theories of failure scet

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


9/30


10/30

4

72844

37

27244

.72443944

.3443344

37

2

7

7

=

=

=

====

=

=

=

y

x

x

B

x

d

d

d

inlbMc

inlbM

dI

dc

I

Mc


11/30

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


14/30

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 .


16/30

400 + 200 lb200 + 500 lb

M


17/30

4

72844

37

27244

.72443944

.3443344

37

2

7

7

=

=

=

==

==

=

=

=

y

x

x

B

x

d

d

d

inlbMc

inlbM

dI

dc

I

Mc


18/30

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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23/30


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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

theories of failure scet

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