E106 Stress and Strain Tensor Summary Page 1

Stress and Strain Tensors

Stress at a point.

Imagine an arbitrary solid body oriented in a cartesian coordinate system. A number

of forces are acting on this body in different directions but the net force (the vector

sum of the forces) on the body is 0. Conceptually slice the body on a plane normal to

the

x

-direction (parallel to the

yz

-plane). Take a small area on this plane and call it

. Calculate the resolved force acting on this small area and call it .

.

Notice that since is the

total

force acting

only

on , the magnitude of will

change as changes.

We can define three scalar quantities,

, , and .

The first subscript refers to the plane and the second refers to the force direction. If

we do the same conceptual experiment at the same location but in the

y

and

z

-direc-

tions, we obtain

, , ,

, , and .

For static equilibrium , , and , resulting in six inde-

pendent scalar quantities. These six scalars arranged in an ordered matrix

forms the

stress tensor

,

.

Ax� Axx�� F�

F� F� x F� y F� z� � F� xx F� yy F� zz� �� �

F� Ax� F�

Ax�

�xxF� x

Ax�-----------

Ax� 0→lim� �xy �xy

F� y

Ax�-----------

Ax� 0→lim� � �xz �xz

F� z

Ax�-----------

Ax� 0→lim� �

�yyF� y

Ay�-----------

Ay� 0→lim� �yx �yx

F� x

Ay�-----------

Ay� 0→lim� � �yz �yz

F� z

Ay�-----------

Ay� 0→lim� �

�zzF� z

Az�-----------

Az� 0→lim� �zx �zx

F� x

Az�-----------

Az� 0→lim� � �zy �zy

F� y

Az�-----------

Az� 0→lim� �

�xy �yx� �yz �zy� �xz �zx�

3 3�

� �ij

�xx �xy �xz

�xy �yy �yz

�xz �yz �zz

� �

E106 Stress and Strain Tensor Summary Page 2

For example, the stress tensor for a cylinder with cross-sectional area in uniaxial

tension from force is

if the cylinder axis and are both parallel to the x-axis,

if the cylinder axis and are both parallel to the y-axis,

and if the cylinder axis and are both parallel to the z-axis.

The sign convention for the stress elements is that a positive force on a positive face

or a negative force on a negative face is positive. All others are negative. As a final

example, a cube oriented so that its faces are perpendicular to the coordinate axes,

with an area per face of has the following forces applied to it: force applied to

the positive x face in the positive x direction, force applied to the positive y face in

A0

F

� �ij

FA0------- 0 0

0 0 00 0 0

� � F

� �ij

0 0 0

0 FA0------- 0

0 0 0

� � F

� �ij

0 0 00 0 0

0 0 FA0-------

� � F

A0 F1

F2

E106 Stress and Strain Tensor Summary Page 3

the positive y direction, and force applied to the positive x face in the negative y

direction. The necessary forces to keep the cube form moving are applied to the other

faces. The resultant stress tensor is

.

Stress on a plane.

It is often necessary to calculate the stress on an arbitrarily-oriented plane with nor-

mal inside a solid. A force balance on the tetrahedron formed by the intersection of

the plane with the coordinate axes provides the needed results. We define a stress

vector, , defined as limit of the net force acting on the plane, , per unit area as the

area shrinks to zero. This vector can be decomposed into the normal stress on the

plane (the force per unit area in the direction normal to the plane), , and the shear

stress on the plane (the force per unit area in a direction lying in the plane), . It

can also be decomposed into components in the three coordinate-axis directions, ,

, and .

In other words, . If we define the direction

cosines of as

, , and ,

then

, , and

,

F3

� �ij

F1

A0-------

F3�

A0------------ 0

F3�

A0------------

F2

A0------- 0

0 0 0

� �

n

s F

sn

ss

sx

sy sz

s sx sy sz� � sxx syy szz� �� �

n

k n x• �cos� � l n y• cos� � m n z• cos� �

sx �xxk �xyl �xzm� �� sy �xyk �yyl �yzm� ��

sz �xzk �yzl �zzm� ��

E106 Stress and Strain Tensor Summary Page 4

or

.

