Instructor’s Manual and Test Bank
For
MACHINE ELEMENTS IN MECHANICAL DESIGN Sixth Edition
Robert L. Mott, University of Dayton Edward M. Vavrek, Purdue University Jyhwen Wang, Texas A&M University
Table of Contents Chapter 1 The Nature of Mechanical Design
1
Chapter 2 Materials in Mechanical Design
3
Chapter 3 Stress and Deformation Analysis
11
Chapter 4 Combined Stresses and Stress Transformations
50
Chapter 5 Design for Different Types of Loading
121
Chapter 6 Columns
158
Chapter 7 Belt Drives, Chain Drives and Wire Rope
178
Chapter 8 Kinematics of Gears
193
Chapter 9 Spur Gear Design
225
Chapter 10 Helical Gears, Bevel Gears and Wormgearing
275
Chapter 11 Keys, Couplings, and Seals
311
Chapter 12 Shaft Design
316
Chapter 13 Tolerances and Fits
354
Chapter 14 Rolling Contact Bearings
360
Chapter 16 Plain Surface Bearings
365
Chapter 17 Linear Motion Elements
372
Chapter 18 Springs
377
Chapter 19 Fasteners
396
Chapter 20 Machine Frames, Bolted Connections and Welded Joints
398
Chapter 21 Electric Motors and Controls
407
Chapter 22 Motion Control: Clutches and Brakes
411
iii
CHAPTER 1 THE NATURE OF MECHANICAL DESIGN Problems 1‐14 require the specification of functions and design requirements for design projects and have no unique solutions. 15.
D
1.75 in 25.4 mm/in
44.5 mm
16.
L
46.0 m 0.3048 m/ft
14.0 m
17.
T
12 500 lbin 0.1130 Nm/lbin
18.
A
4.12 in2 645.2 mm2/in2
19.
Section modulus
20.
Moment of inertia
21.
Given: Amin
S I
1418 Nm
2658 mm2
14.8 in3 1.639104 mm3/in3
2.43105 mm3
88.0 in4 4.162105 mm4/in4
3.66107 mm4
750 mm2: In U.S. units; Amin
1.162 in2
U.S. Angle – From Appendix 15‐1: L223/8; A
1.36 in2 890 mm2
SI Angle – From Appendix 15‐3: Angle 75755; A
P
7.5 hp 745.7 W/hp
22.
Power
23.
Ultimate tensile strength su
24.
Given: Steel shaft; D
35 mm
Volume
0.035 m; L
V
area length
0.035 m 2/4 0.675 m
Mass Weight
876 MPa 0.675 m
9.81 m/s2
7680 kg/m3 From Appendix 3
D2/4 L 6.4910‐4 m3
7680 kg/m3 6.4910‐4 m3
mg
675 mm
mass g ; g
density volume; Density of steel
V
5.59 kW
127 ksi 6.895 MPa/ksi
Find: Weight of the shaft: Weight Mass
5.59103 W
864 mm2
4.98 kg
4.98 kg 9.81 m/s2
48.9 kgm/s2
1
48.9 N
25.
Given: For a torsional spring, Torque Find the scale of the spring
180 lbin for 35 of rotation.
Torque per unit of rotation. Express the result in both
U.S. and SI units.
T
180 lbin 0.1130 Nm/lbin
Angle
26.
20.3 Nm
35 rad/180
0.611 radians 5.14 lbin/degree
Scale
T/
180 lbin/35
Scale
T/
20.3 Nm/0.611 rad
33.3 Nm/rad
Given: 12.5 hp motor operating 16 h/day, 5 days/week Compute the energy, E, used by the motor for one year in both U.S. and SI units.
27.
E
12.5 hp 16h/day 5 days/wk 52 wks/year 550 ftlb/s/hp 3600 s/h
E
1.031011 ftlb/year
E
1.031011 ftlb/year 1.356 J/ftlb 1.0 Nm/J 1.0 W/Nm/s 1h/3600 s
E
38.8106 Wh/year
Given: Viscosity
28.
38.8 MWh/year
3.75 reyn 1.0 lbs/in2 /reyn 144 in2/ft2
3.5 lbs/in2 4.448 N/lb 1.0 in2/645.2 mm2 106 mm2/m2
Given: n
540 lbs/ft2 25.9103 Ns/m2
1750 rpm; 24h/day; 5.0 years. Find life in number of revolutions.
Life
1750 rev/min 24 h/day 60 min/h 365 days/year 5 years
Life
4.60109 revolutions
2
From Eq. 2-5:
3
17) 17)
2
(Table 2-8) SAE
SAE from SAE 1040, 4140, 4340, 4640 (Table 2-9) from SAE 1040, 4140, 4340, 4640 steels (Table (Table 2-9) 2-9)
SAE 1080 steel is a reasonable choice. SAE 5160 OQT 1000
4
SAE1040 1040and and SAE SAE1040 1040WQT WQT SAE
SAE 1015, 1020, [See Appedix A-5.]
ASTM A992 Structural steel is used for most wide-flange beam shapes.
5
Problem 38 (Continued) ASTM A536, Grade 100-70-03 is a ductile iron with a tensile strength of 100 ksi (689 MPa); a yield strength of 70 ksi (483 MPa); 3% elongation (brittle); modulus of elasticity (stiffness) of 24x106 psi (165 GPa). ASTM A47, Grade 32510 is a malleable iron with a tensile strength of 50 ksi (345 MPa); a yield strength of 32.5 ksi (224 MPa); 10% elongation (ductile); modulus of elasticity (stiffnes) of 26x106 (179 GPa.
(Table 2-11)
Aluminum 7178-T6 has the highest strength; tensile strength = 88 ksi (607 MPa); yield strength = 78 ksi (538 MPa).
6
PET, polyurethane
PET PET.
Resins used for composites include polyesters, epoxies, polyimides, phenolics, (all thermosets), and thermoplastics: PE, PA, PEEK, PPS, PVC.
7
Questions 70-73 refer to Figure 2-23 and Table 2-17 in the text. General conclusions from Questions 70-73: The specific strengths of the metals listed range from 0.194x106 to 1.00x106, approximately a factor of 5.0. The specific stiffnesses are very nearly equal for all metals listed, approximately 1.0x108 in. The specific strengths of the composites listed range from 1.87 to 4.88x106 in, much higher than any of the metals. Glass/epoxy has a specific stiffness about 2/3 that of the metals. The other composites listed range from 2.2 to 8.3 times as stiff as the metals. See Section 2-17, Table 2-16, Figures 2-23 and 2-24 for answers to Questions 74 to 100.
8
9
10
CHAPTER 3 STRESS AND DEFORMATION ANALYSIS
Direct Tension and Compression 1.
/ ;
18
12 /4
4500 N/141.4 mm
.
31.8 N/mm
/
. .
.
2.
/
3500 N/
10 /4 mm
3.
/
20
4.
/
860 lb/ 0.40
5.
/
1900 lb/
6.
/ ;
12
.
10 N/ 10 ∙ 30 mm
0.375
.
/4
144 mm ;
5000 N/144 mm
.
mm
/
a)
SAE 1020;
207 GPa
b)
SAE 8650;
207 GPa;
c)
Ductile iron A536(60-40-18);
d)
Aluminum 6061-T6;
e)
Titanium Ti-6AL-4V;
f)
PVC; Tensile Modulus
g)
Phenolic;
10 N/mm ;
207 .
165 GPa;
69 GPa;
7580 MPa;
.
.
.
114 GPa; 2410 MPa
. 2410 N/mm ;
.
Note: The stress is close to the ultimate for f and g.
11
.
7.
/
;
2.25
2.01
in /
.
2.01 in
2.25 in
. .
.
/ 8.
2556 lb/1.02 in
2500 lb; Σ
0
2500 75
2500 75/60
3125 lb
60
Tensile force in
50 A
.
9.
2
1500 lb 1500/ 2
10.
B
45°
/ ; Required
.
/4
. 1500 lb
2898 lb
15°; 2898/18 000
75
/
0.0589 in
/
11.
C
60
.
.
0.161 in ;
12
.
12.
Joint B 10.5 kN 30°
30°
10.5 kN 10.5 kN
5.25 kN
5.25
Joint A
AD sin 30 = 5.25 kN ↓ AD = 10.5 kN = CD AB = AD cos 30 = 9.09 kN = BC →
30°
5.25 kN Stresses: .
AB, BC:
. .
BD: AD, CD:
30
. 20
500 mm
.
13.
.
Figure P3‐13 Σ
0
6000 6
12 000
6000 12 000 12
18
6000 12
18
10 000 lb Σ
0
12 000 6
8000 lb Note: sin
= 8/10 = 0.80; cos
10 6
8
8 6
= 6/10 = 0.60 18
13
6
0.8 /0.8
8000/0.8
8000 lb
10 000 lb Compression
10 000 0.6
6000 lb Tension
0 6000 lb
6000
6000 lb
0 . .
2500 lb Tension 10 000 0.6
1000
0
2500 0.6
7500 lb Compression
12 000 lb /
7500/0.6
12 500 lb Compression
12 000 12 000
12 500 0.8
12 500 0.6
12 000 lb
10 000 lb
14
2000 lb Compression
7500 lb Tension
Areas of Members: Appendix A5, A6 ,
,
2 0.484
,
,
0.484 in
,
,
2 1.21
Stresses:
0.968 in
6000/0.968 7500/0.968
2.42 in 2500/0.484
Note: Compression members must be
2000/0.484
checked for column buckling
10 000/2.42 7500/2.42 12 500/2.42 14.
2.65 1.40 /
15.
52 000
2 1.40 0.5 /2
/4.41
80 40
40 /4
/
10 N/4457 mm
640
4.41 in
4457 mm .
Direct Shear Stress 16.
Pin diameter
75
0.50 in; Double shear 50
2
/4
2
0.50 /4
0.3927 in
/
2500 lb
Member BC: Σ
0
60
2500 75
2500 75/60 And
60
Figure P3 B
3125 lb
3125 lb
15
8
2500 lb Resultant at :
√3125
Pins A and C:
/
Pin B: 17.
/
3125 lb/0.3927 in
4002 lb/0.3927 in
40°:
1167 lb
Assume double shear: /
/
2
/4
2
0.75 /4
15°
1500 lb/ 2
15°
∙
12 45
Torque/Radius
1600 N ∙ m/30 mm
/
98.8 N/mm
0.060
2.50
53 333 N
2.00
1.50
√0.5
2.5
0.060
0.513 in
52 000 lb/0.513 in
Punch – Figure P3‐21. Perimeter 205.7 2.0 /
540 mm
.
Punch – Fig. P3‐20:
/
2898 lb
2898 lb/0.8836 in
See Figure 3‐7: Key is in shear;
8.55
0.8836 in
1500 lb/0.8836 in
Analysis from Problems 9 and 17. Let
/
21.
shear forces on upper pins
1500 lb
1500 lb/ 2
20.
1500 lb/2
1167 lb/0.8836 in
Lower pin:
19.
4002 lb
From Problem 3‐9: Force in each rod For
18.
2500
.
60
2 30
411.4 mm
547 N/mm
16
2 7.5
3
205.7 mm
Torsion ∙
22. 23.
/
:
800 N ∙ m
800
10 N ∙ m:
mm
6.14
10 mm
∙ /
25.
.
/
63 000
/
.
80 GPa
80
0.0138 rad
. .
24.
32.6
/
10 N/mm
.
∙ /
63 000 110 hp /560 rpm
12 375 lb ∙ in
∙ .
26.
/ ∙ /
/
/
622 N ∙ m 8590 mm
∙
27.
72.4 N/mm
:
mm ∙ /
. 1.718
10 mm
0.018 rad
.
17
°
.
°
Noncircular Members in Torsion 28.
Square:
25 mm;
0.208
0.141 /
3250 mm
5.51
230 N ∙ m/3250 mm
70.8 N/mm
∙ /
29.
/
0.60/1.50 0.78;
Figure 3‐10
10 mm
°
0.0339 rad
.
0.40 0.70 3.95 in
0.70 1.50
2.36 in
1.50
∙
Figure P3‐29 ∙
. /
.
.
3.891 in; .
. .
.
.
6.5 ft
/
2.041 in
.
1.604
∙ /
°
78 in
2 0.109 1.891 3.891 6000
.
0.109 in
1.89 in;
2
°
0.0103 rad
4.0 in;
.
°
2.10
0.78 1.50
2.0 in;
.
0.60
.
30.
.
.
1.604 in ∙
0.032 rad
18
°
.
°
Beams 31.
/ : Required
3 ft
/
∙
lb
6.25 in a
Square 3600 lb∙ft
2.43 in
√6
2.50 in; A
Use
0
6.25 in2
Rectangle 1.5 /1.5 3 Use
c
1200 lb
0
/6
b
1200 lb 3 ft
1200 lb
2.40 in
/
1200 lb 4 ft
36
1.17 in
3.50 in 1.25 in;
3.50 in;
4.375 in
Rectangle
3 √2 Use d
1.69 in; 1.75 in;
5.06 in 5.00 in;
8.75 in
Circle
2.90 in Use
3.00 in;
7.07 in 19
e
4
7.7 3.04 in ;
f
15
2.26 in
40 3.37 in ;
g
11.8 in
4 in ‐ Schedule 40 pipe 3.215 in ;
32.
3.17 in
a Volume
2.5
b
1.25 3.50 120
c
1.75 5.00 120
e
g
W
.
40 lb/ft 10 ft
400 lb
3.147
0.283 38.09 33.
a
0.283 1050 0.283 848
38.09 in for 1.0 ft
10.78 lb/ft;
10.78 lb/ft 10
36 in;
120 in; .
4
3
.
3
4
/12
2.50 in /12 . .
750 in
0.283 525
848 in ;
12.0
From Case c ; Appendix 14‐1:
0.283 lb/in
525 in ;
120 in
7.7 lb/ft 10
∙ 12
750 in ;
1050 in ;
.
d
f
120 in
3.26 in :
.
1200 lb;
10 psi
30
in in .
.
.
.
b
/12
1.25 3.50 /12
c
/12
5.00 1.75 /12
4.47 in : 2.23 in :
20
0.511 in;
.
.
.
;
d
3.98 in :
e
6.08 in ;
f
9.23 in ;
g
7.23 in ;
.
. ;
.
.
400 N
600 N
0.3
0.3
. ;
.
;
;
. .
34. 800 N 0.8 m
0.4
0.1
811 N
200 N 0.4
0.5 m
1189 N
811 N
200
11
0 389
0.5 m 0.9 m
989
333
324
217 0.2 m N∙m 0
35.
324
11 0.5
333
389 0.1
80
200 0.2
80
∙ ∙ ∙
For strength: Required Metric shape with smallest
∙
.
/
and
3330 mm : Appendix 15
21
19
Reference g: Mechanical tubing – round:
45 mm;
40 mm;
333.8 mm
3361 mm 36.
a
Simple cantilever – Case b – Appendix 14‐2: 3 4 ft
.
48 in;
6 ft
b
3 72
48
8 ft
96 in
3 96
72
72 in;
3
7.23 in Appendix 15
.
17
.
.
Supported cantilever – Case b – Appendix 14‐3 3 4 ft
48 in;
6 ft
10 ft
120 in;
72 in;
2 ft
4 ft
24 in
72 120 168
1.452
192 168 .
3
48 in
10 in 72 120
4.090 10 48
3 1.452
10
4.09
3 120
48
24
10 in 0.047 in
3
.
22
3 120
72
0.042 in
400
37.
/ ;
20
/ ∙
0.667 in
/
1.637;
30
40
400
Smallest aluminum I‐shape for the beam. 3
500 lb
1.49 in ;
N 100
0
2.24 in
400 4000
N∙m 0.140 in up
.
0 8000
① 0.214 in down
.
Table A14‐
0.140
0.214
.
0.153 in down
0.06415 Note: /
40/70
.
0.112
300 lb 3 ft
.
.
500 lb 3 ft
995
47.6 in
275 0
13.1 in
25
385 885
1245
3810
3480 psi
.
40
1245 lb
995 lb
lb
3195 lb∙in
.
70
in Case g.
6 ft 120 lb/ft
45 720 lb ∙ in
.
30
30
.
3810 lb ∙ ft 12/ft
.
70
Case g
Table A14‐
40
0.112 in up
0.571. Then point C is close to 0.153
38.
0.06415
30 20 40
② .
500
Case b
0
.
23
Beams with Concentrated Bending Moments 4 in
39. 12 in
60 lb 12 in
60
FBD
240 40
20 20 0
lb
0 40 480 240
lb∙in 0 40. 240 lb ∙ in 12 in
12 in
10 lb
FBD
240 lb ∙ in
lb
0
0
10
10 120
lb∙in
10 lb
0 120
24
41. 240 lb ∙ in 12 in
12 in 240 lb ∙ in
10 lb
FBD 10 lb 10 lb
10
0 120
lb∙in
0 120
42.
500 mm
200 175
R 225 N
FBD
225 N
M
39 375 225 N
78.75 78.75
N
0 39 375
N∙mm
0
0 25
300 N
43.
500 N 312.5
187.5
FBD
200 N
500 N 93 750
Shear
93 750
0
0
N
93 750
93 750
N∙mm
0
44.
450 N 25 mm 60 mm
11 250 N ∙ mm 11 250 N ∙ mm
FBD
F F F
N
300 N
11 250 60 187.5 N
187.5 N 450 N
0 187.5
N∙mm
0
11 250
26
450 N
45.
180 6.75 FBD
R
1215 lb ∙ in
460 lb
180 lb
120 lb
180 lb R
264
R 12"
11"
316
5.5"
264 lb
120
0 196 2904 1689
lb∙in 0 660
27
46.
At right end: Net vertical force
2.64
2.64 260
1.80
0.84 kN ↓
1.80 0.260
R
1.154 kN ∙ m
1.24
0.84 kN
FBD 1.20 m
0.40 R
2.08
0.84 N
0 1.24
N∙mm
0 1.154 1.49 693 14
47.
400 22
899 lb ∙ in
22 400
14 693 FBD
M
400 R 496
lb
0
496
0
400 R
693
197
60" 22"
10 912
lb∙in
800
197 11 811
28
48.
Torque
200 4
600 4
3200 lb ∙ in
200 8
600 8
6400 lb ∙ in Not shown
CW – viewed from C
400 lb 12 in
16 in
114.3 lb M
3200 lb ∙ in
FBD R
R
285.7 lb
114.3 lb
0
285.7 4572
1372
lb∙in 0
600 6
49. Net Torque
600 4
200 6 200 10
4800 lb ∙ in 400 lb ∙ in Not shown
800 lb 18
12
FBD R
M
320 lb
4800 lb ∙ in R
480 lb
320 lb
0 480 8640 3840
lb∙in 0
29
1.25 kN 0.22 m
50.
T
1.25 kN 0.30 m
1.25 kN
M
kN
0.275 kN ∙ m
0.375 kN ∙ m Not Shown R
0.30 m
0.275 kN ∙ m
1.18 kN
0.55 m R
2.43 kN
1.18 0 1.25
kN∙m
0
0.275
0.650
Combined Normal Stresses 51.
20 80
1600 mm
/6
20 80 /6
M
21 333 mm
F 400 mm A
400 2.70
F 80 mm
1.69
60.38
F
.
30
3.22
40°
4.2 kN
20
52.
6/12 Σ
0
0.5:
1800 8
1200
12
1200 lb: 2
For W 8 10 Beam: Appendix 15‐9
26.6°
1800
A
600 lb
2.96 in ;
7.81 in
2683 lb
A
8 ft 4 ft
2400
2400
1800
600
1200 0
lb
2400 lb
4800 lb ∙ ft
At B on top of beam
600 57 600 lb ∙ in
lb∙ft .
.
.
.
At B on bottom of beam
53.
At A Top 400 mm /
0.301 At B;
50
20.64
20.94 MPa Tension
430 200
86 000 N ∙ mm
Assume axial stress is small: ∙
Required
.
200 438 N
/ 4107 mm
/
35.1 mm; Let
36 mm;
/6; Note: t
20 mm 720 mm
4320 mm ∙
At C;
430 50
.
301 N
20.33 MPa Ok 21 500 N ∙ mm
1027 mm ;
17.6 mm
31
35°
525 N
Let
18 mm;
20 mm;
20 18 /6
∙
54.
6
2
1
1080 mm ;
18 20
360 mm
.
4 hollow rectangular tube;
Appendix 15‐14
2000 lb
For cross section of beam
4000 lb
30°
6928 lb
6928 lb
3.59 in 4.60 in
lb
At A:
Tension on top surface
2000 lb 2000
0
2000
4 ft
4 ft
2 ft 2 ft
0 ∙ .
.
1930 psi
20 870 psi
lb∙ft
8000 8000
12 000
At B: 55.
From Figure 3‐19;
875 N ∙ mm;
35.0 N 5.0 mm
5.0 2.4
12.0 mm
2.4 5.0 /6
2.4
10.0 mm .
∙
.
56.
60
√50 2.5 in/ Σ
0
on bottom of section
78.1 in 117.15
39.05 in 78.1
117.15/78.1
Σ
.
.
/
1439
0
1152 39.05
25 in
50 in
F
1152
117.15 39.0 F
1728 lb 60 in 78.1
32
7
tan tan
1800 lb
60/50: 50.2° 50/60: 39.8° sin 1152 lb cos 1383 lb
576 lb 1439 For 6
4
1
1383
2822 lb 1728 lb
4 tube: Appendix 15‐14
4.59 in ;
39.05
78.1
7.36 in
576
Just above C:
1152 1152
.
lb
.
Tension
0 576 0
lb∙in
Just below C:
44 986 .
.
Compression 57.
;
/6
4 in
60 lb
/
but assume / is small 240 lb ∙ in
Try
∙
:
1440/
/
0.500 in
1/2 in: ∙
.
58.
240
.
0.25 in ;
60 lb
0.493 in /6
11 520
0.0208 in Ok
FBD from Problem 3‐42: Part from
39 375 N ∙ mm
sees 39 375 N∙mm moment
225 N
500 mm
∙
0.09
1.89
F
50 50 /6
.
On bottom surface between B and C.
33
200
78.75 N 2500 mm 20 833 mm
225 N 50 50
59.
From Problem 3‐43: 500
from
Constant
93 750 N ∙ mm
to
300 N 200 N
500 N
75 75 /6
∙
5625 mm 70 313 mm
75 75
0.089
1.333
.
On top of surface between A and C. 60.
From Problem 3‐47: Point B is where bracket attaches; allowable To right of B;
11 811 lb ∙ in;
/32; /4
32 / 2.41 in ;
Check to left of B: .
/32
166
/
1.69 in: Use
0.526 in
10 912 lb ∙ in; .
∙
/ ; Required
32 0.472 /
20 739
34
25 000 psi
400 lb Ok
1.75 in
0.472 in
Stress Concentrations Stress concentrations:
factors can be obtained from the charts in Appendix 18 or from
the website: www.efatigue.com/constantamplitude/stress‐concentration. While all problems listed here can be solved with the tables in the book, the eFatigue site includes a wider variety of cases and users of this book, both instructors and students, may find it useful. 61.
/
9 mm/6 mm
1.50
/
0.5 mm/6 mm
0.083
.
62.
.
/
Three cases, at each support pad, must be evaluated.
Bottom: / /
2.03 [From Appendix A18-1]
2.0/0.5
4.0
0.08/0.5
0.16
1.86 [From Appendix A18-1]
. .
imum for the entire fixture
/
Middle: /
2.0/0.75
2.67
/
0.08/0.75
0.11
2.08
. /
.
Top: /
2.0/1.0 /
0.08/1.0
2.0
2.22
0.08
. .
/
35
63.
Figure P‐63 Left hole: /
0.72/1.40
0.51
2.15 From Appendix 18‐4 . .
.
.
Middle hole: /
0.40/1.40
0.29 →
2.38
0.50/1.40
0.36 →
2.29
. .
.
.
Right hole: / . .
64.
.
.
/
1.50/0.80
1.875
/
0.12/0.80
0.15
2.10 [From Appendix A182]
. .
65.
.
/
42/30
1.40
/
1.50/30
0.05
.
.
/
66.
/
2.00/1.25
1.60
/
0.10/1.25
0.08
.
68.
/
1.43 [From Appendix A18-
∙ .
67.
2.30 [From Appendix A18-1]
/
2.00/1.25
.
1.60
/
0.06/1.25
0.048
/
1.38/2.00
0.69 →
2.23 From Appendix A18‐6 1.38 .
. .
.
36
.
∙ .
/
Problems of a General Nature 69.
Σ
0
12.5 kN 4.0
12.5 4.0 /2.5
20.0 kN ↑
20 kN
7.5 kN ↓
12.5 kN
18.75 kN ∙ m 18.75
Cross section of
2.5 m
7.5 kN 2.5 m
20 20 mm
∙
&
RB = 20.0 kN D 12.5 kN
lb
12.5
10 N ∙ mm /6
Section Modulus
7.5 0
20.0 /6 lb∙in
1333 mm Area of
20
0 18.75
400 mm
Shear area of pin – double shear .
Tension in
100.5 mm
:
Shear in pin: Bending in
.
: at :
.
/
to make
Suggest change in cross section Required 20 Let
/
.
/6; Required
∙
400 MPa
20
∙ /
6 /20
120 mm: Then
Ok Proposed cross section of
37
70.
/2 40 40 40 40 40 40
/2
/2
20
/2 0 /4
200
/ /6 5.0 kN 20 mm
/2
/2
0.1 for
/2 120
0
100
m
kN ∙ m
mm
mm
MPa
A
0
0
20
1333
0
B
0.040
0.020
0.10
24
1920
52.1
C
0.080
0.040
0.20
28
2613
76.5
D
0.120
0.060
0.30
32
3413
87.9
E
0.160
0.080
0.40
36
4320
92.6
F
0.200
0.100
0.50
40
5333
93.8
G
0.240
0.120
0.60
40
5333
112.5
80 60
MPa 40 20 0 0
71.
From Problem 70,
240
4031 mm
25/40 .
160
mm
0.60 kN ∙ m under load
: /
80
.
0.625;
1.25
40
∙
25 mm diameter
38
72.
Assume lateral bracing a
At C at load;
5143 lb ∙ ft
1.0;
.
/6
.
∙
.
3.20 in
/
.
b
At D at 1.50 in diameter hole 3857 lb ∙ ft;
.
/
0.375
.
1.00 .
.
.
3.03 in
.
1500 2
4 ft
2
2
2
2
. .
c
857
At E at 2.50 in diameter hole .
2572 lb ∙ ft; /
0.625
.
0 V (lb)
1.25 .
.
.
lb∙ft .
/
1.43;
4.00, /
2.80,
0.054;
0.15 2.21
.
1714 lb ∙ ft;
.
. .
39
643
3857 2572
.
At B at step:
5143
2.419 in
.
d
643
1.57 in
1714
1286
73. 3
2
140
12 24 2
2 280
140
2
100
36
100
140
48 8 mm
2
2 N 0
/6 8 12 /6
192 mm
768
380 P
280 P
2
N∙mm 0
1728 3072
180 mm;
80 mm;
40 mm;
200 N
/
/
192
0.25
2.00
1.46
121.7 Maximum
32 000
768
0.125
1.50
1.72
71.67
D
64 000
1728
0.083
1.33
1.88
69.63
E
76 000
3072
‐
‐
1.00
24.70
Point
∙
B
16 000
C
For
: Fillet radius Larger height Smaller height
40
MPa
74.
52
25/2
2500
64.5
161 250 N ∙ mm Along upper part
; At
: /
15/25
.
75.
1.20
0.6 →
NOTE: For Kt at hole in flat plate in bending: If d/W 0.50, use Kt 1.0
.
See also Problem 74. : /
Pivot at
12/25
0.48 →
1.0
.
0
.
200 0
200
1200
76.
.
At fulcrum: /
.
0.50 in
4000 lb ∙ in/0.50 in
At hole: App. 18‐5: / .
8000 psi
1.25/2.0
0.625;
.
0.378 in
. .
20
4000
400 fulcrum
1.25
20 200
. .
77.
Figure P376
Pivot at A .
At Fulcrum:
.
0.50 in
∙
0
260
0 260
1560
.
200
B
3120
At B: 1.25;
0.378 in
6 6
20
Problem 76 460
5200
. .
At C:
26 . .
200 0
41
0 200
140 0 140
Pivot at C At Fulcrum:
20
0.50 in ∙
340
.
2800 14 200 200
78.
0 at , 0.15
0.15 m
0.15
18 kN kN
For fillet
kN∙m
0.044
13 kN
0
1.22
0 0.45 1.575
.
.
42
5
10.5
2.05
∙
0.15
7.5
8946 mm
.
8.5 kN
12.5 kN
10.5 kN
Point B is critical
1.2
0
Lug Joints 79.
One possible design Lug Joint: See Figure P3‐69;
8.0 mm,
20.0 mm
Find: Thickness, , materials for lug and pin. From Problem 3‐69, Let
0.5
Let
20.0 kN
0.5 8.0
/0.40
4.0 mm
8.0 mm/0.40
Hole diameter
20.0 mm
1.002
8.016 mm
: In Figure 3‐34:
416.7 MPa
.
/
8/20
0.40;
3.00
3.00 416.7 Let
5
1250 MPa
; Required
5 1250
6250 MPa
Too high for typical steels in Appendix 3. Redesign for use of SAE 4340 OQT 800,
1450 MPa
290 MPa 3.0
Let
0.40
96.67 MPa
.
Required For
; For /
.
.
.
2.5 ;
0.5 : 2.5
206.9
/0.75
206.9 mm
/
0.5
0.75
16.6 mm
43
206.9 mm
Specify preferred size: Then:
.
18 mm Appendix 2
45.0 mm;
.
0.5
9.0 mm 82.3 MPa
Given design parameters were not feasible. Check shear stress in pin:
/
/
39.3 MPa
/
.
Check
.
217.5 MPa
Hole diameter
1.002
39.3 MPa Ok for pin shear
18.0 mm 1.002
18.036 mm
Curved Beams 81.
ASTM A36 Steel:
36 ksi
/
10
Given:
;
248 MPa – Find F for yield of steel
150 mm;
10
/2
150 10
/
10
100 mm
5
160 mm 155 mm
160/150
0.64539
/
100/0.64539
154.946 mm
∙
325
320
/2
10 mm Cross Section of Hanger
Negative
.
. .
.
. .
Let
.
325 mm At inside surface
Maximum
248 N/mm ;
.
44
126.4 N
82.
Curved beam: Coping saw Find N for 120 N tension in blade SAE 1020 CD Steel; ∙
352 MPa
120
145
22 mm given;
22
22
5
17 400 N ∙ mm Negative 10
32 mm
27 mm
6
/
/
4
32/22
40/1.4987 mm
1.4987 mm 26.688 mm Blade
See Problem 81 for equations. 26.688
22
27.0
26.688
26.688
32.0
.
Cross Section 4 10 40 mm
0.3155 mm 5.3115 mm
231.8 MPa Tension on outside surface
.
1.18 Minimum
120 N
Both low
83.
Curved beam Given:
1.52 Hacksaw
① 10
Figure P3‐83 ②
480 ; Find
SAE 1020 CD;
20;
352 MPa
78.54 mm ;
6
15
F
Consider tube to be a composite section: ① ‐ ② 10 /4
145
140
.
.
4 mm
297.6 MPa Compression on inside surface
.
.
10
4.688 mm
6 /4
50.265 mm
45
28.27 mm
2
/4
/
3.9903 mm
2
/4
/
1.4218 mm
3.9903
1.4218
.
2.5685 mm
19.569 mm
.
See Problem 81 for equations 19.569
25
0.4307 mm
19.569
15
4.569 mm
480
80
∙
38 400 N ∙ mm
. .
540.3 MPa
Compression on inside surface
385.3 MPa
Tension on outside surface
. . .
.
.
.
.
Both indicate failure .
Addition to problem: Find
to achieve
2.0
Stresses are proportional to moment and to F. Stress reduction required:
176 MPa
0.326
.
0.326
0.326 ∙
38 400
12 500 N ∙ mm
.
46
∙
84.
Garden tool Find F for yielding.
Curved beam
207 MPa
Cast aluminum: 356.0‐T6,
12 mm; 16 mm 8 mm
50.265 mm
2 12
12
/
8 /4
.
4.312 mm Figure P3‐84
11.6568 mm
.
12
Let
/
/4
2
11.6568
11.6568
16
11.6568
8
∙ ;
F
0.3431 mm
4
4.343 mm 3.6568 mm
207 MPa positive
∙
Let
207 MPa Compression /
.
.
.
.
Similarly: .
.
.
.
47
Governs
8 4
50.265 mm
81.
Curved beam
Hoop support
Steel: ASTM A53‐Grade B;
35 ksi
∙
230 lb 48 in
= ro
Figure P3‐85
48 in
11 040 lb ∙ in Negative
2.875 in 2.469 in
D1 D2
Figure A15
17
Analysis as in Problem 3‐83
1.704 in
Stress equations in Problem 3‐81 0.61748 .
0.45484
0.1626 in D
10.4769 in
.
1.5231 in; ∙ .
1.35185 in;
.
. ∙ .
.
.
0.08564 in
11 208 psi Compression
.
9.125 in /2 10.563 in
9602.4 psi tension
.
. Satisfactory N 3.645
.
86.
Curved beam
C‐Clamp
Figure P3‐86
Cast zinc ZA‐12 404 MPa; Find:
for
269 MPa 2
3 /2
8 /
8.3 1.5
3 ∙ 11 3
/
/
8/5
3
57/6.355
19/8
1
/2
F
5.5
5.553 mm
;
5
6.355
8.969 mm
48
D
3
8
26
26 .
5.553 .
31.553
0.2776
.
pos.
Part 2:
Tension
134.67 MPa
Let .
/
.
.
0.1846
.
Compression
89.67 N/mm . .
Because
/
14
5
14 19 mm 5 5.553 10.553 mm
8.969 5 3.969 mm 8.969 19 10.0307 mm 10.553 8.969 1.583 mm
.
.
Part 1: 3 ∙ 8 24 mm 3 ∙ 11 33 mm 57 mm
. , the design of the inverted T‐section
uses the material very efficiently.
49
CHAPTER 4 COMBINED STRESSES AND STRESS TRANSFORMATIONS NOTES CONCERNING THE SOLUTIONS PRESENTED FOR MOHR’S CIRCLE PROBLEMS IN CHAPTER 4: 1. The objectives for all problems are to transform the given data for normal and shear stresses, taken from measurements in certain directions on an object, possibly by using strain gaging methods, and then use the equations in this chapter and the drawing of Mohr’s circle to determine the principal stresses and the maximum shear stress. 2.
In general, the procedures follow:
a. The General Procedure for Computing Principal Stresses and Maximum Shear Stresses, given on page 147 of the text. b. The Procedure for Constructing Mohr’s Circle for a 2D Stress System, given on pages 151- 153 of the text. 3. It would be helpful to review Example Problems 4-1 and 4-2 that demonstrate these procedures. 4. Also, Example Problems 4-3 to 4-6 show completed Mohr’s circles for sets of given data that result in the x-axis lying in quadrants 1, 2, 3, and 4. Thus example illustrations resolving the diverse kinds of data that are encountered in practical problems are demonstrated in the chapter. 5.
Problem 4-1 is shown in three forms;
a. The given data are processed using the equations in Section 4-3, particularly equations 4-5 to 4-11 to compute the maximum principal stresses and the maximum shear stresses, along with the angles of orientation on the stress element for those stress values. These calculations are demonstrated in Example Problem 4-1. b. Then the 2D Mohr’s circle is generated manually using the same input data. The three stress elements are then shown; the Original Stress Element, the Principal Stress Element, and the Maximum Shear Stress Element. c.
Then the 3D Mohr’s circle is presented showing the three principal stresses.
6. The remaining problems are shown using software to perform the calculations and to generate the Mohr’s circles for both 2D and 3D analysis. Note for these problems: a.
The software operates only with U.S. units. SI values are converted to U.S. units.
b. Each solution has tabulated results for principal stresses, maximum shear stress, and the angles to orient the stress elements. Those terms are sometimes shown without subscripts: e.g. 1 may be shown as 1; may be shown as . c.
1.
The 2D and 3D Mohr’s circle solutions are shown on separate pages.
:
20 ksi;
0;
10 ksi; x‐axis in the 1st quadrant 50
/2
20
0 /2
20
10
10 ksi;
√
√10
Angle
/ 2
2
/2
10
10 ksi
14.14 ksi 10/10
45.0°/2
45°
2
CW from x
axis to
.
22.5°
90°
90
45
/2
45/2
22.5°
10
14.14
2
10.0 ksi
45° CCW from x
axis to
24.14 ksi 2
2
10
14.14
4.14 ksi
14.14 2
0
R=14.14 2
20
‐4.14 ksi
10
Original Stress Element
(20,10)
10 ksi
10
20
(0,‐10)
Principal Stress Element
Maximum Shear Stress Element
51
Mohr’s Circle
24.14
1.
X-Axis
psi
psi
psi
psi
psi psi psi
52
psi
1.
3D Mohr’s Circle Solution
3D Principal Stresses:
1 24142 psi 2 0 psi 3 4142 psi
X-Axis
psi
53
2.
X-Axis
psi
54
psi
2.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 46827 psi 2 0 psi 3 91827 psi
X-Axis
psi
3
2
55
1
3.
psi
X-Axis
56
3.
3D Mohr’s Circle Solution
3D Principal Stresses:
1 50000 psi 2 0 psi 3 50000 psi
psi
3
2
1
X-Axis
57
4.
psi
psi
X-Axis
psi
psi psi
58
4.
3D Mohr’s Circle Solution
3D Principal Stresses:
1 47082 psi 2 0 psi 3 87082 psi
psi
3
psi
2
X-Axis
59
1
5.
psi
X-Axis
psi
psi psi
60
5.
3D Mohr’s Circle Solution
3D Principal Stresses:
1 42462 psi 2 0 psi 3 122462 psi
3
psi
2
X-Axis
61
1
6.
X-Axis psi
psi
psi
psi psi
62
6,
3D Mohr’s Circle Solution
3D Principal Stresses:
1 42462 psi 2 0 psi 3 122462 psi
X-Axis psi
psi
2
3
63
1
7.
psi
psi
X-Axis
psi
psi psi
64
7.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 71033 psi 2 0 psi 3 51033 psi
psi
3
psi
1
2
X-Axis
65
8.
X-Axis
psi
psi
psi
psi
psi
66
8.
3D Mohr’s Circle Solution
3D Principal Stresses: 1 168062 psi
2 0 psi 3 88062 psi
X-Axis
psi
psi
3
1
2
67
9.
X
‐50.0 MPa
Axis
144.34 MPa
94.34 MPa
44.34 MPa
psi
psi
50.0 MPa
-50.0 MPa = avg psi 144.34 MPa psi
100 MPa psi 80 MPa 80.0 MPa psi
psi
42.3 MPa
68
psi 94.34 MPa
9.
3D Mohr’s Circle Solution
3D Principal Stresses:
1 44.34 MPa 2 0 MPa 3 144.34 MPa
σavg X‐Axis
‐50.0 MPa
‐144.34 MPa σ3 psi
94.34 MPa τmax
42.34 MPa σ1 psi
3
2
1
psi
69
10.
85.0 MPa
198.31 MPa psi
283.31 MPa
113.31 MPa
psi
psi
X-Axis
70
10.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 113.305 MPa 2 0 MPa 3 283.305 MPa
σavg ‐85.0 MPa
198.31 MPa τmax psi
‐283.31 MPa σ2
113.31 MPa σ1 psi
psi
3
2
X-Axis
71
1
11. 11.
15 MPa
76.32 MPa psi
X-Axis 91.32 MPa
61.32 MPa
psi
psi
72
11.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 61.322 MPa 2 0 MPa 3 91.322 MPa
‐15 MPa σavg
76.32 MPa τmax psi
X-Axis ‐91.32 MPa σ2
61.32 MPa σ1 psi
psi
3
2
73
1
12.
35.0 MPa
121.76 MPa
psi
86.76 MPa
156.76 MPa
psi
psi
X-Axis
74
12.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 156.758 MPa 2 0 MPa 3 86.758 MPa
σavg 35.0 MPa
121.76 MPa τmax psi
156.76 MPa σ1 psi
σ2 ‐86.76 MPa psi
3
2
1 X-Axis
75
13.
35.0 MPa
121.76 MPa psi
156.76 MPa psi
86.76 MPa psi
X-Axis
76
13.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 86.758 MPa 2 0 MPa 3 156.758 MPa
σavg ‐35.0 MPa
121.76 MPa τmax psi
‐156.76 MPa σ2
86.76 MPa σ1
psi
3
2
X-Axis
77
1
psi
14.
σx 0, σy 0, τxy 40 MPa
σavg τxy τmax
X-Axis
0, 40 MPa
2ϕσ 90° σ2 ‐40 MPa σ1 40 MPa
τxy
40 MPa
45° σ2 ‐40 MPa
τmax
40 MPa
σ1 40 MPa
78
14.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 40 MPa 2 0 MPa 3 40 MPa
σx 0, σy 0, τxy 40 MPa σavg τxy τmax
X-Axis
0, 40 MPa
2ϕσ 90° σ2 ‐40 MPa
3
2
1 σ1 40 MPa
79
15.
85.0 MPa
; 165.0 MPa psi
80.0 MPa
250 MPa
psi
psi X-Axis
85.0 MPa
psi 80.0 MPa
psi
250 MPa
psi
τ
psi
80
165.0 MPa
15.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 250 MPa 2 0 MPa 3 80 MPa
85.0 MPa σavg; 165.0 MPa τmax psi
‐80.0 MPa σ2
250 MPa σ1
psi
3
psi
1
2
81
X-Axis
16.
15.0 MPa
71.59 MPa psi psi
86.59 MPa
56.59 MPa
psi
psi
X-Axis
psi 80 MPa
psi 15.0 MPa
30 MPa psi
psi 50 MPa
psi 56.59 MPa 86.59 MPa psi
82
psi 71.59 MPa
16.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 56.589 MPa 2 0 MPa 3 86.589 MPa τmax 71.59 MPa
σavg ‐15.0 MPa
psi
σ1 56.59 MPa
σ2 ‐86.59 MPa psi
psi
3
2
1 X‐Axis
83
17.
403.1 MPa
50.0 MPa
psi
X-Axis
353.1 MPa
453.1 MPa
psi
psi
psi
psi
psi
psi psi
psi
84
psi
17.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 453.113 MPa 2 0 MPa 3 353.113 MPa
403.1 MPa τmax
σavg 50.0 MPa
psi
X‐Axis
σ2 353.1 MPa
453.1 MPa σ1 psi
3
psi
2
1
85
18.
30 MPa 170 MPa psi
‐140 MPa
200 MPa
psi
psi
X-Axis
psi
psi psi psi psi 170 MPa
psi
86
30 MPa
18.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 200 MPa 2 0 MPa 3 140 MPa
σavg 30 MPa τmax 170 MPa psi
‐140 MPa psi 3
2
1
200 MPa psi
σ1
σ2
X-Axis
87
19.
5.0 MPa
X-Axis
47.17 MPa psi
-52.17 MPa
42.17 MPa
psi
psi
-5 MPa
psi
psi psi psi psi
τ psi
88
psi 47.17 MPa
σ
19.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 42.170 MPa 2 0 MPa 3 52.170 MPa
σavg ‐5.0 MPa
X‐Axis
τmax 47.17 MPa psi
‐52.17 MPa
42.17 MPa
psi
psi
σ2
3
2
89
1
σ1
20.
50.0 MPa
σ
170 MPa
τ
psi
-120 MPa
220 MPa psi
psi
psi
X-Axis
7252 psi
psi psi
τ 17 405 psi
90
20.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 220 MPa 2 0 MPa 3 120 MPa 50.0 MPa σavg 170 MPa τmax
‐120 MPa
220 MPa psi
σ2
3
psi
2
1
91
σ1
X‐Axis
21.
σ
20.0 ksi τ
20.0 ksi
psi
psi
X-Axis
psi
psi
psi
92
21.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 40 Ksi 2 0 Ksi 3 0 Ksi
Two circles overlap. The third circle is a point at the origin. σavg 20.0 ksi τmax 20.0 ksi psi
1
2 σ2
3
psi
σ1
93
X‐Axis
22.
X-Axis
40.0 ksi psi
psi
psi
40 000 psi
psi
45°
40 000psi 40 000psi
94
22.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 40 Ksi 2 0 Ksi 3 40 Ksi
X‐Axis
τmax 40.0 ksi psi
psi
σ2
psi
3
2
95
1
σ1
23.
psi
psi
X-Axis
psi
psi psi
96
23.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 42.780 Ksi 2 0 Ksi 3 29.780 Ksi
psi
3
psi
1
2
X‐Axis
97
24.
avg,
psi
psi
2
1
98
24.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 55 Ksi 2 0 Ksi 3 0 Ksi
Two circles overlap. The third circle is a point at the origin
psi 2
1
3
99
psi
25.
avg, max
X-Axis
psi
psi
2
1
psi psi
100
25.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 23.932 Ksi 2 0 Ksi 3 1.932 Ksi
X‐Axis
psi
3
psi
2
1
101
26.
X-Axis
psi
psi
2
1
psi
psi
psi
102
26.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 4180.011 psi 2 0 psi 3 5180.011 psi
X‐Axis
psi
3
2
103
psi
1
27.
Both principal stresses are tensile – same sign
164.03 MPa
128.06 MPa
psi psi
71.94 MPa
328.06 MPa psi
psi
104
27.
3D Mohr’s Circle Solution 3D Principal Stresses: Expressed in 3D already
1 328.062 MPa 2 71.938 MPa 3 0 MPa 164.03 MPa τmax psi
3 σ3
128.06 MPa psi
71.94 MPa σ2
328.06 MPa σ1 psi
psi
1
2
105
28.
Both principal stresses are tensile – same sign – 3D Solution
132.02 MPa psi
64.03 MPa psi
135.97 MPa psi
106
264.03 MPa psi
29.
Both principal stresses are compressive – same sign – 3D Solution.
433.53 kPa 279.56 kPa
psi
psi
867.06 kPa =
307.9 kPa psi
psi
107
30.
Both principal stresses are compressive – same sign – 3D Solution
150.02 kPa psi
168.76 kPa 24.48 psi
37.48 kPa 337.5 kPa psi
psi
X‐Axis
108
31.
Use
0.500 in for shaft ABC. Stress element on bottom. 252 lb ∙ in;
From Figure 3-20: /32
0.50
300 lb ∙ in
12 225 psi
20 538 psi
2
0.02454 in
∙
12 225 psi = xy
.
√10 269
12 225
10 269
15 966
26 235 psi
10 269
15 966
5699 psi
2
20 538 psi
0.01277 in
∙ .
/16
/32
0
15 966 psi
15 966 psi 1 = 26 235 psi
2
20 538
10 269
0 2
49.97°
12 225 psi
24.98° CCW 2
2
90°
139.97° -Axis
69.98° 69.98° 24.98° 10 267 psi 15 966 psi
109
31.
3D Mohr’s Circle Solution 3D Principal Stresses:
1 26235 psi 2 0 psi 3 5699 psi τ
τmax 15 966 psi
σ2
2ϕτ
10 269
3
0
2
σAx
σ1
20 538
1
2ϕσ
σ
τ 12 225 psi
x
110
32.
1.500 in for shaft ABC. Stress element on bottom.
Use
4572 lb ∙ in;
From solution for Problem P3-48, /32
0.3313 in ;
/16
2
/
4572 lb ∙ in/0.3313 in
13 800 psi
/
9658 psi
0.6627 in ;
√6900
9658
6900
11 870
18 770 psi
6900
11 870
4970 psi
6400 lb ∙ in
6400/0.6627
11 870 psi
13 800 psi
2
54.46°;
27.2° 11 870 psi
2
2
90°
144.46°;
72.2°
72.2°
0 27.2° 6900 psi .
111
13 800
6900 .
9658 psi
32.
3D Mohr’s Circle Solution
3D Principal Stresses:
1 18770 psi 2 0 psi 3 4970 psi
-4970 psi
0 psi
18770 psi
112
33.
Use
2.25 in for shaft ABC. : Stress element on bottom.
From solution for Problem 3‐49, /32
1.118 in ;
/16
2
/
3863
3863
3867
7730 psi
3863
3867
4 psi
179/3863
400 lb ∙ in
8640 lb ∙ in /1.118 in
2.237 in ;
√179
2
8640 lb ∙ in;
/
7726 psi tension
400 lb ∙ in /2.237 in
3867 psi
3867 psi
2.65° 2
1.33° CW 2
90°
2
179 psi
X-axis
3867
87.35° 3863
3863
2
43.67° CCW 0
1.33°
. 43.67°
7726 psi
3867 psi
179 psi .
113
33.
3D Mohr’s Circle Solution
3D Principal Stresses:
1 7730 psi 2 0 psi 3 4 psi
-4 psi
0 psi
7730 psi
114
34.
50.0 mm for shaft A-B: Stress element on bottom.
Use
0.650 kN ∙ m;
From solution for Problem 3-50, /32
0.375 kN ∙ m
12 272 mm ∙
/ /16
2
52.97 MPa Compression ∙
24 544 mm ;
√15.28
26.48
15.28 MPa
30.57 MPa
26.48
30.57
4.09 MPa
26.48
30.57
57.05 MPa
15.28/26.48
30.0°
2 52.97
26.48 26.48
2
180°
150°
15.28
2
75° CCW 2
90° 60° CW
X-Axis
120° 0
75°
.
52.97 MPa 15.28 MPa CCW
60°
115
34.
3D Mohr’s Circle Solution
3D Principal Stresses:
1 4.09 MPa 2 0 MPa 3 57.05 MPa
-57.05 MPa
0 MPa
116
4.09 MPa
35.
4.00 in;
/4
12.57 in ;
/16
12.57 in 20 000 lb ∙ in
5968 psi
. ∙
75 000 lb
1592 psi
.
2
‐5968
2984
1592 psi
2
y = -5968 psi 2984
0 1592 psi 2
1592 2984
tan
Surface Element 2 √1592 2984
3382
2984
3382
2984
28.08°
3382 psi
398 psi
14.04° CW 2
6366 psi
28.08°
90°
2
61.92°
30.96° CCW
117
35.
3D Mohr’s Circle Solution
3D Principal Stresses:
1 398 psi 2 0 psi 3 6366 psi
-6366 psi
0 psi
118
398 psi
20 mm;
36.
314 mm
450 N ∙ m
114.6 MPa tension 36.0 kN 1571 mm ∙
/
286.48 MPa 0
.
X‐Axis
57.3 MPa √57.3
292.2 MPa 2
286.5
2
‐234.9
57.3
286.5
292.2 MPa 57.3
292.2
57.3
292.2
234.9 MPa
286.5/57.3
2 2
349.5 MPa
90
2
11.3°;
78.7°;
39.35°
5.65°
119
349.5 MPa
36.
3D Mohr’s Circle Solution
3D Principal Stresses:
1 349.5 MPa 2 0 MPa 3 234.9 MPa
- 234.9 MPa
0 MPa
349.5 MPa
120
CHAPTER 5 DESIGN FOR DIFFERENT TYPES OF LOADING
1.
/ :
78.54 mm
3500 N/98.54 mm 500 N/ 3500
.
Stress
500 /2
2000 N/
2000 N Time
. 44.6
/ 2.
25.5
6.37/44.6
/ : 20 8.0
10
30
10
/300 mm
20
8 /2
6
10 N /
/ / :
. . 300 mm
10 N /
. .
6 kN
St
. 66.7
3.
.
Time
20.0
.
26.7/66.7
.
0.40
0.16 in
860 lb/0.16 in
Stress
120 lb/0.16 in 860
120 /2
Time
370 lb
370 lb/
121
5375 / 4.
2313
750/5375
.
.
0.1104 in
/ :
1800 lb/0.1140 in 150 lb/0.1104 in 1800
150 /2
St
975 lb
Time
975 lb/ 16 297 / 5.
8828
1358/16 297 .
/ :
780 N/7.069 mm 360 N/
. 7.069 mm .
. Stress
180
360 /2
570 N/ 110.3
570 N
Time
. 80.6
50.9/110.3
. .
122
6.
/ ;
1.48 in for 4
At B:
2
1
4 Tube:
14 400 lb ∙ in /1.48 in
[Appendix 15‐14] Loading Case I
.
9730
V (lb)
7135 /2 9730
7135 M (lbin)
/
Loading Case II
.
A
B
C
At C V 1320 (lb)
.
.
9730
2595 /2
M (lbin)
9730 /
6162 A D
.
Stress vs. Time diagrams – Same as for Problem 5.
V (lb)
V (lb)
M (lbin) M (lbin)
123
B
C
7.
[Figure above] Beam:
:
3.04 in for S4
500
60
30 000 lb ∙ in
500
10
5000 lb ∙ in
30 000
5000 /2
30 000
17 500
17 500 lb ∙ in
7.7
Stress
12 500 lb ∙ in
Time
∙ .
5000/3.04 17 500/3.04 12 500/3.04 . 8.
Loading case I
Loading case II Beam cross section
V (N)
V (N) M M (Nm) (Nm)
M (Nm)
1440 mm Bending ∙
.
. ∙
. Time
124
107.7
10.4 /2 107.7
/ 9.
. 59.1
10.4/107.7
. .
The spring is a supported cantilever – Case (6) Appendix A14‐3. Deflection is proportional to force. Bending moment is proportional to force. Deflection at load B: 3 .
0.90 mm
V
.
0.30 mm
M
207
10 N/mm General load, shear, and moment diagrams
Solve for P: .
46.78 For
0.25 mm;
46.78 0.25
11.7 : For
0.40 mm,
Moments:
2
25
40
7.617
25
2 40
4.614
Summary: 18.7 N;
7.617 18.7
.
125
∙
Maximum
18.7
.
∙
.
11.7 N;
89.1 N ∙ mm;
475
297 /2
475
386
/ 10.
Find For
Stres s
. 297/475
Time
.
; SAE 1040 CD;
80 ksi;
0.75 in diameter;
31 ksi for cold drawn CD curve
0.90 Figure 5
1.0 Wrought Steel;
12
1.0 Repeated Bending;
0.81 99% Reliability
31 ksi 0.90 1.0 1.0 0.81 11.
Find
′: SAE 5160 OQT 1300;
For 20.0 mm diameter:
795;
0.90
300 MPa Figure 5 5
1.0 Repeated Bending;
.
12 ;
11
1.0 Wrought Steel
0.81 99% Reliability [Table 5‐9]
300 MPa 0.90 1.0 1.0 0.81 12.
Find
: SAE 4130 WQT 1300;
Equation 5‐25: 0.87 Figure 5
676 MPa;
260 MPa Figure 5
0.808√
0.808
12 ;
1.0 Wrought Steel
1.0 Repeated Bending;
60 20
11
28.0 mm
0.81 99% Reliability
260 MPa 0.87 1.0 1.0 0.81 13.
Find
: SAE 301 Stainless Steel, ½ Hard;
150 ksi;
52 ksi: Figure 5
1.0 For axial tensile stress;
1.0 Wrought Steel
0.80 For axial tensile stress.
0.75 for
52 ksi 1.0 1.0 0.80 0.75
126
0.999 .
11
14.
Find
′: ASTM A242 steel;
[Equation (5‐25)];
70 ksi;
0.808√
1.0 Wrought Steel; 27.0
0.808
27.0 ksi, Figure 5 3.5 0.375
1.0 Revolving Bending; 0.883 1.0 1.0 0.81
11
0.926;
0.883
=0.81 (99% Reliability) .
Design and Analysis Problems 15 – 18 are open‐ended design problems for which there is no unique answer. The General Design Procedure from Section 5‐11 should be used. The loading and support conditions should be compared with the cases described in Section 5‐10 to determine the appropriate design stress. A design factor should be specified using the guidelines in Section 5‐9. When needed, the endurance strength should be computed from Equation 5‐21 in Section 5‐6. 15.
The link is subjected to a fluctuating normal stress. See Section 5‐10. See also the solutions to Problem 5‐1.
16.
The rod is subjected to a fluctuating normal stress. See Section 5‐10. See also the solution for Problem 4.
17.
The strut is subjected to a fluctuating normal stress. See Section 5‐10. See also the solution for Problem 2.
18.
The latch part is subjected to a fluctuating normal stress. See Section 5‐10. See also the solution for Problem 5.
127
19.
Figure P5‐8. See also the Problem 8 solution. Fluctuating load – Goodman Criterion: Equation 5 Find . SAE 1020 HR;
28 ;
59.1 MPa;
48.6 MPa
207 MPa;
379 MPa;
140 MPa [Figure 5‐11
Hot Rolled steel] : Rectangle: D
0.808 12 60
1.0 Let
21.7 mm;
,
1.0
0.81
0.99; Then
0.89 1.0 1.0 0.81 140
101 MPa
.
.
.
0.89
.
0.637;
Low
Yield: From [Equation (5‐29)]: 20.
.
.
.
Design Problem – No unique solution Suggestions: Keep 60 mm width for seat application. Consider higher strength material; thinner stock formed into channel shape; e.g. Consider tapering cross section depth – deeper at load – less deep near supports
where moment is smaller. Consider using aluminum. Consider shapes from Appendix 15‐7 or Appendix 15‐8, and the references to suppliers listed there. 21. was
Data is same as in Problem 7 where a S4
7.7
2.26 in
steel beam
proposed. Fluctuating stress. Use Equations 5‐28 and 5‐29 – Goodman Criterion. : Specify ASTM A36 Steel;
;
20 ksi
5
95
0.05 2.26
0.05
0.113/0.0766
36 ksi;
11
1.215 →
0.113 in 0.86 128
0.0776
58 ksi
1.0;
1.0; Let
0.86 0.81 20 .
0.99 →
0.81
13.9 ksi;
5757 psi;
0.395;
4112 psi
Problem 7
2.53
Check Yield: . . 22.
is satisfactory if no unusual conditions or uncertainty of data exists. Data same as Problem 9. Fluctuating normal stress – Goodman Criterion.
386 MPa;
89 MPa. From Appendix 3: SAE 4140 OQT 400 has the highest
with >10% elongation for good ductility. 2000 MPa; Let
99% →
1730 MPa;
5
11
0.81
0.81 450 For rectangle:
450 MPa
364 MPa 0.808√ .
0.808 0.6 5 0.438;
1.40 mm; .
1.0
[Equation 5‐29]
Suggest trying to find an even stronger material from other references. May adjust width or thickness of spring stock. Use analysis from Problem 9 to compute force vs. deflection for spring. Consider moving latch pin farther from fixed end of spring. This is a good spreadsheet problem.
129
23.
Data same as Problem 6. Fluctuating normal stress – Goodman Criterion At B:
8432 psi;
2595 psi
At C:
6162 psi;
3568 psi
For ASTM A500 Grade B: 20 ksi;
0.99 →
3.59 in ;
95
58 ksi; 0.81;
0.05 3.59
1.53 in →
0.369:
Check Yield:
.
2.97
1
4
2.71 Ok; 4
4 tube:
0.0766
0.84 0.81 20
.
At B:
for 4 0.180 in
0.84;
At C:
46 ksi
13.6 ksi
2.71 [Equation 5‐29] Ok
1
4
4 tubing weighs 8.78 lb/ft; Appendix 15
14
A lighter beam can be designed by placing 4 in side vertical and using a thinner wall. In Appendix 15‐15, the source Jorgensen offers 4 with 4.0 in side vertical, reduced by 24.
1.58 in ; original
Fluctuating load.
1.48; so beam is safe. Weight is
0.60 in; A
1500 lb Tensile;
400 /2
550 lb;
1500
/
550 lb/0.283 i
1943 psi
/
950 lb/0.283 in
3357 psi
SAE 4130 WQT 1300: 0.93;
0.134; 5.223 lb/ft
40%.
