biaxial mohrs circle

7/27/2019 Biaxial Mohrs Circle

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Introduction

Stress,by definition, is a vector as zuch in tbe real world one would expect 3

dimensionso be required o fully modela sfess field.Hower/er,sDecan often reduce

complex systemso a 2 dimensional tress ield where it is assumedhat the stressn

one direction is zero. In this handout he classicanatysisof a 2D stress ield will be

examined,n particular he applicationof Mohr's stresscircle to the uo6"1513adingf

2D stress.Further more firndamental xamples f the applicationof 2D stressnnalysis

wiii bepresented.

2D Stress heory

Consider a rectangular element, the

applied forces to the structure (of

which the elements a part) give rise o

two orthogonal direct stresses a

nominally horizontal direct stress, a

nominally vertical direa stressand a

shearstress as llustrated n Figure l).

Note that the orientationof the triangle

ABC may be rotated zuchthat it fits

the criteriaabove, he criteria doesnot

mean hat the triangleABC can only

lie in one direction. Further more,

considera planeat any angle0 relative o horizontalplane(note hat of signifies he

shessacting on the y planeand oy signifies he stressacting hex plane). t shouldbe

noted that o;, o, andrry are sfresseshat havebeendirectly applied o the strucfure,

og and re a.re tresses hicbarise rom that loading on ctDlplane.

If it is assumedhat the triangleABC hasunit width and tlur the hypoteneuss unit

lengtlr, then the area cjf AC is l, the areaof AB is cosOand the areaof BC is sixCI.

Utilising Newton's second aw of motion" and in partiorlar d'Alembert'sprinciple,expressionsor og andr0 may be obtained.

Figure 1 - 2D stressmodel

-rl



Since

l r = o

therq or og

oe = (o, cos9)cosO+(o"in0)sing+2ro cos0sin

=(o ' :o , . ] . [ " - , ] * rze* r ,sn20 . . . . . . . (1 )\ 2 ) [ 2 )

for rg

Te= 6,cos9sin0- orsingcosg- t,n cos'0 + t, sin2

( o , - o , ) .- l ' ' l s i n 2 9 - t co s? A| . , . 2 0 - r c o s 2 A . . . . . . ( 2 )\ z )

(The readershouldattempt o proveboth equations and2)

Upon close exarninatiorgquation1 revealrthat it'is"not"the complex'equationt

seems. ndeed, t is but an eguation or a circle (as llustrated n Figure 2) whose ures

are t and o, whose centre

lies on ((o*+or)/2,0).

Note that within the circle,

the anglss are twice what

they are in the real

structure (2e), further

more note that the angles

are measured from a

different datum to that

normd$ orperienced,hatis they aremeasuredrom

the o* plane.

It is interesting to note

tlat the combinationof the two orthogonal stressanda shearstresshas ncreasedhe

direct stresses ctingon the system.This fact is highlighted y the horizontaloris, this

is the direct stressacting on the syster4 and anyplanealigningwith this aris bas zero

shear stressaaing on it but ma,ximumdirect stresses, ny two orthogonalplanesaligning with this axis are called the PRINCIPAL PIIWES, and the dLect stress o1

xy

-\

/ , i

/ t 'I !

l i il i j

i j-,l-,

2 - Moht's circle of stress

6



and o2) which act on theseplanesare calledPNNCIPAL.S?"ESSES. Tbe angular

orientationof theprincipalplaness easilydeterminedrom equation 2).

( t , \

tzn20 l

' ' '

| . . . . . . . . . . . . . . . . . . (3)\ 6 ' - 6 ' )

AIso, usingFigure2 it is possibleo determine8n expressionor the principlestresses,

that is

c,,.cz=(ry)-+", - or)' +4to (4)

which is ust tle centreof the circleplusand minus he radiusof the circle. t sbouldbe

notedthatwe are aszumin a 2D stress ield, in reality he 3D stress onventionshould

be applied

o1>o2>o3

Where n a 2D stressieldoneof theprincipalstressesszero.

Another important change, hat often goes un-noticed, s the increase n mosmum

shearstress indicatedby tbe growth of the stresscircle).The ma,ximum hearstress s

givenby the radiusof the circle or

I4

L

6 t - c z

2

Note that o2 shouldbe interchangedwith o3 if the stressstate s such hat o2 is zero

and o3 is ess hanzero.

