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8/3/2019 21205749 Theories of Failure 2
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THEORIES OF FAILURE
PRESENTED BY-
PANKAJ SHARMA NAGENDRA PALSINGH100106236 100106218
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Theories of Failure
The material properties are usually determined by simpletension or compression tests.
The mechanical members are subjected to biaxial ortriaxial stresses.
To determine whether a component will fail or not, somefailure theories are proposed which are related to theproperties of materials obtained from uniaxial tension orcompression tests.
Initially we will consider failure of a mechanical membersubjected to biaxial stresses
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The Theories of Failures which are applicable for thissituation are:
Max principal or normal stress theory(Rankines theory)
Maximum shear stress theory(Guests orTrescas theory)
Max. Distortion energy theory(Von Mises &Henckys theory)
Max. strain energy theory
Max. principal strain theory
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Ductile materials usually fail byyielding and hence the limiting strength isthe yield strength of material as determinedfrom simple tension test which is assumed
the same in compression also.
For brittle materials limiting strengthof material is ultimate tensile strength in
tension or compression.
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Max. Principal or Normal stress theory(Rankines Theory):
It is assumed that the failure or yield occursat a point in a member when the max.principal or normal stress in the biaxialstress system reaches the limiting strength ofthe material in a simple tension test.
In this case max. principal stress iscalculated in a biaxial stress case and is
equated to limiting strength of the material.
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Maximum principal stress
2
2
122
xy
yxyx
Minimum principal stress
2
2
222
xy
yxyx
For ductile materials
1 should not exceed in tension,
FOS
Syt
For brittle materials
1 should not exceed in tension
FOS
Sut
FOS=Factor of safety
This theory is basically applicable for brittle materialswhich are relatively stronger in shear and not applicable
to ductile materials which are relatively weak in shear.
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The failure or yielding is assumed to take place at a
point in a member where the max shear stress in abiaxial stress system reaches a value equal to shearstrength of the material obtained from simple tensiontest.
In a biaxial stress case max shear stress developed isgiven by
2.Maximum Shear Stress theory (Guests or Trescastheory):
FOSyt
max
wheremax = FOS2
Sy t
This theory is mostly used for ductile materials.
22
max2
xy
yx
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2
..max
stressdirectMinstressdirectMax
CASE1 (First quadrant )
1 and 2 are +ve
yt
yt
Sei
S
1
1131
max
..
222
0
2
CASE2 (Second quadrant)
1 is -ve and 2 is +ve ,Then
2
2
2
22
)(
1max
11212max
ytS
Then
2max
ytS
CASE3 (Third quadrant)
1 is -ve and 2 is more -ve
,Then
yc
yc
Sei
SThen
. 22
2
0
2
)(
max
223
max
CASE4 (Fourth quadrant)1 is +ve and 2 is -ve ,Then
2
2
2
22
)(
max
2121
max
ytS
Then
Assuming that 1>
2>
3and
3 =0
According to the Maximum shear stress theory,
And also
+1
1=Syt
+2
-1
-2
Syc
Syt
Syc
Syto
1=Syc
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It is assumed that failure or yielding occurs at a pointthe member where the distortion strain energy (alsocalled shear strain energy) per unit volume in a biaxialstress system reaches the limiting distortion energy(distortion energy at yield point) per unit volume asdetermined from a simple tension test.
The maximum distortion energy is the differencebetween the total strain energy and the strain energy dueto uniform stress.
3.Max. Distortion energy theory (Von Mises & Henckystheory):
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3.Max. Distortion energy theory (Von Mises &Henckys theory):
The criteria of failure for the distortionenergy
theory is expressed as
Considering the factor of safety
2132322212
1
FOS
Syt
2
13
2
32
2
212
1 ytS
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3.Max. Distortion energy theory (Von Mises & Henckystheory):
A component subjected to pure shear stresses and thecorresponding Mohrs circle diagram is
Y
X
Element subjected to pure shear stresses
o1-2
Mohrs circle for pure shear stresses
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In the biaxial stress case, principal stress 1, 2 arecalculated based on x ,y & xy which in turn are usedto determine whether the left hand side is more thanright hand side, which indicates failure of the
component.