Piston Rod. Figure P5‐24. Rod Diameter
1500
2
1.0;
0.283 in
400 lb Compression; 550
98 ksi;
89 ksi;
0.80
;
0.93 1.0 0.80 0.81 37 ksi
πD /4
950 lb
99% →
37 ksi Figure 5
11
0.81
22.3 ksi; Use the Goodman Criterion.
130
.
0.170;
Safe but high. Should also check for 0.50 in:
0.196 in ;
. 25.
in final design. If rod diameter is reduced to
2801 psi;
.
.
4838 psi; [Buckling should also be checked.] /
Brittle material – static load: .
/
11 877 psi Compression
/
.
0.25 in/4.00 in
5.87
0.0625;
/
5.00 in/4.00 in
1.25; Then
1.99
. 26.
Brittle material – static load: .
1900 psi;
/
.
/ ame as Problem 25
. 27.
Brittle material – biaxial stress – Section 5‐4 Use modified Mohr method. [See Figures 5‐8 and 5‐9] and
are found from Mohr circle stress element in the fillet area. 11 877 psi
Axial compressive stress from Problem 25: =
Then
+
Torsion:
;
=0
)/2 = (‐11 877 psi + 0)/2 = ‐5939 psi
: For / .
0.0625;
/
1.25;
11 877 psi
12.57 in
1.48 20 000
∙
From Mohr circle: 5939
/12.57 in √2355 6389
1.48 [See Problem 25]
2355 psi
5939
6389 psi
450 psi Tension 131
0 2355 psi
5939
6389
12 328 psi Compression
Graphical solution 4th quadrant Point
450 psi;
at
Line
12 336 psi
Point
is failure point
Line
120 300 psi scaled /
28.
12 328 psi
.
Very Safe
= ‐5939
Ductile Material – Static Load: [See Figure P5‐28] SAE 1137 CD steel; /
565 MPa
565 MPa/3
In middle of shaft,
188 MPa
.
Required
∙
1795 mm
/
/32:
32 /
Use preferred value
M (kNmm) 0
26.3 mm
20 /32 188 /
M = 2.7x
28 mm [Table A2‐1]
Find x which would be safe for /32
V (kN)
337.5 kN ∙ mm
20 mm
785.4 mm 785.4
147 650 N ∙ mm
132
147.7 kN ∙ mm
But
.
2.7 :
.
∙
54.7 mm max
.
Use 125
125 mm
54
54
R d = 20
29.
D = 28
R optional ‐ static stress; Use R = 2.0 mm
Figure P5‐28. See also Problem 28. Repeated and reversed normal stress; SAE 1137 CD;
676 MPa;
0.90;
1.0;
/
182/3 .
∙ .
32 /
/
∶ .
. .
/2.7
20 mm
0.81
337.5 5563 m
38.4 mm: Use
Allowable distance “a” – let ∶
11 ; for
182 MPa
0.83; Recompute:
New
99% →
60.7 MPa:
Required Required
250 MPa Figure 5
1.0;
0.90 0.81 250
′/
: But
785.4 mm
/32
40.0 mm
39.4 mm; 2.0 mm;
10 N ∙ mm at load Problem 28
/
40.0 mm Ok 2.0/20
0.100;
Problem 28 : /
26 485 N ∙ mm 26.5 kN ∙ mm /2.7 kN
133
1.80
40/20
2.00
2.7
Problem 28
26.5 kN ∙ mm 9.81 mm: Use a
.
30.
See Figure P5‐28 and Problems 28 and 29.
2.0 for keyseat
Use Goodman method for shearing stresses:
:
/
; Alternating stress:
; Then
/
SAE 1137 CD:
676 MPa;
0.577
250 MPa [Figure 5‐11]
0.81;
50 mm;
0.81
0.50 0.81 0.81 250
0.75
0.75 676
Required
3 16
.
82.0 MPa
337.5 10 N ∙ mm [Problem 3‐28]
507 MPa .
25 579 mm
/16
50.69 mm: Specify
52.0 mm
.
/
Mohr circle for alternating stress
I
∶ Required
Assume
/2
16 25 579 /
Location of step from 20 mm to 50 mm: /16
20 /16
1571 m
: Solve Equation I for
For
20 mm;
0.90; ′
0.90 0.81 250 0.577
105 MPa
For
: Bending
52 mm
20 mm
1.82
/
∙ /
.
From Problem 3‐28, /2700
2.0 mm 13 143 N ∙ mm
2700 ∙
.
134
very small
x is from middle of bearing to step. Redesign is required, consider larger d or material with higher strength. 31.
30 ksi;
0.5
/
0.5 30 /3
5 ksi
5000 psi
Combined stress – static load, maximum shear stress method. Assume axial compression is small. Then max occurs at front and rear – combined bending and torsion. Use equivalent torque method – Chapter 4, Equation 4‐22 and 4‐23. 200
18
3600 lb ∙ in
400
18
7200 lb ∙ in
Note: Bending due to 200 lb load is only 4000 lb∙in and acts at a different point. √
√7200 / /2
But
1.064 in ;
3600
8050 lb ∙ in ∙
: Then
/
0.805 in → From Appendix A 2.128 in ;
15, Use 2 in schedule 40 pipe
1.704 in 7028/2
Check
1.61 in
.
1693
.
.
135
3903 psi
7002 psi
τ
Ok
32.
Fluctuating shear stress: Use Goodman method for shearing stresses. Ta = 17.5
;
65.0 Tm = 42.5
T (Nm)
30.0
.
2 0.75 0.577
For
1160 MPa;
. SAE 4140 OQT 1000; 400 MPa
708 mm 16
Specify
/
∙
870 MPa 1160 MPa, 5
0.50 0.9 0.81 400 Required
.
/
0.75 1160 MPa
Use
0.0.
∙
Time
11 : Let
1050 MPa 0.90;
146 MPa /16
16 708
/
16.0 mm; Actual
15.3 mm
0.92 Ok;
16 /16
804 mm
Check Yielding ∙
21.77 MPa;
/
/ .
0.81
.
59.06 MPa
.
2.0 Ok
136
33.
Fluctuating normal stress: Goodman Criterion; Equation 5‐28
58 ksi;
Figure 5
11
3;
28 ksi
75 ksi;
Assume machined surface and 0.9 0.81 28
0.9;
20.4 ksi
0.81
20 400 psi
F = 0 to 800 lb
Fx
9600 lb ∙ in V (lb)
4800 lb ∙ in b
/6;
M (lbin)
b
Mmax Ma
/
Mm Mmin = 0
/
I .
Term involving
.
1.75 in → Use
for
1.80 in square.
0.808 1.80
.
.
is small:
1.796 Recheck
.
1.80 in
0.808√
0.808√ .
.
1.454 in. Then
0.84 0.81 28
Preferred size
.
19.05 ksi
I 137
19 050 psi
0.84
0.808 b
Equation
becomes: .
34.
.
.
.
.
.
0.331;
.
.
.
.
.
Fluctuating combined stress: Goodman Criterion
Mohr’s circle – Mean stress Genera l
(m)max = R
τ due to torque: (Figure 3‐10) .
= 5769/b3
.
14 400 b3
From Problem 33, mean stress small
Final
m = 28 800/b3
(m)max = R = 2876
989
Alternating stress:
/2 for bending only
.
Then: 0.75 0.577 For
2700
3;
0.75 75
56.25 ksi
0.577 0.83 0.81 28 1.69 in → use
1.80
138
10 860 psi
5401
Check: .
5401 psi
.
From Mohr circle:
2876 psi
989 psi
.
14 400/ 1.80
2469 psi 0.279 →
.
.
Problems 35, 36 and 37 all ductile materials – steady load 35. 36.
37.
31.8 MPa:
/
/
5000
/ 12
a)
1020 HR:
207/34.7
b)
8650‐OQT 1000;
c)
Ductile Iron, 60‐40‐18:
d)
Aluminum 6061‐T6:
e)
Ti‐6A2‐4V:
f)
PVC:
g)
Phenolic:
/
34.7 MPa:
1070/34.7
/
.
276/34.7
All high ‐ reduce
.
276/34.7
827/34.7
.
. 41/34.7
based on tensile stength; .
.
.
based on tensile strength;
.
290/31.8
11 859 psi:
139
45/34.7 /
.
– Low‐increase .
– Low‐increase
40 000/11 859
.
38.
Ductile material – Steady load: [Problem 38, 39, 40] SAE 1144 CD
90 ksi
90 000 psi
Σ
60 :
75 2500 /60
0
75
/ /
.
3125 lb A +
595 psi
.
90 000/595
151 Very high
Consider a small rod of SAE 1020 HR steel /
30 000/3
10 000 psi;
0.3125 in
A square bar ¾ in on a side would do.
0.44 in
Or from Appendix 15‐15: Rectangular tube – 1.00 39.
Σ
0
1500 lb
1500/2 /
45°
45°
1061 lb
40.
1500/2 .
14 000 psi
15°
2898
.
F = 1500 lb
/
0.0758 in
/
0.0625;
2 FR sin
2
42 000/3
2.00
/4:
:
.
Use
0.207
:
140
41.
Repeated reversed axial load:
1.00;
0.8 Axial load;
0.81
A = d2/4 = (10 mm)2/4 = 78.54 mm2 / /
:
1.5/10
/
.
/
152.8 MPa
.
0.15
12/10
1.60
1.20
1160 MPa →
400 MPa
5
1.00 0.8 400 0.81
259 MPa
259/152.8
→
42.
.
/
Repeated reversed shear stress: ∙
11
0.81,
. 0.81
32.6 MPa
/
SAE 1040 WQT 1000:
780 MPa Figure A4
1 ;
280 MPa Figure 5
11
0.5 0.5 0.81 0.81 280 / 43.
91.9/32.6
91.9 MPa .
Ok
Repeated – One direction shear stress: Fluctuating shear stress T (Nm)
Goodman Criterion:
800 400
400
/16
24 544 mm
400 0.75 .
10 N ∙ mm / 24 544 mm
0.75 780 .
. .
0.205;
Time
0
16.30 MPa
585 MPa .
141
From Problem 42
44.
Ductile Material – Static load:
0.5
/
; Let
3.0 [Design
decision] .
∙
.
7003 psi
/
Required
/0.5
3 7003 /0.5
Aluminum 2024‐T4 has 45. 0.81;
0.82;
Fluctuating shear stress: Goodman Criterion; Assume 3 12 375 lb ∙ in /2 0.75;
6188 lb ∙ in:
0.75 208
156 ksi;
208 ksi →
21.25 ksi
21 250 psi
0.50 0.82 0.81 64
64 ksi Figure 5
: 3 46.
0.993 in ; Steady load: ∙
0.5
mm / 0.5
622 /
/
622
10 N ∙ mm
8590 mm
10 N ∙ mm /8590 mm
:
1.75 in;
/
622 N ∙ m
/
16
/0.5
3 72.4 /0.5
SAE 1040 cold drawn has
142
72.4 MPa
.
11
47.
Fluctuating shear stress – Goodman Criterion (See also data from Problem 46)
Pa = 6.5 kW
/
21 500/45
478 N ∙ m
28 Pm =
/
6500/45
21.5
144 N ∙ m
15 P (kW)
/
478 000 /8590
/
144 000/8590 .
.
55.6 MPa Time
0
16.8 MPa .
After trials: 0.83;
0.81
SAE 1144 CD;
621 MPa;
690 MPa →
0.75
0.75 690
0.50
0.50 0.83 0.81 253
.
.
0.305;
.
253 MPa Figure 5
11
518 MPa
1/0.305
85.0 MPa .
Ok
For Problems 48, 49, 50, and 51: Fluctuating normal stress: Goodman Criterion [Analysis at the section through the hole]
48.
1.50 .
.
/
0.50 0.50
0.50 in
20 000 psi
Pm = 10 000 7500
5000 psi
0.5/1.50
0.333 →
Pa = 2500 12 500
Force, P (lb)
2.31 Time
For SAE 6150 OQT 1300: [Figure 5‐11 for su]
143
Axial
107 ksi,
118 ksi → .
6 /4
49.
0.583 →
.
/
0.5/6
0.083
/
9/6
1.50
.
.
. .
For yield: 50.
0.8 0.81 43
27.9 ksi
.
28.27 mm .
σ
43 ksi;
225 N
36.25 MPa
1250
7.96 MPa
800
1025
P (N) Time 0 2.03
0.126;
779 MPa;
841 MPa
300 MPa
.
0.8 0.81 300
8.68 Ok
Axial From Problem 3‐62, Chapter 3: Maximum stress occurs at bottom. 1.86;
86 000 psi;
0.8 0.81 43 000
121 000 psi;
43 000 psi
27 864 psi:
.
/
Goodman Criterion: .
For yield:
.
0.224;
Ok
144
.
Ok
3056 psi; σ
194.4 MPa
51.
From Problem 3‐63, Chapter 3: Maximum stress occurs at left hole (0.72 diameter) 2.15:
1.40
0.72 0.50
0.34 in 6200
15 000 psi
.
Fm = 5100 4000 F (lb)
3235 psi
.
Goodman Criterion:
For yield:
sy = 125 ksi su = 135 ksi
0.314; N = 3.19 ;
.
3 18 281
51 ksi 0.8 0.81 51 33.05 ksi
.
16 650 psi, including
From Problem 3‐64, Required
Time
0 .
52.
Fa = 1100 lb
:
→
/ ,
53.
Goodman Criterion:
: Note that direct solution is not possible because
both
and are unknown. Also data for endurance for titanium are not directly available
here. As an estimate we will use Figure 5‐11 and the discussion for steel to obtain note from previous problems,
/4. This permits solution for
. Also
. After material
selection, the final “N” can be computed. From Problem 3‐65: Axial tensile force; stepped round bar; 30 /4
707
.
Fa = 5.15 kN 30.3
35.57 MPa
.
Fm = 25.15 20.0 F (kN)
7.28 MPa .
.
Time
.
. /
Then
102.5
Try Titanium Ti 50A;
2.30 for fillet
3 102.5
308 MPa
276 MPa;
145
345 MPa
From Figure 5‐11:
120 MPa Estimate
0.8 0.81 130 .
.
84.2 MPa; Now compute actual value for N:
.
0.302 →
.
54.
.
Ok Specify Titanium Ti 50A
From Problem 3‐66, stepped round bar in torsion: Use Goodman Criterion: For round bar:
Assume .
0.75
;
:
0.75
≅
19 273
/16
0.383 in
57 819 psi /0.75
60 ksi,
87 ksi;
0.75 87 000
65 250 psi
0.405;
2.47 Low
Try SAE 1040 WQT 1000;
113 ksi,
84.75 ksi;
.
0.3175;
53
3 19 273 ≅ 57 819 psi
0.50 0.81 0.85 33 000
.
0.75 113
See Problem 5
/0.75
Try SAE 1137 OQT 1300; 0.50
/4 :
/
55.
1.25
1100 lb ∙ in
1100 lbin /0.383 in = 2868 psi; Also, a = m = 2868 psi
/
But
1.43:
77 092 psi 33 ksi
11 360 psi
88 ksi,
42 ksi
0.50 0.81 0.85 42 .
14.458 ksi
Ok – Specify: SAE 1040 WQT 1000
Figure P3‐67: Stepped round bar in torsion. Specify grade of ASTM A536 ductile iron. Use Maximum Normal Stress method. Higher strength ductile irons are fairly brittle. From Problem 3‐67: Required
32 564 psi, including 3 32 564
;
/ →
146
Grade 100‐70‐03.
56.
From Problem 3‐68: Required
49 323 psi including
3 49 323
:
/
. Specify a grade of stainless steel.
Referring to Figure 5‐11, the required su >> 220 ksi ‐ Excessively high. No practical stainless steel material available. Redesign the member. 57.
Figure P5‐57; Industrial screw‐driver: Load is repeated – one direction – torsional shear stress. Compute the design factor. Fluctuating shear stress – Goodman Criterion:
100 lb ∙ in/2
At fillet: /
0.083:
0.50/0.30
/16
1.67; /
0.30
/16
0.025/0.30
50 lb ∙ in
1.43
0.00530 in
∙ .
SAE 8740 OQT 1000:
167 ksi;
0.50 0.81 1.0 60 ksi
24.3 ksi;
.
58.
175 ksi; Then 0.75
0.627;
.
60 ksi Figure 5 0.75 175 Low
From Figure P5‐58, Rectangular bean; ASTM A48, class 40A cast iron. Steady load – brittle material; Compute the design factor. At middle – between C and D: /6 /
0.75 in 2.25 in /6
0.633 in3
2250 lb ∙ in / 0.633 in
= 3556 psi
At step – point B: /6
0.75 1.25 /6
/
0.20/1.25
0.16
/
2.25/1.25
1.80
0.1953 in Kt = 1.63
V (lb)
M (lbin)
147
11
131.3 ksi
.
12 519 psi
.
/ 59.
40 000/12 519
.
Repeated one direction: Goodman Criterion: Seek 2.5 < N < 3.0 [Design decision] Round tensile member. Specify a material. 0.50 in;
/4 /
0.196 in ; Assume
1500 lb/0.196 in
1st Trial: SAE 1040 CD:
1.0
7639 psi
71 ksi;
Fm = Fa = 1500 lb
3000 F (lb) 1500
80 ksi
Time
30 ksi
5
11 ;
0 0.81 0.8 30
.
0.489;
2nd Trial: SAE 1040 WQT 1000: 42 ksi;
19.4 ksi
2.04 Low
112 ksi;
87 ksi; 23% elongation
27.2 ksi 0.349;
.
Ok
Check Yield: Equation 5‐29, solved for N: . 60.
Ok Specify SAE 1040 WQT 1000 steel.
Repeated – One direction: Use Goodman Criterion:800 Force (lb) 400
Specify a suitable steel. 800 lb;
0;
0.010 in;
25.0 0 in
Consider deflection first: Required √
S .
/ √0.0667
Stress analysis: Assume
Fm = Fa = 400 lb
0.0667 in
.
0.258 in: Try 1.0
148
0.300 in
Cross Section
4444 psi
.
Try SAE 1040 CD:
71 ksi;
From Figure 5‐11:
30 ksi: Let
s
80 ksi; 12% Elongation
1.0 0.80 0.81 30 ksi
1.0;
0.80
;
0.81
19 400 psi .
Equation 5‐28:
0.285;
.
Ok
Check yield; Equation 5‐29, solved for N. N= 61.
.
Ok
Repeated – one direction: a)
1200 lb;
600 lb
[Goodman Criterion]
a=8 in
For illustration use same material As in Problem 60: SAE 1040 CD 80 ksi;
300 V (lb)
30 ksi
0
0.90 1.0 0.81 30 Size Bending At C:
3000 lb ∙ in; /32
2.0
21.9 ksi3000 2400 M (lbin)
1.0 /32
0
0.785 in
Non rotating circle 0.370
.
0.370 2.0
0.74/0.3
Equation 5‐29: /
Fm = Fa = 600 lb
10 in
3000 lb ∙ in/0.785 in 0.222;
.
149
3820 psi .
.
0.74 in 0.91
b)
At section B:
2400 lb ∙ in; /
1.74:
62.
0.10;
/
3.0/2.0
3056 psi
. .
0.20/2.0
0.281;
.
Redesign beam in Problem 61. Note that N is inversely proportional to moment M and distance a. Then a must be reduced by:
Then
Specify 63.
3.54/4.50
0.787
0.787 8.0
300
1875 lb ∙ in;
.
0.220;
6.30 in say 6.25 in /
.
If
.
3.0/2.0
1.50;
8.00 in as given;
0.305;
8.0 0.936
7.25 in 300 lb .
/
2400 lb ∙ in at ;
Dimension a must be reduced to get 4.21/4.50
0.40;
0.40/2.0
1.47
.
Then
2389 psi
Ok Higher than (a)
Refer to Problems 61 and 62: New /
1875/0.785
3.27
4.50 as at .
7.49 in; Let
2175 lb ∙ in;
0.2806;
3056 psi
.
150
7.25 in /
Ok
2771 psi
0.20
1.50
64.
Repeated – one direction: Fluctuating stress: Equation 5‐29‐Goodman Criterion SAE 1040 HR: 1.0
a)
42 ksi;
72 ksi;
direct tension,
0.81; 14.9 ksi
At pin hole:
0.25 in diameter, / .
At fillets: /
5000
1.00 in F 2500
0.25; .
0.02/1.00
(lb)
4.40
Fm = Fa = 2500 lb
4.12; 0.02;
. /
2.0/1.0
From Appendix 18‐2: Point of interest is above chart. . .
.
0
13 333 psi
.
.
b)
0.80 axial load
0.8 0.81 23
Appendix A18‐4:
23 ksi [Figure 5‐11 Hot rolled]
2.0 ≅ 3.7
10 000 psi 2.62;
.
65.
Improvements: 1.) Increase thickness, 2.) Increase fillet radius, 3.) Use
stronger
material, 4.) Increase pin hole size – or – change manner of applying force to
the
part to eliminate hole – or – make part thicker at the holes than in middle of the part. Material may be removed in 2.00 in. section near middle of part to offset added material elsewhere. Could try titanium with lower density than steel. This problem may be too restrictive to permit a practical solution with data in this book. May have to accept lower N < 3.0 or some increase in weight of the component.
151
66.
Fluctuating normal stress – Goodman Criterion – Equation 5‐29. SAE 1040 CD:
71 ksi;
1.0;
0.80 axial;
24.8
3.0 /2
24.8
13.9
0.375 in;
80 ksi;
13.9 kN 1.0 lb/4.448 N
10.9 kN 1.0
.
0.81
19 400 psi
/4.448
2450 lb
0.625 in; /
0.417
Fa = 10.9 kN
24.8
.
.
11 ;
3125 lb
2.84 .
5
0.8 0.81 30 ksi
1.50 in;
At pin holes:
30 ksi
9524 psi
.
Fm = 13.9 F (kN)
7467 psi
.
3.0 .
At middle hole:
1.217;
.
Indicates failure
0.0
Tim e
2.22; same nominal stresses.
.
0.854;
.
Very Low
Part must be redesigned. 67.
Repeated reversed load – bending – SAE 1340 OQT 1300:
517 MPa;
690 MPa;
1.0;
1.0;
0.98 0.81 260
400 N 250/400
250 N
400 N 150/400
150 N
0.98;
At B: /
/
:
260 MPa; 206 MPa 0.808 √ 0.808 12
/6
12 /6
288 mm
2.0/12.0
0.167;
/
0.98
20/12
152
0.81
1.67 [Appendix A18‐7]
=0.808b 9.70 mm
1.60
Load = 400 NReversed .
125 MPa
206/125 At C:
At D:
/
V (N)
1.648
/6
12 20 /6
800 mm
.
46.9 MPa
at B.
14/20
M (Nmm)
1.40 Appendix A18
0.70
5
47.6 MPa; 1.40 47.6 Minimum 68.
.
66.6 MPa
at B.
(Low)
See Problem 67: Specify tempering temperature to achieve 2.5 125 Then
required
312.5 MPa
2.5.
0.98 0.81
1150 MPa (Figure 5‐11)
From Appendix A4‐3: SAE 1340 OQT 900 has 69.
Fluctuating normal stress: 300
700 /2
At fillet: /
300 lb;
500 lb;
0.06/1.25
700
0.048;
/
700 lb; Use Goodman Criterion 500
200 lb
2.0/1.25
1.60 4.0 in
2.35 /32
1.25 /32 ∙ . ∙ .
SAE 1050 HR: Non
49 ksi;
0.192 in F V
10 430 psi
0
4172 psi
0
90 ksi
rotating circular shaft;
0.37 153
4F
M
0.463 in;
0.95
F
Non
rotating circular shaft;
26 ksi
5
0.37
11
;
.
70.
0.463 in;
0.95
0.95 0.81 26
20.0 ksi
0.606;
.
Low
See Problem 69: Increase N to 3.0 or higher by using larger r. Best possible improvement would be .
1.0 with large r.
0.324;
.
Ok
[From Appendix A18‐6] To achieve Kt approaching 1.0, r/d must be > 0.30. Then, r > 0.30d = 0.30(1.25 in) = 0.375 in 71.
See Problem 69: increase material strength to get
3.0.
Try SAE 1340 OQT 900: 158 ksi;
169 ksi
4
Use machined surface:
58 ksi; .
72.
3 17% Elongation 0.95 0.81 58
0.2814;
Repeated reversed stress: Design for Specify material: at : /32 At fillet: /
V (lb)
600 lb ∙ in
0.06/1.0
0.0982 in
0.06;
/
1.38/1.0
1.99 [From Appendix A18‐6] Let Required
.
∙
12 159 psi
3.0 12 159
36 477 psi
.
Ok
3.0.
/ .
1.00 /32
.
44.63 ksi
154
Load = 100 lb
M (lbin)
1.38
600 lbin
0.81;
0.88;
/ 0.88 0.81
51 174 psi
From Figure 5‐11, Required
(Machined surface) ,
One possible solution: SAE 3140 OQT 1000, 73.
%
Fluctuating normal stress – Goodman Criterion; Equation 5‐29 SAE 1340 OQT 700:
197 ksi;
221 ksi
Fa = 3750 lb 16
0.80 0.81 65
Figure 5‐11
Axial 8500
16 000 /2
16 000
12 250
Fm = 12
250 12 250 lb F (lb) 8500
3750 lb
Time
At fillet: /
0.05/0.63
0.079;
/4
0.63 /4
0.312 in
/
/
1.00/0.63
1.59;
2.10
39 300 psi; a = Fa/A = 3250 lb/0.312 in2 = 12 030 psi
. .
=
74.
42.1 ksi 000
0.778;
.
See Problem 73: Try to redesign to achieve Increase fillet radius to
Low
3.0.
0.185 in. Fillet would then just blend with outside of
1.00 in diameter of rod. /
.
0.29;
.
. .
1.59:
r = 0.185 in
1.36
Use stronger material: Try SAE 8650 OQT 700 240 ksi;
222 ksi; 12% elongation
Grind all critical surfaces gently. Then
0.81 0.8 88 .
88 ksi; Use
0.81
57.0 ksi 0.451;
2.22 Still too low
155
Even if
1.0,
.
Diameters may have have to be increased. 75.
Repeated reversed load – Bending moment at fillets = .
.
4.00
800 lb 4.0 in
3200 lb ∙ in
∙
0.5208 in :
6144 psi
.
; Then required for rectangle: 1.25 in SAE 1144 OQT 1000:
2.00 in;
0.808√
112 ksi,
42 ksi; Use
0.85 0.81 42 Then
76.
28.9 ksi
0.808
1.28 in
0.81
28 900 psi
1.57
From Appendix A18‐7: /
2.00/1.25
Then
0.220 in gives
.
Design section at “B” for
3.0.
Σ
2.25
0
800 4.75
800 4.75 25
1.6; For
1689 lb ↓
Let width
2.00 in; same as other sections
Design for
1.50; Specify thickness, h
/
1.57
R1 = 1689 lb 2.25 in in
Multiple solutions possible:
But
1.25 2.00
V (lb)
4.75
Rpin = 2489 lb
Load = 800 lb 800
0 1689
/ /
MB = -1689(1.50) = -2533 lbin
156
Then: .
/
.
0.444 in minimun
.
Let
0.85 in;
Then
0.30
1.50, /
2.00 in: For 0.30 0.85
First estimate by trial → for
0.3
estimated
0.255 in 0.19,
.
Problems 77‐83 are design problems for which multiple solutions are possible.
157
CHAPTER 6 COLUMNS 1.
Radius of gyration:
/4
0.75/4
0.188 in:
42 00 psi: /4
/
1.0 32 0.188
119 → Long column
Euler
0.442 i .
/
2.
[Some data from Problem 1] /
1.0 15 /0.188
79.8
/
3.
L = 15 in → Short
0.442 42 000
Johnson Formula 1
.
[Some data from Problem 1] Use aluminum 6061‐T4. 0.188 in;
/
171;
21 000 psi;
10
10 psi
97 → Long column . /
4.
[Some data from Problem 1] Both ends fixed. /
0.65 32 /0.188
0.442 42 000 5.
111
→ Short
Johnson Formula
1
[Some data from Problem 1] Square: Side dimension = H = 0.65 in
158
171
Square:
6. Let
/√12
0.65/√12
0.188 in Same as round
Problem 1
[Some data from Problem 1] Acrylic plastic: Tensile Strength
5400 psi;
220 000 psi
28.4: .
Long
0.144 in
0.5 in/√12 /
171
.
/
7.
/
1.0 8.5 /0.144
1.00 in
0.50 in
58.9
181 000 psi →
57 Figure 6
5 Long Column
. .
/
8.
/4 /4
1.60
0.529 in:
0.5106 in: /
75/0.529
140 Figure 6
5
30 000 psi; a)
1.382 /4
Pinned Ends:
/
1.0 /
142
6.25 ft
142
Long
Euler
.
b)
Fixed – Fixed:
/
0.5106 30 000 c)
Fixed – Pinned:
/
0.5106 30 000
0.65 /
92.3
1
.
0.8 142
114
1
159
Short
Johnson
Short
Johnson
75 in
d)
Fixed/Free:
/
2.10 142
298
Long
.
152 ksi:
9.
≅ 60
6
5 : Assume column is long:
/
[Equation 6‐11]
/
4.39
1.0
/
1.45 in
Use 1.50 in /4 10.
1.50/4
30 ksi:
0.375:
149
/
1.0 50 /0.375
6
5 : Assume column is long
145 in Same as Problem 9 ; Use
1.50:
/
/
[Equation 6‐12]
133
Long Ok
133
Johnson
.
.
No advantage to 4140 steel over 1020 steel. Only modulus of elasticity involved. 11.
Aluminum 2014‐T4:
42 000 psi; /
/
1.90 in
≅ 69: Assume long column
use
. Ok
1.0 50 /0.50
160
;
/4
0.50 in
12.
Square with side length, S: /12
/12;
: From Equation 6
10
Euler
/ /
For the Johnson Formula:
/12
Equation 6‐9:
1
1
/
: Solve for Divide by
and take √
/
13.
/
Euler: [Equation (6‐10)];
; Express R =d/D; d= RD
D= OD
/
d=
R = ID/OD = d/D d = RD
1
/4
For the Johnson equation: Equation (6‐9) 1
/
161
/
/
14.
/
Assume column is long: From Problem 12: Side length = 0.65 64
41.6 in Fixed ends /
.
Check:
/√12
For 6061‐T6, 15.
/
1.423 in →
1.50/√12
.
0.433:
/
40 ksi: From Figure 6
6
0.8: 1
41.6/0.433
96.1 Ok
0.5904
Assume long; From Problem 13 /
. .
Use
.
;
0.8
/
1.31 in
1.20 in:
0.48 in:
Weight Comparison: Weight proportional to area Square:
Tube:
1.50
2.25 in
1.50
1.20
162
0.636 in
Ok
/ 16.
2.25/0.636
.
Assume column is long: [Equation (6‐11)] [See sketch on following page.] /
/
.
Check:
/4
/
60/0.375
1.46 in Use
.
0.375 in 160
Long Ok
60 [Figure (6‐5)]; Note: Care would have to be used at connections to ensure axial load.
17.
Multiple designs possible. Consider hollow tube – round or square. Cheaper material may also be used.
18.
Multiple designs possible. Cable
/
°
Fv = 9000 lb
Fs 9000 lb 9000 lb
163
Spreader
19.
Multiple designs possible.
°
20.
10.75 ft
129 in:
2.1 129
271 in:
68 ksi;
93
Assume column is long – Euler – [Equation (6‐11)] /
21. 23.
/
.
1.063 in:
Check:
/4
22.
Multiple Solutions Possible Crooked Column: /4
0.188 in;
/
0.08 in;
0.5
8951
Crooked Column:
42 675 lb Problem 6
/2
0.442 in ;
4473 lb
.
1
0.375 in
.
4473
.
10
4 9.226
0.04 in;
10
0.5/2
0.25 in;
7
0.144 in;
0.0208 in ;
181 000 0.50 .
0.5
9.226 8951
Ok
0.75 in,
42 000 0.442 .
.
271/1.063
0.0352 in ;
In Equation 6‐13:
24.
4.21 in
51 220
0.50 in ; 1
.
4.291
10
. .
51 220
181 000 psi 42 675
4 4.291
164
10
51 220
8951
25.
Crooked Column:
7498 lb Problem 6
0.15 in;
0.280 in ;
0.5106 in ; 1
.
.
1.276
10
8677
8677
30 000 0.5106
26.
0.80 in;
8
0.529 in;
0.5
/2
Eccentric Column:
30 000 psi
.
7498
4 1.276
10
.
42 in;
8677
1.25 in;
/2
0.625 in S
/√12
0.361 in;
0.130 in ;
1.563 in ;
0.60 in
/12
0.2035 in ;
13 000 psi for aluminum 6063‐T4
S
From Equation (6‐14):
/
Equation (6‐16): 27.
.
1
.
0.60
Eccentric Column:
3.2 m
3.50 in;
1.16 in 25.4
29.46 mm;
3.02 in 25.4
1.257 207 GPa
.
.
3‐in Schedule 40;
SAE 1020 HR;
. .
.
1
.
3200 mm; /2
30.5 kN
1.75 in 25.4
868 mm;
150 mm
207
207 000 MPa
165
30 500 N
/
2.23 in 25.4
10 mm ; 10 Pa
.
44.5 mm 1439 m
207 000 N/mm
Equation (6‐14): 212 MPa: but
.
207 MPa, then stress is too high
150 28.
1
/
1
.
Eccentric Column:
14.75 in;
.
0.30 in;
if matl. does not yield
0.250 in S
0.0625 in ; 45 lb;
28
/√12
0.25/√
10 psi:
/2
0.0722 in;
0.00521 in S
0.125 in
Equation (6‐14):
Equation (6‐16): 29.
.
1
.
.
50 000 psi; /3
1
.
0.50 in;
3.37 in ;
3; then
.
.
40 in;
From Appendix 15‐14:
Let
.
0.30
Eccentric Column:
ASTM A242:
.
. .
75 000 lb
1.52 in;
30
.
2.31 in;
10 psi
4/2
¼
4
i
16 667 psi
4 [Right side of Eq. (6‐15)]=
Let
30.
Central Load:
:
/
16.0 ft 12
/
From Appendix 15‐9: . .
1
.
:
.
. .
.
/
.
:
192 in;
5.54 in ;
9.13 in ;
119.7
166
/
2.00 in
1.28 in
From Figure 6‐5,
125; short column
.
31.
7.7:
0.65 66
0.65;
2.26 in ;
ASTM A36:
0.581 in;
36 000 psi;
30
32.
0.50
1.60 0.80
1.28 in ;
0.80/2
/
0.90
.
/
9
0.80/2
0.90 in;
0.40 in;
72 in; Rectangle
0.80/√12
0.2309 in
10 psi; [Use Equation 6‐14] 1
0.7955
.
. .
.
1
Using Equation (6‐15):
8315 psi
.
0.386 in [Equation 6‐16]
Specify a material to provide
28 280 psi
3;
.
1
1000 lb;
30
73.8;
10 psi;
Eccentric Load:
Use steel:
42.9 in
/
130 → short column Johnson Equation 6 2.26 36 000 /3
3
.
1
Central Load: Fixed‐End; S4
Johnson formula. Let
3.
.
1
.
. .
.
/ : Then
Specify SAE 1040 WQT 1000,
(Other solutions possible.)
167
33.
Central Load on a steel tube: Specify standare tube size. sy = 36 000 psi; N=3
Assume that the column support ½ of the total load: Pa = 55 000 lb/2 = 27 500 lb ; Cc 130 [Figure 6‐5]
Assume column is long: [Equation 6‐10] Required I = Assume Fixed base; Pinned top: K = 0.80 KL = (0.80)(18.5 ft)(12 in/ft) =178 in > 130 – Long column. .
Required I =
8.79 in
Possible tubes: Square: 4x4x1/2, I = 11.9 34.
Central Load: C5 Pinned ends:
; Rectangular: 6x4x1/4, 11.1
9 Steel Channel: /
112 in;
1.0 112 /0.489
229
2.64 in :
0.489 in;
3
Long column; E = 30x106 psi
.
35.
Central Load on a channel: Data from Problem 34, except ends are fixed rather than
pinned. K = 0.65. KL/r = (0.65)(112)/0.489 = 148.9 30 10 2.64 3 148.9 36.
Same as Problem 34 except load is along back face of channel. Eccentric load. [From Appendix 15‐4] Use Equation (6‐15):
.
1
.
0.478 in and /
36 000/3
. .
.
.
168
. 12 000 psi
By iteration: For
;
The spreadsheet here is for the same data as used in Example Problem 6‐3 in the text. Data from: Example Problem 6-3
CIRCULAR COLUMN ANALYSIS
Cross-section shape: Solid circular section
Refer to Figure 6-7 for analysis logic Enter data for variables in italics in shaded boxes
Use consistent U.S. Customary units.
Data To Be Entered:
Computed Values:
Length and End Fixity: Column length, L = End fixity, K =
25 in 1
------>
Eq. Length, Le = KL =
25.0 in
------>
Column constant, Cc =
140.5
------>
Slender. ratio, KL/r =
81.3
Material Properties: Yield strength, s y =
30000 psi
Modulus of Elasticity, E = 3.00E+07 psi Cross Section Properties: [Note: A and r computed from] [dimensions for circular cross section] [in following section of this spreadsheet] Area, A =
1.188 in2
Radius of Gyration, r = 0.3075 in Properties for round column: Diameter for round column: 1.23 in Area, A = Radius of gyration, r = Design Factor Design factor on load, N =
Column is: short
2
1.188 in 0.3075 in 3
------>
169
Critical Buckling Load =
29,679 lb
Allowable Load =
9,893 lb
The spreadsheet solution below is for Example Problem 6‐4. The user can see the detailed solution in the book and compare it with the spreadsheet shown here. Data from: Example Problem 6-4
CROOKED COLUMN ANALYSIS
Cross-section shape: Solid circular section
Solves Equation 6-13 for Allowable Load Enter data for variables in italics in shaded boxes
Use consistent
Data To Be Entered: Length and End Fixity: Column length, L = End fixity, K = 1
U.S. Customary units.
Computed Values:
32 in ------>
Eq. Length, Le = KL =
32.0 in
------>
Column constant, Cc =
118.7
Material Properties: Yield strength, s y = 42000 psi Modulus of Elasticity, E = 3.00E+07 psi
Euler buckling load =
Cross Section Properties: [ Note: Enter r or compute r = sqrt(I/A ) ] [Always enter Area] [Enter zero for I or r if not used ] Area, A = Moment of Inertia, I =
-9678
C 2 in Eqn. 6-13 = 9.259E+06
0.442 in 2 4 0 in
Radius of Gyration, r = 0.188 in Values for Eqn. 6-11: Initial crook edness = a = 0.125 in Neutral axis to outside = c = 0.375 in Design Factor Design factor on load, N = 3
C 1 in Eqn. 6-13 =
4491 lb
------>
Slenderness ratio, KL/r =
170.7
Column is: long Straight Column 4,491 lb Critical Buckling Load = Crooked Column 1,076 lb ------> Allowable Load =
170
The spreadsheet below is for the analysis of an eccentric column using data as in Example Problem 6‐6. Those that follow are from the Problems section of Chapter 6.
ECCENTRIC COLUMN ANALYSIS
Data from: Example Problem 6-6 Cross section shape: Solid circular section
Solves Equation 6-14 for design stress and Equation 6-16 for maximum deflection Enter data for variables in italics in shaded boxes Use consistent U.S. Customary units. Data To Be Entered:
Computed Values:
Length and End Fixity: Column length, L = 32 in End fixity, K = 1
------>
Eq. Length, Le = KL =
32.0 in
------>
Column constant, Cc =
118.7
Argument for secant = Value of secant =
0.749 for strength 1.3654
Argument for secant = Value of secant =
0.432 for deflection 1.1014
Material Properties: Yield strength, s y =
42000 psi
Modulus of Elasticity, E = 3.00E+07 psi Cross Section Properties: [Note: Enter r or compute r = sqrt(I/A ) ] [Always enter Area] [Enter zero for I or r if not used ] Area, A = 0.785 in 2 4 Moment of Inertia, I = 0 in Radius of Gyration, r = 0.250 in Values for Eqns. 6-14 and 6-16: Eccentricity = e = 0.75 in Neutral axis to outside = c = 0.5 in Allowable load = P a = 1075 lb Design Factor Design factor on load, N =
3
------>
Slenderness ratio, KL/r =
128.0
Column is: long Req'd yield strength = ------>
171
37,764 psi
Must be less than actual yield strength: sy = 42,000 psi Max. deflection, y max = 0.076 in
Data from: Problem 6-37 Cross section shape: Hollow structural section
ECCENTRIC COLUMN ANALYSIS
Solves Equation 6-14 for design stress and Equation 6-16 for maximum deflection Enter data for variables in italics in shaded boxes Use consistent U.S. Customary units. Data To Be Entered:
Computed Values:
Length and End Fixity: Column length, L = 126 in End fixity, K = 1
------>
Eq. Length, Le = KL =
126.0 in
------>
Column constant, Cc =
111.6
Argument for secant = Value of secant =
0.777 for strength 1.4025
Argument for secant = Value of secant =
0.449 for deflection 1.1098
Material Properties: Yield strength, s y =
46000 psi
Modulus of Elasticity, E = 2.90E+07 psi Cross Section Properties: [Note: Enter r or compute r = sqrt(I/A ) ] [Always enter Area] [Enter zero for I or r if not used ] Area, A = 6.020 in 2 4 Moment of Inertia, I = 0 in Radius of Gyration, r = 1.410 in Values for Eqns. 6-14 and 6-16: Eccentricity = e = 3.00 in Neutral axis to outside = c = 2.00 in Allowable load = P a = 17600 lb Design Factor Design factor on load, N =
3
Slenderness ratio, KL/r =
------>
89.4
Column is: short Req'd yield strength = ------>
45,896 psi
Must be less than actual yield strength: sy = 46,000 psi Max. deflection, y max = 0.329 in
172
Data from: Problem 6-38 Cross section shape: Solid Rectangle
ECCENTRIC COLUMN ANALYSIS
Solves Equation 6-14 for design stress and Equation 6-16 for maximum deflection Enter data for variables in italics in shaded boxes Use consistent U.S. Customary units. Data To Be Entered:
Computed Values:
Length and End Fixity: Column length, L = 40 in End fixity, K = 1
------>
Eq. Length, Le = KL =
40.0 in
------>
Column constant, Cc =
70.2
Material Properties: Yield strength, s y =
40000 psi
Modulus of Elasticity, E = 1.00E+07 psi Cross Section Properties: [Note: Enter r or compute r = sqrt(I/A ) ] [Always enter Area] [Enter zero for I or r if not used ] Area, A = 0.600 in 2 4 Moment of Inertia, I = 0 in Radius of Gyration, r = 0.433 in Values for Eqns. 6-14 and 6-16: Eccentricity = e = 1.75 in Neutral axis to outside = c = 0.75 in Allowable load = P a = 685 lb Design Factor Design factor on load, N =
3
Argument for secant = Value of secant =
0.855 for strength 1.5236
Argument for secant = Value of secant =
0.494 for deflection 1.1355
Slenderness ratio, KL/r =
------>
92.4
Column is: long Req'd yield strength = ------>
39,954 psi
Must be less than actual yield strength: sy = 40,000 psi Max. deflection, y max = 0.237 in
NOTE: This calculation was found by iteration to ensure that the required strength was satisfied. It assumes that the shape in Figure P‐38 would buckle in the plane of the page. However, the tendency to buckle about the thinner section (0.40 in thick) must also be checked. See the following page for Problem 6‐38B. The limiting load from that analysis is only 163 lb.
38A.
Note 1:
1.50 .
√
√
0.40 0.433 in;
0.600 in 685 lb [Computed separately]
173
This spreadsheet analyses the shape in Figure P3‐38 for the tendency to buckle about the axis for the thickness of the bar, 0.40 in. The loading is considered to be central without eccentricity. Data from: Problem 6-38B Cross-section shape: Rectangular
RECTANGULAR COLUMN ANALYSIS Refer to Figure 6-7 for analysis logic Enter data for variables in italics in shaded boxes
Analysis of buckling about the thin dimension Use consistent U.S. Customary units.
Data To Be Entered:
Computed Values:
Length and End Fixity: Column length, L = End fixity, K =
40 in 1
------>
Eq. Length, Le = KL =
40.0 in
------>
Column constant, Cc =
70.2
------>
Slender. ratio, KL/r =
347.8
Material Properties: Yield strength, s y =
40000 psi
Modulus of Elasticity, E = 1.00E+07 psi Cross Section Properties: [Note: A and r computed from] [dimensions for circular cross section] [in following section of this spreadsheet] 2 Area, A = 0.600 in NOTE: Buckling about the 0.40 in dimension Radius of Gyration, r = 0.115 in Properties for round column:
Column is: long Critical Buckling Load = Design Factor Design factor on load, N =
38B.
3
------>
489 lb
163 lb Allowable Load = This value governs the design; not solution 38A
Analysis as a straight centrally loaded column that tends to bucke about thin axis, 0.40 in . √
√
0.115 in
From Euler formula with
163 lb
3;
174
Data from: Problem 6-39 Cross section shape: Hollow circular section
ECCENTRIC COLUMN ANALYSIS
Solves Equation 6-14 for design stress and Equation 6-16 for maximum deflection Enter data for variables in italics in shaded boxes Use consistent SI Metric Data To Be Entered:
Computed Values:
Length and End Fixity: Column length, L = 750 mm End fixity, K = 1
------>
Eq. Length, Le = KL =
750.0 mm
200 Gpa ------>
Column constant, Cc =
63.93
Argument for secant = Value of secant =
0.811 for strength 1.4516
Argument for secant = Value of secant =
0.468 for deflection 1.1206
Material Properties: Yield strength, s y =
966 Mpa
Modulus of Elasticity, E =
Cross Section Properties: [Note: Enter r or compute r = sqrt(I/A ) ] [Always enter Area] [Enter zero for I or r if not used ] Area, A = 314 mm 2 4 Moment of Inertia, I = 0 mm Radius of Gyration, r = 7.289 mm Values for Eqns. 6-14 and 6-16: Eccentricity = e = 20 mm Neutral axis to outside = c = 12.5 mm Allowable load = P a = 5200 N Design Factor Design factor on load, N =
3
------>
Slenderness ratio, KL/r =
102.9
Column is: long Req'd yield strength = ------>
389 Mpa
Must be less than actual yield strength: sy = 966 MPa Max. deflection, y max = 2.411 mm Piston rod is safe for applied load of 5200 N.
175
Data from: Problem 6-40A-Straight
HOLLOW CIRCULAR COLUMN
Cross-section shape: Hollow circular section
Refer to Figure 6-7 for analysis logic Enter data for variables in italics in shaded boxes
Use consistent U.S. Customary units.
Data To Be Entered: Length and End Fixity: Column length, L = End fixity, K =
Computed Values:
156 in 1
------>
Eq. Length, Le = KL =
156.0 in
------>
Column constant, Cc =
128.3
------>
Slender. ratio, KL/r =
198.2
Material Properties: Yield strength, s y =
36000 psi
Modulus of Elasticity, E = 3.00E+07 psi Cross Section Properties: [Note: A and r computed from] [dimensions for circular cross section] [in following section of this spreadsheet] Area, A =
1.075 in2
Radius of Gyration, r =
0.787 in
Column is: long Critical Buckling Load = Design Factor Design factor on load, N =
3
------>
176
8,101 lb
2,700 lb Allowable Load = See also Solution 40B for crooked pipe.
Data from: Problem 6-40B-Crooked
CROOKED COLUMN ANALYSIS
Cross-section shape: Hollow circular section
Solves Equation 6-13 for Allowable Load Enter data for variables in italics in shaded boxes
Use consistent
Data To Be Entered: Length and End Fixity: Column length, L = End fixity, K = 1
U.S. Customary units.
Computed Values:
156 in ------>
Eq. Length, Le = KL =
156.0 in
------>
Column constant, Cc =
128.3
Material Properties: Yield strength, s y =
36000 psi
Modulus of Elasticity, E = 3.00E+07 psi
Euler buckling load =
Cross Section Properties: [ Note: Enter r or compute r = sqrt(I/A ) ] [Always enter Area] [Enter zero for I or r if not used ] Area, A = Moment of Inertia, I =
-22074
C 2 in Eqn. 6-13 = 3.483E+07
1.075 in 2 0 in 4
Radius of Gyration, r = 0.787 in Values for Eqn. 6-11: Initial crook edness = a = 1.25 in Neutral axis to outside = c = 1.188 in Design Factor Design factor on load, N = 3
C 1 in Eqn. 6-13 =
8101 lb
------>
Slenderness ratio, KL/r =
198.2
Column is: long Straight Column Critical Buckling Load = 8,101 lb Crooked Column ------> Allowable Load = 1,711 lb Comparisons: Straight (40A) vs. Crooked (40B) Problem # 40A 40B Straight Crooked Allowable Load = 2700 lb 1711 lb Allowable Load for crooked column is limit load
177
CHAPTER 7 BELT DRIVES, CHAIN DRIVES AND WIRE ROPE V-Belt Drives 1.
24.0 in; 2 24
13.95 in; 1.57 13.95
Use 2.
Actual CD from Equation 7‐13: 6.28 13.95
.
4.
.
.
180°
2
.
180°
2
. .
60.0 in;
27.7 in;
[Equation 7‐13]
.
. °
[Equation 7‐14]
.
. °
[Equation 7‐15]
.
8.4 in; 5V Belt [Equation 7‐12]
178.2 in
Actual
[From Table 7‐2]
.
[Equation 7‐13] . °
7.
144 in;
94.8 in;
10.
78.73 in
75 in
.
160.1°;
9.
.
179.4
6.
8.
.
5.25
.
Computed 5.
5.25
[Use Equation 7‐12]
– Standard Length [From Table 7‐2]
4 75
3.
5.25 in; 3V Belt
[Equations 7‐14, 7‐15] 13.8 in; 8V Belt
[Equation 7‐12]
Computed
469.9 in
[From Table 7‐2]
Actual
.
Equation 7‐13
. °;
. °
[Equations 7‐14, 7‐15]
.
/
178
.
11.
.
12. 13.
/
Ratio
/
13.95/5.25
2.66;
Corrected Power 14.
Ratio
27.7/8.4
Ratio
;
0.94;
0.94 1.03 6.25 3.30;
15.5
Corrected Power 15.
≅ 6.25
1.26
1.03
.
16.76 hp;
0.95 1.05 16.76
94.8/13.8
6.87;
Corrected Power
48 hp;
0.95; .
0.904;
0.904 1.09 48
1.05
1.09 .
16.
A 15N belt is a metric size having a top width of 15 mm similar to a 5V belt.
17.
A 17A belt is a metric automotive belt having a top width of 17 mm. Similar to a “3/4 in” automotive belt.
18.
Design: Service Factor
1.5; Design Power
1.5 25
37.5 hp
From Figure 7‐13 – Use 5V Belt Ratio
870/310
For
4000 ft/min;
For
13.1 in;
Rated Power
2.81
22.5
17.56 in 37.4 in;
870
0.94
23.44 hp
13.1
151.5
13.1/37.4
Center Distance: 3 37.4
3 37.4
Try
48 in → Nominal
178 in
Use
180 in → Actual
48.85 in [Equation 7‐13]
179
[Equation 7‐12]
305 rpm Ok
151.2°;
208.8°
0.92;
19.
[Equations 7‐14, 7‐15]
1.06
Corrected Power
0.92 1.06 23.44
Number of Belts
37.5 hp/22.86 hp/belt
1.64 belts → 2 belts
Design: Service Factor
1.2; Design Power
Ratio
1750/72.5 hp
2.41;
For
7.95;
18.95;
103 in → Use 157.6°;
1750
Ratio
1500/550
For
10.2 in;
30 in
10.55 hp/belt
One Belt Required
1.4 40
56 hp → 5V Belt
12 4000 /
1500
1500
10.2/27.7
Rated Power ≅ 25.7 hp by interpolation on Figure 7
10.2 in 552 rpm Ok
15 at 1500 rpm
27.7
114; Use
36 in;
133.6 in → Use
Actual
35.16 in;
151.2°;
208.8°;
Corrected Power Number of Belts 21.
734 rpm Ok
1.09
1.4; Design Power
27.7 in;
8.7 in
28.35 in
0.94;
2.73;
1750
80.7; Try
0.94 1.09 10.3
Design: Service Factor
6.0 hp → 3V Belt
7.95/18.95
100 in; Actual 202.4°;
Corrected Power
1.2 5
12 4000 /
Rated Power ≅ 10.3 hp; 18.95
20.
22.86 hp/belt
0.92 1.01 25.7 56/23.9
Design: Service Factor Ratio
1250/695
For
10.55 in;
Rated Power
0.92;
1.01
23.9 hp/belt
2.35 → Use 3 Belts
1.4; Design Power
1.80;
132 in
12 4000 /
18.95 in;
1250
1.4 20 1250
12.2 in
10.55/18.95
10.4 hp by interpolation on Figure 7
180
28 hp → 3V Belt
695.9 rpm Ok
14 at 1250 RPM
18.95
88.5; Use
Actual
21.43 in;
Corrected Power Number of Belts 22.
20 in; 157.4°;
28/10.46
870/625
For
10.8 in;
Rated Power
136 in → Use 174.9°;
10.46 hp/belt
12 4000 /
14.9 in; 0.77
870
870
17.6 in
10.8/14.9
18.37 hp; 14.9
132 in; Actual
17.6 in
77.1; Use
0.988;
1.01
0.988 1.01 18.37
18.33 hp/belt
200/18.33
Try 8V belt: For
17.8 in;
24.8 in;
624.4 rpm Ok
66 hp; 24.8
12.78;
48 in
163 in → 171.4°; Corrected Power Number of Belts
10.9 → Use 11 Belts Not Acceptable
160 in; Actual 188.6°;
46.43 in
0.98;
0.98 0.94 66 200/60.8
48 in
45.78 in
Number of Belts
Rated Power
1.07
200 hp → 5V Belt
185.1°;
Corrected Power
0.94;
2.68 Belts → Use 3 Belts
1.39;
17.6
90 in
2.0 The crusher may experience choking.
2.0 100
Ratio
202.6°;
0.94 1.07 10.4
Design: Service Factor Design Power
87.2 in → Use
0.94 60.8 hp/belt
3.3 → Use 4 Belts
181
Roller Chain Drives 23.
Chain No. 140: Pitch
24.
Chain No. 60: Pitch
25.
Static Load
14 6
3
8
4 in
1250 lb; Average tensile strength
Use a No. 80 Chain 26.
1 3 4 in
8
.
Load on each chain =
; 2500 lb;
Use No. 120 chain 27.
10
12 500 lb
. .
[Table 7‐12]
. . ;
10
25 000 lb
. .
[Table 7‐12]
Fatigue of link plates; Impact of rollers on sprocket teeth; galling between pins and bushings.
28.
[From Table 7‐15] Given: No. 60 chain, 20 teeth, 750 rpm Rated Power
21.69 hp Using interpolation ; Type B lubrication bath
Service Factor
1.2 for hydraulic drive
Design power rating 29.
30.
21.69/1.2
3 Strands: Factor
2.5
Power Rating
2.5 18.08
.
.
[From Table 7‐14] Given: No. 40 chain, 12 teeth, 860 rpm Rated Power
4.44 hp By interpolation , Type B lubrication bath
Service Factor
1.2: Design Power Rating
31.
4 Strands: Factor
3.3; Power Rating
32.
[From Table 7‐16] Given: No. 80 chain, 32 teeth, 1160 rpm
4.44 hp/1.2
3.3 3.70
. .
Rated Power
78.69 hp By interpolation , Type C lubrication oil stream
Service Factor
1.2; Design Power
78.69 hp/1.2
182
65.57 hp
33.
2 Strands: Factor
1.7; Power Rating
34.
No. 60 chain;
15;
3
4 in
50;
36 in
128 0.75
128 Pitches;
From Equation 7
128
128
47.42 pitches
2 in
0.50 in:
19 47.42 Pitches
0.75 in/pitch 11;
No. 40 chain: 1
19
129.1 Pitches
Use 128 Pitches (Even Number);
36.
Use Equation 7
48 Pitches
2 48
For
111.5 hp
0.75 in
36 in/0.75 in/Pitch
35.
1.7 65.57 hp
.
45;
24 in
24/0.5
48 Pitches
2 48
124.6 Pitches → Use 124 Pitches 124 0.5
37.
124
124
47.69 Pitches 38.
47.69 Pitches
0.5 in/Pitch
23.85 in
Design: 25 hp,
310 rpm;
160 rpm; Nominal Ratio
Service Factor
1.5; Design Power
Use 3 strands: Rating
37.5/2.5
1.5 25
310/160
1.94
37.5 hp
15.0 hp per strand
No. 80 chain; p = 1.00 in; 15 teeth rated > 15.8 hp at 310 rpm; Type B lubrication Ratio
15 1.94
/
310
29.1 → 29 teeth
15/29
160.3 rpm Ok 183
1.00 / Use
180/15
4.810 in;
40 Pitches with
180/29
9.249
[Eq. 7‐20]
1.00 in for No. 80 chain
2 40 102
1.00/
102.1 Pitches → Use 102 Pitches 102
39.94 Pitches
Problems 38‐42 are design problems for chain drives for which there are no unique solutions. The general procedure is illustrated in Example Problem 7‐5 and the solution for Problem 38 above. The example solutions to these and the other design problems are shown on the following pages using spreadsheets. Data for design power per strand from Tables 7‐14, 7‐15 or 7‐16 must be used to ensure that the selected chain design has sufficient capacity. You should also consider using data from commercial chain drive manufacturers such as those listed in the Internet Sites 4, 6, 8, 12, 14, and 15 at the end of Chapter 7.
