SLR 4A11

CHAPTER TWO
Stress and strain relationships
for elastic behaviour

Introduction
• Stress and strain relations are extended from 2D tot 3D
• Equations describing the state of the stress or strain in a
body are applicable to any solid continuum, even if it is a
elastic or plastic solid or a viscous solid ~ continuum
mechanics
• Equations relating stress and strain are called constitutive
equations – depend on material behaviour (Hooke’s law
special case related to elastic materials)

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SLR 4A11

Description of stress at a point

• Stresses at a point associated with a Z

specific plane can be resolved in
normal and shear components  ZX
 ZY
• Normal stress: the subscript
identifies with the direction in
which the stress acts.
– z normal stress in z
direction
– Values greater than zero
implies tension
– Values less than zero
compression compression

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SLR 4A11

Description of stress at a point

z

Plane yz

y
x
• Shear stress
• Need two subscripts
- First – indicates the plane in which the stress acts, refers
to its normal
- Second – indicates direction on that plane in which the
stress acts
- yz – shear stress on a plane perpendicular to the y axis
in the direction of the z axis

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SLR 4A11

Description of stress at a point

• Shear stress is positive when:

– It points in the positive direction on the positive face of a unit cube
– It points in the negative direction on the negative face of a unit
cube
• Shear stress is negative when
– It points in the negative direction on a positive face of a unit cube
– It points in the positive direction on a negative face of a unit cube

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SLR 4A11

Description of stress at a point

• As previously shown stress associated with a specific plane at a
point in a solid may be resolved into a normal and two shear
components
• If this is applied in three dimensions to an infinitesimal cube the
following results:

In essence 18 stress
“values” are required to
fully describe the stress
state on all 6 sides.

Note: Only the three

visible front planes are
shown for now

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SLR 4A11

Description of stress at a point

• The components on the rearward faces are then:
Only rearward plane normal
to Y is shown, X and Z are
similar. Directions per
convention.

Minus sign from positive

gradient aligned with
appropriate axis direction
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SLR 4A11

Description of stress at a point

• For an infinitesimal sized cube ΔX, ΔY and ΔZ ~ 0. Then:
 YX
Y, (etc.)  0
Y

• Nine components of stress are then required to fully describe the

stress at a point.
X,Y,Z, τXY, τXZ, τYX, τYZ, τZX, τZY 7
SLR 4A11

Description of stress at a point

• Still assuming an infinitesimal sized cube ΔX, ΔY and ΔZ ~ 0. Then
taking moments around Z for axes in the centre of cube:

M Z 0

 X X 

 XY  Y Z      Y Z 
2 
XY
 2
 Y Y 
   YX X Z     YX X Z   0
 2 2 

 XY Y Z X    YX X Z Y   0
 XY   YX And in similar manner:  YZ  ZY and  XZ  ZX

• Therefore: state of stress at any point completely described by 6

components:
X, Y, Z, τXY, τXZ, τYZ
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SLR 4A11

Description of stress at a point

• Apply the following stress state on a elemental cube:
σX = -100 MPa, σY = 50 Mpa, τXY = -150 Mpa, τZY = 10 Mpa

Remember that:  XY   YX  YZ  ZY and  XZ  ZX

Therefore:

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SLR 4A11

Description of stress at a point

𝜎𝑥 = 100 Mpa 𝜎𝑥 = 50 Mpa

𝜎𝑦 = 200 Mpa
Z 𝜎𝑦 = 250 Mpa
Z
𝜎𝑧 = 0 Mpa Y 𝜎𝑧 = 0 Mpa
45
τ𝑥𝑦 = -100 Mpa τ𝑥𝑦 = 50 Mpa
Y τ𝑥𝑧 = 50 Mpa τ𝑥𝑧 = 35 Mpa
τ𝑧𝑦 = 0 Mpa τ𝑧𝑦 = -35 Mpa
X X

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SLR 4A11

Stress in two dimensions

Y
 YX

Y  XY

X X
X
 XY

 YX

Y
• 2D simplification for the state of the stress – one dimension is small
compared with the other dimensions
• Stress condition where stress is zero in one of the primary directions
is called plane stress / When stresses act in one plane only

