1. Design of Machine Element
Prepared by: Prof. R. V. Vichare
TSSM’s
Padmabhooshan Vasantdada Patil Institute of Technology, Bavdhan
Pune
Department of Mechanical Engineering
2. 1. Ability to identify and understand failure modes for
mechanical elements and design of machine elements
based on strength.
2. Ability to design Shafts, Keys and Coupling for
industrial applications.
3. Ability to design machine elements subjected to
fluctuating loads.
4. Ability to design Power Screws for various applications.
5. Ability to design fasteners and welded joints subjected to
different loading conditions.
6. Ability to design various Springs for strength and
stiffness.
Course Outcome
3. UNIT 1: Design of Simple Machine Elements
UNIT 2: Design of Shafts, Keys and Couplings
UNIT 3: Design for Fluctuating Load
UNIT 4: Power Screws
UNIT 5: Threaded joints and Welded joints
UNIT 6: Mechanical Springs
..TE-MECH-credit-system-2015-Pattern-
COURSE_SYLLABUS.pdf
Books Required
1)Bhandari V.B., Design of Machine Elements, Tata McGraw Hill Publication Co.
Ltd. 2)Shigley J.E. and Mischke C.R., Mechanical Engineering Design, McGraw
Hill Publication Co. Ltd.
3)Spotts M.F. and Shoup T.E., Design of Machine Elements, Prentice Hall
International. 4)Juvinal R.C., Fundamentals of Machine Components Design, John
Wiley and Sons
Syllabus
4. Unit 1 Design of Simple
Machine Element
Prepared by: Prof. R. V. Vichare
TSSM’s
Padmabhooshan Vasantdada Patil Institute of Technology, Bavdhan
Pune
Department of Mechanical Engineering
5. 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.
Content
6. What is Design????
• Design is to formulate a plan satisfy a particular
need and to create something with physical reality.
• Realization of a concept or idea into a configuration.
• Design is the creation of a plan or
convention for the construction of an
object, system or measurable human
interaction .
7. What is Machine Design???
• Machine is a combination of several machine
elements arranged to work together as a whole to
accomplish specific purpose.
• Machine Design involves designing the elements and
arranging them optimally to obtain some useful work.
• Machine design is the process of engineering
design. A machine is made up of mechanisms that
work together to satisfy the requirements of what the
machine needs to accomplish.
9. Factorsto be considered inMachine
Design…
The method of manufacturing the components and their
assembly.
How wil it operate.
Reliability and safety aspects.
Inspectibilty
Maintenance
Cost and aesthetics of the designed product.
10. STEP: 1
STEP: 2
STEP: 3
STEP: 4
STEP: 5
STEP: 6
STEP: 7
STEP: 8
General Procedure in Machine Design
Analysis of Forces
Selection of Material
Modification
Detailed Drawing
Design of Element (Size, Shapes
and Stresses)
Synthesis (Mechanism)
Production
Define the Need
11. • Standardization is defined as obligatory (or compulsory)
norms, to which various characteristics of a product should
comply (oragree) withstandard.
• The characteristics include materials, dimensions and shape
of the component, method of testing and method of
marking, packing and storingof theproduct.
• A standard is defined as a set of specifications for parts,
materials or processes. The objective of, a standard is to
reduce the variety and limit the number of items to a
reasonable level.
Standardization
12. National Standards:
– India - BIS(Bureau of Indian Standards),
– Germany - DIN (Deutsches Institut für Normung),
– USA - AISI(American Iron and Steel Institute) or SAE (Society of
Automotive Engineers),
– UK -BS(British Standards)
International Standards: These are prepared by the
International Standards Organization (ISO).
Standardization
13. Standards forMaterials,theirchemical 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. The
number indicates ultimate tensilestrength in N/mm2.
Standards for Shapes and dimensions of commonly used Machine
Elements:
The machine elements include bolts, screws and nuts, rivets, belts and
chains, ball
and rollerbearings,wireropes, keysand splines,etc
For example, IS2494(Part 1)specifies dimensions and shape of the cross-
section of endless V-beltsforpower transmission.
