Lecture6-12-MEC223.pptxaegeeeeeeeeeeeeeeeeeeeeeeeeee

MEC223
Design of Machine Element -I

2
Design Against Static Load
• Stress Strain Relationship
• Shear Stress and Shear strain
• Stress due to bending moment
• Stress due to torsion moment
• Eccentric Axial Loading

3
Static Load
A static Load is defined as a force, which is gradually
applied to a mechanical component and which does
not change its magnitude or direction with respect to
time
Three modes of failure
(i) Failure by elastic deflection
(ii) Failure by general yielding; and
(iii) Failure by fracture

4
Static Load
Three modes of failure
a) Failure by Elastic Deflection
b) Failure by General Yielding
c) Failure by fracture
The modules of elasticity and rigidity are the important
properties and the dimensions of the components are
determined by load-deflection equations.
The yield strength of a material is an important
property.
The Ultimate tensile strength of a material is an
important property.

5
Nomenclature used..
• A = Cross sectional area (mm2
)
• E = Young’s Modulus (MPa or N/mm2
)
• G = Shear Modulus or modulus of rigidity (MPa or
N/mm2
)
• I = Moment of interia (mm4
)
• J = Polar Moment of inertia (mm4
)
• K = Bulk Modulus (MPa or N/mm2
)
• Mb = Applied Bending moment (N-mm)
• Mt = Applied Torque (N-mm)

6
Nomenclature used..
• P = External Force (N)
• Ssy= Yield Strength in shear (MPa or N/mm2)
• Sut = Ultimate tensile strength (MPa or N/mm2)
• Syt = Yield Strength of material (MPa or N/mm2)
• U = Strain Energy (N-mm)
• (fs) = Factor of Safety
• K= Stiffness
• 𝛔 = Allowable Stress (MPa or N/mm2)
• 𝛕 = Shear Stress (MPa or N/mm2)

load)
(Actual
component
on the
Load
load)
(Max
component
the
of
Strength
FoS 
Factor of Safety (Safety Factor)
In the calculations,
Material strengths,
Manufacturing process
Is used to provide a design margin over the theoretical design
capacity to allow for uncertainty in the design process.
load
or working
stress
allowable
Load
(stress)
failure
FoS 
While Designing a component, It is necessary to provide sufficient
reserve strength in case of an accident

)
load(
actual
the
to
due
component
in the
Stress
)
(
component
the
of
Strength
FoS

yt
ut,S
S


The magnitude of factor of safety depends upon the
following factors
• Effect of failure
• Type of load
• Degree of accuracy in force analysis
• Material of component
• Reliability of component
• Cost of component
• Testing of machine elements
• Service conditions
• Quality of manufacture

FoS
(Based on
yeild strength)
Application
1.25 – 1.5
Material properties known in detail
Operating conditions known in detail
Load and the resulting stresses and strains are
known to a high degree of accuracy
Low weight is important
2 – 3
For less tried materials or
Brittle materials under average conditions of
environment, load and stress
3 – 4
For untried materials under average conditions of
environment, load and stress
Better known materials under uncertain
environment or uncertain stresses

Stress-Strain Relationship
𝜎 𝑡=
𝑃
𝐴
∈❑=
𝛿
𝑙
𝜎𝑡=𝐸 ∈
𝛿=
𝑃𝑙
𝐴𝐸

15
Assumptions for the Stress strain
analysis
• The material is homogeneous
• The load is applied gradually
• The line of action of force P passes
through the geometric axis of the cross
section.
• The cross section is uniform.
• There is no stress concentration

Shear Stress and Shear Strain
𝜏=
𝑃
𝐴
𝜏=𝐺𝛾
𝐺=80 𝐺𝑃𝑎 𝑓𝑜𝑟 𝑠𝑡𝑒𝑒𝑙𝑠
𝐺=40𝐺𝑃𝑎 𝑓𝑜𝑟 𝑔𝑟𝑒𝑦 𝑐𝑎𝑠𝑡𝑖𝑟𝑜𝑛

Shear Stress and Shear Strain
Relationship between E,G and Poisson's
ration
𝐸=2 𝐺(1+ 𝜇)
For carbon steels, m=0.29
For grey cast iron, m=0.21

Stresses Due to Bending Moment
𝜎 𝑏=
𝑀𝑏 𝑦
𝐼

19
Assumptions for the Stresses due to
bending moment
• The beam is straight with uniform cross
section.
• The forces acting on the beam lie in a
plane perpendicular to the axis of the
beam.
• The material is homogeneous , isotropic
and obey Hook’s law
• Plane cross sections remain plane after
bending.

