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```fields to characterize complex materials.
(b) From eq. 1.2.10 the stress vector t,, acting on the plane whose
normal is fi is just t,, = fi T, where ii = cos 821 + sin O i 2 .
Using matrix multiplication gives
f/a 0 0 f / a cos e
t,, =fi-T=(cose,sine,O) 0 0 0
[ o o J = [ 8 1
= (i) coseil (1.2.26)
The normal stress on the fi plane is just the projection of tn on fi
14 I RHEOLOGY
The shear stress comes from vector subtraction
t,, - Tnni = Tn,i (1.2.28)
where i is the vector in the % I f 2 plane tangent to the plane 6.
Substituting on the left-hand side gives
(~cose ,o ,o ) - ( f . cos3e , -cos28sin8,0 f ) = (f cos 8 sin2 8 , -- f cos2 8 sine, 0
a a a
Since i = sin 812, - cos e i 2 , then
T,,, = (i) cos 8 sin 8 (1.2.30)
This shear stress can be important in failure. For example, if a
certain crystal plane in a material has a lower slip or yield stress,
this stress may be exceeded although the tensile strength between
the planes may not.
Example 1.2.2 Stress on a Surface
Measurements of force per unit area were made on three mutually
perpendicular test surfaces at point P , (Figure 1.2.2a), with the
following results:
Direction of Vector ti
Normal to Test Surface Measured Force/Area { P a = k N / m 2 )
2,
322 - 23
- %* + 3%3
(a) What is the state of stress at P?
(b) Find the magnitude of the stress vector acting on a surface
whose normal is
(1.2.31)
(c) What is the normal stress acting on this interface?
Solution
(a) The state of stress at a point is determined by the stress tensor
(eq. 1.2.24)
T = f i t , + %2t2 + i 3 t 3
ELASTICSOLID / 15
where the measured tractions & are
Thus,
or
(b) From q. 1.2.10 we know that the stress vector tn acting on
the plane whose normal is fi is just
& =ri .T (1.2.10)
Using matrix multiplication (eq. 1.2.14) we find
Remembering the omitted unit vectors, we write
The magnitude of this stress vector is
(c) The normal stress Tnn is just the projection of &, onto the unit
vector
Ten = i ( l , 1, 0)’- L:,] = 2Pa (1.2.35) Jz
1.2.2 Symmetry
Notice that the stress tensor in each of the examples above is sym-
metric; that is, the rows and columns of the matrix for the compo-
nents of T can be interchanged without changing T. The compo-
nents of the traction vectors ti were picked that way intentionally.
The symmetry of the stress tensor can be shown by considering the
16 / RHEOLOGY
Figure 1.2.5.
To balance angular momen-
tum about the 21 axis, the
two shear stress components
T32 and T23 acting on the
tetrahedron must be equal.
shear stresses acting on the small tetrahedron sketched in Figure
1.2.5. The component T23 gives rise to a moment about the XI axis.
To conserve angular momentum, this moment must be balanced
by the one caused by T32. Thus, T32 = T23, and by similar argu-
ments the other pairs of shear components T12 = T21 and TI3 = T31
are equal. Thus, the stress tensor is symmetric with only six in-
dependent components, which when written in matrix form (again
In Gibbs notation, we show that a tensor is symmetric by
writing
T = TT (1.2.37)
where TT is called the rrunspose of T. In the tensor TT the scalar
components of the rows and columns of T have been interchanged.
This interchange may be clearer when TT is written in index nota-
tion
TT = xi ( f i % j T ~ j ) ~ = xi cj 12jkjTji (1.2.38)
The transpose has wider utility in tensor analysis. For exam-
ple, we can use it to reverse the order of operations in the vector
product of eq. 1.2.10
tn = TT .ii (1.2.39)
This result can be verified by using matrix multiplication. Try it
yourself. Follow eq. 1.2.14, switch rows with columns in T, and
make 6 a column vector on the other side. Of course, in the end
this manipulation does not matter for T because it is symmetric,
but we will find the operation useful later.
i 2 i
ELASTICSOLID / 17
Figure 1.2.6.