184
CHAIN DRIVE DESIGN
Application Heavy conveyor driven by a gasoline engine Example Problem 7-5 Initial Input Data: Power input: 15 hp Service factor: 1.4 Table 7-17 Input speed: 900 rpm Desired output speed: 235 rpm Range = 230 to 240 Computed Data: Design power: 21 hp Design speed ratio: 3.83 Design Decisions-Chain Type and Teeth Numbers: Number of strands: 1 1 2 3 4 Strand factor: 1.0 1.0 1.7 2.5 3.3 Required power per strand: 21.00 hp Chain number: 60 Tables 7-7, 7-8, or 7-9 Chain pitch: 0.75 in Nunber of teeth-Driver sprocket: 17 Computed no. of teeth-Driven sprocket: 65.11 Enter: Chosen number of teeth: 65 Check availability from vendor Computed Data: Actual output speed: 235.4 rpm Pitch diameter-Driver sprocket: 4.082 in Pitch diameter-Driven sprocket: 15.524 in Center Distance, Chain Length and Angle of Wrap: Enter: Nominal center distance: 40 pitches 30 to 50 pitches recommended Computed nominal chain length: 122.5 pitches Enter: Specified no. of pitches: 122 pitches Even number recommended Actual chain length: 91.50 in Computed actual center distance: 39.766 pitches Actual center distance: 29.825 in Angle of wrap-Driver sprocket: 157.9 degrees Should be greater than 120 degrees Angle of wrap-Driven sprocket: 202.1 degrees
185
CHAIN DRIVE DESIGN
Application: Hammer mill driven by an electric motor Problem 7-38 Initial Input Data: Power input: 25 hp Service factor: 1.5 Table 7-17 Input speed: 310 rpm Desired output speed: 160 rpm Computed Data: Design power: 37.5 hp Design speed ratio: 1.94 Design Decisions-Chain Type and Teeth Numbers: Number of strands: 3 1 2 3 4 Strand factor: 2.5 1.0 1.7 2.5 3.3 Required power per strand: 15.00 hp Chain number: 80 Tables 7-7, 7-8, or 7-9 Chain pitch: 1.00 in Nunber of teeth-Driver sprocket: 15 Computed no. of teeth-Driven sprocket: 29.06 Enter: Chosen number of teeth: 29 Check availability from vendor Computed Data: Actual output speed: 160.3 rpm Pitch diameter-Driver sprocket: 4.810 in Pitch diameter-Driven sprocket: 9.249 in Center Distance, Chain Length and Angle of Wrap: Enter: Nominal center distance: 40 pitches 30 to 50 pitches recommended Computed nominal chain length: 102.1 pitches Enter: Specified no. of pitches: 102 pitches Even number recommended Actual chain length: 102.00 in Computed actual center distance: 39.938 pitches Actual center distance: 39.938 in Angle of wrap-Driver sprocket: 173.6 degrees Should be greater than 120 degrees Angle of wrap-Driven sprocket: 186.4 degrees
186
CHAIN DRIVE DESIGN
Application Agitator driven by a gasoline engine Problem 7-39 Initial Input Data: Power input: 5 hp Service factor: 1.0 Table 7-17 Input speed: 750 rpm Desired output speed: 325 rpm Computed Data: Design power: 5 hp Design speed ratio: 2.31 Design Decisions-Chain Type and Teeth Numbers: Number of strands: 1 1 2 3 4 Strand factor: 1.0 1.0 1.7 2.5 3.3 Required power per strand: 5.00 hp Chain number: 40 Tables 7-7, 7-8, or 7-9 Chain pitch: 0.5 in Nunber of teeth-Driver sprocket: 16 Computed no. of teeth-Driven sprocket: 36.92 Enter: Chosen number of teeth: 37 Check availability from vendor Computed Data: Actual output speed: 324.3 rpm Pitch diameter-Driver sprocket: 2.563 in Pitch diameter-Driven sprocket: 5.896 in Center Distance, Chain Length and Angle of Wrap: Enter: Nominal center distance: 32 pitches 30 to 50 pitches recommended Computed nominal chain length: 90.8 pitches Enter: Specified no. of pitches: 90 pitches Even number recommended Actual chain length: 45.00 in Computed actual center distance: 31.573 pitches Actual center distance: 15.787 in Angle of wrap-Driver sprocket: 167.9 degrees Should be greater than 120 degrees Angle of wrap-Driven sprocket: 192.1 degrees
187
CHAIN DRIVE DESIGN
Application: Heavy conveyor driven by a 6-cylinder Engine Problem 7-40 Initial Input Data: Power input: 40 hp Service factor: 1.4 Table 7-17 Input speed: 500 rpm Desired output speed: 250 rpm Computed Data: Design power: 56 hp Design speed ratio: 2.00 Design Decisions-Chain Type and Teeth Numbers: Number of strands: 3 1 2 3 4 Strand factor: 2.5 1.0 1.7 2.5 3.3 Required power per strand: 22.40 hp Chain number: 80 Tables 7-7, 7-8, or 7-9 Chain pitch: 1.00 in Nunber of teeth-Driver sprocket: 14 Computed no. of teeth-Driven sprocket: 28.00 Enter: Chosen number of teeth: 28 Check availability from vendor Computed Data: Actual output speed: 250.0 rpm Pitch diameter-Driver sprocket: 4.494 in Pitch diameter-Driven sprocket: 8.931 in Center Distance, Chain Length and Angle of Wrap: Enter: Nominal center distance: 36 pitches 30 to 50 pitches recommended Computed nominal chain length: 93.1 pitches Enter: Specified no. of pitches: 94 pitches Even number recommended Actual chain length: 94.00 in Computed actual center distance: 36.432 pitches Actual center distance: 36.432 in Angle of wrap-Driver sprocket: 173.0 degrees Should be greater than 120 degrees Angle of wrap-Driven sprocket: 187.0 degrees
188
CHAIN DRIVE DESIGN
Application: Centrifugal pump driven by a steam turbine Problem 7-41 Initial Input Data: Power input: 20 hp Service factor: 1.0 Table 7-17 Input speed: 2200 rpm Desired output speed: 775 rpm Computed Data: Design power: 20 hp Design speed ratio: 2.84 Design Decisions-Chain Type and Teeth Numbers: Number of strands: 2 1 2 3 4 Strand factor: 1.7 1.0 1.7 2.5 3.3 Required power per strand: 11.76 hp Chain number: 40 Tables 7-7, 7-8, or 7-9 Chain pitch: 0.50 in Nunber of teeth-Driver sprocket: 25 Computed no. of teeth-Driven sprocket: 70.97 Enter: Chosen number of teeth: 71 Check availability from vendor Computed Data: Actual output speed: 774.6 rpm Pitch diameter-Driver sprocket: 3.989 in Pitch diameter-Driven sprocket: 11.304 in Center Distance, Chain Length and Angle of Wrap: Enter: Nominal center distance: 30 pitches 30 to 50 pitches recommended Computed nominal chain length: 109.8 pitches Enter: Specified no. of pitches: 110 pitches Even number recommended Actual chain length: 55.00 in Computed actual center distance: 30.110 pitches Actual center distance: 15.055 in Angle of wrap-Driver sprocket: 151.9 degrees Should be greater than 120 degrees Angle of wrap-Driven sprocket: 208.1 degrees
189
CHAIN DRIVE DESIGN
Application: Rock crusher driven by a hydraulic drive Problem 7-42 Initial Input Data: Power input: 100 hp Service factor: 1.4 Table 7-17 Input speed: 625 rpm Desired output speed: 225 rpm Computed Data: Design power: 140 hp Design speed ratio: 2.78 Design Decisions-Chain Type and Teeth Numbers: Number of strands: 4 1 2 3 4 Strand factor: 3.3 1.0 1.7 2.5 3.3 Required power per strand: 42.42 hp Chain number: 80 Tables 7-7, 7-8, or 7-9 Chain pitch: 1.00 in Nunber of teeth-Driver sprocket: 21 Computed no. of teeth-Driven sprocket: 58.33 Enter: Chosen number of teeth: 58 Check availability from vendor Computed Data: Actual output speed: 226.3 rpm Pitch diameter-Driver sprocket: 6.710 in Pitch diameter-Driven sprocket: 18.471 in Center Distance, Chain Length and Angle of Wrap: Enter: Nominal center distance: 40 pitches 30 to 50 pitches recommended Computed nominal chain length: 120.4 pitches Enter: Specified no. of pitches: 120 pitches Even number recommended Actual chain length: 120.00 in Computed actual center distance: 39.815 pitches Actual center distance: 39.815 in Angle of wrap-Driver sprocket: 163.0 degrees Should be greater than 120 degrees Angle of wrap-Driven sprocket: 197.0 degrees
Problems 38‐42 are design problems for which no unique solutions exist.
190
Synchronous Belt Drives Note that the data from the ten problems in Tables 7‐27 and 7‐28 can be used for practice problems for synchronous belt drives. However, no unique solutions exist. Problems 43 and 44 that follow illustrate parts of the procedures for those design problems. Example Problems 7‐3 and 7‐4 also demonstrate methods that can be applied to open‐ended design problems.
43
Given: Input speed = n1 = 800 rpm; Output speed = n2 = 600 rpm; Belt 1440‐8MGT (a) Possible synchronous drive sprocket combinations (b) Center distance and belt linear velocity, vb, for each combination Speed ratio = n1/n2: = 800/600 = 1.333; Combinations from Table 7‐7 Example: From Table 7‐7: Driver – 24 teeth; Driven – 32 teeth; CD = 23.93 in (a) Possible synchronous drive sprocket combinations (b) Center distance and belt linear velocity, vb, for each combination From Table 7‐4: Sprocket pitch diameters: D1 = 2.406 in; D2 = 3.208 in Belt speed, using input sprocket: vb = (D1/2)(n1) = (2.406 in/2)(800 rev/min)(2 rad/rev)(1ft/12 in) = 503.9 ft/min Sprocket sets CD (in)
D1 (in)
D2 (in) Belt speed, vb (ft/min)
24:32
2.406
3.208
23.93
503.9
30:40
22.83
3.008
4.010
630.0
36:48
21.72
3.609
4.812
755.9
48:64
19.51
4.812
6.416
1007.8
191
44
Given: Input speed = n1 = 1200 rpm; Output speed = n2 = 600 rpm; Belt 1800‐8MGT‐30
Speed ratio = 1200/600 = 2.000; List possible combinations from Table 7‐7 (a) Possible synchronous drive sprocket combinations (b) Center distance and belt linear velocity, vb, for each combination (c) Required taper‐lock bushing size with Dmin and Dmax Example: From Table 7‐7: Driver – 22 teeth; Driven – 44 teeth; CD = 30.22 in From Table 7‐4: Sprocket pitch diameters and From Table 7‐5: Bushings N1 = 22; D1 = 2.206 in; Bushing 1108; Dmin = 0.500 in; Dmax = 0.875 in N2 = 44; D2 = 4.111 in; Bushing 2012; Dmin = 0.500 in; Dmax = 1.875 in Belt speed, using input sprocket: vb = (D1/2)( n1) = (2.206 in/2)(1200 rev/min)(2 rad/rev)(1ft/12 in) = 693.0 ft/min Sprocket sets
CD (in)
D1 (in)
22: 44
30.22
2.206
24: 48 28: 56 32: 64 36: 72 40: 80 56: 112 72: 144
29.74
D2 (in)
Belt speed, vb (ft/min)
Bushing
Dmin (in)
Dmax (in)
693.0
1108 2012
0.500 0.500
0.875 1.875
755.9
1108 2012 1108 2012 1210 2517 1610 2517 2012 2517 2012 2517 2517 2517
0.500 0.500 0.500 .0500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500
0.875 1.875 0.875 1.875 1.250 2.250 1.500 2.250 1.875 2.250 1.875 2.250 2.250 2.250
4.111 2.406 4.802 28.79
2.807
27.83
3.208
26.87
3.609
881.8 5.614 6.416
1007.8 1133.8
7.218 25.91
4.010
1259.8 8.020
22.03
5.614
1763.7 11.229
18.66
7.598
2387.0 14.437
192
CHAPTER 8 KINEMATICS OF GEARS Gear Geometry 1.
;
a.
/
44/12
3.667 in
b.
/
/12
0.2618 in
c.
Equivalent
25.4/
d.
Nearest standard
25.4/12 2.00 mm
e.
1/
1/12
f.
1.25/
1.25/12
0.1042 in
g.
0.25/
0.25/12
0.0208 in
2.25/
0.1875 in
h. i.
2
2/
j.
/2
0.0833 in
2/12 /2 12
2 /
k. ;
a.
34/24
1.417 in
b.
/24
0.131 in
c.
Equivalent
d.
Nearest standard 1/24
0.1667 in 0.131 in
46/12
2.
e.
2.117 mm
25.4/24
3.833 in
1.058 mm
1.00 mm
0.0417 in
193
f.
.
0.002
0.0502 in
g.
.
0.002
0.0103 in
h.
0.0417
i.
2
j.
/2 24
0.0520
2/24
k.
0.0937 in
0.0833 in
0.0654 in
2 /
36/24
3.
;
a.
45/2
22.500 in
b.
/2
1.571 in
c.
Equivalent
25.4/2
d.
Nearest standard
12.7 mm 12 mm
e.
1/2
f.
1.25/2
0.625 in
g.
0.25/2
0.125 in
h.
2.25/2
1.125 in
i.
2/2
j.
/2 2
0.7854 in
k.
47/2
23.50 in
4.
;
a.
18/8
2.250 in
b.
/8
0.3927 in
c.
Equivalent
1.500 in
0.500 in
1.000 in
25.4/8
3.175 mm
194
d.
Nearest standard
3.0 mm
e.
1/8
0.125 in
f.
1.25/8
0.1563 in
g.
0.25/8
0.0313 in
h.
2.25/8
0.2813 in
i.
2/8
j.
/2 8
0.1963 in
k.
20/8
2.500 in
5.
;
.
a.
22/1.75
12.571 in
b.
/1.75
1.795 in
c.
Equivalent
d.
Nearest standard
e.
1/1.75
f.
1.25/1.75
0.7143 in
g.
0.25/1.75
0.1429 in
h.
2.25/1.75
1.2857 in
i.
2/1.75
j.
/2 1.75
0.8976 in
k.
24/1.75
13.714 in
0.250 in
25.4/1.75
14.514 mm
16 mm
0.5714 in
1.1429 in
195
6.
; a.
20/64
0.3125 in
b.
/64
0.0491 in
c.
Equivalent
d.
Nearest standard
25.4/64
0.397 mm
0.40 mm
e.
1/64
0.0156 in
f.
1.200/64
0.002
0.0208 in
g.
0.200/64
0.002
0.0051 in
h.
0.0364 in
i.
2/64
j.
/2 64
0.0245 in
k.
22/64
0.3438 in
a.
180/80
0.3125 in
b.
/80
c.
Equivalent
d.
Nearest standard
7.
0.0313 in
;
0.0393 in 25.4/80
0.318 mm
0.30 mm
e.
1/80
f.
1.200/80
0.002
0.0170 in
g.
0.200/80
0.002
0.0045 in
h. i.
0.0125 in
0.0295 in 2/80
0.025 in
196
j.
/2 80
0.0196 in
k.
182/80
2.275 in
8.
; a.
28/18
1.556 in
b.
/18
0.1745 in
c.
Equivalent
d.
Nearest standard
25.4/18
1.5 mm
e.
1/18
f.
1.25/18
0.0694 in
g.
0.25/18
0.0139 in
h.
2.25/18
0.125 in
i.
2/18
j.
/2 18
0.0873 in
k.
30/18
1.667 in
a.
28/20
1.400 in
b.
/20
0.1571 in
c.
Equivalent
d.
Nearest standard
9.
1.411 mm
0.0556 in
0.1111 in
;
25.4/20
1.27 mm
1.25 mm
e.
1/20
f.
1.200/20
0.002
0.0620 in
g.
0.200/20
0.002
0.012 in
h.
0.050 in
0.1120 in
197
i.
2/20
j.
/2 20
0.0785 in
k.
30/20
1.500 in
10.
0.1000 in
;
/
a.
3 34
b.
/
c.
Equivalent
102 mm 3
9.425 mm
25.4/
25.4/3
d.
8
e.
3.00 mm
f.
1.25
1.25 3
3.750 mm
g.
0.25
0.25 3
0.750 mm
2.25
6.750 mm
h. i.
2
j.
/2
2
k. 11.
8.47
;
6.00 mm /2
4.712 mm
2
3 36
108 mm
.
a.
1.25 45
56.25 mm
b.
1.25
3.927 mm
c.
25.4/
25.4/1.25 20
d. e.
1.25 mm
f.
1.25
1.25 1.25
1.563 mm
g.
0.25
0.25 1.25
0.313 mm
198
20.3
h.
2.25
i.
2
2 1.25
j.
/2
/2
1.963 mm
2
1.25 47
2.25 1.25
k. 12.
2.813 mm
2.500 mm
58.75 mm
; a.
18 18 12
b.
216 mm
37.70 mm
c.
25.4/12
d.
2
e.
12 mm
f.
1.25 12
15.00 mm
g.
0.25 12
3.00 mm
h.
2.25 12
27.00 mm
i.
2
j.
12 /2
24.00 mm
k. 13.
2.117
18.85 mm 2
12 20
240 mm
; a. b.
20 22 20
62.83 mm 25.4/20
c. d. e. f.
440 mm
1.27
1.25 20 mm 1.25 20
25.00 mm
199
g.
0.25 20
5.00 mm
h.
2.25 20
45.00 mm
i.
2
j.
20 /2
31.42 mm
k.
20 24
480 mm
14.
2 20
40.00 mm
; a.
20.00 mm
b.
20
3.14 mm
c.
25.4/1 24
d.
Nearest standard
e.
1.00 mm
f.
1.25 1
1.25 mm
g.
0.25 2
0.25 mm
h.
2.25 1
2.25 mm
i.
2 1
2.00 mm
j.
1 /2
1.571 mm
k.
1 22
15.
;
25.4
22.00 mm
.
a.
0.4 180
b.
0.4
72.00 mm 1.26 mm
c.
25.4/0.4
d.
Nearest standard
e.
0.40 mm
63.5
64
200
f.
1.25 0.4
0.50 mm
g.
0.25 0.4
0.10 mm
h.
2.25 0.4
0.900 mm
i.
2 0.4
0.80 mm
j.
0.4 /2
0.628 mm
k.
0.4 182
72.80 mm
16.
;
.
a.
1.5 28
b.
1.5
22.40 mm 4.71 mm
c.
25.4/1.5
d.
16
e.
1.50 mm
f.
1.25 1.50
1.875 mm
g.
0.25 1.5
0.375 mm
h.
2.25 1.5
3.375 mm
i.
2 1.5
3.00 mm
j.
1.5 /2
2.36 mm
k.
1.5 30
45.00 mm
17.
16.93
;
.
a.
0.8 28
22.40 mm
b.
0.8
2.51 mm
c.
25.4/0.8
d.
32
31.75
201
e.
0.80 mm
f.
1.25 0.8
1.00 mm
g.
0.25 0.8
0.20 mm
h.
2.25 0.8
1.80 mm
i.
2 0.8
1.60 mm
j.
0.8 /2
1.257 mm
k.
0.8 30
24.00 mm
18.
Backlash – See Section 8‐4 and Table 8‐5
19.
Gear from Problem 1:
12: Backlash 0.006 to 0.009 in
Gear from Problem 12:
12: Backlash 0.52 to 0.82 mm
The amount of backlash depends on center distance.
Velocity Ratio 20.
a.
.
b.
/
64/18 /
c.
. 2450 18/64
d. 21.
/
a.
20
b.
92/20
c.
225 20/92
d.
92 /2 4
.
. . .
/
202
22.
a.
30
68 /2 20
b.
68/30
c.
850 30/68
.
.
d. 23.
/
a.
40
250 /2 64
b.
250/40
c.
3450 40/250
.
d. 24.
/
a.
24
88 /2 12
b.
88/24
c.
1750 24/88
.
.
/
d. 25.
.
a.
/2
b.
/
68
68/22
22 2 /2 .
1750 22/68
c. d.
. 26.
a.
48
b.
48/18
c.
1150 18/48
d.
.
18 0.8 /2
.
.
.
/
203
/
.
27.
a.
45
36 4 /2
b.
45/36
c.
150 36/45
.
d. 28.
.
a.
63
b.
36/15
c.
480 15/36
/
15 12 /2 .
d.
.
/
Errors in statements for Problems 29‐32: 29.
Pinion and gear can’t have different pitches.
30. 31.
8.333 in; Given
is inaccurate by 0.033 in
Too few teeth in the pinion, assuming 20° . . teeth interference would occur.
32.
2.156 in;
cannot be used to find .
Housing Dimensions 33.
Housing must clear the addendum circle of all gears by 0.10 in/side 1/
1/8
8.25 in
0.125 in;
2 0.10 2
2 / .
66/8
8.25 in
in = Y
2 0.10
2 0.1
.
204
2.250
8.00
0.25
0.20
34.
2 / 2
35.
40.0 mm; 2
0.8 48
2 0.8
188 mm: 2
4.0
188
2 2
2 2 2
2 0.8
2 2
.
2 2 0.8 48
0.2
2 2
4.0
Gear Trains – Analysis 37.
15.75 /
1750 rpm/
15.75
38.
8.407 /
1750 rpm/ 8.0407
39.
.
.
.
.
.
.
/
20/16
1.250 in
/
38/16
2.375 in
/
18/12
1.500 in
.
0.0900
40. /
1750/0.0900
205
.
.
40.0
2 2
47 4
0.8 18
3.938 0.20
50 0.8
2
36.
3.938 in:
2 0.1
2
0.8 18
252/64
12.139
.
Helical Gearing 41.
Helical Gear
8;
Helix Angle
30°
Circular Pitch
14 °;
/
/8
45 teeth;
2.00 in
0.3927
30°
8/
30°
.
Normal Circular Pitch Normal Diametral Pitch
/
Axial Pitch
°
Pitch Diameter
/
45/8
.
=
.
=
. .
Normal Pressure Angle 14.5° / 42.
2.00 in/0.680 in
Helical Gear /
30°
12;
20°;
But
12 . .
;
°
43.
.
/ .
Helical Gear / /
8.485
/
/12
.
/
48/8.485
45°;
1.00 in
.
.
36; /6
45°
. °
°
/
1.50 in;
14.5°
; .
/
°=
.
48;
/8.485
.
6;
. 6/
14 °; ;
45°
45°
8.485;
206
°
. .
/ / 44.
36/6 .
.
;
/ .
Helical Gear
.
.
/16.97
.
/
0.1851 in/
Axial pitches in face width (Low)
24;
72;
14 °;
;
/
72/16.97
/ .
.
.
.
. °
°
.
45°
45°
.
. °
/
0.25 in;
0.1851 in 45°
°
Low
Bevel Gears Problems 45 to 51 deal with the geometry of bevel gears. Table 8‐8 in Section 8‐8 gives all of the pertinent equations. Example Problem 8‐3 demonstrated all of the necessary calculations.
Problem 8‐49 has been selected to show the detailed calculations here in the
Solutions Manual.
Problem 8‐46 calls for preparing technical drawings for the bevel gear pair
described in Problem 8‐45.
Problem 8‐48 calls for preparing technical drawings for the bevel gear pair
described in Problem 8‐47.
Solutions are shown in the form of spreadsheets for Problems 8‐45, 8‐47, 8‐49, 8‐50,
and 8‐51. 49.
Given:
;
;
;
°
Computed Values: /
Gear ratio: Pitch diameter – Pinion:
/
72/18 18/12
207
4.000 1.500 in
Pitch diameter – Gear:
/
72/12
Pitch cone angle – Pinion: Pitch cone angle – Gear:
Γ
Outer cone distance:
6.000 in
/
18/72
14.03°
/
72/18
75.96°
0.5 / sin 0.5 6.00
Face width must be specified: Nominal face width:
/ sin 75.96°
3.092 in
.
, based on the following guidelines:
0.30
0.30 3.092
0.928 in
Maximum face width:
/3
3.092
/3
or
10/
10/12
0.833 in
Mean cone distance: Ratio
/
0.5 2.692/3.092
3.092 in
1.031 in
0.5 0.80
2.692 in
0.871
(This ratio occurs in several following calculations) Mean circular pitch:
/
Mean working depth:
2.00/
/
Clearance:
0.125
0.125 0.145
/12 0.871
0.145 in
Mean whole depth: Mean addendum factor:
/
0.228 in
2.00/12 0.871 0.018 in
0.018 in
0.210
0.290/
0.210
0.290/ 4.00
0.145 in
0.163 in
0.228
Gear mean addendum:
0.228 0.145
Pinion mean addendum:
0.145 in
0.033 in
0.112 in
Gear mean dedendum:
0.163 in
0.033 in
0.130 in
Pinion mean dedendum:
0.163 in
0.112 in
0.051 in
Gear dedendum angle:
/
208
0.033 in
0.130/2.692
2.76°
Pinion dedendum angle:
/
Gear outer addendum:
0.051/2.692
1.09°
0.5 0.033
0.5 0.80
1.09°
0.5 0.80
2.76°
0.0406 Pinion outer addendum:
0.5 0.112
0.1313 Gear outside diameter:
2 6.000 in
Pinion outside diameter:
2 0.0406
75.96°
6.020
2 0.1313
14.04°
1.755
2 1.500 in
209
BEVEL GEAR GEOMETRY Computes values for features described in Table 8-8 in Section 8-8 of the text PROBLEM: 8-49 EXAMPLE PROBLEM: 8-3 GIVEN DATA GIVEN DATA No of teeth in pinion = 18 No of teeth in pinion = 16 No of teeth in gear = 72 No of teeth in gear = 48 Diametral pitch = 12 Diametral pitch = 8 Pressure angle = 20 degrees Pressure angle = 20 degrees COMPUTED VALUES Gear ratio 4.000 Pitch diameter: Pinion 1.500 in Pitch diameter: Gear 6.000 in Pitch cone angle: Pinion 14.036 degrees Pitch cone angle: Gear 75.964 degrees Outer cone distance 3.092 in FACE WIDTH Nominal face width 0.928 in Maximum face width (a) 1.031 in Maximum face width (b) 0.833 in Max face width = smallest of (a) or (b) INPUT Face width 0.800 in COMPUTED VALUES Mean cone distance 2.692 in 0.871 Ratio A m /A o Mean circular pitch 0.228 in mean working depth 0.145 in Clearance 0.018 in Mean whole depth 0.163 in mean addendum factor 0.228 Gear mean addendum 0.033 in Pinion mean addendum 0.112 in Gear mean dedendum 0.130 in Pinion mean dedendum 0.051 in Gear dedendum angle 2.767 degrees Pinion dedendum angle 1.090 degrees Gear outer addendum 0.041 in Pinion outer addendum 0.131 in Gear outside diameter 6.020 in Pinion outside diameter 1.755 in
COMPUTED VALUES Gear ratio 3.000 Pitch diameter: Pinion 2.000 in Pitch diameter: Gear 6.000 in Pitch cone angle: Pinion 18.435 degrees Pitch cone angle: Gear 71.565 degrees Outer cone distance 3.162 in FACE WIDTH Nominal face width 0.949 in Maximum face width (a) 1.054 in Maximum face width (b) 1.250 in Max face width = smallest of (a) or (b) INPUT Face width 1.000 in COMPUTED VALUES Mean cone distance 2.662 in Ratio A m /A o 0.842 Mean circular pitch 0.331 in mean working depth 0.210 in Clearance 0.026 in Mean whole depth 0.237 in mean addendum factor 0.242 Gear mean addendum 0.051 in Pinion mean addendum 0.159 in Gear mean dedendum 0.186 in Pinion mean dedendum 0.077 in Gear dedendum angle 3.992 degrees Pinion dedendum angle 1.663 degrees Gear outer addendum 0.065 in Pinion outer addendum 0.194 in Gear outside diameter 6.041 in Pinion outside diameter 2.369 in
210
BEVEL GEAR GEOMETRY Computes values for features described in Table 8-8 in Section 8-8 of the text PROBLEM: 8-45 PROBLEM: 8-47 GIVEN DATA GIVEN DATA No of teeth in pinion = 15 No of teeth in pinion = 25 No of teeth in gear = 45 No of teeth in gear = 50 Diametral pitch = 6 Diametral pitch = 10 Pressure angle = 20 degrees Pressure angle = 20 degrees COMPUTED VALUES Gear ratio 3.000 Pitch diameter: Pinion 2.500 in Pitch diameter: Gear 7.500 in Pitch cone angle: Pinion 18.435 degrees Pitch cone angle: Gear 71.565 degrees Outer cone distance 3.953 in FACE WIDTH Nominal face width 1.186 in Maximum face width (a) 1.318 in Maximum face width (b) 1.667 in Max face width = smallest of (a) or (b) INPUT Face width 1.250 in COMPUTED VALUES Mean cone distance 3.328 in 0.842 Ratio A m /A o Mean circular pitch 0.441 in mean working depth 0.281 in Clearance 0.035 in Mean whole depth 0.316 in mean addendum factor 0.242 Gear mean addendum 0.068 in Pinion mean addendum 0.213 in Gear mean dedendum 0.248 in Pinion mean dedendum 0.103 in Gear dedendum angle 4.257 degrees Pinion dedendum angle 1.774 degrees Gear outer addendum 0.087 in Pinion outer addendum 0.259 in Gear outside diameter 7.555 in Pinion outside diameter 2.992 in
COMPUTED VALUES Gear ratio 2.000 Pitch diameter: Pinion 2.500 in Pitch diameter: Gear 5.000 in Pitch cone angle: Pinion 26.565 degrees Pitch cone angle: Gear 63.435 degrees Outer cone distance 2.795 in FACE WIDTH Nominal face width 0.839 in Maximum face width (a) 0.932 in Maximum face width (b) 1.000 in Max face width = smallest of (a) or (b) INPUT Face width 0.900 in COMPUTED VALUES Mean cone distance 2.345 in Ratio A m /A o 0.839 Mean circular pitch 0.264 in mean working depth 0.168 in Clearance 0.021 in Mean whole depth 0.189 in mean addendum factor 0.283 Gear mean addendum 0.047 in Pinion mean addendum 0.120 in Gear mean dedendum 0.141 in Pinion mean dedendum 0.068 in Gear dedendum angle 3.450 degrees Pinion dedendum angle 1.670 degrees Gear outer addendum 0.061 in Pinion outer addendum 0.148 in Gear outside diameter 5.054 in Pinion outside diameter 2.764 in
211
BEVEL GEAR GEOMETRY Computes values for features described in Table 8-8 in Section 8-8 of the text PROBLEM: 8-50 PROBLEM: 8-51 GIVEN DATA GIVEN DATA No of teeth in pinion = 16 No of teeth in pinion = 12 No of teeth in gear = 64 No of teeth in gear = 36 Diametral pitch = 32 Diametral pitch = 48 Pressure angle = 20 degrees Pressure angle = 20 degrees COMPUTED VALUES Gear ratio 4.000 Pitch diameter: Pinion 0.500 in Pitch diameter: Gear 2.000 in Pitch cone angle: Pinion 14.036 degrees Pitch cone angle: Gear 75.964 degrees Outer cone distance 1.031 in FACE WIDTH Nominal face width 0.309 in Maximum face width (a) 0.344 in Maximum face width (b) 0.313 in Max face width = smallest of (a) or (b) INPUT Face width 0.300 in COMPUTED VALUES Mean cone distance 0.881 in 0.854 Ratio A m /A o Mean circular pitch 0.084 in mean working depth 0.053 in Clearance 0.007 in Mean whole depth 0.060 in mean addendum factor 0.228 Gear mean addendum 0.012 in Pinion mean addendum 0.041 in Gear mean dedendum 0.048 in Pinion mean dedendum 0.019 in Gear dedendum angle 3.113 degrees Pinion dedendum angle 1.227 degrees Gear outer addendum 0.015 in Pinion outer addendum 0.049 in Gear outside diameter 2.007 in Pinion outside diameter 0.596 in
COMPUTED VALUES Gear ratio 3.000 Pitch diameter: Pinion 0.250 in Pitch diameter: Gear 0.750 in Pitch cone angle: Pinion 18.435 degrees Pitch cone angle: Gear 71.565 degrees Outer cone distance 0.395 in FACE WIDTH Nominal face width 0.119 in Maximum face width (a) 0.132 in Maximum face width (b) 0.208 in Max face width = smallest of (a) or (b) INPUT Face width 0.125 in COMPUTED VALUES Mean cone distance 0.333 in Ratio A m /A o 0.842 Mean circular pitch 0.055 in mean working depth 0.035 in Clearance 0.004 in Mean whole depth 0.039 in mean addendum factor 0.242 Gear mean addendum 0.008 in Pinion mean addendum 0.027 in Gear mean dedendum 0.031 in Pinion mean dedendum 0.013 in Gear dedendum angle 5.316 degrees Pinion dedendum angle 2.217 degrees Gear outer addendum 0.011 in Pinion outer addendum 0.032 in Gear outside diameter 0.757 in Pinion outside diameter 0.311 in
212
Wormgearing 52.
Worm Gearing – Given data:
1.250 in;
1;
10;
14.5° 40;
0.625 in
Single Thread
Lead = Axial Pitch = Circular Pitch = /
/10 .
Lead Angle
.
.
.
Addendum
.
2
Worm Root Diameter
2
Gear Pitch Diameter
/
Center Distance
1.250 1.250
0.1157 in
2 0.100
2 0.1157
40/10
.
4.000
1.250 /2
/2 /
°
;
Worm Outside Diameter
Velocity Ratio
.
. .
.
40/1
NOTE: On the following two pages are the results of Problems 52‐57 giving pertinent geometric properties of worms and wormgears and their velocity ratios. The detailed calculations follow the pattern illustrated above for Problem 52. The equations come from Section 8‐10, Equations 8‐20 to 8‐25. See also Example Problems 8‐4 and 8‐5, and the General Guidelines for Worm and Wormgear Dimensions. Compare the results to discern how variations in geometry such as diametral pitch and the number of threads in the worm affect the overall results. This is especially pertinent to Problem 53 in which three different designs (A, B, and C) for worm/wormgear sets provide the same velocity ratio, but the number of worm threads is 1, 2, and 3 respectively. The single threaded worm produces the smallest center distance and overall size of the reducer. But note, also, that it has the smallest lead angle. The lead angle increases as the number of threads is increased. On the positive side, the small lead angle makes the reducer self‐locking. On the negative side, the small lead angle results in lower
213
mechanical efficiency as will be shown in Chapter 10, Sections 10‐10 and 10‐11. The designer must balance these advantages and disadvantages for each application. PROBLEM: 52 WORMGEARING INPUT DATA Worm pitch diameter = 1.250 in Diametral pitch = 10 No. of worm threads = 1 No. of gear teeth = 40 Face width of gear = 0.625 in COMPUTED RESULTS Circular pitch of gear = 0.3142 in Axial pitch of worm = 0.3142 in Lead of the worm = 0.3142 in Lead angle = 4.574 deg Addendum = 0.100 in Dedendum = 0.116 in Worm outside diameter = 1.450 in Worm root diameter = 1.019 in Gear pitch diameter = 4.000 in Center distance = 2.625 in Velocity ratio = 40.00
PROBLEM: 53A WORMGEARING INPUT DATA Worm pitch diameter = 1.000 in Diametral pitch = 12 No. of worm threads = 1 No. of gear teeth = 20 Face width of gear = 0.500 in COMPUTED RESULTS Circular pitch of gear = 0.2618 in Axial pitch of worm = 0.2618 in Lead of the worm = 0.2618 in Lead angle = 4.764 deg Addendum = 0.083 in Dedendum = 0.096 in Worm outside diameter = 1.167 in Worm root diameter = 0.807 in Gear pitch diameter = 1.667 in Center distance = 1.333 in Velocity ratio = 20.00
PROBLEM: 53B WORMGEARING INPUT DATA Worm pitch diameter = 1.000 in Diametral pitch = 12 No. of worm threads = 2 No. of gear teeth = 40 Face width of gear = 0.500 in COMPUTED RESULTS Circular pitch of gear = 0.2618 in Axial pitch of worm = 0.2618 in Lead of the worm = 0.5236 in Lead angle = 9.462 deg Addendum = 0.083 in Dedendum = 0.096 in Worm outside diameter = 1.167 in Worm root diameter = 0.807 in Gear pitch diameter = 3.333 in Center distance = 2.167 in Velocity ratio = 20.00
PROBLEM: 53C WORMGEARING INPUT DATA Worm pitch diameter = 1.000 in Diametral pitch = 12 No. of worm threads = 4 No. of gear teeth = 80 Face width of gear = 0.500 in COMPUTED RESULTS Circular pitch of gear = 0.2618 in Axial pitch of worm = 0.2618 in Lead of the worm = 1.0472 in Lead angle = 18.435 deg Addendum = 0.083 in Dedendum = 0.096 in Worm outside diameter = 1.167 in Worm root diameter = 0.807 in Gear pitch diameter = 6.667 in Center distance = 3.833 in Velocity ratio = 20.00
214
PROBLEM: 54 WORMGEARING INPUT DATA Worm pitch diameter = 0.625 in Diametral pitch = 16 No. of worm threads = 2 No. of gear teeth = 100 Face width of gear = 0.3125 in COMPUTED RESULTS Circular pitch of gear = 0.1963 in Axial pitch of worm = 0.1963 in Lead of the worm = 0.3927 in Lead angle = 11.310 deg Addendum = 0.063 in Dedendum = 0.072 in Worm outside diameter = 0.750 in Worm root diameter = 0.480 in Gear pitch diameter = 6.250 in Center distance = 3.438 in Velocity ratio = 50.00
PROBLEM: 55 WORMGEARING INPUT DATA Worm pitch diameter = 2.000 in Diametral pitch = 6 No. of worm threads = 4 No. of gear teeth = 72 Face width of gear = 1.000 in COMPUTED RESULTS Circular pitch of gear = 0.5236 in Axial pitch of worm = 0.5236 in Lead of the worm = 2.0944 in Lead angle = 18.435 deg Addendum = 0.167 in Dedendum = 0.193 in Worm outside diameter = 2.333 in Worm root diameter = 1.614 in Gear pitch diameter = 12.000 in Center distance = 7.000 in Velocity ratio = 18.00
PROBLEM: 56 WORMGEARING INPUT DATA Worm pitch diameter = 4.000 in Diametral pitch = 3 No. of worm threads = 1 No. of gear teeth = 54 Face width of gear = 2.000 in COMPUTED RESULTS Circular pitch of gear = 1.0472 in Axial pitch of worm = 1.0472 in Lead of the worm = 1.0472 in Lead angle = 4.764 deg Addendum = 0.333 in Dedendum = 0.386 in Worm outside diameter = 4.667 in Worm root diameter = 3.229 in Gear pitch diameter = 18.000 in Center distance = 11.000 in Velocity ratio = 54.00
PROBLEM: 57 WORMGEARING INPUT DATA Worm pitch diameter = 0.333 in Diametral pitch = 48 No. of worm threads = 4 No. of gear teeth = 80 Face width of gear = 0.156 in COMPUTED RESULTS Circular pitch of gear = 0.0654 in Axial pitch of worm = 0.0654 in Lead of the worm = 0.2618 in Lead angle = 14.050 deg Addendum = 0.021 in Dedendum = 0.024 in Worm outside diameter = 0.375 in Worm root diameter = 0.285 in Gear pitch diameter = 1.667 in Center distance = 1.000 in Velocity ratio = 20.00
215
Gear Trains – Analysis For all problems, assume that the input shaft rotates clockwise unless stated otherwise. 58.
Train Value
/
;
3450 rpm 119.1
.
.
Gear H is an idler. It does not affect the TV but changes the direction of the output shaft. 59.
12 200 rpm; Find
:
/ 30 000
. 60.
6840 rpm; Find
:
/ 40 Exactly
61.
2875 rpm; Find
:
/ 5666.7
.
.
216
Kinematic Design of Gear Trains: Solutions to Problems 62-65 at the end of Chapter 8. Enter data in blue-shaded cells. XX shows combination of NP and NG with minimum differential. PROBLEM 62 VELOCITY RATIO FOR GEARS VELOCITY RATIO FOR GEARS PROBLEM 63 DESIRED VR = 3.14159 () DESIRED VR = 1.7321 (3) NP 16 17 18 19 20 XX 21 22 23 24
NG Calc. 50.27 53.41 56.55 59.69 62.83 65.97 69.12 72.26 75.40
NG Actual 50 53 57 60 63 66 69 72 75
VR Differential = Actual Des. VR - VR Act. 3.1250 0.01659 3.1176 0.02395 3.1667 0.02507 3.1579 0.01630 3.1500 0.00841 3.1429 0.00126 XX 3.1364 0.00523 3.1304 0.01116 3.1250 0.01659
Minimum differential =
XX
XX
16 17 18 19 20 21 22 23 24
XX
16 17 18 19 20 21 22 23 24
NG NG VR Differential = Calc. Actual Actual Des. VR - VR Act. 27.71 28 1.7500 0.01795 29.44 29 1.7059 0.02617 31.18 31 1.7222 0.00983 32.91 33 1.7368 0.00479 34.64 35 1.7500 0.01795 36.37 36 1.7143 0.01777 38.11 38 1.7273 0.00478 XX 39.84 40 1.7391 0.00708 41.57 42 1.7500 0.01795 Minimum differential =
0.00126
PROBLEM 64 VELOCITY RATIO FOR GEARS DESIRED VR = 6.16441 (38) NP
NP
0.00478
VELOCITY RATIO FOR GEARS PROBLEM 65 DESIRED VR = 7.42
NG NG VR Differential = Calc. Actual Actual Des. VR - VR Act. 98.63 99 6.1875 0.02309 104.80 105 6.1765 0.01206 110.96 111 6.1667 0.00225 XX 117.12 117 6.1579 0.00652 123.29 123 6.1500 0.01441 129.45 129 6.1429 0.02156 135.62 136 6.1818 0.01740 141.78 142 6.1739 0.00950 147.95 148 6.1667 0.00225 XX [Two equal solutions] Minimum differential = 0.00225
217
NP
XX
16 17 18 19 20 21 22 23 24
NG NG Calc. Actual 118.72 119 126.14 126 133.56 134 140.98 141 148.40 148 155.82 156 163.24 163 170.66 171 178.08 178
VR Differential = Actual Des. VR - VR Act. 7.4375 0.01750 7.4118 0.00824 7.4444 0.02444 7.4211 0.00105 XX 7.4000 0.02000 7.4286 0.00857 7.4091 0.01091 7.4348 0.01478 7.4167 0.00333
Minimum differential =
0.00105
[Ensure that no interference occurs for any pair of gears.] 66.
Design:
1800 rpm;
1800/2
2 rpm; Exact ratio required.
900 Exact: Use factoring:
Factors are: 2
2
5
5
3
150
3
See Table 8‐7 for interference data. For 20° F.D. Teeth, use
16 or 17
‐ Per Pair
Nominal
8.82 Too small
Two Pairs
8.82
77.8
Three Pairs
8.82
687 Too small
Four Pairs
8.82
6061 Four pairs required
Recombine Factors: 16
150/17
Remaining Ratio 900 450 225 45 9 3
Factor 2 2 5 5 3 3
6;
6;
5;
96
80
16
16
5
n in = 1800 rpm
16
6
Too small
n out = 2.00 rpm 3
5
1
A
E
5
D
H 4
6
96
67.
5
2
C F
80
Table 8‐7 says no interference for
16 if
Design:
22;
1800 rpm; 21 1800/21.5
n
83.7;
150/17
101.
150; 20° Full depth teeth
1800/22
From Table 8‐7‐ No interference with er pair
G
B
81.8;
85.7
17 for 20° F. D. teeth.
8.82 Small: Two Pairs
The layout is to be as in Figure 8‐48 in text – Triple reduction. 218
1800/21
8.82
77.9 Low
Try equal reduction ratio:
≅
17: Let
Let 17 4.19
≅ √83.7
5;
83.7/20
3360 rpm
4.19
71
83.53
Design:
.
.
Exact;
12.0 rpm
20° F.D. teeth. From Table 8‐7, let
2
4.37
4; Then
71.2 → Specify
Final 68.
≅
Exact;
Ok 150
17 for no interference.
per pair
150/17
3360/12
280 Exact. Requires 3 pairs similar to Figure 8‐48 in text.
2
Factor 2 2 5 2 7
5
2
7
8.82: 2 Pairs
8.82
280: Recombine 8
Remaining Ratio 280 140 70 14 7
5
7
8:
17;
136
7:
17;
119
5:
17;
85
77.8: 3 Pairs
686
280
(Other designs possible) 69.
Design:
4200 rpm CW: 13.0 .
311.1;
13.5 rpm CW: Positive .
316.98;
.
323.08
From Problem 68, 3 pairs required. Layout in Figure 8‐48 produces a negative TV. Use idler in any pair. Try residual ratio method. Nominal Try
7;
6. Then
Final
√317
6.82 per pair
≅ 317/42 ≅ 7.55: Use 7 6 7.50
315 Ok
It is preferred to place higher ratios early in the train. Let
7.5;
7;
6
219
7.50
Let
18;
7.5 18
135; Let
Let
17,
17 6
Let
17. Place in first pair.
17,
7 17
119
102. Idler needed for positive train. n in = 4200 rpm
Final train value:
Shaft 1
A
5
I
3 E D
315
4 2
.
F
Ok
13.333 rpm
B
Note: Chapter 9 gives information on selection of speed/torque changes,
n out =
C
. Larger
– diametral pitch. Because of gives smaller gears. This is the
reason that larger ratios should be placed earlier in the train. 70.
Design:
5500 rpm Exactly
n in = 5500 rpm
221
225
3 1
A D
Design: Double reduction with all external gears. 5500/221
24.88:
5500/225
2
24.44
C B
5500/223
24.66
Nominal ratio for each par Then
≅ 24.66/5
√24.66
4.93: For
Final
Design:
16;
5 78.6
Use
79
24.6875 .
71.
4.97. Try
Ok
5500 rpm; 13.0 5500/13.5
14.0 rpm
407.4
Two pairs – max TV = 77.85 too low; Three pairs – max TV = 687 – Ok Nominal ratio per pair: √407.4
7.41
220
n out = 222.78 rpm
Try
≅ 8,
For
8:
8 But use hunting tooth approach. 17,
17 8
136: Use
135 n in = 5500 rpm
,
Same for
3 Shaft 1
7.94
A
E
63.06
Residual Ratio: 407.4/63.06
D 4
6.46
2
/
n out =
C F
13.48 rpm
B
Let
17;
6.46 17
408.05
Final Final output speed 72.
Design:
500/408.05
Let
/
Ok
1750 rpm
75,
850
20.24/4.50
n out = 148.2 rpm
C B
Ok 44: Use 2 Pairs
20.24, Let
.
2
51
850 rpm: 40
Residual Ratio
A D
51.06 → 51
.
850/42
1
2.837
18,
1750
3
4.167
18 2.837
18,
Design:
n in =
11.82/4.167
18:
150
11.82 75/18
Residual ratio Let
.
1750 rpm; 146 1750/148
73.
109.82 → Use 110 teeth
/
81/18
4.50
4.50:
18,
81
Ok
n in = 850 rpm 3 1
A D
2
C B
221
n out = 41.98 rpm
74.
Design: Use two pairs: 3000/550 Let
15;
3000 rpm: 548
552
5.4545: Let
15 2.335
2.335
√5.4545
35.03 → 35. Let
3000
15,
35
Ok
(These results used in Problem 9‐76) 75.
3600 rpm; 3.0
Design:
3600/4.0
Factor
Remaining Ratio
2 2 5 5 3
900 450 225 45 9
3
3
5.0
900: Use 4 pairs
n in = 3600 rpm
H 3
Shaft 1
A
E 4
2
C F B
Use:
Let
6
96/16
/
6
96/16
/
5
80/16
/
5
80/16
/
16. Let
95. Let
5.9375;
81/16
178.47. Residual ratio Let
5
D
Alternate solution using hunting tooth:
95/16
n out = 3.984 rpm
81;
81/16
Total
178.47 5.0625
Final
3600/903.5
5.0625
903.5 .
Ok
222
81
5.0625 900/178.47
5.043
G
76.
Design:
3600 rpm; 3.0 3600/4.0
5.0 rpm
900; Use two pairs of worm/wormgears
1 Single thread worms NB = ND = 30
Design: Let
C-worm
1800 rpm; 50
8.0 Exactly
225;
16,
72,
225/50
4.50
1,
50
1800 rpm 78.
Design:
A n in = 3600 rpm
Wormgear Drive.
1800/8 Let
Output shaft
B and D are wormgears
VR1 = VR2 = 30; TV = 302 = 900 77.
A-worm
Input shaft
Output shaft n out = D 8.00 rpm
B C = Worm
. 12.0 rpm Exactly; TV = 3360/12 = 280 = 20 14
3360 rpm;
Use two pairs of wormgear drives as in Problem 76 . Let 79.
20;
14:
2,
40,
4200 rpm; 13.0
Design:
2,
28.
13.5
Use combined helical gear pair with worm/wormgear set as in Problem 77. Let
50. Wormgear Drive, 4200/13.25
Let
316.98;
18.
18 6.34
Final output speed
4200
1,
50 316.98/50
6.34
114.1 → Use 114 .
223
Ok
80.
Design:
5500 RPM 13.0
14.0 RPM
Use two wormgear drives as in Problem 76. 5500/13.5 Let
20. Then
Try
3. 3,
407.4 407.4/20
3 20.37 60,
61
6500
.
62. Then
13.30 rpm
Or 81.
61.11 → Use 61
3,
Final Output Speed Can also use
20.37
60. Then
Ok
13.75 rpm
Given: np = 50 rpm; Pd = 12; NP = 30; Rack length = 7.00 ft = 84 in. a. DP = NP/Pd = 30/12 = 2.500 in b. Distance from pitch line to back of rack = B = 0.917 in. Table 8‐10 for Pd = 12 c. Let center distance = B + DP/2 = B‐C CD = 0.917 in + 2.50 in /2 = 2.167 in d. vrack = Pitch ling speed of pinion vrack = DP/2 nP
vrack = (1.25 in)(50 rev/min)(2 rad/rev)(1 ft/12 in) = 32.72 ft/min e. Time to move rack 7.00 ft: t = s/v = 7.00 ft/(32.72 ft/min) 60 s/min = 12.83 s f. Number of revolutions of pinion for rack to move 7.00 ft (84.0 in): srack = rP
or
= srack/rP = (84 in)/1.25 in = (62.2 rad)(1 rev)/(2 rad)
.
224
CHAPTER 9 SPUR GEAR DESIGN Forces on Spur Gear Teeth 1.
Given:
20°,
7.5 hp,
a.
1750 rpm,
1750 rpm
486.1 rpm 3.600
b.
/
/
72/20
c.
/
20/12
1.667 in;
/
e.
/12
1.667 1750 /12 .
f. .
72/12
12
6.000 in
∙ .
/
764 ft/min
270 lb ∙ in
972 lb ∙ in
.
/
324 lb
Or
.
h.
324
20°
118 lb
324 lb/
20°
345 lb
i.
72,
3.833 in
d.
g.
20,
/
324 lb
A similar method is used for Problems 2‐6. Spreadsheet solution are shown on the following pages. The solution for Problem 1 is also shown for comparison to the solutions shown above.
225
Forces on Spur Gear Teeth Problem 1 Chapter 9 Given data
a. b. c. d. e. f. g. h. i.
Pressure angle = Power = pinion speed = teeth in pinion = teeth in gear = diametral pitch = RESULTS: Gear speed = VR = mG = pinion PD = gear PD = center distance = C = pitch line speed = torque on pinion shaft = torque on gear shaft = tangential force = radial force = normal force =
20 degrees 7.5 hp 1750 rpm 20 72 12 486.1 rpm 3.600 1.667 in 6.000 in 3.833 in 764 ft/min 270 lb in 972 lb in 324 lb 118 lb 345 lb
Forces on Spur Gear Teeth Problem 2 Chapter 9 Given data
a. b. c. d. e. f. g. h. i.
Pressure angle = 20 degrees Power = 50 hp pinion speed = 1150 rpm teeth in pinion = 18 teeth in gear = 68 diametral pitch = 5 RESULTS: Gear speed = 304.4 rpm VR = mG = 3.778 pinion PD = 3.600 in gear PD = 13.600 in center distance = C = 8.600 in pitch line speed = 1084 ft/min torque on pinion shaft = 2739 lb in torque on gear shaft = 10348 lb in tangential force = 1522 lb radial force = 554 lb normal force = 1620 lb
226
Forces on Spur Gear Teeth Problem 3 Chapter 9 Given data
a. b. c. d. e. f. g. h. i.
Pressure angle = Power = pinion speed = teeth in pinion = teeth in gear = diametral pitch = RESULTS: Gear speed = VR = mG = pinion PD = gear PD = center distance = C = pitch line speed = torque on pinion shaft = torque on gear shaft = tangential force = radial force = normal force =
20 degrees 0.75 hp 3450 rpm 24 110 24 752.7 rpm 4.583 1.000 in 4.583 in 2.792 in 903 ft/min 13.7 lb in 62.8 lb in 27.4 lb 10.0 lb 29.2 lb
Forces on Spur Gear Teeth Problem 4 Chapter 9 Given data
a. b. c. d. e. f. g. h. i.
Pressure angle = Power = pinion speed = teeth in pinion = teeth in gear = diametral pitch = RESULTS: Gear speed = VR = mG = pinion PD = gear PD = center distance = C = pitch line speed = torque on pinion shaft = torque on gear shaft = tangential force = radial force = normal force =
25 degrees 7.5 hp 1750 rpm 20 72 12 486.1 rpm 3.600 1.667 in 6.000 in 3.833 in 764 ft/min 270 lb in 972 lb in 324 lb 151 lb 358 lb
227
Forces on Spur Gear Teeth Problem 5 Chapter 9 Given data
a. b. c. d. e. f. g. h. i.
Pressure angle = 25 degrees Power = 50 hp pinion speed = 1150 rpm teeth in pinion = 18 teeth in gear = 68 diametral pitch = 5 RESULTS: Gear speed = 304.4 rpm VR = mG = 3.778 pinion PD = 3.600 in gear PD = 13.600 in center distance = C = 8.600 in pitch line speed = 1084 ft/min torque on pinion shaft = 2739 lb in torque on gear shaft = 10348 lb in tangential force = 1522 lb radial force = 710 lb normal force = 1680 lb Forces on Spur Gear Teeth
Problem 6 Chapter 9 Given data
a. b. c. d. e. f. g. h. i.
Pressure angle = Power = pinion speed = teeth in pinion = teeth in gear = diametral pitch = RESULTS: Gear speed = VR = mG = pinion PD = gear PD = center distance = C = pitch line speed = torque on pinion shaft = torque on gear shaft = tangential force = radial force = normal force =
25 degrees 0.75 hp 3450 rpm 24 110 24 752.7 rpm 4.583 1.000 in 4.583 in 2.792 in 903 ft/min 13.7 lb in 62.8 lb in 27.4 lb 12.8 lb 30.2 lb
228
7.
Given: Pd = 8 for all gears; nA = 1750 rpm NA = 24; NB = 48; NC = 96; ND = 24 Power out: PB = 8 hp; PC = 7 hp; PD = 5 hp Power in = Σ(Power out) = 20 hp a. Pitch diameter of each gear: D = N/Pd in DA = 24/8 = 3.000 in DB = 48/8 = 6.000 in DC = 96/8 = 12.000 in DD = 24/8 = 3.000 in b. Center distance for each mesh: For mesh A‐B: From Table 8‐1: CA‐B = (NA + NB)/2Pd = (24 + 48)/2(8) = 4.500 in For mesh B‐C: CB‐C = (NB + NC)/2Pd = (48 + 96)/2(8) = 9.000 in For mesh C‐D: CC‐D = (Nc + Nd)/2Pd = (96 + 24)/2(8) = 7.500 in c. Rotational speed of each shaft: First compute velocity ratios: VR1 = nA/nB = NB/NA = 48/24 = 2.000; VR2 = nB/nC = NC/NB = 96/48 = 2.000; VR3 = nC/nD = ND/NC = 24/96 = ¼ = 0.250 Given: nA = 1750 rpm nB = nA/VR1 = 1750 rpm/2.00 = 875 rpm nC = nB/VR2 = 875 rpm/2.00 = 437.5 rpm nD = nC/VR3 = 437.5 rpm/0.250 = 1750 rpm d. Torque in each shaft: Using Equation 9‐10: For T in lbin, P in hp, n in rpm:
T = P/n = (63 000)P/n TA = (63 000)PA/nA = (63 000)(20)/1750 = 720 lbin TB = (63 000)PB/nB = (63 000)(8)/875 = 576 lbin TC = (63 000)PC/nC = (63 000)(7)/437.5 = 1008 lbin TD = (63 000)PD/nD = (63 000)(5)/1750 = 180 lbin
229
e. Tangential force on the teeth of each gear: From Equation 9‐8, with P in hp, D in inches, n in rpm, and Wt in lb, Wt = P/[(n)(D/2)] = (126 000)P/(n D) Gear A: WtA = (126 000)(20 hp)/[1750 3.000] = 480 lbin Gear B: WtB = (126 000)(8 hp)/[875 6.000] = 192 lbin Gear C: WtC = (126 000)(7 hp)/[437.5 12.000] = 168 lbin Gear D: WtD = (126 000)(5 hp)/[1750 3.000] = 120 lbin
Gear Quality 8.
Grain harvester:
10.
9.
Printing press:
7.
10.
Auto transmission:
11.
Gyroscope:
12.
Analytical quality measurements include index variation, tooth alignment, tooth
6.
2.
profile, root radius, and runout. 13.
AGMA Standard 2015 is currently used. See Table 9‐4 for the range of quality
numbers in this system and the comparisons with prior systems. For Problems 14‐16: For precision machinery, use the recommendations for machine tool drives in the lower part of Table 9‐5. The choice of quality number is based on the pitch line speed of the gears. 14.
(From Problem 1). Pitch line speed
794 ft/min Use
10.
15.
(From Problem 2). Pitch line speed
1084 ft/min Use
8.
16.
(From Problem 3). Pitch line speed
903ft/ min Use
230
8.
Gear Materials Answers for Problems 17‐25 are found in Sections 9‐7 and 9‐8. Only brief statements are given here. 17.
Bending stresses are created by the tangential force on the gear teeth acting in a manner similar to that on a cantilever. The maximum bending stress occurs in the root of the tooth where it blends with the involute tooth form. High levels of contact stress, called Hertz stress, occur in the face of the teeth near the pitch line as forces are exerted between the pinion and the gear teeth. The probable mode of failure is pitting on the tooth surface.
18.
AGMA standards give allowable bending stress numbers and allowable contact stress numbers related to the hardness of the material of the teeth. See Figures 9‐18 and 9‐19 and Table 9‐9.
19.
Gear steels are typically medium carbon plain or alloy steels that are heat treated by thoroughly hardening using a quenching and tempering process. For examples, see Table 9‐8, Section 9‐7.
20.
The AGMA recommends hardness values from HB 180 to HB 400. See Figures 9‐18 and 9‐19.
21.
Grade 1 steel is typical commercial quality and is recommended for use in this book. Grades 2 and 3 require progressively more stringent quality controls on the alloy content and cleanliness of the materials. Cost increases dramatically for the higher grades. See AGMA Standard 2004‐C08 (R2012) or the latest revision.
231
22.
Grades 2 and 3 may be specified for high‐speed aerospace applications, turbine engine driven systems, ship propulsion drives, and high‐capacity industrial drives such as those in steel rolling mills.
23.
Case hardening by flame hardening, induction hardening, and carburizing are three processes that produce harder surfaces than typical through‐hardening.
24.
See AGMA Standard 2001‐D04 (R2016) or the latest revision.
25.
AGMA Standard 2001‐D04 (R2016) provides data for gray cast iron, ductile iron, and bronze. Table 9‐10.
26.
From Figures 9‐18 and 9‐19: a.
b.
c.
Grade 1; 200 HB:
28.26 ksi;
194.9 MPa;
644.6 MPa
Grade 1; 300 HB:
866.6 MPa
Grade 1; 400 HB:
43.72 ksi;
301.5 MPa;
1088.6 MPa
U. S
125.7 ksi
U. S.
157.9 ksi
U. S.
SI
36.0 ksi;
248.1 MPa;
93.50 ksi
SI
SI
d.
Using HB > 400 HB is not recommended.
e.
Grade 2; 200 HB:
36.80 ksi;
253.7 MPa;
718.5 MPa
f.
g.
Grade 2; 300 HB:
959.5 MPa
Grade 2; 400 HB:
57.20 ksi;
394.3 MPa;
1200.5 MPa
232
U. S.
139.0 ksi
U. S.
173.9 ksi
U. S.
SI
47.0 ksi;
324.0 MPa;
104.1 ksi
SI
SI
27.
From Figure 9‐18: Grade 1: 300 HB. Grade 2: 192 HB
28.
From Table 9‐9: Case hardening by carburizing produces 55‐64 HRC
29.
From Appendix 5: SAE 1020, 4118, 8620, and others
30.
From Table 9‐9: Flame or induction hardening produces 50‐54 HRC with materials
having high hardenability. 31.
SAE 4140, 4340, 6150. All have good hardenability.
32.
ASTM A536, Grade 80‐55‐06 has a minimum hardness of 179 HB.
33.
a.
b.
c.
d. e.
f.
g.
h.
45.0 ksi;
170.0 ksi
U. S.
310 MPa;
1172 MPa
45.0 ksi;
175.0 ksi
310 MPa;
1207 MPa
55.0 ksi;
180.0 ksi
379 MPa;
1241 MPa
SI (Table 9‐9) U. S. SI (Table 9‐9) U. S. SI (Table 9‐9)
Not Listed 55.0 ksi;
180.0 ksi
U. S.
379 MPa;
1240 MPa
5.00 ksi;
50.0 ksi
35.0 MPa;
345 MPa
13.0 ksi;
75.0 ksi
90.0 MPa;
517 MPa
27.0 ksi;
92.0 ksi
186 MPa;
634 MPa
SI (Table 9-9)
U. S. SI (Table 9-10) U. S. SI (Table 9-10) U. S.
233
SI (Table 9-10)
i.
j.
k.
l.
