theories_of_failure.ppt

design engineering

1. Theories of Failure
Failureofamemberisdefinedasoneoftwoconditions.
1.Fractureofthematerialofwhichthememberismade.Thistype
offailureisthecharacteristicofbrittlematerials.
2. Initiationofinelastic(Plastic)behaviorinthematerial.Thistype
offailureistheonegenerallyexhibitedbyductilematerials.
When an engineer is faced with the problem of design using a specific
material, it becomes important to place an upper limit on the state of
stress that defines the material's failure. If the material is ductile, failure
is usually specified by the initiation of yielding, whereas if the material is
brittle it isspecified by fracture.
1

2. Thesemodesoffailurearereadilydefinedifthememberissubjectedto
auniaxialstateofstress,asinthecaseofsimpletensionhowever,ifthe
memberissubjectedtobiaxialortriaxialstress,thecriteriaforfailure
becomesmoredifficulttoestablish.
Inthis section we will discuss four theories that are often used in
engineering practice to predict the failure of a material subjected to a
multiaxial state of stress.
A failure theory is a criterion that is used in an effort to predict the failure
of a given material when subjected to a complex stress condition.
2

3. i.Maximumshearstress(Tresca)theoryforductilematerials.
ii.Maximumprincipalstress(Rankine)theory.
iii.Maximumnormalstrain(SaintVenan’s)theory.
iv.Maximumshearstrain(DistortionEnergy)theory.
Several theories are available however, only four important theories are
discussed here.
3

4. MaximumshearstresstheoryforDuctileMaterials
TheFrenchengineerTrescaproposedthistheory.Itstatesthata
membersubjectedtoanystateofstressfails(yields)whenthe
maximumshearingstress(τ
max)inthememberbecomesequaltothe
yieldpointstress(τ
y)inasimpletensionorcompressiontest(Uniaxial
test).Sincethemaximumshearstressinamaterialunderuniaxialstress
conditionisonehalfthevalueofnormalstressandthemaximum
normalstress(maximumprincipalstress)is
max,thenfromMohr’s
circle.2
max
max


4

5. 5

6. If both of principal stresses are of the same sign tension
compression then
σ
1<σ
y (4)
σ
2<σ
y (5)
IncaseofBiaxialstressstate)3(
22
)2(
22
21
21
max
21minmax
max









y
y
y





6

7. Agraphoftheseequationisgiveninthefigure.Anygivenstateof
stresswillberepresentedinthisfigurebyapointofco-ordinate
1and

2where
1and
2aretwoprincipalstresses.Ifthesepointfallswithin
theareashown,thememberissafeandifoutsidethenmemberfailsasa
resultofyieldingofmaterial.TheHexagonassociatedwiththe
initiationofyieldinthememberisknownas“TrescaHexagon”.(1814-
1885).Inthefirstandthirdquadrant
1and
2havethesamesignsand
7

8. 
maxishalfofthenumericallylargervalueofprincipalstress
1or
2.
Inthesecondandfourthquad,where
1and
2areofoppositesign,

maxishalfofarithmeticalsumofthetwo
1and
2.
Infourthquadrant,theequationoftheboundaryorlimit
(yield/boundarystress)stresslineis

1-
2= 
y
Andinthesecondquadranttherelationis

1-
2= -
y
8

9. y
 
21 y

1 y

1 y

2 y
 
21 y

2 9

10. Problem01:-
ThesolidcircularshaftinFig.(a)issubjecttobeltpullsateach
endandissimplysupportedatthetwobearings.Thematerialhasa
yieldpointof36,000Ib/in
2
•Determinetherequireddiameterofthe
shaftusingthemaximumshearstresstheorytogetherwithasafety
factorof3.
10

