13.021 Marine Hydrodynamics, Fall 2003
Lecture 3
Copyright © 2003 MIT  Department of Ocean Engineering,
All rights reserved.
13.021  Marine Hydrodynamics
Lecture 3
Stress Tensor
. The stress (force per
unit area) at a point in a fluid needs nine components to be
completely specified, since each component of the stress must be
defined not only by the direction in which it acts but also the
orientation of the surface upon which it is acting. The first
index specifies the direction in which the stress
component acts, and the second identifies the orientation of the
surface upon which it is acting. Therefore, the
component of the force acting on a surface whose outward normal
points in the direction is .
Figure 1:
Shear stresses on an
infinitesimal cube whose surface are parallel to the coordinate system.

Figure:
Consider an
infinitesimal body at rest with a surface PQR that is not perpendicular to any of the
Cartesian axis. The unit normal vector to that surface is
. The area of the surface , and
the area of each surface perpendicular to is
, for .

Newton's law:
= (volume
force)
for
i = 1, 2, 3
If is the typical dimension of the body 
: surface forces


: volume forces

An example of surface forces is the shear force and an example of
volumetric forces is the gravity force. At equilibrium, the
surface forces and volumetric forces are in balance. As the body
gets smaller, the mass of the body goes to zero, which makes the
volumetric forces equal to zero and leaving the sum of the surface
forces equal zero. So, as
and
. But the area of each surface
to is
. Therefore
, where
is the notation (represents the sum of all components).
Thus
for i = 1, 2, 3, where
is the component of stress in the direction on
a surface with a normal . We call
the stress vector and we
call the stress matrix or tensor.
In a static fluid, the stress vector cannot be different for
different directions of the surface normal since there is no
preferred direction in the fluid. Therefore, at any point in the
fluid, the stress vector must have the same direction as the
normal vector and the same magnitude for all directions
of
.
Pascal's Law: for hydrostatics 

where is the pressure acting perpendicular to the
surface. If is the pressure acting perpendicular to the
surface PQR, then
, but
.
Therefore
, i = 1, 2, 3 and is
arbitrary.
To prove the symmetry of the stress tensor we follow the steps:
Figure 3:
Material element under tangential
stress.

 The of surface forces body forces
mass acceleration. Assume no symmetry. Balance of the
forces in the direction gives:
since surface forces are
, where the
terms include the body
forces per unit depth. Then, as
.
 The of surface torque body moment angular
acceleration. Assume no symmetry. The balance moment with respect
to gives:
since the body moment is proportional to
. As
,
.
Consider a material volume
and recall that a
material volume is a fixed mass of material. A material volume
always encloses the same fluid particles despite a change in size,
position, volume or surface area over time.
The mass inside the material volume is:
Figure 4:
Material volume
with
surface .

Therefore the time rate of increase of mass inside the
material volume is:
which implies conservation of mass for the material volume
.
The velocity of fluid inside the material volume in the
direction is denoted as . Linear momentum of the material
volume in the direction is
Newton's law of motion: The time rate of change of momentum of the
fluid in the material control volume must equal the sum of all the
forces acting on the fluid in that volume. Thus:
momentum 
body forcesurface force 




Divergence Theorems: 
For vectors:


For tensors:

Thus using divergence theorems:
, 
which gives the conservation of the momentum for the material
volume
.
Consider a flow through some moving control volume
during a small time interval t. Let
be any (Eulerian) fluid property per
unit volume of fluid (e.g. mass, momentum, etc.). Consider the
integral
According with the definition of the derivative, we can write
Figure 5:
Control volume and its
bounding surface at instants and
.

Next, we consider the steps
 Taylor series expansion of f about t.

where
and
is the normal velocity of .
Figure 6:
Element of the surface at instants
and
.

So we have
Kinematic Transport Theorem (KTT)
Leibnitz rule in 3D
If the control volume is a material volume:
and U
, where is
the fluid particle velocity. Then the Kinematic Transport theorem (KTT) assume the form
Using the divergence theorem:

(1) 
1 Kinematic Transport Theorem (KTT)
where f is a fluid property per unit volume.
Let the fluid property per unit volume be mass per unit volume (
f )
since
is arbitrary, so the integrand
everywhere. Therefore, the differential form
of conservation of mass i.e. Continuity equation follows:
Therefore,
In general,
. We consider
the special case of incompressible flow (Note, the
density of the entire flow is not constant when we have more than
one fluid, like water and oil, as illustrated in the picture
above).
Figure 7:
Interface of two fluids (oilwater)

Therefore, for an incompressible flow:
Then
, which is the
Continuity equation for incompressible fluid.
2 Kinematic Transport Theorem ( = 1 KTT + continuity
equation). If = fluid property per unit mass, then
= fluid property per unit volume
and the 2 Kinematic Transport Theorem (KTT) follows:
We consider as the momentum per unit mass ( ).
Then,
But
is an arbitrary material volume,
therefore the integral identity gives Euler's equation
, 
and its Vector Tensor Form
Karl P Burr
20030824