As with most engineeringoolboxes he problem s not how to use the equationbut

where to get the valuesof direct stress and shear stress o put into the equation.

Exa^mplesf 2D stressmodelswill now be erramined.

(s)

q



Example : Shaftunder orsion.

Considera drive shaft,a tran$nissionshaft in a car or a PTO shaft n a tractor for

example. f the shaft has an externaldiarneterof 50 *, is constructed rom thickwalled tubing whereF7 mm and is transmitting 6 kW at 3600 rev/min; determine he

principalstressesn the system.

A circular shaft in torsion may be modelledusing the torsion eguation, his may be

rearrangedo yield :-

t : TrlI

T = P / w =6000 = 15.92Nm

3 6 0 0 x 2 x n / 6 0

. zr(o4-6+)z(.054-.0364)r :

32 r/-

therefore, heappliedshearstresso the zurfaceof the shaft s15 .92.025

t:ff i :0.887 MN/m2

It is noted hat for a systemactingunder orsion only, the systems in pureshear,hatis o" andoy areboth zero. An examination f equation4 with this informationshows

that the principlestresses r€ o1 - '1-!,e'2: 0 ando3 : -1. An examination f equation3 shows hat he principalplanesie at 45o to the x-y planes. hat s as llustratednFieure l . l

Figure l.l

This correspondso a stress ircleas llustrated n FigureBl.2

\--lgrlo.

llv-Zlt'

l l \,/ l l .l l r . \ l l

, / - \ " 2

t o



t

ry

ge450

ry

FigrneE1.2

This alsodemonstrateshe conventionor shearstress,hat is positive f a planecarries

an anti-clockwise hear.Thereforen this problern he horizontalplane s negativeand

the verticalplanepositive.Note that the shears actuallyapplied o the verticalplane,

the shearr the horizontalplane s complementfrT hear.

Example : Ship/boat ropellorshaft.

Consider he sameproblemdiscussedn exanple I but wherethe power is being

transmittedo a propellor.Tbeactionof tbepropellor s to "push" he waterbacwards

in relation to the sbip. Newton's 3rd law states hat every action has an equal and

opposit eaction,aszuch be water mustbe "pushing"back by the samermount.Since

there is nothingretaining he boat it movesforward.However this action inducesa

compressiveoad on the shaft. fis 6edifies the system s llusffated n FigrneE2.l

FigureE2.l

The compressiveoad aaing on the propellorshaft s I kl,I. The compressive tresss

tbereforeF -1000

: -1.06MN/m2y : f =n(.0252-.on21

substiruting heknownvalues nto equation3 yields

t \



S affordsh re Univen ity

School of EngineeringSolid MechanicsGroup : Tutorial

2D stre.ss nalysi.s

1. An a:deof a go-kart s loaded n suchaway that on the surfacea planealigned

circumferentially directcompressiveEess f 100MN/m2 is applied- n association

with the compressive sbess is a shear stess applied from the driving torque

equivalent to -45 MN/m2. On a plane pcrpendicular to the one carrying the

compressive tressonly complementary hearexists.

a) Sketcha portionof the a;dehigblighting he stresssystem.

b) Determine he principal stresses.

c) Deterrrine the maximumshearstress.

d) Sketcha Moh/s stess circle for the systemandcompare be estimateof the

orientationof theprincipal planes rom the circle with that obtained

exactly.(17.3,-11?.3 67.3MN/m2)

2. A component s sfengthenedusinghigh strength ibres. The component s

loadedsuch hat on oneplanea direct stress f 15 MN/m2 anda shearof 5 MN/m2

exists.The fibreshavebeenalignedsuch hat the ong axesof the fibres areat 25o o

the loading. f the maximumstength of tbe fibres is 60 MN/mZ and maximum shear

10 MN/m2deterrninehe factorof safety orthe fibres.

(1 .12 )

3. A thin walled steel tube of 150 mm internaldiameterand wall thicknes2mm

containsa gaspressurisedo I I bar. Determine he tensilestressand shearstress

actingon a helical seem nclinedat25o o the cross ection.(24.4&7.92MN/m2)

4. A shortcolumn s loadedaxiallywith a compressiveorceof l5kN. The column s

also subjectedo a torsional oad of 150 Nm. ffthe column is constnrcted rom ahollow steel ube of 250 mm internal diameterand wall thickness25 mm determine

a) Thecompressivend shearstressesctingon a rectangular lement aligned

with the long axisof the column)on he surfaceof the column.

b) Theprincipal stresses.c) The maximrrm shearstress.