11. 400 + 200 lb
200 + 500 lb
11

12. 12

13. 0
42800
64
2
4200
.42006700
.36006600
64
2
3
4
4









y
x
x
B
x
d
d
d
inlbMc
inlbM
d
I
d
c
I
Mc





 13

14. inlb
OR
inlb
T
d
d
d
J
Tr
xy
xy
.480024200
24)200400(
.480016300
16)200500(
480,24
32
2
4800
3
4










 xy
 yx
 yx
 xy
 3
480,24
d
xy 3
800,42
d
x x 14

15. 
2
2
21
max21
2
2
2
know weAs
xy
yx














 3
36000
2
)(
2
...
...
And
21
21
max




FOS
SOF
SOF
yield
yield
y





 15

16. ''76.1
480,24
2
42800
1036
480,24
2
42800
2
000,12
480,24
2
42800
2
3
000,36
480,24
2
42800
2
2
3
2
3
6
2
3
2
3
2
3
2
3
2
3
2
321





















































d
dd
dd
dd
dd
 16

17. Problem 02
Thestateofplanestressshownoccursatacriticalpointofasteel
machinecomponent.Asaresultofseveraltensiletests,ithasbeenfound
thatthetensileyieldstrengthis
y=250MPaforthegradeofsteelused.
Determinethefactorofsafetywithrespecttoyield,using(a)the
maximum-shearing-stresscriterion,and(b)themaximum-distortion-
energycriterion.
17

18. 18

19. SOLUTION
Mohr's Circle. We construct Mohr's circle for the given state of stress
and find MPaOC
yxave
20)4080()(
2
1
2
1
  MPaFXCFR
m 65)25()60()()(
2222

Principal StressesMPaCAOC
a 856520  MPaCAOC
b 456520 
a.Maximum-Shearing-StressCriterion.Sinceforthegradeofsteelused
thetensilestrengthisay=250MPa,thecorrespondingshearingstressat
yieldisMPaMPa
YY 125)250(
2
1
2
1

19

20. MPaFor
m65 2.19.
65
125
.  SF
MPa
MPa
SF
m
Y

 20

21. b.Maximum-Distortion-EnergyCriterion.Introducingafactorofsafety
intoEq.(7.26),wewrite2
22
.







SF
Y
bbaa


For 
a= +85 MPa, 
b= -45 MPa, and 
y= 250 MPa, we have 
2
22
.
250
45458585 






SF 19.2.
.
250
3.114  SF
SF
21

22. Comment.Foraductilematerialwith
y=250MPa,wehavedrawnthe
hexagonassociatedwiththemaximum-shearing-stresscriterionandthe
ellipseassociatedwiththemaximum-distortion-energycriterion.The
givenstateofplanestressisrepresentedbypointHofcoordinates
a=
85MPaand
b=-45MPa.Wenotethatthestraightlinedrawnthrough
pointsOandHintersectsthehexagonatpointTandtheellipseatpoint
M.Foreachcriterion,thevalueobtainedforF.S.canbeverifiedby
measuringthelinesegmentsindicatedandcomputingtheirratios:  19.2.2.19. 
OH
OM
SFb
OH
OT
SFa
22

23. 23

24. ThesolidshaftshowninFig.10-41ahasaradiusof0.5in.andismade
ofsteelhavingayieldstressof
y=36ksi.Determineiftheloadings
causetheshafttofailaccordingtothemaximum-shear-stresstheory
andthemaximum-distortion-energytheory.
Example 10-12
Solution
Thestateofstressintheshaftiscausedbyboththeaxialforceandthe
torque.Sincemaximumshearstresscausedbythetorqueoccursinthe
materialattheoutersurface,wehave
24

25. ksi
in
ininkip
J
Tc
ksi
in
kip
A
P
xy
x
55.16
2/.)5.0(
.)5.0.(.25.3
10.19
.)5.0(
15
4
2





 Thestresscomponentsareshownactingonanelementofmaterialat
pointAinFig.10-41b.RatherthanusingMohr'scircle,theprincipal
stressescanalsobeobtainedusingthestress-transformationequations,
Eq.9-5.2
2
2,1
2
2
2,1
)55.16(
2
010.19
2
010.19
22





 



















xy
yxyx
25

26. = -9.55 ±19.11
CTI = 9.56 ksi
CT2 = -28.66 ksi
Maximum-Shear-StressTheory.Sincetheprincipalstresseshave
oppositesigns,thenfromSec.9.5,theabsolutemaximumshearstress
willoccurintheplane,andtherefore,applyingthesecondofEq.10-27,
wehave 
362.38
3666.2856.9
?
21