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Transformation of known plane stress state

to new coordinate system
Y Y
 YX X
X '
Y
Y  XY
X X X ' Y '
S

X X
 XY  XY Area A

X
 YX  YX
Y Y

• Derive transformation equations for a known plane stress state to a

new coordinate system
• For a known  x ,  y and xy at point 0 define an oblique plane
(area A) with normal stress x’ and the shear stress τx’y’
• x’ and y’ are appropriate normal and in plane axis for the oblique
plane
• Note that x’ is chosen to be normal to oblique plane 12
SLR 4A11

Y Y
 YX X
Y S
Y  XY SY
X X
 SX
X X
 XY  XY Area A

X
 YX Area 1  YX
Y Y Area 2

• Sx and Sy are the x and y components of the total stress

acting on the oblique plane
• Areas of the sides of the element perpendicular to the x and
y axis are as follows
area 1  A cos 
area 2  A sin 
• Summing the forces in the x and y direction [ F=.A]
S x A   x A cos    xy A sin 
13
S y A   y A sin    xy A cos 
SLR 4A11

Transformation of known plane stress state to new

coordinate system (cont)
Y
• Simplifying the above equations results in:
X
S
Sx   x cos    xy sin Y SYN
SY

Sy   y sin   xy cos  X

 SX
• The components of Sx and Sy in the
 XY
direction of the normal stress are then: SXN
SxN  Sx cos  X
 YX
SyN  Sy sin Y

• Normal stress acting on the oblique plane

is then given by
 x '  S XN  S YN  S x cos   S y sin 
(2-2)
 x '   x cos    y sin   2 xy sin  cos 
2 2

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SLR 4A11

Transformation of known plane stress state to new

coordinate system (cont)
Y
X
• The shear stress on the oblique SY S
Y
plane is given by summing the
tangential or shear X  
components: SX
SYT
 XY
x ' y '  S YT  S XT SXT

• But  YX
S XT  S X sin  Y
S YT  S Y cos 
• Substituting back

 x ' y '  Sy cos   Sx sin

 x ' y '   xy cos   sin     y   x sin cos 
2 2 (2-3)
SLR 4A11

Transformation of known plane stress state to new

coordinate system (cont)

• We therefore have:
Y
 XY
X
X
 XY
Y Y
Y Y
X

X X

• But we do not have a relation for  Y 

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SLR 4A11

Transformation of known plane stress state to new

coordinate system (cont)

• To obtain y’ substitute  with ( + /2) in Eq. 2-2 since y’ is

orthogonal to x’


 y '   x cos 2   
2
 y sin2

2
 2 xy sin     2  cos    2 
• and since sin(+/2 ) = cos  and cos(+/2 )= - sin 
 y '   x sin2    y cos2   2 xy sin cos (2-4)

• Eq. 2-2~2-4 are the transformation of the stress equations which give
the stresses in the x’y’ coordinate system if the stresses in an xy
coordinate system and the angle  are known

 Y  may also be found in a more general way, as follows (similar to

previous derivation):

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SLR 4A11
SY
Y X
 XY
Assume oblique Y   XY
plane where Y SX
normal aligns X
with Y 
X
 XY Y
• Summing the forces in the x and y direction [ F=.A] and dividing by A
Sx   x sin   xy cos   0
Sy   y cos    xy sin  0
• But:
S xN  S x sin 
S y N  S y cos 
• Therefore:
 y '  S xN  S y N  S x sin   S y cos 
 y '   x sin 2    y cos 2   2 xy sin  cos  (2-4)
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SLR 4A11

Transformation of known plane stress state to new

coordinate system (cont)

• Using a double angle 2 and the following identities, eq. 2-2~2-4 can
be simplified:
cos 2   1 cos 2  1
2
sin2   1 1  cos 2 
2
2 sin cos   sin 2
cos 2   sin2   cos 2
• Transforming the equations gives the following equations [2-5~2-7]:
x  y x  y
x'   cos 2   xy sin 2 (2-5)
2 2
x  y x  y
y'   cos 2   xy sin 2 (2-6)
2 2
y  x
 x'y '  sin 2   xy cos 2 (2-7)
2
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SLR 4A11

