LECTURE

3

Analysis of Stresses:

at
Considera pointq in somesortofstructural
memberlikeas showningure below.Assuming
thatat pointexist.q a planestateofstressexist.i.e.thestateofstate
stressis todescribebya parameters
ax,cyandrxyThesestressescouldbe indicate
a on thetwodimensional
diagramas shownbelow:

Thisis a commenwayof representing
thestresses.it mustbe realizea thatthematerialis unawareofwhatwe havecalledthexand yaxes.i.e.the materialhasto
resisttheloadsirrespective
lessof howwe wishto namethemorwhetherthey
are horizontal,
vertical
orotherwise
furthermore,thematerialwillfailwhenthestresses
exceedbeyonda permissible
value.Thus,a fundamental
problemin engineering
designis to determine
the maximumnormalstressor maximumshearstressatany
particular
pointin a body.Thereis no reasonto believeapriorithatox,rsyandrxyarethe maximumvalue.Ratherthe maximumstressesmayassociatesthemselves
withsomeotherplaneslocatedat9. Thus,itbecomesimperative
to determine
thevaluesof69andre.inorderttoachievethisletus considerthe
following.

Shear
stress:
P
L

Theseareparallel
M

lftheappliedloadP consistsoftwoequalandoppositeparallelforcesnotinthesameline,thanthereis a tendencyfor
onepartofthebodytoslideoverorshearfrom
the otherpartacrossanysectionLM.ifthecrosssectionat LMmeasuredparalleltothe loadisA,thentheaveragevalueof shearstress1.= P/A.The shearstressis
tangential
totheareaoverwhichitacts.

lfthe
shear
stress
varies
then
atapointthen
I may
bedened
asT= LimE

5A>D59"-K

Complementary
shearstress:

:
')b >
A

>>
B

x

5'
D

C
*<d(-(

I

<

4(-

LetABCDbe a smallrectangular
elementofsidesx,yandzperpendicuiarto
theplaneofpaperlettherebeshearstressactingon planesABandCD
it is obvious
thatthesestresseswillfroma couple( T. xz)ywhichcanonlybe balancedbytangential
forceson planesADandBC.Theseareknownas complementary
shearstresses.i.e.theexistence
ofshearstressesonsidesABandCD oftheelementimpliesthattheremustalsobe complementaryshear
stresseson to maintain

equilibrium.

Thus,everyshearstress is accompaniedbyan equalcomplementaryshearstress.
Stresses on obligue glanezTill now we havedealt with either pure normal direct stress or pure shear stress. in many instances,howeverboth direct and shear
stressesacts andthe resultantstressacrossanysectionwill be neithernormalnortangentialto the plane.
Aplane stse ofstress is a 2 dimensionalstae ofstress in a sense thatthe stress componentsin one directionare all zeroi.e
62= Tyz= rzx= 0
examplesof planestateofstress includesplatesand shells.
Conslderthegeneralcase ofa barunderdirectload F givingrise to a stress cy vertically
F

\°~«

Thiclrnessof the
aliamzarnl
in 2-r.li:r
is thin

andiskhansmity.

unitdepth

5

Tr; C

Thestress actingat a pointis representedbythe stressesactingon the facesofthe eiementenclosingthe point.
The stresseschangewith the inclinationofthe planespassingthroughthat pointi.e.the stress on the faces ofthe elementvaryas the angularpositionofthe element
changes.
Letthe blockbe ofunitdepth nowconsideringthe equilibriumof forceson the triangleportionABC
ResolvingforcesperpendiculartoBC,gives
G9.BC.1
= GySln9.AB .1
butAB/BC= sineorAB = BCsin6
Substitutingthis valuein the aboveequation,we get

o9.BC.1
=oysln6
.BCsin6.1
or

(1)

Now resolvingthe forces parallelto BC
r9.BC.1=crycose. ABsine.1
againAB= BCcos6
1:9.BC.1
= oycose. BCsin6.1 or 19= cysinecose
<2>

lfe = 90°theBCwillbeparalleltoABandr9= 0,i.e.therewillbeonlydirect
stressor normalstress.
Byexaminingthe equations(1) and (2),the followingconclusionsmaybe drawn