Sign conventions for Bending Moment

Stresses due to Torsional Moment
t
M l
JG
 
t
M r
J
 
𝐽=
𝜋 𝑑4
32
𝐽=
𝜋(𝑑𝑜
4
− 𝑑𝑖
4
)
32

22
Assumptions for Torsional moment
• The Shaft is straight with circular cross section.
• The plane transverse section remains plane
after twisting
• The material is homogeneous , isotropic and
obey Hook’s law
Calculation of torque from the power transmitted
and the speed of rotation
𝑘𝑊 =
2𝜋 𝑁 𝑀𝑡
60 𝑋 10
6

Eccentric Axial Loading
P Pey
A I
  

Principle Stresses (Mohr’s Circle)
sx
sy
txy
Some Part
A particular point
on the part
x
y
x & y  orientation

Mohr’s Circle t (CW)
s
x-axis
y-axis
sx
sy
txy
(sx, txy)
sx
txy
sy
txy
Center of Mohr’s
Circle
(sy, txy)

Requirement of Theories of Failures
• Theories of failure are used to determine the
safe dimension of a component when it is
subjected to combined stresses due to various
loads.
• Theories of failure are used in design by
establishing a relationship between stresses
induced under combined loading conditions
and properties obtained from tension test like
Syt & Sut

Various Theory Of Failure
• Maximum principal stress theory (Rankine’s
Theory)
•Maximum shear stress theory (Guest and
Treska’s Theory)
• Maximum Distortion energy theory (Von-
mises and Henky’s Theory)
•Maximum principal strain theory (St. Venant’s
theory)
• Maximum total strain energy theory (Haigh’
Theory)

Maximum Principle Stress Theory:
• Condition For Failure):
σ1 Syt or Sut
• Condition for safe design:
σ1 < Syt/N or Sut/N
Where N is factor of safety.
• For Tensile Stresses:
For Compressive stresses:
The Theory states that the failure of the mechanical component
subjected to bi-axial or tri-axial stresses occurs when maximum
principal stress reaches the yield or ultimate strength of the material.

Maximum Principle Stress Theory
• This theory is suitable for the safe design of
machine component made up of brittle material,
because brittle material are weak in tension
• This theory is not good for design of ductile
material because shear failure may occur

Maximum shear stress theory
The Theory states that the failure of the mechanical
component subjected to bi-axial or tri-axial stresses occurs
when maximum shear stress at any point in the component
becomes equal to the maximum shear stress in the standard
specimen of the tension test, when yield starts.
𝜏𝑚𝑎𝑥=
𝜎1
2
𝜏𝑚𝑎𝑥=
𝑆𝑦𝑡
2

Maximum shear stress theory
• Condition For Failure:
• Condition for safe design:
𝑆𝑠𝑦 (𝜏¿¿𝑚𝑎𝑥)≥
𝑆𝑦𝑡
2
¿
𝑆𝑠𝑦 (𝜏¿¿𝑚𝑎𝑥)<
𝑆𝑦𝑡
2
¿

For tri-axial state of stress condition
𝜏12=[𝜎1 − 𝜎2
2 ] 𝜏23=[𝜎2 − 𝜎3
2 ] 𝜏31=[𝜎3 − 𝜎1
2 ]
The largest of the three stresses is equated to or
[𝜎1 − 𝜎 2
2 ]¿
𝑆𝑦𝑡
2 𝑓 𝑠 ❑
[𝜎 1 − 𝜎2 ] ¿
𝑆𝑦𝑡
𝑓 𝑠 ❑

Region of safety
[𝜎 1− 𝜎2]¿± 𝑆𝑦𝑡❑
For bi axial system (=0)
𝜎1=±𝑆𝑦𝑡
𝜎2=± 𝑆𝑦𝑡
Used for safe
design for
ductile material

Distortion Energy theory
• For biaxial stress
The Theory states that the failure of the mechanical
component subjected to bi-axial or tri-axial stresses occurs
when strain energy of distortion per unit volume at any
point in the component becomes equal to the strain energy
of distortion per unit volume in the standard specimen of
the tension test, when yield starts.
𝑆𝑦𝑡
𝑓𝑠
=√(𝜎1
2
− 𝜎1 𝜎2+𝜎2
2
)
𝑆𝑠𝑦 =0.577 𝑆𝑦𝑡

37
Summary Theories of failures
Maximum Principle Stress Theory Used for
brittle materials with design criteria as
Maximum shear stress theory is used for safe
design for ductile materials with design criteria
as
Distortion energy theory is used for safe design
for ductile materials with design criteria as
(more accurate and complicated than Max.
Shear stress theory)

An Example
Two Plates, Subjected to a tensile force of 50kN, are fixed
together by means of three rivets as shown in fig (a). The
plates and rivets are made of plain carbon steel 10C4 with
a tensile yield strength of 250 N/mm2
. Using maximum
shear stress theory and factor of safety 2.5. Neglecting
stress concentration, determine (i) the diameter of rivets;
and (ii) the thickness of plates.