Transformation of known plane stress state to new

coordinate system (cont)

The sum of the normal stresses for plane stress is an invariant quantity
(does not change with direction)

x y x y
 x'   cos 2   xy sin 2
2 2

x y x y
 y'   cos 2   xy sin 2
2 2

 x'   y '   x   y

Convenient quantity that may be used for a simple comparative

description of stress at a point
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SLR 4A11

Transformation of known plane stress state to new

coordinate system (cont)
Variation of normal stress and shear stress
on oblique plane (eq. 2-5, 2-7)

• The variation of stress is in the

form of a sine wave with a
period of 180 
• Max and min normal stress
occurs where shear stress = 0
• The max and min values of
normal and shear stress is 90
apart.
• Max shear occurs halfway
between min and max normal
stress.
• Minimum and maximum normal
stresses are your principal
stresses.
• These relationships are valid for
any state of stress
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SLR 4A11

Transformation of known plane stress state to new

coordinate system (cont)
Principal stresses
• It is possible to define a new coordinate system which has axes
perpendicular to the plane on which the maximum normal stresses
act, and on which no shear stress acts. These are referred to as
Principal planes – The stresses normal to these planes are the
Principal stresses
• The plane of principal stress is found by setting x’y’ = 0 for eq. 2-7:
y  x
sin2   xy cos 2  0
2
Rearrange:
2 xy
tan 2  (2-8)
x y
• Note that tan 2  tan   2  therefore at least two perpendicular
planes free from shear are defined. These are associated for 2D with
the two principal stresses σ1 & 2 which are 90 apart
• For 3D – σ1, σ2 & σ3 with σ1 - greatest value, σ3 – lowest value, also
22
90 apart
SLR 4A11

Transformation of known plane stress state to new

coordinate system (cont)
Principal stresses (cont)
Principal stresses may be found by substituting values for 2ϴ from eq. 2-8
into eq. 2.5-6. Values for sin 2ϴ and cos 2ϴ may also be found by
Pythagorean relationships:
2 xy
If: tan 2 
x y
   y   2 xy  
1
2 2 2
2 xy x
Then:

2
x y
Then:

sin 2  
 xy
cos 2  
 x   y / 2

   y  / 4   xy   
1

  y  / 4   xy
1
2 2 2 2 2 2
x x
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SLR 4A11

Transformation of known plane stress state to new

coordinate system (cont)
Principal stresses (cont)
Substituting these values for sin 2ϴ and cos 2ϴ into eq. 2-5,6 the principal
stresses for plane stress is then:

x y x y

2

 max   1       xy
2

2  2  (2-9)

x y x y 

2

 min   2       xy
2

2  2 

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SLR 4A11

Transformation of known plane stress state to new

coordinate system (cont)
Maximum shear stress
Maximum shear stress is found by differentiating eq. 2-7:

d x ' y '
  y   x  cos 2  2 xy sin 2  0
d

y x x y
therefore tan 2s    (2-10)
2 xy 2 xy

When compared to equation for angle for normal principal stresses

eq. 2-8:

x y 1
tan 2s   
2 xy tan 2 n
Noting above:

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Transformation of known plane stress state to new

coordinate system (cont)
Maximum shear stress (cont)
90  2 n

2 xy
If: tan 2 n 
x y Then: 2 xy

2 n
x y
x y
But: tan 2s   Therefore:  2  s
2 xy

Therefore: 90  2 n   2 s or  s   n  45
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Transformation of known plane stress state to new

coordinate system (cont)
Maximum shear stress (cont)
The maximum shear stress is found by substituting eq. 2-10 into eq. 2-7
after first obtaining values for sin2ϴ and cos2ϴ :

x y 2s
   x   2 xy  
1
2 2
tan 2s  
2
If:
2 xy
Then: 2 xy y

Then: y x
sin 2 s  
   x   4 xy 
1

y
2 2 2 y x
2 xy
cos 2 s  
   x   4 xy 
1
2 2 2
y

y x y x 2 xy
Then:  max    xy
   x   4 xy     x   4 xy 
1 1
2 2 2 2 2 2 2
y y
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Transformation of known plane stress state to new

coordinate system (cont)
Maximum shear stress (cont)
Rearrange:
1
 y   x   2 xy
2 2
1     2
 4  2

 max  2  y x xy


 y   x   4 xy  
 y   x 2  4 xy 2
1 1
2 2 2 2 2

1 1

  y   x  2
   x   y 
2
2 2

then: max      xy   
2
   xy 
2 (2-11)
 2    2  

In terms of the principal stresses 2-11 simplifies to (because τXY= 0):