We can then find the normal component of the stress, , by

.

The shear component can be determined but requires a little more work.

sx

sy

sz

�ijn�xx �xy �xz

�xy �yy �yz

�xz �yz �zz

kl

m

� �

sn

sn s n• sx sy sz

kl

m

� �

E106 Stress and Strain Tensor Summary Page 5

Coordinate transformations and stress invariants.

It is often useful to know the stress tensor in a coordinate system that has been

rotated and/or translated with respect to the original coordinate system. We can

transform the coordinates, into the coordinates, by use of a

transformation matrix, , where has the property . In particular, a

rotation about the -axis through an angle is given by

.

To get all of the elements of the stress tensor in the new coordinate system,

.

The above relationship is often used to define a tensor of rank 2. Several properties

of the stress tensor remain unchanged by a change in coordinates. These properties

are called invariants. These invariants are closely related to important quantities.

The first invariant, , is the trace of the matrix,

.

The hydrostatic component of (the part due to uniform pressure on all exterior

surfaces of the solid) is equal to .

The second invariant, is given by

.

The third invariant is

.

A common question in stress problem is: Is there a coordinate system for which all of

the shear stresses disappear, and the remaining stresses are purely tensile or com-

x y zT

x� y� z�T

Q Q Q 1� QT�

z �

x�

y�

z�

�cos � sin 0 �

sin

�

�

cos 0

0 0 1

xyz

�

�� Q�QT�

I1

I1 �xx �yy �zz� ��

�ij

I1 3�

I2

I2 �xy2 �yz

2 �zx2 �xx�yy� �yy�zz� �zz�xx�� ��

I3 �xx�yy�zz 2�xy�yz�zx �xx�yz2� �yy�zx

2� �zz�xy2���

E106 Stress and Strain Tensor Summary Page 6

pressive? It turns out that there is. The resultant stresses are called the

principal

stresses

, the planes on which they occur are the

principal planes

, and the directions

of the resultant force components are the

principal directions

or

principal axes

. If we

call the principal stresses , , and , then the problem appears as: Are there

values of for which

?

The principal stresses are the eigenvalues and the principal directions are the eigen-

vectors. The eigenvalue problem can be rewritten in terms of the three invariants as

.

For any stress tensor, three real (but possibly not distinct) roots will result.

The Von Mises yielding criterion.

In a complex stress field it is not easy to determine if the stress has exceeded the

yield stress in the body. Von Mises proposed the following criterion: Yielding occurs

when the second invariant of the stress deviator exceeds some critical value.

The stress deviator is the stress tensor with the hydrostatic component removed,

i.e.

,

where .

The second invariant of is

,

which must be greater than some constant for yielding to occur. In terms of the

yield stress, , the criterion is

�1 �2 �3

�

Det�xx �� �xy �xz

�xy �yy �� �yz

�xz �yz �zz ��

0�

�3 I1�2� I2�� I3� 0�

�ijD �ij hI�

�xx h� �xy �xz

�xy �yy h� �yz

�xz �yz �zz h�

� � h �xx �yy �zz� �

3------------------------------------------�

�ijD

I2�xx �yy�( )2 �yy �zz�( )2 �zz �xx�( )2 6 �xy

2 �yz2 �xz

2� �( )� � �

6-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------�

k2

�y

E106 Stress and Strain Tensor Summary Page 7

for yielding to occur.

The Strain Tensor.

Normal strain is the change in length in a given direction divided by the initial

length in that direction. Shear strain is the complement of the angle between two

initially perpendicular line segments. If you apply a force to a solid object you may

end up simultaneously translating, rotating and deforming the object. The vector

function which describes the difference between the initial position and the final

position of each point in the object is

.

If we take the gradient of u we end up with a tensor

.

For small strains we find that where

.

The diagonal terms are the normal strains in the x, y, and z directions respectively.