(-0.85,0. 83,0. 08,{. 86,0.43MN/m2)

t3



f - I . 0 6 \ 1 -01,62

[.fJt7r/(t.oo)'+4x 0 8872

01= 0.5302:0 & o3 = -1.54MN/m2.

Which when plotted on a stresscircle becomes

FigveE2.2

The angleof the principalplanesn relation he thex-plane s givenby equation3

,(2 x 0.882\2e: t*-t[,:fr6:J = 141-11167):59.1o therefore = +29.550

Themaximum hearstress ctingonthe system s

/o.sr-(-t .srt \tmax:

W)=1.03MNim2.

t

l-t.up.*nz1

xy

C'

\

o? Y

r FD€a /

v(0.4.887)

l ' \



StaffordshireUniversitySchoolof Enginccring

Module 3.2 EnginceringMcchanics

ExperimentalStrain Analysis"'

JEarnLrauges

In 1856 I-ord Kelvin found that when a wire is extcnded s rcsistance hanges,

indeed that the change n rcsistancc s proportional to thc change in the wile's

dimensions.n. the 1930's his fact was developcd or usc in cxperimentalstrain

analysis, mallgaugesmade rom fine wire bcing attachedo the surfaceof an object

such hatwhen he objectdeforms hewire gaugedcformswirh t.

Bas€

Wres/Folls

Figure A TypicalStrainGauge

Srain gauges avedevelopedurther,modernsrain gaugesseeFigure l) aremade

from foil andcomeanachedo a basewhichcanbereadilyadheredo most materials.

Their designhasevolvedso that axial strains anbc measured ith confidencecross

sensitiviry cing imitedto at east ess han27o).

t 4



Since t is commonly impossible o know the exactdirectionof the principat stresses

(andhenccprincipal strains) t is commonpractice o mount straingaugesas groups.

Figure 2 illustrates somecommonconfigurations, hesearc lnown as strain gauge

rosenes.Although, in theory,any configuration(basedon a commoncentre) can be.

utilised as a strain gaugerosene;ttuec strain gaugesconfigured as 120o and 45o

rosenesare usual(see heoreticalanalysisater). nd€ed he useof tbrecstraingauges

also facilitates a graphical solution to thc problem of detcrmining principal

strains.Indeed traingauges rcnot solelyuscd or srain analysisbut they alsoplay a

major role in expcrimentalequipmentastheprimary components f load cells.

-J tll LIJ+I,"W-#N".rllfl3!'

r^crsd'd.nskr\

1 2 3 . 5 6

WNry.:

Figure2 Typical StrainGaugeRosenes

t 5



Mohr's Strain Circle

Figure 3 is a graphicalrcprcscntationof a 2-D strain systemwhich exists within a

material specimn. As with the stresscircle, the horizontal axis dcpicts direct strain

and the vertical ods depictsshearstrain (asopposed o direct and shearstress);one

subtlediffercnce,however, s that he verticalaxis s infactTl2.

Figure3 Mohr's StrainCircle

Anothercommon earurewith the stress ircle s that ncluded angleswithin thecircle

are nfact 20.It is therforepossibleo determinehe shearstress nd direct stress t

any anglularorientationwithin the material.Arguably, the more important faciliry is

that the magnitude and directionsof the kincipal Strains can be found- Principal

Strains are direct strainswhich lie on a planeon which there s no shear similar to

principal stresses),hey are the maximum and minimum direct strains n the system

and are denotedEt Ezas ndicatedon Figure3.

If three direct strains and their respective orientations are known then the

constnrctionof a strain circle by geomebrys a fairly simple task. For example

Lb



consider he system llustrated n Figrrrc 4, three gaugesare scparatedby included

angles of 45o . The strainsmeasuredat a particular nstant are -200, 150 and 200

x106e respectively. From this information alone a circle can be constructedby

adhereingo the following guidelines:-

Figure4 Strain GaugeConfiguration

i) On graphpaperdraw vertical lines qpaced uch hat they representvaluesof -200,

0, 150 and200.The0line is theshear trainaxis.

ii) Choose omepointon the"middle" ine, n thiscase150 s in the "middle"of-200

and 200. Redraw the strain gaugeorientationon this line such that the "middle"

gauge s alignedwith the "middle" line. Extrapolatehe other gaugesso that their

orientationsntersectwith theirrespective alue ines.

iii) Bisect the two gauge ines produced n ii) andmark where they intersect, his is

thecentreof thestraincircle.

iv) Using the centre and the intersectionpoints deterrrined in ii) draw the circle.