Y
26

27. Thus,shearfailureofthematerialwilloccuraccordingtothistheory.
Maximum-Distortion-EnergyTheory.ApplyingEq.10-30,wehave
Using this theory, failure will not occur.    
12961187
)36(66.2866.2856.956.9
)(
2
?
22
2
221
2
1



y
27

28. Maximum Principal Stress theory or
(Rankine Theory)
Accordingtothistheory,itisassumedthatwhenamemberis
subjectedtoanystateofstress,fails(fractureofbrittlematerialor
yieldingofductilematerial)whentheprincipalstressoflargest
magnitude.(
1)inthememberreachestoalimitingvaluethatisequal
totheultimatenormalstress,thematerialcansustainwhensubjectedto
simpletensionorcompression.
The equations are shown graphically as)1(
1 ult
 )2(
2 ult

28

29. ult These equations are shown graphically if the point obtained by plotting
the values of 
1+ 
2falls within the square area the member is safe.
29

30. Itcanbeseenthatstressco-ordinate
1and
2atapointinthematerial
fallsontheboundaryforoutsidetheshadedarea,thematerialissaidto
befracturedfailed.Experimentally,ithasbeenfoundtobeinclose
agreementwiththebehaviorofbrittlematerialthathavestress-strain
diagramsimilarinbothtensionandcompression.Itcabbefurther
noticedthatinfirstandthirdquad,theboundaryisthesaveasfor
maximumshearstresstheory.
Problem
Athin-walledcylindricalpressurevesselissubjecttoaninternal
pressureof600lb/in
2
.themeanradiusofthecylinderis15in.Ifthe
materialhaayieldpointof39,000lb/in
2
andasafetyof3isemployed,
30

31. determinetherequiredwallthicknessusing(a)themaximumnormal
stresstheory,and(b)theHuber-vonMises-Henckytheory.
P=600psi.
r=15"

y=39000psi
F.O.S=3
t=?
Accordingto
c
Circumferential stress / girth stresswhere
Pr
t
c
 31

32. By comparing 
1= 
c.
Thus, According to the normal stress theory, maximum Principal stress
should be equal to yield stress/FOS i.e.
|
1| ≤ 
ult |
2| ≤ 
ult
f = maximum (|
1|, |
2|, |
3| -
ult= 0)tt
t
stressallengitudinFor
tt
c
l
c
4500
2
"15600
2
Pr
9000"15600










32

33. As 
y= 39000psi.69.0
9000
13000
9000
3
39000
Anst
t
t



Problem02:-
ThesolidcircularshaftinFig.18-12(a)issubjecttobeltpullsat
eachendandissimplysupportedatthetwobearings.Thematerialhas
ayieldpointof36,000Ib/in
2
•Determinetherequireddiameterofthe
shaftusingthemaximumnormalstresstheorytogetherwithasafety
factorof3.
33

34. 400 + 200 lb
200 + 500 lb
34

35. 35

36. 0
42800
64
2
4200
.42006700
.36006600
64
2
3
4
4









y
x
x
B
x
d
d
d
inlbMc
inlbM
d
I
d
c
I
Mc





 36

37. inlb
OR
inlb
T
d
d
d
J
Tr
xy
xy
.480024200
24)200400(
.480016300
16)200500(
480,24
32
2
4800
3
4










 xy
 yx
 yx
 xy
 3
480,24
d
xy 3
800,42
d
x x 37

38. 

2
2
2
2
2
1
22
22
xy
yxyx
xy
yxyx





























 2
3
2
331
20.24446
2
42800
2
42800













ddd
 According to maximum normal stress theory .2
3
2
33
1
20.24446
2
42800
2
42800
3
36000














ddd
ult

38

39. "48.1
51.10
10514.1
10144
1062.5971096.457
2
1092.915
10144
20.24446
2
42800
2
42800
12000
6
6
9
6
6
6
6
6
3
6
6
2
3
2
33























d
d
d
ddd
ddd 39

40. Mohr’s Criterion:-
Insomematerialsuchascastiron,havemuchgreaterstrength
incompressionthantensionsoMohrproposedthatis1
st
andthird
quadrantofafailurebrokes,amaximumprincipalstresstheorywas
appropriatebasedontheultimatestrengthofmaterialsintensionor
compression.Thereforein2
nd
and4
th
quad,wherethemaximumshear
stresstheoryshouldapply.
Pureshearisoneinwhich
xand
yareequalbutofoppositesense.
40

41. c
ult Failure or strength envelope
Pure shear
Smaller tension strength
Largest compressive strength
41

