Stress Transformation GENERALEQUATIONSOFPLANESTRESSTRANSFORMATION

SignConvention Positivenormalstressesx andy actsoutwardfromallfaces. Positiveshearstressxy actsupwardontherighthandfaceoftheelement.

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Theorientationoftheinclinedplaneisdeterminedusingtheangle. Establishapositivexandyaxesusingtherighthandrule. Angleispositiveifitmovescounterclockwisefromthe+x axistothe+xaxis.

x

y

2

GENERALEQUATIONSOFPLANESTRESSTRANSFORMATIONNormal and shear stress components Sectionelementasshown. AssumesectionedareaisA. Freebodydiagramofelementisshown.

3

Applyequationsofforceequilibriumtodetermineunknownstresscomponents

4

Simplifytheabovetwoequationsusingtrigonometricidentities:sin2= 2 sincos

sin2= (1 cos2)/2

cos2=(1+cos2)/2

Normalandshearstresscomponentsare:

Thenormalstresscomponentycanbeobtainedbysubstituting=+90o intoxexpression,then:

y

(1)

(2)

(3)

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The stress transformation equations show that if the state of stress (x, y and xy)at a point is known, we can calculate the stresses that act on any plane passingthrough that point.From the expressions of normal stress components, one obtains:

x + y = x + y

How to obtain the Inclined Angle

x

y

xy

=30o=135o

(4)

6

Example (1): State of stress at a point is represented by the element shown.Determine the state of stress at the point on another element orientated 30clockwisefrom the position shown.

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x = -80 MPa

y = 50 MPa

xy = -25 MPa

=30o (or330o)

80 50 sin 60 ( 25) cos( 60) 68.802x y

MPa

80 50 80 50 cos 60 ( 25)sin( 60) 25.852 2x

MPa 80 50 80 50 cos 60 ( 25)sin( 60) 4.15

2 2yMPa

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Problem 1: Determine the normal stress and shear stress acting on the inclined planeAB. Solve the problem using the stress transformation equations. Show the result onthe sectioned element.

Answer

9

Trytosolvethisproblem

InPlanePrincipalStressesandPrincipalPlanes

The maximum and minimum normal stresses (1 and 2) at a point are called theprincipal stresses at that point. The planes on which the principal stresses act arereferred to as the principal planes. The directions that are perpendicular to theprincipal planes are called the principal directions. The values of the angle thatdefine the principal directions are found from the condition dx/d=0. If we usethe expression for x from equation (1), this condition becomes:

Equation (5) yields two solutions for 2p that differ by 180o. If we denote onesolution by 2p1, the second solution is 2p2 = 2p1+180o. Then, the two principaldirections differ by 90o.

2 sin 2 22

tc an 2os2 02

x yxxy

xyp

x y

dd

(5)

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sin2p andcos2pcanbeobtainedas:

Substitutingtheaboveexpressionsintoequation(1),yields:

wherethemaximumandminimuminplaneprincipalstresses1 (ormax )and2(ormin )

actontheprincipalplanesdefinedbyp1 andp2,respectively,notethat1 2.

2x y

xy2 p

22

2x y

xy 2 2sin 2

2x y

p xy xy

22cos 2

2 2x y x y

p xy

11

(9)

(7)

(8)2

2cos 22 2

x y x yp xy

22

1,2 2 2x y x y

xy

12

The shear stresses acting on the principal planes are obtained by substituting equations (78)into equation (2). The result is:

Itcanbeseenthatxy=0.Thatmeansthereisnoshearstressactsontheprincipalplanes (p1andp2).

2 22 2

2sin2 cos2 0

2 2

2 2

x y

x y x y xyx y xy xy

x y x yxy xy

MaximumInPlaneShearStressThe largest magnitude of xy at a point, denoted by max, is called the maximum inplaneshear stress. The values of that define the planes of maximum inplane shear are foundfrom the equation dxy/d=0, where xy is given in equation (2). Setting the derivative equalto zero and solving for the angle give:

Thetworootsofthisequation,S1andS2canbedeterminedusingtheshadedtrianglesasshown.

Theplanesformaximumshearstresscanbedeterminedbyorientinganelement45fromthepositionofanelementthatdefinestheplaneofprincipalstress.

cos2 2 sin

n

2

ta 22

0x y x y xy

x yS

xy

dd

(10)

2x y

xy 12 S

xy2

x y

22 S

13

14

Fromtriangleshown:

and

Substitutingtheaboveexpressionsintoequation(2),themaximuminplaneshearstressis:

Thenormalstressesactingontheplanesofmaximuminplaneshearstressarefoundbysubstitutingtheexpressionsofsin2 andcos2intoequation(1)as:

2 2sin22

x yS x y xy

2

2cos22

x yS xy xy

(11)

2x y

avg (12)

22 1 2

maxin plane 2 2

x yxy

maxin plane

avg

avg

ExamplesExample(1): FindtheprincipalstressesandtheprincipaldirectionsforthestateofplanestressgiveninFig(1).Showtheresultsonasketchofanelementalignedwiththeprincipaldirections.Solution:x=8000,y=4000andxy=3000.

