theories_of_failure.ppt

7/25/2019 theories_of_failure.ppt

1/77

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. 1


2/77

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.

2


3/77

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.

3


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5/77


6/77

If both of principal stresses are of the same sign tension

compression then

1y (3)

2y (4)

In case of 5iaial stress state

)6(

22

)2(22

21

21

ma

21minmama


7/77

% graph of these e/uation is given in the figure. %ny given state of

stress will be represented in this figure by a point of co7ordinate 1and

2where 1and 2are two principal stresses. If these point falls withinthe area shown" the member is safe and if outside then member fails as

a result of yielding of material. The 8eagon associated with the

initiation of yield in the member is nown as 9Tresca 8eagon:. (1;137

1;;4). In the first and third /uadrant 1and 2have the same signs and

7


8/77

mais half of the numerically larger value of principal stress 1or 2. In

the second and fourth /uad" where 1and 2are of opposite sign" mais

half of arithmetical sum of the two 1and 2.In fourth /uadrant" the e/uation of the boundary or limit

(yield


9/77

y = 21

y =1

y =1

y =2

y

= 21

y =2

9


10/77

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 6>"??? Ib


11/77

400 + 200 lb200 + 500 lb

11


12/77


13/77

?

32;??

>3

232??

.32??>A??

.6>??>>??

>3

2

6

3

3

=

=

=

====

=

=

=

y

x

x

B

x

d

d

d

inlbMc

inlbM

dI

dc

I

Mc

13


14/77

inlb

OR

inlb

T

d

d

d

J

Tr

xy

xy

.3;??232??

23)2??3??(

.3;??1>6??

1>)2??4??(

3;?"23

62

23;??

6

3

==

=

==

==

=

=

xy

yx

yx

xy

6

3;?"23

dxy= 6

;??"32

dx=

x

14


15/77

( )22

21

ma21

22

2

nowwe%s

xy

yx

+

=

=

6

6>???

2

)(2

...

...

%nd

21

21

ma

==

=

=

FOS

SOF

SOF

yield

yield

y

15


16/77

##A>.1

3;?"23232;??1?6>

3;?"23

2

32;??

2

???"12

3;?"23

2

32;??2

6

???"6>

3;?"23

2

32;??2

2

6

2

6

>

2

6

2

6

2

6

2

6

2

6

2

621

=

+

=

+

=

+

=

+

=

d

dd

dd

dd

dd

16


17/77

Problem 02

The state of plane stress shown occurs at a critical point of a steel

machine component. %s a result of several tensile tests" it has been found

that the tensile yield strength is y= 24? &Pa for the grade of steel used.

,etermine the factor of safety with respect to yield" using (a) themaimum7shearing7stress criterion" and (b) the maimum7distortion7

energy criterion.

17


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18

SOLUTION


19/77

SOLUTION

&ohr#s Bircle. !e construct &ohr#s circle for the given state of stress

and findMPaO yxa!e 2?)3?;?()( 2

121 ==+==

MPaF"FRm >4)24()>?()()(2222 =+=+==

Prini!al S"resses

MPa#Oa ;4>42? +=+=+=MPa#Ob 34>42? ===

a. &aimum7hearing7tress Briterion.ince for the grade of steel used

the tensile strength is ay = 24? &Pa" the corresponding shearing stress at

yield is

MPaMPa$$ 124)24?(21

21

=== 19


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MPaF%r m >4= 2.1C.>4124

. === SFMPa

MPaSF

m

$

20

b i i i B i i d i f f f


21/77

b. &aimum7,istortion7-nergy Briterion.Introducing a factor of safety

into -/. (A.2>)" we write

222

.

=+

SF

$bbaa

For a= D;4 &Pa" b= 734 &Pa" and y= 24? &Pa" we have

( ) ( )( ) ( )2

22

.

24?3434;4;4

=+

SF

1C.2..

24?6.113 == SF

SF

21


22/77


23/77

23


24/77

The solid shaft shown in Fig. 1?731a has a radius of ?.4 in. and is made

of steel having a yield stress of y= 6> si. ,etermine if the loadings

cause the shaft to fail according to the maimum7shear7stress theory

and the maimum7distortion7energy theory.