Thedimensions of the trapezoidal cross-section of the belt, viz.width, height
and
included angle are specified in thisstandard
Standards are Used in Mechanical Engineering System
14. Standards forFits,Tolerances and SurfaceFinishof
Component:
For example, selection of the type of fitfor different applications isillustratedinIS
2709on 'Guide forselection offits'.
Thetolerances orupper and lower limitsforvarioussizesof holes and shaftsare
specified inIS919on 'Recommendations forlimitsand fitsforengineering'.
IS10719explainsmethod for indicating surface texture on technical drawings.
StandardsforTesting of Products:
Thesestandards,sometimescalled 'codes', give procedures to testthe
products
suchas pressurevessel,boiler, crane and wirerope, wheresafety
of the operatorisan important consideration.
Forexample, IS807isa code of practice fordesign, manufacture,
erection and testingof cranes and hoists.
15. Reductions intypes and dimensions of identical components
(inventorycontrol).
Reduction inmanufacturing facilities.
Easy to replace (Interchangeability).
No need to design ortestthe elements.
Improvesqualityand reliability.
Improves reputation of the company which manufactures
standard
components.
Sometimesitensuresthe safety.
It resultsinoverallcost reduction.
Benefits ofStandardization
16. With the acceptance of standardization, there is a need to
keep the standard sizes or dimensions of any component or
product indiscrete steps.
The sizes should be spread over the wide range, at the same time
these should be spaced properly.
For example, if shaft diameters are to be standardized between 10
mm and 25 mm, then sizesshould be like :10 mm, 12.5mm, 16mm,
20mm,25mmand not like:10mm,11mm,13mm,18mm,25mm.
This led to the use of geometric series known as series of preferred
numbersorpreferred series.
Preferred series are series of numbers obtained by
geometric progression and rounded off.
Preferred Numbers
17.
18. • The difference in two successive terms has a fixed
percentage.
• It provides small steps for small quantities and
large steps forlarge quantities.
• The product range is covered with minimum
number of size without restricting the choice of the
customers.
AdvantagesPreferred Numbers
19. While designing any machine part, it is necessary to keep the stress lower than
the maximum or ultimate stress at which failure of the material takes place.
This stress is called as Design Stress or Working stress.
Hence, FOS is defined as the ration of the maximum stress to the working
stress.
Factor of Safety (FOS)
20. 2 Variable or Live Load
3 Shock or Sudden Load
4 Impact Load
The mechanical components or structural members are subjected to
different external loads during the operations.
These external loads may be either forces, torques etc.
Types loads are as follows:
Static or Steady or Dead Load
1
Types of loads
21. Variable or Live Load
The load which does not
change in magnitude, direction
and point of application with
respect to time.
The load which varies in
magnitude, direction and point
of application with respect to
time.
Static or Steady or Dead Load
22. Impact Load
It is suddenly applied
load with some velocity
There is rapid build up
od stresses.
When a load is applied
from a certain height or a
distance, its intensity
increases
Shock or Sudden Load
23. • When amaterialissubjectedto anexternalforce,aresistingforceis
setupwithin thecomponent,this internalresistanceforceperunit
areaiscalledstress.
• SI unit isN/m²(Pa).1kPa=1000Pa, 1MPa=10^6Pa,1
Gpa=10^9Pa,1TerraPascal=10^12Pa
• In engineeringapplications,we usethe
theoriginalcrosssectionareaof thespecimen
and it is known as conventional stress or
Engineeringstress
Stress
25. Direct Stress
• Direct stress may be normal stress orshear stress
• Normal stress (σ)is the stress which acts in direction
perpendicular to the area. Normal stress is further
classifiedinto tensilestress
• Tensile stress is the stress induced in a body, when it
is subjected to two equal and opposite pulls (tensile
forces) as a result of which there is a tendency in
increasein length
• It acts normal to the areaand pulls on the area
28. Shear Stress
• Shear stress :- Stress Induced in a body, when subjected to
two equal and opposite forces which are acting tangentially
acrosstheresistingsectionasa resultof whichthebodytends
to shearoff acrossthat section
29. Bending Shear stress
• When amemberisbeingloadedsimilarto that infigureone
bendingstress(orflexurestress)will result. Bendingstressis
a more specific type of normal stress. When a beam
experiencesloadlikethat shownin figureonethetop fibers
of thebeamundergoanormalcompressivestress. Thestress
at the horizontal plane of the neutral is zero. The bottom
fibersof thebeamundergoa normaltensilestress. It canbe
concludedthereforethat the valueof thebendingstresswill
varylinearlywith distancefromtheneutralaxis.