1  2 1  2
max  min 
2 2 28
SLR 4A11

Transformation of known plane stress state to new

coordinate system (cont)
Mohr’s circle of stress in two dimensions

• Stress at a point (transformation equations) on an oblique plane can be

represented graphically → Mohr’s circle
• Transformation equations eq. 2-5 and 2-7 can be rearranged
x y x y
 x'   cos 2   xy sin 2
2 2
y x
 x 'y '  sin 2   xy cos 2
2
x y  y x
2 2
square each of these equations and then 
    
sum noting that:  2   2 

 y  y
2 2
  
then:   x '  x    x2' y '   x    xy
2 (2-12)
 2   2 
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Transformation of known plane stress state to new

coordinate system (cont)
Mohr’s circle of stress in two dimensions (cont)
•  y  y
2 2
Equation 2-12 is the equation of a circle   
  x '  x    x2' y '   x    xy
2

x  h 2  y 2  r 2    
2 2

• Mohr’s circle
• σx’ and x’y’ as the two axis
• Radius equal to max
• Centre is displaced (σx + σy )/2 to the right of the
origin
• Basic rules for using Mohr’s circle
• An angle  of the physical element is
represented by 2 on circle
• Same sense of rotation should be used
• Shear stress causing a clockwise rotation about
a point is plotted above the horizontal axis
• A point on the Mohr’s circle gives the magnitude
and direction of the normal and shear stress on
any plane [Figure 2-6]
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State of stress in three dimensions

Special states of 3D stress
• A 3D state of stress consisting of three unequal principal stresses acting at
a point → tri axial state of stress
• If 2 of the 3 principal stresses are equal → cylindrical stress
• If all 3 principal stresses are equal → hydrostatic / spherical stress
Assume and elemental free body with
diagonal plane JKL of area A
• JKL is assumed to be a principal plane
cutting through the unit cube
• σ is assumed to be the principal stress
acting normal to the plane JKL
• Let l, m and n be the direction cosines
of σ
• Forces acting on each of its faces must
be in balance
• Components of σ along each of the axis
are as follows 31
SLR 4A11

State of stress in three dimensions (cont)

Components of σ along each of the axis are as follows

Sx=σl Sy= σm Sz= σn

Area KOL=Al Area JOK=Am Area JOL=An

Sum forces in x direction:

lA   x Al   y xAm   zx An  0
This reduces to:
   x l   yxm   zx n  0 (2-13a)
Summing forces in other directions yields:

  xy l     y m   zy n  0 (2-13b)

  xzl   yzm     z n  0 (2-13c)

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State of stress in three dimensions (cont)

Compiling in matrix formation:

   x   yx   zx   l 
 
   xy    y   zy  m  0
   xz   yz    z   n 

Only non trivial solution is when determinant is zero:

   x   yx   zx
  xy    y   zy  0
  xz   yz    z

Solution of the determinant results in a cubic equation in σ:

 3   x   y   z  2   x y   y  z   x z   xy
2
  yz
2
  xz
2


  x y  z  2 xy yz xz   x yz
2
  y xz
2 2

  z xy 0 (2-14)
33
SLR 4A11

34
SLR 4A11

State of stress in three dimensions (cont)

• Three roots of eq. 2-14 are the 3 principal stress σ1, σ2, σ3
• To determine the directions of each principal stress Eq. 2-13a-c must be
solved simultaneously for l, m and n by substituting σ1, σ2, σ3 back in turn
and with the help of an auxiliary equation (sum of direction cosines of
vector)
l 2  m2  n 2  1
• The three coefficients of the cubic equation 2-14 are invariant:
 x   y   z  I1
 x y   y  z   x z   xy
2
  xz
2
  yz
2
 I2
 x y  z  2 xy yz xz   x yz
2
  y xz
2
  z xy
2
 I3