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22

1,2

1 2

2 2

9605.6 2394.4

x y x yxy

and

1 112

2 2 30001tan tan 28.2

2 8000 4000

90 28.2 118.2

xy op

x y

o o op

Todeterminewhichofthetwoanglesisp1 (associatedwith1)andwhichisp2 (associatedwith2),equation(2)isused:

whichisequalto1.Thentheprincipalplaneanglep1 isassociatedwithmaximumprincipalstress1.

Notethatthereisnoshearstressontheprincipalplanes.

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1

cos2 sin22 2

8000 4000 8000 4000cos 2 28.2 3000sin 2 28.2 9605.6

2 2

x y x yx y

UseMohrsCircleforPlaneStress:x:(x,xy)=(8000,3000)y:(y,xy)=(4000,3000)

17

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Example (2): For the state of plane stress shown in the figure, determine (i) the principal stresses and theprincipal directions, (ii) the maximum inplane shear stress and the planes on which it acts. Show the results onsketches of elements aligned with the planes of principal stresses and maximum inplane shear stress.Solution: x=40MPa,y=100MPaandxy=50MPa.

Tocheckwhichofthetwoanglesisp1 (associatedwith1)andwhichisp2 (associatedwith2),equation(2)isused:

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1,2

1 2

2 2

56 116

x y x yxy

and

1 112

2 2 ( 50)1tan tan 17.8

2 40 100

90 ( 17.8) 72.2

xy op

x y

o o op

Sincex =56.0MPa=1,thentheprincipalplaneanglep1=17.8o isrelatedtomaximumprincipalstress1.

MaximumInplaneShearStress:

Theorientationofplanesthatcarrythemaximuminplaneshearstressareas:

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1

cos2 sin22 2

40 100 40 100cos 35.6 50sin 35.6 56.0

2 2

x y x yx y

y

xp1=17.8o

2=-116MPa

1=56.0MPa

p2=72.2o

1 2m ax

in plane

56 11686 M Pa

2 2

1 11

2

40 1001 1tan tan 27.23

2 2 2 100

90 27.23 117.23

x y oS

xy

o o oS

Todeterminethedirectionsofthemaximuminplaneshearstressesonthesidesoftheelement,wemustfindthesignoftheshearstressononeoftheplanes.BysubstitutingS1=27.23o intoequation(2):

Thenegativesignindicatesthattheshearstressonthepositivexfaceactsinthenegativeydirection,asshowninthefigure.

Thenormalaveragestressesactingontheelementarecomputedas:

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sin2 cos22

40 ( 100)sin 2 27.23 50 cos 2 27.23 86.0MPa

2

x yx y xy

40 ( 100)30.0MPa (comp)

2 2x y

avg

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UseMohrsCircleforPlaneStress:x:(x,xy)=(40,50)y:(y,xy)=(100,50)

ProblemsProblem(1):Thestateofstressatapointisshownwithrespecttothexyaxes.Determinetheequivalentstateofstresswithrespecttothexyaxes.Showtheresultsonasketchofanelementalignedwiththexyaxes.

Problem(2):Forthestateofstressshown,determinetheprincipalstressesandtheprincipaldirections.Showtheresultsonasketchofanelementalignedwiththeprincipaldirections.

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Answera:)

Answerb:)

(a) (b)

(a) (b)

Answera:)Answerb:)

Problem(3):Forthestateofstressshown,determinethemaximuminplaneshearstress.Showtheresultsonasketchofanelementalignedwiththeplanesofmaximuminplaneshearstress.

Problem(4):Forthestateofstressshown,usingtheMohrscircledetermine(a)theprincipalstresses;and(b)themaximuminplaneshearstress.Showtheresultsonproperlyorientedelements.

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Answera:Answerb:

(a) (b)

(i) (ii)

Answeri:

Answerii:

GoodLuck,anyquestion,pleasedonothesitatetocontactme. PreparedbyDr.MohammedAliHjajiMechanicalandIndustrialEngineeringDepartment,OfficeL105;Emails:mahjaji@hotmail.com orm.hjaji@me.uot.edu.ly

SolvedProblemsExample(1):Thestateofplanestressatapointwithrespecttothexyaxesisshowninthefigure.Determinetheequivalentstateofstresswithrespecttothexyaxes.Showtheresultsonasketchofanelementalignedwiththexandyaxes.

UseMohrscircletochecktheseresults

Example(2):Forthestateofstressshown,determine(a)theprincipalstresses;and(b)themaximuminplaneshearstress.Showtheresultsonproperlyorientedelements.

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