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7/26/2019 Theories of Failure Scet
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Theories of FailureFailure of a member is defined as one of two conditions.
1. Fracture of the material of which the member is made. This type
of failure is the characteristic of brittle materials.
2. Initiation of inelastic (Plastic) behavior in the material. This
type of failure is the one generally ehibited by ductile materials.
!hen an engineer is faced with the problem of design using a specific
material" it becomes important to place an upper limit on the state ofstress that defines the material#s failure. If the material is ductile, failure
is usually specified by the initiation ofyielding, whereas if the material
is brittle it isspecified byfracture.
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These modes of failure are readily defined if the member is sub$ected to
a uniaial state of stress" as in the case of simple tension however" if the
member is sub$ected to biaial or triaial stress" the criteria for failure
becomes more difficult to establish.
Inthis section we will discuss four theories that are often used in
engineering practice to predict the failure of a material sub$ected to a
multiaxial state of stress.
% failure theory is a criterion that is used in an effort to predict the
failure of a given material when sub$ected to a comple stress condition.
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i. &aimum shear stress (Tresca) theory for ductile materials.
ii. &aimum principal stress ('anine) theory.
iii. &aimum normal strain (aint *enan+s) theory.
iv. &aimum shear strain (,istortion -nergy) theory.
everal theories are available however" only four important theories are
discussed here.
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&aimum shear stress theory for ,uctile &aterials
The French engineer Tresca proposed this theory. It states that a
member sub$ected to any state of stress fails (yields) when the
maimum shearing stress (ma)in the member becomes e/ual to the
yield point stress (y)in a simple tension or compression test (0niaial
test). ince the maimum shear stress in a material under uniaial stress
condition is one half the value of normal stress and the maimum
normal stress (maimum principal stress) is ma" then from &ohr+scircle.
2
mama
=
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In case of iaial stress state
)(
22
)2(22
21
21
ma
21minmama
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Problem 01:-
The solid circular shaft in Fig. (a) is sub$ect to belt pulls at each
end and is simply supported at the two bearings. The material has a
yield point of 3"444 Ib5in26 ,etermine the re/uired diameter of the
shaft using the maimum shear stress theory together with a safety
factor of .
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400 + 200 lb200 + 500 lb
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4
72844
37
27244
.72443944
.3443344
37
2
7
7
=
=
=
====
=
=
=
y
x
x
B
x
d
d
d
inlbMc
inlbM
dI
dc
I
Mc
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inlb
OR
inlb
T
d
d
d
J
Tr
xy
xy
.784427244
27)244744(
.78441344
13)244:44(
784"27
2
27844
7
==
=
==
==
=
=
xy
yx
yx
xy
784"27
dxy=
844"72
dx=
x
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( )22
21
ma21
22
2
nowwe%s
xy
yx
+
=
=
3444
2
)( 2...
...
%nd
21
21
ma
==
=
=
FOS
SOF
SOF
yield
yield
y
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##93.1
784"27272844143
784"27
2
72844
2
444"12
784"27
2
728442
444"3
784"27
2
728442
2
2
3
2
2
2
2
2
2
21
=
+
=
+
=
+
=
+
=
d
dd
dd
dd
dd
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Maximum Principal Stress theor or
!"an#ine Theor$
%ccording to this theory" it is assumed that when a member is
sub$ected to any state of stress" fails (fracture of brittle material or
yielding of ductile material) when the principal stress of largest
magnitude. (1) in the member reaches to a limiting value that is e/ual
to the ultimate stress"
)1(1 ult = )2(2 ult =
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Problem 02:-
The solid circular shaft in Fig. 1 (a) is sub$ect to belt pulls at
each end and is simply supported at the two bearings. The material has
a yield point of 3"444 Ib5in26 ,etermine the re/uired diameter of the
shaft using the maimum Principal stress theory together with a safety
factor of .
