namas chandra advanced mechanics of materials chapter 2-1 egm 5653 chapter 2 theories of stress and...

Namas ChandraAdvanced Mechanics of Materials Chapter 2-1

EGM 5653

CHAPTER 2

Theories of Stress and Strain

EGM 5653Advanced Mechanics of Materials


EGM 5653

Objectives

Definition of stress/strain as tensorStress/strain on arbitrary planesPrincipal quantities, Mohr’s circle in 2-D and 3-DSmall displacement theoriesStrain rosettes

Sections

2.1,2.2 Stress Definition

2.3 Stress tranformation, 2.4 Principal quanities

2.5 Equilibrium equation, 2.6 strain in any direction

2.7 strain transformation, 2.8 strain theory

2.9 Strain measurment


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2.1 Stress at a point

0LtA

Ft

A

0NLt

N NA

Ft

A

0SLt

S SA

Ft

A


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2.2 Stress Notation (Tensor)

PA

xx xy xz

xy yy yz

xz yz zz

T

Body force vector is

, ,x y zB B B


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2.2 Alternate Stress Notation

PA

xx xy xz

xy yy yz

xz yz zz

T

xx xy xz

xy yy yz

xz yz zz

T

11 12 13

12 22 23

23 23 33

T

Diagonal –Normal

Off-diagonal-Shear

Eng. Shear= 2*Tensor Shear


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2.2 Stress on arbitrary planes

PA

ˆˆ ˆ ˆcos cos cos

ˆˆ ˆ

ˆˆ ˆx Y Z

N i j k

li mj nk

N i N j N k

xx xy xz

xy yy yz

xz yz zz

T

t T NStress on arbitrary plane is


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2.4 Stress Transformation (3-D to another 3-D)

1e

1e

2e2e

3e

3e

Consider two systems

1 2 3 1 2 3X , , and X , ,e e e e e e ,

ie is obtained from ieusing a rigid body rotation.

We are interested to relate the unit vectors

ie in X from that of X

.

1 11 1 21 2 31 3

2 12 1 22 1 32 3

3 13 1 23 1 33 3

ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ

e Q e Q e Q e

e Q e Q e Q e

e Q e Q e Q e

We can see that,

ˆ

cos( ,

ˆcos( , )

)

ij i j

old

Q e e

new

Transformation matrix component


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2.4 Stress Transformation (Vector)

1e

1e

2e2e

3e

3e

,

.

Vector from old to new system, '

'

11 21 311 1

2 12 22 32 2

3 313 23 33

[ ] [ ] [ ]

i mi m

T

a Q a

a Q a

Q Q Qa a

a Q Q Q a

a aQ Q Q

12a e Find in terms of '

{ }ieGiven,

0 1 0

1 0 0

0 0 1

Q

'

0 1 0 2 0

1 0 0 0 2

0 0 1 0 0

Ta Q a


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2.4 Stress Transformation (Tensor)

1e

1e

2e2e

3e

3e

,

.

Tensor from old to new system,

'

'

.

( . )

m nij mi nj

m nmi nj

ij mi nj mn

T Q e TQ e

Q Q e Te

T Q Q T

' ' '11 12 13 11 12 13 11 12 13 11 12 13

' ' '21 22 23 12 22 32 21 22 23 12 22 32

' ' '31 32 33 13 23 33 31 32 33 13 23 33

'

Thus

. .

T

T T T Q Q Q T T T Q Q Q



T Q T Q


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2.4 Stress Transformation (Tensor)

,

.

0 1 0

1 2 0

0 0 1

T

0 1 0

1 0 0

0 0 1

Q

'

2 1 0

1 0 0

0 0 1

TT Q T Q


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2.4 Stress Transformation (All tensors)

,

.

0 1 0

1 0 0

0 0 1

Q

'

'

'

'

th... ...

Scalar

Vector

Second Order Tensor

Third Order Tensor..... n Order Tensor

nn

i mi m

ij mi nj mn

ijk mi nj rk mnr

ijk mi nj ok sp mno

a Q a

T Q Q T

T Q Q Q T

T Q Q Q Q T


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2.4.2 Principal Stresses

ˆ ˆ ˆn n In ˆ 0I n

11 1 12 2 13 3

21 1 22 2 23 3

31 1 32 2 33 3

( ) 0

( ) 0

( ) 0

1 2 3, , are the direction cosines 2 2 21 2 3 1

is the one of the principal values11 12 13

21 22 23

31 32 33

0

This is a cubic equation in Solution to this equation gives thethree principal stress values.