5.70 ksi;
30.0 ksi
U. S.
39.0 MPa;
207 MPa
23.6 ksi;
65.0
163 MPa;
448 MPa
SI (Table 9-10)
U. S. SI (Table 9-10)
12.0 ksi;
Not listed
U. S.
83.0 MPa;
Not listed
SI (Table 9-14)
9.0 ksi; 62.0 MPa;
Not listed
U. S.
Not listed (Table 9-14)
34.
Depth
0.027 in. (Figure 9-20)
35.
Depth
0.90 mm. (Figure 9‐20)
On the following pages, the solutions to Problems 36 ‐83 are shown in spreadsheet form. Please study the example problems throughout the chapter for details of the solution procedure, reviewing figures, tables, and equations that present pertinent data. Many sources used are cited on the spreadsheets, and several employ equations for analytically acquiring data from tables and graphs. Uniform color schemes are used in the spreadsheets to guide in interpreting the data.
Items that are to be inserted manually are shaded in BLUE.
Sections, such as ‘Initial Input Data’ and ‘Factors in Design Analysis’ are in RED.
Results, typically stresses, are in LIGHT VIOLET.
WHITE SPACES show calculated data and general information.
Most of the problems are true design problems and only examples of appropriate data and results are shown. These are not to be taken as the only solutions, or even the best solutions. Users are urged to critique the given solutions and, perhaps, seek to improve on them.
234
Note that for Problems 36 to 59, the problem statements call for parts of complete problem solutions, while the spreadsheets show the complete design. For example, Problems 36, 42, 48, and 54 all deal with parts of the same problem:
Problem 36 calls for computing only the Bending Stresses in the Gear Teeth.
Problem 42 calls for determining the Required Allowable Bending Stress Number.
Problem 48 calls for calculating the Contact Stress Number.
Problem 54 calls for determining the Required Allowable Contact Stress Number.
All of these calculations use the same data set and are included in the same spreadsheet. Other sets of data are problems:
37, 43, 49, 55;
38, 44, 50, 56
39, 45, 51, 57;
41, 47, 53, 59
235
236
Pd = NP = nG = NG =
ber of Pinion Teeth: ired Output Speed: mber of gear teeth: No. of Gear Teeth:
C= vt = Wt =
Center Distance: Pitch Line Speed: Transmitted Load:
Prob 48: s c = 175207 psi
35253 psi 27514 psi
Prob 36: s t =
Pinion
Gear
Pinion
Fig. 9-10 Fig. 9-17
Gear: J G = 0.410 g Geometry Factor: I = 0.108 REF: m G = 4.72 Prob 36: s t =
Fig. 9-10
J P = 0.320
Geometry Factors: Pinion:
Table 9-4
A v = 11
Table 9-7
1.333
Max
r: Quality Number:
Recommended ratio F/D P < 2.00 Cp = 2300 Elastic Coefficient:
0.83
1.000 1.250 in
F=
Nom
0.667
th/pinion diameter: F/D P =
dth Guidelines (in): Enter: Face Width:
Min
480.19 lb
4.292 in 687 ft/min
7.083 in
1.00
1.00
1.35
Bending Stress Cycle Factor: Y NP =
0.083
0.051
Ex. Prec.
28,963 psi
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.92
0.88
0.95
0.93
<107
stress levels. Could use Grade 2 carburized steel.
s ac too high for any form of Grade 1 steel. Recommend redesign to lower
Gear: Required s ac = 190,443 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
Stress Analysis: Pitting Pinion: Required s ac = 199,099 psi
37,906 psi Gear: Required s at =
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.88 0.92
Pitting Stress Cycle Factor: Z NP = Pitting Stress Cycle Factor: Z NG =
0.95
0.93
0.83
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.147
Commer. Precision
Gear - Number of load cycles: N G = 4.4E+08 107 cycles Bending Stress Cycle Factor: Y NG =
F/D P =
[0.50 < F/DP < 2.00] Figure 9-12
0.061
If F >1.0
Enter: Design Life: 20000 hours Pinion - Number of load cycles: N P = 2.1E+09
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
1.00
1.500 in
DP =
Gear Rim Thickness Factor: K BG =
1.00
1.50
1.208
0.147
0.268
Open
1.00
DG =
Secondary Input Data:
0.058 0.061
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Industrial conveyor driven by an electric motor Problems 36, 42, 48, 54 Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0
4.72
370.6 rpm
Diameter - Pinion:
Gear Ratio:
85
85.1
370 rpm
18
h Diameter - Gear:
nG = mG =
ual Output Speed:
uted data:
1750 rpm
nP =
Input Speed: Diametral Pitch: 12
10 hp
P=
APPLICATION:
Input Power:
nput Data:
UR GEARS
O
237
Pd = NP = nG = NG =
Diametral Pitch: er of Pinion Teeth: ed Output Speed: ber of gear teeth: No. of Gear Teeth:
8.000 in
DP = DG = C= vt = Wt =
Diameter - Pinion: Diameter - Gear: Center Distance: Pitch Line Speed: Transmitted Load:
0.68
32743 psi 26940 psi 172128 psi
Prob 37: s t = Prob 49: s c =
Pinion
Gear
Pinion
Fig. 9-10 Fig. 9-17
Gear: J G = 0.395 Geometry Factor: I = 0.095 REF: m G = 2.40 Prob 37: s t =
Fig. 9-10
J P = 0.325
eometry Factors: Pinion:
Table 9-4
A v = 11
Table 9-7
in
2.667
Max
Quality Number:
Recommended ratio F/D P < 2.00 Cp = 2300 lastic Coefficient:
2.250
2.000
F=
1.333
Nom
1315 lb
h/pinion diameter: F/D P =
th Guidelines (in): ter: Face Width:
Min
Secondary Input Data:
3.333 in
mG =
Gear Ratio:
5.667 in 1004 ft/min
2.40
nG =
479.2 rpm
48
48.0
479 rpm
al Output Speed:
ed data:
6
nP =
Input Speed: 20
1150 rpm
P=
40 hp
APPLICATION:
Input Power:
put Data:
R GEARS
1.00
1.00
1.42
1.00
1.00
1.00
1.75
1.22
0.162
0.284
Open
0.058
0.043
0.061
Ex. Prec.
28,063 psi
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.93
0.91
0.96
0.95
<107
Recommend redesign to lower stress levels.
Contact stress requires case hardened Grade 2 steel.
Gear: Required s ac = 185,084 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
Stress Analysis: Pitting Pinion: Required s ac = 189,152 psi
34,466 psi
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Gear: Required s at =
0.93
Pitting Stress Cycle Factor: Z NG =
0.096
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.91
0.96 Pitting Stress Cycle Factor: Z NP =
Bending Stress Cycle Factor: Y NG =
0.95
0.68
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.162
Commer. Precision
Gear - Number of load cycles: N G = 2.3E+08 107 cycles Bending Stress Cycle Factor: Y NP =
F/D P =
[0.50 < F/DP < 2.00] Figure 9-12
0.058
If F >1.0
Enter: Design Life: 8000 hours Pinion - Number of load cycles: N P = 5.5E+08
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Cement k iln driven by an electric motor Problems 37, 43, 49, 55 Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0
238
Pd = NP = nG = NG =
of Pinion Teeth: Output Speed: er of gear teeth: of Gear Teeth:
C= vt = Wt =
enter Distance: ch Line Speed: ansmitted Load:
0.500 in 0.67
pinion diameter: F/D P =
0.500
Max
Prob 50: s c = 73669 psi
8071 psi 6603 psi
Prob 38: s t =
Pinion
Gear
Pinion
Fig. 9-10 Fig. 9-17
Gear: J G = 0.440 eometry Factor: I = 0.118 REF: m G = 5.00 Prob 38: s t =
Fig. 9-10
J P = 0.360
ometry Factors: Pinion:
Table 9-4
Av = 7
Quality Number:
Table 9-7
0.375
Guidelines (in): 0.250 r: Face Width: F = ommended ratio F/D P < 2.00 stic Coefficient: Cp = 2300
Nom
24.36 lb
Min
Secondary Input Data:
3.750 in
DG =
iameter - Gear: 2.250 in 677 ft/min
0.750 in
DP =
5.00
690.0 rpm
120
120.0
690 rpm
24
32
3450 rpm
0.5 hp
APPLICATION:
ameter - Pinion:
Gear Ratio: m G =
Output Speed:
nG =
nP =
Input Speed: Diametral Pitch:
d data:
P=
Input Power:
t Data:
GEARS
0.042
1.50
1.25
1.11
1.00
1.00
1.00
1.50
1.12
0.074
0.255
Open
0.042
0.043
Fig. 9-14: Use 1.00 if solid blank
13,033 psi
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.91
0.88
0.95
0.92
<107
Gear: Carburized case hardened, 55-64 HRC, s ac = 180,000 psi
Pinion: Carburized case hardened, 55-64 HRC, s ac = 180,000 psi
Gear: Required s ac = 151,791 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
Stress Analysis: Pitting Pinion: Required s ac = 156,966 psi
16,448 psi
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Gear: Required s at =
0.91
Pitting Stress Cycle Factor: Z NG =
0.074
Ex. Prec.
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.88
0.95
Bending Stress Cycle Factor: Y NG = Pitting Stress Cycle Factor: Z NP =
0.92
0.67
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.135
Commer. Precision
Gear - Number of load cycles: N G = 5.0E+08 107 cycles Bending Stress Cycle Factor: Y NP =
F/D P =
[0.50 < F/DP < 2.00] Figure 9-12
0.035
If F >1.0
Enter: Design Life: 12000 hours Pinion - Number of load cycles: N P = 2.5E+09
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Small machine tool driven by an electric motor Problems 38, 44, 50, 56 Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0
O
<
239
NG =
Output Speed: er of gear teeth: of Gear Teeth:
Wt =
ansmitted Load:
0.50
pinion diameter: F/D P =
1.600
Max
Prob 51: s c = 37758 psi
2285 psi 1983 psi
Prob 39: s t =
Pinion
Gear
Pinion
Fig. 9-10 Fig. 9-17
Gear: J G = 0.530 eometry Factor: I = 0.130 REF: m G = 2.93 Prob 39: s t =
Fig. 9-10
J P = 0.460
ometry Factors: Pinion:
Table 9-4
Av = 5
Quality Number:
Table 9-7
1.200 1.500 in
Guidelines (in): 0.800 r: Face Width: F = ommended ratio F/D P < 2.00 stic Coefficient: Cp = 2300
Nom
97.0 lb
Min
Secondary Input Data:
C= vt =
enter Distance: ch Line Speed:
5.900 in 5105 ft/min
3.000 in 8.800 in
DP = DG =
iameter - Gear:
2.93
2215.9 rpm
88
88.0
2216 rpm
30
10
6500 rpm
15 hp
APPLICATION:
ameter - Pinion:
Gear Ratio: m G =
Output Speed:
nG =
nG =
of Pinion Teeth:
d data:
Pd = NP =
Diametral Pitch:
P= nP =
Input Power: Input Speed:
t Data:
GEARS
1.50
1.00
1.00
1.00
1.00
1.00
1.50
1.08
0.053
0.272
Open
0.031
0.025
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.91
0.89
0.95
0.93
<107
Stresses are quite low for steel gears. Recommend redesign
Gear: Required s ac = 62,239 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
63,637 psi
3,131 psi
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
3,685 psi Stress Analysis: Pitting Pinion: Required s ac =
0.053
Fig. 9-14: Use 1.00 if solid blank [Computed: See Fig. 9-16]
Gear: Required s at =
0.91
Pitting Stress Cycle Factor: Z NG =
0.086
Ex. Prec.
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.89
0.95 Pitting Stress Cycle Factor: Z NP =
0.93
Bending Stress Cycle Factor: Y NP =
0.50
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.150
Commer. Precision
Gear - Number of load cycles: N G = 5.3E+08 107 cycles Bending Stress Cycle Factor: Y NG =
F/D P =
[0.50 < F/DP < 2.00]
Figure 9-12
0.031
Enter: Design Life: 4000 hours Pinion - Number of load cycles: N P = 1.6E+09
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Aircraft actuator driven by a universal electric motor Problems 39, 45, 51, 57 Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0 If F >1.0
O
<
240
er of gear teeth: of Gear Teeth:
788 lb
Wt =
ansmitted Load:
3.000 1.500 in 0.50
Guidelines (in): 2.000 r: Face Width: F = pinion diameter: F/D P =
4.000
Pinion Gear Pinion
Prob 40: s t = 14,850 psi Prob 52: s c = 95,948 psi
Fig. 9-10 Fig. 9-17
Gear: J G = 0.520 eometry Factor: I = 0.124 REF: m G = 2.38 Prob 40: s t = 13,280 psi
Fig. 9-10
J P = 0.465
ometry Factors: Pinion:
Table 9-4
Av = 9
Quality Number:
mended ratio 0.5 < F/D P < 2.00 stic Coefficient: Cp = 2300 Table 9-7
Entered
Nom
Min
Max
13.500 in 5236 ft/min
C= vt =
enter Distance: ch Line Speed:
Secondary Input Data:
8.000 in 19.000 in
DP = DG =
iameter - Gear:
2.38
1052.6 rpm
76
76.2
1050 rpm
32
ameter - Pinion:
Gear Ratio: m G =
Output Speed:
nG =
NG =
Output Speed:
d data:
nG =
of Pinion Teeth:
4
Pd = NP =
Diametral Pitch:
2500 rpm
P= nP =
Input Power:
125 hp
APPLICATION:
Input Speed:
t Data:
GEARS
1.00
1.00
1.56
1.00
1.00
1.05
1.70
1.18
0.15
0.272
Open
0.031
0.025
0.053
Fig. 9-14: Use 1.00 if solid blank
13,979 psi
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.91
0.90
0.95
0.93
<107
Gear: SAE 1040 WQT 1000, 269 HB, 22% elongation
Pinion: SAE 1040 WQT 1000, 269 HB, 22% elongation
Gear: Required s ac = 105,438 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
Stress Analysis: Pitting Pinion: Required s ac = 106,609 psi
15,968 psi
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Gear: Required s at =
0.91
Pitting Stress Cycle Factor: Z NG =
0.086
Ex. Prec.
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.90
0.95 Pitting Stress Cycle Factor: Z NP =
0.93
Bending Stress Cycle Factor: Y NP =
0.50
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.150
Commer. Precision
Gear - Number of load cycles: N G = 5.1E+08 107 cycles Bending Stress Cycle Factor: Y NG =
F/D P =
[0.50 < F/DP < 2.00]
Figure 9-12
0.031
Enter: Design Life: 8000 hours Pinion - Number of load cycles: N P = 1.2E+09
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Portable indusrial water pump driven by a gasoline engine Problems 40, 46, 52, 58 Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0 If F >1.0
O
<
[For Problems 60 to 83, single sample solutions are shown for each set of data.]
241
Pd = NP = nG = NG =
of Pinion Teeth: Output Speed: er of gear teeth: of Gear Teeth:
C= vt = Wt =
iameter - Gear: enter Distance: ch Line Speed: nsmitted Load:
1.250 in 0.52
pinion diameter: F/D P =
1.600
Max
Prob 53: s c = 78263 psi
9458 psi 8134 psi
Prob 41: s t =
Pinion
Gear
Pinion
Fig. 9-10 Fig. 9-17
Gear: J G = 0.500 eometry Factor: I = 0.122 REF: m G = 2.58 Prob 41: s t =
Fig. 9-10
J P = 0.430
ometry Factors: Pinion:
Table 9-4
A v = 11
Quality Number:
Table 9-7
1.200
Guidelines (in): 0.800 r: Face Width: F = ommended ratio F/D P < 2.00 stic Coefficient: Cp = 2100
Nom
Min
Secondary Input Data:
193.09 lb
4.300 in 427 ft/min
6.200 in
DG =
ameter - Pinion: 2.400 in
2.58
263.2 rpm
62
62.1
263 rpm
24
10
680 rpm
2.5 hp
APPLICATION:
DP =
Gear Ratio: m G =
Output Speed:
nG =
nP =
Input Speed: Diametral Pitch:
d data:
P=
Input Power:
t Data:
GEARS
0.85
1.25
1.28
1.00
1.00
1.00
1.75
1.18
0.147
0.268
Open
0.03
0.027
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.97
0.95
1.00
0.98
<107
Gear: Ductile iron 100-70-03 Q&T, s ac = 92 000 psi, s at = 27 000 psi
Pinion: Ductile iron 100-70-03 Q&T, s ac = 92 000 psi, s at = 27 000 psi
Gear: Required s ac = 85,727 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
87,531 psi
8,642 psi
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
10,254 psi Stress Analysis: Pitting Pinion: Required s ac =
0.051
Fig. 9-14: Use 1.00 if solid blank
Gear: Required s at =
0.97
Pitting Stress Cycle Factor: Z NG =
0.083
Ex. Prec.
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.95
1.00 Pitting Stress Cycle Factor: Z NP =
Bending Stress Cycle Factor: Y NG =
0.98
0.52
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.147
Commer. Precision
Gear - Number of load cycles: N G = 3.2E+07 107 cycles Bending Stress Cycle Factor: Y NP =
F/D P =
[0.50 < F/DP < 2.00] Figure 9-12
0.030
If F >1.0
Enter: Design Life: 2000 hours Pinion - Number of load cycles: N P = 8.2E+07
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Lawn and garden tractor with fluid power drive Problems 41, 47, 53, 59 Factors in Design Analysis: Alignment Factor, K m =1.0+C pf +C ma If F <1.0
O
<
242
18 387.5 rpm
3.11 1.800 in 5.600 in 3.700 in 565 ft/min 292 lb
Pd = NP = nG = NG = nG = mG = DP = DG = C= vt = Wt =
f Pinion Teeth: Output Speed: r of gear teeth: of Gear Teeth:
Output Speed: Gear Ratio: meter - Pinion: ameter - Gear: enter Distance: ch Line Speed: nsmitted Load:
F=
0.800 0.69
1.600
Max
127348 psi 127348 psi
sc = sc =
Gear
Pinion
Gear
Pinion
17245 psi 13796 psi
st =
Fig. 9-10 Fig. 9-17
Gear: J G = 0.400 ometry Factor: I = 0.100 REF: m G = 3.11 stresses: s t =
Fig. 9-10
J P = 0.320
metry Factors: Pinion:
Table 9-4
A v = 11
uality Number:
Table 9-7
1.250 in
1.200
Nom
385.7 rpm
56
55.7
ommended ratio F/D P < 2.00 Cp = 2300 tic Coefficient:
nion diameter: F/D P =
Guidelines (in): : Face Width:
Min
Secondary Input Data:
data:
10
nP =
Input Speed: ametral Pitch:
1200 rpm
P=
5 hp
APPLICATION:
Input Power:
Data:
GEARS
0.044
1.00
1.00
1.32
1.00
1.00
0.048
0.051
Ex. Prec.
Fig. 9-14: Use 1.00 if solid blank
14,522 psi
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.92
0.89
0.95
0.93
<107
Gear: Requires HB 340: SAE 4140 OQT 1000; HB 340, 18% elongation
Pinion: Requires HB 354: SAE 4140 OQT 900; HB 388, 16% elongation
Gear: Required s ac = 138,422 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
Stress Analysis: Pitting Pinion: Required s ac = 143,088 psi
18,543 psi
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Gear: Required s at =
0.92
Pitting Stress Cycle Factor: Z NG =
0.083
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.89
0.95
Bending Stress Cycle Factor: Y NG = Pitting Stress Cycle Factor: Z NP =
0.93
0.69
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.147
Commer. Precision
Gear - Number of load cycles: N G = 4.6E+08 107 cycles Bending Stress Cycle Factor: Y NP =
F/D P =
[0.50 < F/DP < 2.00] Figure 9-12
Enter: Design Life: 20000 hours Pinion - Number of load cycles: N P = 1.4E+09
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
1.00
1.50
Overload Factor: K o = Size Factor: K s =
1.20
0.147
0.268
Open
0.048
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Reciprocating compressor driven by an electric motor Problem 60 Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0 If F >1.0
243
NG =
f gear teeth: Gear Teeth:
DP = DG = C= vt = Wt =
Gear Ratio: eter - Pinion: meter - Gear: er Distance: Line Speed: mitted Load:
Pinion Gear
20033 psi 16694 psi 119624 psi
esses: s t = st = sc =
Pinion
Fig. 9-10 Fig. 9-17
Gear: J G = 0.420 etry Factor: I = 0.108 REF: m G = 2.96
Table 9-4
Fig. 9-10
Av = 9
Table 9-7
2.667
Max
J P = 0.350
etry Factors: Pinion:
ity Number:
mended ratio F/D P < 2.00 Cp = 2300 Coefficient:
0.50
2.000 2.000 in
F=
Nom
1146 lb
1.333
on diameter: F/D P =
delines (in): Face Width:
Min
econdary Input Data:
11.833 in
mG =
7.917 in 576 ft/min
2.96 4.000 in
nG = 185.9 rpm
71
71.4
put Speed:
ata:
nG =
tput Speed: 185 rpm
6 24
Pd = NP =
inion Teeth:
550 rpm
20 hp
APPLICATION:
metral Pitch:
P= nP =
nput Power: nput Speed:
ata:
ARS
1.25
1.00
1.20
1.00
1.00
1.00
1.50
1.13
0.093
0.280
Open
0.038
0.025
0.093
0.058
Ex. Prec.
21,737 psi
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.93
0.91
0.96
0.94
<107
Gear: SAE 4140 OQT 800; HB 352; 21% elong.-core; Case harden to HRC 50 min
Pinion: SAE 4140 OQT 800; HB 352; 21% elong.-core; Case harden to HRC 50 min
Gear: Required s ac = 160,785 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
Stress Analysis: Pitting Pinion: Required s ac = 164,319 psi
26,640 psi Gear: Required s at =
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.91 0.93
Pitting Stress Cycle Factor: Z NP =
0.96 Pitting Stress Cycle Factor: Z NG =
0.94
Bending Stress Cycle Factor: Y NP =
0.50
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.158
Commer. Precision
Gear - Number of load cycles: N G = 2.2E+08 107 cycles Bending Stress Cycle Factor: Y NG =
F/D P =
[0.50 < F/DP < 2.00] Figure 9-12
0.038
If F >1.0
Enter: Design Life: 20000 hours Pinion - Number of load cycles: N P = 6.6E+08
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Milling machine driven by an electric motor Problem 61 Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0
244
24 227.5 rpm
Pd = NP = nG = NG =
ametral Pitch: f Pinion Teeth: Output Speed: r of gear teeth: of Gear Teeth:
1167 lb
DG = C= vt = Wt =
ameter - Gear: nter Distance: ch Line Speed: nsmitted Load:
100597 psi
sc =
Pinion
Gear
Pinion
14539 psi 12462 psi
st =
Fig. 9-10 Fig. 9-17
Gear: J G = 0.420 ometry Factor: I = 0.114 REF: m G = 3.96 stresses: s t =
Fig. 9-10
J P = 0.360
metry Factors: Pinion:
Table 9-4
A v = 11
uality Number:
ommended ratio F/D P < 2.00 Cp = 2300 tic Coefficient:
4.000
Table 9-7
F= 0.50
3.000 3.000 in
2.000
nion diameter: F/D P =
Guidelines (in): Face Width:
Min
Max
23.750 in 14.875 in 1414 ft/min
DP =
meter - Pinion:
Nom
6.000 in
mG =
Gear Ratio:
Secondary Input Data:
3.96
nG =
227.4 rpm
95
94.9
Output Speed:
data:
900 rpm
nP =
Input Speed: 4
50 hp
P=
APPLICATION:
Input Power:
Data:
GEARS
1.25
1.00
1.50
1.00
1.00
1.05
1.75
1.22
0.173
0.296
Open
0.050
0.025
Bending Stress Cycle Factor: Y NP =
0.068
Ex. Prec.
16,226 psi
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.93
0.90
0.96
0.94
<107
Gear: Requires HB 330: SAE 1040 WQT 800; HB 352; 21% elong.
Pinion: Requires HB 344; SAE 1040 WQT 800; HB 352; 21% elong.
Gear: Required s ac = 135,211 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
Stress Analysis: Pitting Pinion: Required s ac = 139,718 psi
19,333 psi
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Gear: Required s at =
0.93
Pitting Stress Cycle Factor: Z NG =
0.105
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.90
Pitting Stress Cycle Factor: Z NP =
0.96
0.94
0.50
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.173
Commer. Precision
Gear - Number of load cycles: N G = 2.7E+08 107 cycles Bending Stress Cycle Factor: Y NG =
F/D P =
[0.50 < F/DP < 2.00] Figure 9-12
0.050
If F >1.0
Enter: Design Life: 20000 hours Pinion - Number of load cycles: N P = 1.1E+09
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Punch press driven by an electric motor Problem 62 Factors in Design Analysis: Alignment Factor, K m =1.0+C pf +C ma If F <1.0
245
NG =
r of gear teeth: of Gear Teeth:
nsmitted Load:
66573 psi
sc =
Pinion
Gear
Pinion
6457 psi 4880 psi
st =
Fig. 9-10 Fig. 9-17
Gear: J G = 0.430 ometry Factor: I = 0.116 REF: m G = 12.00 stresses: s t =
Fig. 9-10
J P = 0.325
metry Factors: Pinion:
Table 9-4
A v = 12
uality Number:
ommended ratio F/D P < 2.00 Cp = 2100 tic Coefficient:
2.000
Table 9-7
F= 1.00
1.500 2.250 in
1.000
nion diameter: F/D P =
Guidelines (in): Face Width:
Min
Max
156 lb
Wt =
nter Distance: ch Line Speed:
Nom
14.625 in 530 ft/min
C= vt =
ameter - Gear:
Secondary Input Data:
2.250 in 27.000 in
DP = DG =
meter - Pinion:
12.00
nG = mG =
Gear Ratio:
75.0 rpm
216
Output Speed:
data:
75 rpm
Output Speed: 216.0
nG =
8 18
Pd = NP =
900 rpm
ametral Pitch:
nP =
Input Speed:
2.5 hp
APPLICATION:
f Pinion Teeth:
P=
Input Power:
Data:
GEARS
1.00
1.00
1.38
1.00
1.00
1.00
2.00
1.38
0.284
0.284
Open
0.091
0.075
0.061
Fig. 9-14: Use 1.00 if solid blank
4,929 psi
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.97
0.92
0.99
0.95
<107
NOTE: The large gear can be conveniently cast and affixed to the drum of the mixer.
Gear: Grey cast iron, ASTM A48,Class 40
Pinion: Requires approximately HB 134: SAE 1040 CD; HB 160; 12% elong.
Gear: Required s ac = 68,632 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification: Steel pinion driving cast iron gear
72,362 psi
6,796 psi
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Gear: Required s at = Stress Analysis: Pitting Pinion: Required s ac =
0.096
Ex. Prec.
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.92 0.97
Pitting Stress Cycle Factor: Z NP =
0.99 Pitting Stress Cycle Factor: Z NG =
0.95
Bending Stress Cycle Factor: Y NP =
1.00
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.162
Commer. Precision
Gear - Number of load cycles: N G = 3.6E+07 107 cycles Bending Stress Cycle Factor: Y NG =
F/D P =
[0.50 < F/DP < 2.00] Figure 9-12
0.091
If F >1.0
Enter: Design Life: 8000 hours Pinion - Number of load cycles: N P = 4.3E+08
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Cement mixer driven by a gasoline engine Problem 63 Factors in Design Analysis: Alignment Factor, K m =1.0+C pf +C ma If F <1.0
246
NP = nG = NG =
f Pinion Teeth: Output Speed: r of gear teeth: of Gear Teeth:
5.083 in 3936 ft/min 629 lb
DG = C= vt = Wt =
meter - Pinion: ameter - Gear: nter Distance: h Line Speed: nsmitted Load:
22,166 psi 25,524 psi 152,339 psi
Prob 38: s t = Prob 50: s c =
Pinion
Gear
Pinion
Fig. 9-10 Fig. 9-17
Gear: J G = 0.330 ometry Factor: I = 0.096 REF: m G = 2.05 Prob 38: s t =
Fig. 9-10
J P = 0.380
metry Factors: Pinion:
Table 9-4
A v = 11
uality Number:
ommended ratio F/D P < 2.00 Cp = 2300 tic Coefficient:
2.667
Table 9-7
F= 0.44
2.000 3.000 in
1.333
nion diameter: F/D P =
Guidelines (in): Face Width:
Min
Max
6.833 in 3.333 in
DP =
Nom
2.05
mG =
Gear Ratio:
Secondary Input Data:
4510.0 rpm
nG =
20
19.8
4550 rpm
Output Speed:
data:
6
Pd =
ametral Pitch: 41
2200 rpm
P= nP =
Input Power:
75 hp
APPLICATION:
Input Speed:
Data:
GEARS
1.00
1.00
1.81
1.00
1.00
1.00
2.75
1.35
0.296
0.296
Open
0.05
0.025
26,868 psi
Gear: Required s at =
1.00
1.00
1.00
1.00
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.91
0.90
0.95
0.94
<107
Note: Equation for contact stress adjusted to use D = 3.333 in; smaller gear
Gear: SAE 4140 OQT 800; HB 352, 21% elong-core; Case harden to HRC 54 min
Pinion: SAE 4140 OQT 800; HB 352, 21% elong-core; Case harden to HRC 54 min
Gear: Required s ac = 167,405 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification: Requires case hardening
Stress Analysis: Pitting Pinion: Required s ac = 169,266 psi Note
23,581 psi
0.91
Pitting Stress Cycle Factor: Z NG = Stress Analysis: Bending Pinion: Required s at =
0.90
0.95 Pitting Stress Cycle Factor: Z NP =
0.94
Bending Stress Cycle Factor: Y NP =
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Gear - Number of load cycles: N G = 5.2E+08 107 cycles Bending Stress Cycle Factor: Y NG =
0.068
Ex. Prec.
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.105
Commer. Precision 0.173
0.50 [0.50 < F/DP < 2.00]
Figure 9-12
0.050
Enter: Design Life: 8000 hours Pinion - Number of load cycles: N P = 1.1E+09
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Wood chipper driven by a gasoline engine Speed increaser Problems 64 Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0 If F >1.0 F/D P =
247
NP = nG = NG =
ametral Pitch: f Pinion Teeth: Output Speed: r of gear teeth: of Gear Teeth:
1592 N
DG = C= vt = Wt =
ameter - Gear: nter Distance: ch Line Speed: nsmitted Load:
F=
48.0
901 psi
sc =
Pinion
Gear
Pinion
140 psi 112 psi
st =
Fig. 9-10 Fig. 9-17
Gear: J G = 0.415 ometry Factor: I = 0.104 REF: m G = 3.40 stresses: s t =
Fig. 9-10
J P = 0.330
metry Factors: Pinion:
Table 9-4
A v = 12
Table 9-7
mm
uality Number:
ommended ratio F/D P < 2.00 Cp = 191 tic Coefficient:
0.67
36.0 40.0
24.0
nion diameter: F/D P =
idelines (mm): Face Width:
Min
Max
204.0 mm 132.0 mm 1.88 m/s
DP =
meter - Pinion:
Nom
60.0 mm
mG =
Gear Ratio:
Secondary Input Data:
3.40
nG =
176.5 rpm
68
68.6
175 rpm
20
3.00 mm
600 rpm
3.0 kw
APPLICATION: SI-Metric data
Output Speed:
data:
P= nP = m =
Input Power: Input Speed:
Data:
GEARS
1.00
1.00
1.32
1.00
1.00
1.00
2.00
1.32
0.274
0.793
Bending Stress Cycle Factor: Y NP =
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.96
0.94
0.99
0.97
<107
ear: Requires HB 332: SAE 4340 OQT 1000; HB 363
Pinion: Requires HB 341: SAE 4340 OQT 1000; HB 363
Gear: Required s ac = 939 MPa Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
959 Mpa
113 MPa
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
Read from Fig. 9-16
Fig. 9-14: Use 1.00 if solid blank
145 Mpa Stress Analysis: Pitting Pinion: Required s ac =
0.314
Ex. Prec.
Fig. 9-14: Use 1.00 if solid blank
Gear: Required s at =
0.96
Pitting Stress Cycle Factor: Z NG =
0.67
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Stress Analysis: Bending Pinion: Required s at =
0.94
Pitting Stress Cycle Factor: Z NP =
0.99
0.97
0.431 Read from Figure 9-13
0.584
Commer. Precision
Gear - Number of load cycles: N G = 5.3E+07 107 cycles Bending Stress Cycle Factor: Y NG =
F/D P =
[0.50 < F/DP < 2.00] Read from Figure 9-12
0.529
Enter: Design Life: 5000 hours Pinion - Number of load cycles: N P = 1.8E+08
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Open
0.049
Enter: C pf = Type of gearing:
0.042
Pinion Proportion Factor, Cpf =
Reciprocating compressor driven by an electric motor Problem 65 Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0 If F >1.0
248
3600 rpm 1.25
nG = mG =
Output Speed: Gear Ratio:
nsmitted Load:
F=
32.0 0.52
50.0
48.0
Fig. 9-10 Fig. 9-17 Pinion
Gear: J G = 0.365 ometry Factor: I = 0.084 REF: m G = 1.25 86.5 psi 82.3 psi 737 psi
stresses: s t = st = sc =
Pinion
Gear
Fig. 9-10
J P = 0.347
metry Factors: Pinion:
Table 9-4
Av = 7
Table 9-7
mm
64.0
uality Number:
ommended ratio F/D P < 2.00 Cp = 191 tic Coefficient:
nion diameter: F/D P =
idelines (mm): Face Width:
Min
Max
3316 N
Wt =
nter Distance: ch Line Speed:
Nom
120.0 mm 108.0 mm 22.62 m/s
C= vt =
ameter - Gear:
Secondary Input Data:
96.0 mm
DP = DG =
30
30.0
meter - Pinion:
data:
NG =
r of gear teeth: of Gear Teeth:
24
nG = 3600 rpm
NP =
f Pinion Teeth:
4.00 mm
4500 rpm
75.0 kw
APPLICATION: SI-Metric data
Output Speed:
ametral Pitch:
P= nP = m =
Input Power: Input Speed:
Data:
GEARS
1.00
1.00
1.26
1.00
1.00
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.84
0.83
0.89
0.88
<107
Gear: Requires HB 305: SAE 4340 OQT 1100; HB 321
Pinion: Requires HB 310: SAE 4340 OQT 1100; HB 321
Gear: Required s ac = 878 MPa Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
888 Mpa
92.4 MPa
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
Read from Fig. 9-16
Fig. 9-14: Use 1.00 if solid blank
98.3 Mpa Stress Analysis: Pitting Pinion: Required s ac =
0.343
Ex. Prec.
Fig. 9-14: Use 1.00 if solid blank
Gear: Required s at =
0.84
Pitting Stress Cycle Factor: Z NG =
0.52
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Stress Analysis: Bending Pinion: Required s at =
0.83
0.89
Bending Stress Cycle Factor: Y NG = Pitting Stress Cycle Factor: Z NP =
0.88
0.476 Read from Figure 9-13
0.644
Commer. Precision
Gear - Number of load cycles: N G = 2.2E+10 107 cycles Bending Stress Cycle Factor: Y NP =
F/D P =
[0.50 < F/DP < 2.00] Read from Figure 9-12
0.640
Enter: Design Life: 100000 hours Pinion - Number of load cycles: N P = 2.7E+10
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
1.00
1.20
Overload Factor: K o = Size Factor: K s =
1.20
0.158
0.891
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Open
0.040
Enter: C pf = Type of gearing:
0.027
Pinion Proportion Factor, Cpf =
Electric power generator driven by a water turbine Problem 66 Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0 If F >1.0
249
nG = NG =
tput Speed: f gear teeth: Gear Teeth:
244 lb
DG = C= vt = Wt =
meter - Gear: er Distance: Line Speed: mitted Load:
Pinion
14974 psi 11605 psi 113445 psi
esses: s t = st = sc =
Pinion
Gear
Fig. 9-10 Fig. 9-17
Gear: J G = 0.400 etry Factor: I = 0.106 REF: m G = 4.72
Table 9-4
Fig. 9-10
Av = 9
J P = 0.310
etry Factors: Pinion:
ity Number:
mended ratio F/D P < 2.00 Cp = 2300 Coefficient:
1.600
Table 9-7
F= 0.69
1.200 1.250 in
0.800
on diameter: F/D P =
delines (in): Face Width:
Min
Max
5.150 in 1626 ft/min
DP =
Gear Ratio: eter - Pinion:
Nom
8.500 in
mG =
econdary Input Data:
4.72 1.800 in
nG = 730.6 rpm
85
85.1
put Speed:
ata:
18 730 rpm
10
Pd = NP =
inion Teeth:
3450 rpm
12 hp
APPLICATION:
metral Pitch:
P= nP =
nput Power: nput Speed:
ata:
ARS
1.00
1.00
1.33
1.00
1.00
1.00
1.50
1.20
0.147
0.268
Open
0.048
0.044
0.083
0.051
Ex. Prec.
12,088 psi
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.92
0.89
0.96
0.93
<107
Comment: It would be reasonable to specify the same heat treatment
Gear: Requires HB 293: SAE 4340 OQT 1200; HB 293, 20% elongation
Pinion: Requires HB 305: SAE 4340 OQT 1100; HB 321, 18% elongation
Gear: Required s ac = 123,310 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification: SAE 4340 specified; Fig. A4-5
Stress Analysis: Pitting Pinion: Required s ac = 127,467 psi
16,101 psi Gear: Required s at =
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.89 0.92
Pitting Stress Cycle Factor: Z NP =
0.96 Pitting Stress Cycle Factor: Z NG =
0.93
Bending Stress Cycle Factor: Y NP =
0.69
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.147
Commer. Precision
Gear - Number of load cycles: N G = 3.5E+08 107 cycles Bending Stress Cycle Factor: Y NG =
F/D P =
[0.50 < F/DP < 2.00] Figure 9-12
0.048
Enter: Design Life: 8000 hours Pinion - Number of load cycles: N P = 1.7E+09
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Commercial band saw driven by an electric motor Problem 67 Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0 If F >1.0
250
18
4.72 1.286 in 6.071 in 3.679 in 1161 ft/min 341 lb
NP = nG = NG = nG = mG = DP = DG = C= vt = Wt =
f Pinion Teeth: Output Speed: r of gear teeth: of Gear Teeth:
Output Speed: Gear Ratio: meter - Pinion: ameter - Gear: enter Distance: ch Line Speed: nsmitted Load:
156310 psi
sc =
Pinion
Gear
Pinion
28427 psi 22031 psi
st =
Fig. 9-10 Fig. 9-17
Gear: J G = 0.400 ometry Factor: I = 0.106 REF: m G = 4.72 stresses: s t =
Fig. 9-10
J P = 0.310
metry Factors: Pinion:
Table 9-4
Av = 7
Table 9-7
1.143
Max
uality Number:
ommended ratio F/D P < 2.00 Cp = 2300 tic Coefficient:
0.88
0.857 1.125 in
F=
Nom
730.6 rpm
85
85.1
730 rpm
0.571
nion diameter: F/D P =
Guidelines (in): : Face Width:
Min
Secondary Input Data:
data:
14
Pd =
ametral Pitch:
3450 rpm
P= nP =
Input Power:
12 hp
APPLICATION:
Input Speed:
Data:
GEARS
1.00
1.00
1.15
1.00
1.00
Bending Stress Cycle Factor: Y NP =
1.00
1.00
1.00
1.00
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.92
0.89
0.96
0.93
<107
Gear: Use same material as for pinion.
Pinion: SAE 4620 DOQT 300; Case harden to HRC 62; Core has 22% elongation
Gear: Required s ac = 169,902 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification: Specified-Case hardened; Appendix 5
Stress Analysis: Pitting Pinion: Required s ac = 175,629 psi
30,567 psi 22,949 psi
Gear: Required s at =
0.92
Pitting Stress Cycle Factor: Z NG = Stress Analysis: Bending Pinion: Required s at =
0.89
Pitting Stress Cycle Factor: Z NP =
0.96
0.93
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Gear - Number of load cycles: N G = 3.5E+08 107 cycles Bending Stress Cycle Factor: Y NG =
0.049
Ex. Prec.
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.082
Commer. Precision 0.145
0.88 [0.50 < F/DP < 2.00]
Figure 9-12
0.064
Enter: Design Life: 8000 hours Pinion - Number of load cycles: N P = 1.7E+09
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
1.00
1.50
Overload Factor: K o = Size Factor: K s =
1.21
0.145
0.266
Open
0.064
0.063
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Commercial band saw driven by an electric motor Problem 68 - Same data as in Problem 67. Use case hardening Factors in Design Analysis: F/D P = Alignment Factor, K m =1.0+C pf +C ma If F <1.0 If F >1.0
251
22 112.5 rpm
NP = nG = NG =
f Pinion Teeth: Output Speed: r of gear teeth: of Gear Teeth:
DP = DG = C= vt = Wt =
meter - Pinion: ameter - Gear: enter Distance: ch Line Speed: nsmitted Load:
155711 psi
sc =
Pinion
Gear
Pinion
30981 psi 24292 psi
st =
Fig. 9-10 Fig. 9-17
Gear: J G = 0.440 ometry Factor: I = 0.106 REF: m G = 5.82 stresses: s t =
Fig. 9-10
J P = 0.345
metry Factors: Pinion:
Table 9-4
Av = 7
Table 9-7
2.000
Max
uality Number:
ommended ratio F/D P < 2.00 Cp = 2300 tic Coefficient:
0.73
1.500 2.000 in
F=
Nom
1410 lb
1.000
nion diameter: F/D P =
Guidelines (in): : Face Width:
Min
Secondary Input Data:
2.750 in 16.000 in
mG =
Gear Ratio:
9.375 in 468 ft/min
5.82
nG =
111.7 rpm
128
Output Speed:
data:
8
Pd =
ametral Pitch:
127.1
20 hp 650 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
Data:
GEARS
1.00
1.00
1.10
1.00
1.00
1.00
Bending Stress Cycle Factor: Y NP =
0.058
Ex. Prec.
Fig. 9-14: Use 1.00 if solid blank
25,043 psi
Gear: Required s at =
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.94
0.90
0.97
0.94
<107
Gear: Use same material as for pinion.
Pinion: SAE 4620 DOQT 300; Case harden to HRC 62; Core has 22% elongation
Gear: Required s ac = 165,650 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification: Case hardening required; Appendix 5
Stress Analysis: Pitting Pinion: Required s ac = 173,012 psi
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
32,958 psi
0.94
Pitting Stress Cycle Factor: Z NG =
0.093
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.90
Pitting Stress Cycle Factor: Z NP =
0.97
0.94
0.73
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.158
Commer. Precision
Gear - Number of load cycles: N G = 1.7E+08 107 cycles Bending Stress Cycle Factor: Y NG =
F/D P =
[0.50 < F/DP < 2.00] Figure 9-12
0.060
Enter: Design Life: 25000 hours Pinion - Number of load cycles: N P = 9.8E+08
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
1.15 1.50
Overload Factor: K o =
0.093
0.280
Open
0.06
0.048
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Special purpose machine tool-Milling machine Problem 69 - Small size is expected; Use case hardening Factors in Design Analysis: Alignment Factor, K m =1.0+C pf +C ma If F <1.0 If F >1.0
[Note that Problems 71, 72, and 73 compute the Power Transmitting Capacity.]
252
17
NP = nG = NG =
f Pinion Teeth: Output Speed: r of gear teeth: of Gear Teeth:
DP = DG = C= vt = Wt =
meter - Pinion: ameter - Gear: enter Distance: ch Line Speed: nsmitted Load:
128839 psi
sc =
Pinion
Gear
Pinion
19710 psi 13844 psi
st =
Fig. 9-10 Fig. 9-17
Gear: J G = 0.420 ometry Factor: I = 0.109 REF: m G = 5.65 stresses: s t =
Fig. 9-10
J P = 0.295
metry Factors: Pinion:
Table 9-4
Av = 7
Table 9-7
2.667
Max
uality Number:
ommended ratio F/D P < 2.00 Cp = 2300 tic Coefficient:
0.92
2.000 2.600 in
F=
Nom
1202 lb
1.333
nion diameter: F/D P =
Guidelines (in): : Face Width:
Min
Secondary Input Data:
2.833 in 16.000 in
mG =
Gear Ratio:
9.417 in 686 ft/min
5.65
nG =
163.8 rpm
96
96.5
Output Speed:
data:
6
Pd =
ametral Pitch: 163 rpm
25 hp 925 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
Data:
GEARS
1.00
1.00
1.11
1.00
1.00
1.00
Bending Stress Cycle Factor: Y NP =
14,421 psi
Gear: Required s at =
1.00
1.00
1.00
1.00
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.92
0.89
0.96
0.93
<107
Comment: It is reasonable to specify the same heat treatment
Gear: Requires HB 345: SAE 8650 OQT 1000; HB 363, 14% elongation
Pinion: Requires HB 359: SAE 8650 OQT 1000; HB 363, 14% elongation
Gear: Required s ac = 140,043 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification: SAE 4340 specified; Fig. A4-5
Stress Analysis: Pitting Pinion: Required s ac = 144,763 psi
21,194 psi
0.92
Pitting Stress Cycle Factor: Z NG = Stress Analysis: Bending Pinion: Required s at =
0.89
Pitting Stress Cycle Factor: Z NP =
0.96
0.93
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Gear - Number of load cycles: N G = 3.1E+08 107 cycles Bending Stress Cycle Factor: Y NG =
0.064
Ex. Prec.
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.100
Commer. Precision 0.167
0.92 [0.50 < F/DP < 2.00]
Figure 9-12
0.087
Enter: Design Life: 31200 hours Pinion - Number of load cycles: N P = 1.7E+09
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
1.25 1.50
Overload Factor: K o =
0.167
0.290
Open
0.087
0.067
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Crane drum drive driven by an electric motor Problem 70 - Fit gear pair within 24-in inside diameterof drum Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0 If F >1.0 F/D P =
253 6.000 in
DP = DG = C= vt = Wt =
Gear Ratio: h Diameter - Pinion: ch Diameter - Gear: Center Distance: Pitch Line Speed: ad at Pmin Capacity:
1.00
1.00
1.28
1.00
1.00
1.00
Table 9-12 Use 1.00 for R
Use 1.00 if no unusual con
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid
Fig. 9-14: Use 1.00 if solid
Table 9-2: Use 1.00 if P d >
Table 9-1
smission Capacity: 13.07 hp Elastic Coefficient: Cp = 2300 er: Quality Number: A v = 9
Gear: g Geometry Factor:
Fig. 9-10 J G = 0.415 I = 0.104 Fig. 9-17 REF: mG = 2.40
Geometry Factors: Press. angle = 20 deg Pinion: J P = 0.363 Fig. 9-10
25
Pitting Stress Cycle Factor: Z NG =
REF: NP, NG =
Pitting Stress Cycle Factor: Z NP =
on Contact Stress: 13.07 hp
Table 9-4
Bending Stress Cycle Factor: Y NG =
on Contact Stress: 14.50 hp
0.89
0.94
0.93
1.00
1.00
0.89
0.94
0.93
>107
Fig
Fig
Fig
<
37,000 psi
Gear: s at =
Table 9-9
See Fig. 9-18 o
Table 9-9
Pinion material: AISI 4140 OQT 1000, HB 340 Gear material: AISI 4140 OQT 1100, HB 311
Material specification: Steel pinion; Steel gear: through hard
Gear: s ac = 130,000 psi
Allowable Contact Stress Numbers: (Input --Use known hardness) Pinion: s ac = 140,000 psi See Fig. 9-19 o
39,000 psi
Pinion: s at =
0.91 1.00 0.91 Fig Allowable Bending Stress Numbers: (Input - Use known hardness)
Bending Stress Cycle Factor: Y NP =
27.42 hp
60
1.00
on Bending Stress:
Gear - Number of load cycles: NG = 6.5E+08 10 cycles
7
Table 9-7
0.
Enter: Design Life: 15000 hours See Table 9-12 Guidelines: YN , ZN Pinion - Number of load cycles: NP = 1.6E+09
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: Kv =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
1.50
Figure 9-13 1.18
[Computed]
0.147
Enter: C ma =
0.083
Commer. Precision Ex.
Alignment Factor: Km = Overload Factor: K o =
F/DP = 0.50 [0.50 < F/DP <
0.147
Open
0.028
Figure 9-12
on Bending Stress: 25.01 hp
nsmission Capacity: (Using Eq. 9-35, 9-37)
382 lb
4.250 in 1129 ft/min
2.40 2.500 in
mG =
718.8 rpm
nG =
tual Output Speed:
uted data:
0.025 0.028
If F>1.0
0.268
Mesh Alignment Factor, Cma =
Type of gearing:
25 60
NP = NG =
Enter: C pf =
Pinion Proportion Factor, Cpf =
Centrifugal pump driven by an electric motor Problem 71 Factors in Design Analysis: If F<1.0 Alignment Factor, Km =1.0+Cpf +Cma
10
1725 rpm
ber of Pinion Teeth:
Pd =
1.250 in
APPLICATION:
mber of Gear Teeth:
nP =
Input Speed: Diametral Pitch:
SMISSION CAPACITY nput Data: Enter: Face Width: F =
254 5.833 in 16.667 in 11.250 in
mG = DP = DG = C= vt = Wt =
Gear Ratio: Diameter - Pinion: Diameter - Gear: Center Distance: Pitch Line Speed: at Pmin Capacity:
35
100
Gear: Geometry Factor:
Fig. 9-10 J G = 0.450 I = 0.114 Fig. 9-17 REF: mG = 2.86
Geometry Factors: Press. angle = 20 deg Pinion: J P = 0.410 Fig. 9-10
REF: NP, NG =
1.00 1.00
0.89
1.00 0.95
0.93
0.89
0.95
0.93
>107
Fig. 9
Fig. 9
Fig. 9
<10
8,500 psi
40,000 psi
Table 9-10
See Fig. 9-18 or
65,000 psi
Table 9-10
Pinion material: SAE 1040 WQT 800, HB 352 Gear material: Gray cast iron, ASTM A48-83, Class 30
Material specification: Steel pinion; Gear: Gray cast iron gear
Gear: s ac =
Allowable Contact Stress Numbers: (Input --Use known hardness) Pinion: s ac = 144,000 psi See Fig. 9-19 or
Gear: s at =
Pinion: s at =
0.92 1.00 0.92 Fig. 9 Allowable Bending Stress Numbers: (Input - Use known hardness)
Pitting Stress Cycle Factor: Z NP = Pitting Stress Cycle Factor: Z NG =
n Contact Stress: 19.26 hp mission Capacity: 19.26 hp Elastic Coefficient: Cp = 2100 Quality Number: A v = 11
Bending Stress Cycle Factor: Y NP =
Table 9-4
7
Gear - Number of load cycles: NG = 4.7E+08 10 cycles Bending Stress Cycle Factor: Y NG =
Table 9-7
Table 9-12 Use 1.00 for R = .
Use 1.00 if no unusual condit
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid bla
Fig. 9-14: Use 1.00 if solid bla
Table 9-2: Use 1.00 if P d >=
Table 9-1
0.05
Enter: Design Life: 15000 hours See Table 9-12 Pinion - Number of load cycles: NP = 1.4E+09 Guidelines: YN , ZN
1.00
1.00
Service Factor: SF = Reliability Factor: K R =
1.63
1.00
1.00
1.00
2.00
Dynamic Factor: Kv =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Figure 9-13 1.20
[Computed]
0.158
Enter: C ma =
0.093
Commer. Precision Ex. P
Alignment Factor: Km =
21.63 hp
Bending Stress:
[0.50 < F/DP < 2.0 Figure 9-12
0.038
0.158
Open
0.038
0.025
0.280
Mesh Alignment Factor, Cma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Heavy duty conveyor for crushed rock driven by a gasoline engine Problem 72 Factors in Design Analysis: If F<1.0 If F>1.0 Alignment Factor, Km =1.0+Cpf +Cma F/DP = 0.50
n Contact Stress: 88.46 hp
n Bending Stress: 90.79 hp
smission Capacity: (Using Eq. 9-35, 9-37)
277 lb
2291 ft/min
2.86
nG =
al Output Speed:
525.0 rpm
100
NG =
ber of Gear Teeth:
ted data:
6 35
Pd = NP =
Diametral Pitch:
1500 rpm
nP =
2.000 in
APPLICATION:
er of Pinion Teeth:
Input Speed:
MISSION CAPACITY put Data: nter: Face Width: F =
Note that Problems 74 to 79 call for the design of Double‐Reduction Drives.
255 5.833 in 16.667 in 11.250 in
mG = DP = DG = C= vt = Wt =
Gear Ratio: Diameter - Pinion: Diameter - Gear: Center Distance: Pitch Line Speed: at Pmin Capacity:
35
100
face widty from 2.00 to 2.42 in. y number from 11 to 10.
Gear: Geometry Factor:
Fig. 9-10 J G = 0.450 I = 0.114 Fig. 9-17 REF: mG = 2.86
Geometry Factors: Press. angle = 20 deg Pinion: J P = 0.410 Fig. 9-10
REF: NP, NG =
1.00 1.00
0.89
1.00 0.95
0.93
0.89
0.95
0.93
>107
Fig. 9
Fig. 9
Fig. 9
<10
8,500 psi
40,000 psi
Table 9-10
See Fig. 9-18 or
65,000 psi
Table 9-10
Pinion material: SAE 1040 WQT 800, HB 352 Gear material: Gray cast iron, ASTM A48-83, Class 30
Material specification: Steel pinion; Gear: Gray cast iron gear
Gear: s ac =
Allowable Contact Stress Numbers: (Input --Use known hardness) Pinion: s ac = 144,000 psi See Fig. 9-19 or
Gear: s at =
Pinion: s at =
0.92 1.00 0.92 Fig. 9 Allowable Bending Stress Numbers: (Input - Use known hardness)
Pitting Stress Cycle Factor: Z NP = Pitting Stress Cycle Factor: Z NG =
n Contact Stress: 25.08 hp mission Capacity: 25.08 hp Elastic Coefficient: Cp = 2100 Quality Number: A v = 10
Bending Stress Cycle Factor: Y NP =
Table 9-4
7
Gear - Number of load cycles: NG = 4.7E+08 10 cycles Bending Stress Cycle Factor: Y NG =
Table 9-7
Table 9-12 Use 1.00 for R = .
Use 1.00 if no unusual condit
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid bla
Fig. 9-14: Use 1.00 if solid bla
Table 9-2: Use 1.00 if P d >=
Table 9-1
0.06
Enter: Design Life: 15000 hours See Table 9-12 Pinion - Number of load cycles: NP = 1.4E+09 Guidelines: YN , ZN
1.00
1.00
Service Factor: SF = Reliability Factor: K R =
1.50
1.00
1.00
1.00
2.00
Dynamic Factor: Kv =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Figure 9-13 1.21
[Computed]
0.164
Enter: C ma =
0.098
Commer. Precision Ex. P
Alignment Factor: Km =
28.17 hp
Bending Stress:
[0.50 < F/DP < 2.0 Figure 9-12
0.043
0.165
Open
0.043
0.025
0.287
Mesh Alignment Factor, Cma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Heavy duty conveyor for crushed rock driven by a gasoline engine Problem 73 - Redesign of system in Problem 72 to get capacity > 25 hp Factors in Design Analysis: If F<1.0 If F>1.0 Alignment Factor, Km =1.0+Cpf +Cma F/DP = 0.50
n Contact Stress: 115.19 hp
n Bending Stress: 118.23 hp
smission Capacity: (Using Eq. 9-35, 9-37)
361 lb
2291 ft/min
2.86
nG =
al Output Speed:
525.0 rpm
100
NG =
ber of Gear Teeth:
ted data:
6 35
Pd = NP =
Diametral Pitch:
1500 rpm
nP =
2.420 in
APPLICATION:
er of Pinion Teeth:
Input Speed:
MISSION CAPACITY put Data: nter: Face Width: F =
Two separate, successive spreadsheets comprise the full solution to each problem. The division of the total reduction ratio into that for the first pair and for the second pair are described in each problem solution. Often, the larger part of the reduction is taken in Pair 1 because the speeds of rotation are higher, resulting in lower forces on the gear teeth on the teeth of that pair, resulting in generally lower stress levels. Thus, a higher value for diametral pitch, Pd, can be used, producing smaller teeth, diameters, and center distance. For Pair 2, the diametral pitch is typically smaller resulting in generally larger gears. An attempt was made to seek a design for the full double‐reduction train for which the sizes of the resulting pairs are approximately the same; being well balanced. As stated before, the given solutions are only examples and other designs are certainly possible. The designer’s judgment should be used to determine what would be appropriate results for any given problem. Note: Having spreadsheets such as those shown here facilitates iteration, helping the designer to hone in on the most desirable design for a given application.
Problems 74, 75, and 76 are typical double‐reduction drives, designed to be made from steel gears. The design for each pair is done in the same manner at those in Problems 36 to 70. o Note, however, that the input speed for Pair 2 is taken to be the output speed from Pair 1. Then the final output speed for Pair 2 is within the desired range called for in the problem statement. Problem 77 calls for using plastic gears. See Section 9‐14 for data and the procedures used for such gear drives. Problem 78 involves a different type of data in that the double‐reduction gear drive in turn drives a rack‐and‐pinion drive that is to have a given linear velocity. Therefore, the analysis of the relationship between the rack speed and the rotational speed of the output of the gear drive must be completed before the gear drive is designed. Problem 79 calls for the design of a double‐reduction gear drive for an industrial lift truck that will travel at a stated design speed and that will travel on wheels of a given size. Therefore, the relationship between the truck speed and the rotational speed of the output of the gear drive must be calculated first.
256
257
NP = nG = NG =
f Pinion Teeth: Output Speed: r of gear teeth: of Gear Teeth:
320 lb
Wt =
nsmitted Load:
Fig. 9-10 Fig. 9-17 Pinion
Gear: J G = 0.410 ometry Factor: I = 0.106 REF: m G = 4.17 13894 psi 10675 psi 110157 psi
stresses: s t = st = sc =
Pinion
Gear
Fig. 9-10
J P = 0.315
metry Factors: Pinion:
Table 9-4
A v = 11
uality Number:
ommended ratio F/D P < 2.00 Cp = 2300 tic Coefficient:
2.000
Table 9-7
F= 0.67
1.500 1.500 in
1.000
nion diameter: F/D P =
Guidelines (in): : Face Width:
Min
Max
5.813 in 1031 ft/min
C= vt =
enter Distance: ch Line Speed:
Nom
9.375 in
Secondary Input Data:
2.250 in
DP = DG =
ameter - Gear:
mG =
Gear Ratio: meter - Pinion:
4.17
nG =
420.0 rpm
75
74.1
425 rpm
Output Speed:
data:
8
Pd =
ametral Pitch: 18
10 hp 1750 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
Data:
GEARS
1.00
1.00
1.43
1.00
1.00
0.086
0.053
Ex. Prec.
11,237 psi
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.92
0.89
0.95
0.93
<107
Comments: A larger part of the total reduction (4.17) is taken on this first pair.