• I1 – sum of the normal stresses for any orientation in the coordinate system
is equal to the sum of the normal forces for any other orientation (similar to
the 2D version previously introduced):
 x   y   z   x '   y '   z'   1   2   3
35
SLR 4A11

State of stress in three dimensions (cont)

General-including shear
s2   2   2
• Previous example excluded shear stresses
• Consider a stress S acting on an oblique S σ
plane. S may be resolved into a normal (σ)
and a shear () component

• The total stress can be expressed as:
S 2  Sx2  Sy2  Sz2

• Summing the force in the x, y and z direction

(and dividing by the area), we arrive at the
orthogonal components of the total stress:
Sx   x l   yxm   zx n (2-17a)
Sy   xy l   y m   zy n (2-17b)

Sz   xzl   yzm   z n (2-17c)

36
SLR 4A11

State of stress in three dimensions (cont)

General-including shear
• To obtain the normal stress, it is necessary to determine the
components of Sx, Sy , Sz in the direction of the normal of the
oblique plane therefore: Remember that l, m and n
are the unit vectors of a normal
  Sx l  Sy m  Sz n on the plane

• Substitute eq.’s 2-17 into above and simplifying with xy= yx , etc.
   x l 2   y m2   z n 2  2 xy lm  2 yzmn  2 zx nl (2-18)

• Magnitude of the shear stress on the oblique plane can be found from

 2  S2   2

• Since plastic flow involves shearing stress, it is important to identify the

plane on which the maximum or principal shear stress occurs. This
easiest to define in terms of the principal stresses. This is given by:
 2   1   2  l 2m2   1   3  l 2n 2   2   3  m2n 2
2 2 2
(2-19)
37
SLR 4A11

State of stress in three dimensions (cont)

General-including shear
• The maximum or principal shear stresses occur on planes bisecting
the normal principal stress planes at 45 ̊ . Their magnitude and
directions in terms of their directional cosines relative to the normal
principal stress planes are then given by (assume principal normal
stress are along the main axis):
l m n 
1 1 2 3
0   1 
2 2 2
(2-20)
1 1 1   3
 0  2 
2 2 2
1 1  2
  0 3  1
2 2 2

38
SLR 4A11

State of stress in three dimensions (cont)

General-including shear

• This is graphically illustrated by:

• If σ1 is the largest normal principal stress

and σ3 is the smallest (per convention)
then 2 is the greatest shear stress →
maximum shear stress

1  3
max  (2-21)
2
• Maximum shear stress is important in the theories of
yielding and metal forming

39
SLR 4A11

Stress tensor

General definition of a tensor

• Tensors are arrays of numbers or functions that transform according to
certain rules under change of coordinates (Physicslink).
• Tensors provide a convenient and natural framework for formulating
and solving problems in areas of physics and engineering (Wolfram
research).
• Tensors are geometric objects that describes the linear relationship
between scalars, vectors and other tensors (Wikipedia).

• Tensors may have different rank or orders:

• Tensor of rank zero is a scalar
• Tensor of rank one is a vector
• Tensor of rank two is a matrix
• Stress and strain are second rank tensors

40
SLR 4A11

Stress tensor (cont)

• Consider a practical application of a tensor, the transformation of a vector

from one coordinate system to another. Define the vector S that is to be
transformed as:
S  S1x1  S2 x 2  S3 x 3

• If we now want to find the components of S with reference to x1’, x2’ and x3’
axes. These refer to angle
between these axis

    
S1'  S1 cos x1 x1'  S 2 cos x 2 x1'  S3 cos x 3 x1' 
• Let a11 be the direction cosine between x1’ and x1
etc then,

S1'  a11S1  a12S 2  a13S3

S 2'  a21S1  a22S 2  a23S3 2-22(a-c)
S3'  a31S1  a32S 2  a33S3
41
SLR 4A11

Stress tensor (cont)

In terms of matrix notation eq. 2.22(a-c) then becomes:

3
S   a1 j S j
3 3
S   a2 j S j S   a3 j S j
' ' '
1 2 3
j 1 j 1 j 1

They can also be combined:

3
S   aij Si i  1,2,3  ai1S1  ai 2 S 2  ai 3 S3
i
'

j 1

Finally, the Einstein suffix notation can also be used

Si'  aij S j (2-24)

• aij is the transformation tensor between S’ and S and is of rank 2

• j is dummy suffix implying summation (occurs more than once in argument)
• i is free suffix that denotes expansion
42
SLR 4A11

Stress tensor (cont)

Product of two vectors A and B, with components (A1, A2, A3) and (B1, B2,
B3) results in a second-rank tensor Tij

 A1   A1B1 A1B2 A1B3  T11 T12 T13 

 A B B B3    A2B1 A2B2 A2B3   T21 T22 T23   Tij
 2 1 2

 A3   A3B1 A3B2 A3B3  T31 T32 T33 

or Ai B j  Tij

If A and B are now transformed to a new coordinate system:

(both A and B are transformed from x,y,z to x’,y’, z’ where directional cosines are aij )

Ai '  aij A j and Bk '  akl Bl

43
SLR 4A11

Stress tensor (cont)

Taking the product of the transformed vectors:

Ai ' Bk '  aij A j akl Bl 

Tik '  aij aklT jl (2-25)

Since stress is a second rank tensor, the components of the

stress tensor can be written as:

 11  12  13   x  xy  xz 
   
 ij   21  22  23    yx  y  yz 
 31  32  33   zx  zy  z 

Transformation of the stress tensor σij from x1, x2, x3 to x’1, x’2, x’3 is
then given by:
 'kl  aki alj  ij (2-26)

where i and j are the dummy suffixes, and k and l are the free suffixes 44
SLR 4A11

Stress tensor (cont)

Kronecker delta ij – second rank unit isotropic tensor:

 1 0 0
  1 i j (2-28)
 ij  0 1 0  
i j
0 0 1 
0

If we use the Kronecker delta with a stress tensor, the following

contraction in terms is observed:
 ij  ij   ii   11   22   33

Since stress tensor is symmetric (σ12 = σ21 etc. ) the first invariant of
the stress tensor is the sum of the main diagonals:
I1   11   22   33

45
SLR 4A11

Stress tensor (cont)

Also, second invariant is then the sum of the principal
minors (determinant of the next lower order of matrix):
 22 23 11 13 11 12
I2   
32 33 31 33 21 22
The third invariant is the determinant of entire matrix:
 11  12  13
I3   21  22  23
 31  32  33
The invariants may then be written in tensor notation as:

I1  ii

I2  iikk  ik ki 

1
2
I3  2 ij jk ki  3 ij jikk  ii jjkk 
1 46
6
SLR 4A11

Description of strain at a point

• Displacement of points in a continuum may be due to translation,

rotation (part of dynamics) or deformation
• Deformation is made up of:
• Dilatation: Change in volume
• Distortion: Change in shape
• Deformations can be:
– Small: Described by the theory of elasticity
– Large: Described by the theory of plasticity and hydrodynamics
amongst others.

47
SLR 4A11

Description of strain at a point (cont)

• A body subjected to a load may induce displacement of a point Q to
Q’
• If the displacement vectors (components given by u,v,w) for all points in
the body are the same there is no strain
• If not, strain has to occur.

• Displacement may cause strain and rigid-body rotation

• For the current case only strain component is of concern 48
SLR 4A11

Description of strain at a point (cont)

du
X X

dx
u
Total normal strain is then:  XX  u,v,w are the components of
x displacement vector of a point

Y
Y X
du
u
 XY y
dy v
 XY x
dv
X
dx
Y X
u v
Total shear strain is then :  XY     XY   YX
y x
But:  XY   Y X “Engineering” shear strain
1 1  u v 
Then:  XY  2  XY Or:  XY   XY     49
2 2  y x 
SLR 4A11

Description of strain at a point (cont)

For normal and angular (shear) strain the strain tensor is defined in a
general sense as:
u 1  u v  1  u w 
    
x 2  y x  2  z x 
 xx  xy  xz
1  u v  v 1  v w 
ij   y x  y y  y z       
2  y x  y 2  z y 
 zx  zy  zz (2-42)
1  u w  1  v w  w
     
