2001, W. E. Haisler
1
Chapter 8: Stress, Strain and Deformation in Solids
Chapter 8 - Stress, Strain and Deformation in Solids
For a beam in bending, we are often interested in determining the
transverse displacement along the beam as shown below.
y
x
u y ( x)
u y ( x)  displacement in y direction
To do that, we would start with Conservation of Linear
Momentum. Unfortunately, COLM is in terms of stress,  .
Consequently, we must have some way to relate stress to
deformation. We will need additional equations as follows:
Constitutive relations - relate stress to strain
Kinematic relations - relate strain to displacement (gradients)
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
2
In the study of the motion of a solid or fluid, we will find it
necessary to describe the kinematic behavior of a continuum
body by defining expressions called strains in terms of the
gradients of displacement components. In the example below, we
consider an elastic bar of length L. If the bar is subjected by an
axial force F, it will stretch
an amount  as shown in
figure b). The quantity /L
L
is a measure of the change
a) undeformed
in length relative to the
original length and is
F
F
defined to be the axial
L+
strain for the bar.
b) stretched (deformed)
2001, W. E. Haisler
In figure d), a shear load is applied that
is parallel to the top surface as shown.
The angle  measures the amount the
original right angle in figure c) has
changed from a right angle, and the
angle  is related to the shear strain.
What causes strain?
3
Chapter 8: Stress, Strain and Deformation in Solids
o
90
c) undeformed

F
 Mechanical loads (forces,
pressures, etc.)
 Temperature change (thermal
expansion)
d) sheared (deformed)
 Moisture absorption
 ** Stress (relationship between stress and strain is the
constitutive relationship)
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
4
In this chapter we will mathematically formalize these simple
ideas to develop expressions for strains in terms of displacement
components. We will consider two approaches:
1) mathematically precise approach and
2) a simpler geometrical approach.
2001, W. E. Haisler
5
Chapter 8: Stress, Strain and Deformation in Solids
Deformations in solids are characterized by displacements of
points and by elongations and rotations of line segments in
the solid.
z
Initial State of Body
u(r+dr)
Q
P
Deformed State
Q*
dr
u(r)
dr*
P*
r
r*
x
y
r =position vector of point P (initial)
r* =position vector of point P* (deformed)
u(r) =displacement vector of point P
dr = vector line segment between P and Q
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
Note the following geometry quantities:
P moves to P*,
Q moves to Q*,
line segment P-Q deforms (stretches/rotates) to P*-Q*:
= position vector of point P (initial state)
= position vector of point P (in deformed state)
u (r ) = displacement vector of point P (from initial to
deformed state)
dr
= vector line segment between P and Q (initial state)
dr * = vector line segment between P and Q (deformed
state)
u (r  dr ) = displacement vector of point Q (from initial to
deformed state)
r
r*
6
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
7
From geometry of the vectors, we can write two equations:
displacement vector of point P: u (r )  r *  r
and
dr  u (r  dr )  u (r )  dr * or dr *  dr  u (r  dr )  u (r )
The last two terms represent the change (gradient) in
displacement u with respect to position r , i.e.,
u (r  dr )  u (r )  du  dr  (u )
ux x u y x uz x displacement gradient

u  ux y u y y uz y matrix (3x3). Note: not
symmetric!
ux z u y z uz z






















2001, W. E. Haisler
8
Chapter 8: Stress, Strain and Deformation in Solids
Definition of strain. Strain is a measure of the deformation
and rotation of line segments. Consider two material
elements dr & dr , which undergo deformations that will
1
2
bring them into new locations dr * & dr *; respectively.
1
2
z
Initial State of Body
dr
1
dr 2
u(r)
r
Deformed State
dr*
1
dr*2
r*
y
x
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
9
Previously, we obtained for a line segment dr *:
dr *  dr  u (r  dr )  u (r )  dr  du  dr  dr  (u )
We will obtain identical expressions for the two line
segments dr * and dr * defined in the figure above:
1
2
dr *  dr  dr  (u )
1
1
1
and
dr *  dr  dr  (u )
2
2
2
Consider the dot product of these two vectors (a scaler result!)
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
10
dr *  dr *  [dr  dr  (u )]  [dr  dr  (u )]
1
2
1
1
2
2
 dr  dr  dr  [dr  (u )]

2
1
2

 1
 [dr  (u )]  dr  [dr  (u )]  [dr  (u )] 