Through the centrepoint drawa horizontal ine (directstrainaxis).

v) To obtain the true gaugeorientationfirst frnd the true position of the "middle"

gauge.This is done by ftnding the intersectionof the circle with the "middle" value

line oposite o that where the constructiongaugewas initially placed-Join this point

with thecentre,t rcpresentshe arm of t}16middle" gauge.

t-?



vi) The two remeining gaugescan thus bc found by joining their rcspectivecircle/value ntersectionswith thestraincircle centre.

vii) The orientation of the principal strainswith any of the threegauges an thusbe

found by measuirng he included angle benveen he gaugearms and the horizontal

axis.

It

(Draw StrainCircle Here)



NumericalAnalysis

Strain GaugeOrientation

Since heorientation f the hreegauges re nown then:-

eL=+(e1 e) * i (. , - e/cos o

€m=* (r, * E2) i Cr, e/cos2(0+ cr)

en=i(e1+ e) * j C., e/cos2(0+ a+ P)

Thesehold rr:e for an three-gaugeosettewith any angularorientation. f the gauges

are either 45o or 1200 then the following generalsolutions or the principal srains

and heangularorientationmaybeutilised:-

rn



. +e { + E n ) _.

- 2 \

tan2g

and

Et Ez

tan 20 =

and

q - 2 e - + e nQ - e n

S(r- - rn)(2q -e - -en )

45o rosette

Er E2=i(.t * €m+ en)t r=3{t(e{

- e6)z+ (em en)2+ q - en)21

For otherst'aingaugeorientationshe fundarnentalquationsescribed arliermay

besolved singnumericalechniqueso obtaina solution or theprincipalstrains nd

t(q-em)2+(En-em)21

12@rosctte

Le



their associated rientation.Furtherto the use of threegauges osettesmade of four

gaugesare alsounlised, he additionof anothergaugecan mproveaccuracy.

1t



S affo r dsh r e Un v ers tyFaculty of Computing, Engineering and Technology

2-D StressStrain Manipulation

Figure 1PlaneStress-Planeu'ainRelationships

The results obtained from experimental echniquessuch as strain analysis can be

mademore useful if they cal be transformed nto valuesof stress.Considera 3-D

stresssystemas shown n Figure 1(a), he orthogonalstrainscan be determined orm

an nvestigation f thePoisson'satio effect:-

I

er= E [o1 -v (o2+o3)J1

ez= E[o2-v(o1+o3)J1

es=E[o3-v (o2+or ) ]

E-:_



If the system s in plane stress the body is thin in the z axis then the stresssystem s

2-D as shown n Figue 1(b)) then 03 = 0. If however, he body is consideredo be

tong in the z-anis, as shown in Figure 1(c), then the strain in that ocis can bc

considcredo be zero(e3= 0), this is calledplanestrain; he strcsss not rcro.

Removing 6i3 from the equationsabove and rcarrangingyields two cquations(for

planestress)which yield principalstrcssesrom principal strains:-

E(e1+ ve2)or=-?i]'vzr

E(e2 ve1)oz=-lF

The values of principal strainscan be deterrrinedfrom an analysisof experimental

strainsusingMohr's straincircle or any otherapplicablemethod.

13



Ex4At a point on the surfaceof a component' a 60o osettestraingauge

positioned,,noJi'ilfi;, l:tmeasurestrainsfer:0'00046'

r- = 0.0002 nds,,: -0.00016.-UseMohr'ssttt'incircle o determine

themagnirude"#

dir".tion of th-e rincipal trains ndhence he

princip-altresses. = 208GN/m2, = A'29'

€r: 0.000525 and €2: -0'00019

The angle between el and tr on the circle is 34" and betwe€r €; and e2,

214".Therefore

0 t :17" and 0z: LW"

The principal stresses re given by208x loe

or:ffi(o.0o0s2s +0.29(-0.00019))

208x loeor:

-ffi(-0'000 19+ 0'29 0'00025)

from which

or = 106.3MN/m2

oz: -86.0 MN/m2

P

o,,='o 000{6Lo.omto

( r =

2tr



STAFFORDSH RE UNIVERSITY

School of Engineering and Advanced Technology

PROBLEMS ON TWO-DIMENSIONAL STRAINANALYSIS

1. Astateof two-dimensionalstra in ise*=70010{,er= -600 10-6 ndy"y 300x 10-6.Calculateheprincipalstrainsn magnitude nddirection.Check he results singMohr'scircle construction.