2  z x  2  z y  z

This becomes: Note that the shear

components are ½ of the

1  ui u j 
engineering shear strain

 ij   
2  x j xi 
50
SLR 4A11

Description of strain at a point (cont)

• Since strain tensor is tensor of rank two same rules apply as for
stress tensor
• Transformation of strain tensor to new coordinate system is then:
 'kl  aki alj  ij (2-46)

• Analogues to principal stress, planes exist with no shear strain

• For an isotropic body the direction of the principal strains coincide
with the principal stress direction
• Element will undergo pure extension or contraction without any
rotation or shear
• Once again it is the roots of the cubic equation:
 3  I1 2  I 2  I 3  0 Remember
engineering
Where: I1   x   y   z shear strain
 ij  2 ij
1 2
I 2   x  y   y  z   z x 
4

 xy   zx2   yz2 
1
4
1
4

I3   x  y  y   yx zx  yz   x  yz
2
  y  zx2   z xy
2
 51
SLR 4A11

Description of strain at a point (cont)

• Directions of the principal strains
2l  x     m xy  n xz  0
l xy  2m y     n yz  0
(Calculate principal
strains, substitute into
eq’s then solve for l,m,n)
l xz  m yz  2n z     0 Analogous to eq 2-13 for
stress
• Principal shearing strains are then:
1   2  3
 max   2  1   3 (analogous to 2.20 for (2-48)
stress)
 3  1   2
• Deformation is a combination of volume and shape change
• Volume strain – change in volume per unit volume for small
strains is equal to first invariant of the strain tensor:

   x   y   z  1   2   3 52
SLR 4A11

Description of strain at a point (cont)

Mean or hydrostatic (spherical) strain is given by:
 x   y   z  kk 
m    (2-50)
3 3 3
Deviatoric strain (part of strain involved in shape change rather than volume change)
is then:
ij  ij   m ij
'

2 x  y  z
 xy  xz
 x  m  xy  xz 3
2 y  z  x
 ij'   yx y  m  yz   yx  yz
3
 zx  zy z  m 2 z  x  y
 zx  zy
3
The total strain tensor divided into the deviatoric and dilatational
components are then given by:
   
ij     m ij   ij  ij   ij
' (2-52)
 3  3
ij
53
SLR 4A11

Hydrostatic and deviator components of stress

• Stress tensor can now also be divided into a component that produces pure
tension and compression (hydrostatic or mean stress tensor σm) only
and a component that represents the shear stresses (deviator stress
tensor σ’ij)
• Hydrostatic component produces only elastic volume changes, does not
cause plastic deformation (experimentation has shown that yield stress of metals
are largely independent of hydrostatic component)
• Fracture strain is strongly influenced by hydrostatic stress
• Stress deviator involves shear stresses → important in causing plastic
deformation
• Hydrostatic or mean stress is given by
 kk  x   y   z  1   2   3 (2-54)
m   
3 3 3
The decomposition of the stress tensor is then:
1
 ij   ij'   ij kk (2-55)
3
54
SLR 4A11

Hydrostatic and deviator components of stress

Therefore:
 ij'   ij   m ij (2-56)

 2 x y z 
  xy  xz 
 3 
2 y   z   x
 ij'    yx  yz 
(2-57)
 3 
 2 z   x   y 
  zx  zy 
 3 
Stress deviator involves shear stress. For system of principle stresses:

2 1   2   3  1   2    1   3 
 1'  
3 3
2       2
 1'   1 2  1 3    3   2 
3 2 2  3
From previous
derivation eq 2-20
where 2 and 3 are the principal shearing stresses
55
SLR 4A11

Elastic stress-strain relationships

• To relate the stress tensor with the strain tensor, we must introduce
the properties of the material → constitutive equations
• Will consider only constitutive equations for elastic solids →
isotropic elastic solids
• Elastic stress is linearly related to the elastic strain by means of the
modulus of elasticity (Hooke’s law)

x  E x (2-62)

• E is the modulus of elasticity in tension or compression

• Tensile force in x direction produces
• extension along the x-axis
• contraction in the transverse y and z directions
• Transverse strain is usually a constant fraction of the strain in the
longitudinal direction → Poisson’s ratio
 x
 y   z   x   (2-63)
E 56
SLR 4A11

