FME 461: ENGINEERING DESIGN II
EXAMPLE-DESIGN OF INPUT GEAR-SHAFT
IDENTIFICATION OF EXTERNAL LOAD CARRIED BY GEAR SHAFT
The loading diagram of the gear shaft is shown below:
The specification of performance requirements for shaft is given below in terms of power
to be transmitted, and the speed of transmission:
Performance specification required=30 Kw*1450 r.p.m.
SELECTION OF MATERIAL FOR PART
Material selected is medium carbon steel, to British Standard specification BS
970:080M040(H&T), whose mechanical properties are:
(a) Tensile yield strength S y  385 Mpa
(b) Ultimate tensile strength Sut  625  775 Mpa
1
(1)
(c) Elongation %=16%
(d) Hardness number=179-229 HB
The material is a ductile material, having an elongation % of 16 %>5%.
DETERMINATION OF EXTERNAL LOADS CARRIED BY GEAR SHAFT
Torque required to transmit power at the speed specified
The relationship between power and torque transmitted is given by the equation:
Power  T * watts
Where
T  Torque transmitted in N-m
  Angular velocity in radians/sec
Consequently
Power  T *
Where
2N
60
watts
and T  Power *
60
2N
N-m
(2)
N  Angular velocity of shaft in revs/min
Substituting for performance specification required from (1) into (2)
T  30 * 1000 *
60
2 * 1450
N-m
(3)
Torque to be transmitted becomes
T  197.6 N  m
(4)
TRANSVERSE LOAD ON GEAR SHAFT ARISING FROM THE TORQUE
TRANSMITTED
TANGENTIAL FORCE ON GEAR TEETH F REQUIRED TO TRANSMIT
TORQUE SPECIFIED
As shown in the gear set diagram, the tangential force on the gear teeth, is the gear tooth
force operating at the pitch diameter d of the driving gear on the input shaft.
The torque transmitted is then given as function of the tangential tooth force and the pitch
diameter of the gear as below:
Torque T  F *
2
d
Therefore F  T * Newtons
d
2
2
Substituting for the pitch diameter of the input gear d  40 mm. and T=197.6 N-m
F  197 .6 *
2
Newtons
0.04
F  9880 Newtons
(5)
RESULTANT FORCE ON GEAR TOOTH REQUIRED TO TRANSMIT
TORQUE SPECIFIED
The tangential force F is the component of the resultant gear tooth force that gives rise
to the transmitted torque. It acts at the pitch point, along the tangent to the pitch circle
diameter of the gear tooth. The resultant gear tooth force is normal to the tooth surface,
and therefore inclined to the tangent to pitch circle diameter at an angle equal to the
pressure angle of the gear tooth.
The resultant force on the gear tooth is given by the equation
Re sul tan t R 
F
cos
where   pressure angle of gear tooth
R  F * sec 
Substituting for F =9880 N and  =20o
R  9880* sec 20 =10154 N
R  10154 Newtons
(6)
DETERMINE BENDING MOMENT LOADS ON THE INPUT SHAFT
LOADING DIAGRAM OF THE INPUT SHAFT-CONSIDERED AS A SIMPLY
SUPPORTED BEAM SUBJECT TO TRANSVERSE POINT LOADS
The loading diagram is shown at Appendix A and reproduced below:
Reactions at the simple supports are then given by
3
R1 
Rb
l
R2 
and
R1  R *
35
85
Ra
l
R2  R *
N , and
50
85
N
Substituting for R=10154 Newtons
R1  10154 *
35
85
N , and
R1  4181 N , and
R2  10154 *
R2  5972
50
85
N
(7)
N
SHEAR FORCE DIAGRAM
The shear force diagram is shown at appendix a. however, the direct shear stress induced
by the shear force reaches its maximum value at the centre of the shaft, while the stresses
caused by bending and torsion reach their maximum values the surface of the shaft. the
effect of the direct shear stress is therefore ignored.
BENDING MOMENT DIAGRAM FOR INPUT SHAFT
The bending moment diagram for a straight beam with intermediate load and simple
supports is then as shown below
The maximum bending moment is then given by
M AB 
Rbx
l
M max  M a  R *
and
35
* 50
85
M BC 
N  mm
Ra
l  x 
l
and
occurs at x  50 mm.
4
M max  10154 *
M max  209052
35
* 50
85
N  mm  209052
N  mm
N  mm
(8)
EXTERNAL LOAD ON THE GEAR SHAFT
The external load on the input gear shaft then reduces to combined torsion and bending,
where the torsion and bending loads are:
(9)
T  197.6 N  m  197600 N  mm
M max  209052
N  mm
(10)
Determination of stresses induced by the external loads
STRESSES DUE TO COMBINED TORSION AND BENDING OF SHAFT
In this situation, there is a plane stress at the location of maximum bending moment as
shown below
The stress elements are:
 x  normal stress due to bending 
y  0
 xy  shear stress due to torque 
32 M
d 3
16T
d3
Simplifying the loading situation of the input shaft into a static load which remains
constant in spite of the rotation of the shaft, determine the significant stress at the
location of highest stresses in terms of principal and maximum shear stress arising
from the loads on the member
Applying maximum shear stress theory of failure
5
The MAXIMUM SHEAR STRESS theory of failure states:
When Yielding occurs in any material, the maximum shear stress at the point of failure
equals or exceeds the maximum shear stress when yielding occurs in the tension test
specimen.
STRESS ELEMENTS IN THE PLANE STRESS SITUATION
The plane stress situation is the stress situation in which the stress elements are
 x ,  y , and  xy , and the stresses on the z-axis are zero,
MAXIMUM SHEAR STRESS IN TERMS OF PLANE STRESS ELEMENTS
The maximum shear stress is the significant stress in this situation and is given by the
expression
2
x  y 
   xy 2