1
2
1
2
T
T
=dr  dr  dr  [(u )  (u )  (u )  (u ) ]  dr
1
2
1
2
Recall that (u ) is a 3x3 matrix. The “T” in (u )T means
that the matrix (u ) is transposed. The underlined term is
defined as 2E so that dr *  dr *  dr  dr  dr  (2 E )  dr
1
2
1
2
1
2
and
E  12[(u )  (u )T  (u )  (u )T ] Finite Strain Tensor
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
11
Note that the expression for the finite strain tensor
E  12[(u )  (u )T  (u )  (u )T ]
contains two distinctive terms:
1 [(u )  (u )T ]  linear in displacement gradients
2
1 [(u )  (u )T ]  quadratic in displacement gradients
2
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
12
Infinitesimal Strain Tensor
If the higher order terms are neglected from the finite strain
tensor [E] (i.e., keep only linear displacement gradient
terms), we obtain the infinitesimal strain tensor, []:
  12[(u )  (u )T ]
where [] is a 3x3 matrix given by:
2001, W. E. Haisler
13
Chapter 8: Stress, Strain and Deformation in Solids




 xx
xy
xz 


    yx  yy  yz 



 

zx
zy
zz 

u u

u
1
y
x

(
 x)
x
2 x y


u
1  ux  u y
y
 (

)
x
y
2  y


u

u

u
 1 ( x  z ) 1 ( y   uz )
2  z
x
2 z
y

1  uz  ux 
(

)
2 x z 
1  uz
(

2 y
Note: Both [E] and [] are symmetric matrices.
u
z
z

y 
)
z 




u
2001, W. E. Haisler
14
Chapter 8: Stress, Strain and Deformation in Solids
A 2-D geometrical look at  xx (defined to be the change in
length of a line segment dx which is originally oriented in
the x direction) and undergoes displacements ux and uy:
y
Q*
dx*
P*
ux(x)
P
x
u (x)
y
dx
uy(x+dx)
u (x+dx)
x
Q
(x+dx)
x
2001, W. E. Haisler
15
Chapter 8: Stress, Strain and Deformation in Solids
dx * dx
 xx 
= (change in length)/(original length) .
dx
2 1/ 2
2
dx*  dx  u x ( x  dx)  u x ( x)  u y ( x  dx)  u y ( x) 




u
u
 dx(1 
)  dx(
)
x
x
u
u
u
u
 dx 1 

 dx 1  2


x
x
x
x

 ux
  dx  dx
x




2
  uy 
 dx

 x



2
x



2
x



y
y



2 1/ 2



2 1/ 2






2 1/ 2






x



x
   uy 
 

  x 
2
Note: 1  a  1  (1/ 2)a (for small a). Thus, the last result is
approximately:
2 1/ 2



2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
16
2

2

u
 y

u

u


1
1
 
dx*  dx 1  x   x   
 x 2   x  2   x  


The strain now becomes
 xx
2

2

u
 y

u

u


1
1
   dx
dx 1  x   x   
 x 2   x  2   x  


dx * dx


dx
dx
2001, W. E. Haisler
17
Chapter 8: Stress, Strain and Deformation in Solids
After canceling the dx terms, the finite strain term becomes
 ux 1   ux  1   u y 
 xx 
 

  
 x 2  x  2  x 
2
2
 finite strain
For small gradients, we assume that the squared terms are
small compared to the un-squared terms and we obtain the
linear, or small strain expression:
 ux
 xx 
x
 small strain
2001, W. E. Haisler
18
Chapter 8: Stress, Strain and Deformation in Solids
Shear strain (rotation of line segments):
dux
y
Note that
1  u y  ux
xy = (

).
2 x y
dy
dy*
dx*
Geometrically, each of the
dy
two terms is an angle as
dx
shown at the left. xy is

called a shear strain and
dx
geometrically is 1/2
x
(average) of the angular
rotation of line segments dx & dy which originally form a right

duy
angle. In contrast, the engineering shear strain xy is defined as
u
y  ux
the sum of these two angles, ie, xy = 2 xy =
.

x y
2001, W. E. Haisler
19
Chapter 8: Stress, Strain and Deformation in Solids
The definition of the engineering shear strain xy from a
graphical viewpoint is an approximation (similar to the square
root approximation made in xx ). From the geometry above,
dx * dy*  cos * dx * dy* . Define xy to be the sum of the
u
y  ux
angular rotations, ie, xy =
. (engineering shear strain)

