q
For AISI 8620, carburized, 20o
FD
750=gK for 1010
cycles
( )( )
22
350,712
30cos
750583.11530
ddd
w
PPP
F =











=
By trial and error, dw FF ≥
dP dF wF
5 4433 28,494
4 5454 44,522
6 3817 19,788
8 3173 11,130
9 3008 8794
For carefully cut gears, 001.0=e
fpmv 1400max = (Fig. AF 9)

7. SECTION 12 – HELICAL GEARS
Page 7 of 14
dP
d
m
P
v
781,11
=
5 2356.2
4 1963.5
6 1683
8 1473
9 1309 fpm
use 9=dP
lbFd 3008=
dw FlbF >= 5794
in
P
b
d
3.3
9
3030
===
use inb 0.3=
To check for strength
dfdnf
s
PK
sbY
PK
sbY
F
ψcos
==
ψ3
cos
p
ep
N
N =
15=pN
23
30cos
15
3
==epN
Table AT 24, Load near middle
23=epN , FDn
o
20=φ
565.0=Y
assume 7.1=fK
8620, SOQT 450, ksisu 167=
3
u
n
s
s =
5.83
2
167
2
=== u
n
s
s
( )( )( )
( )( )
( )lbFlbF ds 30088011
97.1
30cos565.00.3500,83
=>==
Designed Data:
9=dP
inb 0.3=
15=pN
( )( ) 57158.3 === pwg NmN

8. SECTION 12 – HELICAL GEARS
Page 8 of 14
in
P
N
D
d
p
p 67.1
9
15
===
in
P
N
D
d
g
g 33.6
9
57
===
CHECK PROBLEMS
707. The data for a pair of carefully cut gears are: 5=dnP , o
20=nφ , o
12=ψ ,
.5.3 inb = , 18=pN , 108=gN teeth; pinion turns 1750 rpm. Materials: pinion,
SAE 4150, OQT to BHN = 350; gear, SAE 3150, OQT to BHN = 300. Operation
is with moderate shock for 8 to 10 hr./day. What horsepower may be transmitted
continuously?
Solution:
d
p
p
P
N
D =
( ) 89.415cos5cos === ψdnd PP
inDp 681.3
89.4
18
==
Wear load
ψ2
cos
gp
w
QKbD
F =
.5.3 inb =
( ) 7143.1
10818
10822
=
+
=
+
=
gp
g
NN
N
Q
Table AT 26, o
20=nφ
Sum of BHN = 350 + 300 = 650
233=gK
( )( )( )( ) lbFw 5379
12cos
2337143.1681.35.3
2
==
Strength of gear
lb
PK
sbY
F
dnf
s =
For gear: SAE 3150, OQT to BHN = 300
ksisu 151=
( ) ksiss un 5.751515.05.0 ===
116
12cos
108
cos 33
===
ψ
g
eg
N
N

9. SECTION 12 – HELICAL GEARS
Page 9 of 14
Table AT 24, Load near middle, o
20=nφ
763.0=Y
( )( ) 6.57763.05.75 ==Ysn
For pinion: SAE 4150, OQT to BHN = 350
( ) ksiBHNsu 1753505.05.0 ===
( ) ksiss un 5.871755.05.0 ===
19
12cos
18
cos 33
===
ψ
p
ep
N
N
Table AT 24, Load near middle, o
20=nφ
534.0=Y
( )( ) 7.46534.05.87 ==Ysn
Therefore use pinion as weak
Assume 7.1=fK
( )( )( )
( )( )
lbFs 240,19
57.1
534.05.3500,87
==
For moderate shock, 8 to 10 hr./day
Use 5.1=sfN
dsfs FNF ≥
dF5.1240,19 =
lbFd 827,12≤
Therefore use lbFF wd 5379==
( )
( )
lb
CbFv
CbFv
FF
tm
tm
td
2
1
2
2
cos05.0
coscos05.0
ψ
ψψ
++
+
+=
Fig. AF 20, carefully cut gears, 5=dnP , ine 001.0=
Table AT 25, steel and steel, 20o
FD
1660=C
( )( ) fpm
nD
v pp
m 1686
12
1750681.3
12
===
ππ
( ) ( )[ ]
( ) ( )[ ]
lb
F
F
FF
t
t
td
2
1
2
2
12cos5.31660168605.0
12cos12cos5.31660168605.0
++
+
+=
[ ]
[ ]
lb
F
F
FF
t
t
td 5379
55593.84
555946.82
2
1
=
++
+
+=
Trial and error
lbFt 1800=
( )( ) hp
vF
hp mt
92
000,33
16861800
000,33
===