2


 max  
(11)
Maximum shear stress in the case of plane stress situation with  y  0
THE GENERAL CASE OF PLANE STRESS SITUATION WITH  y  0
Substituting for  y  0 into the equation for maximum shear stress yields
2
x  0
2
   xy
 2 
 max  
 max 
x 2
4
  xy 2
=
2

 2
=  x    xy 2 = x   xy 2
 2 
4
1
 x 2  4 xy 2
2
(12)
Stresses induced in gear shaft by the external loads
SOLID CIRCULAR SHAFT SUBJECT TO BENDING AND TORSION
In the case of combined torsion and bending, the stress elements in the plane stress
situation are:
 x  normal stress due to bending 
y  0
 xy  shear stress due to torque 
32 M
d 3
16T
d3
MAXIMUM SHEAR STRESS IN ELEMENT IN TERMS EXTERNAL LOADS
6
Substituting for  x and  xy in equation for maximum shear stress yields the expression
for maximum shear stress in terms of load and dimension of element as shown below:
 max 
1
1
 x 2  4 xy2 
2
2
 max 
16
d
3
2
 32 M 
 16T 
 3   4 3 
 d 
 d 
2
M2 T2
(13)
SHEAR STRENGTH OF CHOSEN (DUCTILE) MATERIAL
The yield strength in shear of ductile materials such steel is predicted to be half the
tensile yield strength by the maximum shear stress theory of failure, and the shear yield
strength of such materials can therefore be derived from the tensile yield strength
Therefore S sy 
Sy
(14)
2
Where
S sy  Shear strength of the material
S y  Yield strength of the material in tension
COMPARE SIGNIFICANT STRESS WITH STRENGTH: DESIGN EQUATION
Design equation then becomes
 max 
16
d
3
M2 T2 =
S sy
f .s.

Sy
2 * f .s.
OR
d3 
16

M2 T2 *
2 * f .s.
Sy
Where
f .s.  factor of safety , and
ssy
f .s

Sy
2 * f .s
  d , the design or allowable shear stress ,
The design equation then becomes
d3 
16
M2 T2 *

2 * f .s.
Sy
(15)
SOLVING DESIGN EQUATION
Substituting the TORQUE and BENDING MOMENT loads into design equation
(16)
T  197.6 N  m
M max  209052
d3 
32

N  mm
209052 2  197600 2 *
(17)
f .s.
Sy
Substituting for yield strength of chosen material and the factor of safety
Factor of safety =2.5 and Tensile yield strength S y  385 Mpa
7
d3 
32

209052 2  197600 2 * 2.5
385
d  26.7
(18)
mm.
SELECT SHAFT SIZE FROM PREFERRED METRIC RANGE
Select the shaft size to be used form the nearest size in the range of preferred metric
sizes1 (1,1.2,1.6,2,2.5,3,4,5,6,8,10,12,16,20,25,30,35,40,45,50,55,60,65,70,75,80,90,100
mm.)
Nearest shaft size selected is 30 mm.
REVIEW DESIGN
Determine the actual factor of safety resulting from the use of the selected standard
shaft size
Rewriting the design equation in terms of the factor of safety
f .s. 
S sy
 max
But  max 

Sy
2 *  max
16
d
M2 T2
3
Substituting
T  197.6 N  m
(16)
M max  209052
N  mm
(17)
And d  30 mm
 max 
16
209052 2  197600 2  54.3 Mpa
 30 3
Substituting S y  385 Mpa and  max  54.3 Mpa
f .s. 
Sy
2
*
1
 max

