ME 6201: Applied Elasticity and Plasticity

3-D stress-Strain

Prof. S.K.Sahoo

xx xz

yy

x

y

z

zx

xzxx

xy

yy

yxzz yz xy

zy

yz

zy

A

B

C

O

N(l1,m1,n1)

The stresses (normal and shear) on a inclined plane• Let consider the general 3D state of stress

at a point specified by,• Similar to 2D case, let find out the

normal and shear stress on aarbitrarily doubly inclined plane ABC.

zxyzxyzyx ,,,,,

• The plane ABC isspecified by its normal N.It is represented by itsdirection cosines(l1,m1,n1).

• If N makes angles , Φ, with x,y,z axes,then l1 = cos,m1 =cos Φ,n1 =cos .

• Values of l1,m1,n1 are not independent.Having relations, l12+m1

2+n12=1

zz

zx

yx

zx

xx xz

yy

x

y

z

zx

xzxx

xy

yy

yxzz yz xy

zy

yz yx

A

B

C

O

N(l1,m1,n1)

The stresses (normal and shear) on a inclined plane• As the elemental cube is in static equilibrium, the

tetrahedron ABCO will be in equilibrium by thestress act in faces OAC, OAB, OBC and the

resultant stress/traction on face ABC.

zz

zy

• Let Area of the plane ABC is A.

So, Area of OAB=An1

Area of OBC=Am1

Area of OAC=Al1

x

y

z

zx zy

zz

xx xy

xz

yy

yx yz

A

B

C

O

Sx

Sy

Sz

The stresses (normal and shear) on a inclined plane• Let the resultant stress has components Sx, Sy &

Sz in x, y, z-directions. Applying force balanceequations, we have:

0xF

,0 yF

,0 zF

0111 AnAmAlAS yzyxyy

0111 AnAmAlAS zyzzxz

0111 AnAmAlAS zxxyxx

x

y

z

zx zy

zz

xx xy

xz

yy

yx yz

A

B

C

O

Sx

Sy

Sz

The stresses (normal and shear) on a inclined plane

• Simplifying,111 nmlS yzyxyy

111 nmlS zyzzxz

111 nmlS zxxyxx

111 nSmSlS zyxn

• Adding normalcomponents of Sx, Sy,Sz, we have,

n3

n2

n1

321 nnnn

x

y

z

zx zy

zz

xx xy

xz

yy

yx yz

A

B

C

O

n

s

11111121

21

21 222 lnnmmlnml zxyzxyzyxn

2222

222

nzyx

ns

SSS

• Putting values of Sx,Sy, Sz, we have

The stresses (normal and shear) on a inclined plane

x

y

z

zx

zy

zz

xx xy

xz

yy

yx yz

A

B

C

O

= n

Sx

Sy

Sz

Principal stresses and Principal planes in 3D stress system• Let the plane ABC (direction cosine of l,m,n) is oriented in such a

way that the resultant stress σ has only normal component, ie, isnormal to plane ABC, ie, a principal stress.

nmlS zzyzzxz

• Also Sx, Sy, Sz can befound from stress terms

nmlS yzyyxyy

nmlS zxxyxxx

N(l,m,n)

nS z mS y lSx

• As σ is itself normal toplane ABC its componentsin x,y,z-directions Sx, Sy, Szcan be found bymultiplying it with itsdirection cosines l, m, n.

Principal stresses and Principal planes in 3D stress system• Subtracting first set from second ones,

we have,

0)( nml zzyzzx

0)( nml yzyyxy

0)( nml zxxyxx

• These are three homogeneous linearequations in l,m,n. To give a non-zerosolution it is necessary that thedeterminant should be zero.

0)(

)()(

nml

zzyzxz

yzyyxy

xzxyxx

0)(

)()(

zzyzxz

yzyyxy

xzxyxx

• Expanding this determinant we get acubic equation in σ as,

0)2(

)(

)(

222

222

23

xyzzzxyyyzxxzxyzxyzzyyxx

zxyzxyxxzzzzyyyyxx

zzyyxx

• The three roots of the cubic equation (designated as σI, σII , σIII ) are threeprincipal stresses and for each Principal stress associated by a principal plane.It can be obtained by putting the value of σ (σI, σII , σIII ) in the equations andsolved separately. These planes are orthogonal to each other.