30.
31. Torsional Shear stress
• When a machinememberis subjected to the action of two
equalandoppositecouplesactingin parallelplanes(ortorque
ortwistingmoment),thenthemachinememberissaidto be
subjectedto TorsionThestresssetup bytorsionisknown as
torsionalshearstress.
32. • The Service factor or application factor or overload
factor is defined as the ration of maximum torque
(load) to the average torque.
• It denoted as Ka.
• Ka = Max. Torque/ Avg or Mean Torque
= Tmax/ Tavg
• It is greater than or equal to one
Service Factor
33. A cotter joint is used to connect two co-axial rods, which are subjected to either
axial tensile force or axial compressive force.
Applications of cotter joint are as follows:
(i) Joint between the piston rod and the crosshead of a steam engine
(ii) Joint between the slide spindle and the fork of the valve mechanism
(iii) Joint between the piston rod and the tail or pump rod
(iv) Foundation bolt
Design of Cotter Joint
36. 1. Tensile Failure of Rod
Each rod of diameter d is subjected to a tensile force P. The tensile
stress in the rod is given by,
37. 2. Tensile Failure of Spigot
Figure shows the weakest cross-section at XX of the spigot end,
which is subjected to tensile stress.
Therefore, tensile stress in the spigot is given by,
Empirical relationship,
t = 0.31d
38. 3. Tensile Failure of Socket
Figure shows the weakest section at YY of the socket end, which is
subjected to tensile stress. The area of this section is given by,
The tensile stress at section YY is given by,
From the above equation, the outside diameter of socket
(d1) can be determined.
39. 4. Shear Failure of Cotter
The cotter is subjected to double shear as illustrated in Fig. The area of
each of the two planes that resist shearing failure is (bt). Therefore,
shear stress in the cotter is given by,
where tou is
permissible shear
stress for the cotter.
From Eq. , the
mean width of the
cotter (b) can be
determined.
40. 5. Shear Failure of Spigot End
The spigot end is subjected to double shear as shown in Fig. The
area of each of the two planes that resist shear failure is (ad2).
Therefore, shear stress in the spigot end is given by,
where tou is the
permissible shear
stress for the
spigot. From Eq.
the dimension a
can be determined.
41. 6. Shear Failure of Socket End
The socket end is also subjected to double shear as shown in Fig. The
area of each of the two planes that resist shear failure is given by,
area = (d4 – d2) c
Therefore, shear stress in the socket end is given by
From the above equation, the
dimension c can be determined.
42. 7. Crushing Failure of Spigot End
As shown in Fig., the force P causes compressive stress on a narrow
rectangular area of thickness t and width d2 perpendicular to the plane
of the paper.
The compressive stress is given by,
43. 8. Crushing Failure of Socket End
As shown in Fig. the force P causes compressive stress on a narrow
rectangular area of thickness t. The other dimension of rectangle,
perpendicular to the plane of paper is (d4 – d2).
Therefore, compressive stress in the socket end is given by,
44. 9. Bending Failure of Cotter
When the cotter is tight in the socket and spigot, it is
subjected to shear stresses. When it becomes loose,
bending occurs. The forces acting on the cotter are
shown
in Fig.
45. Numerical on Cotter Joint
It is required to design a cotter joint to connect two steel
rods of equal diameter. Each rod is subjected to an axial
tensile force of 50 kN. Design the joint and specify its
main dimensions.
46.