#$am!le 10-12

Solu"ion

The state of stress in the shaft is caused by both the aial force and the

tor/ue. ince maimum shear stress caused by the tor/ue occurs in the

material at the outer surface" we have

24


25/77

'iin

inin'iJTc

'iin

'i

#

P

xy

x

44.1>2


26/77

= 7C.44 G 1C.11

BTI = C.4> si

BT2 = 72;.>> si

Maximum-Shear-Stress Theory. ince the principal stresses have

%%ite ign, then from ec. C.4" the absolute maimum shear stress

will occur in the plane" and therefore" applying the second of -/. 1?72A"

we have

( )

6>2.6;

6>>>.2;4>.CH

21

>

$

26


27/77

Thus" shear failure of the material will occur according to this theory.

Maximum*+it%rti%n*nergy T-e%ry. %pplying -/. 1?76?" we have

0sing this theory" failure will not occur.

( ) ( ) ( ) ( )[ ]12C>11;A

)6>(>>.2;>>.2;4>.C4>.C

)(

2

H

22

2

221

2

1

+

+ y

27


28/77

%a$imum Prini!al S"ress "heor& 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 normal stress" the material can sustain when sub$ected to

simple tension or compression.

The e/uations are shown graphically as

)1(1 ult = )2(2 ult =28


29/77

ult

These e/uations are shown graphically if the point obtained by plotting

the values of 1D 2falls within the s/uare area the member is safe.29

I b h di d i i h i l


30/77

It can be seen that stress co7ordinate 1and 2at a point in the material

falls on the boundary for outside the shaded area" the material is said to

be fractured failed. -perimentally" it has been found to be in close

agreement with the behavior of brittle material that have stress7strain

diagram similar in both tension and compression. It cab be further

noticed that in first and third /uad" the boundary is the save as formaimum shear stress theory.

Problem

% thin7walled cylindrical pressure vessel is sub$ect to an internal

pressure of >?? lb


31/77

determine the re/uired wall thicness using (a) the maimum normal

stress theory" and (b) the 8uber7von &ises78ency theory.

P = >??psi.

r = 14

y= 6C???psi

F.E. = 6t = H

%ccording to

=c Bircumferential stress < girth stress

where

Prt

c=

31

C???1>??


32/77

5y comparing 1= c.

Thus" %ccording to the normal stress theory" maimum Principal stressshould be e/ual to yield stress??

2

Pr

C???14>??

==

=

=

=

32

% 6C??? i


33/77

%s y= 6C???psi

.>C.?

C???16???

C???

6

6C???

#nt

t

t

=

=

=

Problem 02:-

The solid circular shaft in Fig. /*0(a) is sub$ect to belt pulls at

each end and is simply supported at the two bearings. The material hasa yield point of 6>"??? Ib


34/77

400 + 200 lb200 + 500 lb

34


35/77

35

Mc


36/77

?

32;??

>3

232??

.32??>A??

.6>??>>??

>3

2

6

3

3

=

=

=

==

==

=

=

=

y

x

x

B

x

d

d

d

inlbMc

inlbM

dI

dc

I

Mc

36

Tr


37/77

inlb

OR

inlb

T

d

d

d

J

Tr

xy

xy

.3;??232??

23)2??3??(

.3;??1>6??

1>)2??4??(

3;?"23

62

23;??

6

3

==

=

==

=

=

=

=

xy

yx

yx

xy

6

3;?"23

dxy= 6;??"32

dx=

x

37

2


38/77

( )

( )22

2

2

2

1

22

22

xy

yxyx

xy

yxyx

+

+=

+

+

+=

2

6

2

661

2?.2333>

2

32;??

2

32;??

++= ddd

%ccording to maimum normal stress theory .

2

6

2

66

1

2?.2333>

2

32;??

2

32;??

6

6>???

+

+=

=

ddd

ult

38

22


39/77

I3;.1

41.1?

1?413.11?133

1?>2.4CA1?C>.34A2

1?C2.C141?133

2?.2333>

2

32;??