Gear: Requires HB 281: SAE 4340 OQT 1200; HB 293, 20% elongation
Pinion: Requires HB 294: SAE 4340 OQT 1100; HB 321, 19% elongation
Gear: Required s ac = 119,736 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
Stress Analysis: Pitting Pinion: Required s ac = 123,772 psi
14,940 psi Gear: Required s at =
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.89 0.92
Pitting Stress Cycle Factor: Z NP =
0.95 Pitting Stress Cycle Factor: Z NG =
0.93
Bending Stress Cycle Factor: Y NP =
0.67
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.150
Commer. Precision
Gear - Number of load cycles: N G = 3.8E+08 107 cycles Bending Stress Cycle Factor: Y NG =
F/D P =
[0.50 < F/DP < 2.00] Figure 9-12
0.048
Enter: Design Life: 15000 hours Pinion - Number of load cycles: N P = 1.6E+09
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
1.00
1.50
Overload Factor: K o = Size Factor: K s =
1.20
0.15
0.272
Open
0.048
0.042
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Assembly conveyor driven my an electric motor Problem 74 - First pair of a double reduction drive Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0 If F >1.0
258
nG = NG =
Output Speed: er of gear teeth: of Gear Teeth:
DP = DG = C= vt = Wt =
ameter - Pinion: iameter - Gear: enter Distance: ch Line Speed: nsmitted Load:
21567 psi 17199 psi 135967 psi 135967 psi
st = sc = sc =
Gear
Pinion
Gear
Pinion
Fig. 9-10 Fig. 9-17
Gear: J G = 0.395 eometry Factor: I = 0.108 REF: m G = 2.83 stresses: s t =
Fig. 9-10
J P = 0.315
ometry Factors: Pinion:
Table 9-4
A v = 11
Table 9-7
2.667
Max
Quality Number:
ommended ratio F/D P < 2.00 Cp = 2300 stic Coefficient:
0.67
2.000 2.000 in
F=
Nom
1000 lb
1.333
pinion diameter: F/D P =
Guidelines (in): r: Face Width:
Min
Secondary Input Data:
3.000 in 8.500 in
mG =
Gear Ratio:
5.750 in 330 ft/min
2.83
nG =
148.2 rpm
51
51.1
148 rpm
18
6
420 rpm
10 hp
APPLICATION:
Output Speed:
d data:
Pd = NP =
of Pinion Teeth:
nP =
Input Speed: Diametral Pitch:
P=
Input Power:
t Data:
GEARS
1.00
1.00
1.25
1.00
1.00
1.00
1.50
1.21
0.158
0.280
Open
0.054
0.042
Bending Stress Cycle Factor: Y NP =
0.093
0.058
Ex. Prec.
17,731 psi
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.94
0.92
0.97
0.95
<107
A lower diametral pitch - 6 is used, compared with 8 for the first pair.
Comments: A smaller part of the total reduction (2.83) is taken on this second pair.
Gear: Requires HB 359: SAE 4340 OQT 1000; HB 363, 17% elongation
Pinion: Requires HB 369: SAE 4340 OQT 900; HB 388, 15% elongation
Gear: Required s ac = 144,645 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
Stress Analysis: Pitting Pinion: Required s ac = 147,790 psi
22,702 psi Gear: Required s at =
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.92 0.94
Pitting Stress Cycle Factor: Z NP = Pitting Stress Cycle Factor: Z NG =
0.97
0.95
0.67
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.158
Commer. Precision
Gear - Number of load cycles: N G = 1.3E+08 107 cycles Bending Stress Cycle Factor: Y NG =
F/D P =
[0.50 < F/DP < 2.00]
Figure 9-12
0.054
Enter: Design Life: 15000 hours Pinion - Number of load cycles: N P = 3.8E+08
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Assembly conveyor driven my an electric motor Problem 74 - Second pair of a double reduction drive Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0 If F >1.0
O
<
259
18
NP = nG = NG =
f Pinion Teeth: Output Speed: r of gear teeth: of Gear Teeth:
3.094 in 250 ft/min 66 lb
DG = C= vt = Wt =
meter - Pinion: ameter - Gear: enter Distance: ch Line Speed: nsmitted Load:
107497 psi
sc =
Pinion
Gear
Pinion
13025 psi 10043 psi
st =
Fig. 9-10 Fig. 9-17
Gear: J G = 0.415 ometry Factor: I = 0.106 REF: m G = 4.50 stresses: s t =
Fig. 9-10
J P = 0.320
metry Factors: Pinion:
Table 9-4
Av = 9
uality Number:
ommended ratio F/D P < 2.00 Cp = 2300 tic Coefficient:
1.000
Table 9-7
F= 0.44
0.750 0.500 in
0.500
nion diameter: F/D P =
Guidelines (in): : Face Width:
Min
Max
5.063 in
DP =
Nom
1.125 in
mG =
Gear Ratio:
Secondary Input Data:
4.50
nG =
188.9 rpm
81
80.5
Output Speed:
data:
16
Pd =
ametral Pitch: 190 rpm
0.5 hp 850 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
Data:
GEARS
1.00
1.00
1.14
1.00
1.00
1.00
Bending Stress Cycle Factor: Y NP =
0.043
Ex. Prec.
Fig. 9-14: Use 1.00 if solid blank
10,248 psi
Gear: Required s at =
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.95
0.92
0.98
0.95
<107
Comments: Equal reduction ratios used for both pairs. P d = 16 for both pairs.
Gear: Requires HB 260: SAE 4340 OQT 1200; HB 293, 21% elongation
Pinion: Requires HB 272: SAE 4340 OQT 1200; HB 293, 21% elongation
Gear: Required s ac = 113,155 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
Stress Analysis: Pitting Pinion: Required s ac = 116,845 psi
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
13,710 psi
0.95
Pitting Stress Cycle Factor: Z NG =
0.074
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.92
Pitting Stress Cycle Factor: Z NP =
0.98
0.95
0.50
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.135
Commer. Precision
Gear - Number of load cycles: N G = 9.1E+07 107 cycles Bending Stress Cycle Factor: Y NG =
F/D P =
[0.50 < F/DP < 2.00] Figure 9-12
0.019
Enter: Design Life: 8000 hours Pinion - Number of load cycles: N P = 4.1E+08
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
1.16 1.50
Overload Factor: K o =
0.135
0.255
Open
0.025
0.025
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Food waste grinder driven by an electric motor Problem 75 - First pair of a double reduction drive Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0 If F >1.0
260
18 42 rpm
NP = nG = NG =
er of Pinion Teeth: red Output Speed: mber of gear teeth: No. of Gear Teeth:
5.063 in 3.094 in 56 ft/min 297 lb
DG = C= vt = Wt =
Diameter - Pinion: h Diameter - Gear: Center Distance: Pitch Line Speed: Transmitted Load:
149581 psi
sc =
Pinion
Gear
Pinion
25219 psi 19446 psi
st =
Fig. 9-10 Fig. 9-17
Gear: J G = 0.415 Geometry Factor: I = 0.106 REF: m G = 4.50 ted stresses: s t =
Fig. 9-10
J P = 0.320
Geometry Factors: Pinion:
Table 9-4
Av = 9
r: Quality Number:
Recommended ratio F/D P < 2.00 Cp = 2300 Elastic Coefficient:
1.000
Table 9-7
F= 1.02
0.750 1.150 in
0.500
th/pinion diameter: F/D P =
dth Guidelines (in): nter: Face Width:
Min
Max
1.125 in
DP =
Nom
4.50
mG =
Gear Ratio:
Secondary Input Data:
42.0 rpm
nG =
81
ual Output Speed:
ted data:
16
Pd =
Diametral Pitch:
81.0
0.5 hp 188.9 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
nput Data:
UR GEARS
1.00
1.00
1.07
1.00
1.00
1.00
Bending Stress Cycle Factor: Y NP =
0.050
Ex. Prec.
Fig. 9-14: Use 1.00 if solid blank
19,253 psi
Gear: Required s at =
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.98
0.95
1.01
0.98
<107
Comments: Pinion and gear made from same material and heat treatment.
Gear: Requires HB 384: SAE 4340 OQT 800; HB 415, 12% elongation
Pinion: Requires HB 399: SAE 4340 OQT 800; HB 415, 12% elongation
Gear: Required s ac = 152,634 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
Stress Analysis: Pitting Pinion: Required s ac = 157,454 psi
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
25,734 psi
0.98
Pitting Stress Cycle Factor: Z NG =
0.082
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.95
Pitting Stress Cycle Factor: Z NP =
1.01
0.98
1.02
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.145
Commer. Precision
Gear - Number of load cycles: N G = 2.0E+07 107 cycles Bending Stress Cycle Factor: Y NG =
F/D P =
[0.50 < F/DP < 2.00] Figure 9-12
0.079
Enter: Design Life: 8000 hours Pinion - Number of load cycles: N P = 9.1E+07
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
1.22 1.50
Overload Factor: K o =
0.145
0.266
Open
0.079
0.077
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Food waste grinder driven by an electric motor Problem 75 - Second pair of a double reduction drive Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0 If F >1.0
261
1285.7 rpm 2.33 0.625 in 1.458 in 1.042 in 491 ft/min 17 lb
NP = nG = NG = nG = mG = DP = DG = C= vt = Wt =
f Pinion Teeth: Output Speed: r of gear teeth: of Gear Teeth:
Output Speed: Gear Ratio: meter - Pinion: ameter - Gear: nter Distance: ch Line Speed: nsmitted Load:
F=
0.333
15
0.40
0.667
Max
Fig. 9-10 Fig. 9-17 Pinion
Gear: J G = 0.355 ometry Factor: I = 0.088 REF: m G = 2.33 13288 psi 9358 psi 115385 psi
stresses: s t = st = sc =
Pinion
Gear
Fig. 9-10
J P = 0.250
metry Factors: Pinion:
Table 9-4
Av = 9
uality Number:
Table 9-7
0.250 in
0.500
Nom
35
34.6
1300 rpm
ommended ratio F/D P < 2.00 Cp = 2300 tic Coefficient:
nion diameter: F/D P =
Guidelines (in): Face Width:
Min
Secondary Input Data:
data:
Pd =
ametral Pitch:
24
0.25 hp 3000 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
Data:
GEARS
1.00
1.00
1.19
1.00
1.00
0.041
Ex. Prec.
9,851 psi
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.92
0.90
0.95
0.94
<107
Comments: Equal reduction ratios and P d used for both pairs.
Gear: Requires HB 297: SAE 4340 OQT 1100; HB 321, 19% elongation
Pinion: Requires HB 306: SAE 4340 OQT 1100; HB 321, 19% elongation
Gear: Required s ac = 125,418 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
Stress Analysis: Pitting Pinion: Required s ac = 128,205 psi
14,137 psi
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Gear: Required s at =
0.92
Pitting Stress Cycle Factor: Z NG =
0.071
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.90
0.95 Pitting Stress Cycle Factor: Z NP =
Bending Stress Cycle Factor: Y NG =
0.94
0.50
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.131
Commer. Precision
Gear - Number of load cycles: N G = 3.9E+08 107 cycles Bending Stress Cycle Factor: Y NP =
F/D P =
[0.50 < F/DP < 2.00] Figure 9-12
0.016
Enter: Design Life: 5000 hours Pinion - Number of load cycles: N P = 9.0E+08
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
1.00
1.50
Overload Factor: K o = Size Factor: K s =
1.16
0.131
0.251
Open
0.025
0.025
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Small hand drill driven by an electric motor Problem 76 - First pair of a double reduction drive Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0 If F >1.0
262
nG = NG =
sired Output Speed: umber of gear teeth: n No. of Gear Teeth:
1.042 in 210 ft/min 39 lb
C= vt = Wt =
Center Distance: Pitch Line Speed: Transmitted Load:
116826 psi 116826 psi
9593 psi
st = sc =
Pinion
13622 psi
puted stresses: s t =
sc =
Fig. 9-10 Fig. 9-17
Gear: J G = 0.355 ng Geometry Factor: I = 0.088 REF: m G = 2.33
Gear
Pinion
Gear
Fig. 9-10
J P = 0.250
g Geometry Factors: Pinion:
Table 9-4
Av = 9
er: Quality Number:
Recommended ratio F/D P < 2.00 Cp = 2300 Elastic Coefficient:
0.667
Table 9-7
F= 0.90
0.500 0.560 in
0.333
dth/pinion diameter: F/D P =
Width Guidelines (in): Enter: Face Width:
Min
Max
1.458 in
DG =
ch Diameter - Gear:
Nom
0.625 in
DP =
h Diameter - Pinion:
Secondary Input Data:
2.33
mG =
551.0 rpm
35
35.1
550 rpm
Gear Ratio:
ctual Output Speed:
nG =
NP =
mber of Pinion Teeth:
puted data:
24
Pd =
Diametral Pitch: 15
0.25 hp 1285.7 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
Input Data:
PUR GEARS
1.00
1.00
1.12
1.00
1.00
1.00
1.50
1.20
0.136
0.256
Open
0.065
0.065
0.044
Ex. Prec.
Fig. 9-14: Use 1.00 if solid blank
9,890 psi
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.94
0.92
0.97
0.95
<107
Larger face width used for second pair. Same material + heat treatment used
Comments: Equal reduction ratios and P d used for both pairs.
Gear: Requires HB 296: SAE 4340 OQT 1100; HB 321, 19% elongation
Pinion: Requires HB 304: SAE 4340 OQT 1100; HB 321, 19% elongation
Gear: Required s ac = 124,283 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
Stress Analysis: Pitting Pinion: Required s ac = 126,985 psi
14,339 psi
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Gear: Required s at =
0.94
Pitting Stress Cycle Factor: Z NG =
0.075
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.92
0.97 Pitting Stress Cycle Factor: Z NP =
0.95
Bending Stress Cycle Factor: Y NP =
0.90
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.136
Commer. Precision
Gear - Number of load cycles: N G = 1.7E+08 107 cycles Bending Stress Cycle Factor: Y NG =
F/D P =
[0.50 < F/DP < 2.00]
Figure 9-12
0.059
Enter: Design Life: 5000 hours Pinion - Number of load cycles: N P = 3.9E+08
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Small hand drill driven by an electric motor Problem 76 - Second pair of a double reduction drive Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0 If F >1.0
7
0
O
<
DESIGN OF PLASTIC SPUR GEARS Connect blade wheel to output shaft Application: Band saw driven by electric motor of gear reducer in Problem 76 Problem 77 - Produce a linear speed of blade = 375 ft/min Blade speed = perpheral speed of 9.0 in wheel v t of wheel = D w n w /12 Initial Input Data: P =
0.25 hp
n w = 12v tw /(D w ) =159.2 rpm
Input Speed: n P =
551 rpm
From Problem 76
Input Power:
Diametral Pitch: P d =
12
Number of Pinion Teeth: N P =
18
Desired Output Speed: n G =
159.2 rpm
To produce blade
62.32
speed of 375 ft/min
Computed number of gear teeth: Enter: Chosen No. of Gear Teeth: N G = Computed data: Actual Output Speed: n G = Gear Ratio: m G =
62 160.0 rpm 3.444
Pitch Diameter - Pinion: D P =
1.500 in
Pitch Diameter - Gear: D G = Center Distance: C= Pitch Line Speed: v t =
5.167 in 6.667 in 216.4 ft/min
Transmitted Load: W t =
38.11 lb
Secondary Input Data - Pinion: Tooth Form: 20 degree full depth Lewis Form Factor: Y = 0.521 Safety Factor: SF = 1.50 Unfilled Nylon Material: Allowable Bending Stress: s at = 6000 psi Required Face Width: Enter: Specified Face Width: Actual Bending Stress in Pinion:
F= F= st =
Table 9-15 Ref: Table 9-1 (K o ) Table 9-14 or Fig. 9-26
0.219 in 0.220 in 5985 psi
Secondary Input Data - Gear: Tooth Form: 20 degree full depth Lewis Form Factor: Y = 0.716 Safety Factor: SF = 1.50 Unfilled Acetal Material: 5000 psi Allowable Bending Stress: s at = Face Width - Gear: F = 0.220 in 4355 psi Actual Bending Stress in Gear: s t =
263
Same as for pinion Table 9-15 Same as for pinion Table 9-14 or Fig. 9-26 Same as for pinion Must be < s at
78.
Rack drives furnace door.
2.0 ft/s
120 ft/min
/12
Shaft for gears B and C
.
Possible values for DD:
/ 10.0 in
45.84 rpm
32.72
9.0 in
50.93
29.45
12.167
37.76
39.72 (Used This)
Pair 1:
1500 rpm;
172.3 rpm
17, Pair 2:
148, 172.3 rpm,
16,
73,
8.71 37.76 4.56
See two spreadsheet solutions on later pages: Design Life: In each cycle, rack moves 6 ft each way, a total of 12 ft for full cycle. At 2.0 ft/s, the cycle time is: .
12 ft
6.0 sec/cycle
.
1314 hr
Use 1500 hr design life for the gear drive.
264
Summary:
.
;
.
;
.
;
Pair 1:
10
17
172
1.700
14.800
1.200
8.250
Pair 2:
6
16
73
2.667
12.167
2.500
7.417
Well balanced in size. All gears made from SAE 4340: Pair 1 – OQT 1100; HB 321 Pair 2 – OQT 900; HB 388 Two spreadsheets follow for the double‐reduction gear drive:
265
266
Pd = NP = nG = NG =
Diametral Pitch: Number of Pinion Teeth: Desired Output Speed: d number of gear teeth: osen No. of Gear Teeth:
247 lb
Wt =
Transmitted Load:
1.200 in
F=
1.600
Pinion Gear
st =
Pinion Gear
s c = 118,498 psi s c = 118,498 psi
11,679 psi
Computed basic stresses: s t = 16,826 psi
Fig. 9-10 Fig. 9-17
Gear: J G = 0.425 itting Geometry Factor: I = 0.110 REF: m G = 8.71
148 Fig. 9-10
17
J P = 0.295
ding Geometry Factors: Pinion:
. of teeth, Pinion, Gear:
ommended range of ratio: 0.50 < F/D P < 2.00 nter: Elastic Coefficient: Cp = 2300 Table 9-7 Enter: Quality Number: A v = 11 Table 9-4
0.71
1.200
Min 0.800
e width/pinion diameter: F/D P =
e Width Guidelines (in): Enter: Face Width:
Max
8.250 in 668 ft/min
C= vt =
Center Distance: Pitch Line Speed:
Nom
14.800 in
DG =
Pitch Diameter - Gear:
Secondary Input Data:
1.700 in
DP =
8.71
mG =
Gear Ratio: Pitch Diameter - Pinion:
172.30 rpm
nG =
148
148.3
172 rpm
17
10
1500 rpm
5 hp
APPLICATION:
Actual Output Speed:
omputed data:
P= nP =
Input Power: Input Speed:
tial Input Data:
F SPUR GEARS
1.00
1.00
1.35
1.00
1.00
1.00
1.50
1.19
0.146
0.267
Open
0.048
0.046
F/D P =
0.71
[0.50 < F/D P < 2.00]
0.083
0.050
Bending Stress Cycle Factor: Y NP =
Table 9-11 Use 1.00 for R = .99
11,564 psi
1.00
1.00
1.00
1.00
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.99
0.94
1.01
0.97
Gear: Required s ac = 119,695 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification: Pinion: Through hardened, 301 HB min, SAE 4340 OQT 1100, HB = 321 Pinion: Through hardened, 281 HB min, SAE 4340 OQT 1100, HB = 321
Stress Analysis: Pitting Pinion: Required s ac = 126,062 psi
17,347 psi Gear: Required s at =
0.99
Pitting Stress Cycle Factor: Z NG = Stress Analysis: Bending Pinion: Required s at =
0.94
1.01
0.97 Pitting Stress Cycle Factor: Z NP =
Bending Stress Cycle Factor: Y NG =
Fo
Oth
<---
5
Use 1.00 if no unusual conditions 0.8
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.146
Commer. Precision Ex. Prec.
Figure 9-12
0.048
If F>1.0
Enter: Design Life: 1500 hours See Table 9-12 Guidelines: Y N , Z N Pinion - Number of load cycles: N P = 1.4E+08 7 >107 <107 Gear - Number of load cycles: N G = 1.6E+07 10 cycles
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, C pf =
Rack and pinion driven by a fluid power motor Problem 78 - First reduction Factors in Design Analysis: Alignment Factor, K m =1.0+C pf +C ma If F<1.0
267
NP = nG = NG =
Number of Pinion Teeth: Desired Output Speed: d number of gear teeth: osen No. of Gear Teeth:
C= vt = Wt =
Center Distance: Pitch Line Speed: Transmitted Load:
2.667
Max
Gear
s c = 151,889 psi
Gear Pinion
17,793 psi s c = 151,889 psi
st =
Pinion
Fig. 9-10 Fig. 9-17
Computed basic stresses: s t = 26,858 psi
Fig. 9-10
73 J P = 0.265
16
Gear: J G = 0.400 itting Geometry Factor: I = 0.102 REF: m G = 4.56
ding Geometry Factors: Pinion:
. of teeth, Pinion, Gear:
ommended range of ratio: 0.50 < F/D P < 2.00 nter: Elastic Coefficient: Cp = 2300 Table 9-7 Enter: Quality Number: A v = 11 Table 9-4
0.94
2.000 2.500 in
F=
Nom
1372 lb
1.333
e width/pinion diameter: F/D P =
e Width Guidelines (in): Enter: Face Width:
Min
Secondary Input Data:
12.167 in
DG =
Pitch Diameter - Gear: 7.417 in 120 ft/min
2.667 in
4.56
37.76 rpm
DP =
Gear Ratio:
73
72.2
38.2 rpm
Pitch Diameter - Pinion:
nG = mG =
Actual Output Speed:
omputed data:
6
Pd =
Diametral Pitch: 16
172.3 rpm
P= nP =
Input Power:
5 hp
APPLICATION:
Input Speed:
tial Input Data:
F SPUR GEARS
1.00
1.00
1.15
1.00
1.00
1.00
1.50
F/D P =
0.94 [0.50 < F/D P < 2.00]
0.099
0.063
Bending Stress Cycle Factor: Y NP =
Table 9-11 Use 1.00 for R = .99
17,109 psi
1.00
1.00
1.00
1.00
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
1.03
0.99
1.04
1.01
Same material used for both Pair 1 and Pair2; Different tempering temperatures.
Same material and heat treatment used for pinion and gear.
Comments:
Gear: Required s ac = 147,465 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification: Pinion: Through hardened, 386 HB min, SAE 4340 OQT 900, HB = 388 Pinion: Through hardened, 368 HB min, SAE 4340 OQT 900, HB = 388
Stress Analysis: Pitting Pinion: Required s ac = 153,423 psi
26,592 psi Gear: Required s at =
1.03
Pitting Stress Cycle Factor: Z NG = Stress Analysis: Bending Pinion: Required s at =
0.99
1.04
1.01 Pitting Stress Cycle Factor: Z NP =
Bending Stress Cycle Factor: Y NG =
Fo
Oth
<---
5
Use 1.00 if no unusual conditions 0.
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.166
Commer. Precision Ex. Prec.
Figure 9-12
0.088
If F>1.0
Enter: Design Life: 1500 hours See Table 9-12 Pinion - Number of load cycles: N P = 1.6E+07 Guidelines: Y N , Z N 7 >107 <107 Gear - Number of load cycles: N G = 3.4E+06 10 cycles
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
1.25
0.166
Enter: C ma = Alignment Factor: K m =
0.288
Open
0.088
0.069
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, C pf =
Rack and pinion driven by a fluid power motor Problem 78 - Second reduction Factors in Design Analysis: Alignment Factor, K m =1.0+C pf +C ma If F<1.0
79.
Gear drive for a lift truck. Rolling wheel is the inverse of a pinion driving a rack. at centerline of wheel equals speed of lift truck. /12
Wheel /
v = 20 mi/hr
560.23 rpm
Motor rotational speed = 3000 rpm Train value This
3000/560.23
Floor
5.355
could be produced by one pair of gears. However, for the power required, the gear
would be too large to attach to axle with 12 in wheel. Use double reduction. 2.94 Motor
Whee l
1.81 Gear drive 3000
563.7 rpm ‐ Slightly higher than target
Floor 6 for Pair 1 and
Notes: Gears have large.
2.50 in,
5 for Pair 2. Face widths are relatively
3.00 in. Redesign with helical gears should allow a
smaller system. Also all gears use same material and heat treatment. Life Calculation: 99 840 hr Use L Design decisions:
1.00
1.50
268
100 000 h
One failure in 10 000;
0.9999
269
er of gear teeth: . of Gear Teeth:
5.583 in 2225 ft/min 297 lb
C= vt = Wt =
Center Distance: tch Line Speed: ansmitted Load:
71442 psi 71442 psi
sc = sc =
Gear
Pinion
Gear
Pinion
5430 psi 4187 psi
st =
Fig. 9-10 Fig. 9-17
Gear: J G = 0.380 eometry Factor: I = 0.097 REF: m G = 2.94 d stresses: s t =
Fig. 9-10
J P = 0.293
ometry Factors: Pinion:
Table 9-4
Av = 7
Quality Number:
commended ratio F/D P < 2.00 Cp = 2300 astic Coefficient:
2.667
Table 9-7
F= 0.88
2.000 2.500 in
1.333
pinion diameter: F/D P =
Guidelines (in): er: Face Width:
Min
Max
8.333 in
DG =
Diameter - Gear:
Nom
2.833 in
DP =
ameter - Pinion:
Secondary Input Data:
2.94
mG =
1020.0 rpm
50
51.0
Gear Ratio:
Output Speed:
nG =
NG =
d Output Speed:
d data:
nG =
of Pinion Teeth: 1000 rpm
6 17
Pd = NP =
Diametral Pitch:
20 hp 3000 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
ut Data:
R GEARS
1.50
1.00
1.19
1.00
1.00
1.00
1.50
1.25
0.166
0.288
Open
0.082
0.063
Bending Stress Cycle Factor: Y NP =
0.063
Ex. Prec.
Fig. 9-14: Use 1.00 if solid blank
6,901 psi
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.86
0.84
0.91
0.89
<107
Same material used for pinion and gear because contact stresses are similar.
Comments:
Gear: Requires HB 297: SAE 4340 OQT 1100; HB 321, 19% elongation
Pinion: Requires HB 306: SAE 4340 OQT 1100; HB 321, 19% elongation
Gear: Required s ac = 124,609 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
Stress Analysis: Pitting Pinion: Required s ac = 127,576 psi
9,152 psi
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Gear: Required s at =
0.86
Pitting Stress Cycle Factor: Z NG =
0.099
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.84
Pitting Stress Cycle Factor: Z NP =
0.91
0.89
0.88
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.166
Commer. Precision
Gear - Number of load cycles: N G = 6.1E+09 107 cycles Bending Stress Cycle Factor: Y NG =
F/D P =
[0.50 < F/DP < 2.00] Figure 9-12
0.082
Enter: Design Life: 100000 hours Pinion - Number of load cycles: N P = 1.8E+10
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Industrial lift truck drive to wheels - Driven by a DC motor Problem 79 - First pair Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0 If F >1.0
[Problems 80 to 83 are single‐reduction gear pairs made from plastic gears.]
270
er of gear teeth: . of Gear Teeth:
5.900 in 1122 ft/min 588 lb
C= vt = Wt =
Center Distance: tch Line Speed: ansmitted Load:
75676 psi 75676 psi
sc = sc =
Gear
Pinion
Gear
Pinion
6338 psi 5504 psi
st =
Fig. 9-10 Fig. 9-17
Gear: J G = 0.380 eometry Factor: I = 0.092 REF: m G = 1.81 d stresses: s t =
Fig. 9-10
J P = 0.330
ometry Factors: Pinion:
Table 9-4
Av = 7
Quality Number:
commended ratio F/D P < 2.00 Cp = 2300 astic Coefficient:
3.200
Table 9-7
F= 0.71
2.400 3.000 in
1.600
pinion diameter: F/D P =
Guidelines (in): er: Face Width:
Min
Max
7.600 in
DG =
Diameter - Gear:
Nom
4.200 in
DP =
ameter - Pinion:
Secondary Input Data:
1.81
mG =
563.7 rpm
38
Gear Ratio:
Output Speed:
nG =
NG =
d Output Speed:
d data:
560.23 rpm
nG =
of Pinion Teeth: 38.2
5 21
Pd = NP =
Diametral Pitch:
20 hp 1020 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
ut Data:
R GEARS
1.50
1.00
1.14
1.00
1.00
1.00
1.50
1.24
0.173
0.296
Open
0.071
0.046
Bending Stress Cycle Factor: Y NP =
0.068
Ex. Prec.
Fig. 9-14: Use 1.00 if solid blank
8,974 psi
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.87
0.86
0.92
0.91
<107
Same material used for pinion and gear because contact stresses are similar.
Comments:
Gear: Requires HB 315: SAE 4340 OQT 1100; HB 321, 19% elongation
Pinion: Requires HB 320: SAE 4340 OQT 1100; HB 321, 19% elongation
Gear: Required s ac = 130,475 psi Table 9-9 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification:
Stress Analysis: Pitting Pinion: Required s ac = 131,992 psi
10,447 psi
1.00
1.00
1.00
1.00
>107
See Table 9-12 Guidelines: Y N , Z N
Table 9-11 Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Gear: Required s at =
0.87
Pitting Stress Cycle Factor: Z NG =
0.105
Fig. 9-14: Use 1.00 if solid blank
Stress Analysis: Bending Pinion: Required s at =
0.86
Pitting Stress Cycle Factor: Z NP =
0.92
0.91
0.71
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.173
Commer. Precision
Gear - Number of load cycles: N G = 3.4E+09 107 cycles Bending Stress Cycle Factor: Y NG =
F/D P =
[0.50 < F/DP < 2.00] Figure 9-12
0.071
Enter: Design Life: 100000 hours Pinion - Number of load cycles: N P = 6.1E+09
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Industrial lift truck drive to wheels - Driven by a DC motor Problem 79 - Second pair Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F <1.0 If F >1.0
DESIGN OF PLASTIC SPUR GEARS Application: Small band saw driven by an electric motor Problem 80 - One possible design Initial Input Data: Input Power: P = 0.5 hp Input Speed: n P = 860 rpm Diametral Pitch: P d =
12
Number of Pinion Teeth: N P =
18
Desired Output Speed: n G =
266 rpm
Computed number of gear teeth: Enter: Chosen No. of Gear Teeth: N G = Computed data: Actual Output Speed: n G = Gear Ratio: m G =
58.2 58 266.9 rpm 3.222
Pitch Diameter - Pinion: D P =
1.500 in
Pitch Diameter - Gear: D G = Center Distance: C= Pitch Line Speed: v t =
4.833 in 6.333 in 337.7 ft/min
Transmitted Load: W t =
48.84 lb
Secondary Input Data - Pinion: Tooth Form: 20 degree full depth Lewis Form Factor: Y = 0.521 Table 9-15 Safety Factor: SF = 1.50 Ref: Table 9-1 (K o ) Material: Nylon 6000 psi Table 9-14 or Fig. 9-26 Allowable Bending Stress: s at = Required Face Width: Enter: Specified Face Width: Actual Bending Stress in Pinion:
F= F= st =
0.281 in 0.300 in 5624 psi
Secondary Input Data - Gear: Tooth Form: 20 degree full depth Lewis Form Factor: Y = 0.709 Safety Factor: SF = 1.50 Material: Acetal Allowable Bending Stress: s at = 5000 psi Face Width - Gear: F = 0.300 in 4133 psi Actual Bending Stress in Gear: s t =
271
Same as for pinion Table 9-15 Same as for pinion Table 9-14 or Fig. 9-26 Same as for pinion Must be < s at
DESIGN OF PLASTIC SPUR GEARS Application: Paper feed roll driven by an electric motor Problem 81 - One possible design Initial Input Data: Input Power: P = 0.06 hp Input Speed: n P = 88 rpm Diametral Pitch: P d =
20
Number of Pinion Teeth: N P =
16
Desired Output Speed: n G =
21 rpm
Computed number of gear teeth: Enter: Chosen No. of Gear Teeth: N G = Computed data: Actual Output Speed: n G = Gear Ratio: m G =
67.05 67 21.0 rpm 4.188
Pitch Diameter - Pinion: D P =
0.800 in
Pitch Diameter - Gear: D G = C= Center Distance: Pitch Line Speed: v t =
3.350 in 4.150 in 18.4 ft/min
Transmitted Load: W t =
107.4 lb
Secondary Input Data - Pinion: 20 degree stub Tooth Form: Y = 0.578 Lewis Form Factor: Safety Factor: SF = 1.25 Material: Nylon-glass filled Allowable Bending Stress: s at = 12000 psi Required Face Width: Enter: Specified Face Width: Actual Bending Stress in Pinion:
F= F= st =
Table 9-15 Ref: Table 9-1 (K o ) Table 9-14 or Fig. 9-26
0.387 in 0.400 in 11612 psi
Secondary Input Data - Gear: 20 degree stub Tooth Form: Y = 0.782 Lewis Form Factor: Safety Factor: SF = 1.25 Material: Nylon-glass filled Allowable Bending Stress: s at = 12000 psi F = 0.400 in Face Width - Gear: Actual Bending Stress in Gear: s t = 8583 psi
272
Same as for pinion Table 9-15 Same as for pinion Table 9-14 or Fig. 9-26 Same as for pinion Must be < s at
DESIGN OF PLASTIC SPUR GEARS Application: Wheels of remote control car-Electric motor drive Problem 82 - One possible design Initial Input Data: Input Power: P = 0.025 hp 430 rpm Input Speed: n P = Diametral Pitch: P d =
48
Number of Pinion Teeth: N P =
14
Desired Output Speed: n G =
121 rpm
Computed number of gear teeth: Enter: Chosen No. of Gear Teeth: N G = Computed data: Actual Output Speed: n G = Gear Ratio: m G =
49.75 50 120.4 rpm 3.571
Pitch Diameter - Pinion: D P =
0.292 in
Pitch Diameter - Gear: D G = Center Distance: C= Pitch Line Speed: v t =
1.042 in 1.333 in 32.8 ft/min
Transmitted Load: W t =
25.1 lb
Secondary Input Data - Pinion: Tooth Form: 20 degree stub Lewis Form Factor: Y = 0.54 Safety Factor: SF = 1.25 Nylon-unfilled Material: 6000 psi Allowable Bending Stress: s at = Required Face Width: Enter: Specified Face Width: Actual Bending Stress in Pinion:
F= F= st =
Table 9-15 Ref: Table 9-1 (K o ) Table 9-14 or Fig. 9-26
0.465 in 0.470 in 5938 psi
Secondary Input Data - Gear: Tooth Form: 20 degree stub Lewis Form Factor: Y = 0.758 Safety Factor: SF = 1.25 Material: Nylon-unfilled Allowable Bending Stress: s at = 6000 psi Face Width - Gear: F = 0.470 in 4230 psi Actual Bending Stress in Gear: s t =
273
Same as for pinion Table 9-15 Same as for pinion Table 9-14 or Fig. 9-26 Same as for pinion Must be < s at
DESIGN OF PLASTIC SPUR GEARS Application: Food-chopping machine driven by electric motor Problem 83 - One possible design Initial Input Data: Input Power: P = 0.65 hp Input Speed: n P = 1560 rpm Diametral Pitch: P d =
16
Number of Pinion Teeth: N P =
18
Desired Output Speed: n G =
469 rpm
Computed number of gear teeth: Enter: Chosen No. of Gear Teeth: N G = Computed data: Actual Output Speed: n G = Gear Ratio: m G =
59.87 60 468.0 rpm 3.333
Pitch Diameter - Pinion: D P =
1.125 in
Pitch Diameter - Gear: D G = C= Center Distance: Pitch Line Speed: v t =
3.750 in 4.875 in 459.5 ft/min
Transmitted Load: W t =
46.7 lb
Secondary Input Data - Pinion: Tooth Form: 20 degree full depth Y = 0.521 Lewis Form Factor: Table 9-15 Safety Factor: SF = 1.75 Ref: Table 9-1 (K o ) Nylon-unfilled Material: 6000 psi Table 9-14 or Fig. 9-26 Allowable Bending Stress: s at = Required Face Width: Enter: Specified Face Width: Actual Bending Stress in Pinion:
F= F= st =
0.418 in 0.420 in 5971 psi
Secondary Input Data - Gear: Tooth Form: 20 degree full depth Y = 0.713 Lewis Form Factor: Safety Factor: SF = 1.75 Acetal-unfilled Material: 5000 psi Allowable Bending Stress: s at = F = 0.420 in Face Width - Gear: 4363 psi Actual Bending Stress in Gear: s t =
274
Same as for pinion Table 9-15 Same as for pinion Table 9-14 or Fig. 9-26 Same as for pinion Must be < s at
CHAPTER 10 HELICAL GEARS, BEVEL GEARS AND WORMGEARING 1.
Helical Gears:
a)
5.0 hp;
8;
14 °;
45;
.
1250 rpm; Torque
/2
45 8
; 252 ∙ 5.625 /2
W b)
2.00 in;
252 lb ∙ in
5.625 in
89.6 lb
89.6
30°
89.6
1 14 ° 2
15; Drive to a reciprocating pump;
51.7 lb 23.2 lb
1.50
; Assume Approximate
1.0
= 0.38 from Figure 10‐5(a) for a 75‐tooth mate. Modifying K‐factor = 0.97
from Figure 10‐5(b).
Data are for
15° 14 °
Actual
0.38 0.97 /
15 8
1.875 in;
1.0
11: The
/
30°
0.075
5.625 1250 /12
1.55 (Figure 9‐16) 275
12.6°
0.369
1.867; 1.00
/12 Specify
30°
0.075; 0.155
1.23
1841 ft/min
0.155
89.6 8 1.50 1.0 1.23 1.0 1.55 2.00 0.369
89.6 1.50 1.0 1.23 1.55 2.00 1.875 0.20
1960 36 228 psi: c)
2778 psi
estimated from Table 10
1.
Specify cast iron because of low stresses
Gray cast iron – Class 20:
;
2.
16;
48;
.
90.0 lb ∙ in
Helical Gears:
2.50 hp;
1.50 in; Torque
12;
20°;
1750 rpm a)
[From results of Problem 8‐42:
b)
. /
.
8.485;
31.8 lb
45°
31.8 lb
27.2°
1.25
16.4 lb 1.00;
5.657 1750 /12
Centrifugal blower: Let 1.50/1.887
9;
2592 ft/min
0.05; .
. .
0.21;
≅ 0.30
1.40
0.795; .
Pitting:
1.887 in
31.8 lb
; /12
/
27.2°
31.8 lb
/
16 8.485
c)
5.657 in;
. .
1960 for Cast iron
276
0.15; .
.
1.20
1259 psi
45°
.
1960
.
.
.
.
: 3.
Helical Gears:
15 hp;
12;
.
14 °;
45°;
1.00 in
6.00 in;
2.00 in
143 lb
/
;
143
45°
143
143
1 14 ° 2
37.0
1.00
/12
6.00 2200 /12 1.44;
1.00 2.00
;
0.15;
9;
6;
430 lb ∙ in
2.00
b)
,
36;
From the results of Problem 8‐43:
/
.
;
2200 rpm;
a)
. .
0.50;
1.18 3456 ft/min
0.30
,
0.190 Estimated
143 6 1.0 1.18 1.0 1.44 2.0 1.00 0.30 Use nodular (Ductile) iron:
0.025
9720 psi
2050
2050
143 2.0 1.0 1.18 1.44 1.0 2.00 0.19
277
73 300 psi
c)
Specify: Ductile iron ASTM A536 60‐40‐18;
Or Cast Iron, Class 40:
,
Note: Problem 8‐43 gives
Axial Pitch
0.5236 in
Should be
2.0
.
Then 4.
1.91
.
low
Helical Gears:
0.50 hp;
45°;
72;
16;
From Problem 8‐44;
.
a)
/
.
3450 rpm;
24;
0.25 in; Winch
Moderate shock
16.97; .
/12 9;
1.50
0.943 in
.
9.13 lb ∙ in
4.30 lb
/
4.30
Let
14 °
4.243 in;
4.30 lb
b)
,
20°
1.57
4.243 3450 / 12
1.48;
1.00
0.25 0.943
0.265;
;
45° ;
4.30 lb
20.0°
44
3832 ft/min 0.32
≅ 0;
8
;
0.14;
0.25 Estimated
1.14
4.30 16.97 1.50 1.0 1.14 1.0 1.48 0.25 0.32
2308 psi
Try Cast iron: 1960
4.30 1.50 1.0 1.14 1.48 0.25 0.943 0.25
278
26 633 psi
Specify class 20 cast iron; Note: Problem 8‐44 gives Then
0.25 0.1815
, Axial Pitch
0.1815 in
1.35
Should be
Low
2.0
.
SPREADSHEETS FOR SOLUTIONS TO CHAPTER 10 PROBLEMS: The following pages give sample designs for helical gears, bevel gears, and wormgearing. The methods used to create the spreadsheets follow generally materials presented in the pertinent sections of the book and demonstrated in Example Problems in Chapter 10. The spreadsheets are shown in U.S. units.
Helical gearing is used for Problems 5 to 11.
Bevel gearing is used for Problems 14‐17.
Wormgearing is used for Problems 19‐24.
Because many design decisions are required, other designs are possible. The reader is encouraged to work toward a particular goal of material type, center distance, overall size or other application‐specific goals, creating improved designs. Notes concerning the spreadsheets: 1.
All design approaches involve gearing geometry, force analysis, stress analysis, and
surface durability analysis. 2.
Blue highlighting is used for data that are input by the designer.
3.
Red highlighting is used to identify the various sections of the solution.
4.
Green highlighting is used for the efficiency of the gearing.
279
5.
Light purple highlighting is used for key results for stresses and surface durability
calculations. 6.
Items without highlighting are calculated data.
7.
Care should be used to not change cells where calculations are performed.
8.
Spreadsheets are shown in landscape form to enhance the size of the images. It may
be desirable to print these pages for greater ease of reading. HELICAL GEARING The first spreadsheet shown here is for Example Problem 10‐3 so the user can compare the data in Chapter 10 with the presentation of those data and calculations in spreadsheet form.
280
281
NG =
uted number of gear teeth: Chosen No. of Gear Teeth:
C= vt = Wt =
Center Distance: Pitch Line Speed: Transmitted Load:
15.0
= px =
Helix angle: Axial Pitch:
Fig 10-5,6,7 Tab. 10-1,2
I = 0.202
307 lb 432 lb
Radial Force:
Wr =
3.13
Fig 10-5,6,7
Table 9-5
0.162
0.096
0.061
Table 9-11: Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
38,163 psi
Gear: Required s at =
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
<107
Table 9-9
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.91
0.88
0.94
0.93
>107
For Kv :
Red: HB > 4
HB 457
HB 469
HB 328
HB 381
Grade
Through
B = 0.6303 C = 70.71
Gear
Computed contact stress no. = s c = 128,343 psi
Gear Pinion
28,699 psi Computed contact stress no. = s c = 128,343 psi
Computed bending stress no. = s t =
Computed stresses without modifying factors for reliability and life: Pinion Computed bending stress no. = s t = 31,449 psi
Gear requires HRC 58 min.: SAE 4320 SOQT 450; HRC 59; Carburized
Pinion requires HRC 58 min.: SAE 4320 SOQT 450; HRC 59; Carburized
One possible material specification: Steel pinion and gear: Carburized, Grade 1 for through h
Specify materials, alloy and heat treatment, for most severe requirement.
Gear: Required s ac = 176,295 psi
Stress Analysis: Pitting Pinion: Required s ac = 180,257 psi
42,270 psi
1.00
1.00
1.00
1.00
7
Stress Analysis: Bending Pinion: Required s at =
0.91
Pitting Stress Cycle Factor: Z NG =
0.94
0.93
in
J P = 0.480
75
Bending Stress Cycle Factor: Y NP = Bending Stress Cycle Factor: Y NG =
0.89
Table 9-7
F/DP = 1.09 [0.50 < F/DP < 2.00]
Commer. Precision Ex. Prec.
Figure 9-12
0.099
If F>1.0
Gear - Number of load cycles: NG = 6.6E+08 10 cycles
Pitting Stress Cycle Factor: Z NP =
J G = 0.526
24
1.25
1.00
1.35
1.00
1.00
1.00
1.50
1.26
0.162
0.284
Open
0.099
0.084
Enter: Design Life: 10000 hours See Table 9-7 Pinion - Number of load cycles: NP = 2.1E+09 Guidelines: YN , ZN
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: Kv =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: Km =
Enter: C ma =
Mesh Alignment Factor, Cma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Milling machine driven by an electric motor Example Problem 10-3 Factors in Design Analysis: If F<1.0 Alignment Factor, Km =1.0+Cpf +Cma
in
Axial Force: W x =
REF: mG =
Gear: : Pitting Geometry Factor:
ending Geometry Factors: Pinion:
REF: NP, NG =
e Width (2 x Axial Pitch): F min = 2.023 Enter: Face Width: F = 2.250 Enter: Elastic Coefficient: Cp = 2300 Enter: Quality Number: A v = 9
1.012 in
deg
20.65 deg
t =
1147 lb
ransverse pressure angle:
Secondary Input Data:
6.471 in
DG =
Pitch Diameter - Gear: 4.271 in 1870 ft/min
2.071 in
DP =
3.13
1104.0 rpm
75
75.3
Pitch Diameter - Pinion:
Gear Ratio: m G =
Actual Output Speed:
nG =
nG =
Desired Output Speed:
Computed data:
24
Number of Pinion Teeth: 1100 rpm
11.59
Pd = NP =
verse Diametral Pitch, P d :
65 hp 3450 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
nitial Input Data:
OF HELICAL GEARS-U.S.
NG =
uted number of gear teeth: Chosen No. of Gear Teeth:
282
3.111 in 2.056 in 314 ft/min 525 lb
DG = C= vt = Wt =
Pitch Diameter - Gear: Center Distance: Pitch Line Speed: Transmitted Load:
in
0.749 1.200
Axial Pitch: ce Width (2 x Axial Pitch): F min = Enter: Face Width: F=
Fig 10-5,6,7 Tab. 10-1,2
I = 0.205
191 lb
Wr =
Radial Force:
245 lb
3.11
Fig 10-5,6,7
J P = 0.453 J G = 0.486
56
Table 9-5
A v = 11 18
Table 9-7
Cp = 2300
in
Axial Force: W x =
REF: mG =
Gear: r: Pitting Geometry Factor:
ending Geometry Factors: Pinion:
REF: NP, NG =
Enter: Elastic Coefficient: Enter: Quality Number:
in
25.0 0.374
= px =
Helix angle:
deg
20.00 deg
t =
Transverse pressure angle:
Secondary Input Data:
1.000 in
DP =
3.11
385.7 rpm
56
Pitch Diameter - Pinion:
Gear Ratio: m G =
Actual Output Speed:
nG =
387.5 rpm
nG =
Desired Output Speed:
Computed data:
18 55.7
18
Pd = NP =
1200 rpm
Number of Pinion Teeth:
nP =
Input Speed:
5 hp
verse Diametral Pitch, P d :
P=
Input Power:
Initial Input Data:
1.00
1.00
1.24
1.00
1.00
F/DP = 1.20 [0.50 < F/DP < 2.00]
0.083
0.050
Table 9-11: Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.146
Commer. Precision Ex. Prec.
Figure 9-12
0.098
If F>1.0
39,050 psi
Gear: Required s at =
1.00
1.00
1.00
1.00
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
<107
Table 9-9
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.92
0.90
0.96
0.94
>107
For Kv
Red: HB
HB 455
HB 467
HB 340
HB 388
Gra
Throu
B = 0.8 C = 59
Gear
Computed contact stress no. = s c = 161,614 psi
Gear Pinion
37,488 psi Computed contact stress no. = s c = 161,614 psi
Computed bending stress no. = s t =
Computed stresses without modifying factors for reliability and life: Pinion Computed bending stress no. = s t = 40,219 psi
Gear requires HRC 58 min.: SAE 4118 DOQT 300; HRC 62; Carburized
Pinion requires HRC 58 min.: SAE 4118 DOQT 300; HRC 62; Carburized
One possible material specification: Steel pinion and gear: Carburized, Grade 1 for throug
Specify materials, alloy and heat treatment, for most severe requirement.
Gear: Required s ac = 175,668 psi
Stress Analysis: Pitting Pinion: Required s ac = 179,572 psi
42,786 psi
0.92
Pitting Stress Cycle Factor: Z NG = Stress Analysis: Bending Pinion: Required s at =
0.90
0.96
0.94
Pitting Stress Cycle Factor: Z NP =
Bending Stress Cycle Factor: Y NG =
Bending Stress Cycle Factor: Y NP =
Gear - Number of load cycles: NG = 3.5E+08 10 cycles
7
Enter: Design Life: 15000 hours See Table 9-12 Pinion - Number of load cycles: NP = 1.1E+09 Guidelines: YN , ZN
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: Kv =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
1.00
1.50
Overload Factor: K o = Size Factor: K s =
1.24
0.146
0.267
Open
0.098
0.095
Alignment Factor: Km =
Enter: C ma =
Mesh Alignment Factor, Cma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Factors in Design Analysis: If F<1.0 Alignment Factor, Km =1.0+Cpf +Cma
Problem 10-5
Problem 10‐5 – Helical gearing.
NG =
d number of gear teeth: osen No. of Gear Teeth:
283
C= vt = Wt =
Center Distance: Pitch Line Speed: Transmitted Load:
in
2.345 2.500
Axial Pitch: Width (2 x Axial Pitch): F min = Enter: Face Width: F=
Fig 10-5,6,7 Tab. 10-1,2
I = 0.220
695 lb
Wr =
Radial Force:
512 lb
3.00
Fig 10-5,6,7
J P = 0.480 J G = 0.526
72
Table 9-5
Av = 9 24
Table 9-7
Cp = 2300
in
Axial Force: W x =
REF: mG =
Gear: itting Geometry Factor:
ding Geometry Factors: Pinion:
REF: NP, NG =
nter: Elastic Coefficient: Enter: Quality Number:
in
15.0 1.172
= px =
Helix angle:
deg
20.00 deg
t =
1910 lb
nsverse pressure angle:
Secondary Input Data:
7.200 in
DG =
Pitch Diameter - Gear: 4.800 in 346 ft/min
2.400 in
DP =
3.00
183.3 rpm
72
71.4
185 rpm
24
10
Pitch Diameter - Pinion:
Gear Ratio: m G =
Actual Output Speed:
nG =
nG =
Desired Output Speed:
omputed data:
Pd = NP =
se Diametral Pitch, P d : Number of Pinion Teeth:
20 hp 550 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
tial Input Data:
F HELICAL GEARS-U.S.
1.25
1.00
1.16
1.00
1.00
1.00
1.50
F/DP = 1.04 [0.50 < F/DP < 2.00]
0.099
0.063
Table 9-11: Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.166
Commer. Precision Ex. Prec.
Figure 9-12
0.098
If F>1.0
38,931 psi
Gear: Required s at =
1.00
1.00
1.00
1.00
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
<107
Table 9-9
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.94
0.91
0.97
0.95
>107
For Kv :
Red: HB >
HB 431
HB 448
HB 338
HB 398
Grad
Through
B = 0.63 C = 70.7
Gear
Computed contact stress no. = s c = 126,177 psi
Gear Pinion
30,210 psi Computed contact stress no. = s c = 126,177 psi
Computed bending stress no. = s t =
Computed stresses without modifying factors for reliability and life: Pinion Computed bending stress no. = s t = 33,105 psi
Gear requires HRC 58 min.: SAE 4320 SOQT 450; HRC 59; Carburized
Pinion requires HRC 58 min.: SAE 4320 SOQT 450; HRC 59; Carburized
One possible material specification: Steel pinion and gear: Carburized, Grade 1 for through
Specify materials, alloy and heat treatment, for most severe requirement.
Gear: Required s ac = 167,788 psi
Stress Analysis: Pitting Pinion: Required s ac = 173,320 psi
43,560 psi
0.94
Pitting Stress Cycle Factor: Z NG = Stress Analysis: Bending Pinion: Required s at =
0.91
0.97
0.95
Pitting Stress Cycle Factor: Z NP =
Bending Stress Cycle Factor: Y NG =
Bending Stress Cycle Factor: Y NP =
Gear - Number of load cycles: NG = 1.7E+08 10 cycles
7
Enter: Design Life: 15000 hours See Table 9-12 Guidelines: YN , ZN Pinion - Number of load cycles: NP = 5.0E+08
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: Kv =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
1.20
0.099
Enter: C ma = Alignment Factor: K m =
0.288
Open
0.098
0.079
Mesh Alignment Factor, Cma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Milling machine driven by an electric motor Problem 10-6 Factors in Design Analysis: If F<1.0 Alignment Factor, K m =1.0+Cpf +Cma
Problem 10‐6 – Helical gearing
C= vt = Wt =
Center Distance: Pitch Line Speed: Transmitted Load:
284 in
2.246 2.500
Axial Pitch: Width (2 x Axial Pitch): F min = Enter: Face Width: F=
Fig 10-5,6,7 Tab. 10-1,2
I = 0.220
637 lb
Wr =
Radial Force:
816 lb
3.96
Fig 10-5,6,7
J P = 0.465 J G = 0.496
95
Table 9-5
Av = 9 24
Table 9-7
Cp = 2300
in
Axial Force: W x =
REF: mG =
Gear: Pitting Geometry Factor:
ding Geometry Factors: Pinion:
REF: NP, NG =
nter: Elastic Coefficient: Enter: Quality Number:
in
25.0 1.123
= px =
Helix angle:
deg
20.00 deg
t =
1751 lb
nsverse pressure angle:
Secondary Input Data:
15.833 in
DG =
Pitch Diameter - Gear: 9.917 in 942 ft/min
4.000 in
DP =
3.96
227.4 rpm
95
Pitch Diameter - Pinion:
Gear Ratio: m G =
Actual Output Speed:
nG =
NG =
omputed data:
ed number of gear teeth: osen No. of Gear Teeth:
94.9
227.5 rpm
nG =
Desired Output Speed:
6 24
Pd = NP =
se Diametral Pitch, P d : Number of Pinion Teeth:
50 hp 900 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
tial Input Data:
F HELICAL GEARS-U.S.
1.25
1.00
1.26
1.00
1.00
1.00
1.75
1.22
0.166
0.288
Open
0.056
0.038
F/DP = 0.63 [0.50 < F/DP < 2.00]
0.099
0.063
Table 9-11: Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.166
Commer. Precision Ex. Prec.
Figure 9-12
0.056
If F>1.0
29,607 psi
1.00
1.00
1.00
1.00
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
<107
Table 9-9
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.93
0.90
0.96
0.94
>107
Gear
Computed contact stress no. = s c = 106,299 psi
Gear Pinion
22,738 psi Computed contact stress no. = s c = 106,299 psi
Computed bending stress no. = s t =
Computed stresses without modifying factors for reliability and life: Computed bending stress no. = s t = 24,254 psi Pinion
Gear requires HB 353 min.: SAE 4140 OQT 900:HB 388
Pinion requires HB 368 min.: SAE 4140 OQT 900:HB 388
One possible material specification: Steel pinion and cast iron gear
Specify materials, alloy and heat treatment, for most severe requirement.
Gear: Required s ac = 142,875 psi
Stress Analysis: Pitting Pinion: Required s ac = 147,637 psi
Gear: Required s at =
32,253 psi
0.93
Pitting Stress Cycle Factor: Z NG = Stress Analysis: Bending Pinion: Required s at =
0.90
0.96
0.94
Pitting Stress Cycle Factor: Z NP =
Bending Stress Cycle Factor: Y NG =
Bending Stress Cycle Factor: Y NP =
Gear - Number of load cycles: NG = 2.0E+08 10 cycles
7
Enter: Design Life: 15000 hours See Table 9-12 Pinion - Number of load cycles: NP = 8.1E+08 Guidelines: YN , ZN
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: Kv =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: Km =
Enter: C ma =
Mesh Alignment Factor, Cma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Punch press driven by an electric motor Problem 10-7 Factors in Design Analysis: If F<1.0 Alignment Factor, Km =1.0+Cpf +Cma
For Kv :
for through
Red: HB >
HB 353
HB 368
HB 217
HB 252
Grad
Through
B = 0.63 C = 70.7
Problem 10‐7 – Helical gearing
NG =
puted number of gear teeth: Chosen No. of Gear Teeth:
285
20.000 in 10.833 in 393 ft/min 210 lb
DG = C= vt = Wt =
Pitch Diameter - Gear: Center Distance: Pitch Line Speed: Transmitted Load:
in
1.123 1.750
Axial Pitch: ace Width (2 x Axial Pitch): F min = Enter: Face Width: F=
Fig 10-5,6,7 Tab. 10-1,2
I = 0.260
98 lb 76 lb
Radial Force:
Wr =
12.00
Fig 10-5,6,7
J P = 0.451 J G = 0.512
240
Table 9-5
A v = 12 20
Table 9-7
Cp = 2100
in
Axial Force: W x =
REF: mG =
Gear: er: Pitting Geometry Factor:
Bending Geometry Factors: Pinion:
REF: NP, NG =
Enter: Elastic Coefficient: Enter: Quality Number:
in
25.0 0.561
= px =
Helix angle:
deg
20.00 deg
t =
Transverse pressure angle:
Secondary Input Data:
1.667 in
DP =
12.00
75.0 rpm
240
Pitch Diameter - Pinion:
Gear Ratio: m G =
Actual Output Speed:
nG =
75 rpm
nG =
Desired Output Speed:
Computed data:
20 240.0
12
Pd = NP =
Number of Pinion Teeth:
900 rpm
nP =
Input Speed: sverse Diametral Pitch, P d :
2.5 hp
P=
APPLICATION:
Input Power:
Initial Input Data:
N OF HELICAL GEARS-U.S.
1.00
1.00
1.33
1.00
1.00
1.00
2.00
1.37
0.276
0.276
Open
0.089
0.080
F/DP = 1.05 [0.50 < F/DP < 2.00]
0.090
0.056
Table 9-11: Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.154
Commer. Precision Ex. Prec.
Figure 9-12
0.089
If F>1.0
Gear: Required s at =
68,583 psi
72,310 psi
10,295 psi
1.00
1.00
1.00
1.00
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
<107
Table 9-9
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.97
0.92
0.99
0.95
>107
10,192 psi 66,525 psi 66,525 psi
Computed bending stress no. = s t = Computed contact stress no. = s c = Computed contact stress no. = s c =
Gear
Pinion
Gear
Computed stresses without modifying factors for reliability and life: Pinion Computed bending stress no. = s t = 11,571 psi
Gear: Gray cast iron-Grade 40; s at = 13 ksi; s ac = 75 ksi (Table 9-10)
Pinion requires HB 134 min.: SAE 1020 CD:HB 160 - or almost any steel
One possible material specification: Steel pinion and cast iron gear
Specify materials, alloy and heat treatment, for most severe requirement.
Gear: Required s ac =
Stress Analysis: Pitting Pinion: Required s ac =
12,180 psi
0.97
Pitting Stress Cycle Factor: Z NG = Stress Analysis: Bending Pinion: Required s at =
0.92
0.99
0.95
Pitting Stress Cycle Factor: Z NP =
Bending Stress Cycle Factor: Y NG =
Bending Stress Cycle Factor: Y NP =
Gear - Number of load cycles: NG = 3.6E+07 10 cycles
7
Enter: Design Life: 8000 hours See Table 9-12 Guidelines: YN , ZN Pinion - Number of load cycles: NP = 4.3E+08
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: Kv =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: Km =
Enter: C ma =
Mesh Alignment Factor, Cma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Small cement mixer driven by a gasoline engine Problem 10-8 Factors in Design Analysis: If F<1.0 Alignment Factor, Km =1.0+Cpf +Cma
For Kv :
for through
Red: HB >
HB 123
HB 134
HB -32
HB -8
Grad
Through
B = 0.91 C = 54.7
Problem 10‐8 – Helical gearing
nG = NG =
Desired Output Speed:
puted number of gear teeth: Chosen No. of Gear Teeth:
286
9.000 in 6.667 in 2496 ft/min 992 lb
DG = C= vt = Wt =
Pitch Diameter - Gear: Center Distance: Pitch Line Speed: Transmitted Load:
in
2.246 2.500
Axial Pitch:
ace Width (2 x Axial Pitch): F min = Enter: Face Width: F=
Fig 10-5,6,7 Tab. 10-1,2
I = 0.196
462 lb 361 lb
Radial Force:
Wr =
2.08
Fig 10-5,6,7
J P = 0.426 J G = 0.492
54
Table 9-5
Av = 9 26
Table 9-7
Cp = 2300
in
Axial Force: W x =
REF: mG =
Gear: er: Pitting Geometry Factor:
Bending Geometry Factors: Pinion:
REF: NP, NG =
Enter: Elastic Coefficient: Enter: Quality Number:
in
25.0 1.123
= px =
Helix angle:
deg
20.00 deg
t =
Transverse pressure angle:
Secondary Input Data:
4.333 in
DP =
2.08
4569.2 rpm
54
53.8
Pitch Diameter - Pinion:
Gear Ratio: m G =
Actual Output Speed:
nG =
26
NP =
Number of Pinion Teeth:
Computed data:
6
Pd =
sverse Diametral Pitch, P d : 4550 rpm
75 hp 2200 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
Initial Input Data:
N OF HELICAL GEARS-U.S.