1 2 3

For each of the values, find the direction corresponding to that principal value,


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2.4.2 Principal Stresses-example

2 0 0

0 3 4

0 4 3

T

Given: Find principal values and directions.

2

2 0 0

0 3 4 (2 )( 25) 0

0 4 3

T I

1

2

3

5

2

5

1 5 For 1 1 2 3

1 2 3

2 1 3

2 0 0 0

0 (3 ) 4 0

0 4 ( 3 ) 0

n n n

n n

n n

1 13 0 0n n

2 3

2 3

2 4 0

4 8 0

n n

n n

2 2 21 2 3 3 2

1 21 and

5 5n n n n n

1 2 3

1ˆ ˆ ˆ2

5n e e

2 1ˆ ˆn e 3 2 3

1ˆ ˆ ˆ2

5n e e


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2.4.2 Stress Invariants

1

2

2 2 2

3

xx yy zz

xx xy yy yz xx zx

xy yy yz zz zx zz

xx yy yy zz zz xx xy yz zx

xx xy zx

xy yy yz

zx yz zz

I

I

I

1 1 2 3

2 1 2 2 3 3 1

3 1 2 3

I

I

I

In terms of principalStresses,


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2.4.4 Octohedral Stresses

2 2 2cos cos cos 1 13

ˆˆ ˆˆ ( )n i j k

1 1oct 1 1 2 33 3I

2 2 21oct 1 2 1 2 1 23

1 1oct 13 3 xx yy zzI

2 22 2 2 21oct 3 6( )xx yy zz xx yy xx xy yz zx


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2.4.5 Mean and Deviatoric Stresses

1 2 31

3 3 3xx yy zz

m

I

ˆm d mT T T T T

13

ˆij kk ij ijT T T

0 0

0 0

0 0

m

m m

m

T

xx m xy xz

d xy yy m yz

xz yz zz m

T


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2.4.6 Coordinate Transformation in 2-D

PA

TT Q T Q

2 2

2 2

2 2

cos sin 2 sin cos

sin cos 2 sin cos

sin cos (cos sin )

xx xx yy xy

yy xx yy xy

xy xx yy xy

1 12 2

1 12 2

12

cos 2 sin 2

cos 2 sin 2

sin 2 cos 2

xx xx yy xx yy xy

yy xx yy xx yy xy

xy xx yy xy


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2.4.7 Mohr’s Circle in 2-D

12 ,0xx yy Center

2 214 xx yy xyR Radius

2 21 11 2 4

2 21 11 2 4

2 21max 4

xx yy xx yy xy

xx yy xx yy xy

xx yy xy


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2.4.7 Mohr’s Circle in 3-D

•The best way to draw Mohr’s circle in 3-D is first to find the principal stresses.

•Using the three principal stresses, the three circles can then be drawn.

•The horizontal direction then indicates the principal direction.


EGM 5653

Problem 2.5 (page72)

Given: Stress components 80Mpa; 60Mpa; 20Mpa

20Mpa; 40Mpa; 10Mpa

xx yy zz

xy xz yz

Find: Stress vector on a plane normal to ˆˆ ˆ2i j k

1ˆ (1,2,1)

6n

80 20 40

20 60 10

40 10 20

T

Stress vector ˆˆ ˆˆ 65.32 61.24 32.66Tn i j k


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20Mpa; 40Mpa; 10Mpa

xx yy zz

xy xz yz

Find: Invariants

First Invariant 1 160Mpaxx yy zzI

Second Invariant 2I 2 2 2 5500xx yy yy zz zz xx xy yz zx

3 0xx xy zx

xy yy yz

zx yz zz

I

Third Invariant


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20Mpa; 40Mpa; 10Mpa

xx yy zz

xy xz yz

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

Characteristic Equation

3 2

2

1

2

1

160 5500 0 0

( 160 5500) 0

110Mpa

50Mpa

0

1max 1 32

2 2 22 1oct 1 2 2 3 3 19

55Mpa

44.97Mpa


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2.5 Equation of Motion

yxxx zxx

xy yy zyy

yzxz zzz

B 0x y z

B 0x y z

B 0x y z


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2.2 Equation of Motion (deformation)