Stress Invariants• The cubic equations of σ can be written as, 032

21

3 JJJ

zzyyxxJ 1

zzxz

xzxx

zzyz

yzyy

yyxy

xyxx

zxyzxyxxzzzzyyyyxxJ

2222

zzyzzx

yzyyxy

zxxyxx

xyzzzxyyyzxxzxyzxyzzyyxxJ

)2( 2223

• Where,

• J1, J2, J3 are known as the first,second, and third invariants of stressrespectively.

• J1, J2, J3 are called invariants as theydo not depend on reference axisorientation.

When Principal axes are the Referred axes

IIIIIIJ 1

IIIIIIIIIIIIJ 2

IIIIIIJ 3

• Stress matrix will be,

• Stress invariant are:

III

II

I

ij

000000

• Let find the stress on ainclined plane having directioncosine l2,m2,n2

• Components Sx, Sy & Sz in x, y, z-directionsof resultant stress will be,

2nS IIIz 2mS IIy 2lS Ix 22

22

22 nml IIIIIIn

222

22

22

22

222

222

2

2222

222

)( nmlnml

SSS

IIIIIIIIIIII

nzyx

ns

• Normal stress is:

• Shear stress is:

22

22

22

22

22

22

1,

1

mlnSonml

n

s

• When plane l3,m3,n3 has the maximumshear stress, conditions to satisfy are:

• The solutions are:

Planes & Values of Maximum Shear Stress

0)()(

3

2

3

2

mlss

221,

21,0 333

IIIIIstressshearMaximumnml

IIIIII • Assuming:

• Each one of the three planes of maximum shear is inclined at 450

to two of the principal directions and parallel to the third one.• When,

221,0,

21

333IIIIstressshearMaximumnml

20,

21,

21

333IIIstressshearMaximumnml

IIIIII • No shear stress exists on any plane and

state of stress is hydrostatic, where IIIIIIS

• The mean stress, also called hydrostatic stress is,

Mean and Deviatoric Stresses

)(31)(

31

31

31

1 IIIIIIzzyyxxiim JS

• In tensor notation,

SS

S

000000

• It can be expressed as,

jiwhenjiwhendeltaKronecij

0

1ker

ij

zzyzzx

yzyyxy

zxxyxx

zzyzzx

yzyyxy

zxxyxx

zzyzzx

yzyyxy

zxxyxx

SSSSSSSSSS

SS

S

SS

S

000000

• This tensor indicate, on any arbitraryplane the stress resultant will benormal and equal to S

ijijij SS

Where,

• When the mean/hydrostatic stress components are subtracted from thestress tensor components, deviatoric stress are obtained

Significance of Hydrostatic stress & Deviatoric Stress

Hydrostatic stress Deviatoric Stress

• Responsible forvolume change.

• K, εv attached to it.

• Responsible forShape change.

• Deformation ordistortion take place.

Octahedral Stress

• So, l=m=n, as l2+m2+n2=1, we have

We can get 8 such planes

• A plane that makes equal angles with theprincipal planes is called an octahedral plane.

31

nml

• The normal and shear stresses on theseplanes are called the octahedral normalstress and octahedral shear stress.

2222 )()()(91)( IIIIIIIIIIIIocts

stressmeanJzzyyxxoctn 131)(

31)(

)(6)()()(91)( 2222222

zxyzxyxxzzzzyyyyxxocts

• In stress terms, we have,• If on a plane, mean stress is zero, then

normal stress on octahedral plane is zero& only shear stress will act.

3I

Ix lS

stressmeanJnml IIIIIIIIIIIIoctn

1

222

31

3)(

octnoctocts )()()( 222

3III

IIIz nS 3II

IIy mS

333

2222222 IIIIIIzyxoct SSS

21

22

1 )3(32)( JJocts

Principal StrainsThe procedure is analogous to stress

zzyzzx

yzyyxy

zxxyxx

ij

0

nml

zzyzzx

yzyyxy

zxxyxx

0322

13

zzxz

xzxx

zzyz

yzyy

yyxy

xyxx

2

zzyzzx

yzyyxy

zxxyxx

3

zzyyxx 1

• 1, 2, 3 are known as the first,second, and third invariants of strainrespectively.

• The three roots of the cubic equation (designated as I, II , III ) are threeprincipal strains and for each Principal strain associated by a principal plane. Itcan be obtained by putting the value of ( I, II , III ) in the equations andsolved separately. These planes are orthogonal to each other.

