1. Notes on Lesson Faculty Name : V. KIRAN KUMAR Code AU28 Subject Name
DESIGN OF MACHINE ELEMENTS
ME2303
Year
III
Semester V
Degree & Branch
B.E. – AUTOMOBILE
Section -

2. Failure Theories
Stress in machine components should be accurately computed.
Designer must understand material limits to ensure a safe design.

3. Design Factor Factor of Safety (N)
Suitable values depend on inherent danger, certainty of calculations, certainty of material properties, etc.

4. Static Stresses - Brittle Materials
Percent elongation < 5%
for parts in tension
for parts in compression
for parts with general stress

5. Example
The Gray Cast Iron (Grade 40) cylinder carries an axial compressive load of 75,000 lbs and a torque of 20,000 in lbs. Compute the resulting design factor.
Ø4.00”
Ø5.00”
R0.25”

6. Static Stresses - Ductile Materials
Percent elongation > 5%
Distortion Energy Theory
Define von Mises Stress
For nominal stress
For localized stress

7. Static Stresses - Ductile Materials
Percent elongation > 5%
Maximum Shear Stress Theory
For nominal stress
For localized stress

8. Example
Specify a diameter for the middle portion of the rod, if it is to be made from AISI 1040-hot rolled steel.
5000 lbs
450

9. Example
For the seat support shown, specify a standard structural tube to resist static loads shown. The tube has properties similar to AISI 1020 hot-rolled steel. Use a design factor of 3.
200 lb
400 lb
20”
14”

10. Repeated Loads
salt
smean

11. Example
The notched bar is machined from AISI 1020 steel. This bar is subjected to a load that varies from 2000 lb to 3000 lb. Determine the mean and alternating nominal stresses.
0.1” R 1”
1.25”
.75”

12. Fatigue Strength R.R. Moore Test Motor Alternating Stress, sa
103
104
105
106
107
108
Endurance Strength, sn
Cycles to Failure, N (log)
Alternating Stress, sa

13. Endurance Strength sn = Endurance strength Listed in tables
If no information is available, use
sn  0.5 su (Steel)
sn  0.4 su (Aluminum)

14. Adjusted Endurance Strength
The data from the standard R.R. Moore test is adjusted for a particular application.
sn’ = Adjusted endurance strength
= (Cs) (Cm) (Cst) (CR) (sn)

15. Size and Stress Type Factors
Cs = Size Factor
D< 0.4 in Cs = 1.0
0.4 < D 2.0 in Cs = (D/0.3)-0.068
2.0 < D 10.0 in Cs = D-0.19
For rectangular sections, D=.808(h b)1/2
Cst = Stress Type Factor
= 1.0 for bending
= 0.80 for axial tension
= 0.50 for torsion

16. Material and Reliability Factor
Cm = Material Factor
= 1.0 for wrought steel
= 0.80 for cast steel
= 0.70 for cast iron
CR = Reliability Factor
50% CR = 1.0
90% CR = 0.90
99% CR = 0.81
99.9% CR = 0.75

17. Example
The notched bar is machined from AISI 1020 steel. This bar is subjected to a load that varies from 2000 lb to 3000 lb. Determine the endurance limit of the material.
0.1” R 1”
1.25”
.75”

18. Repeated Stresses - Ductile Materials
Distortion Energy Theory
Define repeated von Mises Stress
Solderberg criterion

19. Repeated Stresses - Ductile Materials
Maximum Shear Stress Theory
ssy = 0.5 sy
s’sn = 0.5 sn

20. Example
The notched bar is machined from AISI 1020 steel. This bar is subjected to a load that varies from 2000 lb to 3000 lb. Comment on the robustness of the design.
0.1” R 1”
1.25”
.75”

21. Example
Comment on the robustness of a 1-1/4” round bar made from AISI 1213 C-D steel. It carries a constant tensile load of 1500 lbs, a bending load that varies from 0 to 800 lbs at the senter of the 48” length and a constant torque of 1200 in lbs.
48”

22. Shafts Connect power transmission components.
Inherently subjected to transverse loads and torsion.

23. Shaft Forces
Gears
As before Wr Wt T

24. Shaft Forces
Chains
Fslack = 0
Ftight D T

25. Shaft Forces
V-belts
Ftight D T
Fslack

26. Shaft Forces
Flat belts
Ftight D T
Fslack

27. Material Properties For steady load (torsion) sys=.5sy
For fatique load ( bending)
sn’=cs cR sn
cT = 1 (bending)
cm = 1 (wrought steel)

28. Stress Concentrations
Keyseats
Sled Runner Kt = 1.6
Profile Kt = 2.0
Woodruff Kt = 1.5

29. Stress Concentrations
Shoulders
Sharp, Bearing (r/d .03) Kt = 2.5
Round, Gear Bore (r/d .17) Kt = 1.5
Grooves
Retaining Rings Kt = 1.5
Try not to let Kt’s overlap.
Leave ” between

30. Strength Analysis Bending stress Torsion stress For round sections

31. Strength Analysis Mohr’s circle and Solderberg
Suggested Design Factors:
N=2 smooth operation
N=3 typical industrial operation
N=4 shock or impact loading

32. Minimum Acceptable Diameter
The designer must size the shaft.
Solve for appropriate diameters

33. Example
Determine a suitable diameter for a shaft made from AISI 1144 OQT It is subjected to a reversing bending moment of 3000 ft lbs and a steady torque of 1800 ft lbs. The shaft has a profile keyway.