[Ans: 717x 10-6,617x 10-6]

2. The followingstrainsare ecorded y wo straingauges,heiraxesbeingat right angles:

E*= 390x 10-6; v= -!20 x 10{. Find he values f the stresses* ando, actingalongthese xes f the elevant lastic onstantsreE = 208GN/m2andv = 0.3.

[Ans: 80.9MNVm2;-0.69MN/m2]

3. The followingstrainswere ecorded n a rectangulartrain osette:-

E " = 4 5 0 x10 -6 , u=230x1 0 { ;e = 0

Determine:-

(a) theprincipa-ltrains nd hedirections f theprincipal trainaxes;

(b) theprincipalstressesf E = 200GN/m2 ndv = 0.3

[Ans: 450.05x 10-6 t i" clockwiserom A, -0.055 105at 91" clockwiserom A; 98

MN/m2l

4. Thevaluesof straingiven n problem were ecorded n a 60" delta osettegauge.Whata-re ow thevaluesof theprincipal tninsand heprincipal [esses.Check heanalyticalsolutionby means f Mohr's circles.

[Ans: 488.22 10-6, 31.56 10-6; c5.22MN/m2 25.25MN/m2)

2 5



5. A circularbar 50 mm diameter s subjectedo an

axial tensionand an applied orque. A rectangular

strain gaugerosette s attached o the shaft with

gauge B alignedparallel o the lon-eitudinal xis.

Themeasuredtrains re€e= 350x 105,€s= 250x

10-6; . = -100x 10-6. f E = 70 GN/mz nd = 0.3

calculate the maximum tensile stess and themaximumshearstressand the valueof the applied

torque.

[Ans: 26.36MN/m2,13.86MN/m2,340.15Nm]

Write a program o determineheprincipal trainmagnitudesnddirectionsor any strain

gauge os.ti". Determine lso heprincipal tressesor prescribedtressesor prescribed

uutu.. of Young'smodulus ndPoisson'satio. Test he program sing he tollowingdataobtained y rectangular nddeltastraingauge osettes.

(a) Rectangular traingauge osette:-

I. €"= 450x 10-6, ,"-- 420 10{, €"= 1100 10-6;

2. t "= 200x 10-6, u=900x 106,%= 600 106;

3 - € " = 1100x 0 -6 ,u=200x 0 - 6 ,L=200x10 -6

(b) Deltastraingaugeosette:-

t. e"= 2000 10-6, u=1000 106, L= -500x 10-6;

2. r" = 1.100 10-6, u 400x 10-6, "= -550x 106;

3. r" = 1100 10-6, u=200x 10-6, = -600x 10-6

7. Extend the prograrn for a graphicalsolution option by meansof Mohr's circle

constructions.

6 .

2 6



Staffordshire University

Schml of Engineering

Module 3.2 EngineeringMechanics

Yield Criteria

Introduction

A srudy of Biaxial Stress heory illustrates hat the stresswhich exists n a system s

more complex then just a force/areascenario. n fact both direct stressand shearstress lay a major role in defining he stress t aparticularpoint.

If one considers he failure of materials,and in panicular the failure of a simple

tensile est specimen, ne can magine he complexnatureof material ailure and n

par-ticular ield. A round tensile estspecimenas llustratedn Figure 1) commonly

fails such that a cup and coneareproduced.The failure indicates he existence f

shear tresseset only tensile tresses ere nduced!Biaxialstressheorysolveshe

dilemaof co-existence, ut evenso thematerialwas oaded n tensionandyet failedeffectively n shear.

Theabovedescriptionllustrateshecomplexnarure f yield failureanddemonstrates

the need for yield failurc critcria. Indecd the dcvelopmentof an cffective yield

critcria has bccn a major topic of rcscarch,n this scction hrecof the morc common

n\7

gFigure TensileSpecimen ailure

biaxial mohrs circle

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