42. tension
compression
42

43. Problem3:-
Inacast-ironcomponentthemaximumprincipalstressistobelimited
toone-thirdofthetensilestrength.Determinethemaximumvalueof
theminimumprincipalstresses,usingtheMohrtheory.Whatwouldbe
thevaluesoftheprincipalstresses"associatedwithamaximumshear
stressof390MN/m
2
?Thetensileandcompressivestrengthsofthecast
ironare360MN/m
2
and1.41GN/m
2
respectively
43

44. stress Shearingfourthand2
stress Normalquadrant thirdand1
criterionsMohr'perAs
?
nd
st
2



quadrant
 2
)(
1
/120
3
360
3
thatindicatedAs
mMN
tult



 26
2
/101414
/360
mMN
mMN
cult
tult



 44

45. Byusingprincipalstresstheory2
2
2
1
/414
/360
)(
)(
mMN
mMNAs
b
a
cult
tult
ult
ult







 2
1 /120 mMN
Maximumprincipalstress=360/3=120MN/m
2
(tension).According
toMohr'stheory,inthesecondandfourthquadrant
From(a)and(b)
45

46. Therefore 2
2
2
/9401
1410360
120
mMNand 

 

TheMohr'sstresscircleconstructionforthesecondpartofthisproblem
isshowninFig.13.7.Ifthemaximumshearstressis390MN/m
2
,a
circleisdrawnofradius390unitstotouchthetwoenvelopelines.The
principalstressescanthenbereadoffas+200MN/m
2
and-580
MN/m
2
•1
11
21
21


culttult
culttult
and








46

47. cult
 tult
 360
tult
 1410
cult
 390
max 47

48. Fig 13.7 48

49. Now from second state21
21
max
21
21
21cult
21tult
2
max
2
2
1414
380
quadfourth in and
theorystress shearingBy
/390
















R
Also
Thus
mMN
49

50. Example 10-11
Thesolidcast-ironshaftshowninFig.10-40aissubjectedtoatorque
ofT=400Ib.ft.Determineitssmallestradiussothatitdoesnotfail
accordingtothemaximum-normal-stresstheory.Aspecimenofcast
iron,testedintension,hasanultimatestressof(σ
ult)
t=20ksi.
Solution
Themaximumorcriticalstressoccursatapointlocatedonthesurface
oftheshaft.Assumingtheshafttohavearadiusr,theshearstressis34max
..8.055
)2/(
)/.12)(.400(
r
inlb
r
rftinftlb
J
Tc 



50

51. 51

52. 52

53. Mohr'scircleforthisstateofstress(pureshear)isshowninFig.10-
40b.SinceR=
max,then
Themaximum-normal-stresstheory,Eq.10-31,requires
|
1|≤
ult3max21
..8.3055
r
inlb
  2
3
/000,20
..8.3055
inlb
r
inlb

Thus, the smallest radius of the shaft is determined from ..535.0
/000,20
..8.3055
2
3
Ansinr
inlb
r
inlb


53

54. ThesolidshaftshowninFig.10-41ahasaradiusof0.5in.andismade
ofsteelhavingayieldstressof
y=36ksi.Determineiftheloadings
causetheshafttofailaccordingtothemaximum-shear-stresstheory
andthemaximum-distortion-energytheory.
Example 10-12
Solution
Thestateofstressintheshaftiscausedbyboththeaxialforceandthe
torque.Sincemaximumshearstresscausedbythetorqueoccursinthe
materialattheoutersurface,wehave
54

55. ksi
in
ininkip
J
Tc
ksi
in
kip
A
P
xy
x
55.16
2/.)5.0(
.)5.0.(.25.3
10.19
.)5.0(
15
4
2





 Thestresscomponentsareshownactingonanelementofmaterialat
pointAinFig.10-41b.RatherthanusingMohr'scircle,theprincipal
stressescanalsobeobtainedusingthestress-transformationequations,
Eq.9-5.2
2
2,1
2
2
2,1
)55.16(
2
010.19
2
010.19
22





 



















xy
yxyx
55

56. = -9.55 ±19.11
CTI = 9.56 ksi
CT2 = -28.66 ksi
Maximum-Shear-StressTheory.Sincetheprincipalstresseshave
oppositesigns,thenfromSec.9.5,theabsolutemaximumshearstress
willoccurintheplane,andtherefore,applyingthesecondofEq.10-27,
wehave 
362.38
3666.2856.9
?
21



