Introduction to Design of Machine Elements Levers Bell Crank Lever Eccentric Loading

1. Unit 1
Design of Simple Machine Elements
Prepared By
Prof. M.C. Shinde [9970160753]
Mech. Engg. Dept., JSCOE, Hadapsar

2. SPPU Syllabus Content
Machine Design, Design cycle, Design considerations - Strength,
Rigidity, Manufacture, Assembly and Cost, Standards and codes, Use
of preferred series, Factor of safety, Service factor. Design of Cotter
joint, Knuckle joint, Levers - hand / foot lever, lever for safety valve,
bell crank lever, and components subjected to eccentric loading.

3. What is Machine?

4. What is Machine Design?
Machine design is defined as the use of scientific principles, technical
information and imagination in the description of a machine or a
mechanical system to perform specific functions with maximum
economy and efficiency.
A design is created to satisfy a recognised need of customer.

5. Basic Procedure of Machine Design

6. Basic Procedure of Machine Design

7. Basic Requirements Of Machine Elements
1. Strength:
2. Rigidity:
3. Wear Resistance
4. Minimum Dimensions and Weight:
5. Manufacturability:
6. Safety:
7. Conformance to Standards:
8. Reliability:
9. Maintainability:

8. Basic Procedure Of The Design Of Machine
Elements

9. Unit 1
Design of Simple Machine Elements
Session 1.2 Use of Standards in Design
Prepared By
Prof. M.C. Shinde
Mech. Engg. Dept., JSCOE, Hadapsar

10. Use of Standards in Design
Standardization is defined as obligatory norms, to which various
characteristics of a product should conform

11. Standards are used in Mechanical
Engineering Design
(i) Standards for Materials, their Chemical Compositions, Mechanical
Properties and Heat Treatment .
For example, Indian standard IS 210 specifies seven grades of grey cast iron
designated as FG 150, FG 200, FG 220, FG 260, FG 300, FG 350 and FG 400
(ii) Standards for Shapes and Dimensions of Commonly used Machine
Elements The machine elements include bolts, screws and nuts, rivets, belts
and chains, ball and roller bearings, wire ropes, keys and splines, etc.
For example, IS 2494

12. Standards are used in Mechanical
Engineering Design
(iii) Standards for Fits, Tolerances and Surface Finish of Component For
example, selection of the type of fi t for different applications is illustrated in
IS 2709 on ‘Guide for selection of fits’.
(iv) Standards for Testing of Products These standards, sometimes called
‘codes’, give procedures to test the products such as pressure vessel, boiler,
crane and wire rope, where safety of the operator is an important
consideration.
For example, IS 807 is a code of practice for design, manufacture, erection
and testing of cranes and hoists.

13. Standards are used in Mechanical
Engineering Design
(iii) Standards for Fits, Tolerances and Surface Finish of Component For
example, selection of the type of fit for different applications is illustrated in
IS 2709 on ‘Guide for selection of fits’.
(v) Standards for Engineering Drawing of Components For example, there is
a special publication SP46 prepared by Bureau of Indian Standards on
‘Engineering Drawing Practice for Schools and Colleges’ which covers all
standards related to engineering drawing.
(vi) International standards These are prepared by the International
Standards Organization (ISO).

14. Selection Of Preferred Sizes
The ‘size’ of the product is a general term, which includes different
parameters like power transmitting capacity, load carrying capacity, speed,
dimensions of the component such as height, length and width, and volume
or weight of the product.
There are five basic series, denoted as R5, R10,
R20, R40 and R80 series

15. Numerical on Preferred Sizes
Example 1.1 Find out the numbers of the R5 basic series from 1 to 10.
Solution:-
Step I Calculation of series factor
Step II Calculation of numbers
First number = 1
Second number = 1 *(1.5849) = 1.5849 = 1.6
Third number = (1.5849)*(1.5849) = 2.51 = 2.5
Fourth number = (1.5849)2 * (1.5849) = 3.98 = 4
Fifth number = (1.5849)3 * (1.5849) = 6.3
Sixth number = (1.5849)4 * (1.5849) = 10