34. Example The shaft shown is part of a grain drying system
At A, a 34 lb. propeller-type fan requires 12 hp when rotating at 475 rpm.
A flat belt pulley at D delivers 3.5 hp to a screw conveyor handling the grain.
All power comes to the shaft through the v-belt at C.
Using AISI 1144 cold drawn steel, determine the minimum acceptable diameter at C.

35. Example
12”
10” 4” A B C D E
Sheave C
150
Sheave D

36. Shafts Accessories
Components used to securely mount power transmitting elements on a shaft.
Axial
Rotational

37. Keys
Allow torque to be transferred from a shaft to a power transmitting element (gear, sprocket, sheave, etc.)

38. Key Design Use a soft, low strength material (ie, low carbon steel)
Standard size H=W=1/4 D
Design length
based on strength H L W

39. Standard Key Sizes
Shaft Dia. (in)
W (in) T S H W

40. Key Design
Key Shear
Failure Theory
Length

41. Example
Specify a key for a gear (grade 40, gray cast iron) to be mounted on a shaft (AISI 1144, hot rolled) with a 2.00 in. diameter. The gear transmits lb-in of torque and has a hub length of 4 in.

42. Retaining Rings Also known as snap rings
Provides a removable shoulder to lock components on shafts or in bores.
Made of spring steel, with a high shear strength.
Stamped, bent-wire, and spiral-wound.

43. Retaining Ring Selection
Based on shaft diameter & thrust force

44. Set Screws
Setscrews are fasteners that hold collars, pulleys, or gears on shafts.
They are categorized by drive type and point style.

45. Standard Set Screw Sizes

46. Set Screw Holding

47. Pins A pin is placed in double shear Holds torsion and axial loads
Hole is made slightly smaller than the pin (FN1 fit)

48. Example
Specify a pin for a gear (grade 40, gray cast iron) to be mounted on a shaft (AISI 1144, hot rolled) with a 2.00 in. diameter. The gear transmits lb-in of torque and has a hub length of 4 in.

49. Roll Pins
Easier disassembly

50. Collars Creates a shoulder on shaft without increasing stock size.
Held with either set screw or friction (clamped)

51. Mechanical Couplings Couplings are used to join two shafts
Rigid couplings are simple and low cost. But they demand almost perfect alignment of the mating shafts.
Misalignment causes undue forces and accelerated wear on the shafts, coupling, shaft bearings, or machine housing.

52. Mechanical Couplings
In connecting two shafts, misalignment is the rule rather than the exception. It comes from such sources as bearing wear, structural deflection, thermal expansion, or settling machine foundations.
When misalignment is expected, a flexible coupling must be used.

53. Mechanical Couplings Selection factors include:
Amount of torque (or power & speed)
Shaft Size
Misalignment tolerance

54. Fasteners, Powers Screws, Connections
Helical thread screw was an important invention.
Power Screw, transmit angular motion to liner motion
Transmit large or produce large axial force
It is always desired to reduce number of screws

55. Definition of important Terminologies
Major diameter d, Minor diameter dr Mean dia or pitch diameter dp
Lead l, distance the nut moves for one turn rotation

56. Single and Double threaded screws
Double threaded screws are stronger and moves faster

57. Screw Designations United National Standard UNS
International Standard Organization
Roots and crest can be either flat or round
Pitch diameter produce same width in the thread and space,

58. Coarse thread Designated by UNC
Fine Thread UNF, is more resistance to loosening, because of its small helix angle.
They are used when Vibration is present
Class of screw, defines its fit, Class 1 fits have widest tolerances, Class 2 is the most commonly used
Class three for very precision application
Example:1in-12 UNRF-2A-LH, A for Ext. Thread and B for Internal, R root radius
Metric M10x diameter mm major diameter,1.5 pitch

59. Some important Data for UNC, UNF and M threads
Lets Look at the Table 8-1 on Page 398

60. Square and Acme Threads are used for the power screw
Preferred pitch for Acme Thread
d, in
1/4
5/16
3/8
1/2
5/8
3/4
7/8 1
1 1/4
p,in
1/16
1/14
1/12
1/10
1/8
1/6
1/5

61. Mechanics of Power Screws

62. Used in design to change the angular motion to linear motion, Could you recall recent failure of power screw leading to significant causalities

63. What is the relationship between the applied torque on power screw and lifting force F

64. Torque for single flat thread
If the thread as an angle α, the torque will be
Wedging action, it increases friction

65. Stresses in the power Screw
Shear stress in the base of the screw
Bearing stress
Bending stress at the root of the screw
Shear stress in the thread
nt number of engaged thread