0.033
1.00
1.00
1.44
1.00
1.00
1.00
2.75
1.34
0.288
0.288
Open
0.051
0.051
[0.50 < F/DP < 2.00]
0.099
0.063
Table 9-11: Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.166
Commer. Precision Ex. Prec.
Figure 9-12
27,011 psi
Gear: Required s at =
1.00
1.00
1.00
1.00
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
<107
Table 9-9
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.91
0.90
0.95
0.94
>107
Gear
Computed contact stress no. = s c = 114,456 psi
Gear Pinion
25,660 psi Computed contact stress no. = s c = 114,456 psi
Computed bending stress no. = s t =
Computed stresses without modifying factors for reliability and life: Computed bending stress no. = s t = 29,638 psi Pinion
Gear requires HB 300 min.: SAE 3140 OQT 1000 steel; HB = 311
Pinion requires HB 305 min.: SAE 3140 OQT 1000 steel; HB = 311
One possible material specification: Steel pinion and gear: Through hardened
Specify materials, alloy and heat treatment, for most severe requirement.
Gear: Required s ac = 125,776 psi
Stress Analysis: Pitting Pinion: Required s ac = 127,173 psi
31,529 psi
0.91
Pitting Stress Cycle Factor: Z NG = Stress Analysis: Bending Pinion: Required s at =
0.90
0.95
0.94
Pitting Stress Cycle Factor: Z NP =
Bending Stress Cycle Factor: Y NG =
Bending Stress Cycle Factor: Y NP =
Gear - Number of load cycles: NG = 5.1E+08 10 cycles
7
Enter: Design Life: 8000 hours See Table 9-12 Guidelines: YN , ZN Pinion - Number of load cycles: NP = 1.1E+09
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: Kv =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: K m =
Enter: C ma =
Mesh Alignment Factor, Cma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Wood chipper driven by a gasoline engine Problem 10-9 Speed increaser - cells changed Factors in Design Analysis: If F<1.0 If F>1.0 Alignment Factor, K m =1.0+Cpf +Cma F/DP = 0.58
For Kv :
for through
Red: HB >
HB 300
HB 305
HB 184
HB 242
Grade
Through
B = 0.63 C = 70.7
Problem 10‐9 – Helical gearing: Note that this design is for a speed increaser. Modest adjustments in the manner if inputting data are used.
NG =
Desired Output Speed: puted number of gear teeth: Chosen No. of Gear Teeth:
6
287
20.333 in 11.917 in 412 ft/min 1601 lb
DG = C= vt = Wt =
Pitch Diameter - Gear: Center Distance: Pitch Line Speed: Transmitted Load:
25.0
= px =
Helix angle: Axial Pitch:
1.00
1.00
1.27
1.00
1.00
1.00
Bending Stress Cycle Factor: Y NP =
REF: NP, NG =
Fig 10-5,6,7 Tab. 10-1,2
I = 0.240
746 lb 583 lb
Radial Force:
Wr =
5.81
Fig 10-5,6,7
J P = 0.444 J G = 0.494
122
Axial Force: W x =
REF: mG =
Gear: er: Pitting Geometry Factor:
21
0.061
Table 9-11: Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
7
37,194 psi
Gear: Required s at =
1.00
1.00
1.00
1.00
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
<107
Table 9-9
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.97
0.93
0.99
0.96
>107
Pinion Gear
Gear Computed contact stress no. = s c = 138,176 psi
36,822 psi Computed contact stress no. = s c = 138,176 psi
Computed bending stress no. = s t =
Computed stresses without modifying factors for reliability and life: Pinion Computed bending stress no. = s t = 40,969 psi
Gear requires HB 352 min.: SAE 4340 OQT 1000; HB 363
Pinion requires HB 371 min.: SAE 4340 OQT 900:HB 388
One possible material specification: Steel pinion and gear: Through hardened
Specify materials, alloy and heat treatment, for most severe requirement.
Gear: Required s ac = 142,449 psi
Stress Analysis: Pitting Pinion: Required s ac = 148,576 psi
42,676 psi
0.97
Pitting Stress Cycle Factor: Z NG =
in
Bending Geometry Factors: Pinion:
0.096
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Stress Analysis: Bending Pinion: Required s at =
0.93
Pitting Stress Cycle Factor: Z NP =
0.99
0.96
in
Bending Stress Cycle Factor: Y NG =
ace Width (2 x Axial Pitch): F min = 2.246 Enter: Face Width: F = 2.250 Enter: Elastic Coefficient: Cp = 2300 Enter: Quality Number: A v = 11 Table 9-5
0.162 Figure 9-13
Gear - Number of load cycles: NG = 3.7E+07 10 cycles
in
Table 9-7
F/DP = 0.64 [0.50 < F/DP < 2.00]
Commer. Precision Ex. Prec.
Figure 9-12
0.055
If F>1.0
Enter: Design Life: 8000 hours See Table 9-12 Guidelines: Y N , ZN Pinion - Number of load cycles: NP = 2.2E+08
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
1.22 2.75
Overload Factor: K o =
0.162
0.284
Open
0.055
0.039
Alignment Factor: Km =
Enter: C ma =
Mesh Alignment Factor, Cma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Small tractor driven by a gasoline engine Problem 10-10 Factors in Design Analysis: If F<1.0 Alignment Factor, Km =1.0+Cpf +Cma
1.123
deg
20.00 deg
t =
Transverse pressure angle:
Secondary Input Data:
3.500 in
DP =
5.81
77.5 rpm
122
121.9
77.5 rpm
21
Pitch Diameter - Pinion:
Gear Ratio: m G =
Actual Output Speed:
nG =
nG =
Number of Pinion Teeth:
Computed data:
Pd = NP =
sverse Diametral Pitch, P d :
20 hp 450 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
Initial Input Data:
N OF HELICAL GEARS-U.S.
For K v :
for through
Red: HB >
HB 352
HB 371
HB 316
HB 386
Grad
Through
B = 0.82 C = 59.7
Problem 10‐10 – Helical gearing
NG =
d number of gear teeth: osen No. of Gear Teeth:
288
2.083 in 1.875 in 1963 ft/min 252 lb
DG = C= vt = Wt =
Pitch Diameter - Gear: Center Distance: Pitch Line Speed: Transmitted Load:
1.25
1.00
1.36
1.00
1.00
1.00
1.20
1.20
0.147
0.268
Open
0.053
0.050
Fig 10-5,6,7 Tab. 10-1,2
I = 0.150
118 lb 92 lb
Radial Force:
Wr =
1.25
Fig 10-5,6,7
J P = 0.418 J G = 0.432
25
Table 9-5
Axial Force: W x =
REF: mG =
Gear: itting Geometry Factor:
ding Geometry Factors: Pinion:
20
0.84
Pitting Stress Cycle Factor: Z NG =
in
REF: NP, NG =
0.83
Pitting Stress Cycle Factor: Z NP =
in
Width (2 x Axial Pitch): F min = 1.123 Enter: Face Width: F = 1.250 ter: Elastic Coefficient: Cp = 2300 Enter: Quality Number: A v = 9
0.083
0.051
Table 9-11: Use 1.00 for R = .99
Use 1.00 if no unusual conditions
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
7
15,397 psi
Gear: Required s at =
1.00
1.00
1.00
1.00
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
<107
Table 9-9
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.84
0.83
0.89
0.88
>107
10,962 psi 91,382 psi 91,382 psi
Computed bending stress no. = s t = Computed contact stress no. = s c = Computed contact stress no. = s c =
Gear
Pinion
Gear
Computed stresses without modifying factors for reliability and life: Computed bending stress no. = s t = 11,329 psi Pinion
Gear requires HB 332 min.: SAE 4340 OQT 1000; HB 363
Pinion requires HB 337 min.: SAE 4340 OQT 1000:HB 363
One possible material specification: Steel pinion and gear: Through hardened
Specify materials, alloy and heat treatment, for most severe requirement.
Gear: Required s ac = 135,985 psi
Stress Analysis: Pitting Pinion: Required s ac = 137,623 psi
16,093 psi
0.89
Stress Analysis: Bending Pinion: Required s at =
Bending Stress Cycle Factor: Y NG =
0.88
Axial Pitch:
Bending Stress Cycle Factor: Y NP =
in
25.0 0.561
= px =
Helix angle:
Table 9-10
0.147 Figure 9-13
Gear - Number of load cycles: NG = 2.2E+10 10 cycles
20.00 deg
t = deg
F/DP = 0.75 [0.50 < F/DP < 2.00]
Commer. Precision Ex. Prec.
Figure 9-12
0.053
Enter: Design Life: 100000 hours See Table 9-12 Guidelines: YN , ZN Pinion - Number of load cycles: NP = 2.7E+10
Reliability Factor: K R =
Service Factor: SF =
Dynamic Factor: Kv =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
Overload Factor: K o =
Alignment Factor: Km =
Enter: C ma =
Mesh Alignment Factor, Cma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, Cpf =
Electric power generator driving by a water turbine Problem 10-11 Factors in Design Analysis: If F<1.0 If F>1.0 Alignment Factor, Km =1.0+Cpf +Cma
nsverse pressure angle:
Secondary Input Data:
1.667 in
DP =
1.25
3600.0 rpm
25
25.0
3600 rpm
20
12
Pitch Diameter - Pinion:
Gear Ratio: m G =
Actual Output Speed:
nG =
nG =
Desired Output Speed:
mputed data:
Pd = NP =
se Diametral Pitch, P d : Number of Pinion Teeth:
15 hp 4500 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
ial Input Data:
F HELICAL GEARS-U.S.
For Kv :
for through
Red: HB >
HB 332
HB 337
HB 34
HB 43
Grad
Through
B = 0.63 C = 70.7
Problem 10‐11 – Helical gearing
7.765 in
DG = C= vt = Wt =
Pitch Diameter - Gear: Center Distance: Pitch Line Speed: Load at P min Capacity:
289 75
REF: mG =
Gear: ing Geometry Factor:
Tab. 10-1,2
I = 0.200 3.75
Fig. 10-5,6,7
J G = 0.521
ng Geometry Factors: Press. angle = 20 deg Pinion: J P = 0.465 Fig. 10-5,6,7
20
0.92
Pitting Stress Cycle Factor: Z NG =
nsmission Capacity: 22.59 hp er: Elastic Coefficient: Cp = 2300 nter: Quality Number: A v = 12 REF: NP, NG =
0.89
Pitting Stress Cycle Factor: Z NP =
ed on Contact Stress: 24.14 hp
Table 9-5
0.099
0.063
Use 1.00 if no unusual conditions Fig. 9-25 or 9-26; Gear only Table 9-11 Use 1.00 for R = .99
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.166
Commer. Precision Ex. Prec.
39,200 psi
1.00
1.00
1.00
1.00
Fig. 9-24
Fig. 9-24
Fig. 9-22
Fig. 9-22
Table 9-9
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.92
0.89
0.95
0.93
29,165 psi 29,792 psi 98,897 psi 102,230 psi
Computed bending stress no. = s t = Computed bending stress no. = s t = Computed contact stress no. = s c = Computed contact stress no. = s c =
Pinion material: AISI 4140 OQT 1000 Gear material: AISI 4140 OQT 1000
gear
pinion
gear
pinion
341 HB 341 HB
Material specification: Steel pinion; Steel gear, Through HT
Gear: s ac = 138,900 psi
Allowable Contact Stress Numbers: (Input) Pinion: s ac = 138,900 psi
Gear: s at =
Allowable Bending Stress Numbers: (Input) Pinion: s at = 39,200 psi
0.95
0.93
Bending Stress Cycle Factor: Y NP = Bending Stress Cycle Factor: Y NG =
Table 9-7
1.25
1.00 1.00
1.50
1.00
1.00
1.00
F/D P = 1.21 [0.50 < F/D P < 2.00]
Figure 9-12
0.114
If F>1.0
Enter: Design Life: 15000 hours See Table 9-12 Pinion - Number of load cycles: N P = 1.6E+09 Guidelines: Y N , Z N 7 >107 <107 Gear - Number of load cycles: N G = 4.1E+08 10 cycles
Reliability Factor: K R =
Service Factor: SF = Hardness Ratio Factor: C H =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
47.41 hp
d on Bending Stress:
1.28
Alignment Factor: K m = 1.25
0.166
Enter: C ma = Overload Factor: K o =
0.288
Open
0.114
0.096
Mesh Alignment Factor, C ma =
Type of gearing:
Enter: C pf =
Pinion Proportion Factor, C pf =
Centrifugal pump driven by an electric motor Problem 10-12 Factors in Design Analysis: Alignment Factor, K m=1.0+C pf +C ma If F<1.0
ed on Contact Stress: 22.59 hp
ed on Bending Stress: 41.43 hp
ransmission Capacity: (Using Eq. 9-35, 9-37)
797 lb
4.918 in 935 ft/min
2.071 in
DP =
3.75
460.0 rpm
tch Diameter - Pinion:
Gear Ratio: m G =
Actual Output Speed:
nG =
75
NG =
Number of Gear Teeth:
mputed data:
20
9.659
Pd = NP =
Diametral Pitch:
1725 rpm
2.500 in
APPLICATION:
mber of Pinion Teeth:
nP =
Input Speed:
ARS ANSMISSION CAPACITY al Input Data: Enter: Face Width: F =
138.9 ks
138.9 ks
39.2 ks
39.2 ks
Gra
Throu
0 54
For K B A
Problem 10‐12 – Helical gearing: Note that this problems calls for rating the transmission capacity of the gear set. Modest adjustments in the spreadsheet are used.
nG =
7.765 in
DG = C= vt = Wt =
Pitch Diameter - Gear: Center Distance: Pitch Line Speed: Load at Pmin Capacity:
290 75
REF: mG =
Gear: ing Geometry Factor:
Tab. 10-1,2
I = 0.200 3.75
Fig. 10-5,6,7
J G = 0.521
ng Geometry Factors: Press. angle = 20 deg Pinion: J P = 0.465 Fig. 10-5,6,7
20
0.92
Pitting Stress Cycle Factor: Z NG =
nsmission Capacity: 37.94 hp er: Elastic Coefficient: Cp = 2300 nter: Quality Number: A v = 12 REF: NP, NG =
0.89
Pitting Stress Cycle Factor: Z NP =
ed on Contact Stress: 40.54 hp
Table 9-5
[0.50 < F/D P < 2.00]
0.099
0.063
Use 1.00 if no unusual conditions Fig. 9-25 or 9-26; Gear only Table 9-11 Use 1.00 for R = .99
[Computed: See Fig. 9-16]
Fig. 9-14: Use 1.00 if solid blank
Fig. 9-14: Use 1.00 if solid blank
Table 9-2: Use 1.00 if P d >= 5
Table 9-1
[Computed]
Figure 9-13
0.166
Commer. Precision Ex. Prec.
55,000 psi
1.00
1.00
1.00
1.00
Fig. 9-22
Fig. 9-22
Fig. 9-21
Fig. 9-21
Table 9-9
See Fig. 9-19 or
Table 9-9
See Fig. 9-18 or
0.92
0.89
0.95
0.93
40,920 psi 41,800 psi 128,160 psi 132,480 psi
Computed bending stress no. = s t = Computed bending stress no. = s t = Computed contact stress no. = s c = Computed contact stress no. = s c =
Pinion material: SAE 4620 DOQT 300 Gear material: SAE 4620 DOQT 300
gear
pinion
gear
pinion
62 HRC 62 HRC
Material specification: Steel pinion+gear, Carburized case hard.
Gear: s ac = 180,000 psi
Allowable Contact Stress Numbers: (Input) Pinion: s ac = 180,000 psi
Gear: s at =
Allowable Bending Stress Numbers: (Input) Pinion: s at = 55,000 psi
0.95
0.93
Bending Stress Cycle Factor: Y NP = Bending Stress Cycle Factor: Y NG =
Table 9-7
1.25
1.00 1.00
1.50
1.00
1.00
1.00
0.114 Figure 9-12
Enter: Design Life: 15000 hours See Table 9-12 Guidelines: Y N , Z N Pinion - Number of load cycles: N P = 1.6E+09 7 >107 <107 Gear - Number of load cycles: N G = 4.1E+08 10 cycles
Reliability Factor: K R =
Service Factor: SF = Hardness Ratio Factor: C H =
Dynamic Factor: K v =
Gear Rim Thickness Factor: K BG =
Pinion Rim Thickness Factor: K BP =
Size Factor: K s =
66.52 hp
d on Bending Stress:
1.28
Alignment Factor: K m = 1.25
0.166
Enter: C ma = Overload Factor: K o =
0.288
Open
ed on Contact Stress: 37.94 hp
ed on Bending Stress: 58.12 hp
ransmission Capacity: (Using Eq. 9-35, 9-37)
1338 lb
4.918 in 935 ft/min
2.071 in
DP =
3.75
460.0 rpm
tch Diameter - Pinion:
Gear Ratio: m G =
Actual Output Speed:
mputed data:
0.096 0.114
Mesh Alignment Factor, C ma =
Type of gearing:
20 75
NP = NG =
mber of Pinion Teeth: Number of Gear Teeth:
Enter: C pf =
Pinion Proportion Factor, C pf =
1725 rpm 9.659
Pd =
Diametral Pitch:
Centrifugal pump driven by an electric motor Problem 10-13; Same as Problem 10-12 except case hardened steel Factors in Design Analysis: F/D P = 1.21 Alignment Factor, K m=1.0+C pf +C ma If F<1.0 If F>1.0
nP =
2.500 in
APPLICATION:
Input Speed:
ARS ANSMISSION CAPACITY al Input Data: Enter: Face Width: F =
49.1 ks
49.1 ks
17.6 ks
17.6 ks
Gra
Throu
0. 54
For K B= A=
Problem 10‐13 – Helical gearing: Note that this problems calls for rating the transmission capacity of the gear set. Modest adjustments in the spreadsheet are used. Also, this is a modest change from Problem 12 with case hardened steel.
8 16
Pd = NP = nG = NG = nG =
ametral Pitch: Pinion Teeth: Output Speed: of gear teeth: of Gear Teeth: mputed data: utput Speed:
291 3.162 in
= = Ao = vt =
angle - Pinion: one distance: h Line Speed: ress analysis: Max
A v = 11 J P = 0.230
ality Number: actor - Pinion:
312 lb 35.9 lb 108 lb
- Radial load: W rG = r - Axial load: W xG =
787.5 lb-in
TG = ngential load: W tG =
haft - Torque:
108 lb 35.9 lb
- Radial load: W rP = n - Axial load: W xP =
0.842 in 20 degrees 312 lb [Eq. 10-10]
rm =
essure angle: = ngential load: W tP =
dius of pinion:
Factor - Gear: J G = 0.187 Fig. 10-15 metry Factor: I = 0.076 Fig. 10-19 ue for shaft and bearing load analysis: haft - Torque: T P = 262.5 lb-in
Fig. 10-15
Table 9-5
1.000 in Cp = 2300 Table 9-7
1.250
ic Coefficient:
Max 1.054
Nom
263 lb
314 ft/min
uidelines (in): 0.949 Face Width: F =
dary Input Data:
Wt =
71.57 degrees
DG =
meter - Gear: angle - Gear:
6.000 in
DP = 18.43 degrees
3.00 2.000 in
mG =
Gear Ratio: meter - Pinion:
200.0 rpm
48
48.0
200 rpm
2.5 hp 600 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
l Input Data:
EARS
0.948
2.700
107,938 psi
7,502 psi
9,227 psi
Figure 9-21
Figure 9-17
Figure 9-17
1.109
Gear: HB 247 required: SAE 6150 OQT 1300; HB = 241
Gear: Required s ac = 107,938 psi Figure 9-21 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification: Pinion: HB 247 required: SAE 6150 OQT 1300; HB = 241
Pinion: Required s ac =
Stress Analysis: Pitting
Gear: Required s at =
Stress Analysis: Bending Pinion: Required s at =
>104
5.599E-01
9.479E-01
>103but<3x106 >3x106 cyc.
number of cycles.
Input value from pertinent
See Table 9-12
Table 9-11; ;Use 1.00 for R = .99
Figure 10-16 <103 Cyc.
1.8E+08
F >3.15 0.83 Use 1.00 if no unusual conditions
15000 hours 5.4E+08
0.5<F<3.15 0.563
Use 0.50 if P d > 15
Computed: Table 9-6
F <0.5 0.50
0.513
Table 9-1
Neither gear straddle mounted
4 Stress cycle factor-Pitting Figure 10-20 <10 Cyc. Enter Stress cycle factor-Pitting, C L = 1.109 2.000
Enter Stress cycle factor-Bending, K L =
Stress cycle factor-Bending
Gear - Number of load cycles: N G =
Enter: Design Life: Pinion - Number of load cycles: N P =
1.00
1.00
Reliability Factor: K R =
1.239
Service Factor: SF =
0.56
Dynamic Factor: K v =
Fig. 10-18-Gear Size Factor: C s =
0.513
1.50
Figure 10-13-Pinion Size Factor: K s =
1.004
Enter K m =
1.254
1.104
1.004
Overload Factor: K o =
Neither gear straddle mounted:
One gear straddle mounted:
Both gears straddle mounted:
Load with moderate shock driven by an electric motor Example Problems 10-6, 7, 8 Both gears straddle mounted Factors in Design Analysis: Load distribution factor, K m: From Figure 10-14 and Equation 10-16
F
Bevel Gearing ‐ Example Problems – This is the first of a set of five spreadsheets dealing with bevel gearing. This page uses data from Example Problems 10‐6, ‐7, and ‐8. Following pages give solutions to the four end‐of‐chapter problems.
292
15
3.00 2.500 in 7.500 in 71.57 degrees 3.953 in
NP = nG = NG = nG = mG = DP = DG = = = Ao = vt = Wt =
Pinion Teeth: Output Speed: of gear teeth: of Gear Teeth: puted data: utput Speed: Gear Ratio: meter - Pinion: meter - Gear: ngle - Pinion: angle - Gear: one distance: h Line Speed: ess analysis: Max
Av = 8 J P = 0.228
ality Number: actor - Pinion:
599 lb 68.9 lb 207 lb
- Radial load: W rG = r - Axial load: W xG =
1890 lb-in
haft - Torque:
TG = ngential load: W tG =
207 lb 68.9 lb
- Radial load: W rP = n - Axial load: W xP =
1.052 in 20 degrees 599 lb [Eq. 10-10]
rm =
essure angle: = ngential load: W tP =
dius of pinion:
Factor - Gear: J G = 0.190 Fig. 10-15 metry Factor: I = 0.074 Fig. 10-19 ue for shaft and bearing load analysis: haft - Torque: T P = 630 lb-in
Fig. 10-15
Table 9-5
1.250 in Cp = 2300 Table 9-7
1.667
c Coefficient:
Max 1.318
Nom
504 lb
196 ft/min
18.43 degrees
100.0 rpm
45
45.0
100 rpm
uidelines (in): 1.186 Face Width: F =
ary Input Data:
6
Pd =
ametral Pitch:
300 rpm
P= nP =
Input Power:
3 hp
APPLICATION:
Input Speed:
l Input Data:
EARS
1.006
Use 1.00 if no unusual conditions
1.007
2.700
15,075 psi
131,709 psi
12,563 psi
Figure 9-21
Figure 9-17
Figure 9-17
1.274
>104
8.397E-01
Gear: HB 238 required: SAE 6150 OQT 1300; HB = 241
1.007E+00
>103but<3x106 >3x106 cyc.
number of cycles.
Gear: Required s ac = 131,709 psi Figure 9-21 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification: Pinion: HB 295 required: SAE 6150 OQT 1200; HB = 295
Pinion: Required s ac =
Stress Analysis: Pitting
Gear: Required s at =
Stress Analysis: Bending Pinion: Required s at =
See Table 9-12 Input value from pertinent Figure 10-16 <103 Cyc.
6.0E+06
F >3.15 0.83
Table 9-11; ;Use 1.00 for R = .99 1000 hours 1.8E+07
0.5<F<3.15 0.594
Use 0.50 if P d > 15
Computed: Table 9-6
F <0.5 0.50
0.522
Table 9-1
Neither gear straddle mounted
4 Stress cycle factor-Pitting Figure 10-20 <10 Cyc. Enter Stress cycle factor-Pitting, C L = 1.274 2.000
Enter Stress cycle factor-Bending, K L =
Stress cycle factor-Bending
Gear - Number of load cycles: N G =
Enter: Design Life: Pinion - Number of load cycles: N P =
1.00
1.00
Reliability Factor: K R =
1.091
Service Factor: SF =
0.594
Dynamic Factor: K v =
Fig. 10-18-Gear Size Factor: C s =
0.522
2.00
Overload Factor: K o = Figure 10-13-Pinion Size Factor: K s =
1.256
1.256
1.106
Enter K m =
Neither gear straddle mounted:
One gear straddle mounted:
Both gears straddle mounted:
Concrete mixer with moderate shock driven by a gasoline engine Problem 10-14 Neither gear straddle mounted Factors in Design Analysis: Load distribution factor, K m: From Figure 10-14 and Equation 10-16
F
Problem 10‐14 – Bevel gearing
293 63.43 degrees 2.795 in
nG = NG = nG = mG = DP = DG = = = Ao = vt =
ed Output Speed: mber of gear teeth: No. of Gear Teeth: Computed data: al Output Speed: Gear Ratio: Diameter - Pinion: h Diameter - Gear: one angle - Pinion: cone angle - Gear: ter cone distance: Pitch Line Speed:
Max
A v = 10 J P = 0.228
: Quality Number: ry Factor - Pinion:
Fig. 10-15
Table 9-5
Table 9-7
rm =
161 lb 26.3 lb 53 lb
ear - Radial load: W rG = Gear - Axial load: W xG =
352.8 lb-in
TG = - Tangential load: W tG =
ar Shaft - Torque:
53 lb 26.3 lb
nion - Axial load: W xP =
20 degrees 161 lb [Eq. 10-10]
1.093 in
nion - Radial load: W rP =
r: Pressure angle: = - Tangential load: W tP =
n radius of pinion:
etry Factor - Gear: J G = 0.218 Fig. 10-15 Geometry Factor: I = 0.083 Fig. 10-19 orque for shaft and bearing load analysis: on Shaft - Torque: T P = 176.4 lb-in
Cp = 2300
Elastic Coefficient:
1.000 0.700 in - Low
Max 0.932
Nom
141 lb
818 ft/min
26.57 degrees
625.0 rpm
50
50.0
625 rpm
25
10
th Guidelines (in): 0.839 nter: Face Width: F =
condary Input Data:
Wt =
5.000 in
NP =
er of Pinion Teeth:
d: stress analysis:
2.00 2.500 in
Pd =
Diametral Pitch:
1250 rpm
P= nP =
Input Power:
3.5 hp
APPLICATION:
Input Speed:
nitial Input Data:
EL GEARS
1.002
0.936
2.700
115,859 psi
15,005 psi
15,694 psi
Figure 9-21
Figure 9-17
Figure 9-17
0.993
Gear: HB 293 required: SAE 4140 OQT 1100; HB = 311
Gear: Required s ac = 115,859 psi Figure 9-21 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification: Pinion: HB 309 required: SAE 4140 OQT 1100; HB = 311
Pinion: Required s ac =
Stress Analysis: Pitting
Gear: Required s at =
Stress Analysis: Bending Pinion: Required s at =
>104
0.513
0.936
>103but<3x106 >3x106 cyc.
number of cycles.
Input value from pertinent
See Table 9-12
Table 9-11; ;Use 1.00 for R = .99
Figure 10-16 <103 Cyc.
5.63E+08
F >3.15 0.83 Use 1.00 if no unusual conditions 15000 hours 1.13E+09
0.5<F<3.15 0.525
Use 0.50 if P d > 15
Computed: Table 9-6
F <0.5 0.50
0.508
Table 9-1
Neither gear straddle mounted
4 Stress cycle factor-Pitting Figure 10-20 <10 Cyc. Enter Stress cycle factor-Pitting, C L = 0.993 2.000
Enter Stress cycle factor-Bending, K L =
Stress cycle factor-Bending
Gear - Number of load cycles: N G =
Enter: Design Life: Pinion - Number of load cycles: N P =
1.00
1.00
Reliability Factor: K R =
1.305
Service Factor: SF =
0.525
Dynamic Factor: K v =
Fig. 10-18-Gear Size Factor: C s =
0.508
2.00
Overload Factor: K o = Figure 10-13-Pinion Size Factor: K s =
1.252
1.252
1.102
Enter K m =
Neither gear straddle mounted:
One gear straddle mounted:
Both gears straddle mounted:
Conveyor with moderate shock driven by a gasoline engine Problem 10-15 Neither gear straddle mounted Factors in Design Analysis: Load distribution factor, K m: From Figure 10-14 and Equation 10-16
F
Problem 10‐15 – Bevel gearing
294 70.56 degrees 3.380 in
nG = NG = nG = mG = DP = DG = = = Ao = vt =
ed Output Speed: mber of gear teeth: No. of Gear Teeth: Computed data: al Output Speed: Gear Ratio: Diameter - Pinion: Diameter - Gear: ne angle - Pinion: one angle - Gear: ter cone distance: Pitch Line Speed:
Max
A v = 10 J P = 0.240
: Quality Number: ry Factor - Pinion:
395 lb 47.9 lb 136 lb
ear - Radial load: W rG = Gear - Axial load: W xG =
1050 lb-in
ar Shaft - Torque:
TG = - Tangential load: W tG =
136 lb 47.9 lb
ion - Radial load: W rP = nion - Axial load: W xP =
0.938 in 20 degrees 395 lb [Eq. 10-10]
rm =
r: Pressure angle: = - Tangential load: W tP =
n radius of pinion:
etry Factor - Gear: J G = 0.192 Fig. 10-15 Geometry Factor: I = 0.078 Fig. 10-19 orque for shaft and bearing load analysis: on Shaft - Torque: T P = 370.59 lb-in
Fig. 10-15
Table 9-5
1.125 in Cp = 2300 Table 9-7
1.250
Elastic Coefficient:
Max 1.127
Nom
330 lb
501 ft/min
19.44 degrees
300.0 rpm
51
51.0
300 rpm
18
8
th Guidelines (in): 1.014 nter: Face Width: F =
ondary Input Data:
Wt =
6.375 in
NP =
er of Pinion Teeth:
: stress analysis:
2.83 2.250 in
Pd =
Diametral Pitch:
850 rpm
P= nP =
Input Power:
5 hp
APPLICATION:
Input Speed:
nitial Input Data:
EL GEARS
1.005
Use 1.00 if no unusual conditions
0.942
2.700
136,032 psi
10,615 psi
13,268 psi
Figure 9-21
Figure 9-17
Figure 9-17
1.016
>104
0.537
Gear: HB 330 required: SAE 1340 OQT 1100; HB = 331
0.942
>103but<3x106 >3x106 cyc.
number of cycles.
Gear: Required s ac = 136,032 psi Figure 9-21 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification: Pinion: HB 330 required: SAE 1340 OQT 1100; HB = 331
Pinion: Required s ac =
Stress Analysis: Pitting
Gear: Required s at =
Stress Analysis: Bending Pinion: Required s at =
See Table 9-12 Input value from pertinent Figure 10-16 <103 Cyc.
2.7E+08
F >3.15 0.83
Table 9-11; ;Use 1.00 for R = .99 15000 hours 7.7E+08
0.5<F<3.15 0.578
Use 0.50 if P d > 15
Computed: Table 9-6
F <0.5 0.50
0.513
Table 9-1
Neither gear straddle mounted
4 Stress cycle factor-Pitting Figure 10-20 <10 Cyc. Enter Stress cycle factor-Pitting, C L = 1.016 2.000
Enter Stress cycle factor-Bending, K L =
Stress cycle factor-Bending
Gear - Number of load cycles: N G =
Enter: Design Life: Pinion - Number of load cycles: N P =
1.00
1.00
Reliability Factor: K R =
1.241
Service Factor: SF =
0.578
Dynamic Factor: K v =
Fig. 10-18-Gear Size Factor: C s =
0.513
2.00
Overload Factor: K o = Figure 10-13-Pinion Size Factor: K s =
1.005
1.255
1.105
Enter K m =
Neither gear straddle mounted:
One gear straddle mounted:
Both gears straddle mounted:
Heavy-duty conveyor with moderate shock driven by a gasoline engine Problem 10-16 Both gears straddle mounted Factors in Design Analysis: Load distribution factor, K m: From Figure 10-14 and Equation 10-16
F
Problem 10‐16 – Bevel gearing
295 75.17 degrees 1.759 in
nG = NG = nG = mG = DP = DG = = = Ao = vt =
ed Output Speed: mber of gear teeth: No. of Gear Teeth: Computed data: al Output Speed: Gear Ratio: Diameter - Pinion: h Diameter - Gear: cone angle - Gear: ter cone distance: Pitch Line Speed: : stress analysis: Max
A v = 11 J P = 0.252
: Quality Number: ry Factor - Pinion:
68 lb 6.3 lb 24 lb
ear - Radial load: W rG = Gear - Axial load: W xG =
99.167 lb-in
TG = - Tangential load: W tG =
ar Shaft - Torque:
24 lb 6.3 lb
ion - Radial load: W rP = nion - Axial load: W xP =
0.386 in 20 degrees 68 lb [Eq. 10-10]
rm =
r: Pressure angle: = - Tangential load: W tP =
n radius of pinion:
etry Factor - Gear: J G = 0.208 Fig. 10-15 Geometry Factor: I = 0.086 Fig. 10-19 orque for shaft and bearing load analysis: on Shaft - Torque: T P = 26.25 lb-in
Fig. 10-15
Table 9-5
0.500 in Cp = 2300 Table 9-7
0.500
Elastic Coefficient:
Max 0.586
Nom
58 lb
424 ft/min
14.83 degrees
476.5 rpm
68
68.2
475 rpm
18
th Guidelines (in): 0.528 nter: Face Width: F =
condary Input Data:
Wt =
3.400 in
NP =
er of Pinion Teeth:
ne angle - Pinion:
3.78 0.900 in
Pd =
Diametral Pitch: 20
0.75 hp 1800 rpm
P= nP =
Input Power:
APPLICATION:
Input Speed:
nitial Input Data:
EL GEARS
Use 1.00 if no unusual conditions
0.930
2.700
119,148 psi
9,196 psi
11,142 psi
Figure 9-21
Figure 9-17
Figure 9-17
0.971
>104
0.491
Gear: HB 280 required: SAE 1340 OQT 1200; HB = 293
0.930
>103but<3x106 >3x106 cyc.
number of cycles.
Gear: Required s ac = 119,148 psi Figure 9-21 Specify materials, alloy and heat treatment, for most severe requirement. One possible material specification: Pinion: HB 280 required: SAE 1340 OQT 1200; HB = 293
Pinion: Required s ac =
Stress Analysis: Pitting
Gear: Required s at =
Stress Analysis: Bending Pinion: Required s at =
See Table 9-12 Input value from pertinent Figure 10-16 <103 Cyc.
4.3E+08
F >3.15 0.83
Table 9-11; ;Use 1.00 for R = .99 15000 hours 1.6E+09
0.5<F<3.15 0.500
Use 0.50 if P d > 15
Computed: Table 9-6
F <0.5 0.50
0.497
Table 9-1
Neither gear straddle mounted
4 Stress cycle factor-Pitting Figure 10-20 <10 Cyc. Enter Stress cycle factor-Pitting, C L = 0.971 2.000
Enter Stress cycle factor-Bending, K L =
Stress cycle factor-Bending
Gear - Number of load cycles: N G =
Enter: Design Life: Pinion - Number of load cycles: N P =
1.00
1.00
Reliability Factor: K R =
1.277
Service Factor: SF =
0.500
Dynamic Factor: K v =
Fig. 10-18-Gear Size Factor: C s =
0.500
1.75
Overload Factor: K o = Figure 10-13-Pinion Size Factor: K s =
1.001
1.251
1.101
1.001
Enter K m =
Neither gear straddle mounted:
One gear straddle mounted:
Both gears straddle mounted:
Reciprocating saw driven by an electric motor Problem 10-17 Both gears straddle mounted Factors in Design Analysis: Load distribution factor, K m: From Figure 10-14 and Equation 10-16
F
Problem 10‐17 – Bevel gearing
WORMGEARING 18.
[Shown here are data from the solution to Problem 8‐52 that has the same
requirements]
924 lb ∙ in,
30 rpm ∙
Forces:
462 lb
.
/
/
Pitch Line Speed of Gear Sliding Velocity
/
From Figure 10‐25;
31.4/
4.57
.
.
.
.
.
.
.
.
.
.
.
Input Power Efficiency
. .
Friction Power Loss
100%
0.44
. .
Pitting Because
0.185
0.625 hp
70.4% 30 40
Input Speed Stress
15.5 lb (Equation 10‐32)
0.185 hp
0.185 /
27
120 lb (Equation 10‐31)
. .
Friction Force
394 ft/min
53 lb (Equation 10‐30)
.
. .
31.4 ft/min
0.0323 Computed from Equation 10 .
462
4.00 30 /12
.
1000 4.00
.
. .
1200 rpm 24 235 psi
0.625 0.814 0.427
659 lb
– Ok for pitting
(The spreadsheet solution to Problem 10‐18 is on the following page.)
296
1 40 40
NG = NG =
n =
Required No. of gear teeth: Specify No. of gear teeth: Normal pressure angle:
297 L= =
[Used given face width]
0.625 in
Effective gear face width:
Fe =
Sliding velocity:
0.735 in 0.8375 in
Nominal gear face width:
F eG =
Enter: C m = 4.200 in
D tG =
Gear Ratio: m G =
1000
<2.5 in
6 to 20
0.427
0.427
<700
0.814
#NUM!
1143 1
>25 in <
0.439 0.642
700-3000 >3000
Actual v s = 39
>76 Actual m G = 40
0.814 0.885
20 to 76
<8 in
1137 1000
>8 in
1000 Sand Cast
903
>2.5 in
NOTE: For the Ratio correction factor, if #NUM! appears, the argument of equation for C m is negative resulting in an invelid result.
*
659 lb 462 lb Satisfactory Suggest using larger face width; Approx. 0.75 in, to reduce bending stress
Rated tangential load: W tR = Must be > W t =
Enter: C v =
Velocity factor: C v =
Ratio correction factor: C m =
Gear throat diameter:
*
1.789 in
Enter: Materials factor: C s =
Centrifugally cast: C s =
Chill cast or forged: C s =
DG → Sand cast: C s =
Surface Durability: [Hardened steel worm; bronze gear] Type of bronze ↓
D RW =
Max effective gear face width: 0.67*D W =
0.100 --------> 0.100 0.125 0.150 0 0.313 in 0.974 [Stress slightly high]
14.5 20 25 Lewis form factor, y
Normal pressure angle
Bending stress on gear: 24223 psi [Using effective gear face w Allowable stresses-Bronze: Manganese = 17000 psi; Phosphor = 24000 p
Normal circular pitch: Dynamic factor: K v =
Bending Stress on Gear: Enter: Lewis form factor: y =
Stresses:
70.3 %
0.626 hp
0.440 hp
Power input: Efficiency:
F Wnom =
1.019 in
Worm 53 120 462
15.6 lb
Gear 462 120 53
0.186 hp
Worm root diameter:
1.450 in
D oW =
Worm outside diameter:
394 ft/min 0.032 If v s > 10 ft/min
Power loss - friction:
Power output from gear:
Power:
Forces: (lb) Tangential: Radial: Axial: Friction force, W f =
Coefficient of friction:
Sliding velocity v s =
Additional Computed Results: Pitch line speed - Gear: 31.42 ft/min
Nominal worm face length:
0.116 in
0.100 in
a= b=
Dedendum:
4.574 deg
0.314 in
0.314 in
Addendum:
Lead angle:
p xW =
Lead of the worm:
Should be >1.6 and <3.0 pG = 0.314 in
Axial pitch of worm:
Circular pitch of gear:
1.250 in DW = C= 2.625 in C 0.875 /D W = 1.86
4.000 in
DG =
Gear pitch diameter: Specify worm diameter: Actual center distance:
40
VR =
Actual velocity ratio:
Computed Results and Additional Inputs: nW = 1200 rpm Actual input speed:
14.5 degrees
10
NW =
30 rpm 40
No. of worm threads:
924 lb-in
To =
Problem: 10-18
Output speed: nG = Velocity Ratio: VR = Design Decisions: Diametral pitch: Pd =
Desired output torque:
Input Data:
Wormgearing - Design
Spreadsheet solution to Problem 10-18. Small differences from the solution on the preceding page are due to rounding differences and reading of data from graphs, rather than calculations.
Comments on the following Problems 19 and 20:
These problems have three distinct solutions for the three configurations described in the Problem 19 statement. Also, Problem 20 uses the same data and calls for calculating the efficiency. These results are integrated in the spreadsheet. Each configuration is presented in a separate spreadsheet called: o
19a and 20a
o
19b and 20b
o
19c and 20c
The three solutions are followed by a summary and comparison of the results.
You are advised to review the text coverage of Wormgearing in Chapters 8 and 10, evaluate each solution yourself, and compare your observations with the summary presented here. Furthermore, you are advised to work toward improving the performance of any particular version of the solutions by changing the given data and any design decisions that were used in the given solutions. As an example, following the summary page presented after Problems 19c and 20c is a modified version of problems 19a and 20a. o Note that the solution to the given Problem 19a calls for using a worm diameter, DW = 1.00 in. This resulted in a value for the center distance that is below the range defined in Equation 8‐27: 1.6 < C0.875/DW < 3.0 Thus the worm diameter is too large. o The modified solution reduces the worm diameter to DW = 0.700 in, bringing the center distance into the recommended range. This is an example of how the spreadsheet can be used to iterate on design decisions to work toward a desirable solution.
298
1 20 20
NG = NG =
n =
No. of gear teeth: No. of gear teeth: al pressure angle:
299 0.096 in 1.167 in
a= b=
Addendum: Dedendum:
F G2 = Fe =
ce width-Eq. 8-30: d gear face width:
0.500 in
0.500 in
0.601 in
1.833 in
1.054 in
0.807 in
Design decision
D tG = F G1 =
ce width-Eq. 8-28:
F Wnom =
worm face length: ar throat diameter:
D oW = D RW =
outside diameter: orm root diameter:
0.083 in
4.764 deg
0.262 in
L= =
Lead angle:
Lead of the worm:
0.262 in
p xW =
xial pitch of worm:
DW = 1.000 in center distance: C= 1.333 in C 0.875 /D W = 1.29 LOW r worm diameter Should be >1.6 and <3.0 pG = cular pitch of gear: 0.262 in
D G = 1.66667 in
ar pitch diameter: fy worm diameter:
20
VR =
tual velocity ratio:
ted Results and Additional Inputs: nW = ctual input speed: 1800 rpm
14.5 degrees
12
NW =
90 rpm 20
of worm threads:
202 lb-in
To =
Problem: 10-19a and 20a
Output speed: nG = Velocity Ratio: VR = Design Decisions: Diametral pitch: Pd =
Trial output torque:
Input Data:
aring - Design ve procedure] 473 ft/min Worm 28 63 242
72.9 %
0.396 hp
0.107 hp
0.289 hp
7.5 lb
Gear 242 63 28
14.5 20 25 30 Lewis form factor, y
Normal pressure angle, n
0.030 If v s > 10 ft/min - Eq. 10-27
1000
<2.5 in
<8 in
1311 1000
>8 in
1211 1000
>25 in <25 in
1000 Chilled cast phosphor bronze
1084
0.392
0.392
<700
0.819
0.820
0.395 0.557
700-3000 >3000
0.819 1.017 Actual v s = 473
242 lb Eq. 10-42 242 lb Adjusted output torque until limits reached on either bending or surface durab Using chilled phosphur bronze, surface durability controls this design.
Rated tangential load: W tR = Must be > W t =
Enter: C v =
Velocity factor: C v =
Sliding velocity:
Enter: C m =
Ratio correction factor: C m =
Gear Ratio: m G = 6 to 20 20 to 76 >76 Actual m G = 20
Enter: Materials factor: C s =
Centrifugally cast: C s =
Chill cast or forged: C s =
Sand cast: C s =
Type of bronze:/D G ---> >2.5 in
Surface Durability: [Hardened steel worm; bronze gear]
0.100 --------> 0.100 0.125 0.150 0.175 Normal circular pitch: 0.261 in Dynamic factor: K v = 0.968 Bending stress on gear: 19190 psi- Using specified gear face width Allowable stresses-Bronze: Manganese = 17000 psi; Phosphor = 24000 psi
Bending Stress on Gear: Enter: Lewis form factor: y =
Stresses:
Efficiency:
Power input:
Power loss - friction:
Power output from gear:
Power:
Forces: (lb) Tangential: Radial: Axial: Friction force, W f =
Coefficient of friction:
Sliding velocity v s =
Additional Computed Results: Pitch line speed - Gear: 39.27 ft/min
Problems 10‐19a and 10‐20a – Wormgearing
2 40 40
NG = NG =
n =
No. of worm threads: ed No. of gear teeth: ify No. of gear teeth: rmal pressure angle:
300 D G = 3.33333 in
Gear pitch diameter: ecify worm diameter:
L= = a=
Lead of the worm: Addendum:
0.807 in 1.491 in
D RW = F Wnom = D tG =
Worm root diameter: al worm face length: Gear throat diameter: inal gear face width:
0.096 in
84.1 %
14.5 20 25 30 Lewis form factor, y
Normal pressure angle, n
1000
<2.5 in
<8 in
1173 1000
>8 in
1157 1000
>25 in <25 in
1000 Chilled Cast - Phosphor bronze
940
Sliding velocity: 0.670 in 0.500 in
ctive gear face width:
0.390
0.390
<700
0.819
0.820
0.393 0.553
700-3000 >3000
0.819 1.017 Actual v s = 478
418 lb 290 lb Adjusted output torque until limits reached on either bending or surface durab Bending stress controls this design
Rated tangential load: W tR = Must be > W t =
Enter: C v =
Velocity factor: C v =
0.601 in
F eG =
Ratio correction factor: C m =
Gear Ratio: m G = 6 to 20 20 to 76 >76 Actual m G = 20
Enter: Materials factor: C s =
Centrifugally cast: C s =
Chill cast or forged: C s =
Sand cast: C s =
Type of bronze:/D G ---> >2.5 in
Surface Durability: [Hardened steel worm; bronze gear]
0.100 --------> 0.100 0.125 0.150 0.175 Normal circular pitch: 0.258 in Dynamic factor: K v = 0.939 Bending stress on gear: 23963 psi [Using effective gear face width Allowable stresses-Bronze: Manganese = 17000 psi; Phosphor = 24000 psi
Bending Stress on Gear: Enter: Lewis form factor: y =
Stresses:
Efficiency:
0.822 hp
0.691 hp 0.131 hp
ctive gear face width: 0.67*D W = Fe =
Worm 58 77 290
9.1 lb
Gear 290 77 58
Power input:
Enter: C m =
[Used given face width] e width is small; Could use F > 0.601 in
478 ft/min 0.030 If v s > 10 ft/min
Power loss - friction:
Power output from gear:
Power:
Forces: (lb) Tangential: Radial: Axial: Friction force, W f =
Coefficient of friction:
Sliding velocity v s =
Additional Computed Results: Pitch line speed - Gear: 78.54 ft/min
3.500 in
1.167 in
b= D oW =
Dedendum:
0.083 in
9.462 deg
0.524 in
0.262 in
rm outside diameter:
Lead angle:
p xW =
Axial pitch of worm:
1.000 in DW = ual center distance: C= 2.167 in OK C 0.875 /D W = 1.97 Should be >1.6 and <3.0 pG = 0.262 in Circular pitch of gear:
VR =
Actual velocity ratio:
20
puted Results and Additional Inputs: nW = Actual input speed: 1800 rpm
14.5 degrees
12
90 rpm 20
NW =
484 lb-in
To =
Problem: 10-19b and 20b
Output speed: nG = Velocity Ratio: VR = Design Decisions: Diametral pitch: Pd =
esired output torque:
Input Data:
aring - Design
Problems 10‐19b and 10‐20b – Wormgearing
4 80 80
NG = NG =
n =
No. of worm threads: ed No. of gear teeth: ify No. of gear teeth: rmal pressure angle:
301 D G = 6.66667 in
Gear pitch diameter: ecify worm diameter:
L= = a=
Lead of the worm: Addendum:
0.807 in 2.108 in
D RW = F Wnom = D tG =
Worm root diameter: al worm face length: Gear throat diameter: inal gear face width:
0.096 in
90.8 %
14.5 20 25 30 Lewis form factor, y
Normal pressure angle, n
1000
<2.5 in
<8 in
1036 1000
>8 in
1103 1000
>25 in <25 in
1000 Chilled Cast - Phosphor bronze
797
Sliding velocity: 0.670 in 0.500 in
ctive gear face width:
0.382
0.382
<700
0.819
0.820
0.384 0.537
700-3000 >3000
0.819 1.017 Actual v s = 497
714 lb 263 lb Adjusted output torque until limits reached on either bending or surface durab Bending stress controls this design
Rated tangential load: W tR = Must be > W t =
Enter: C v =
Velocity factor: C v =
0.601 in
F eG =
Ratio correction factor: C m =
Gear Ratio: m G = 6 to 20 20 to 76 >76 Actual m G = 20
Enter: Materials factor: C s =
Centrifugally cast: C s =
Chill cast or forged: C s =
Sand cast: C s =
Type of bronze:/D G ---> >2.5 in
Surface Durability: [Hardened steel worm; bronze gear]
0.100 --------> 0.100 0.125 0.150 0.175 Normal circular pitch: 0.248 in Dynamic factor: K v = 0.884 Bending stress on gear: 23987 psi [Using effective gear face width Allowable stresses-Bronze: Manganese = 17000 psi; Phosphor = 24000 psi
Bending Stress on Gear: Enter: Lewis form factor: y =
Stresses:
Efficiency:
1.381 hp
1.254 hp 0.127 hp
ctive gear face width: 0.67*D W = Fe =
Worm 97 73 263
8.4 lb
Gear 263 73 97
Power input:
Enter: C m =
[Used given face width] e width is small; Could use F > 0.601 in
497 ft/min 0.029 If v s > 10 ft/min
Power loss - friction:
Power output from gear:
Power:
Forces: (lb) Tangential: Radial: Axial: Friction force, W f =
Coefficient of friction:
Sliding velocity v s =
Additional Computed Results: Pitch line speed - Gear: 157.08 ft/min
6.833 in
1.167 in
b= D oW =
Dedendum:
0.083 in
18.435 deg
1.047 in
0.262 in
rm outside diameter:
Lead angle:
p xW =
Axial pitch of worm:
1.000 in DW = ual center distance: C= 3.833 in HIGH C 0.875 /D W = 3.24 er worm diameter Should be >1.6 and <3.0 pG = 0.262 in Circular pitch of gear:
VR =
Actual velocity ratio:
20
puted Results and Additional Inputs: nW = Actual input speed: 1800 rpm
14.5 degrees
12
90 rpm 20
NW =
878 lb-in
To =
Problem: 10-19c and 20c
Output speed: nG = Velocity Ratio: VR = Design Decisions: Diametral pitch: Pd =
esired output torque:
Input Data:
aring - Design
Problems 10‐19c and 10‐20c – Wormgearing
SUMMARY OF RESULTS OF THE THREE PARTS OF PROBLEMS 19 AND 20:
Problems 10-19 and 10-20 Summary Given data:
COMPARISON OF THREE PROPOSED DESIGNS See details on the preceding three spreadsheets
Diametral pitch, Pd = Velocity ratio, VR = Output speed (Gear) =
12 20 90
rpm
Worm pitch diameter, DW = Gear face width, F =
1.000 0.500
in in
See comment See comment
Normal pressure angle, n = 14.5 degrees Assumed gear is made from chilled cast phosphor bronze Allowable bending stress = 24,000 psi DESIGN Results: A B Number of threads in worm: 1 2
C 4
Output torque (lb-in), To = Output power (hp), Po =
202 0.289
484 0.691
878 1.254
Gear bending stress (psi) Allowable bending stress (psi) Rated load for surface durability (lb) Gear transmitted load (lb) Efficiency (%) [Problem 20] Power input (hp) Lead angle (degrees) Gear pitch diameter (in) Center distance (in)
19190 24000
23963 24000
23987 24000
242 242 72.9 0.396 4.76 1.667 1.333
418 290 84.1 0.822 9.46 3.333 2.167
714 263 90.8 1.381 18.4 6.667 3.833
Limits in Bold Limits in Bold
Comments on results: The given face width is small. Could use F > 0.601 in to maximize effective face width. Worm diameter is too large for Design A. See Equations 10-52 and 10-53 Worm diameter is too small for Design C. See Equations 10-52 and 10-53 Design A is limited by surface durability Designs B and C are limited by bending stress in gear teeth. As number of threads in worm increases: Lead angle increases Efficiency increases Torque and power capacity increase BUT: Gear size and center distance increase
The following page shows the results for a modified design for the data for Problems 10‐ 19a and 10‐20a. Note that the initial design for the worm diameter was 1.00 in. That was considered to be too large. The new worm diameter is 0.700 in, putting the center distance term in the proper range. Bending stress and surface durability remain acceptable.
302
1 20 20
NG = NG =
n =
No. of worm threads: equired No. of gear teeth: Specify No. of gear teeth: Normal pressure angle:
303 DG =
Gear pitch diameter: Specify worm diameter:
1.667 in
20
L= =
Lead of the worm:
332 ft/min
0.289 hp
76.4 %
0.378 hp
0.089 hp
1.833 in
D tG = F G1 = F G2 = Fe =
Gear throat diameter: Gear face width-Eq. 8-28: Gear face width-Eq. 8-30: Spedified gear face width:
Design decision
0.500 in
0.500 in
0.511 in
1.054 in
0.507 in
D RW = F Wnom =
Worm root diameter:
0.867 in
D oW =
Worm outside diameter: ominal worm face length:
0.096 in
0.083 in
a=
<8 in
1311 1000
>8 in
1211 1000
>25 in <25 in
1000 Chilled cast phosphor bronze
1000
<2.5 in
0.392
0.457
<700
0.819
0.820
0.484 0.733
700-3000 >3000
0.819 1.017 Actual v s = 332
242 lb Eq. 10-42 242 lb Adjusted output torque until limits reached on either bending or surface durab Using chilled phosphur bronze, surface durability controls this design.
Rated tangential load: W tR = Must be > W t =
Enter: C v =
Velocity factor: C v =
Sliding velocity:
Enter: C m =
Ratio correction factor: C m =
Gear Ratio: m G = 6 to 20 20 to 76 >76 Actual m G = 20
Enter: Materials factor: C s =
Centrifugally cast: C s =
Chill cast or forged: C s =
1084
Sand cast: C s =
6.789 deg
b=
14.5 20 25 30 Lewis form factor, y
Normal pressure angle, n
Surface Durability: [Hardened steel worm; bronze gear] Type of bronze:/D G ---> >2.5 in
Dedendum:
Worm 38 63 242
8.9 lb
Gear 242 63 38
0.035 If v s > 10 ft/min - Eq. 10-27
0.100 --------> 0.100 0.125 0.150 0.175 Normal circular pitch: 0.260 in Dynamic factor: K v = 0.968 Bending stress on gear: 19259 psi- Using specified gear face width Allowable stresses-Bronze: Manganese = 17000 psi; Phosphor = 24000 psi
Bending Stress on Gear: Enter: Lewis form factor: y =
Stresses:
Efficiency:
Power input:
Power loss - friction:
Power output from gear:
Power:
Forces: (lb) Tangential: Radial: Axial: Friction force, W f =
Coefficient of friction:
Sliding velocity v s =
Additional Computed Results: Pitch line speed - Gear: 39.27 ft/min
0.262 in
0.262 in
Addendum:
Lead angle:
p xW =
Axial pitch of worm:
DW = 0.700 in Actual center distance: C= 1.183 in OK C 0.875 /D W = 1.66 Should be >1.6 and <3.0 pG = 0.262 in Circular pitch of gear:
VR =
Actual velocity ratio:
Computed Results and Additional Inputs: nW = 1800 rpm Actual input speed:
14.5 degrees
12
90 rpm 20
NW =
202 lb-in
To =
Problem: 10-19a and 20a Modified-changed D w
Output speed: nG = Velocity Ratio: VR = Design Decisions: Diametral pitch: Pd =
Trial output torque:
Input Data:
ormgearing - Design [Iterative procedure]
Modified Problems 10‐19a and 10‐20a – New worm diameter, DW = 0.700 in.
80 rpm 7.5 5 2 15 15
NW = NG = NG =
n =
No. of worm threads: ed No. of gear teeth: ify No. of gear teeth: rmal pressure angle:
304 DG =
Gear pitch diameter: ecify worm diameter:
3 in
7.5
L= = a=
Lead of the worm: Addendum:
0.737 in 0.737 in
inal gear face width:
ctive gear face width: 0.67*D W = ctive gear face width:
Used Max effective face width
Fe =
1.020 in
D tG = F eG =
Gear throat diameter:
3.400 in
2.191 in
0.637 in
D RW = F Wnom =
al worm face length:
1.500 in
0.231 in
0.200 in
19.983 deg
1.257 in
0.628 in
Worm root diameter:
b= D oW =
Dedendum: rm outside diameter:
Lead angle:
p xW =
Axial pitch of worm:
1.1 in DW = ual center distance: C= 2.050 in OK. C 0.875 /D W = 1.70 Should be >1.6 and <3.0 pG = Circular pitch of gear: 0.628 in
VR =
Actual velocity ratio:
puted Results and Additional Inputs: nW = 600 rpm Actual input speed:
25 degrees
984 lb-in
esired output torque:
To =
Problem: 10-21 One possible solution
Output speed: nG = Velocity Ratio: VR = Design Decisions: Diametral pitch: Pd =
Input Data:
aring - Design
Worm 276 331 656
86.5 %
1.445 hp
0.195 hp
1.250 hp
35.0 lb
Gear 656 331 276
14.5 20 2 Lewis form fac
Normal pressure
0.045 If v s > 10 ft/min
184 ft/min
1000
962
1000
<2.5 in
<8 in
1194 1000
>8 in
116
>25
Rated tangential load: W tR = Must be > W t =
Enter: C v =
Velocity factor: C v =
Sliding velocity:
Enter: C m =
Ratio correction factor: C m =
0.794 1.099
0.678 1.158
700-3000 >3000
687 lb 656 lb
0.538
0.538
<700
0.719
0.719
Actual v s
Gear Ratio: m G = 6 to 20 20 to 76 >76 Actual m G
Enter: Materials factor: C s =
Centrifugally cast: C s =
Chill cast or forged: C s =
Sand cast: C s =
Type of bronze:/D G ---> >2.5 in
Surface Durability: [Hardened steel worm; bronze gear]
0.150 --------> 0.100 0.125 0.1 Normal circular pitch: 0.590 in Dynamic factor: K v = 0.950 Bending stress on gear: 10575 psi [Using effective gear fa Allowable stresses-Bronze: Manganese = 17000 psi; Phosphor = 24
Bending Stress on Gear: Enter: Lewis form factor: y =
Stresses:
Efficiency:
Power input:
Power loss - friction:
Power output from gear:
Power:
Forces: (lb) Tangential: Radial: Axial: Friction force, W f =
Coefficient of friction:
Sliding velocity v s =
Additional Computed Results: Pitch line speed - Gear: 62.83 ft/min
Spreadsheet solution to Problem 10-21.