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

Cotter Joint
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(1 in 24))
from one end to another for an easy adjustment.

Notations Used
P= tensile Load
d= diameter of each rod
d1= OD of socket
d2= OD of spigot or ID of socket
d3= diameter of spigot collar
d4= diameter of socket collar
a= distance from end of slot to the end
of spigot
b=mean width of cottar
t= thickness of the cottar
l= length of the cottar (all in mm)

Possible Failure of Cotter Joint
Component Tensile
Failure
Shear
failure
Crushing Failure
(compressive)
Bending
Failure
Rod
Spigot
Socket
Cotter
Component Tensile
Failure
Shear
failure
Crushing Failure
(compressive)
Bending
Failure
Rod YES NO NO NO
Spigot YES YES YES NO
Socket YES YES YES NO
Cotter NO YES NO YES

Tensile Failure of Rods
𝜎 =
𝑃
[𝜋
4
𝑑
2
] 𝑑=
√4𝑃
𝜋 𝜎𝑡
= permissible tensile stress

Tensile failure of Spigot
𝜎 =
𝑃
[𝜋
4
𝑑2
2
− 𝑑2 𝑡]
The thickness of the cottar
is usually as t= .31 d 𝑃=𝜎
[𝜋
4
𝑑2
2
− 𝑑2 𝑡
]

Shear Failure of Spigot
𝜏=
𝑃
2(𝑎 𝑑2)
𝑃=2𝑎𝑑2 𝜏

Crushing/Compressive Failure of Spigot
𝜎𝑐=
𝑃
𝑡 𝑑2
𝑃=𝑡 𝑑2 𝜎𝑐

Tensile Failure of Socket
𝑃 = 𝜎 ¿

Shear Failure of Socket
𝜏=
𝑃
2 (𝑑4 − 𝑑2 )𝑐
𝑃=2(𝑑4 − 𝑑2 )𝑐 𝜏

Crushing Failure of Socket
𝜎𝑐=
𝑃
𝑡(𝑑¿¿4−𝑑2)¿
𝑃=𝑡(𝑑¿¿4 −𝑑2)𝜎𝑐 ¿

Shear Failure of Cotter
𝜏=
𝑃
2(𝑏𝑡)
𝑃=2𝑏𝑡 𝜏

Bending Failure of Cotter
𝑀𝑏=
𝑃
2 [𝑑2
2
+𝑥]−
𝑃
2
( 𝑧)
¿
𝑃
2 [𝑑2
2
+
𝑑4 − 𝑑2
6 ]−
𝑃
2 [𝑑2
4 ]
¿
𝑃
2 [𝑑2
4
+
𝑑4 − 𝑑2
6 ]
𝜎𝑏=
𝑀𝑏 𝑦
𝐼
𝜎 𝑏=
𝑃
2 [𝑑2
4
+
𝑑4 − 𝑑2
6 ]𝑏
2
(𝑡 𝑏
3
12 )

Standard Proportions
• d1= 1.75d d2= 1.21d d3= 1.5d d4= 2.4d
• a=c= 0.75d b= 1.6d
• t= 0.31d t1= 0.45d
• Clearance= 1.5 to 3mm
• Taper for cotter= 1 in 32

Procedure for design
• Calculate diameter of each rod (d)
• Calculate thickness of cotter by empirical relation as t=
0.31d
• Calculate d2 of the spigot on the basis of tensile stress.
• Calculate OD d1 of the socket on the basis of tensile
stress
𝑑=
√ 4 𝑃
𝜋 𝜎𝑡
𝑃=𝜎
[𝜋
4
𝑑2
2
− 𝑑2 𝑡
]
𝑃 = 𝜎 ¿

Procedure for design
• Calculate the diameter of spigot collar as(d3= 1.5d) and
the socket collar (d4= 2.4d)
• Calculate a=c= 0.75d
• Calculate width b of cotter by considering shear and
bending. Take the value larger from above
considerations
• Check for the crushing and shear stresses in spigot and
socket
• Calculate t1= 0.45d

58
• 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

Knuckle joint
Two or more rods
subjected to tensile and
compressive forces are
fastened together
Their axes are not in
alignments but meet in a point
The joint allows a small
angular moment of one rod
relative to another
It can be easily connected
and disconnected, simple
design and manufacuring
Applications: tie bars in
roof trusses, Chains,
valve rods, Hand Pump
etc