16. Numerical on Preferred Sizes
Example 1.2 A manufacturer is interested in starting a business with five
different models of tractors ranging from 7.5 to 75 kW capacities. Specify
power capacities of the models. Specify the power capacities of the models.
Solution:-
Calculation of ratio factor
Rating of first model = (7.5) kW
Rating of second model = 7.5* (1.7783) = 13.34 = 13 kW
Rating of third model = 7.5 * (1.7783)2 = 23.72 = 24 kW
Rating of fourth model = 7.5*(1.7783)3 = 42.18 = 42 kW
Rating of fifth model = 7.5*(1.7783)4 = 75.0 = 75 kW

18. Assignment 1.2
Q.1 It is required to standardize eleven shafts from 100 to 1000 mm diameter.
Specify their diameters.
Q.2 Find out the numbers of R20/4(100, …, 1000) derived series
Q.3 What is standardization?

19. Factor Of Safety
The factor of safety is defined as
fs = failure stress / allowable stress

20. Factors affecting magnitude of factor of safety
(i) Effect of Failure
(ii) Type of Load
(iii) Degree of Accuracy in Force Analysis
(iv) Material of Component
(v) Reliability of Component
(vi) Cost of Component
(vii)Testing of Machine Element
(viii)Service Conditions
(ix) Quality of Manufacture

21. Unit 1
Design of Simple Machine Elements
Session 1.3 Design of Cotter Joint & Knuckle Joint
Prepared By
Prof. M.C. Shinde
Mech. Engg. Dept., JSCOE, Hadapsar

22. COTTER JOINT
A cotter joint is used to connect
rigidly two co-axial rods or bars which
are subjected to axial tensile or
compressive forces .
It is a temporary fastening.
A cotter is a flat wedge shaped piece
of rectangular cross section and its
width is tapered (either on one side or
on both sides) from one end to
another for an easy adjustment.

23. APPLICATIONS OF COTTER
1. Connection of the piston rod with
the cross heads
2. Joining of tail rod with piston rod of
a wet air pump
3. Foundation bolt
4. Connecting two halves of fly wheel
(cotter and dowel arrangement)

24. APPLICATIONS OF COTTER

25. Design of Cotter Joint

26. Design of Spigot and Socket Cotter

27. Design of Spigot and Socket Cotter

28. Design of Spigot and Socket Cotter

29. 1. Failure of the rod (Spigot) in tension

30. 2. Failure of spigot in tension across the
weakest section (or slot)

31. 3. Failure of the rod (spigot) or cotter in
crushing
P=d2 x t x σc

32. 4. Failure of the socket in tension across the
slot

33. 5. Failure of cotter in shear

34. 6. Failure of the socket collar in crushing

35. 7. Failure of socket end in shearing

36. 8. Failure of rod(Spigot) end in shear

37. 9. Failure of spigot collar in crushing

38. 10. Failure of the spigot collar in shearing

40. Applications
• Joints Between The Tie Bars In A Roof Trusses.
• Between The Links Of Suspension Bridge.
• In Valve Mechanism Of A Reciprocating Engine.
• Fulcrum For The Levers.
• Joints Between The Links Of A Bicycle Chain.

43. Design of Knuckle Joint

44. 1. Failure of rod in tension

45. 2. Failure of knuckle pin in shear

46. 3. Failure of single eye in shear

47. 4. Failure of single eye in tension

48. 5. Failure of single eye in crushing

49. 6. Failure of forked end in tension

50. 7. Failure of forked end in shear

51. 8. Failure of forked end in crushing

52. Assignment 1.3
Q.1 What is Cotter Joint? Write any 5 steps for design of Cotter Joint.
Q.2 What are applications of Knuckle Joint. Write down any 4 steps in design of
Knuckle Joint.

