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Transcript of Machine Design
![Page 1: Machine Design](https://reader036.fdocument.pub/reader036/viewer/2022081715/5451ce8bb1af9f04078b47f2/html5/thumbnails/1.jpg)
SECTION 7 – SHAFT DESIGN
Page 1 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
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SECTION 7 – SHAFT DESIGN
Page 2 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=
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SECTION 7 – SHAFT DESIGN
Page 3 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
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SECTION 7 – SHAFT DESIGN
Page 4 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
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SECTION 7 – SHAFT DESIGN
Page 5 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
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SECTION 7 – SHAFT DESIGN
Page 6 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==
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SECTION 7 – SHAFT DESIGN
Page 7 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
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SECTION 7 – SHAFT DESIGN
Page 8 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=
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SECTION 7 – SHAFT DESIGN
Page 9 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
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SECTION 7 – SHAFT DESIGN
Page 10 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==
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SECTION 7 – SHAFT DESIGN
Page 11 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 ==
=−=+=
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SECTION 7 – SHAFT DESIGN
Page 12 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
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SECTION 7 – SHAFT DESIGN
Page 13 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:
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SECTION 7 – SHAFT DESIGN
Page 14 of 76
Vertical:
lbFB BV 40545cos57345cos ===
[ ]∑ = 0DM
( ) VV EB 24815006 =+
( ) ( ) VE24405815006 =+
lbEV 510=
[ ]∑ = 0VF
VVV BDE +=+1500
4055101500 +=+ VD
lbDV 1605=
Shear Diagram
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SECTION 7 – SHAFT DESIGN
Page 15 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
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SECTION 7 – SHAFT DESIGN
Page 16 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
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SECTION 7 – SHAFT DESIGN
Page 17 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
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SECTION 7 – SHAFT DESIGN
Page 18 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
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SECTION 7 – SHAFT DESIGN
Page 19 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=
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SECTION 7 – SHAFT DESIGN
Page 20 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=
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SECTION 7 – SHAFT DESIGN
Page 21 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
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SECTION 7 – SHAFT DESIGN
Page 22 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
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SECTION 7 – SHAFT DESIGN
Page 23 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 ==
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SECTION 7 – SHAFT DESIGN
Page 24 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=
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SECTION 7 – SHAFT DESIGN
Page 25 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+=+=+=
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SECTION 7 – SHAFT DESIGN
Page 26 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
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SECTION 7 – SHAFT DESIGN
Page 27 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=
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SECTION 7 – SHAFT DESIGN
Page 28 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).
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SECTION 7 – SHAFT DESIGN
Page 29 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
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SECTION 7 – SHAFT DESIGN
Page 30 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
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SECTION 7 – SHAFT DESIGN
Page 31 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)
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SECTION 7 – SHAFT DESIGN
Page 32 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
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SECTION 7 – SHAFT DESIGN
Page 33 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
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SECTION 7 – SHAFT DESIGN
Page 34 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 =
+=
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SECTION 7 – SHAFT DESIGN
Page 35 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=
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SECTION 7 – SHAFT DESIGN
Page 36 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
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SECTION 7 – SHAFT DESIGN
Page 37 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 ==
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SECTION 7 – SHAFT DESIGN
Page 38 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?
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SECTION 7 – SHAFT DESIGN
Page 39 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
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SECTION 7 – SHAFT DESIGN
Page 40 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
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SECTION 7 – SHAFT DESIGN
Page 41 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
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SECTION 7 – SHAFT DESIGN
Page 42 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=+
=
Octahedral shear theory
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
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SECTION 7 – SHAFT DESIGN
Page 43 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
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SECTION 7 – SHAFT DESIGN
Page 44 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
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SECTION 7 – SHAFT DESIGN
Page 45 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
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SECTION 7 – SHAFT DESIGN
Page 46 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
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SECTION 7 – SHAFT DESIGN
Page 47 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=+
=
Octahedral shear theory
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
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SECTION 7 – SHAFT DESIGN
Page 48 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=+
=
Octahedral shear theory
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=+
=
Octahedral shear theory
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SECTION 7 – SHAFT DESIGN
Page 49 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=+
=
Octahedral shear theory
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
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SECTION 7 – SHAFT DESIGN
Page 50 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=+
=
Octahedral shear theory
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.)
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SECTION 7 – SHAFT DESIGN
Page 51 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
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SECTION 7 – SHAFT DESIGN
Page 52 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
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SECTION 7 – SHAFT DESIGN
Page 53 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
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SECTION 7 – SHAFT DESIGN
Page 54 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
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SECTION 7 – SHAFT DESIGN
Page 55 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=+
=
Octahedral shear theory
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 +=
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SECTION 7 – SHAFT DESIGN
Page 56 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=+
=
Octahedral shear theory
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:
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SECTION 7 – SHAFT DESIGN
Page 57 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
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SECTION 7 – SHAFT DESIGN
Page 58 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=
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SECTION 7 – SHAFT DESIGN
Page 59 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.
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SECTION 7 – SHAFT DESIGN
Page 60 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
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SECTION 7 – SHAFT DESIGN
Page 61 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=
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SECTION 7 – SHAFT DESIGN
Page 62 of 76
EI
M, Scale inper
Ds
EIM 4
4102 −×=
Slope θ , Scale inradDs
44104 −×=θ
y deflection, Scale ininDsy
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 −− ×=×=
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SECTION 7 – SHAFT DESIGN
Page 63 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
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SECTION 7 – SHAFT DESIGN
Page 64 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
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SECTION 7 – SHAFT DESIGN
Page 65 of 76
y deflection, Scale ininDsy
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
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SECTION 7 – SHAFT DESIGN
Page 66 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 −×=
Slope θ , Scale inradDs
4032.0=θ
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SECTION 7 – SHAFT DESIGN
Page 67 of 76
y deflection, Scale ininDsy
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:
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SECTION 7 – SHAFT DESIGN
Page 68 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:
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SECTION 7 – SHAFT DESIGN
Page 69 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 ++=
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SECTION 7 – SHAFT DESIGN
Page 70 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 ++=
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SECTION 7 – SHAFT DESIGN
Page 71 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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SECTION 7 – SHAFT DESIGN
Page 72 of 76
(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=
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SECTION 7 – SHAFT DESIGN
Page 73 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
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SECTION 7 – SHAFT DESIGN
Page 74 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 =
use ksisud 3.1518.0 ==τ
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SECTION 7 – SHAFT DESIGN
Page 75 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:
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SECTION 7 – SHAFT DESIGN
Page 76 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 =
use ksisud 54.1818.0 ==τ
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 -