66. Loading to the fasteners and their Failure considerations

67. Bolts are used to clamp two or more parts It causes pre tension in the bolt Grip length is the total thickness of parts and washers d t

68. ld h t2
lt=L’- ld
L’ effective grip= h+t2 if t2<d
=h=d/2 for t2 d

69. Failure of bolted or riveted joints

70. Type of Joints
Lap Joint (single Joint) But Joint

71. Example 1

72. Example 2

73. Example 2

74. Example 3

75. Weld

77. Weld under Bending

81. Springs
Flexible machine elements
Used to:
Exert force
Store energy

82. Spring Rate Effective springs have a linear deflection curve.
Slope of the spring deflection curve is the rate
Force
Deflection k 1

83. Example
A compression spring with a rate of 20 lb/in is loaded with 6 lbs and has a length of 1.5 in. Determine the unloaded spring length (free length)

84. Geometry Wire diameter, Dw (Standard gages) Mean Diameter, DmDo Di Dw L
Wire diameter, Dw (Standard gages)
Mean Diameter, Dm
Dm = Do - Dw

85. Spring Parameters Spring index C > 5 (manufacturing limits)
Active coils, Na
= N for plain ends
= N-1 for ground ends
= N-2 for closed ends

86. Deflection Deflection for helical springs
G = Shear modulus
Spring rate for helical springs

87. Example
A helical compression spring is formed from 35 gage music wire with 10-1/4 turns and an O.D. if in. It’s ends are squared. The free length is 2 inches. Determine the force to press the spring solid.

88. Stress Analysis Spring wire is in torsion Wahl factor, KV T
Wahl factor, K
Accounts for the curvature of the wire

89. Example
A helical compression spring is formed from 35 gage music wire with 10-1/4 turns and an O.D. if in. It’s made from A228 and the ends are squared. The free length is 2 inches.
If the spring is repeatedly compressed to 1.3 in, do you expect problems?

90. Design Procedure Select a material Compute required spring rate
Estimate Dm based on size constraints
Determine required Dw (use K=1.2)
Select standard wire
Verify actual stress is satisfactory.
Compute number of coils required.

91. Example
Design a helical compression spring to exert a force of 22 lbs when compressed to a length of 1.75 in. When its length is 3.0 in, it must exert 5 lb. The spring will be cycled rapidly. Use ASTM A401 steel wire.

92. Rolling Element Bearings
Provides support for machine elements, while allowing smooth motion. m=

93. Types Single-row Radial Roller Radial Ball Angular Contact Ball

94. Types
Tapered Roller
Spherical
Roller
Needle
Thrust

95. Ball Bearings

96. Stress Analysis Contact Stress sc=300,000 is not unusual
Balls, rollers and races are made from extremely high strength steel
ex. AISI 52100
sy = 260,000 psi
su=322,000 psi

97. Empirical relationship:
Bearing Load/Life
Test (fatigue) data
Empirical relationship:
L10 Life (cycles)
Radial Load (lbs)
k=3.0 (ball)
k=3.33 (roller)

98. Example
A bearing is mounted on a shaft rotating at 1200 rpm. The bearing has been tested to have a L10 life of 300 hrs, when loaded with 500 lbs. Determine the expected L10 life, if the load is increased to 700 lbs.

99. Manufacturer’s Data
Vendors publish the Basic Dynamic Load rating (C) of a bearing at an L10 life of 1 million cycles.

100. Bearing Selection Determine the design life (in cycles)
Determine the design load
Pd = V R
Calculate the required basic dynamic load
Select a bearing with (C > Creq’d) and a bore that closely matches the shaft diameter.
V=1 for inner race rotation
V=1.2 for outer race rotation

101. Example
Specify suitable bearings for a shaft used in an grain dryer. The shaft rotates at 1700 rpm. The required supporting loads at the bearing are
and the minimum acceptable diameter is 2.16”.
RBx=589 lb
RBy=164 lb

102. Mounting of Bearings Shaft/bearing bore has a light interference fit.
Housing/outer race has a slight clearance fit.
Check manufacturers catalog
Match maximum permissible fillet radius.
Shaft or housing shoulders not to exceed 20% of diameter.

103. Mounted Bearings Pillow block
Bearing is inserted into a cast housing, with base or flange slots, which can be readily attached to a machine base.

104. Bearings with Varying Loads
Compute a weighted average load based on duty cycle.
Fm=equivalent load
Fi= load level for condition i
Ni= cycles for condition i
p = exponent for load/life

105. Example
Bearing 6211 is carrying the following load cycle, while rotating at 1700 rpm.
Stage Load (lbs) Time (min)
Compute the bearing L10 life in minutes.

106. Radial & Thrust Loads Calculate an equivalent load P=VXR +YT
T=thrust load
X factors depending
Y on bearing =

107. Thrust factors, Y Deep -groove, ball bearings
X = 0.56 for all values of Y

108. Example
A bearing is to carry a radial load of 650 lb and a thrust load of 270 lb. Specify a suitable single-row, deep-groove ball bearing if the shaft rotates at 1150 rpm and the design life is 20,000 hrs.