x y
u
y  ux
cos *  cos( / 2   )  sin    

 2
xy
xy
x y
xy
xy
As in the square root approximation made for xx (for the
geometrical interpretation of strain), an assumption of “small”
rotations has been made in defining the shear strain xy .
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
20
Some examples
u(x)
Consider a bar of lenght L,
P
fixed at the left end, and
with a force P applied at the
L
right end. The bar stretches
uL
x
an amount uL at the right
end. If the “engineering” strain in the x direction is defined to
be (change in length)/(original length), then xx = uL/L.
We could also find the expresssion for u(x) and apply the
mathematical definition of xx . We have two boundary
conditions on u(x): u(0)=0 and u(L)= uL. Assume
u ( x)  C1  C2 x where C1 and C2 are constants. Applying B.C.s
gives C1 = 0 and C2 = uL/L so that u(x) = (uL/L) x. Thus the
 ux
strain is given by  xx 
= (uL/L).
x
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
Example: At the point "A" shown on the wheel, the
displacement field has been determined
Point A
to be as follows (using the finite
element method):
u x ( x, y )  (0.38 x 2  1.31xy  0.045 y 2
 2.22 x  8.41 y )104 in
u y ( x, y )  (0.25 x 2  1.62 xy  1.65 y 2
+3.44 x  7.03 y )104 in
u z ( x, y )  0
Determine the strains at point "A" which is located at x=4in,
y=5in.
21
2001, W. E. Haisler
22
Chapter 8: Stress, Strain and Deformation in Solids
u x ( x, y )  (0.38 x 2  1.31xy  0.045 y 2
 2.22 x  8.41 y )104 in
u y ( x, y )  (0.25 x 2  1.62 xy  1.65 y 2
+3.44 x  7.03 y )104 in
 xx
u x

x
 yy 
 xy
 xz
 (0.76 x  1.31 y  2.22)104 in / in
x  4", y 5"
u y
y
 7.37 x104 in / in
x  4", y 5"
 2.01x104 in / in
x  4", y 5"
1  u x u y 
4
 


3.27
x
10
in / in , Note:  xy  2 xy

2  y
x 
x  4", y 5"
  yz   zz  0
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
23
Now, do again, but consider finite strain.
Comparing the two results, is it OK to assume small strains for
this problem?
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
24
Some Thought Exercises
Suppose we have an 8x8 square area outlined on a larger chunk of
planar material (like a plate). Suppose the plate is loaded in the x-y
plane so that the square is displaced and deformed to a rectangle as
shown below [new x-y coordinates starting at lower left corner and
going CCW are: (8,4), (20,4),
(20,12) and (8,12)]. Picture shows
“deformed” and “undeformed”
area. What are your guesses for
undeformed
the strains? There is stretching in
the x direction, so  xx  0 . There
is no stretching in the y direction,
deformed
so  yy  0. All right angles
remain right angles, hence there is
no shear strain and  xy  0 .
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
25
We can actually solve for the displacements and strains since we
know the initial and final positions of the 4 corner points.
Assume that the displacements are given by (a 2-D curve fit):
u x ( x, y )  C1  C2 x  C3 y  C4 xy
u y ( x, y )  C5  C6 x  C7 y  C8 xy
Four constants are chosen because we know information at 4
points. Now apply know conditions for the four corner points:
u x (0,0)  8  0  C1  C2 (0)  C3 (0)  C4 (0)(0)
u x (8,0)  20  8  C1  C2 (8)  C3 (0)  C4 (8)(0)
u x (8,8)  20  8  C1  C2 (8)  C3 (8)  C4 (8)(8)
u x (0,8)  8  0  C1  C2 (0)  C3 (8)  C4 (0)(8)
u y (0,0)  4  0  C5  C6 (0)  C7 (0)  C8 (0)(0)
u y (8,0)  4  0  C5  C6 (8)  C7 (0)  C8 (8)(0)
u y (8,8)  12  8  C5  C6 (8)  C7 (8)  C8 (8)(8)
u y (0,8)  12  8  C5  C6 (0)  C7 (8)  C8 (0)(8)
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
26
In each of the above, the displacement at a point is set equal to
the “final position” – “initial position” of that point. Solve for
the constants and substitute back into u x and u y to obtain:
u x ( x, y )  8  0.5 x  0 y  0 xy  8  0.5 x
u y ( x, y )  4  0 x  0 y  0 xy  4
Last equation says that displacement of all points in y direction
( u y ) is a constant, which is consistent with the “picture” of
deformed and undeformed area. Now calculate the strains:
u x
 xx 
 0.5 in / in (a positive value indicates stretching)
x
u y
(no stretching in y direction)
 yy 
0
y
1  u x u y 
 xy  