53. Unit 1
Design of Simple Machine Elements
Session 1.4 Levers, Types and Design of Hand /Foot Lever
Prepared By
Prof. M.C. Shinde
Mech. Engg. Dept., JSCOE, Hadapsar

54. Lever is rigid rod or bar pivoted at a point called
fulcrum and used to overcome load by application
of small effort, to facilitate application of effort in
desired direction
A lever consists of a rod or bar that rests and turns
on a support called a fulcrum.
Levers

56. Terminology
• Simple Machine- a machine with one moving part
• Lever –a beam, bar, rod that turns or rotates on or around a fixed point
• Fulcrum – a fixed point that allows the beam to rotate around it.
• Work – the use of force to move an object
• Force –any kind of push or pull on an object
• Effort – the force that is used to do the work
• Resistance – the force (load) that works against the effort
• Load –the object or resistance being moved by the effort
• Friction – the force that is caused when 2 surfaces rub together as an object
moves
• Mechanical Advantage – how the simple machine increases the effort

57. Mechanical Advantage
Mechanical Advantage=Load/Effort

58. Types of levers

59. 1.Lever with fulcrum between load and effort
Mechanical Advantage>1
Mechanical Advantage=Load/Effort

60. 2.Lever with load between fulcrum and effort
Mechanical Advantage>1
Mechanical Advantage=Load/Effort

61. 3.Lever with effort between fulcrum and load
Mechanical Advantage<1
Mechanical Advantage=Load/Effort

62. 4.Angular Lever
Rocker arms in IC engines & brake levers

63. 5. Bell Crank Levers
Levers in Railway Signals & Governors

64. Design of Bell Crank Lever

65. Design of one arm Lever
(Hand Lever or Foot Lever)

66. Design of one arm Lever
(Hand Lever or Foot Lever)

67. A foot lever has 1m length between the centre of the shaft and point of application of force.the
maximum force exerted by the foot is 800N.the lever has a rectangular cross section with its
depth as 3 times its thickness. The allowable tensile stress is 73N/mm2, while allowable shear
stress is 70 N/mm2,determine 1) the diameter of shaft, 2) the dimensions of the key, 3) the
dimensions of foot lever cross section at 60 mm from the centre of shaft

68. 1.Shaft in boss under Torsional shear stress
• To find diameter of shaft in boss of lever

69. 2.Size of Boss of Lever

70. 3. Shear & Crushing failure of key
For Square key, w=d/4, w=t

71. 4. Shaft in bearing under Torsion & Bending

72. 5. Lever Cross section Bending Stress

73. Eccentric Loading

74. • Centric loading:
• The load is applied at the centroid of the cross section.
The limiting allowable stress is determined from
strength (P/A) or buckling.
• Eccentric loading:
• The load is offset from the centroid of the cross
section because of how the beam load comes into the
column. This offset introduces bending along with
axial stress.

75. Eccentric Loading
C Clamp Hacksaws Blade Bracket

76. Ex. A ‘C’ frame subjected to a load of 10kN is shown in fig. it is
made of grey cast iron with allowable stress of 120 N/mm2.
determine the dimensions of cross section of frame.
• Resultant stress is maximum at inner most section and is
given by
Ans….t=12.9mm & b=64.5 mm

77. Ex. The hacksaw frame to be made of plain carbon steel 30C8 (Syt=300 N/mm2) is
shown in fig. the initial tension ‘P’ in the blade should be 300N. The cross section
of the frame is rectangular with depth to thickness ratio of 3.if required factor of
safety is 2.5. determine dimensions of the cross section.
• Resultant stress is maximum at inner most section and is
given by
Ans….t=6.3mm & d=18.9 mm

78. Ex. A bracket shown in fig. is subjected to a pull of 5kN acting at an
angle of 45 to vertical. The bracket has a rectangular section whose
depth is two times its thickness. If the permissible tensile stress is 55
N/mm2, determine the cross section of the bracket.
Assignment 1.4

79. Ex. A bracket shown in fig. is subjected to a pull of 5kN acting at an angle of 45 to
vertical. The bracket has a rectangular section whose depth is two times its
thickness. If the permissible tensile stress is 55 N/mm2, determine the cross
section of the bracket.
Ans….t=34.2mm & d=68.4 mm

80. Ex. Fig. shows a hanger with rectangular cross section .the force P acting on the
hanger is 6kN and acts at 30 to the vertical as shown. If the permissible stress in
the hanger material is 60MPa, determine the size of the cross section.
Ans….t=33mm & d=66 mm
Note : assume d=2t