Lecture 9 Plane Stress of materials .pdf

Mechanics of materials

1. 9. Two Dimensional State of Stress: Transformation of
Stress…Cont’d
Introduction
•Thisisachangingthecomponentsofthestateof
stressgiveninonesetofcoordinateaxestoany
othersetofrotatedaxes.
•Thepossibilityoftransformingagivenstateof
stressinvolvingbothnormalandshearstressesto
anyothersetofrotatedcoordinateaxespermits
anexaminationoftheeffectofsuchstressesona
material.
•Inpreviouschapters,stressescausedbyseparate
actionscausingeithernormaland/orshear
stresseswereconsidered.
1

2. 9. Two Dimensional State of Stress: Transformation of
Stress
2
Representations of stresses acting on an element

3. 9. Two Dimensional State of Stress: Transformation of Stress
•Superpositionofnormalstressesactingonthe
sameelement,whenaxialforcesandbending
momentoccursimultaneously,wasalsodiscussed
inlecture8.
•Ofteninstressanalysis,amoregeneralproblem
arises,suchasshowninfigure1.
•Intheillustratedcase,elementAissubjectedtoa
normalstressσ
x
duetoaxialpullandbending
moment,andsimultaneouslyexperiencesadirect
shearstressτ
xy
.
3

4. 9. Two Dimensional State of Stress: Transformation of Stress



x
x
x
x
y
y
y
y

 



x
x
x
x
x
xx
y
y
y
y
y
y
y




Fig. 1: A Two Dimensional Stress System
4

5. Equations for Transformation of Two Dimensional Stress
•Givenarethestresses
x,
yand
xyactingonthe
sidesofaplaneelement.fig.2(a)
•Requiredarethenormalstress()andthe
shearingstress()actingonanyplaneintermsof
thegivenstressesandtheangleofinclination(q
o
)
oftheplanebeinginvestigated.Fig2(b)



x
x
x
x
xx
y
y
y
y
y
y



 


x
x
x
y
y
y



q

Fig. 2(a)
Fig. 2(b)
5

6. Equations for Transformation of Two Dimensional Stress
•Considerthestressesandactingonaninclined
planeofuniformlystressedrectangularblock.fig3(a).
•Theinclinedplanemakesanangleq
o
withtheyaxis.
Thelengthofthehypotenuseiscandthethicknessof
theblockistunitsoflength.Seefig3(a)andfig3(b)
Fig. 3(a)
Fig. 3(b)


x
x
x
y
y
y



q

c
c.sinq
c.cos
q 



x
x
x
x
x
x
x
y
y
y
y
y
y
y





+u
+v

A
B
q
q
q




x
x
x
x
y
y
y
y


tunits
6

7. Equations for Transformation of Two Dimensional Stress
•Consider the forces acting on the wedge. Fig.4(a)
Fig. 4(a)
Fig. 4(b)q
-A.cosq
+A.sinq
 q
-A.cosq
-A.sinq

Fig. 4(c)


+u
+v
x
x
y
y
y
x



q

c
c.sinq
.t.c.sinq
.t.c.sinq
.t.c.cosq
.t.c.cosq
c.cos
q
.t.c .t.c
7

8. •Equilibrium of the forces In the u-and v-directions:
Sum of forces in u-direction =0:
.t.c-
x.t.c.cosq.cosq-
y.t.c.sinq.sinq-
xy .t.c.cosq.sinq-
yx.t.c.sinq.cosq=0
Sum of forces in v-direction =0:
.t.c+
x.t.c.cosq.sinq-
y.t.c.sinq.cosq-
xy .t.c.cosq.cosq+
yx.t.c.sinq.sinq=0
=
x.cos
2
q.+ 
y.sin
2
q+2. 
xy.cosq.sinq
Hence
= (
y–
x.) cosq.sinq+ 
xy(cos
2
q-sin
2
q)
Hence
[1]
[2]
8