47. Given
P = (50 × 103) N
Step I Diameter of rods
Step II Thickness of cotter t = 0.31 d = 0.31(32) = 9.92 or 10
S
m
te
m
p III Diameter (d2) of spigot
48. Step IV Outer diameter (d1) of socket
Step V Diameters of spigot collar (d3) and socket collar (d4)
d3 = 1.5d = 1.5(32) = 48 mm
d4 = 2.4d = 2.4(32) = 76.8 or 80 mm
49. Step VI Dimensions a and c
a = c = 0.75d = 0.75(32) = 24 mm
Step VII Width of cotter
50. Step VIII Check for crushing and shear stresses in spigot end
Step IX Check for crushing and shear stresses in socket end
The stresses induced in the spigot and
the socket ends are within limits.
51. Step X Thickness of spigot collar
t1 = 0.45d = 0.45(32) = 14.4 or 15 mm
The taper for the cotter is 1 in 32.
Dimensioned sketch of cotter joint
52. Introduction
A knuckle joint is a mechanical joint used to connect two rods which are
under a tensile load, when there is a requirement of small amount
of flexibility, or angular moment is necessary. There is always axial or
linear line of action of load.
Design of knuckle Joint
53. 1. Knuckle joint can withstand large tensile loads.
2. It has good mechanical rigidity.
3. It is easy to manufacture and set up.
4. It can be easily dismantled and assembled.
5. Design is simple and easy.
54. 1. The joint cannot withstand large
compressive loads.
2. It permits angular movement in only one plane.
3. It is not as flexible as universal joint.
55.
56. Parts of Knuckle joint
A typical knuckle joint has the following parts:
⦁ Fork end
⦁ Eye end
⦁ Knuckle pin
⦁ Collar
⦁ Taper pin
57.
58. Notations used in design
⦁ P = Tension in rod ( Load on the joint)
⦁ D = Diameter of rod
⦁ D1= Enlarged diameter of rod
⦁ d = Diameter of pin
⦁ d1 = Diameter of pin head
⦁ d0 = Outer diameter of eye or fork
⦁ T1 or b = thickness of eye end
⦁ T2 or a= thickness of forked end (double eye)
⦁ x= distance of the Centre of fork radius R from the
eye
59. Step 1 : Design of Rods (D,D1)
Tensile failure of rod
Using basic strength equation
Load = Stress * Area
Empirical relations
Using Empirical relation the enlarged diameter of rod D1 is determined
Design of Knuckle Joint
60. Step 2 : Decide the thickness of eye end and forked end (t1,t2)
Empirical relations
Both these dimensions are decided on the basis of empirical relations,
t1= 1.25 D and t2= 0.75 D
61. Step 3 : Decide the dimensions of pin (d,d1)
Double shear Failure
The pin may get sheared off into three pieces as shown below, since the pin breaks
at two places it is called double shear. Both areas are taken as resisting areas.
Using basic strength equation
Load = Stress * Area
where tou is the permissible shear stress for the pin.
The standard proportion for the diameter of the pin
is as follows,
d = D
62. Step 4: Bending failure of pin
When the pin is tight in the eye and the fork, failure occurs due to shear.
On the other hand, when the pin is loose, it is subjected to bending
moment
63. Step 5: Check Stresses in Eye end
Tensile failure of eye end
The single eye may fail in tension as shown below { please note that when the
plane of failure is perpendicular to the direction of force then the failure is either
tensile or compressive}
Using the basic equation for stress
Empirical relation for outside diameter of
eye and fork
d0=2d
64. Shear failure of eye end
The single eye may fail in shear as shown below { please note that when the plane
of failure is parallel to the direction of force then the failure is Shear failure}
Using the basic equation for stress
The area of each of the two planes resisting
the shear failure is [b (d0 – d)/2]
approximately. Therefore, shear stress is
given by,
Standard proportion for outside diameter of the eye or the fork is
given by the following relationship,
d0 = 2d
65. ⦁ Step 6 : Check Stresses fork end
Fork end is also subjected to same failures as that of eye end, the only difference is
that it has two eyes. So we get the same equations except multiplied by 2.
The equations for tensile, shear and crushing failures are given below
Tensile failure of fork end
Fork is a double eye and as such, Fig. 4.23 is applicable to a fork except for dimension
b which can be modified as 2a in case of a fork.
Shear failure of fork end
Each of the two parts of the fork is subjected to double shear.