(i) Thevalue
ofdirectstress
:59
ismaximum
andisequal
tocywhen9=900.
(ii) Theshearstress
rehasa maximum
value
of0.5crywhen6=450
(iii) Thestresses0&#39;9
and G9are notsimplytheresolutionofoy
Materialsubjectedto pure shear:
considerthe elementshownto whichshearstresses havebeenappliedto the sides ABand DC

A

-1,
i

ty,

B

1.

l.pry
Complementaryshearstresses of equalvaluebut of oppositeeffectare then set up on the sides AD and BC in orderto preventthe rotationof the element.Sincethe
appliedand complementary
shearstressesare of equalvalueon the xand yplanes. Therefore,theyare bothrepresentedbythe symbolrxy.
Nowconsidertheequilibriumof portionof PBC

Assumingunitdepthand resolvingnormalto PCor in the directionofce
c9.PC.1= TXy.PB.COS9.1+
tXy.BC.sin6.1
= &#39;cXy.PB.cos9
+ rxy.BC.sine
NowwritingPBand BC in terms of PCso that it cancelsoutfrom the two sides
PB/PC= sine BC/PC= cos6

c9.PC.1
=rXy.cosesin6PC+
rxy.cose.sin6PC
<59
=Zrxysinecose
69= rXy.2.sinBcos6
05 =1 It _sin2t?

(1)

Nowresolvingforcesparallelto PCor in the direction19.thenTXYPC
.1 = 1Xy.PBsin9 1:Xy.BCcos9
ve sign has beenput becausethis componentisin the same directionas thatofre.
againconvertingthe variousquantitiesin terms of PCwe have

rxypc. 1=1xy.PB.sin26
- 1Xy.PCcos26
=[rXy(cos29 sin20)]
=1.&#39;xyCOS29
or T3= - T 00328

(2)

the negativesign meansthatthe sense ofre is oppositeto thatofassumed one. Letus examinethe equations(1) and (2) respectively
Fromequation(1) Le,
69= Txysin26

Theequation
(1) represents
thatthemaximum
value
ofceisrxywhen9=45°.
Letus take intoconsiderationthe equation(2)which statesthat
19= Cxy
cos26

itindicates
thatthe
maximum
value
ofreisrxywhen6=00or900.ithasavalue
zerowhen9=450.
From
equation
(1)itmaybenoticed
thatthenormal
componentce
hasmaximum
andminimum
values
of+rXy
(tension)
and~rXy
(compression)
onplane
at: 45°tothe
appliedshearand on these planesthe tangentialcomponentreis zero.

Hencethesystemofpureshearstressesproduces
andequivalentdirect
stresssystem,onesetcompressive
andonetensileeachlocatedat45°totheoriginalshear
directionsas depictedin the gure below:

YXJ

x.

N

V

M

531:

45:3}:
&#39;

Tn

m :9"
1

g_

,3, T

DP.

TN? are

x

M

Nowconsidera rectangularelementofunitdepth,subjectedto a systemoftwo directstressesbothtensile,6x and cyactingrightanglesto eachother.

for equilibriumofthe portionABC,resolvingperpendiculartoAC
G9.AC.1=6ysin e.AB.1+cXcos 9. BC.1
convertingABand BC in terms ofAC so thatACcancelsoutfrom the sides

63=crysinze4-rsxcosze
Futher,recallingthatcos26 sin29= cos26or (1 ~ cos26)/2= sin20
Simiiarly(1+ cos26)/2= coszq
Hencebythesetransformationsthe expressionfor <59
reducesto
=1/26y(1 COS29)+1/2GX(1
+ cos29)
On rearrangingthe varioustermswe get

(3)

Nowresolvingparallelto AC
sq.AC.1=~1:Xy..cos9.AB.1+
rXy.BC.sin9.1
The ve sign appearsbecausethis componentis in the same directionas thatofAC.
Againconvertingthe variousquantitiesin terms ofAC so thatthe ACcancelsoutfrom the two sides.

:rB.r-\C.1
= [1100ssin
crysinEcost?]AC
T5= [0, oyjsincns

L Q0*)
sin2E

<4>

Conclusions:
Thefollowingconclusionsmaybe drawnfrom equation(3)and (4)

(i) Themaximum
directstress
wouldbeequal
to(ixorcywhich
everisthegreater,
when6=00or900
(ii) Themaximum
shearstressintheplaneoftheappliedstressesoccurswhen6=45°

Trnax
: (U1
; 03)
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