9. 2
2cos1
sin
2
12cos
cos
2sin
2
1
cossin
2cossincos
1sincos
2
2
22
22
q
q
q
q
qqq
qqq
qq
-




-
 from
It followsqq

 2sin2cos
22
xy
yxyx

-


 qq

 2cos2sin
2
xy
yx

-
-
[3]
[4]
9

10. From eq. 4 it follows that the shear stresses varnish when 02cos2sin
2

-
- qq


xy
yx
[5]
That is when:( )
( )
( )







-


-


-


-
yx
xy
yx
xy
yx
xy
or
whenei


q


q


q
q
2
tan
2
1
2
2tan..
2
2cos
2sin
1
Thus, in a two dimensional stress system there are two planes
separated by 90
o
on which the shearing stress is zero.
10

11. Notes on the Equations for the Transformation of Stress
 Thederivativesofthealgebraicexpressionsforthenormaland
shearingstresswithrespecttotheangleofinclination,whenset
equaltozero,locatetheplanesonwhicheitherthenormalor
theshearingstressreachesamaximumorminimumvalue.
 Thenormaltensilestressesarepositiveandthecompressive
stressesarenegative.
 ShearstressesthattendtorotatetheelementANTICKLOWISE
arePOSITIVE.



x
x
x
x
x
xx
y
y
y
y
y
y
y




Fig. 5: Sign Convention
1.The Normal Stresses are positive
2.Theshearstressesontherightand
leftfacesarePOSITIVE
3. The shear stresses on the top and
bottom faces are NEGATIVE
11

12. PRINCIPAL STRESSES 02cos22sin2
2

-
- qq

q

xy
yx
d
d
That is the planes with maximum or minimum normal stress are given by( )
( )
( )






-


-


-


-
yx
xy
yx
xy
yx
xy
or
ei


q


q


q
q
2
tan
2
1
2
2tan..
2
2cos
2sin
1
1
1
1
1
TheseplanesarecalledPRINCIPALPLANES.The
correspondingvaluesofarecalledPRINCIPALSTRESSES.
To find the plane of maximum or minimum normal stress Eq. 3 is differentiated
with respect to qand the derivative is set equal to zero, i.e.,
[6]
[7]
12

13. PRINCIPAL STRESSES…Cont’d ( )
yx
xy
p


q
-


2
2tan
1
 Notethatequations[5]and[7]aresimilar.Thereforeonplanesonwhich
maximumorminimumnormalstressesoccur,therearenoshearing
stresses.
 Themagnitudesoftheprincipalstressescanbeobtainedbysubstituting
valuesofthesineandcosinefunctionscorrespondingtothedouble
anglegivenbyeq.[6]intoequation[3]






-








-








-


2
2
2cos
2
2sin
2
1
2
1
yx
xy
yx
p
yx
xy
xy
p



q



q
13

14. From eq.[3], i.e.,
It followsqq

 2sin2cos
22
xy
yxyx

-


 2
2
2,1minmax/
2
2
2
2
minmax/
2
2
2
2
minmax/
22
2
2
2
22
2
22







-










-








-











-









-








-








-



yx
xy
yx
yx
xy
yx
xy
yx
yx
xy
xy
xy
yx
xy
yx
yxyx



















[8]
and eq.[6]
14

15. MAXIMUM SHEARING STRESSESxy
yx
yx
xy
xy
yx
or
d
d


q


q
q
qq

q


-
-
-

-
-
-
-
2
)(
2tan
2
2sin
2cos
02sin22cos2
2
To find the plane of maximum or minimum shearing stress Eq. 4 is
differentiated with respect to qand the derivative is set equal to zero, i.e.,2
2
1
2
2
1
1
2
2cos
2
2
)(
2sin
2
)(
2tan







-









-

-
-

-
-
yx
xy
xy
s
yx
xy
yx
s
xy
yx
s



q



q


q
[9]
15

16. MAXIMUM SHEARING STRESSESqq

 2cos2sin
2

-
-
xy
yx
Tofindthemagnitudesofthemaximumandtheminimumshearing
stresses,thevaluesofthesineandcosinefunctionscorrespondingtothedouble
anglegivenbyeq.[9]aresubstitutedintoequation[4]2
2
2
2
minmax/
22
2
)(
2







