Namas Chandra Advanced Mechanics of Materials Chapter 2-1 EGM 5653 CHAPTER 2 Theories of Stress and Strain EGM 5653 Advanced Mechanics of Materials.
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Namas ChandraAdvanced Mechanics of Materials Chapter 2-1
EGM 5653
CHAPTER 2
Theories of Stress and Strain
EGM 5653Advanced Mechanics of Materials
Namas ChandraAdvanced Mechanics of Materials Chapter 2-2
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-3
EGM 5653
2.1 Stress at a point
0LtA
Ft
A
0NLt
N NA
Ft
A
0SLt
S SA
Ft
A
Namas ChandraAdvanced Mechanics of Materials Chapter 2-4
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-5
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-6
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-7
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-8
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-9
EGM 5653
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 T T Q Q Q T T T Q Q Q
T T T Q Q Q T T T Q Q Q
T Q T Q
Namas ChandraAdvanced Mechanics of Materials Chapter 2-10
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-11
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-12
EGM 5653
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,
Namas ChandraAdvanced Mechanics of Materials Chapter 2-13
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-14
EGM 5653
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,
Namas ChandraAdvanced Mechanics of Materials Chapter 2-15
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-16
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-17
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-18
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-19
EGM 5653
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.
Namas ChandraAdvanced Mechanics of Materials Chapter 2-20
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-21
EGM 5653
Problem 2.5 (page72)
Given: Stress components 80Mpa; 60Mpa; 20Mpa
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-22
EGM 5653
Problem 2.5 (page72)
Given: Stress components 80Mpa; 60Mpa; 20Mpa
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-23
EGM 5653
2.5 Equation of Motion
yxxx zxx
xy yy zyy
yzxz zzz
B 0x y z
B 0x y z
B 0x y z
Namas ChandraAdvanced Mechanics of Materials Chapter 2-24
EGM 5653
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 )
Namas ChandraAdvanced Mechanics of Materials Chapter 2-25
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-26
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-27
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-28
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-29
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-30
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-31
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-32
EGM 5653
Problem 2.55 (contd)
Namas ChandraAdvanced Mechanics of Materials Chapter 2-33
EGM 5653
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
Namas ChandraAdvanced Mechanics of Materials Chapter 2-34
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
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