6 18 18
NG = NG =
n =
ed No. of gear teeth: ify No. of gear teeth: rmal pressure angle:
305 DG =
Gear pitch diameter: ecify worm diameter:
1.5 in
3
L= =
Lead of the worm:
0.335 in 0.335 in
D tG = F eG =
Gear throat diameter: inal gear face width:
ctive gear face width: 0.67*D W = ctive gear face width:
Fe =
0.441 in
F Wnom =
al worm face length:
1.667 in
1.000 in
0.307 in
0.667 in
D oW = D RW =
0.096 in
0.083 in
Worm root diameter:
Dedendum:
45.000 deg
1.571 in
0.262 in
rm outside diameter:
a= b=
Addendum:
Lead angle:
p xW =
Axial pitch of worm:
0.5 in DW = ual center distance: C= 1.000 in OK. C 0.875 /D W = 2.00 Should be >1.6 and <3.0 pG = Circular pitch of gear: 0.262 in
VR =
Actual velocity ratio:
puted Results and Additional Inputs: nW = Actual input speed: 1800 rpm
25 degrees
12
NW =
600 rpm 3
No. of worm threads:
52.5 lb-in
To =
Problem: 10-22
Output speed: nG = Velocity Ratio: VR = Design Decisions: Diametral pitch: Pd =
esired output torque:
Input Data:
aring - Design
Worm 76 48 70
14.5 20 Lewis form fac
Normal pressure
1000
1106
1000
<2.5 in
<8 in
1331 1000
>8 in
122
>25
Rated tangential load: W tR = Must be > W t =
Enter: C v =
Velocity factor: C v =
Sliding velocity:
Enter: C m =
Ratio correction factor: C m =
0.779 1.129
0.483 0.731
700-3000 >3000
122 lb 70 lb
0.457
0.457
<700
0.578
0.578
Actual v s
Gear Ratio: m G = 6 to 20 20 to 76 >76 Actual m G
Enter: Materials factor: C s =
Centrifugally cast: C s =
Chill cast or forged: C s =
Sand cast: C s =
Type of bronze:/D G ---> >2.5 in
Surface Durability: [Hardened steel worm; bronze gear]
0.150 --------> 0.100 0.125 0.1 Normal circular pitch: 0.185 in Dynamic factor: K v = 0.836 Bending stress on gear: 9003 psi [Using effective gear fa Allowable stresses-Bronze: Manganese = 17000 psi; Phosphor = 24
Bending Stress on Gear: Enter: Lewis form factor: y =
Stresses:
92.6 %
0.540 hp
Efficiency:
0.040 hp
Power input:
0.500 hp
4.0 lb
Gear 70 48 76
0.035 If v s > 10 ft/min
333 ft/min
Power loss - friction:
Power output from gear:
Power:
Forces: (lb) Tangential: Radial: Axial: Friction force, W f =
Coefficient of friction:
Sliding velocity v s =
Additional Computed Results: Pitch line speed - Gear: 235.62 ft/min
Spreadsheet solution to Problem 10-22.
80 80
NG = NG =
n =
uired No. of gear teeth: ecify No. of gear teeth: Normal pressure angle:
10 in
DG =
Gear pitch diameter: Specify worm diameter:
306 L= =
Lead of the worm:
10.250 in 1.090 in 1.5075 in
D tG = F eG =
Gear throat diameter: ominal gear face width:
ffective gear face width: 0.67*D W =
1.090 in
3.162 in
F Wnom =
minal worm face length:
Fe =
1.961 in
ffective gear face width:
2.500 in
D oW = D RW =
0.145 in
0.125 in
Worm root diameter:
Dedendum:
6.340 deg
0.785 in
0.393 in
Worm outside diameter:
a= b=
Addendum:
Lead angle:
p xW =
Axial pitch of worm:
2.25 in DW = Actual center distance: C= 6.125 in C 0.875 /D W = 2.17 OK Should be >1.6 and <3.0 pG = 0.393 in Circular pitch of gear:
40
VR =
Actual velocity ratio:
mputed Results and Additional Inputs: nW = Actual input speed: 1800 rpm
14.5 degrees
8 2
NW =
45 rpm 40
No. of worm threads:
4200 lb-in
To =
Problem: 10-23
Output speed: nG = Velocity Ratio: VR = Design Decisions: Diametral pitch: Pd =
Desired output torque:
Input Data:
gearing - Design 1067 ft/min Worm 111 219 840
84.0 %
3.570 hp
0.570 hp
3.000 hp
17.6 lb
Gear 840 219 111
14.5 20 25 Lewis form factor
Normal pressure ang
0.020 If v s > 10 ft/min
1000
<2.5 in
956
>8 in
713 Sand Cast
713
1000
<8 in
1072
>25 in
0.814
0.248
0.204
<700
0.248 0.297
700-3000 >3000
0.814 0.885 Actual v s =
990 lb 840 lb NOTE: For the Ratio correction factor, if #NUM! appears, the argument equation for Cm is negative resulting in an invalid result.
Rated tangential load: W tR = Must be > W t =
Enter: C v =
Velocity factor: C v =
Sliding velocity:
Enter: C m =
Ratio correction factor: C m = #NUM!
Gear Ratio: m G = 6 to 20 20 to 76 >76 Actual m G =
Enter: Materials factor: C s =
Centrifugally cast: C s =
Chill cast or forged: C s =
Sand cast: C s =
Type of bronze:/D G ---> >2.5 in
Surface Durability: [Hardened steel worm; bronze gear]
0.100 --------> 0.100 0.125 0.150 Normal circular pitch: 0.390 in Dynamic factor: K v = 0.911 Bending stress on gear: 21689 psi [Using effective gear face Allowable stresses-Bronze: Manganese = 17000 psi; Phosphor = 2400
Bending Stress on Gear: Enter: Lewis form factor: y =
Stresses:
Efficiency:
Power input:
Power loss - friction:
Power output from gear:
Power:
Forces: (lb) Tangential: Radial: Axial: Friction force, W f =
Coefficient of friction:
Sliding velocity v s =
Additional Computed Results: Pitch line speed - Gear: 117.81 ft/min
Spreadsheet solution to Problem 10-23.
30 30
NG = NG =
n =
quired No. of gear teeth: pecify No. of gear teeth: Normal pressure angle:
DG =
Gear pitch diameter: Specify worm diameter:
5 in
30
307 1.614 in 2.582 in 5.333 in 1.202 in 1.340 in 1.000 in
F Wnom = D tG = F eG =
minal worm face length: Gear throat diameter: ominal gear face width:
ffective gear face width: 0.67*D W = ffective gear face width:
[Used given face width] face width is small; Could use F > 1.202
Fe =
2.333 in
D oW = D RW =
Worm root diameter:
0.193 in
0.167 in
4.764 deg
0.524 in
0.524 in
Worm outside diameter:
a= b=
Lead angle: Addendum:
L= =
Lead of the worm:
Dedendum:
p xW =
Axial pitch of worm:
2.000 in DW = Actual center distance: C= 3.500 in LOW C 0.875 /D W = 1.50 maller worm diameter Should be >1.6 and <3.0 pG = 0.524 in Circular pitch of gear:
VR =
Actual velocity ratio:
omputed Results and Additional Inputs: nW = 600 rpm Actual input speed:
14.5 degrees
6 1
NW =
20 rpm 30
No. of worm threads:
1200 lb-in
To =
Problem: 10-24A
Output speed: nG = Velocity Ratio: VR = Design Decisions: Diametral pitch: Pd =
Desired output torque:
Input Data:
gearing - Design 315 ft/min Worm 58 125 480
69.1 %
0.552 hp
0.171 hp
0.381 hp
17.9 lb
Gear 480 125 58
14.5 20 25 Lewis form factor
Normal pressure ang
0.036 If v s > 10 ft/min
1000
<2.5 in
<8 in
1093 1000
>8 in
857 Sand cast
857
1126
>25 in
0.466
0.466
<700
0.824
0.759
0.498 0.763
700-3000 >3000
0.824 0.951 Actual v s =
1193 lb OK For sand cast bro 480 lb Can use Manganese bronze based on bending stress in gear. Worm diameter is too large.
Rated tangential load: W tR = Must be > W t =
Enter: C v =
Velocity factor: C v =
Sliding velocity:
Enter: C m =
Ratio correction factor: C m =
Gear Ratio: m G = 6 to 20 20 to 76 >76 Actual m G =
Enter: Materials factor: C s =
Centrifugally cast: C s =
Chill cast or forged: C s =
Sand cast: C s =
Type of bronze:/D G ---> >2.5 in
Surface Durability: [Hardened steel worm; bronze gear]
0.100 --------> 0.100 0.125 0.150 Normal circular pitch: 0.522 in Dynamic factor: K v = 0.979 Bending stress on gear: 9400 psi [Using effective gear face Allowable stresses-Bronze: Manganese = 17000 psi; Phosphor = 2400
Bending Stress on Gear: Enter: Lewis form factor: y =
Stresses:
Efficiency:
Power input:
Power loss - friction:
Power output from gear:
Power:
Forces: (lb) Tangential: Radial: Axial: Friction force, W f =
Coefficient of friction:
Sliding velocity v s =
Additional Computed Results: Pitch line speed - Gear: 26.18 ft/min
Spreadsheet solution to Problem 10-24A.
Noting that the worm diameter, DW = 2.00 in, is too large, a modified solution follows with a smaller diameter, DW = 1.75 in. A successful design is achieved.
30 30
NG = NG =
n =
quired No. of gear teeth: pecify No. of gear teeth: Normal pressure angle:
DG =
Gear pitch diameter: Specify worm diameter: 5 in
30
Problem 10‐24B follows.
308 L= = a=
Lead of the worm: Addendum:
0.524 in
1.173 in 1.130 in
ffective gear face width: 0.67*D W = ffective gear face width:
5.333 in
[Used nominal face width]
Fe =
1.130 in
D tG = F eG =
Gear throat diameter: ominal gear face width:
2.582 in
1.364 in
D RW = F Wnom =
Worm root diameter:
2.083 in
0.193 in
0.167 in
5.440 deg
0.524 in
minal worm face length:
b= D oW =
Dedendum: Worm outside diameter:
Lead angle:
p xW =
Axial pitch of worm:
DW = 1.750 in Actual center distance: C= 3.375 in C 0.875 /D W = 1.66 OK Should be >1.6 and <3.0 pG = 0.524 in Circular pitch of gear:
VR =
Actual velocity ratio:
omputed Results and Additional Inputs: nW = Actual input speed: 600 rpm
14.5 degrees
6 1
NW =
20 rpm 30
No. of worm threads:
1200 lb-in
To =
Problem: 10-24A - Revised Larger worm diameter
Output speed: nG = Velocity Ratio: VR = Design Decisions: Diametral pitch: Pd =
Desired output torque:
Input Data:
gearing - Design 276 ft/min Worm 65 125 480
14.5 20 25 Lewis form factor
Normal pressure ang
1000
<2.5 in
<8 in
1093 1000
>8 in
857 Sand cast
857
1126
>25 in
0.486
0.486
<700
0.824
0.759
0.537 0.845
700-3000 >3000
0.824 0.951 Actual v s =
1406 lb OK For sand cast bro 480 lb Can use Manganese bronze based on bending stress in gear. Changed worm diameter to 1.75 in; Changed face width to 1.13 in
Rated tangential load: W tR = Must be > W t =
Enter: C v =
Velocity factor: C v =
Sliding velocity:
Enter: C m =
Ratio correction factor: C m =
Gear Ratio: m G = 6 to 20 20 to 76 >76 Actual m G =
Enter: Materials factor: C s =
Centrifugally cast: C s =
Chill cast or forged: C s =
Sand cast: C s =
Type of bronze:/D G ---> >2.5 in
Surface Durability: [Hardened steel worm; bronze gear]
0.100 --------> 0.100 0.125 0.150 Normal circular pitch: 0.521 in Dynamic factor: K v = 0.979 Bending stress on gear: 8324 psi [Using effective gear face Allowable stresses-Bronze: Manganese = 17000 psi; Phosphor = 2400
Bending Stress on Gear: Enter: Lewis form factor: y =
Stresses:
70.6 %
0.540 hp
Power input: Efficiency:
0.159 hp
0.381 hp
19.0 lb
Gear 480 125 65
0.038 If v s > 10 ft/min
Power loss - friction:
Power output from gear:
Power:
Forces: (lb) Tangential: Radial: Axial: Friction force, W f =
Coefficient of friction:
Sliding velocity v s =
Additional Computed Results: Pitch line speed - Gear: 26.18 ft/min
Spreadsheet solution to Problem 10-24A – Modified: DW = 1.75 in.
2 60 60
NG = NG =
n =
No. of worm threads: uired No. of gear teeth: ecify No. of gear teeth: Normal pressure angle:
DG =
Gear pitch diameter: Specify worm diameter: 6 in
30
309 L= =
Lead of the worm:
1.019 in 2.191 in 6.200 in 0.735 in 0.838 in 0.625 in
D RW = F Wnom = D tG = F eG =
Worm root diameter: minal worm face length: Gear throat diameter: ominal gear face width:
ffective gear face width: 0.67*D W = ffective gear face width:
[Used given face width] face width is small; Could use F > 0.735
Fe =
1.450 in
D oW =
Worm outside diameter:
0.100 in 0.116 in
a= b=
Addendum:
9.090 deg
0.628 in
0.314 in
Dedendum:
Lead angle:
p xW =
Axial pitch of worm:
DW = 1.250 in Actual center distance: C= 3.625 in OK C 0.875 /D W = 2.47 Should be >1.6 and <3.0 pG = 0.314 in Circular pitch of gear:
VR =
Actual velocity ratio:
mputed Results and Additional Inputs: nW = 600 rpm Actual input speed:
14.5 degrees
10
20 rpm 30
NW =
1200 lb-in
To =
Problem: 10-24B
Output speed: nG = Velocity Ratio: VR = Design Decisions: Diametral pitch: Pd =
Desired output torque:
Input Data:
gearing - Design 199 ft/min Worm 82 106 400
77.6 %
0.491 hp
0.110 hp
0.381 hp
18.3 lb
Gear 400 106 82
14.5 20 25 Lewis form factor
Normal pressure ang
0.043 If v s > 10 ft/min
1000
<2.5 in
<8 in
1057 1000
>8 in
819 Sand cast
819
1111
>25 in
0.530
0.530
<700
0.824
0.759
0.648 1.090
700-3000 >3000
0.824 0.951 Actual v s =
937 lb OK For sand cast br 400 lb Can use Phosphor bronze based on bending stress in gear.
Rated tangential load: W tR = Must be > W t =
Enter: C v =
Velocity factor: C v =
Sliding velocity:
Enter: C m =
Ratio correction factor: C m =
Gear Ratio: m G = 6 to 20 20 to 76 >76 Actual m G =
Enter: Materials factor: C s =
Centrifugally cast: C s =
Chill cast or forged: C s =
Sand cast: C s =
Type of bronze:/D G ---> >2.5 in
Surface Durability: [Hardened steel worm; bronze gear]
0.100 --------> 0.100 0.125 0.150 Normal circular pitch: 0.310 in Dynamic factor: K v = 0.974 Bending stress on gear: 21171 psi [Using effective gear face Allowable stresses-Bronze: Manganese = 17000 psi; Phosphor = 2400
Bending Stress on Gear: Enter: Lewis form factor: y =
Stresses:
Efficiency:
Power input:
Power loss - friction:
Power output from gear:
Power:
Forces: (lb) Tangential: Radial: Axial: Friction force, W f =
Coefficient of friction:
Sliding velocity v s =
Additional Computed Results: Pitch line speed - Gear: 31.42 ft/min
Spreadsheet solution to Problem 10-24B.
COMPARISON OF THE RESULTS FOR PROBLEMS 10‐24A, 10‐24A (REV.), AND 10‐24B
Wormgearing Comparisons for:
10‐24A
Problems 10‐24A (Rev)
Common data: Output Torque Output speed Pressure angle Input speed
1200 lb‐in 20 rpm 14.5 deg. 600 rpm
1200 lb‐in 20 rpm 14.5 deg. 600 rpm
1200 lb‐in 20 rpm 14.5 deg. 600 rpm
6 1 30 2.00 in 5.00 in 3.5 in 1.00 in
6 1 30 1.75 in 5.00 in 3.375 in 1.13 in
10 2 60 1.25 in 6.00 in 3.625 in 0.625 in
Forces (on gear): Tangential Radial Axial
480 lb 125 lb 58 lb
480 lb 125 lb 65 lb
400 lb 106 lb 82 lb
Efficiency:
69.10%
70.60%
77.60%
Bending Stress Rated tangential load
9400 psi 1193 lb
8324 psi 1406 lb
21171 psi 937 lb
10‐24B
Variable data: Diametral pitch, P d No. of worm threads No. of gear teeth Worm diameter Gear diameter Center distance Gear face width Computed results:
Observations: Bending stress for Design 24B are much higher Design 24B has highest efficiency Forces are nearly equal for all three designs All rated tangential loads are well above the actual loads
310
Too large
OK
CHAPTER 11 KEYS, COUPLINGS, AND SEALS
General Notes for key design problems 1‐15: For SAE 1018 CD steel:
54 000 psi. If key material is weakest –
.
.
9000 psi 18 000 psi
1.
2.00 in; Use in square key, SAE 1018 CD. ∙ /
But hub length
2.
0.5 77 /3
77 000 psi
.
= 3.27 in; Use L = 3.75 in to more nearly match hub length.
.
3.60 in; Use in square key; SAE 1018 CD
.
3.
4.667 in
.
4.00 in. Use SAE 1045 CD;
0.5 / .
.
.
1.48 in Use L = 3.50 in to more nearly match hub length.
1.75 in; Use in square key; SAE 1018 CD; For hub; Class 20 CI,
20 000 psi
For bearing
6667 psi Conservative
9000 psi
Because compressive strength of CI is much greater than tensile strength. Shear of Key:
.
.
311
.
Bearing on hub:
.
.
.
Use L = 1.50 in to more nearly match hub length. 4.
63 000 110 /1700
.
Use 5.
.
2.50 in; Use square key.
0.580 in
.
to more nearly match hub length.
Express data from Table 11‐6 as Required a)
/
b)
2.00 in,
21 000 lb ∙ in;
.
3.60 in;
;
;
1112 lb ∙ in;
1.75 in;
4076 lb ∙ in;
2.50 in;
At 220 hp:
63 000 220 /1700
8153 lb ∙ in
At 110 hp:
63 000 110 /1700
4076 lb ∙ in 0.75
;
Pin should not yield at
;
Section 2
0.5
For a given pin d and Shaft D in Equation 11‐18: 0.5
;
4.00 in
1.75 in
3.25 in
; Use either 4 (K = 219) or 6 splines (K = 208)
.
Pin should shear at
4.00 in
; Use 6 splines; K = 208
.
Problem 4 data: .
Fit
Too high for any spline in Table 11‐6.
.
Problem 3 data: .
d)
21 000 lb ∙ in;
Problem 2 data: .
c)
Hub length Use B
Problem 1 data: .
6.
4076 lb ∙ in;
0.75
312
2
Ratio
.
0.75
.
Most cold drawn steels have
54 ksi;
For SAE 1018 CD;
.
7.
.
.
Minimum
At
.
:
.
64 ksi 0.294 in
.
4076 lb ∙ in; τ
.
.
From example Problem 12‐3:
4168 lb ∙ in
2 in; Sprocket:
in square key; SAE 1018 CD
.
Wormgear:
.
8.
1 in
0.823 in
.
.
Square key 1.647 in
.
.
Woodruff key 204: Nominal
: Nominal
Actual dimensions in Table 11‐3. 9.
Woodruff key 1628: Nominal
: Nominal
10, 11, 12 Are Drawings 13.
/2;
2 / ; No. 1210 Woodruff key
From Table 11‐3: Key
in
; .
.
.
∙
313
0.75
.
14.
.
No. 406 Woodruff key: 15.
No. 2428 key: W
19.
a
139
139 1.50
b
326
326 3.50
c
688
688 2.500
0.75 in;
.
∙ in .
.
∙
.
/ ∙ ∙
∙
Problem 16 – 4 Splines
/
Problem 17 – 10 Splines
/
Problem 18 – 16 Splines
NOTE: Problems 20‐46 call for narrative answers for which the proper information can be found in the text. Guidance is provided below for sections in which additional information can be found. 20.
Section 11‐6 includes discussion of applying set screws to transmit torque. A table of approximate holding force capacity vs. set screw size is provided.
21.
Press fit is described briefly in Section 11‐6. More discussion follows in Chapter 13.
22.
Section 11‐7 describes both rigid and flexible couplings and compares their performance. Examples of commercially available couplings are shown and described.
23.
Section 11‐8 contains general information about universal joints.
314
24.
Section 11‐9 contains general information about retaining rings and other means of locating machine elements axially on shafts and in other devices. Included are collars, shoulders, spacers, and locknuts.
25.
to 38. Section 11‐10 contains general information about seals.
39.
to 46. Section 11‐11 contains general information about seal materials, including elastomers.
40.
to 45. A list of 14 elastomers in included Section 11‐11. Following the list of 14 elastomers, their general performance capabilities are described.
46.
The required conditions for shafts on which elastomeric seals operate are discussed in the last part of Section 11‐11. Examples are: Steels, hardened to HRC 30 with tolerances of less than 0.005 in 0.13 mm are typically used for shafts on which seals operate. The surface must be free of burrs with a surface finish of 10 to 20 microinches is recommended. Lubrication is recommended.
315
CHAPTER 12 SHAFT DESIGN GENERAL NOTES CONCERNING SOLUTIONS TO SHAFT DESIGN PROBLEMS Design values for stress concentrations as given in Section 12‐4 are used for the initial calculations. These values must be checked once final design details are specified for diameters, fillet radii, and other features. Estimates are originally used for the size factors used in calculations because they depend on the shaft sizes that are unknown at the start of a design problem. These values must be checked once final design decisions have been specified. The choice of the reliability factor, CR, is a design decision. Other values may be preferred. In most cases, the proposed final values for diameters are expected to be safe because trial values are typically conservative and because final specified diameters are typically made to the next larger preferred size according to Appendix Table A‐2. Final specifications for diameters where bearings are to be mounted must await the selection of suitable bearings that can accommodate the radial and thrust loads applied to them. This process is described in Chapter 14. Use of commercially available software is an excellent tool for making those decisions. The computed ‘minimum required diameter’ from the shaft design process should be used as to limit the bearing selection to only feasible sizes. Shafts with Only Radial Loads Applied to Them: Problems P1 through P30 relate to one of the Figures P12‐1 through P12‐17 showing shafts carrying a variety of combinations of gears, belt sheaves, chain sprockets, and a few other items such as a flywheel and a propeller‐type fan. All of these elements apply only radial loads to the shafts on which they are mounted. Problems 22‐30 are comprehensive design problems that use the same shaft assemblies that are used for Problems 1‐21. The solutions to these problems include the analyses of forces and torques and should be used as the solutions for problems 1‐21. Problems 1‐11 include only force and torques exerted by gears on shafts. No separate solutions for these problems are included here. Problems 12‐21 include only forces and torques exerted by belt drives and chain drives on shafts. No separate solutions are included here.
316
The parts of the solutions for torques and forces give discrete, single‐answers. The remaining parts of the comprehensive shaft designs include many design decisions and multiple solutions are possible. The given solutions should be considered examples only. There are multiple ways in which the problems P1 through P30 may be assigned. The following table may help instructors decide how to assign the problems for student solution and may help students comprehend how the sets of problems lead to the more general shaft designs. Any combination of problems may be chosen. Torques and Forces Acting Radial to Shaft
Comprehensive
Figure P12‐1: P1 – Gear B; P14 – Sheave D
P22
Figure P12‐2: P2 – Gear C; P12 – Sprocket D; P13 – Pulley A
P23
Figure P12‐3: P3 – Gear B; P15 – Sprocket C; P16 – Sheaves D, E
P24
Figure P12‐4: P4 – Gear A; P19 – Sprockets C, D
P25
Figure P12‐5: P5 – Gear D; P20 – Sheave A; P21 – Sprocket E
P26
Figure P12‐6: P6 – Gear E; (No separate analysis of Sheave A)
P27 (Includes Sheave A)
Figure P12‐7: P7 – Gear C; P8 – Gear A
P28 (Includes Sheaves D, E)
Figure P12‐9: P9 – Gear C; P10 – Gear D; P11 – Gear F
P29 (Includes Sheave B)
Figure P12‐17: P17‐ Sheave C; P18 – Pulley D
P30 (Includes Fan A)
Shafts with both Radial and Axial Loads Applied to Them: Problems P31 to P34 deal with shafts carrying helical gears and wormgears that produce forces directed axially in addition to radial forces. Solutions are only shown for the comprehensive problems (12‐32 and 12‐34) in which the details of the analyses of torques and forces are included. Torques and Forces Acting Radial and Axial to Shaft
Comprehensive
Figure P12‐31: P31 – Helical Gear B
P32
Figure P12‐33: P33 – Wormgear C
P34 (Includes Sheave A)
Other Comprehensive Design Problems Problems 35 to 41 contain a variety of loading situations for which the general solution procedure must be adapted. Some of the problems involve more than one shaft, considering shafts for mating gears and multiple reductions.
317
Figure P12‐35: P35 – Double reduction helical drive Figure P10‐8 in Chapter 10: P36 – Bevel gear drive Figure P12‐37: P37 – Bevel gear drive with two chain sprockets Figure P12‐38: P38 – Double reduction spur gear drive; design three shafts. Figure P12‐39: P39 – Drive system consisting of an electric motor, a V‐belt drive, a double reduction spur gear type reducer, and a chain drive. Figure P12‐40: P40 – Shaft with three spur gears Figure P12‐41: P41 – Shaft for windshield wiper mechanism with two levers 22.
Figure P12‐1 Torque on gear B:
63 000 30 /550
Same torque on sheave D: Torque in shaft:
3436 lb ∙ in
3436 lb ∙ in
0;
3436 lb ∙ in
Forces on gear B: ∙
430 lb ←
20°
156 lb ↓
F
Forces on sheave D: ∙ .
687 lb 1.5 687
1.5 40°
790
→
40°
663
↓
318
1031 lb
Bending Moment Diagrams
Vertical plane: Horizontal plane:
V (lb)
V (lb)
M (lbin) M (lbin)
√4520
1210
4679 lb ∙ in
√4740
3980
6189 lb ∙ in
3436 lb ∙ in Assume
2.5 at left of
SAE 1040
:
80 ksi;
30 ksi
.5
Bearin g seat
71 ksi
11
0.81,
0.80 0.81 30 000
19 440 psi
Design geometry at C Use N = 3
/
.
23.
0.80
2.90 in → Specify D
3.00 in, C
0.78 Ok
Figure P12‐2 Torque on Pulley A:
63 000 10 /200
Torque on Gear C:
63 000 6 /200
Torque on Sprocket D:
3150 lb ∙ in 1890 lb ∙ in
63 000 4 /200
Torque Distribution in Shaft:
1260 lb ∙ in
3150 lb ∙ in;
319
1260 lb ∙ in;
0
Forces on Pulley A: 315 lb 2.0 315
2.0
630 lb ↓
;
0
Forces on Gear C: ∙
378 lb →
.
20°
Gear C drives Q
138 lb ↑
Forces on Sprocket D: 420 lb ↑ 0 Bending Moment Diagrams Vertical plane:
Horizontal plane:
V (lb)
V (lb)
M (lbin)
M (lbin)
3780 lb ∙ in √1588
3732
4056 lb ∙ in
√454
2272
2317 lb ∙ in
3150 lb ∙ in To left; 1260 lb ∙ in To right;
Ring groov e
3.0 Ring Groove 2.0 Key
320
Design at C
69 ksi;
SAE 1170 CD;
57 ksi
3.00 in
Design at C Specify 26 ksi;
0.78 Ok
0.85 0.81 26
17.9 ksi
/ .
2.75 in 1.06 2.75
Increase by 6% at groove 24.
2.92 in
Figure P12-3 63 000 5 /480
Torque on Gear B:
656 lb ∙ in
63 000 3 /480
Torque on Sheaves D and E:
63 000 11 /480
Torque on Sprocket C:
394 lb ∙ in
1444 lb ∙ in
Torque Distribution in Shaft: 0;
656 lb ∙ in
T (lbin)
788 lb ∙ in 394 lb ∙ in;
0 Torque diagram for shaft
Forces on Gear B: 437 lb →
.
20°
159 lb ↑
Forces on Sprocket C: 289 lb
.
15°
289
15°
75 lb ←
15°
289
15°
279 lb ↓
321
Forces on Sheave D: .
197 lb
1.5
1.5 197
;
296 lb ↑
0
Forces on Sheave E: 296 lb 30°
296
30°
256 lb → ;
296
30°
148 lb ↑
Vertical plane
Horizontal plane
V (lb)
V (lb)
M (lbin)
M (lbin) 0
Worst Case at D: 1254
√1014 788 lb ∙ in
1613 lb ∙ in &
SAE 1137 OQT 1300;
87 ksi;
0.9 0.81 33
24.1 ksi
;
3.0 Retaining Ring
60 ksi;
33 ksi Figure 5
8
0.90 Estimate /
.
Increase by 6%: Check:
0.81;
1.83 in 1.06 1.83
1.94 in; Specify
21 200 psi;
2.00 in
2.01 in; Acceptable to use
322
D
.
25.
Figure P12-4 Torque on Gear A:
63 000 40 /120
21 000 lb ∙ in
63 000 20 /120
Torque on Sprockets C & D: 21 000 lb ∙ in;
Torque in Shaft:
10 500 lb ∙ in
10 500 lb ∙ in;
0
Forces on Gear A: 2625 lb ↓ 20°
255 lb ←
Forces on Sprockets C & D: 1500 lb ← Vertical plane
Horizontal plane
V (lb)
V (lb)
M (lbin)
M (lbin)
At B: 21 000 lb ∙ in √7640 SAE 1020 CD: Use
0.75
21 000
22 347 lb ∙ in
51 000 psi;
61 000 psi;
Large Diameter;
22 000 psi Figure 5
0.81 0.75 22 000
323
13 400 psi
11
D DBR Small radius, Kt
To the right of B: ≅ 2.5
25
DBL
small fillet radius / .
Well-rounded radius; Kt 1.5
.
;
.
;
.
Ok Bearing seat
To the left of B: / .
. 26.
;
.
;
.
Ok
Figure P12-5 63 000 10 /240
Torque on Sheave A:
63 000 15 /240
Torque on Gear D:
2625 lb ∙ in 3938 lb ∙ in
63 000 5 /240
Torque on Sprocket E:
In shaft
1313 lb ∙ in
In shaft
Forces on Sheave A: 438 lb;
0
1.5
1.5 438
Forces on Gear D: 985 lb 20°
358
985
30° ←
358
985
30°
358
30° → 30°
324
182 lb ←
1032 lb ↑
657 lb ↓
Forces on Sprocket E: 438 lb 30°
379 lb →;
30°
Horizonta l plane
219 lb ↑ Vertical plane
V (lb)
V (lb)
M (lbin)
M (lbin)
C
At C: 2625 lb ∙ in: Use √4980
Kt 1.5
4 Shock
9708
10 911 lb ∙ in Kt 2.5
Choose 1137 CD; 0.8 0.81 37
82 000 psi,
98 ksi,
37 ksi Bearing seat
24.0 ksi /
.
. With
; 1.5;
3.59 in .
Ok
3.03 in;
.
325
;
.
Ok
27.
Figure P12-6 63 000 20 /310
Torque on Sheave A and Gear E:
4065 lb ∙ in
Forces on Sheave A: 370 lb 554 lb
1.5 20°
521 lb →
20° ↑
120 lb ↓
189 lb ↑
120 lb ↑
69 lb ↑
Forces on Gear E: 1355 lb ↑ 20° 45
493 lb →
1310 lb ↑
Vertical plane
Horizontal plane
V (lb)
V (lb)
M (lbin) M (lbin)
At C: Profile keyseat
4065 lb ∙ in 2662
9782
/
10 138 lb ∙ in
326
Light interference fit
2.0;
Use
3 84 ksi;
SAE 1050 CD: 0.8 0.81 38
100 ksi;
38 ksi
24.6 ksi /
.
Specify 28.
2.93 in
.
;
.
Ok
63 000 30 /480
Torque on Gear A:
63 000 50 /480
Torque on Gear C:
6563 lb ∙ in
63 000 10 /480
Torque on Sheaves D and E: 3938 lb ∙ in;
Torque in Shaft:
3938 lb ∙ in
1313 lb ∙ in
2626 lb ∙ in;
Figure P12-7 T (lbin)
Forces on Gear A:
Torque diagram
1575 lb ↑
.
20°
1575
20°
573 lb →
20°
478 lb ↑
Forces on Gear C: 1313 lb → 20°
1313
327
1313 lb ∙ in;
0
Forces on Sheave D: 438 lb 1.5
1.5 438
657 lb ↑:
20°
225 lb ↓
0
Forces on Sheave E: 657 lb 20°
617 lb →; Vertical plane
Horizonta l plane
V (lb)
V (lb)
M (lbin
M (lbin
At B: 3938 lb ∙ in;
3
2292
6300
SAE 3140 OQT 1000; 0.85 0.81 52
Wellrounded fillet, Kt 1.5
6704 lb ∙ in
133 ksi;
152 ksi;
35.8 ksi
328
52 ksi
Sharp fillet Kt 2.5
/ .
Specify
.
2.43 in
;
.
2.05 in with 29.
Ok
1.5;
.
;
.
Ok
Figure P12-9 63 000 2.5 /220
Torque on Sheave B:
716 lb ∙ in
63 000 5 /220
Torque in Shaft Gears C and F: Torque on Gear D:
63 000 12.5 /220
Torque in Shaft:
0;
1432 lb ∙ in
3580 lb ∙ in
716 lb ∙ in;
2148 lb ∙ in;
Forces on Sheave B: 239 lb 1.5 30°
1.5 239 179 lb ←;
358 lb 30°
310 lb ↓
Forces on Gear C: 477 lb ← 20°
477
20°
174 lb ↓
20°
217 lb ↑
Forces on Gear D: 597 lb ← 20°
597
Forces on Gear F: 477 lb
174 lb
329
1432 lb ∙ in
477
45° ←
477
45°
174 174
45° → 45°
214 lb ←
460 lb ↓
Horizonta l plane
Vertical plane
V (lb)
V (lb)
M (lbin
M (lbin
At C: 2148 lb ∙ in to right;
3.0;
716 lb ∙ in to left; 57 ksi;
SAE 1020 CD;
0.85 0.81 22
2.0;
3428 lb ∙ in 3428 lb ∙ in
61 ksi;
22 ksi 3.0
Retaining ring groove, Kt
C
Profile keyseat Kt 2.0
15.1 ksi /
.
2.40 in
/ .
Increase by 6%: Specify
.
2.75 in Critical 1.06 2.75
2.98 in
;
Ok
.
330
Well-rounded fillet Kt 1.5
Shaft at C where Gear C is mounted
30.
Figure P12-17 Torque at Fan:
63 000 12 /475
1592 lb ∙ in
Torque on Pulley D:
63 000 3.5 /475
Torque on Sheave C:
63 000 15.5 /475
Torque in Shaft:
1592 lb ∙ in;
464 lb ∙ in 2056 lb ∙ in
464 lb ∙ in;
0
Forces on A: 0;
34 lb ↓
The fan would also produce an axial thrust force but its effect on shaft diameter is small. Forces on C: (V-Belt sheave) 411 lb
.
1.5
1.5 411
617 lb ← ;
Forces on D: (Flat belt pulley) 155 lb 2.0
2.0 155
60°
155 lb →
60°
268 lb ↑
310 lb
Vertical plane
Horizontal plane
V (lb)
V (lb)
M (lbin)
M (lbin)
331
0
At C: 1592 lb ∙ in;
464 lb ∙ in
688
√3340
0.85 0.81 38
100 ksi;
38 ksi
26.2 ksi
2.0 keyseat
At C & to left:
/ .
2.00 in 3.0
At right of C:
Ring groove /
.
Increase by 6%: Specify 31.
.
2.29 Critical 1.06 2.29
2.42 in
;
Ok
.
Figure P12-31 And
Torque on Gear B and in Shaft from B to Coupling: 63 000 7.5 /650
32.
Use SAE 1020 CD: 0.9 0.81 22
C
Retaining ring groove, Kt 3.0
3410 lb ∙ in
90 ksi;
SAE 1144 CD:
Profile keyseat; Kt 2.0
727 lb ∙ in 51 ksi;
61 ksi;
22 ksi
16.0 ksi
Vertical plane
Horizontal plane
V (lb)
V (lb)
M (lbin)
M (lbin) 332
Wellrounded Fillet, Kt 1.5
At B: 275
421
503 lb ∙ in; Use
3;
2.0 Keyseat
/ .
1.25 in;
33.
Figure P12-33: Torque at A:
63 000 7.5 /1750
And
Forces on Sheave at A:
1.5
270 lb ∙ in
.
. .
Vertical plane
34.
0.9 Ok
162 lb ↓
Horizontal plane
V (lb)
V (lb)
M (lbin)
0 M (lbin)
MC = (1131)2 + (398)2 = 1199 lbin
Assume Torque is Negligible, Use N = 4 . .
4357 psi
/
0.83 0.81
0.67
4 4357
Required
17428 psi
Required
/0.67
26 000 psi
From Figure 5
11;
68 ksi
Specify SAE 1040 CD; 30 ksi;
0.67 30
80 ksi;
0.67
70 ksi
20.1 ksi
333
and .
Check using Equation 12-24 with
/ .
. .
This is less than actual 35.
Ok
Figure P12-35 Torques on Gears and in Shafts: 63 000 5 hp /1800 rpm
Gear P, Shaft 1:
63 000 5 hp /900 rpm
Gears B & C, Shaft 2:
63 000 5 hp /300 rpm
Gear Q, Shaft 3:
14 °; Ψ
Shaft 2: Given:
175 lb ∙ in 350 lb ∙ in
1050 lb ∙ in
45° (See Chapter 10)
Gear B: .
233 lb → 233
(Equation 10-2)
45°
233 lb
20.1° or
0.366 (Equation 10-1)
233 lb 0.366
85 lb
Gear C: .
350 lb ← 350 lb
(Equation 10-9)
45°
350 lb
350 lb 0.366
128 lb
334
(Equation 10-8)
114 ksi;
For SAE 4140 OQT 1200; 0.9 0.81 46
130 ksi;
46 ksi Figure 5
11
33.5 ksi
Vertical plane
Horizonta l plane
V (lb)
V (lb)
M (lbin)
M (lbin)
At C: 350 lb ∙ in;
470
√592
.
Specify
.
756 lb ∙ in: Use
2.0 Key
1.11 in
;
.
Ok
Shafts 1 and 3 can be designed similar to Problem 33. 36.
Data from Figure 10-8 to 10-12 and Example Problems 10-4 and 10-5. Pinion: 263 lb ∙ in; SAE 1040 OQT 1200;
√220
96.6
63 ksi;
335
240 lb ∙ in; 93 ksi;
3, 36 ksi
2.0 Key
0.86 0.81 36
25.1 ksi /
.
.
Gear: 788 lb ∙ in;
404 lb ∙ in;
3;
2.0 Key
/ .
.
63 000 4.0 hp /600 rpm
37.
420 lb ∙ in;
42 lb .
V (lb)
420/2 Γ
108 lb 108 lb
20°
71.6°
12 lb
108
20°
71.6°
37 lb
See Figure P12-37 Bottom
;
Top
V (lb)
M (lbin) 0
0 M (lbin)
336
.
Ok
210 lb ∙ in each 45/15
71.6°
15/45
18.4°
At Bearing D: 351
√105
2.5 Fillet
3;
Use
366 lb ∙ in
SAE 4140 OQT 1000
152 ksi;
0.95 0.81 58
44.6 ksi
168 ksi;
58 ksi
/ .
1.00 in;
Specify 38.
0.86 in
15.0 hp;
0.88 Ok
1725 rpm;
575 rpm;
287.5 rpm
Torque on Shaft 1
63 000 15 /1725
548 lb ∙ in
Torque on Shaft 2
63 000 15 /575
1643 lb ∙ in
Torque on Shaft 3
63 000 15 /287.5
Figure P12-38
3287 lb ∙ in
609 lb ←
.
609 lb Reactions act on A:
20°
222 lb ↓
609 lb →;
222 lb ↑
Gear A drives Gear B
Shaft 1
Horizontal
Vertical
V, M
V, M
At middle of Shaft: 337
333
√913
972 lb ∙ in
548 lb ∙ in; From Coupling to Gear
Couplin g
Gear A
Bearin g
Bearing
Shaft 1
Equation 12‐24 Used to compute all diameters. Diameter D1:
548 lb ∙ in;
Use SAE 1040 CD steel:
0; Design for 0.999 Reliability
71 000 psi;
30 000 psi; Then
0.75
80 000 psi; 12% Elongation
30 000 0.9 0.75
20 250 psi; Let
3
0.589 in
Diameter D2:
ame conditions as
;
0.589 in
Diameter D3: Depends on D4 Diameter D4:
972 lb ∙ in
548 lb ∙ in; At shoulder At Shoulder:
1.54 in
At Ring Groove:
0;
3.0:
Increase by 6% for Diameter D5:
0;
2.5 ; At keyseat
:
2.0
1.64 in ≅ 1.06 1.64
1.74 in Governing value
0: Very small diameter required to resist shear.
Bearing Seats: ,
: Assume bearings with bore
0.7874 in 20 mm
can be found to carry radial loads. Chapter 14, Table 14‐3; Bearing No. 6204 Specifications:
338
0.750 in 0.7874 in and right end of
Relief provided on left side of
of bearing does not contact rotating shaft. 0.969 in Shaft shoulder 2.00 in 1.80 in checked for all diameters
Ok
Shaft 2 1643 lb ∙ in ∙ .
822 lb ←
822 lb → 822
20°
299 lb ↑
299 lb ↓ Vertical
Horizontal
V (lb)
M (lbin)
V (lb)
M (lbin)
339
to ensure that outer race
Gear C drives Gear D
At A:
0;
At B:
√2001
240
2015 lb ∙ in
At C:
√2291
470
2339 lb ∙ in
At D:
0;
0
√667
0
√764
80
641 lb
157
DC = 4.00 in
780 lb DD = 8.00 in
Shaft 2
Bearing
Gear B
SAE 1040 CD steel: 30 000 psi;
Bearin g
Gear
71 000 psi;
80 000 psi; 12% Elongation
20 250 psi Same as Shaft 1
: 2015 lb ∙ in; 1.968 in At ring groove:
164 lb ∙ in; Keyseat K
2.0 ; Shoulder
At shoulder 1.06 2.089
.
: Governs
: 2339 lb ∙ in;
1643 lb ∙ in; At shoulder
2.068 in
340
2.5
2.5
At ring groove:
0;
1.06 2.1959
3.0;
2339 lb ∙ in
.
Governs
0;
780 lb;
: Right Bearing:
0;
2.94 :
/
671 lb →
Let:
. and
2.5
2.94 2.5 780 3 / 20 250
0.922 in
0.855 in ;
.
;
.
depend on bearing selection
Shaft 3 Horizontal
Vertical
V (lb)
V (lb)
M 0 (lbin)
M (lbin)
√1233
449
1312 lb ∙ in
3287 lb ∙ in from keyseat to right end of shaft. Use same material as for
341
Shafts 1 and 2. 0.80;
71 000 psi;
18 000 psi
Shaft 3
Bearing
Bearin g
Gear D
Diameter D2: 1312 lb ∙ in;
3287 lb ∙ in;
2.5 at shoulder
1.79 in at shoulder At Groove:
1.88 in for
Increase by 6%: Diameters
;
3.0;
1.06 1.88
.
:
3287 lb ∙ in;
0;
Specify: .
;
. . .
0
;
. Ok
342
Governs 1.07 in
39.
12.0 hp;
1150 rpm; Figure P12
39
Motor Shaft:
V-belt drive
Net driving force 63 000
/
/
63 000 12
Bending force
/1150
657 lb ∙ in View from left end
1.5
352 lb
35°
352 lb
35°
202 lb ←
35°
352
35°
288 lb ↑
Forces act on motor shaft. Directions as viewed from behind the left end of the motor. Reducer Input Shaft: Forces shown as viewed from right end for consistency with views of gear system in Figure P12‐38. Shaft speed:
1150 rpm
63 000 12 hp / 767 rpm Forces at sheave : 1.5
/
352 lb;
. .
767 rpm
986 lb ∙ in 986
∙
202 lb ←;
/ 8.4
/2
235 lb
288 lb ↓
See motor shaft analysis
V-belt drive
From Problem 38: Gear :
∙ .
1096 lb →
1096
20°
343
399 lb ↑
View from right end
Position of sheave on Shaft 1 Assumed Horizontal plane
Vertical plane
Bearing D
Sheave Bearing Gear V (lb)
V (lb)
M (lbin)
M (lbin)
Resultant Moments: √606
864
√1947
1032
1055 lb ∙ in 2204 lb ∙ in
Bearing Forces: √245
232
358 lb
√649
344
735 lb
Design of Shaft 1: Completed in a manner similar to that shown in Problem 38. Shaft 2: See Problem 38 for analysis and design procedure. Forces on Gear :
1096 lb ←;
344
399 ↓
Shaft 2 speed
256 rpm
63 000 12 /256 /
2958 lb ∙ in
2958 lb ∙ in/2.00 in 20°
128 rpm
1479 lb → Gear C
538 lb ↓
Shaft 3: See Problem 38 for analysis and design procedure. Forces:
1479 lb ←;
538 lb ↑
Chain sprocket at end of Shaft 3: Speed:
256 rpm
128 rpm
63 000 12 /128
5917 lb ∙ in
∙
2817 lb ↑
.
20°
/
2817
20°
345
Chain drive
964 lb →
20°
2817
20°
2647 lb ↑ Vertical plane
Horizontal plane
Bearin g
Bearing Gear
Gea r Bearing
Bearing
Sprocket Sprocket V (lb)
V (lb)
M (lbin)
M (lbin)
Resultants: √3666
3096
4798 lb ∙ in
√2892
7941
8451 lb ∙ in
√1222
1032
1600 lb
√707
4262
4320 lb
Bearing Forces:
Conveyor Shaft Forces: 964 lb →;
2647 lb ↓
346
Speed:
4.2/10.6
128
63 000 12 hp / 50.6 rpm 40.
Figure P12‐40: Shaft 2: Power out at
50.6 rpm
14 932 lb ∙ in
480 rpm; Power in at
15 kW; Power out at
Coal Crusher – Use
4.2/10.6
22.5 kW
7.5 kW
4 because of impact and shock. 50.27 rad/s
∙ / . .
298 N ∙ m
/ ∙ /
.
Torque on Gear
TC = 447 Nm
T (Nm)
149 N ∙ m
/ .
∙ / .
/
Torque in shaft
447 N ∙ m Gear Q drives Gear C
Forces: ∙
Gear : 20°
5960
20°
∙
Gear : 20° Gear :
5960 N ← 2169 N ↑
2980 N ←
2980
20°
∙
1065 N ↓
4806 N ↑
347
Geat A drives Gear R
Geat E drives Gear S
Horizontal plane
V (N)
Vertical plane
V (N)
M 0 (Nm)
M (Nm)
Resultants: √596
217
634 N ∙ m
√389
249
462 N ∙ m
√481
175
512 N ∙ m
√8027
1846
8237 N
√5719
1006
5807 N
Bearing Forces:
348
Proposed Shaft Design: Gear A
Bearing B
D1
D2
D3
Gear C
D4
Bearing D
D5
D6
D7
2.5, except
, ; Use
Gear E
D8
D9 Assume all fillets are small radius with at ring grooves with shaft diameter Use
1.5 at
and
1.06
3.0
groove diameter.
Well rounded
Material Selection: SAE 4140 OQT 1000,
1160 MPa,
1050 MPa
17% Elongation – Good strength and ductility. From Figure 5‐11: Select
0.80
400 MPa 65
;
0.80 0.81 400
349
0.81 0.99 Reliability
259 MPa
259 N/mm
Kt
Location
2.5
N ∙ mm 0
N ∙ mm 298 000
2.156
Specified D 50.0 mm
1.5
634 000
298 000
53.13
60.0 mm
2.5
634 000
298 000
62.96
65.0 mm
3.0
525 000*
298 000
62.82
2.5
462 000
298 000
56.69
3.0
462 000
149 000
60.19
2.5
512 000
149 000
58.62
60.0 mm
1.5
512 000
149 000
49.45
55.0 mm
2.5
0
149 000
17.11
45.0 mm
Solutions for diameters using Equation 12‐24 – Summary: N
1.06 D5
D4
66.6 mm D6
1.06
80.0 mm
63.8 mm
4
Notes:
and
*Moment at
80.0 mm used for
are standard bearing bores from Table 14‐3. estimated between points B and C. ,
, and
to provide shoulder for bearing at B and D.
and
provide ease of installation for bearings.
and
made larger than required for compatibility with adjacent diameters
and to withstand moments at shoulders.
350
Final stresses must be checked after specifying fillet radii, groove geometry, and bearing selection.
41.
Figure P12‐41 Design shaft and levers.
20 N
60
60
/ 40
Use SAE 1137 CD steel 0.9,
Let
0.75;
20 N: Torque 676 MPa;
1200 N ∙ mm from B to D.
565 MPa;
0.9 0.75 260
260 MPa Figure 5
175 MPa
V (N)
M 0 (Nmm)
Wiper mechanism has an oscillating motion. Both bending and torsion will be varying. Shaft restrained in bearing at A but can float in bearing C.
Lever 1
Lever 2
351
11
Assume
2.5 at fillet to right of A. Assume
1.0 at C.
Assume levers are installed with a light press fit and tack welded in position. Use 3.0 for weld area. Design Equation 12‐24 must be modified for varying torque. Add
for torsion. Substitute
for
. Then: /
At C: / .
.
.
.
5.94 mm
At B: /
At
.
8.18 mm
, , .
.
.
Lever Design: Use flat stock, 4.0 mm thick
b
Same material as shaft:
175 MPa;
20
1200 N ∙ mm:
Required
60 ∙ .
/
20.57 mm
352
58.3 MPa
Required
6 /
6 20.57
/4.0
5.55 mm at shaft
Because shaft is 10.0 mm diameter, lever is adequate.
Cross section of lever
353
CHAPTER 13 TOLERANCES AND FITS Clearance Fits: Refer to Table 13‐3 for tolerance grades; data in thousandths of an inch. 1.
.
Limits: Hole 2.
.
;
.
Shaft
;
.
7 ; Clearance 10.5
7 15.5
Clearance 0.0070 to 0.0155
Basic size = 0.500 in; Precision fit – RC2: Hole
0.4 ; Shaft 0 .
Limits: Hole 3.
5 ; Shaft 0
Basic size = 3.500 in; Loose fit – RC8: Hole
;
.
0.25 ; Clearance 0.55 .
Shaft
0.25 0.95
; Clearance 0.00025 to 0.00065
.
Basic size = 5/8 in – 0.625 in; Loose fit – RC8: Hole
2.8 ; Shaft 0
3.5 ; Clearance 5.1
3.5 7.9
Adjust tolerances for basic shaft system – Add 3.5 6.3 ; 3.5
Hole Limits: Hole 4.
; Shaft
.
.
Clearance
3.5 7.9
; Clearance 0.0035 to 0.0079
.
Basic size = 0.800 in; Close fit – Reliable motion – RC5: Hole
1.2 ; Shaft 0
Limits: Hole 5.
.
0 ; 1.6
Shaft
. .
1.6 ; Clearance 2.4 ;
.
Shaft
;
.
1.6 3.6 Clearance 0.0016 to 0.0036
Basic size = 1.250 in; Loose fit – RC8: Hole Limits: Hole
. .
;
Pin
. .
;
4.0 ; Pin 0
5.0 ; Clearance 7.5
Clearance 0.0050 to 0.0115
354
5.0 11.5
6.
Basic size = 4.000 in; Loose fit – RC8: Hole
7.0 ; Clearance 10.5
5.0 ; Pin 0
7.0 15.5
Adjust tolerance for basic shaft system – Add 7.0 12.0 ; Pin 7.0
Hole Limits: Hole 7.
.
.
Shaft
.
0 ; Clearance 3.5
7.0 15.5
Clearance 0.0070 to 0.0155
.
Basic size = 0.750 in; Precision with wide temperature variations would typically
call for RC3 or RC4. RC2 – probably too tight for temperature change. RC5 probably too loose for required precision. RC3 or RC4 not available in Table 13‐4. Illustrate limits with RC5: Hole Limits: Hole 8.
1.2 ; Pin 0 . .
;
1.6 ; Clearance 2.4 Pin
.
;
.
1.6 3.6
Clearance 0.0016 to 0.0036
Basic size = 1.500 in; Loose – RC8: Hole
4.0 ; Shaft 0
5.0 ; Clearance 7.5
Adjust tolerances for basic shaft system – Add 5.0 9.0 ; Shaft 5.0
Hole Limits: Hole
. .
;
Shaft
. .
0 ; Clearance 2.5 ;
5.0 11.5
Clearance 0.0050 to 0.0115
355
5.0 11.5
9.
Tolerance diagrams for Problems 1 to 8 shown below:
Problem number B.H. B.H. B.S. B.H. B.H. B.S. B.H. B.S. B.H. = Basic Hole System; B.S. = Basic Shaft
356
Force or Shrink Fits: See Table 13‐4 10.
Basic ID = 3.25 in; OD = 4.00 in: Both are steel – E = 30 X 106 psi 0;
3.25/2
1.625 in;
4.000/2
2.000 in
Use FN5 – Heavy force fit 8.4 4.8 ; Interference: 7.0 8.4
2.2 ; Shaft 0
Hole
.
Limits: Hole
.
Shaft
.
0.0084 in
.
.
[Eq. 13‐2]
.
.
. .
. .
13 175 psi [Eq. 13‐4]
13 175
Not acceptable for 1020 HR; 11.
1.750 in;
4.0/2
FN3: Hole
1.4 Shaft 0
4.9 2.6 Int. 4.0 4.9 .
1575 psi Using
30
.
. .
64 363 psi
.
55 ksi
3.50/2
Limits: Steel sleeve
.
2.000 in;
4.50/2
.
: Bronze bushing
Equation 13
3
10 psi;
15
10 psi;
2.250 in
.
0.27;
:
0.0049 in
0.27
13 438 psi
Equation 13
4 At inner surface of steel sleeve
11 869 psi
Equation 13
5 At outer surface of bronze bushing
Note: Appendix A‐13 – Bearing bronze has a yield strength of 18 000 psi. 1.52 Low Evaluate FN5 fit for maximum interference. Hole tolerance:
0.0018 ; Shaft: 0
0.0072 0.0042 ; Interference: 0.0060 0.0072 357
Maximum interference = 4.0000 Stress in aluminum Let
4.0072
0.0072 [Smallest hole, Largest shaft]
18 901 psi Tension
2. Required
2 18 901
37 802 psi
Specify any aluminum alloy with
37 800 psi
42 ksi or 6061
Examples: 2014‐T4,
T6,
40 ksi
Any steel for rod would be satisfactory, provided 2 8834
17 788 psi
Any carbon or alloy steel with
18 000 psi
From Appendix 3: Examples:
13.
AISI 1020 HR,
30 ksi
AISI 1040 HR,
42 ksi
Steel sleeve on aluminum tube 2.00
2 0.065 /2
0.935 in;
[Equation 13‐5] 8500 psi;
For
2.00/2 .
.
.
.
1.000 in;
14.898
570 psi Maximum allowable pressure
.
From Equation 13‐3: Solve for 2 30
10 psi;
10
3.00/2
10 psi;
.
358
0.27;
0.33
1.50 in
14.
Nominal Diameter = 3.250 in; Assume maximum interference = 0.0084 in For final clearance = 0.0020 in; Change in diameter .
∆ 15.
492°F:
.
°
.
Bronze – shrink from 75 to ∆
10.0
10
75°
20°; ∆ 4.00
0.0084
0.0020
492°F
°
0.0104
95°F 95
0.0038 in
Max interference = 0.0049 in 0.0049 in
0.0038 in
0.0011in
0.0040 in Desired Clearance
∆ 209°F 16.
. .
.
75°F
°
0.0011 in 0.0051 in (Required expansion of steel)
209°F .
Use Equation 13‐7 to find decrease in diameter: .
.
.
.
.
0.27
0.00269 in Final inside diameter
3.500 00 in
0.002 69 in
359
.
CHAPTER 14 ROLLING CONTACT BEARINGS
1.
Equation 14 – 2:
2.
10
20 000
880
/
Equation 14 – 3: 3.
Use
0.343 1150
/
/
.
10
10 1500
.
1250 1.06
Equation 14 – 2 (a)
/10 5.
60
/10
(b) 4.
10
10
10
/
1450 1.04
10 /10 .
. 60
/
1.04
10 rev
/
10 /10
Shaft for an industrial blower. Reactions from Figure 12‐12: Reaction at B:
√458
4620
4643 lb
Reaction at D:
√1223
1680
2078 lb
From Example Problem 12‐1; Dia. At B (min.) 1.09 in
Dia. At D (min.) Use
10 000
Required
600
3.55 in
/
60
value at :
3.6
10 /10
/
4643 7.114
33 029 lb
:
3.6
10 /10
/
2078 7.114
14 782 lb
From Table 14‐3:
360
/
3.60
10 rev
;
.
;
.
6.
Data of Example Problem 12‐2: From Figure 12‐16 we find, √589
164
611 lb;
2.02 in;
188
√393
436 lb
1.98 in; From Table 12‐2
Table 14‐4: Agricultural Equipment – Let
5000 hr
5000 hr 1700 rev/min 60 min/hr Required C value at B:
611 5.1
5.1
10 rev
/
4882 lb
10 /10
At B: From Table 14‐3: Bearing 6011 has
6317 lb and a bore of 2.1654 in.
C is higher than required but specify Bearing 6011 for both B and D. 7.
Data of Example Problem 12‐3 and Figures 12‐17 and 12‐18. Conveyor shaft. √507
41
509 lb;
√1697
393
1742 lb Radial
purely radial; Bearing C also carries 265 lb thrust load 0.59 in; Use
20 000
Required
value at :
Bearing 6302 has
2.26 in [From Table 12‐3] 101
/
60
509 1.2
/
10 /10
1.2 /
10 rev
2519 lb
2563 lb and a bore of 0.5906 in [Figure 12‐16] to provide a shoulder for the
Should be compatible with diameter
bearing. A lighter bearing with a larger bore may be preferred. Bearing C: Combined radial & thrust load. [Equation 14‐5] Assume
1.5:
1.0 0.56 1742
1373 1.2 Bearing 6212 has Check: /
/
10 /10
10 679 lb, bore
265/7307
1.5 265
0.0363 →
1373 lb
6772 lb 2.3622 in, ≅ 0.24
361
7307 lb
/
265/1742 1742 1.2
19.
0.152
10 /10
use Equation 14 /
Varying loads: (Bearing 6324) 1.
4500 lb
25 min
2.
2500 lb
15 min
600 rpm;
47 783 lb
25 4500
1630
Bearing 6314,
600 rpm;
23 381 lb
25 min
2.
1500 lb
15 min
/
Bearing 6209,
1700 rpm,
7464 lb
480 min
2.
200 lb
115 min
3.