10. SECTION 12 – HELICAL GEARS
Page 10 of 14
708. Two helical gears are used in a single reduction speed reducer rated at 27.4 hp at
a motor speed of 1750 rpm; continuous duty. The rating allows an occasional 100
% momentary overload. The pinion has 33 teeth. 10=dnP , .2 inb = , o
20=nφ ,
o
20=ψ , 82.2=wm . For both gears, the teeth are carefully cut from SAE 1045
with BHN = 180. Compute (a) the dynamic load, (b) the endurance strength;
estimate 7.1=fK . Also decide whether or not the 100 % overload is damaging.
(c) Are these teeth suitable for continuous service? If they are not suitable
suggest a cure. (The gears are already cut.)
Solution:
d
p
p
P
N
D =
( ) 66.915cos10cos === ψdnd PP
inDp 42.3
66.9
33
==
( )( ) fpm
nD
v pp
m 1567
12
175042.3
12
===
ππ
( ) lb
v
hp
F
m
t 577
1567
4.27000,33000,33
===
(a) Dynamic load
( )
( )
lb
CbFv
CbFv
FF
tm
tm
td
2
1
2
2
cos05.0
coscos05.0
ψ
ψψ
++
+
+=
Fig. AF 20, carefully cut gears, 10=dnP , ine 001.0=
Table AT 25, steel and steel, 20o
FD
1660=C
inb 2=
( ) ( )[ ]
( ) ( )[ ]
lbFd 2578
15cos21660577156705.0
15cos15cos21660577156705.0
577
2
1
2
2
=
++
+
+=
(b) Endurance strength
lb
PK
sbY
F
dnf
s =
For SAE 1045, BHN = 180
( ) ksiBHNsu 901805.05.0 ===
( ) ksiss un 45905.05.0 ===
37
15cos
33
cos 33
===
ψ
p
ep
N
N
Table AT 24, Load near middle, o
20=nφ

11. SECTION 12 – HELICAL GEARS
Page 11 of 14
645.0=Y
7.1=fK
( )( )( )
( )( )
lb
PK
sbY
F
dnf
s 3415
107.1
645.02000,45
===
For 100 % overload
( ) lbFt 11545772 ==
( )
( )
lb
CbFv
CbFv
FF
tm
tm
td
2
1
2
2
cos05.0
coscos05.0
ψ
ψψ
++
+
+=
( ) ( )[ ]
( ) ( )[ ]
lbFd 3475
15cos216601154156705.0
15cos15cos216601154156705.0
1154
2
1
2
2
=
++
+
+=
Since ds FF ≈ , 100 % overload is not damaging
(c)
ψ2
cos
gp
w
QKbD
F =
.2 inb =
( ) 476.1
182.2
82.22
1
2
=
+
=
+
=
w
w
m
m
Q
Table AT 26, o
20=nφ
Sum of BHN = 2(180) = 360
5.62=gK
( )( )( )( ) ( )lbFlbF dw 2578676
15cos
5.62476.142.32
2
=<==
Therefore not suitable for continuous service.
Cure: Through hardened teeth
For Bhn
( ) 2385.62
676
2578
==gK
min Bhn = 0.5(650) = 325
709. Two helical gears are used in a speed reducer whose input is 100 hp at 1200 rpm,
from an internal combustion engine. Both gears are made of SAE 4140, with the
pinion heat treated to a BHN 363 – 415, and the gear to 321 – 363; let the teeth
be F.D.; 20o
pressure angle in the normal plane; carefully cut; helix angle
o
15=ψ ; 22=pN , 68=gN teeth; 5=dP , inb 4= . Calculate the dynamic load,
the endurance strength load, and the limiting wear load for the teeth. Should these
gears have long life if they operate continuously? (Data courtesy of the Twin
Disc Clutch Co.)
Solution:

12. SECTION 12 – HELICAL GEARS
Page 12 of 14
in
P
N
D
d
p
p 4.4
5
22
===
( )( ) fpm
nD
v pp
m 1382
12
12004.4
12
===
ππ
( ) lb
v
hp
F
m
t 2388
1382
100000,33000,33
===
Dynamic load
( )
( )
lb
CbFv
CbFv
FF
tm
tm
td
2
1
2
2
cos05.0
coscos05.0
ψ
ψψ
++
+
+=
Fig. AF 20, carefully cut gears, 5=dnP , ine 001.0=
Table AT 25, steel and steel, 20o
FD
1660=C
inb 4=
( ) ( )[ ]
( ) ( )[ ]
lbFd 5930
15cos416602388138205.0
15cos15cos416602388138205.0
2388
2
1
2
2
=
++
+
+=
Endurance strength load
lb
PK
sbY
F
df
s
ψcos
=
Assume 7.1=fK
Pinion
( ) ksiBHNsn 75.9036325.025.0 ===
25
15cos
22
cos 33
===
ψ
p
ep
N
N
Table AT 24, Load near middle, o
20=nφ
580.0=Y
( )( )( )
( )( )
lb
PK
sbY
F
df
s 925,23
57.1
15cos580.04750,90cos
===
ψ
Gear
( ) ksiBHNsn 25.8032125.025.0 ===
75
15cos
68
cos 33
===
ψ
p
ep
N
N
Table AT 24, Load near middle, o
20=nφ
735.0=Y
( )( )( )
( )( )
lb
PK
sbY
F
df
s 811,26
57.1
15cos735.04250,80cos
===
ψ
use lbFs 925,23=

13. SECTION 12 – HELICAL GEARS
Page 13 of 14
Limiting Wear Load
ψ2
cos
gp
w
QKbD
F =
Table AT 26, o
20=nφ
Sum of BHN = 684 to 778 use 700
270=gK
( ) 511.1
6822
6822
=
+
=
+
=
gp
g
NN
N
Q
( )( )( )( ) lbFw 7696
15cos
270511.14.44
2
==
Since ( ) ( )lbFlbF dw 59307696 =>= these gears have long life if they operate
continuously.
CROSSED HELICAL
710. Helical gears are to connect two shafts that are at right angles
( 201 =N , 402 =N , 10=dnP , o
4521 ==ψψ ). Determine the center distance.
Solution:
11
11
1 cos
cos
ψ
ψπ
DP
P
D
N dn
cn
==
( )( ) 45cos1020 1D=
inD 83.21 =
222 cosψDPN dn=
( )( ) 45cos1040 2D=
inD 66.52 =
( ) ( ) inDDC 25.466.583.22
1
212
1
=+=+=
712. Two shafts that are at right angles are to be connected by helical gears. A
tentative design is to use 201 =N , 602 =N , 10=dnP , and a center distance of 6
in. What must be the helix angles?
Solution:
o
9021 =+=Σ ψψ
1
1
1
cosψdnP
N
D =
2
2
2
cosψdnP
N
D =
( )212
1
DDC +=

14. SECTION 12 – HELICAL GEARS
Page 14 of 14
2
2
1
1
coscos
2
ψψ dndn P
N
P
N
C +=
( )
21 cos10
60
cos10
20
62
ψψ
+=
21 cos
6
cos
2
12
ψψ
+=
21 cos
3
cos
1
6
ψψ
+=
By trial and error method
11 sin
3
cos
1
6
ψψ
+=
o
5.391 =ψ
o
5.502 =ψ
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