Elastic stress-strain relationships (cont)

• For 3D stress-strain relationship, consider a unit cube

• subjected to normal stresses σx ,σy , σz
• shearing stresses xy, yz, zx
• Assume elastic stresses are small and the material is isotropic
• normal stress σx produce no shear strain on the x, y or z planes
• shear stress xy produce no normal strains on the x, y or z planes
• Then:

57
SLR 4A11

Elastic stress-strain relationships (cont)

By superposition of the components of strain in the x, y and z are then:

 x   x    y   z 
1
E
 y   y    z   x 
1 (2-64)
E
 z   z    x   y 
1
E
Shear stresses acting on the unit cube are simply written as:

 xy  G xy  yz  G yz  xz  G xz (2-65)

But:
E  ij
G and  ij 
2 1    2 58
SLR 4A11

Elastic stress-strain relationships (cont)

Therefore:

1
       1     1     
 E 11 22 33
E
12
E
13 
 1   1  
 ij    21
1
 22    33   11   23 
 E E E 
 1     1     1
      
 E
31
E
32
E
33 11 22 

And then in tensor notation

1  
 ij   ij   kk  ij (2-69)
E E

59
SLR 4A11

Calculation of stresses from elastic strain

• Since for small elastic strains there is no coupling between the

expressions for normal stress and strain and the equations for shear
stress and shear strain it is possible to invert Eqs. (2-64) and (2-65)
to solve for stress in terms of strain
• Sum equations 2-64;

 x   x    y   z 
1
E




 y   y    z   x 
1
E




 z   z    x   y 
1
E


E
 x   y   z    x   y   z (2-70)
1  2 60
SLR 4A11
Calculation of stresses from elastic strain
Rewrite eq. 2.64 for σx ;

x 
1
E
  
E E
 
E

E

 x   y  z   x   y  z   x   x
E
1  
 x   x   y   z 
(2-71)
x 
E E
Substitute eq. (2-70) into eq. (2-71);
E
x 
E
x   x   y   z 
1  1   1  2 
Other components along with the shear components are similar
therefore in tensor notation;
E E
 ij 
 ij   kk ij
1  1   1  2  (2-73)

This can be further simplified using Lame’s E E

 and  2G
constant (λ); 1   1  2  1 

 ij  2G ij   kk ij (2-74)

61
(general case of Hooke’s law):
SLR 4A11

Strain energy

• Elastic strain energy is the energy expended by external forces that

causes deformation in a material
• Neglecting damping effects it is usually fully recoverable with the
removal of the forces
• In essence energy is force times displacement
• For an elemental cube subjected to a force P in the x direction:
Z

dU  Pdu   x A x dx 
1 1 x 
du
X
2 2 dx

dU   x x Adx 
1
Y
2
X
• Note that the contribution of the lateral X
strains are not included as there is no force
in these directions.
• A.dx is the volume of the element therefore dU describes the total
strain energy by the element 62
SLR 4A11

Strain energy

• The strain energy per unit volume is then:

1
U 0   x x
2
• With the same reasoning energy expended for simple shear along a
single set of parallel planes is then:
1
U 0   xy xy
2
• The elastic strain energy for a three dimensional stress distribution
is then obtained by superposition:

U 0   x x   y  y   z z   xy xy   yx yx   xz xz   zx  zx   yz yz   zy  zy 

1
2
• In tensor notation this is:
1
U 0   ij  ij (2-83)
2

63
SLR 4A11

Strain energy

• Equation (2-83) may now be simplified by using engineering strain:

U0 
1
 x x   y  y   z z   xy xy   xz xz   yz yz 
2
• This is then further simplified by substituting Hooke’s law
1  
 ij   ij   kk  ij
E E

U0 
2E
 E

 x   y2   z2   x y   y z   x z  
1 2 1 2
2G
 xy   xz
2

  yz
2
 (2-84)

• With appropriate substitution (and noting Lame’s constant) it may also be

written in terms of the strain components only:

1
2
 2
  G 2
U 0    x   y   z  G  x2   y2   z2   xy
2

  xz
2

  yz
2
 (2-85)

64