2

32;??12???

>

>

C>

>

>

>

>

6

>

>

2

6

2

66

=

=

=

++=

+

+=

d

d

d

ddd

ddd

39

% h , i" i


40/77

%ohrs ,ri"erion:-

In some material such as cast iron" have much greater strength

in compression than tension so &ohr proposed that is 1st and third

/uadrant of a failure broes" a maimum principal stress theory was

appropriate based on the ultimate strength of materials in tension or

compression. Therefore in 2ndand 3th/uad" where the maimum shear

stress theory should apply.

Pure shear is one in which and yare e/ual but of opposite sense.

40


41/77

( )cult

Failure or s"ren"h en.elo!e

Pure shear

Smaller "ension s"ren"h

Lares" om!ressi.e s"ren"h

41


42/77

tension

compression

42

P bl /


43/77

Problem /:-

In a cast7iron component the maimum principal stress is to be limited

to one7third of the tensile strength . ,etermine the maimum value ofthe minimum principal stresses "using the &ohr theory. !hat would be

the values of the principal stresses associated with a maimum shear

stress of 6C? &L? &L


44/77

stresshearingfourthand2

stressLormal/uadrantthirdand1

criterions&ohr#per%s

H

nd

st

2

=

=

=

1uadrant

2

)(

1

?

6

thatindicated%s

mM2

tult

==

=

2>

2


45/77

5y using principal stress theory

2

2

2

1


46/77

Therefore

22

2 ?

12? mM2and ==+

The &ohr#s stress circle construction for the second part of this problem

is shown in Fig. 16.A. If the maimum shear stress is 6C? &L


47/77

cult tult

6>?=tult131?=cult

6C?ma=

47


48/77

Fi 1/ 48

Low from second state


49/77

Low from second state

21

21ma

21

21

21cult

21tult

2

ma

2

2

1313

6;?

/uadfourthinand

theorystressshearing5y


50/77

#$am!le 10-11

The solid cast7iron shaft shown in Fig. 1?73?ais sub$ected to a tor/ue

of T = 3?? Ib . ft. ,etermine its smallest radius so that it does not fail

according to the maimum7normal7stress theory. % specimen of cast

iron" tested in tension" has an ultimate stress of (ult)t= 2? si.

Solu"ion

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

63ma

..;.?44

)2


51/77

51


52/77

52

&ohr#s circle for this state of stress (pure shear) is shown in Fig. 1?7


53/77

(p ) g

3?b. inceR = ma" then

The maimum7normal7stress theory" -/. 1?761" re/uires

J1J K ult

6ma21..;.6?44

rinlb===

2

6


54/77

The solid shaft shown in Fig. 1?731a has a radius of ?.4 in. and is made

of steel having a yield stress of y= 6> si. ,etermine if the loadings

cause the shaft to fail according to the maimum7shear7stress theory

and the maimum7distortion7energy theory.

#$am!le 10-12

Solu"ion

The state of stress in the shaft is caused by both the aial force and the

tor/ue. ince maimum shear stress caused by the tor/ue occurs in the

material at the outer surface" we have

54

'iP 14


55/77

'iin

inin'iJTc

'iin

'i

#

P

xy

x

44.1>2


56/77

BTI = C.4> si

BT2 = 72;.>> si

Maximum-Shear-Stress Theory. ince the principal stresses have

%%ite ign, then from ec. C.4" the absolute maimum shear stress

will occur in the plane" and therefore" applying the second of -/. 1?72A"

we have

( )

6>2.6;

6>>>.2;4>.CH

21

>

$

56

Thus shear failure of the material will occur according to this theory


57/77

Thus" shear failure of the material will occur according to this theory.

Maximum*+it%rti%n*nergy T-e%ry. %pplying -/. 1?76?" we have

0sing this theory" failure will not occur.