Y
56

57. Thus,shearfailureofthematerialwilloccuraccordingtothistheory.
Maximum-Distortion-EnergyTheory.ApplyingEq.10-30,wehave
Using this theory, failure will not occur.    
12961187
)36(66.2866.2856.956.9
)(
2
?
22
2
221
2
1



y
57

58. Maximum Normal Strain or Saint Venant’s
Criterion
Inthistheory,itisassumedthatamembersubjectedtoanystateof
stressfails(yields)whenthemaximumnormalstrainatanypoint
equals,theyieldpointstrainobtainedfromasimpletensionor
compressiontest(
y=σ
y/).
Principalstrainoflargestmagnitude|
max|couldbeoneoftwo
principalstrain
1and
2dependinguponthestressconditionsactingin
themember.ThusthemaximumPrincipalstraintheorymaybe
representedbythefollowingequation. 58

59. )1(
2max
1max








y
y

 Asstressinonedirectionproducesthelateraldeformationintheother
twoperpendiculardirectionsandusinglawofsuperposition,wefind
threeprincipalstrainsoftheelement.

x=

y=

z=
σ
x
σ
x / E
-μσ
x / E
-μσ
x / E
σ
y
μσ
y / E
σ
y / E
-μσ
y / E
σ
z
–μσ
z / E
-μσ
z / E
σ
z / E

x=

y=

z=

x=

y=

z= 59

60. 
x = σ
x/ E -μσ
y/ E -μσ
z/ E
= σ
x/ E -μ/ E(σ
y+ σ
z)

y = σ
y/ E -μσ
x/ E -μσ
z/ E
= σ
y/ E -μ/ E(σ
x+ σ
z)

z = σ
z/ E -μσ
y/ E -μσ
x/ E
= σ
z/ E -μ/ E(σ
x+ σ
y)
(2)
Thus)3(
)(
)(
)(
12
3
3
31
2
2
32
1
1
























E
E
E
E
E
E
60

61. Also )6(
)5(
Then
and
4and1Equating
)4()2(
12y
21y
12
2
21
1
21
1














EE
EEE
dFor
EE
y
y
61

62. TheyieldsurfaceABCDisthestraightunderbiaxialtensionorbiaxial
compressionIndividualprincipalstressesgreaterthanσ
ycanoccur
withoutcausingyielding.
62

63. Maximum Shear Strain Energy (Distortion
Energy Criterion (Von MISES Criterion)
Accordingtothistheorywhenamemberissubjectedtoanystateof
stressfails(yields)whenthedistortionenergyperunitvolumeata
pointbecomesequaltothestrainenergyofdistortionperunitvolumeat
failure(yielding)inuniaxialtension(orcompression).
Thedistortionstrainenergyisthatenergyassociatedwithachangein
theshapeofthebody.
Thetotalstrainenergyperunitvolumealsocalledstrainenergy
densityistheenergyinabodystoredinternallythroughoutitsvolume
duetodeformationproducedbyexternalloading.Iftheaxialstress
63

64. arisinginatensiontestis“”andthecorrespondingaxialstrainisε
thentheworkdoneoraunitvolumeofthetestspecimenistheproduct
ofmeanvalueoftheforceperunitarea(/2)timesthedisplacementin
thedirectionoftheforce,orε.Theworkisthus)1(
2

U
Thisworkisstoredasinternalstrainenergy.
WhenanelasticelementsubjectedtotriaxialloadingasshowinFigthe
stressescanberesolvesintothreeprincipalstressesσ
1,σ
2andσ
3.where
1,2and3aretheprincipalaxes.Thesethreeprincipalstresseswillbe
accompaniedbytheseprincipalstrainrelatedtothestressesby
equations3,4and5
64

65. Ifitisassumedthatloadsareappliedgraduallyandsimultaneousthen
stressesandstrainwillincreaseinthesamemanner.Thetotalstrain
energyperunitvolumeinthesumofenergiesproducedbyeachofthe
stresses(asenergyisascalarquantity)E
u
E
)(
321
1



 E
u
E
)(
312
2



 E
u
E
)(
213
3




(2)
(3)
(4)
65

66. Whereε
1,ε
2,andε
3,arethenormalstraininthedirectionofprincipal
stressesrespectivelyofstrainareexpressedintermofstressesthan
equation(5)takenthefollowingform)5(,
2
1
,,
2
1
,,
2
1
3322  U 






