The off-diagonal terms are equal to one-half of the engineering shear strain, e.g.,

. In terms of ,

�xx �yy�( )2 �yy �zz�( )2 �zz �xx�( )2 6 �xy2 �yz

2 �xz2� �( )� � �

2-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

1 2/�y�

u x y z, ,( )u x y z, ,( )v x y z, ,( )w x y z, ,( )

�

�u

x��u

y��u

z��u

x��v

y��v

z��v

x��w

y��w

z��w

�

�u �uT� 2��

� �kl

�xx �xy �xz

�yx �yy �yz

�zx �zy �zz

� �

�xy xy 2�� u�

E106 Stress and Strain Tensor Summary Page 8

, , and .

As was the case with stress , , and . We can determine

the strains in a rotated coordinate system in the same way as for stresses. We can

transform the coordinates, into the coordinates, by use of a

transformation matrix, , where has the property . To get all of the

elements of the strain tensor in the new coordinate system, .

Relationship between stress and strain.

Every member of will cause a corresponding stress in . The relationship can

be written as . Writing out the first term explicitly should suffice to

explain the notation.

.

Fortunately only 21 of the 81 -terms are unique. To simplify the notation, the

stress and strain tensors are rewritten as vectors. The simplified notation is known

as contracted notation. First the off-diagonal strain terms are converted to engineer-

ing shear strains.

.

The resulting matrix is no longer a tensor because it doesn’t follow the coordinate-

transformation rules. Then the elements are renumbered.

�xy12---

y��u

x��v�� �xz

12---

z��u

x��w�� �yz

12---

z��v

y��w��

�yx �xy� �yz �zy� �xz �zx�

x y zT

x� y� z�T

Q Q Q 1� QT�

��[ ] Q �[ ]QT�

�kl �ij

�ij Cijkl�kl�

�xx Cxxxx�xx Cxxxy�xy Cxxxz�xz� ��

Cxxyx�yx Cxxyy�yy Cxxyz�yz� � �

Cxxzx�zx Cxxzy�zy Cxxzz�zz� � �

Cijkl

�xx 2�xy 2�xz

2�yx �yy 2�yz

2�zx 2�zy �zz

�xx xy xz

yx �yy yz

zx zy �zz

�

E106 Stress and Strain Tensor Summary Page 9

, .

Then the matrices are written as vectors,

, .

Finally the relationships between the stress vector and the strain vector is

expressed.

.

The materials-property matrix with all of the Q’s is known as the stiffness matrix.

Unfortunately is used for both the stiffness matrix and the coordinate transfor-

mation matrix. Don’t get them confused. The stiffness matrix is used when all of the

strains are known and the values of the stresses are to be determined. In the more

common case of the stresses being known and the strains to be determined, the

inverse of the stiffness matrix, called the compliance matrix, , must be used.

The relationship between and is that . There are a number of simpli-

fied cases for the stiffness and compliance matrices.

�xx �xy �xz

�xy �yy �yz

�xz �yz �zz

�1 �6 �5

�6 �2 �4

�5 �4 �3

�

�xx xy xz

yx �yy yz

zx zy �zz

�1 �6 �5

�6 �2 �4

�5 �4 �3

�

�1 �6 �5

�6 �2 �4

�5 �4 �3

�1

�2

�3

�4

�5

�6

⇒

�xx

�yy

�zz

�yz

�xz

�xy

�

�1 �6 �5

�6 �2 �4

�5 �4 �3

�1

�2

�3

�4

�5

�6

⇒

�xx

�yy

�zz

yz

xz

xy

�

�1

�2

�3

�4

�5

�6

Q11 Q12 Q13 Q14 Q15 Q16

Q12 Q22 Q23 Q24 Q25 Q26

Q13 Q23 Q33 Q34 Q35 Q36

Q14 Q24 Q34 Q44 Q45 Q46

Q15 Q25 Q35 Q45 Q55 Q56

Q16 Q26 Q36 Q46 Q56 Q66

�1

�2

�3

�4

�5

�6

�

Q

S

Q S S Q 1��

E106 Stress and Strain Tensor Summary Page 10

.