Notations Used in Knuckle Joint
• D= Diameter of each rod (mm)
• D1 = Enlarged diameter of each
rod (mm)
• d = diameter of knuckle pin
(mm)
• do= outside diameter of eye or
fork (mm)
• a= thickness of each eye of fork
• b= thickness of eye end rod-B
(mm)
• d1= diameter of pin head (mm)
• X = distance of the center of
fork radius R from the eye
(mm)

FBD for Knuckle Joint and Tension
Failure of Rod

Tensile Failure of Rods
𝜎 =
𝑃
[𝜋
4
𝐷2
] 𝐷=
√4𝑃
𝜋 𝜎𝑡

Shear Failure of Pin
𝜏=
𝑃
2(𝜋
4
𝑑
2
)
𝑑=
√2𝑃
𝜋𝜏

Crushing Failure of Pin in Eye
𝜎𝑐=
𝑃
𝑏𝑑

Crushing Failure of Pin in Fork
𝜎𝑐=
𝑃
2𝑎𝑑

Bending Failure of Pin
𝑀𝑏=
𝑃
2 [𝑏
2
+𝑥
]−
𝑃
2
(𝑧)
¿
𝑃
2 [𝑏
2
+
𝑎
3 ]−
𝑃
2 [𝑏
4 ]
¿
𝑃
2 [𝑏
4
+
𝑎
3 ]
𝜎𝑏=
𝑀𝑏 𝑦
𝐼
𝜎 𝑏=
𝑃
2 [𝑏
4
+
𝑎
3 ]𝑑
2
(𝜋 𝑑
4
64 )
𝜎𝑏=
32
𝜋 𝑑
3
𝑃
2 [𝑏
4
+
𝑎
3 ]

Tensile Failure of Eye
𝜎𝑡=
𝑃
𝑏(𝑑¿¿0−𝑑)¿

Shear Failure of Eye
𝜏 =
𝑃
2 ¿ ¿
𝜏=
𝑃
𝑏(𝑑¿¿0−𝑑)¿

Tensile and Shear failure of Fork
• Fork is a double eye
• Replace b with 2a to find corresponding equations
𝜎𝑡=
𝑃
2𝑎(𝑑¿¿0−𝑑)¿
𝜏=
𝑃
2𝑎(𝑑¿¿0−𝑑)¿

Design Procedure for Knuckle Joint
• Calculate diameter of each rod
• Calculate enlarged diameter of rod end D1 = 1.1 D
• Calculate a and b as a = 0.75D and b = 1.25D
• Calculate diameters of pin by shear and bending and
take maximum value from these
𝐷=
√4 𝑃
𝜋 𝜎𝑡
𝑑=
√2𝑃
𝜋𝜏
𝑑=
3
√16 𝑃
𝜋 𝜎𝑏
[𝑏
4
+
𝑎
3 ]

Design Procedure for Knuckle Joint
• Calculate d0 = 2d and d1 = 1.5d
• Check tensile, crushing and shear stresses for eye
• Check tensile, crushing and shear stresses for eye and
fork
• The distance x is usually taken as 10mm
𝜎𝑡=
𝑃
𝑏(𝑑¿¿0−𝑑)¿
𝜎𝑐=
𝑃
𝑏𝑑
𝜏=
𝑃
2𝑎(𝑑¿¿0−𝑑)¿
𝜎𝑐=
𝑃
2𝑎𝑑
𝜎𝑡=
𝑃
2𝑎(𝑑¿¿0−𝑑)¿
𝜏=
𝑃
𝑏(𝑑¿¿0−𝑑)¿

73
Problem
It is required to design a knuckle joint to
connect two circular rods subjected to an
axial tensile force of 80kN. The rods are co-
axial and a small amount of angular
movement between their axes is
permissible. Design the joint and specify
the dimensions of its components. Material
used for two rods and pin is Grade 30C8
( Syt = 400 N/mm2
).

74
Problem
It is required to design a knuckle joint to connect two circular
rods subjected to an axial tensile force of 80kN. The rods are co-
axial and a small amount of angular movement between their
axes is permissible. Design the joint and specify the dimensions
of its components. Material used for two rods and pin is Grade
30C8( Syt = 400 N/mm2
).
Given P= 80 X 103
N
I. Select factor of safety
II. Calculation of permissible stresses
III. Diameter of rods
IV. Enlarged diameter of the rod(D1)
V. Dimensions a and b
VI. Diameter of pin
VII. Dimensions d0 and d1
VIII.Check for stress in eye and fork

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