EEEEEEEEE
U
213
3
312
2
321
1
2
1
2
1
)(
2
1 











Whichcanbereducedto  )6()(2
2
1
323121
3
3
2
2
2
1  
E
U
66

67. Asthedeformationofamaterialcanbeseparatedintotwoparts
(a)Changeinvolume,(b)changeinshapeordistortion.
Similarlythetotalstrainenergycanbebrokenintotwoparts.Onepart
representingtheenergyneededtocausevolumechangeoftheelement
withnochangeinshape(U
v)andtheotherpartrepresentingtheenergy
neededtodistorttheelement(U
d).)7(
dvUUU 
Theprincipalstressesσ
1,σ
2andσ
3ofFig.(01)canberesolvedintotwo
statesofstressesinFig.(02)b&c.thestateofstressshowninFig.b
representsahydrostaticstressconditioninwhichalltheseprincipal
stressareequaltothequantityσ. 67

68. 
1

3

1
(a)



(b)

1-

3-

2-
(c)
68

69. Somematerialsweresubjectedtohydrostaticpressureresultingin
appreciablechangesinvolumebutnochangeinshapeandnofailureby
yielding.Thereforetheremainingportionofthestress(
1-),(
2-)
and(
3-)willresultindistortiononly(Novolumechange)andthe
algebraicsumofthethreeprincipalstrainsproducedbythethree
principalstresses(
1-
ave),(
2-
ave)and(
3-
ave)mustbeequalto
zero.i.e
(ε
1+ε
2+ε
3)d=0 (9)
Expressing strains in term of stresses.
69

70.      
    
    0
213
132
321321





 dE    
   
   


2
2
2
213
132
321


 

2
22
213
132321

  63
333222111   63222
332211  70

71. a)One part that causes Volumetric (U
v)
b)Change and one causes distortion an (U
d) )4(
dvUUU 
a)As according to theory of distortion only energy due to
distortion is responsible for failure. Some experimental evidence
supports this assumption some materials were subjected 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 
1
2and 
3
known as average stress. )5(
3
321
max




71

72. Thisprincipalstrainsεinthematerial.
DuetotherePrincipalstresses(
1,
2,
3)threeprincipalstrainsε
1,ε
2,
andε
3,areproduced.ThestateofstressinFigbelowwillresultin
distortiononly(Novolumechange)ifsumofthreenormalstrainsis
zero.Thatis    
  
)6(0
2
221
)(
213
3123221
321 












u
E
E
Which eh reduces to  
)5(3/
03)21(
321
321




72

73. Thenormalstrainscorrespondingtothesestressesarefoundfromthree
dimensionfromofHook’slawgivenbythefollowingequation.)7()21(
E


Strainenergyresultingfromthesestressesorstrainscanbeobtainedby
subjectthroughthesevaluesis)8(.
2
3
.
2
1
.
2
1
.
2
1
,
321321




Uv
73

74. (5)3
)()21(
3
)(
.
2
3
321321
 

E
u
U
v 2
321 )(
6
)21(
 


E
u
U
v
(9)
Strain energy due to distortion
U=U
y+ U
d
U
d= U-U
y    













2
321323121
2
3
2
2
2
1
6
21
2
2
1

E
u
u
E
U
d
(10)   
2
321323121
2
3
2
2
2
1 21)(6)(3
6
1
  uu
E
U
d
74

75. Whenthethirdforceinthe    
2
113
2
3
2
332
2
2
2
221
2
1 222
6
1
 


E
u
U
d     
2
13
2
32
2
21
6
1
 


E
u
U
d
(12)
(13)
Forbiaxialstresssystem,σ
3=0 
2
221
2
12
3
1
 


E
u
U
d
(14)
Foruniaxialstresssystem
2
1
3
1

E
u
U
d


(15)
75

76. ForDistortiontheorythefailureoccurswhendistortionenergyofthe
memberhoweverequaltothestrainenergyofdistortionatfailure
(yielding)inuniaxialtorsion(orequilibrium).Sosubstitutingσ
1=σ
yin
equ(15)
2
2
6
1
yd
E
u
U 


(16)
Andequatingitwith(13)    
2
13
2
32
2
21
2
6
1
)2(
6
1
 



E
u
E
u
y    
2
13
2
32
2
21
2
2  
y
(17)
76

77. Forbiaxialstresssystem2
221
2
1
2
2  
y
(18)
77