Linear elastic isotropic materials:

The simplest materials are ones in which the properties do not vary with direction,

or linear elastic isotropic materials. In a linear elastic isotropic material character-

ized by Young’s modulus, , Poisson’s ratio, , and the shear modulus,

, the following relationships between the stress tensor and the strain

tensor hold:

,

,

and

,

also

,

,

and

�1

�2

�3

�4

�5

�6

S11 S12 S13 S14 S15 S16

S12 S22 S23 S24 S25 S26

S13 S23 S33 S34 S35 S36

S14 S24 S34 S44 S45 S46

S15 S25 S35 S45 S55 S56

S16 S26 S36 S46 S56 S66

�1

�2

�3

�4

�5

�6

�

E �

G E2 1 ��( )----------------------�

�xx1E---- �xx � �yy �zz�( )�[ ]�

�yy1E---- �yy � �xx �zz�( )�[ ]�

�zz1E---- �zz � �xx �yy�( )�[ ]�

xy 2�xy�xy

G--------� �

xz 2�xz�xz

G--------� �

yz 2�yz�yz

G--------� �

E106 Stress and Strain Tensor Summary Page 11

It is left to the reader (usually on an exam) to wade through the alphabet soup and

determine the stiffness and compliance matrices for a linear elastic isotropic mate-

rial.

Linear elastic orthotropic materials:

For an orthotropic material (one in which the properties in the y- and z-directions are

the same but different in the x-direction, such as a composite material with the fibers

all oriented in the x-direction) the stiffness matrix has the form

,

and the compliance matrix has the form

.

Since the properties in the y and z directions are equal, the two-dimensional x,y-case

is often considered when determining materials properties. In this case the stiffness-

matrix relationship becomes

and the compliance-matrix relationship becomes

Q

Q11 Q12 Q13 0 0 0

Q12 Q22 Q23 0 0 0

Q13 Q23 Q33 0 0 0

0 0 0 Q44 0 0

0 0 0 0 Q55 0

0 0 0 0 0 Q66

�

S

S11 S12 S13 0 0 0

S12 S22 S23 0 0 0

S13 S23 S33 0 0 0

0 0 0 S44 0 0

0 0 0 0 S55 0

0 0 0 0 0 S66

�

�xx

�yy

�xy Q11 Q12 0

Q12 Q22 0

0 0 Q66

�xx

�yy

xy

�

E106 Stress and Strain Tensor Summary Page 12

.

First a uniaxial tensile test is performed in the x-direction. is applied and ,

and are measured. As always in a uniaxial tensile test, , and

, where is the elastic modulus in the longitudinal or x direction, and

is Poisson’s ratio for stress in the longitudinal direction and strain in the trans-

verse direction. By comparison with the compliance matrix, it is seen that

, and .

Next a uniaxial tensile test is performed in the y-direction. is applied and ,

and are measured. As before,

, and , where is the elastic modulus in the transverse

or y direction, and is Poisson’s ratio for stress in the transverse direction and

strain in the longitudinal direction. By comparison with the compliance matrix, it is

seen that

, and .

A final test with pure shear gives the relationship

.

The interested reader should be able to determine the relationship between and

�xx

�yy

xy S11 S12 0

S12 S22 0

0 0 S66

�xx

�yy

�xy

�

�xx �xx

�yy �xx Ecl�xx�

�lt�yy

�xx--------�� Ecl

�lt

S111

Ecl--------� S12

�lt

Ecl--------��

�yy �xx

�yy

�yy Ect�yy� �tl�xx

�yy--------�� Ect

�tl

S221

Ect--------� S12

�tl

Ect--------��

S661G----�

�lt

E106 Stress and Strain Tensor Summary Page 13

from the relationships and .

The structure of is not nearly so simply related to , , , and as is ,

but, as before, it can be determined from , or .

�tl S12�lt

Ecl--------�� S12

�tl

Ect--------��

Q Ecl Ect �lt �tl S

SQ I� S Q 1��