Let ij is the designated strain tensor• To give a non-zero solution for required

principal plans of strain, it is necessarythat the determinant should be zero.

• The solved cubic equations can bewritten as,

Where,

Example:Given: Find principal values and directions.

For

340430002

ij

0)25)(2()3(40

4)3(000)2(

2

Answer:

525 IIIIII

5I 0)53(40

4)53(000)52(

nmnm

l0

)()(

)(

nml

Izzyzxz

yzIyyxy

xzxyIxx

003 ll

1222 nml

08400420

0003

nmlnmlnml

084042

nmnm

51

52

n

m

Similarly, putting the values for σII , σIII , we canfind the other two associated principal planes

Given: Stress components

80Mpa; 60Mpa; 20Mpa

20Mpa; 40Mpa; 10Mpaxx yy zz

xy xz yz

Find normal stress on a plane normal to ˆˆ ˆ2i j k

Putting values, σn =95.3 MPa

Example:

Find: Invariants, Principal Values, max. shear and oct. shear stress.

Answer: First InvariantSecond Invariant 2 2 2 5500xx yy yy zz zz xx xy yz zx

Third Invariant

Characteristic Equation:3 2

2

1

2

1

160 5500 0 0( 160 5500) 0

110Mpa50Mpa0

1max 1 32

2 2 22 1oct 1 2 2 3 3 19

55Mpa

44.97Mpa

11111121

21

21 222 lnnmmlnml zxyzxyzyxn

,ie, direction cosines are:

61,

62,

61

σ3= 0

Mohr’s Circle for the Three-Dimensional Stress System

• Limitation: If the principal stresses σI , σII , σIII are given, then the normal stressσn and σs on a arbitrary plane having direction cosine (l,m,n) can be found out.

222 nml IIIIIIn

1222 nml

σ is the resultant stress

2222222222

2222

222

)( nmlnml

SSS

IIIIIIIIIIII

nzyx

ns

222 1 nlm

So, 2222 1 nnll IIIIIIn IIIII

IIIIIn ln

22

Substituting,

22222222222IIIIIIIIIIIIIIIIIIIIs nlnl

2sSubstituting value of n2 in we can get,

IIIIIIIII

nIInIs

IIIIIIII

nInIIIs nm

2

22

2 ,

We know that,

IIIIIII

nIIInIIsl

2

2

Similarly, we can get,

sn &

n

s

Thus, if l,m,n are given for a particular plane,

specified by equation given above. When

the ordinate, the circle having a centre at

0,

2IIIII

and a radius 22

2

IIIII

IIIIIIIl

is the abscissa and

From first equation of l2 IIIIIIIsIIInIIn l 22

2

222

22

IIIII

IIIIIIIsIIIII

n l

or

rradiusofaatcentreryax )0,()( 222 It is a equation of a circle

lie on the circle

Instruction to draw Mohr’s Circle

1. Mark the points P1, P2, P3 on the n axis where .,, 321 IIIIII OPOPOP

2. Draw circles on diameter P1P2, P2P3 & P1P3 having centres at C1, C2, C3 at

0,2

&0,2

,0,2

IIIIIIIIIIII

s cosl

3. Through P1, P2, P3 draw lines parallel to theaxis, say P1T1, P2T2, P3T3. From P1 draw aline at angle to P1T1,, Where

which cut the circles at Q2 & Q3 respectively to circle P1P3 & P1P2.The coordinates of Q3 are sincos,cos2

IIIIIIII

222

232 1

22 lllQC III

IIIIIIII

So,

IIIIIIIIIIII l 2

2

2

sn &

)m

n

s n

i.e., It is the radius of the circle & C2Q3 = C2Q2 . So, Draw curve with centre C2.In the same way, starting from the expression for m2 & n2, lie on a circle

•From P3 draw a line at angle (Where cos=n) to cut the circles P3P2, P3P1 on S3 & S2. Draw acurve with centre C1. From P2 draw line at angle (where cos

to P2P3 & P2P1 cross on R2 & R3 & draw a curve with centre C3. All three curves meets at point P.•Draw perpendicular PN to axis

•PN = & ON =

.

III

P1

T2

n

s

T1T3

P3 P2 C3 C1C2

Q2S2

R2 Q3S3

R3

O

P

N

Mohr’s Circle for the 3-D Stress System

I

IIn