A hydrostatic state of stress
on a tetrahedron of fluid.
The possibility of a nonsymmetrical stress tensor is discussed
by Dahler and Scriven (1961, 1963). Asymmetry has not been
observed experimentally for amorphous liquids. Body torques do
exist on suspension particles, but these can be treated by calculating
the stress distribution over the particle surface for each orientation
(see Chapter 10).
1.2.3 Pressure
One particularly simple stress tensor is that of uniform pressure. A
fluid is a material that cannot support a shear stress without flowing.
When a fluid is at rest, it can support only a uniform normal stress,
TI I = T22 = T33, as indicated in Figure 1.2.6. This normal stress
is called the hydrostatic pressure p . Thus, for a fluid at rest, the
stress tensor is
0 1 0 0
0 0 1
Tij = [ -p - p :] = - p [ 0 1 0 1 (1.2.40)
0 -P
where the minus sign is used because compression is usually con-
sidered to be negative.
The matrix or tensor with all ones on the diagonal given in
eq. 1.2.40 has a special name. It is called the identity or unit tensor.
When multiplied by another tensor, it always generates the same
tensor back again:
T . L = T (1.2.41)
The Gibbs notation for the identity tensor is I. Its components are
18 / RHEOLOGY
Thus, the stress tensor for a fluid at rest is
T = -PI (1 -2.43)
When dealing with fluids in motion, it is convenient to retain
p . Thus, we write the total stress tensor as the sum of two parts
T = - p I + r (1.2.44)
where r is known as the extra or viscous stress tensor.* Often T is
referred to as the toral stress tensor and r as just the stress tensor.
In rheology we generally assume that a material is incom-
pressible, The deviations from simple Hookean or Newtonian be-
havior due to nonlinear dependence on deformation or deformation
history are usually much greater than the influence of compress-
ibility. We discuss the influence of pressure briefly in Chapters 2
and 6. For incompressible materials the overall pressure cannot
influence material behavior. In other words, increasing the baro-
metric pressure in the room should not change the reading from a
rheometer. For incompressible materials the isotropic pressure is
determined solely by the boundary conditions and the equations of
motion (see Sections 1.7 and 1.8).
Thus, it makes sense for incompressible materials to subtract
p . The remaining stress tensor I contains all the effects of defor-
mation on a material. Constitutive equations are usually written in
terms of z. However, experimentally we can measure only forces
which, when divided by the area, give components of the total stress
T. Since T includes p and r , we would like to remove the pressure
term from r . This presents no problem for the shear stress compo-
nents (because T12 = 712, etc.), but the normal stress components
will differ by p; T;; = - p + ~ i ; . As we said, determination of
p requires boundary conditions for the particular problem. Thus,
normal stress difSerences are used to eliminate p since
TII - T22 = t i 1 - 722
T22 - T33 =r22 - 733
(1.2.45)
As an example of how we use the normal stress difference,
consider the simple uniaxial extension shown in Figure 1.1.2b. The
figure shows that tension acting on the f 2 l faces of the rubber cube
will extend it. However, as shown in Figure 1.2.7, a compression,
-T22 = -T33, on the f22 and f23 faces could generate the same
deformation. The rubber cube does not know the difference. The
deformation is caused by Ti, - T22, the net difference between
tension in the 21 direction and the compression in the 22 direction.
Figure 1.2.7.
Uniaxial extension generated
by a uniform compression.
T22
- - J 7-22
t
*r is an exception to the general rule for using boldface Latin capital letters for
tensors. Howeves it is in such common use in rheology that we retain it here,
I
T22
ELASTICSOLID / 19
For the special case of simple shear we use
We call NI thejrs t normul stress```