* *

* *

* *

x x (x, y, z)

y y (x, y, z)

z z (x, y, z)

* * *

* * *

* * *

x x(x , y , z )

y y(x , y , z )

z z(x , y , z )


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2.2 Strain of a Line Element

Engineering strain: Definition

*

E

ds ds

ds

Finite or Green Strain

2 2 2

xx

2 2 2

yy

2 2 2

zz

u 1 u v w

x 2 x x x

v 1 u v w

y 2 y y y

w 1 u v w

z 2 z z z


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2.7.3 Shear Strain (Angle Change between elements)

1 1 1 2 2 2

* * * * * * * * * *1 1 1 2 2 2

PA : (l ,m ,n ) PB : (l ,m ,n )

P A : (l ,m ,n ) P B : (l ,m ,n )

* * * * * * *1 2 1 2 1 2

cos2

cos l l m m n n

Let PA & PB be along x and y axes12 xy xy

23 yz yz

31 zx xz

2

2

2

*12 E1 E

*

2(1 )(1 )c

2

os


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2.7.4 Principal Strains

This theory is analogous to stress theory

3 21 2 3

2

0

1

2

xx xy xz

xy yy yz

xz yz zz

E E

M

M

M

or

M I M I M I

M

1

2

2 2 2

3

,

xx yy zz

xx xy yy yzxx xz

xz zzxy yy yz zz

xx yy xx zz yy zz xy xz yz

xx xy xz

xy yy yz

xz yz zz

where

I

I

I


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2.8 Small Displacement Theory

Here the strains are infinitesimal and rigid body motion is negligible.Then,

, ,

1 1 1, ,

2 2 2

xx yy zz

xy xz yz

u v w

x y z

v u w u w v

x y x z y z

and the Magnification factor is,

EM


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2.9 Strain Measurements

The direction cosines or arms a,b,and c are

2 2

2 2

( , , ) (1,0,0), ( , , ) (cos ,sin ,0), ( , , ) (cos 2 ,sin 2 ,0)

(cos ) (sin ) 2 (cos )(sin )

(cos 2 ) (sin 2 ) 2 (cos 2 )(sin 2 )

a a a b b b c c c

a xx

b xx yy xy

c xx yy xy

l m n l m n l m n

Theextensional strains are


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2.9 Strain Measurements contd.

Therefore,

2

2 2 2 2 2 2

2

( 2 )sin 4 2 sin 2

4sin sin 2

2 (sin cos 2 sin 2 cos ) 2( sin 2 sin )

4sin sin 2

60

2( ), ,

3 3

45

1, , ( )

2

xx a

a b cyy

a b cxy

o

b c a b cxx a yy xy

o

xx a yy c xy b a c

When

When


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Solved Problem 2.55

2.55 A Square panel in the side of a ship is loaded so that the panel is in a state of plane strain ( ) a. Determine the displacements for the panel given that the deformations shown and the strain components for the (x,y) coordinate axes. b. Determine the strain components for the (X,Y) axes.

0zz zx zy


EGM 5653

Problem 2.55 (contd)


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Problem 2.74

2.74 Show that for Ex 2.12 when θ=600 the principal strains are given by

1 2

1/ 22 2

1 1( ) ; ( )

3 31

(2 ) 3( )3

a b c a b c

a b c b c

R R

R

and the maximum strain is located at angle Φ ccw from x axis

3( )tan 2

2( )b c

a b c


EGM 5653

Problem set in chapter 2 (HW2)

NoHomework(due date)

Suggested Problems In-class and others

1

2.2, 2.6, 2.7, 2.21,2.25, 2.45,2.54, 2.57, 2.58, 2.67

(9 Feb 2006)

2.4, 2.10,2.13, 2.18, 2.26, 2.42, 2.56, 2.59, 2.64, 2.71,

2,77, 2.73,2.76

namas chandra advanced mechanics of materials chapter 2-1 egm 5653 chapter 2 theories of stress and...

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