385
1
*
 3.85
2
54 .3
Factor of safety =3.5
1
Shigley Joseph, Mechanical Engineering Design, First Metric Edition, , McGraw Hill, 1986, page 660
8
APPENDIX A: LOADING, SHEAR FORCE, AND BENDING MOMENT
DIAGRAMS
LOADING DIAGRAM
R1 
Rb
l
and
R2 
Ra
l
SHEAR FORCE DIAGRAM
VAB  R1
VBc   R2
and
BENDING MOMENT DIAGRAM
M AB 
Rbx
l
and
9
M BC 
Ra
l  x 
l
APPENDIX B2: MECHANICAL PROPERTIES OF SOME STEELS
Material
British
Standard3
Productio
n process
0.20C
070M20
HR4
CD5
0.30C
080M30
HR
CD
Maximum
section
size, mm.
152
254
13
76
152
254
13
63
63
150
63
63
150
63
150
100
29
150
29
152
102
64
29
64
0.40C
080M40
0.50C
080M50
1Cr
530M40
H&T6
HR
CD
H&T
HR
CD
H&T
H&T
1.5MnMo
605M36
H&T
1.25NiCr
640M40
H&T
3NiCr
653M31
H&T
1CrMo
708M40
H&T
3CrMo
722M24
H&T
150
13
152
2.5NiCrMo
826M40
H&T
150
3NiCrMo
830M31
H&T
1.5MnNiCrMo
945M38
H&T
254
152
64
152
64
29
2
Yield
Strength
Mpa
215
200
385
340
245
230
470
385
385
280
430
385
310
510
430
525
680
525
755
525
585
680
755
755
680
525
940
680
755
755
850
1020
650
680
940
525
680
850
Tensile
Strength,
Mpa
430
400
530
430
490
460
600
530
550-700
550
570
625-775
620
650
625-775
700-850
850-1000
700-850
925-1075
700-850
770-930
850-1000
930-1080
930-1080
850-1000
700-850
1075-1225
850-1000
930-1080
925-1075
1000-1150
1150-1300
850-1000
850-1000
1080-1240
690-850
850-1000
1000-1160
Shigley, Joseph E., Mechanical Engineering Design, pp. 664, McGraw-Hill Inc., 1986
British Standards Institution, BS 970: Part 1: 1983
4
HR-Hot rolled and normalised
5
CD-Cold drawn
6
H&T-Hardened and tempered
3
10
Elong
ation
%
22
20
12
14
20
19
10
12
13
16
10
16
14
10
11
17
13
17
12
17
15
13
12
12
12
17
12
13
12
12
12
10
13
12
11
17
13
12
Hardness
Number,
HB
126-179
116-170
154
125
143-192
134-183
174
154
152-207
152-207
165
179-229
179-229
188
179-229
202-255
248-302
202-255
269-331
202-255
223-277
248-302
269-331
269-331
248-302
201-255
311-375
269-331
269-331
269-331
293-352
341-401
248-302
248-302
311-375
201-255
248-302
293-352
APPENDIX C: STEEL APPLICATION AND HEAT-TREATING GUIDE7
USE
OR
PART
Arbors
Armature shafts
Axles
Ball races
Bolts and studs
Bushings
Cams
Camshaft
Cant dogs
Chain Links
Chain Pins
Chuck Jaws
Chuck screws
Clutches
Collets
Connecting Rods
Crankshafts
Drift Pins
Engine bolts
Gears
Guide Pins
Mandrels
Pinions
Pins
Pistons
Pump Shafts
Rollers
Rolls
Lead Screws
Set Screws
Spindles
Stay Bolts
Thrust washers
Turbine Shafts
Turnbuckles
U bolts
Universal Joint Pins
Universal joint bodies
Worm Gears
Low-Carbon
Plain
Alloy
Carbon
Or
Lean
Alloy
C 1020
A2315-20
C 1117
3115-20
4615-20
5120
8620
C
C
C
C
C
C
C
C
Medium-Carbon
Plain
Medium
Carbon
Alloy
Or
Lean
Alloy
C1040-50
A3140-50
4140-50
5145
8640-50
8740-50
6145
N,T
T
T
T
N,T,A,
S,T,
S
T
T,A
T
T
High-Carbon
Rich
Alloy
A 4340
3250
Oil
Hardening
Tool
Steel
Water
Hardening
Tool
Steel
T
T
T
T
T
T
T
T
T
T
T
T
T
C
C
C
T
T
N,A
T
T
C
C
C
C
C
C
C
C
C
N,S,A
N
N,T
N,S,T,A
N,S,T
T
S,T
T
T
S,T
S,T
C
C
C
N
C
A
C
C
C
C
C
N,T,A
T
S
N,A
T
S,T,A,
S,T
T
T
S,T,
N,T,A
T
T
T
T
T
N,T,A,
S,T
T
S,T
T
T
S,T
S,T
S,T
C
C
C
T
T
T
T
T
T
T
T
T
T
S,T
T
T
S,T
T
T
T
T
N=Normalised; C= Case-hardened; S= Surface-hardened; T=
Through-hardened; A= As-rolled
7
pp. 10, ASME Handbook, Metals Engineering-Processes, McGraw-Hill Book Company, 1958
11