100 lb
45 min
/
40
23 381/2226
600 lb
15 1500
25 2500
/
1.
/
7464
480 600
/547
/
362
2226 lb
/
115 200 640
640 min
/
3975 lb
10 rev
40 min
21.
15 2500 40
46 783/3975
2500 lb
1742 lb
Ok
/
1.
1.0
8592 →
40 min
20.
4;
/
45 100
/
547 lb
22.
Bearing 6209,
1700 rpm,
1.
450 lb
480 min
2.
180 lb
115 min
3.
50 lb
45 min
7464 lb
480 450
45 50
/
0.85 100
/
115 180 640
411 lb
640 min
23.
/
7464/ 411
Bearing 6205,
101 rpm,
1.
500 lb
6.75 hr
2.
800 lb
0.40 hr
3.
100 lb
0.85 hr
/
/
3147 lb 0.40 800 8.0
6.75 500
508 lb
8.00 hr
. /
24.
Bearing 6211,
101 rpm,
/
9802 lb,
6502 lb
Combined radial and thrust loads: Compute equivalent load
as in Section 14 – 10.
R
T
T/R
/
e
Y
P
1.
6.75 hr
1750 lb
350 lb
0.20
0.653
0.26
‐
1750 lb
2.
0.40 hr
600 lb
250 lb
0.417
0.0383
0.234
1.89
809 lb
3.
0.85 hr
280 lb
110 lb
0.393
0.017
0.20
2.24
403 lb
8.00 hr
0.56
363
Use equivalent loads to compute .
.
: /
.
1658 lb
. . /
25.
1450 lb, Actual
1150 rpm,
10 rev
1.04
10 /0.62 /
60
20 000 hr,
20000 101 60
1.21
/
436 lb,
1.21 509
5.10
5.10 436
1250 lb,
880 rpm,
0.21
10 rev
0.97 →
0.44
10 rev 1.16
10 rev
/
.
20 000 hr,
0.95 →
0.62
/
1.06
10 rev
1.70
10 rev
20 000
880
60
Adjusted
/
1.06
10 /0.62
1250
5.77
10 /0.44
Actual
[Equation 14 – 3]
0.99 →
/
.
5000 1700 60
[Equation 14 – 3]
10 rev /
.
10 /0.21
5000 hr,
/
1.04
10 rev
1700 rpm,
Adjusted
1.68
1450
101 rpm,
[Equation 14 – 3]
28.
/
/
509 lb,
Actual
0.62 [Table 14‐6]
/
Adjusted
27.
0.95 →
1150
[Equation 14 – 3]
Actual
15 000 hr,
15 000
Adjusted
26.
/
.
/
364
CHAPTER 16 PLAIN SURFACE BEARINGS All of the problems in this chapter are design problems with no unique solutions. A sample of each type of design problem is shown here. Plain Surface Bearings 1. Let
225 lb;
3.00 in;
1750 rpm
1.5
1.5 3.00
4.50 in
.
16.67 psi
. .
1374 ft/min
16.67 psi 1374 ft/min Required
Rating
22 900 psi ∙ fpm
2
2 22 900
∙
Porous bronze bearing material/oil impregnated or BU or DU dry lubricated bearing. 4.
75 lb; Let
1.5
0.50 in; 1.5 0.50 in
600 rpm 0.75 in
75 lb/ 0.75 in 0.50 in 0.50 in 600 rpm /12 200 psi 78.5 fpm Required
200 lb/in 78.5 ft/min
15 700 psi ∙ fpm
2 15700
∙
Porous bronze or BU bearing
365
7.
800 lb; Let
3.00 in;
1.50
350 rpm
1.50 3.00 in
.
4.50 in
59.3 psi
.
3.00 in 350rpm /12
275 fpm
16 290 psi ∙ fpm Required
2 16 290
∙
Porous bronze or BU bearing 8.
60 lb; Let
0.75 in;
1.00 in; / .
750 rpm; Try 1.00/0.75
1.33 Ok
0.75 in 750 rpm /12
80 psi 147.3 fpm Required
1.25 0.75
80 psi
.
/12
1.25
rating
147.3 ft/min
11 784 psi ∙ fpm
2 11 784
∙
Use Babbitt – high tin Hydrodynamically Lubricated Bearings 9.
1250 lb;
2.60 in;
Let
2.75 in,
For
300 psi;
/
1.515/2.75 .
0.0036 in;
/2
1750 rpm; Electrical motor
1.375 in /
.
0.55: Let
1.515 in
.
0.5
1.375 in; /
331 psi Ok 0.0018 in;
. .
366
754
0.50
0.938 in
Figure 16-4 Surface Finish: 16
32 μ in Avg.
0.000 25 2.75 /
0.0069 in: Use
0.0007/0.0018 /60
1750/60
0.389 →
0.29 Fig. 16
6.5
10
S is proprotional to : From Fig 16 /
5.79
.
/
SAE 50 oil has
9:
.
∙
500 lb; 1.200 in;
For
200 psi;
/2
2500 rpm; Precision spindle 0.600 in /
1.73; Let
.
18.2 lb ∙ in
0.506 hp
1.15 in;
Let
.
0.326
8.0
0.0106 1250 1.375
.
reyns
0.0106
/
13.
10
reyns @ 160°F
0.29 6.5/5.79
/
.
8
29.17 rev/s .
Required
0.0007 in
1.50;
2.08 in
.
1.50
1.5 1.200 in
1.800 in
232 psi Ok
.
0.0014 in;
0.0007 in;
Surface finish: 8
16 μ in Avg.
.
0.00025 1.200 in
0.00030 in;
0.11;
41.67 rev/s
2500/60
857
.
367
/
0.0003/0.0007
0.429
.
Required
/
SAE 5W has
0.91
10
S is proportional to μ: /
0.832
.
reyns
@ 160°
0.11 0.91/0.832
0.120
2.80 From Figure 16-9: /
.
0.0033
/
0.0033 500 .
0.60
∙
100 mm;
0.98 lb ∙ in
0.039 hp 100 mm;
18.7 kN;
16.
10
500 rpm; Conveyor
50 mm
For
2.0 MPa
Let /
1.0;
.
2.0 N/mm ;
.
93.5 mm
/
100 mm
.
1.87 N/mm
0.15 mm;
1.87 MPa Ok
0.075 mm; /
667
.
(Large clearance desired) 1.0
Then 8 μin
0.40 μm; 32 μ in
63 μin
0.20 μm; 16 μm
0.0254 μm 0.80 μm
1.60 μm
Specify surface finish
0.8 to 1.6 μm Avg.
0.00025 100 mm 0.096 Required
10 in
.
Surface Finish: Note: 1.0 μ in
16
0.025 mm; 8;
/60
. /
/
. .
0.025/0.075
500/60
8.33 rev/s
0.0485 Pa ∙ s
/
368
0.333
At 70°C, SAE 50 has .
0.096
0.091 →
.
/
0.046 Pa ∙ s;
.
0.0039 18.7 3.65 N ∙ m
/2
0.036 mm; .
2.0 MPa;
Let
2
∙
191 watts
2200 RPM; Machine Tool
12.5 mm
.
50 mm; /
45 mm
/
2.0
1.80 N/mm
Surface Finish: 16
0.00625 mm ≅ 0.006 mm
0.006/0.018 2200/60
0.333 →
/
SAE 5W has μ
0.057
36.7 rev/s .
Required
1.80 MPa Ok
32 μin Avg. 0.4 to 0.8 μm Avg.
0.00025 25
. .
0.0058 Pa ∙ s
/
0.0066 Pa ∙ s @ 70°C
0.057 0.0066/0.0058 /
3.65 N ∙ m
191
.
.
/
10
0.018 mm
For
/60
50
694
.
/
2.6
8.33
25 mm;
/
10
25 mm;
2.25 kN; Let
/
0.025 mm
0.0039
/
17.
Slightly
.
0.065 →
0.0023
369
/
1.6
0.0023 2.25 0.065
10
12.5
10
.
∙
0.065 N ∙ m ∙
15.0
15.0 Watts
Hydrostatic Bearings 19. Let
1250 lb; Supply pressure
p
/
1.40 [Figure 16-13]
0.50;
0.55;
.
4 /2 Let
9.09 in
/
/
4 9.09 1.70 in;
.
3500 lb; Let
0.5 1.7
.
/
.
/
0.85 in
0.60;
0.62;
4
/2
1.60 [Figure 16-13]
4 16.13 /
4.53 in
/4
4.50 /4
15.90 in
2.25 in;
∙
350 psi
/
.
lb ∙ s/in
16.13 in
/
.
10
725 lb ∙ in/s
/
2.90
8.3 2.90 in /s
∙ /
500 psi: Let
.
Let
/4
0.005 in; SAE 30 oil @ 120°F →
250
250 psi
3.40 in Ok 0.5
.
21.
300 psi: Let Recess pressure
Use
4.50 in
355 psi Ok 0.60
1.35 in
0.008 in: SAE 40 oil @ 146°F;
370
7.0
10
lb ∙ s/in
∙ /
0.11 hp
.
.
.
.
355 25.
25.8 in /s
∙ /
/
25.8
∙
/
∙
/
1.39 hp
/ /
Hydrostatic Lubrication – Metric Data 22.5 kN; /
20 MPa: Let
0.60;
0.62;
4
.
/
/2
2.27
/
4 2.27 85 mm;
10 / 0.6
.
51.0 mm
.
1.20 kN;
26. Let
9.91
750 kPa
0.75 MPa
0.62;
.
4
/
/2 Let
.
10
32 mm;
∙
m /s 159 Watts
1.60 [Figure 16-13]
64 mm Ok
0.6
0.6 32
0.10 mm: at 60° , SAE 10W oil has .
. .
10
159
10
3226 mm
/
4 3226 /
0.60
0.054 Pa ∙ s 9.91
∙ /
10
0.60 MPa; .
1.60 N/mm
170 mm Ok
.
.
1.60
10 N/m
10 mm
0.15 mm; at 50°C, SAE 30 oil has
Let
1.6
1.60 [Figure 16-13]
. .
1.60 MPa
0.014 Pa ∙ s 4.25
∙ /
4.25
19.2 mm
10
371
25.5
10 ∙
m /s 25.5 watts
CHAPTER 17 LINEAR MOTION ELEMENTS 5.
Required Use 2 1/2
6.
3.0 in
/
3 Acme thread
Required
. 5.0 in
/
For an Acme thread: 2 ½ ‐3: Required
5.0/4.075
4.075 in per in of length .
minimum .
7.
0.0463
.
2.65°
tan 17
10 ;
14.5° .
0.968
.
.
.
.
. .
∙ 8.
.
Use Problem 7 data:
.
. .
. .
.
∙ 9.
Square thread: ¾ ‐6;
4000 lb;
0.1667 in .
Equation 17‐2:
.
. .
. .
.
∙ 10.
.
Equation 17‐4:
. .
372
.
. .
.
∙
11.
.
Equation 17‐3: 0.083
12.
.
.
°
→ .
Equation 17‐7:
.
%
10 yrs
.
.
13.
.
. 14.
Travel At 600 lb; ¾‐2 screw required;
15.
.
.
.
∙
.
. .
17.
.
Equation 17‐13:
16.
.
.
.
For the ¾ ‐2 screw at 600 lb, travel life = . 1.04
10 in
.
Metric trapezoidal power screws – Problems 18‐29 – Table 17‐1 M 18.
Required Use
75 MPa
Specify a size: Load = 125 kN;
/
1667 mm –
; Dp = 50.5 mm; Nominal Lead
Problems 18 to 23 use same data
Pitch
.
373
55 mm;
19.
Find torque to raise 125 kN load for
0.15
.
.
.
.
.
658 559 N ∙ mm 20.
.
125 kN,
2
0.15
.
.
. .
21.
Equation 17
.
∙
Find torque to lower load
291 905 N ∙ mm
.
.
Equation 17
.
4
∙
Find power to raise 125 kN – 4250 mm in 7.5 s; M55
9 Screw
658.6 N ∙ m From prob. 19 /
Define:
395.6 rad/s
.
658.6 N ∙ m
395.6 rad/s
260 531 N ∙ m/s
P = 260.5 kW ‐ (Very high) 22.
Find lead angle : For λ
°
Equation 17
3
; or
. %
5°
23.
Find efficiency [Equation 17‐7]:
24.
Specify a power screw size:
8500 ;
Required
77.3 mm
. 25.
.
.
/
;
.
Find torque to raise load; 17 12 167 N ∙ mm
.
∙
;
110 MPa
.
0.15 .
2
.
. .
.
∙
374
. .
26.
Find torque to lower load;
8500 ; .
0.15; [Use Equation 17‐4]
.
. .
3956 N ∙ mm 27.
.
.
∙
Find power to raise load; 240 mm in 3.5s ;
12.167 N ∙ m From Problem 25 / .
143.62 rad/s
.
12.167 N ∙ m 143.62
/
1747 N ∙ m/s
1747 W
. .
.
28.
Lead angle
29.
[Use Equation 17‐7] Efficiency
30.
Ball screw from problem 14: ¾ ‐2, Length = 28.0 in
.
° [Borderline self‐locking] . ∙
0.334
. %
Find estimate of critical speed. [Equation 17‐15] .
minor diameter of screw. This value not available in this book, or on Internet site 10. As an estimate, use minor diameter for a ¾ acme screw – Machinery’s handbook, 28th ed., page 1830. 28.0 in
between single bearings at ends
1.00 Let
1.0 to estimate critical speed. .
. .
.
3339 RPM
375
≅ 0.55 in – from
Safe operating speed ‐
3.0
376
CHAPTER 18 SPRINGS 1.
∆ /∆
12.0/ 2.75
2.
;
;
3.
:
∆ /∆
0.830
/
. .
37.1 mm
21.4
29.4
134/8.95
.
2
1.100
2 0.085
.
1.100
0.085
; 8.
2
/
134 N
.
.
/
0.563/0.085
0.560
0.059
.
.
0.501 in:
2
19
2
/
. .
0.501/0.059
17 Active coils
.
Tan 19 0.059
.
.
: .
32.2
.
/
8.95 29.4
1.015/0.085
0.626
.
. .
.
/
.
/
/
Tan
32.2/76.7
39.47 mm
6.
7.
/
1.25
99.2 / 63.5
5.
.
76.7 0.830 /
4.
1.85
.
.
377
°
.
9.
.
From Equation 18‐6:
.
.
.
Equation 18‐4:
.
.
10.25 lb
74 500 psi
.
.
.
.
.
.
.
1.17
.
From Figure 18‐9: Music wire: 10.
/
4.22/0.501 0.18
Actual 11.
8.42: From Figure 18
0.18 4.22
4.22
.
3.00
.
15, Curve A,
0.18
0.76 in
.
0.76
– Buckling should occur
(Equation 18‐3)
8.40 4.22
12.
0.501
.
1.12
.
.
From Problem 9: 74 500 psi
.
0.059
.
8.40 lb/in
.
.
; Too high
.
From Figure 18‐9, Light service,
.
147 000 psi
approximate
Notes concerning Problems 13‐35: Most of these problems are design problems. No single unique solutions exist. Sample solutions are shown. Problems 13 through 24 are compression springs. Problem 13 is done by both methods 1 and 2 as outlined in the text. Others are done by one or the other method. In some problems only summary results are shown. Problems 25 through 31 are extension springs. Problem 25 is worked out in detail. Problems 26 through 30 were designed with the aid of a computer program using the same procedure. Only summary results are shown. Problem 31 is the stress analysis of the ends of the spring.
378
Problems 32 through 35 are torsion springs. In each case, the design of each end is required. It was assumed that ends would be straight with length L1 and L2 as shown. The effects of the ends on the spring rate were then included in the analysis. 13.
220 lb;
Method 1
180 lb; ∆
Step 1
ASTM A229; Severe service;
Steps 2‐5
Let
3.00 in; ∆ /∆
5.25
2.50
2.75 in
(Equation 18‐7)
80 000 psi
/
.
.
/
3.00/0.3065 .
.
.
Step 12
2
.
.
0.3065 5.72
2
14
2.365 in
2365
231 lb
67 126 231/220
;
1.15 Fig. 18
5.72 coils
80 5.25
Step 13
104 ksi
67 126 psi Ok
.
/
0.293 in
78 ksi; max
9.79:
.
Step 11
/
.
0.3065 in U.S. Gage 0;
Select
Step 10
Summary:
10 psi
5.25 in
2.50 in
Step 7
Step 9
180/80 0.50
Assume
0.50 in.
80 lb/in
3.00
Step 6
40 lb; ∆
3.00 in Design Decisions 3.00
∆
180
11.2
40 lb/0.50 in /
Step 8
220
70 420 psi
3.00
0.3065
3.3065 in
3.00
0.3065
2.694 in
.
;
379
.
;
Ok
.
13.
Method 2
Process starts the same as method 1 without
Step 2
(Eq. 18‐10)
21.4
Step 3
Try
Step 4
(Equation 18‐11)
/
21.4 220 / 80 000
0.2625 in U. S. Gage 2;
2
/
Step 5
Use
Step 6
(Equation 18‐12)
.
2.50
0.243 in
80 000 psi
2 0.2625 0.2625
7.5 coils
6.0 /
.
.
/
.
9.15
1.16 (Figure 18‐4) (Equation 18‐4; Note: 8/ = 2.546) .
Step 8
.
.
86 262 psi Too high
.
Repeat from step 3: Try
0.2830 in U. S. Gage 1,
79 000 psi
104 000 psi Step 4 Step 5
2.50
2 0.2830 /0.2830
6.83
6 coils
Use .
Step 6
/
. . .
Operating stress
.
9.38;
.
1.155 75 770 psi Ok
.
Solid length & Stress at solid length 2
0.2830 6.0
2
2.264 in
80 lb/in 5.25
2.264 in
Stress at solid height Final geometry
/
239 lb
75 770 239/220
9.38 0.2830
380
2.655 in
82 272 psi Ok
2.655
0.2830
2.938 in
2.655
2.830
2.372 in
2.50
/
14.
Summary:
.
Method 2
1.75 in;
2.264 /6.0 ,
.
.
.
3.368
21.4 22 /125 000
2
/
/
.
/
.
.
.
6.85 →
.
13.6 3.368
1.375
Buckling:
/
1.75
1.225
27.1 lb 148 245 psi Ok
0.428 in
0.491; /
20
1.375 in
120 325 psi 27.1/22.0 6.85 0.0625
26; Try
120 325 psi Ok
.
0.0625 22
188 ksi
2 0.0625 /0.0625
.
/
5.0 lb
0.061 in
130 ksi; τ
1.75
.
2
3.00 in;
.
3.368 in
0.0625 in → τ
.
,
1.618 in
5.0/13.6
/
Try: 16 Ga.;
.
125 ksi [125 000 psi]
1.75
3.00
21.4
Ok
13.6
.
/
,
22.0 lb;
ASTM A401; Severe service; .
0.0393 in
0.366 in 1.375 /20
3.368/0.428
7.87:
381
0.019 in /
/10 Ok 0.21 Fig. 18
15
0.21
0.21 3.368
0.71 in. : Actual
will buckle
After several iterations this design was produced which will not buckle.
15.
0.072 in;
129 ksi; τ
22.3;
12;
108 200 psi;
1.008 in;
185 ksi 8.52; 32.1 lb;
0.062 in
: Buckling ⇾
0.53
1.78 in; Actual
Method 2
1.25 in;
.
/
5.49; .
.
2.00 in;
21.4 Try 16 Ga.;
0.0625 in; 2
115 ksi;
/
1.25
/
/
.
.
.
.
0.0625 18 16.67 2.09
6.64 0.0625 /
2.09/0.415 /
2.09 in
0.055 in 128 ksi
6.64 Then
18.0 → Use 1.23
1.125 in 1.125
74 540 16.09/14.0
/
1.50 lb
75 540 psi Ok
.
2
.
2 0.0625 /0.0625
. .
.
2.00
21.4 14.0 /100 000
/
.
100 ksi
16.67 lb/in:
.
/
Ok
14.0 lb;
ASTM A313, Type 302, Avg. Service; .
.
16.09 lb 85 650 psi Ok
0.415 in
5.04 No buckling 1.25
1.125 /16
382
0.008 in
/10 Ok
16
16.
0.415
0.0625
0.478 in
0.415
0.0625
0.353 in
4.00 in;
Method 2 ASTM A231:
250 lb,
21.4
29.23 lb/in:
.
/
10.50
21.4 250 /90 000
Trials with gages 2 and 1 failed 0.3065 in 0 2
/
/
.
90 000 psi,
4.00
2 0.3065 /0.3065 /
.
.
/
.
3.678 in
29.23 12.55
3.678
259/250
11.37 0.3065 12.55/3.485 / 17.
Method 2 .
4.00
259 lb
89 900 psi
Ok
3.485 in
3.678 /10
0.032 in
0.68 in;
0;
/10 Ok 1.75 in;
13.08 lb/in; Music wire; Average service; τ
21.4 14 /120 000 135 ksi;
1.125
3.60 No buckling
14.0 lb; .
11.05; Try
86 680 psi Ok
0.3065 12
86 680
130 ksi
11.37; Then
.
2
12.55 in
0.244 in
,
.
.
60 lb/29.23 lb/in
too high
.
/
60 lb
90 ksi For severe service .
For
10.50 in,
0.050 in; Use
150 ksi
383
120 ksi
0.055 in, 24 Gage
10
2 .
/
. .
.
.
0.68
/
9.20 →
. .
2
/
8.0
1.16
0.055 10
0.550 in
13.08 1.75
0.55
15.70 lb
125 750 15.7/14.0 9.20 0.055
1.75/0.506 0.68
8.9 coils; Use
125 750 psi Ok
.
/
2 0.055 /0.055
141 000 psi Ok
0.506 in
3.46 No buckling
0.55 /8
0.016 in
/10 Ok
0.561 in 0.451 in 18.
Method 2 .
2.75;
1.75 in;
1.00 in;
8.00 in
4.57 lb/in
.
ASTM A313, Type 316, Average service; 21.4 8.0 /90 000 0.85 122 ksi
.
/
.
.
.
.
.
0.0475 17 /
1.00
/
. .
0.044 in → Use
104 ksi;
2
2
90 ksi
0.85 135
0.0475 in; 18 gage 115 ksi
2 0.0475 /0.0475
9.14 →
19.05 → Use
17
1.16
95 740 psi Ok 0.9025 in;
4.57 lb/in 2.75
95 740 8.44/8.0
101 050 psi Ok
384
0.9025 in
8.44 lb
9.14 0.0475 / 19.
0.434 in;
0.32;
0.32 2.75 0.625 in
Same as 18 except Let
0.75 in; /
.
0.85 115 ksi
.
.
.
.
.
2
1.00
130 ksi; /
0.743
9.17 lb 80 400 psi Ok
0.743 /9.89
0.026 in Good
0.75
0.0625
0.8125 in
0.75
0.0625
0.6875 in
Ok
0.75 in Method 1
0.625 in; .
1.12
3.67 Ok No buckling
Same as 17 except
.
109 ksi
0.743 in
70 100 9.17/8.0
/
.
0.057 → Use
0.85 128
0.0625 11.89
2.75/0.75
Let
/
.
9.89 coils
4.57 2.75
20.
6.33 High 0.88 Will buckle
Use method 1
12.0 →
.
/
1.75
70 100 psi Ok
.
/
0.88 in;
97.8 ksi;
0.75/0.0625 .
2.75/0.434
100 ksi
.
/
/
120 ksi /
0.061 in → Use
0.063 in
145 ksi 0.625/0.063
9.92 →
385
1.15
0.0625 in
.
.
.
.
102 500 psi Ok
. .
. .
7.31 coils;
.
13.08 1.75 / /
21.
0.063 7.31
0.586
2
0.586 in
15.22 lb
111 500 psi Ok
1.75/0.625
2.80 Ok No buckling
0.68
0.586 /7.31
0.625
0.063
0.013 in
/10 Ok
0.688 in
Clearance with hole = 0.75
0.688
0.062 in Ok
/10
Method 2
45 lb;
3.50 in;
22.0 lb
.
3.05 in; 51.1 lb/in;
.
/
21.4 45 /90 000
0.103 in → Use
109 ksi;
ASTM A231: 2
3.05
.
.
/
.
.
5.44 →
71 950 psi;
.
2 /
0.1055 20
1.285
51.1 lb/in 3.93
148 700 psi
21.1
Ok
0.574 in
3.93/0.574
6.85:
0.27 3.93
1.061 in 3.93
26.9 coils → Use
Ok
2.11 in;
71 950 93.0/45.0 5.44 0.1055
/
3.93 in
0.1055 in
2 0.1055 /0.1055
. .
22/51.1
157 ksi; Severe service
/
/
3.50
3.05
/
0.27
0.88 in
386
Ok No buckling
93.0 lb
18
3.05
2.11 /18
0.052 in
/10 Ok
0.680 in 0.469 in 22.
Compression Spring: Method 1 ASTM A227 Steel wire;
11.5
10 psi Table 18
4
Stress from Figure 18‐8. 2.50 in,
20.0 lb,
2.10 in,
35 lb
Install spring around 1.50 in. diameter shaft. Spring scale
.
/
Free length Try
37.5 lb/in
.
2.50
20.0/37.5
1.75 in;
Severe service: ≅
1.75
/
.
Try U.S. wire gage 9: 87 000
0.25 in
/
.
Severe service ;
.
.
.
.
2
.
116 000 psi Light service
11.80;
1.125 Figure 18
14
53 800 psi Ok
.
.
0.130 in
0.1483 in
1.75/0.1483
3.45 coils
0.1483 3.45 37.5 3.033
/
1.50
≅ 85 000 psi
.
/
3.033 in
2
0.809 in
0.809
83.4 lb
53 800 psi 83.4/35
387
128 200 psi Too high
Try
0.162 in 8 gage ;
86 000 psi;
114 000 psi
Summary of results: 10.80;
1.13
41 455 psi Ok 4.93 coils;
1.122 in;
71.7 lb
84 900 psi Ok 1.75 23.
0.162
1.588 in Ok
Compression Spring: Method 1: Metric ASTM A227 Steel wire; 60 mm,
79.3 GPa
90 N,
79.3
50 mm,
10 MPa Table 18
155 N
6.5 N/mm /
60 mm
73.85 mm
/
Spring installed on 38 mm diameter shaft: Try Figure 18
8 Severe service;
Trial
.
Try
600 MPa 87 ksi ;
/
80 ksi 550 MPa
/
3.80 mm Table 18
11.84;
1.125 Fig. 18
364 MPa Ok
. .
2
3.39 mm
2 0.150 in
.
.
/
.
800 MPa 116 ksi
45/3.80
.
45 mm
3.49 coils
3.80 3.49
2
20.86 mm
388
14
4
6.5 N/mm 73.85 /
364 MPa 45
24.
3.80
0.054 in;
41.22 mm Ok
For
11.2
106 ksi Severe ;
0.531 in; /
344 N
809 MPa slightly high
Compression Spring: Analysis: ASTM A229, 17Ga. ;
0.531
0.477/0.054
128 ksi Avg ;
0.054
8.83;
10 psi
1.17;
141 ksi Light
0.477 in 7.0 coils;
7.0
2
5.0
10.0 lb: .
.
90 219 psi Ok for severe service
. .
.
.
.
1.25 in
0.456
. .
0.456 in 0.794;
/
0.378
Extension Spring
172 500 psi at solid length
141 000 psi Too high
7.75 lb;
2.75 in;
2.25 in,
0.300 in; Music wire, Severe service 0.250 in; /
Use 16 Ga. ,
0.378 in
19.12 lb
90 219 psi 19.12 lb/10.0 lb For light service
But
0.054 7
21.93 lb/in
.
21.93 1.25
25.
20.86 mm
120 000 psi; Trial K = 1.20 .
.
0.037 in;
.
/
0.0367 in
120 000 psi
0.287 in Ok;
0.213 in
389
5.25 lb installed
/
0.25/0.037
6.76 →
.
.
1.225
.
119 320 psi Ok
. ∆
.
.
∆
.
.
5.0 lb/in . .
.
35.5 Coils
1
0.037 36.5
.
Body length ≅
1.350 in
. .
Assume full loop at each end . . 2 Deflection to
1.350
2 0.213
:
1.777 in
2.750
Initial force
7.75
Initial stress
/
1.777
0.973 in
5.0 0.973
2.88 lb
119 320 2.88/7.75
From Figure 18‐21 – Stress should be
14
44 340 psi High
21 ksi
Use smaller end loops – say 0.06 each . . 2 0.06 2.750 7.75
1.35 1.47
5.0 1.28 /
0.12
1.47 in
1.28 in
1.35 lb
119 320
1.35/7.75
20 800 psi Ok
Summary: . .
. ;
.
;
.
35.5 Coils; 2 Loops @ 0.06 in each .
@ .
.
@ .
;
390
.
@ .
26.
Extension: Music wire, Average service, 15.0 lb;
5.20 lb;
130 ksi
5.00 in;
7.84 lb/in;
0.60 in;
132 000 psi;
9.52;
3.75 in 0.0596 in;
1.153;
0.063 26 Ga.
105 600 psi Ok,
Body length = BL = 0.931 in; For end loop size = Change ELS 0.663 in 27.
6.80 lb/in;
0.65 in;
110 ksi;
0.709 in;
9.61 Coils;
0.626;
1.20 in →
0.591 in
;
,
0.591; 0.591 in
,
,
;
, , ,
6.00 in;
:
100 ksi
25.14 Coils;
10.0 lb;
1.542 in
6.00 in;
100 ksi
0.0545 in; 0.6325 in;
1.58 in;
1.0 in;
12.5;
113 ksi (Ok) 10.80 in;
1.50 in:
0.210 in →
81 ksi;
1.2747 in;
6.66;
391
1.114
0.6325 in;
0.96 lb; 162 lb;
0.055 in,
80 ksi Avg. service ; 1.7253 in;
17.3 ksi High
13 500 psi (Ok Fig. 18‐21)
0.7425 in;
Extension: ASTM A313: Type 302:
91.1 ksi Ok
1.89 lb;
10.0 lb;
0.6875 in 11/16 ;
0.14 lb: Change
1.31;
10.95 ksi Ok (Fig. 18‐21)
1.48 lb;
27.81 Coils;
3.00 in,
0.059 25 Ga.
11.02;
Extension: Music wire, Average service;
117.3 ksi;
1.0 lb;
0.058 in → Use
1.92 lb;
135 ksi; 24 Ga. ;
30.
100 ksi;
Extension: Music wire, Severe service;
130 ksi;
8.48 lb
0.75 in Ok
(Same as Problem 24 for
29.
0.537 in = Di;
13 500 psi Ok (Fig. 18‐21)
Extension: Music wire, Severe service;
Change 28.
1.92 lb;
1.20 in →
13.78 Coils
1.225
38.0 lb/in 0.225 in 4 Ga.
66.3
;
25.11 Coils;
71.7 lb;
1.27 in
,
29 300 psi Too high
Change 31.
5.88 in:
43.6 lb;
0.90 in;
17.8 ksi (Ok Fig. 18‐21)
Extension Spring Ends 19 Ga. :
0.041 in:
0.28 in;
146 ksi;
0.094 in;
5.0 lb
130 ksi For average service
Bending Equation 18 2
0.25 in;
/
13
Figure 18
22
2 0.25 /0.041
12.20
1.065 .
.
.
.
.
114 000 psi Ok
.
Torsion (Equation 18‐15) 2
/
2 0.094 /0.041
4.59
1.209 .
.
.
62 600 psi Ok
.
32.
Torsion Spring: ASTM A313, Type 302, Average service Assume
0.420 in;
140 000 psi;
/
.
.
/
1.15;
180°
0.0437 in; Use 18 Ga. ;
0.50 rev 0.0475 in 160 000 psi
/
0.420/0.0475
Actual 1.0 lb ∙ in/0.50 rev
.
. .
8.84: 103 800 psi Ok
2.00 lb ∙ in 392
1.092
. .
.
.
Ends: Specify
0.75 in .
.
0.38 Coil Ends
.
.
Operating
. .
Minimum
0.408 in
.
0.408 in
0.408
Make rod 90% of
0.9 0.360
0.0475
0.360 in
0.324 in
0.300 in Standard size
Use
1
0.0475 16.64 0.420 in
33.
16.64 Coils Total
.
0.0475
1
0.5
0.862 in If coils touch
0.4675 in Ok
Torsion Spring: ASTM A313, Type 302; Severe service; Mo
12.0 lbin
Assume
0.75 rev
1.125;
110 ksi;
/
.
1.15; /
.
270°
0.1085 in; Use 11 Ga. ;
0.1205 in 118 000 psi
/
1.125/0.1205 .
.
12.0 lb ∙ in/0.75 rev .
.
Ends: Let Body Operating
.
1.087
75 920 psi Ok
.
/
9.34;
.
16.0 lb ∙ in/rev 32.15 Coils total
.
1.50 in;
0.28 Coil Ends
32.15
0.28
31.89 Coils
.
1.100 in
. .
.
393
1.100 Use
0.1205
7/8 in 4.08 in;
34.
0.9 0.9795
0.9795 in;
0.875 in; 1.125
1
0.1205
0.1205 32.15
140 ksi; /
1
1.0 rev
1.15
.
/
.
0.0549 in; Use 26 Ga. ;
0.063 in
156 000 psi /
0.625/0.063
.
.
9.92;
.
2.50 lb ∙ in/rev;
.
Specify ends; . .
0.51 coil:
0.604
0.063
28.15 coils
0.541 in
0.487 in → Use
7/16 in
0.4375 in
0.688 in
1
1.0
1.93 in
0.038;
0.368 in;
1.125 in; 401 Steel;
.
. .
180°
9.50 coils 0.50 rev
:
From Eq. 18‐22, Solve for .
0.51 0.604 in
.
0.063
Torsion Springs: 0.50 in;
. .
0.063 28.66 35.
28.66 coils
.
28.66 .
0.9 0.514 0.625
.
1.50 in
Minimum Minimum
1.081
110 100 psi Ok
.
.
0.75
1.246 in
Torsion Spring: Music wire; Severe service; 0.625 in;
0.882 in
. .
394
0.91 lb ∙ in
Where:
0.368 .
Ends:
.
. .
Where
/
0.330 in
.
0.52 coil
0.52
10.02 coils
.
9.50
0.038
184 800 psi
0.330/0.038
8.68
1.094 For severe service
200 000 psi;
184 800 psi Ok
395
CHAPTER 19 FASTENERS 4.
Load per bolt = 6000 lb/4 0.75
1500 lb; Select SAE Grade 2 bolts, Table 9‐1
0.75 55 000 psi
41 250 psi
0.0364 in
/
Use ¼‐28 UNF or 5/16‐18 UNC For 5/16‐18:
0.15 0.3125 in 1500 lb
5.
From Tables 19‐1 and 19‐4:
6.
M4
.5:
85 000
800 MPa = T.S.
.
Yield strength
0.6
. .
0.6 800 MPa
480 MPa
Proof strength
0.9
. .
0.9 480 MPa
432 MPa
9.79 7/8‐14 UNF;
432 /
4229 N
25.4 mm/in
22.2 mm
0.875 in
. .
.
8. 1/
1/14
0.0714 in
25.4 mm/in
1.81 mm
Nearest metric thread = M24
2 (Table 19‐5)
Difference in
1.8 mm
24.0
22.2
0.071 in
Metric thread is approximately 8% larger in this size 9.
Pitch
∙ /
9.79 mm ; Grade 8.6 (Section 19‐2; Tables 19‐3 and 19‐5)
Tensile strength
7.
0.0140
.
0.630/20
0.196 in
0.0315 in
25.4 mm/in
Closest standard thread is
25.5 mm/in 4.98 mm .
396
0.800 mm
American standard thread #10‐32 is similar but not as close to measured dimensions. 10.
M10
1.5:
58.0 mm 0.75 TS
Nylon 6/6: TS = 146 MPa: Max. 58.0 11.
¼‐20 UNC:
109.5 /
0.0318 in ;
6351 N
SAE grade 2:
0.5 74
37 ksi;
b)
SAE grade 5:
0.5 120
60 ksi;
c)
SAE grade 8:
0.5 150
75 ksi;
d)
ASTM A307:
0.5 60
30 ksi;
e)
ASTM A574:
0.5 180
90 ksi;
f)
Metric grade 8.8;
0.0318 37000
0.5 830
0.0318
415 MPa
60 200
4.448 N/lb
/
8513 N 0.5 68
.
Aluminum 2024‐T4;
h)
S43 000, Ann.;
i)
Ti‐6AI‐4V, Ann.;
j)
Nylon 66 dry;
0.5 21
10.5 ksi;
k)
Polycarbonate;
0.5 9
4.5 ksi;
l)
ABS (high impact);
0.5 75
34 ksi;
37.5 ksi;
0.5 130
65 ksi;
0.5 5
2.5 ksi;
Alternate for h): Screw may be full hard – App.6 90 ksi,
0.5 90
45 ksi;
397
. .
60.2 ksi
1914 lb Similar to SAE grade 5
g)
S43 000
.
0.5
a)
1914 lb
109.5 MPa
CHAPTER 20 MACHINE FRAMES, BOLTED CONNECTIONS AND WELDED JOINTS 1.
Direct shear, 5 bolts, A307 lb
24 000 lb/bolt single shear
/ lb
Required
0.240 in2
lb/in2
/4 ∶
4
/
4 0.240 /
Figure P20-1
0.55 in
Use five 9/16‐12 UNC bolts, 1 ¾ in long 2.
Direct shear; Each bar supports 4 hangers 4 750 lb
3000 lb
Total load on joint
4 bolts in double shear, A307 lb
Total As required 2 4
0.30 in2
lb/in2
2
/ 2
0.30
/ 2
0.219 in
Figure P20-2
Use – ¼‐20 UNC Bolts, 2.00 in long 3.
4 Bolts Horizontal direct shear:
5196 lb/4
Vertical direct shear:
3000 lb/4
5196 lb 13 in
67 548 lb ∙ in
√1.50
1.00
1299 lb/bolt
750 lb
1.80 in ‐ all bolts 398
Figure P20-3
4
13.0 in2
4 1.80 ∙
.
Forces on bolt at upper left:
9353 lb/bolt
.
Total
7794 lb
1299 lb
9093 lb
Total
5196 lb
750 lb
5940 lb
Total
√9093
5946
10 865 lb
Total F = 10 865 lb
A490 bolts, double shear lb
0.494 in2
lb/in2
Required
2
/
2
/4
2 0.494
/
/2 0.561 in
Use – 9/16 – 12 UNC, 3 ½ in long 4.
Fixed‐end beam – Case e; Appendix 14‐3. .
Each side: Direct shear
Figure P20-4
32 873 lb∙in
32 873 lb∙in
l between centroids of bolt patterns (See dwg. on next page)
lb Bolts
Bolt
Use 4 bolts on each face of column through flange and plate. Force on each bolt due to moment: 16.00 in2
8 1.414
1.414 in; ∙
lb
.
Bolt
.
Forces on bolt at upper right:
FR = 3041 lb
399
Solution continued on following page.
l = 87.66 in
Total
2054 lb
188 lb
2054
√2242
Tapere d washer
2242 lb
3041 lb
Specify ASTM A325 high strength bolts 17500 psi
Top view
Side view
/
Required
lb
/
lb/in2
/4;
4 /
0.174 in2
4 0.174
/ l = 84 + 2(1.83) l = 87.66 in
0.470 in Use – ½ ‐ 13 UNC bolts. No threads in the shear plane. 5.
6 Bolts, A307, 1400 lb/side of bracket lb
Shear:
1400 √2
Figure P205
233 lb/Bolt 13 in
3
18 200 lb∙in
3.61 in;
4 3.61 in
2.00 in 60 in2
2 2.00 in
13 in to load
Bolt ① ∙
Total
.
1
1095 lb
Fh = 608 lb
1296 lb (Resultant of Fv and Fh)
As = F/d =
/ .
= 0.130 in =
/4
0.406 in
Use six 7/16‐14 Bolts on each side of bracket
Fv = 1144 lb F = 1296 lb
400
6.
Vertical component of force in each cable must be 2000 lb 25°
Figure P20‐6
2000 lb
2000 lb/
25°
sin 25°
2207 lb
2207 sin 25°
933 lb Circular pattern; 8 Bolts, r = 3.00 in; A325 8 3.0
72 in2
On each bolt, Fi =
=
∙
.
= 2500 lb
Shear‐horizontal
933
/8
117 lb ←
Shear‐vertical
2000 lb /8
250 lb ↓
√2750
117 /
4 0.157
One of four brackets
2753 lb 0.157 in2 /
/4
0.448 in
Use eight ½‐13 Bolts
Moment about center of bolt pattern: M = (2000 lb)(30 in) = 60 000 lb∙ in
2750 lb = Net vertical force
401
Welded Joints 7.
Figure P20‐1: Direct shear;
/
; V = 12 000 lb
Weld both vertical sides, 4.0 in long: lb
2 4 in
1500 lb/in
. in
For A36 steel and E60 electrode; / /
8.
8.0 in
9600 lb/in/in of leg [Table 20‐3]
0.156 in;
/
Use 3/16 in leg
From problem 4 –At each end of the beam, V = 1500 lb T = 32 873 lb∙in – Shared equally on front and back of plates that are welded to S7x153 column. Weld along top and bottom of plate. Case 3 – Fig. 20‐8: 3 37.45 in
2
/6
2 3.66 in
ch
7.32 in
3 3.66 4.00
3.66
1.83 in
A
/6
cv = 2.0 d = 4.00 in
Torsion
On each weld pattern: V = 750 lb, T = 16 436 lb ∙ in lb
At Ⓐ:
.
102
in ∙
in
.
878
. . .
Vector sum of forces:
b= 3.66 in
↓
804
→ 87 8
↓
1211
804
Use E60 electrode: fa = 9600 lb/in/in
906
/
102
/ /
0.131 in
Use w = 3/16 in = 0.188 in – minimum for ½ in plate.
402
fR = 1211 lb/in
9.
Figure P20‐5: Trim sides so that 2.00 in. extend onto rigid structure. b= 2
W Weld top and bottom only. Each side carries 1400 lb Torque on weld:
1400 8
/2 = 12 600 lbin d = 8.0 in
2
2 2 in
4.00 in 65.3 in 350
. ∙
↓
.
771
. .
193
.
Case 3 – Figure 20-8 →
↓
E60 Electrode
771 lb/in 193
/ / /
543
0.098 in
943 lb/in
350
[T Use w = 3/16 in = 0.188 in (min) 10.
Figure P20‐6 2 10
A
20 in
667 in
Weld back side
At Ⓐ
Weld
Figure 20-8 – Case 2
Shear
933 lb/20 in
46.7 lb/in ←
Shear
2000 lb/20 in
100 lb/in ↓
Torsion
450
403
↓
Torsion
450
497
497 lb/in
550
ft = 741 lb/in
550 lb/in 741 lb/in 0.077 in – Use w = 3/16 in (min) 11.
Figure P20‐11; Case ⑥, Figure 20‐8 2
9
2 6.5
.
b= 90
22.0 in 1.92 =
1.92 in
.
.
.
22.3 in
.
887
461
bottom
A
426 in
At Ⓐ: Shear
191
.
.
Bending
.
↓ 4056
into wall .
Torsion .
Torsion 191
209
Required
213 209
← ↓
400 lb/in
Resultant = 4081 lb/in
.
.
√4056 Use E70 electrode 0.364 in;
Use 3/8 in = w
404
6.5 = d
ch = 4.5 0
cv = 4.58
213
400
12.
Figure P20‐11; With shear
0; No torsion 136
.
↓
.
bending
2892
.
2892
√136
(Into wall)
2896
Use E60 electrode: /
0.302 in – Use 5/16 in = 0.312 in
/
13.
Figure P20‐1
P = 4000 lb; 6061 Aluminum
Direct shear – data from Table 19‐2 Throat width
.
Shear area
0.707
For 4043 filler alloy, Let
2 4.00
Required
Length of weld 5000 psi
8.00 in Vertical sides of bracket .
.
.
/
0.141 in
Use 3/16 in weld (min) 14.
Figure P20‐14
Direct shear, 6061 Aluminum, 4043 Filler alloy
Length of weld = As in Problem 13:
4.50 .
14.1 in .
.
/
0.030 in
Use w = 3/16 in. Could also consider partial weld. 15.
Figure P20‐15 Let
3/16 in
Direct shear, 6063 Aluminum, 4043 Filler 0.188 in
Solve for L
405
.
.
1.20 in
.
Distribute partial welds totaling 1.20 in equally on both sides of tab. 16.
Figure P20‐16
Direct shear, 3003 Aluminum, 4043 Filler alloy
As in Problem 14: Required
.
5000 psi 2
2 .
2.0
12.57 in
.
Weld all around
0.225 in
/
Use w = ¼ in = 0.250 in 17.
Material Comparison F = 4800 lb; N = 2; Required
/
/2 ;
4 /
Compute weight per inch of length 1.0 in
ksi
ksi
in
in
in
lb/in
lb
a) 1020 HR
30
15
0.320
0.638
0.32
0.283
0.0906
b) 5160 OQT 1300
100
50
0.096
0.350
0.096
0.283
0.0272
c) Aluminum 2014‐T6
60
30
0.160
0.451
0.16
0.100
0.0160
d) Aluminum 7075‐T6
73
36.5
0.132
0.409
0.132
0.100
0.0132
e) Ti‐6Al‐4V (annealed)
120
60
0.080
0.319
0.080
0.160
0.0128
f) Ti‐3Al‐13V‐11Cr
175
87.5
0.055
0.264
0.055
0.160
0.0088
Material
406
CHAPTER 21 ELECTRIC MOTORS AND CONTROLS Questions 1‐8: see Sections 21‐2 and 21‐3. 9.
Standard frequency for AC power in the U.S. 60 hertz.
10.
Standard frequency for AC power in Europe is 50 hertz.
11.
Single phase AC power at 115 and 230 volts.
12.
Two conductors plus a ground wire.
13.
480V, three‐phase is preferred because the current would be lower and the size of the motor would be smaller.
14.
Synchronous speed is the speed at which an AC motor tends to run at zero load. ns = 120 (f)/p, where f is the frequency of the power and p is the number of poles in the motor.
15.
Full‐load speed is the speed of the motor when it is delivering its rated torque.
16.
In U.S. :
120
In France: 17.
/
120
120 60 /4 /
1800 rpm
120 50 /4
1500 rpm
2‐pole motor. Zero‐load speed approximately 3600 rpm.
18.
120
/
120 400 /4
1200 rpm
19.
Two speed motor; 1725 rpm and 1140 rpm
20.
Variable frequency control
21, 22, 24.
See Section 21‐8.
23.
National Electrical Manufacturers Association
25.
TEFC – Totally enclosed – fan‐cooled. See Section 21‐8.
407
26.
TENV – Totally enclosed – non‐ventilated. Section 21‐8.
27.
TEFC – XP – Explosion proof motor; Enclosure – NEMA Design 9 ‐ Hazardous locations (Table 21‐6). Flour can explode.
28.
TENV because motor may be bathed in water during cleaning and to protect food from contaminants from the motor.
30.
Locked rotor torque is the torque that a motor can exert when the rotor is at rest. Also called starting torque.
31.
A poorer speed regulation means that the motor would slow down more when subjected to an increase in torque.
32.
Breakdown torque is the maximum torque a motor can develop during the increase in speed after start or the torque at which a motor would be stalled if the torque is increased after it is running.
33.
Split‐phase; capacitor‐start; permanent‐split capacitor; shaded pole.
34.
a)
Single phase, split‐phase AC motor because of the moderate starting torque and the change of torque when the switch cuts out the starting winding.
b)
From Table 21‐2, full‐load speed = 1140 rpm 63000
c)
63000 0.75 /1140
Starting torque = 150% (F.L. Torque) 1.5 41.4
d)
/
62.2 lb∙in approximate
Breakdown = 350% (F.L. Torque) 3.5 41.4
145 lb∙in approximate
408
41.4 lb∙in
35.
2‐pole; 1.50 kW rated power. b)
From Table 21‐2, full‐load speed = 3450 rpm 3450 rev/ min 2π rad/rev 1 min/60 s /
36.
1.5 103 N∙m/s / 361 rad/s 1.5 4.15 N∙m
c)
Starting torque
d)
Breakdown torque
361 rad/s
4.15 N∙m
6.23 N∙m
3.5 4.15 N∙m
14.5 N∙m
Fan requires about 18 lb∙in of torque at 1725 rpm; low starting torque; assume fan cools motor. Recommend 4‐pole, single phase permanent split capacitor AC motor. Power
37.
/63 000
18 1725 /63 000
0.49 hp use 1/2 hp
Full load torque about 0.5 N∙m at 3450 rpm; high starting torque (about 2.8xF.L.T.) due to starting compressor against high pressure in the system. Recommend capacitor start, single phase, 2‐pole, AC motor. 3450 rev/ min
2π rad/rev 1 min /60 s
0.5 N∙m 361 rad/s 38.
181 N∙m/s
361 rad/s 181 watts
Speed is adjusted by varying the resistance in the rotor circuit through an external resistance control.
39.
F.L. speed = synchronous speed = 720 rpm. (Table 21‐2)
40.
Pull‐out torque is the torque that would disengage the motor from its synchronous speed and cause it to stop.
41.
Universal motors are very small and light weight for a given power rating. Some vacuum cleaners, appliances, and hand tools utilize the high speed of rotation effectively.
409
42.
A universal motor can operate on DC or almost any frequency of AC voltage when operating near its full‐load point.
43.
Batteries, generators, rectified AC, hydrogen fuel cells.
44.
See Table 21‐7.
45.
SCR – Silicon controlled rectifier. Used to produce DC power from AC.
46.
SCR controls do not produce pure DC power; it has some variation, called ripple due to the AC input. A low‐ripple control would produce a nearly true DC power.
47.
Could use a 90V DC motor powered from a NEMA Type K SCR power supply to convert 115V AC to 90V DC power.
48‐50.
See Section 21‐11.
51.
The motor would speed up without limit and may fail catastrophically.
52.
Speed is proportional to torque.
15.0 N∙m 3000/2200 53, 56‐61.
/ 20.5 N∙m
See Sections 21‐9, 21‐11 and 21‐12.
54.
NEMA Size 2 motor starter for 10 hp, 220V AC, 3‐phase.
55.
NEMA Size 1 motor starter for 1.0 kW, 110V AC, single phase.
410
CHAPTER 22 MOTION CONTROL: CLUTCHES AND BRAKES 1.
From EQ. 22‐1:
/
63 025 5.0 2.75 /1750
495 lb∙in C and K from
Section 22‐4. Summary for problems 2‐7.
8.
Problem
C
P
K
m
Torque
2
5252
75 hp
5.0
2500
788 lb∙ft
3
63 025
0.50 hp
1.5
1150
41 lb∙in
4
63 025
5.0 hp
2.75
180
4814 lb∙in
5‐1
63 025
5.0 hp
1.0
1750
180 lb∙in
5‐2
5252
75 hp
1.0
2500
158 lb∙ft
5‐3
63 025
0.50 hp
1.0
1150
27.4 lb∙in
5‐4
63 025
5.0 hp
1.0
180
1751 lb∙in
6
9549
20 kW
2.75
3450
152 N∙m
7 (a)
9549
50 kW
4.0
900
2122 N∙m Clutch
7 (b)
9549
50 kW
1.0
900
531 N∙m Brake
Disk: D = 24.0 in; R = 12.0 in; L = 2.50 in; R2 = 0; Steel .
.
.
.
∆ .
lb∙ft
160 lb ∙ ft
. lb∙ft
411
9. R1
R2
L
Wk2
Shaft
0.625
0
16.0
0.007 54
Coupling
1.50
0.625
2.25
0.0341
Bearing 1
1.00
0.625
1.80
0.0047
Hub
2.00
0.625
1.00
0.0489
Gear
6.00
0.625
3.00
12.0023
Bearing 2
1.00
0.625
1.80
0.0047
Total
12.102 lb∙ft2
∆
.
. lb∙ft
.
10.
Neglect clutch and short shaft between clutch and gear A. Speed of Shaft 2: In EQ. 22‐4
.
1750
/
.
/
466.7 rpm 466.7/1750
0.0711
R1
R2
L
Wk2
W
Gear A
2.00
0
3.00
0.1482
0.1482
Gear B
7.50
1.25
3.00
29.2833
2.0824
Shaft 2
1.25
0
54.0
0.4070
0.0289
Hub 1
2.50
1.25
5.00
0.5653
0.0402
Hub 2
2.50
1.25
5.00
0.5653
0.0402
End Plates
9.00
1.25
2.00
40.4974
2.8798
Hollow Cyl.
9.00
7.50
30.0
314.6284
22.3736
Total
27.5933 lb∙ft2
∆ 308
27.5933 1750 308 1.50
. lb∙ft
412
11.
Load Speed
50 ft/ min v
Drum Speed
50 ft min
.
rad min
rpm of drum
rev 2π rad
150
rad min
ω
23.87 rpm
R1
R2
L
wk2
Shaft
0.75
0
24.0
0.0234
End Plates
4.00
0.75
3.0
2.3682
Hollow Cyl.
4.00
3.00
13.0
7.0238
Total
9.4154 lb∙ft2
600 lb
50 ft/min 150 rad/min
66.6667 lb∙ft
66.6667
76.08 lb∙ft 2
Load: Total;
9.4154
Braking Torque
∆ 308
Additional Torque to Hold Load Total Brake Torque 12.
12 in ft
(a)
76.08 23.87 308 0.25
23.6 lb∙ft
600 lb 4 in / 12in /1ft
200 lb∙ft
223.6 lb∙ft
Clutch on Motor Shaft: Motor Speed = Clutch Speed = 1150 rpm = nc Barrel Speed = 38 RPM = n (n/nc)2 = (38/1150)2 = 0.001092
413
[Solution continued on following page.]
R1
R2
L
Wk2
Worm
1.75
0
8.00
0.2316
0.2316
Worm Gear
8.00
0
2.50
31.6147
0.0344
2 Hubs
4.00
2.00
4.00
2.9639
0.0032
End Plates
14.00
2.00
4.00
474.2204
0.5178
Hollow Cyl.
14.00
12.00
18.00
982.5255
1.0728
Shaft
2.00
0
36.0
1.7783
0.0019
Total
1.8617 lb∙ft2
∆
.
3.48 lb∙ft
.
(b)
Clutch on worm gear shaft. Only barrel and hubs are accelerated to 38 rpm = clutch speed. Wk2 2 Hubs
2.9639
End Plates
474.2204
Hollow Cyl.
982.5255
Shaft
1.7783
Total
1461.5 lb∙ft2
∆
.
. lb∙ft [Note: Tb/Ta = 90.2/3.48 = 25.9]
.
13.
;
Using Eq. 22‐8:
Let: A
hp
1.37 hp [Equation 22‐10]
0.10 hp/in2
/
1.37 hp / 0.10 hp/in2
/
1.5
1.5 1.60 Similarly, For 1.45 in;
13.7 in2
/2
: 2
1.5
2.5
2.40 in;
2.40
1.75 2.55 in;
2.00 in
[Equation 22‐11]
; Try:
75 lb∙in / 0.25 150 lb
/
.
in
414
2 .
;
.
1.60
.
10.05 low
1.60 in
14.
From Problem 9:
64 lb∙in; Try .
.
1.50 in;
0.25
170 lb
.
0.79 hp For:
0.10
Try:
1.50 2
7.9 in
; ;
/2
1.5
2.5
/1.25
1.50 in / 1.25 5.65 in
Try:
2.0
; 2
1.5 1.2
3.0
2.0 in;
2.0
; 1.0
/1.5 .
in2
From Equation 22‐13: sin α
cos α
∙
sin .
°
cos
.
°
.
in ft
lb 16.
64 lb∙in; sin
17.
0.25; °
.
cos
.
°
12 in/ft
2
2 1800 12.0 . .
1800 lb∙in
.
/2
300 lb in
lb
.
18.
12°
lb
150 lb∙ft /
2.0 in;
.
For self‐actuation; W<0. For
0;
0
1.80 in
low
2.0
2.0 1.0 15.
1.20 in;
/
4.0 in / 0.25
16.0 in For self‐actuation 415
16.0 in
1.0 in
19.
100 lb∙ft Try:
12 in/ft
10.0 in;
1200 lb∙in 0.25
240 lb
In Fig. 22‐17 (c); Let:
2.0 in;
6.00 in; .
.
.
100 lb ∙ ft
20.
18 in
. lb
12 in/ft
1200 lb∙in
Select woven asbestos; In Figure 22‐19:
0.25;
0.25;
6.0 in;
30 psi 9.00 in;
20.0 in;
45°;
135°
From Equation 22‐18: lb∙in cos
Use:
cos
.
.
3.25 in;
30 psi
135°
90°
45°
/ .
cos 45° cos 135°
3.14 in
29.0 psi
.
1.57 rad
°
(Equation 22‐20) 0.25 29.0 3.25 6.0 9.0 2 1.57
sin 270°
sin 90°
lb∙in (Equation 22‐21) 0.25 29.0 3.25 6.0 6.0 cos 45°
cos 135°
0.25 9.0 cos 270°
lb∙in (Equation 22‐19) /
6540
1200 /20
267 lb Actuation Force
[Solution continues on following page.]
416
cos 90°
Check wear ratio /63 000
1200 lbin 480 rpm /63 000
.
hp
(Equation 22‐23) 2
sin hp
.
hp/in2
.
. in2
21.
Use woven asbestos,
25.0 psi;
375 lb 5.50 in;
.
.
275 lb
2.5 in 375 lb
150 lb 6.0 in
1350 lb∙in High
210°;
2.0 in
5.50 in 2.0 in
275 lb
110 lb 110 lb 5.5 in
980 lb∙in Ok
For a simple band brake: Let /
0.25
150 lb
.
25.0 lb/in
350 RPM
6.0 in 2.5 in
25.0 lb/in .
900 lb∙in ;
210° 3.67 rad ;
6.0in;
Try:
Somewhat high – intermittent service only
75 lb∙ft 12 in/ft
Band Brake:
Try:
27.6 in2
2 3.25 6.0 sin 45°
5.50 in;
12.0 in
110 lb 5.5 in/12 in
. lb
2π 5.5 in 2.0 in
40.3 in2
Wear Ratio: 2
∙
/
5.04 hp / 40.3 in
5.04 hp .
417
hp/in2 Ok
22
From Figure P22‐22: Solid steel cylinder a. Mass moment of inertia,
w = density volume =
;
= 0.283 lb/in3 ; V = Area Length; g = 32.2 ft/s2
Cylinder has three different diameters; 4.00 in, 8.00 in, 16.00 in. I ‐ Total length of 4.00 in diameter sections = LI = 12.00 in II ‐ Total length of 8.00 in diameter sections = LII = 16.00 in III ‐ Total length of 16.00 in diameter sections = LIII = 80.00 in Area = D2/4; Then w = (D2/4)L WI = [(4.00 in)2/4](12.00 in)(0.283 lb/in3) = 42.17 lb WII = [(8.00 in)2/4](16.00 in)(0.283 lb/in3) = 227.6 lb WIII = [(16.00 in)2/4](80.00 in)(0.283 lb/in3) = 4 552.0 lb Total weight = 4 822.3 lb II =
.
=
III =
=
IIII =
=
.
. .
/ . .
/
0.018 lb ∙ ft ∙ s = 0.22 lbins2
/
0.40 lb ∙ ft ∙ s = 4.70 lbins2 31.4 lb ∙ ft ∙ s = 377 lbins2
Itotal = 381.9 lbins2 b. Required deceleration rate, , to stop the cylinder from 800 rpm to zero in 5.0 s: =
∆ ∆
= ‐160 rpm/s
= ‐16.76 rad/s2 =
c. Required braking torque to sop the cylinder in 5.0 s or less: Torque = Tbrake = Itotal = (381.9 lbins2) (16.7 rad/s2) = 6399.1 lbin = Tbrake
418