( ) ( ) ( ) ( )[ ]12C>11;A

)6>(>>.2;>>.2;4>.C4>.C

)(

2

H

22

2

221

2

1

+

+ y

57


58/77


59/77

)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=

N=

< -

7O < -

7O < -

y

Oy < -

y < -

7Oy < -

N

ON < -

7ON < -

N < -

=y=

N=

=

y=

N= 59

= < - 7Oy< - 7ON< -


60/77

y N

= < - 7O < - (yD N)

y = y< - 7O< - 7ON< -= y< - 7O < - (D N)

N = N< - 7Oy< - 7O< -

= N< - 7O < - (D y)

(2)

Thus

)6(

)(

)(

)(

126

6

612

2

621

1

+=

+=

+=

60

%lso


61/77

)>(

)4(

Then

and

3and1-/uating

)3()2(

12y

21y

122

21

1

211

=

=

=

=

=

=

,,

,,,

dF%r,,

y

y

61

The yield surface %5B, is the straight under biaial tension or biaial


62/77

compression Individual principal stresses greater than y can occur

without causing yielding.

62


63/77

arising in a tension test is 9: and the corresponding aial strain is Q


64/77

then the wor done or a unit volume of the test specimen is the product

of mean value of the force per unit area (


65/77

If it is assumed that loads are applied gradually and simultaneous then

stresses and strain will increase in the same manner. The total strain

energy per unit volume in the sum of energies produced by each of the

stresses (as energy is a scalar /uantity )

,,

)( 6211 =

,

u

,

)( 6122

+=

,u

,)( 216

6

+=

(2)

(6)

(3)

65

)4(111

3


66/77

!here Q1" Q2" and Q6" are the normal strain in the direction of principal

stresses respectively of strain are epressed in term of stresses than

e/uation (4) taen the following form

)4("2

""2

""2

6622 ++=3

++

++

+=

,,,,,,,,,3 2166

6122

6211

2

1

2

1)(

2

1

!hich can be reduced to

[ ] )>()(22

1626121

6

6

2

2

2

1 ++++=,

366


67/77


68/77

1

/

1

'a*

'b*

1-

/-

2-

'*

68

ome materials were sub$ected to hydrostatic pressure resulting in


69/77

appreciable changes in volume but no change in shape and no failure by

yielding. Therefore the remaining portion of the stress (17 )" (27 )

and (6 7 ) will result in distortion only (Lo volume change) and the

algebraic sum of the three principal strains produced by the three

principal stresses (17 ave)" (27 ave) and (67 ave) must be e/ual toNero. i.e

(Q1DQ2DQ6)d=? (C)

-pressing strains in term of stresses.

69

( ) ( ) ( ) ( )[ ]621621 +=++ d,


70/77

( ) ( ) ( ) ( )[ ]

( ) ( ) ( )[ ]

( ) ( ) ( )[ ] ?216

162

621621

=++++

( ) ( )[ ]

( ) ( )[ ]

( ) ( )[ ]

2

2

2

216

162

621

+++++=

2

22

216

162621

++++++=

>6666222111 +++=

>6222 662211 +++=

70

a) Ene part that causes *olumetric (0v)


71/77

b) Bhange and one causes distortion an (0d)

)3(d! 333 +=a) %s according to theory of distortion only energy due to

distortion is responsible for failure. ome eperimental

evidence supports this assumption some materials were

sub$ected to hydrostatic pressure result in appreciable changes

in volume but no change in shape and no failure by yielding.

The hydrostatic pressure is the average of three principal

stresses 12and 6nown as average stress.

)4(6

621

ma

++= 71

This principal strains Q in the material.


72/77

,ue to there Principal stresses (1" 2" 6) three principal strains Q1" Q2"

and Q6" are produced. The state of stress in Fig below will result in

distortion only (Lo volume change) if sum of three normal strains is

Nero. That is

( ) ( ) ( ) ( )( ) ( )

)>(?2

221)(

216

6126221

621 =

+++++=++

u

,,

!hich eh reduces to

( )

)4(6

1yd ,

u

3

+

=(1>)

%nd e/uating it with (16)

( ) ( ) ( )[ ]2

162

622

212

>1)2(

>1 +++=+

,u

,u

y

( ) ( ) ( )2162

62

2

21

22 ++=y (1A)76

For biaial stress system


77/77

2

221

2

1

22 +=y (1;)

theories_of_failure.ppt

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