212221 FOS
Syt
From the figure, 1 = -2 = and 3=0Substituting the values in the equation
We get
Replacing by Ssy, we get
3ytS
yt
yt
sy SS
S 577.03
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+1
+2
-1
-2
Syc
Syt
Syc
Syto
Boundary for distortion energy theory under bi axial stresses
Case 1 (First quadrant)
1 and 2 are +ve and equal
to , then
FOS
SFOS
S
yt
yt
212
2
2
1
Case 4 (Fourth quadrant)
1 is +ve and 2 is -ve and equalto , then
FOS
SFOS
SFOS
SFOS
S
yt
yt
yt
yt
577.0
33 2
21
2
2
2
1
21
2
2
2
1
Case 2 (Second quadrant)
1 is -ve and 2 is +ve and equalto , then
FOS
SFOS
SFOS
SFOS
S
yt
yt
yt
yt
577.0
332
21
2
2
2
1
21
2
2
2
1
Case 3 (Third quadrant)
1 is -ve and 2 is +ve andequal to , then
FOS
SFOS
S
yt
yt
212
2
2
1
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Failure is assumed to take place at a point in a memberwhere strain energy per unit volume in a biaxial stresssystem reaches the limiting strain energy that is strainenergy at yield point per unit volume as determinedfrom a simple tension test.
Strain energy per unit volume in a biaxial system is
The limiting strain energy per unit volume for yieldingas determined from simple tension test is
mEU 21
2
2
2
11
2
2
1
4. Max. Strain energy theory (Heighs Thoery):
2
2 2
1
FOS
S
EU
yt
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Equating the above two equations then we get
In a biaxial case 1
, 2
are calculated based as x
, y
& xy
2
212
2
2
1
2
FOS
S
m
yt
It will be checked whether the Left Hand Side ofEquation is less than Right Hand Side of Equation ornot. This theory is used for ductile materials.
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EFOS
S
mE
E
Ey t21
max
It is assumed that the failure or yielding occurs at a point
in a member where the maximum principal (normal) strainin a biaxial stress exceeds limiting value of strain (strain atyield port) as obtained from simple tension test.
In a biaxial stress case
One can calculate 1 & 2 given x , y & xyand checkwhether the material fails or not, this theory is not usedin general as reliable results could not be detained in
variety of materials.
5.Max. Principal Strain theory (Saint Venants Theory):
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Example :1 The load on a bolt consists of an axial pull of
10kN together with a transverse shear force of5kN. Find the diameter of bolt requiredaccording to
1. Maximum principal stress theory2. Maximum shear stress theory
3. Maximum principal strain theory
4. Maximum strain energy theory
5. Maximum distortion energy theoryPermissible tensile stress at elastic limit =100MPa
and Poissons ratio =0.3
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Solution 1
Cross sectional area of the bolt,
Axial stress,
And transverse shear stress,
227854.0
4ddA
2
221/
73.12
7854.0
10mmkN
ddA
P
2
2/365.6
7854.0
5mmkN
dA
Ps
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According to maximum principal stress theory
Maximum principal stress,
According to maximum principal stresstheory, Syt = 1
22
122
xyxx
2
2
122
xy
yxyx
2
21
2
2
2
221
/15365
365.6
2
73.12
2
73.12
mmNd
ddd
mmd
d
4.1215365
1002
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According to maximum shear stress theory
Maximum shear stress,
According to maximum shear stress,
mmdd
Syt
42.132
1009000
2 2max
2
2
max2
xy
yx
2
2
2
2
2
2
2
2
2
2
max
/9000
/9365.673.12
2
mmNd
mmkNddd
xyx
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According to maximum principal strain theory
The maximum principal stress,
And minimum principal stress,
2
2
1 22 xyyxyx
2
2
222
xy
yxyx
2
2
2
1
15365
22 dxy
xx
2
22
2
2
2
22
22
2
/2635
365.6
2
73.1273.12
22
mmNd
dddxy
xx
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And according to maximum principal strain theory,
mmddd
7.12
1003.0263515365
Sm
E
S
mE
E
22y t2
1
y t21
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According to maximum strain energy theory
According to maximum distortion theory
mmd dddd
Sm
yt
78.12
1003.0263515365
2263515365
2
2
22
2
2
2
2
22122
21
mmd
dddd
Syt
4.13
263515365263515365100
22
2
2
2
2
21
2
2
2
1
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