  0  0  0 (no shear strain)
2  y
x 
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
27
Suppose we have an 8x8 square area outlined on a larger chunk
of planar material (like a plate). Suppose the plate is loaded in
the x-y plane so that the square is displaced and deformed to a
parallelogram as shown below [new x-y coordinates starting at
lower left corner and going CCW are: (8,4), (16,4), (19,12) and
(11,12)]. Picture shows
“deformed” and “undeformed”
area. What are your guesses for
the strains? There is no
undeformed
deformed
stretching in the x direction, so
 xx  0. There is no stretching in
the y direction, so  yy  0.
However, original right angles
are no longer right angles, hence
there is some shear strain and
 xy  0 .
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
28
Suppose we have an 8x8 square area outlined on a larger chunk
of planar material (like a plate). Suppose the plate is loaded in
the x-y plane so that the square is displaced and deformed to a
quadrilateral as shown below [new x-y coordinates starting at
lower left corner and going
CCW are: (9,4), (16.8,6),
(17.5, 14.5) and (10.8,12.5)].
Picture shows “deformed” and
“undeformed” area. What are
your guesses for the strains? It
appears that there is shortening
or stretching in the x and y
direction; hence,  xx  0 and
 yy  0 . Original right angles
are no longer right angles, hence there is some shear strain and
 xy  0 .
2001, W. E. Haisler
29
Chapter 8: Stress, Strain and Deformation in Solids
Note on Tensor Stress & Strain Transformation
For 2-D, Cauchy's formula provided the following relation:
t( n)  n  
The component of t( n ) in the
direction of the unit outward
normal n of a surface (or in x'
direction is:
 n   x ' x '  t( n)  n
(3.31)
y’
t (n )
n’
 xx
n

x’
s
t( n)
n
 xy
 yx
 yy
Substitute Cauchy's formula
into the above, and write in both vector and matrix notation:
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
30
 n   x'x'  n   n
 [n] [ ] [ n]
(1x 2) (2 x 2) (2 x1)
The expression  n  [n][ ][n] gives the component of stress,
 n , in the direction of the unit normal n (or in the direction of
the x'-axis which makes an angle  CCW from the x-axis).
Lets do both the vector and matrix operations to show that
they are the same. First, Cauchy's formula is

t( n)  n    (cos i  sin  j )   xx ii   yx ji   xy ij   yy jj
 ( xx cos   xy sin  )i  ( yx cos   yy sin  ) j
 ( xx cos   xy sin  )i  ( xy cos   yy sin  ) j

2001, W. E. Haisler
31
Chapter 8: Stress, Strain and Deformation in Solids
Note  yx   xy . Now do the remaining vector operation for
 n to obtain:
 ( xx cos   xy sin  )i 
 n   x ' x '  t( n)  n  
 cos i  sin  j 
  ( xy cos   yy sin  ) j  


  cos2   2 sin  cos   sin 2 
xx
xy
yy
If we do this in matrix notion we obtain:
 xx  xy  cos 
 n  [n][ ][n]  cos sin   




sin


yy  
 yx
  xx cos2   2 xy sin  cos   yy sin 2 
(5.3)
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
32
The above is called the stress transformation equation (see
eq. 5.3 in the notes). The stress transformation transforms
stresses from an (x,y) coordinate system to an (x',y') system
where x' is rotated by an angle  CCW from the x-axis.
We can similarly show that the strain transformation is
given by:
n   x' x'  n    n
 [n] [ ] [ n]
(1x 2) (2 x 2) (2 x1)
The quantity  n is the component of strain in the direction of
a unit normal n .  n is often called the unit elongation in the
n direction (just as  xx is the unit elongation in the i or xcoordinate direction).
2001, W. E. Haisler
Chapter 8: Stress, Strain and Deformation in Solids
33
Notes:
1. Both [ ] and [ ] are second order tensors.
2. All second order tensors follow the same transformation
form in transforming from (x,y,z) to another orthogonal
coordinate system (x',y',z'), i.e.,
 n   x'x'  n   n
 [n][ ][n]
and
n   x' x'  n    n
 [n][ ][n]
2001, W. E. Haisler
34
Chapter 8: Stress, Strain and Deformation in Solids
3. The same transformation applies to
moments of inertia of a cross-section
A (which is also a second order
tensor):
y’
y
x’

I xx   y 2dA , I xy   xydA, I yy   x 2dA
A
A
A
With respect to the x'-y' coordinate
system at some angle , we have:
I x ' x '   ( y ')2 dA
A
Can also get I x ' x ' by applying the coordinate transformation
to the x-y moments of inertia written as a matrix: ( n is unit
vector in x' direction):
 I xx
[I ]  
 I yx
I xy 
 , then I n  I x ' x '  [n][ I ][n]
I yy 
x