-









-

-
-
-
-
yx
xy
xy
xy
yx
xy
yxyx








i.e. from eq.4 i.e.
One gets 2
2
2
2
2
2
minmax/
2
2
2







-








-








-


yx
xy
yx
xy
yx
xy







That is
Note: The maximum shear stress differ from the minimum shear stress only in sign
[10]
and eq.[9]
16

17. From eq.[8], i.e.
It follows2
2
2
2].[].[
].[
22
].[
22
21
minmax/
2
2
minmax/
2
2
21
2
2
2
2
2
1














-








-








-
--







-
-









-



yx
xy
yx
xy
yx
xy
yx
yx
xy
yx
But
givesiiEqiEq
iiEq
iEq
[11]2
2
2/1
22







-



yx
xy
yx




17

18. From eq.[3], i.e.,
It followsqq

 2sin2cos
22
xy
yxyx

-


 stressshearwithplaneonstressNormalei
yx
yx
xy
yx
xy
yx
xy
xyyxyx
minmax/
2
..
2
2
2
22
2
2
2
2










-

-








-









-
-












Themagnitudesofthenormalstressesactingontheplaneswith
maximumandminimumshearingstressescanbeobtainedby
substitutingvaluesofthesineandcosinefunctionscorrespondingto
thedoubleanglegivenbyeq.[9]intoequation[3]
Normal Stresses on Planes with Max/Min Shear Stresses
Therefore a normal stress acts with the maximum/ minimum shear stress.
[12]
and eq.[9]
18

19. From eq.[3]qq

 2sin2cos
22
xy
yxyx

-


-
Mohr’s Circle of Stress
[15]qq

 2cos2sin
2
xy
yx

-
-
From eq.[4]
[13]
[14]
Squaring Eqs.[13] and [14] and adding gives2
2
2
2
22
xy
yxyx




 







-








 
- ()
222
ryax -
We know that a circle with radius rand center at x=a , y=0on the x,y planeis given by
[16]
Hence, eq [15] is an equation of a circle on the ,plane with its center at







0,
2
yx

And radius2
2
2
xy
yx










-
19

20. 



x
x
x
x
x
x
x
y
y
y
y
y
y
y






A
B
q 


x
x
y
y Fig.6(c):Positive Normal Stresses
Fig.6(d) Element for fig 4(a)
Fig.6(a) Mohr’s Circle of Stress
Fig.6(b)




 
 



max
xy
yx
112
x
x
y
,
,
xy y
min
O CJ
+ -
2 2
(
(
(
(
A
B-
2q
20

21. •PointAonthecirclecorrespondstothestressesonthe
rightfaceofthegivenelement.
•NotethatAJ/CJ=tan2q
1andthisisthesameasEq.[6].Therefore
angleACJistwicetheangleofinclinationoftheprincipleplane.
•Thefollowingconclusionsregardingthestateofstressatapointcan
bedrawnfromtheMohr’scircle
 
1isthelargestnormalstressand
2isthesmallest
normalstress
 TheradiusoftheMohr’scircleisnumericallyequal
to
maxandalsoto(
1-
2)/2
 Thenormalstressontheplanesofmaximum
shearingstressis(
1+
2)/2
21

22. 22

23. Example 1
Thestateofstressat
apointjustbefore
failureoftheshaftis
shownintheFig.
Determine;
(a)Principalplanes
(b)Principalstresses
andshowtheresults
onproperlyorientated
element.
23

24. Example 2
Thestateofplanestressat
apointonabodyis
representedontheelement
shown intheFig.
Determine;
(a)Planes with
maximum/minimum shear
stress (b) Maximum and
minimum shear stresses
(c) Associated normal
stress.
Show the results on
properly oriented element
24