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400 + 200 lb200 + 500 lb
M
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4
72844
37
27244
.72443944
.3443344
37
2
7
7
=
=
=
==
==
=
=
=
y
x
x
B
x
d
d
d
inlbMc
inlbM
dI
dc
I
Mc
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inlb
OR
inlb
T
d
d
d
J
Tr
xy
xy
.784427244
27)244744(
.78441344
13)244:44(
784"27
2
27844
7
==
=
==
=
=
=
=
xy
yx
yx
xy
784"27
dxy= 844"72
dx=
x
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( )
( )22
2
2
2
1
22
22
xy
yxyx
xy
yxyx
+
+=
+
+
+=
2
2
1 24.27773272844
272844 ++= ddd
%ccording to maimum normal stress theory .
2
2
1
24.27773
2
72844
2
72844
3444
+
+=
=
ddd
ult
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;78.1
:1.14
14:17.114177
1432.:
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%xample 0&
The solid cast=iron shaft shown in Fig. is sub$ected to a tor/ue of T >
744 Ib . ft. ,etermine its smallest radius so that it does not fail
according to the maimum=Principal=stress theory. % specimen of cast
iron" tested in tension" has an ultimate stress of (?ult)t> 24 si.
Solution
The maimum or critical stress occurs at a point located on the surface
of the shaft. %ssuming the shaft to have a radius r, the shear stress is
7ma
..8.4::
)25(
)5.12)(.744(
r
inlb
r
rftinftlb
J
Tc 3===
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&ohr#s circle for this state of stress (pure shear) is shown in Fig. . ince
R > ma" then
The maimum=Principal=stress theory"" re/uires
@1@ A ult
ma21
..8.4::
r
inlb===
2
5444"24
..8.4::inlb
r
inlb
Thus" the smallest radius of the shaft is determined from
..::.4
5444"24..8.4:: 2
n!inr
inlbr
inlb
=
=
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Maximum 'ormal Strain or Saint (enant)s
*riterion
In this theory" it is assumed that a member sub$ected to any state of
stress fails (yields) when the maimum normal strain at any point
e/uals" the yield point strain obtained from a simple tension or
compression test (y> ?y5-).
Principal strain of largest magnitude @ma@ could be one of two principal
strain 1 and 2 depending upon the stress conditions acting in the
member . Thus the maimum Principal strain theory may be
represented by the following e/uation.
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)1(
2ma
1ma
==
==
y
y
%s stress in one direction produces the lateral deformation in the other
two perpendicular directions and using law of superposition" we find
three principal strains of the element.
>
y>
B>
?
? 5 -
=C? 5 -
=C? 5 -
?y
=C?y 5 -
?y 5 -
=C?y 5 -
?B
=C?B 5 -
=C?B 5 -
?B 5 -
>
y>
B>
>y>
B>
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> ?5 - =C?y5 - =C?B5 -
> ?5 - =C 5 - (?yD ?B)
y > ?y5 - =C?5 - =C?B5 -> ?y5 - =C 5 - (?D ?B)
B > ?B5 - =C?y5 - =C?5 -
> ?B5 - =C 5 - (?D ?y)
(2)
Thus
)(
)(
)(
)(
12
12
2
21
1
+=
+=
+=
""
""
""
l
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%lso
)3(
):(Then
and
7and1-/uating
)7()iaial(
12y
21y
122
21
1
211
=
=
=
=
=
=
""
"""
F#r""
yield
yield
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Maximum Shear Strain %ner !,istortion
%ner *riterion !(on MS%S *riterion$
%ccording to this theory when a member is sub$ected to any state of
stress fails (yields) when the distortion energy per unit volume at a
point becomes e/ual to the strain energy of distortion per unit volumeat failure (yielding).
The distortion strain energy is that energy associated with a change in
the shape of the body.The total strain energy per unit volume also called strain energy
density is the energy in a body stored internally throughout its volume
due to deformation produced by eternal loading. If the aial stress
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train energy due to distortion per unit volume for biaial stress system
,istortion energy per unit volume is given by
( )2
23
1yd "
u
$ +
=
[ ]2
221
2
1
1 +
+= "
u
$d
%ccording to distortion energy theory
[ ]
2
221
2
1
2
1)2(
3
1
+
+
=
+
"
u
"
u
y
2
221
2
1
2 +=y
o" e/uation becomes