we have learned that the stress state of a mature point can be captured by tensor
with nine components now let's have a closer look of these nine values
in fact the nine values of a stress tensor are not absolute with a different coordinate
system the values will be different but still can represent the same stress state
let's have a look at this simple torsion bar problem a pair of torsional force
applied to the ends of the bar the stress state is uniform along the axis which means if we cut
the bar at any location the stress distribution over the cross section will be always the same
if we pick an infinitesimal cube with position a on the bar the stress state will only have values
for x y and y x components as shear stresses we say the material is in pure shear under such
torsional loads so no normal stress value exists for the stress tension however if we rotate the
cube by 45 degree and reach position b the shear forces on the cube will be transformed to normal
forces the stress tensor will only have components for normal x and y directions these two tensors
also they look very different they're representing the same stress state of the chosen material point
the only difference between them is the coordinate systems they're in now you might wonder why such
coordinate transformation is needed why not just stick with one coordinate system well we want the
stress result to serve engineering purposes instead of just numbers without any physical
meanings let's go back to the torsion bar problem again in real life if the torsion bar is made of
ductile material for example it's made of steel with excessive torsion load the material will
break and you will find the failure plan will be almost perpendicular to the axis direction
this is because ductile material fail when shear stress reaches a certain value in this
case naturally engineers want to use a coordinate system that can reflect maximum shear stress so we
prefer to review the stress with the infinitesimal cube in position a however if the torsion bar
is made of brittle material for example it's a piece of chalk or as a concrete column you will
find the failure plan is usually 45 degree to the axis this is because brittle material fail when
normal stress reaches a certain value so it makes sense to evaluate stress out of brittle materials
in a coordinate system that maximum normal stress value is reflected just like position b
the coordinate transformation leads to another very important concept principle stress so what
is principle stress and how do you find them a figurative way to describe principle stress
if we are given a stress tensor we change the coordinate system of it in a certain way
that the shear stress components become zero and the three normal stresses left
in the tensor are the principal stresses for the stress state more specific let's
have a look of the three stress components on the top face of the infinitesimal cube
there are two shear stresses sigma z y and sigma zx and one normal stress sigma's easy on yet
the three stress components are represented by three vectors with directions and lengths now
if we do vector addition on these three vectors we will create a new vector let's call it sigma p in
a reserved way if we decompose the new vector sigma p in the x y and z direction it gives
back the original three stress components right next let's rotate the cube surface to a position
that the phase is normal to sigma p and identify a new coordinate system x prime y prime and z
prime in this new coordinate system sigma p is the only stress component and is normal to the surface
such sigma p is a principle of stress of the original stress state similarly we can expand
principle stresses on the other two phases we use index one two and three to denote the three
principle stresses another look at principle stress so we start with a cube with nine stress
components to a rotated cube with three stress components the tensor expressions are as shown
here for the principal stress the three components are located along the diagonal of the tensor
remember that although they look very different these two stress expressions represent exactly the
same stress state the reason they look different is they're written in different coordinate system
one is in the original x y z axis the other is in the principal axis x prime y prime
and z prime another way to understand this we know that a vector is independent of the coordinate
system right which means vectors components are not the same in different coordinate system
same thing for tensor tensor is just a high order object compared to vector and the principle
stress is obtained at a very special coordinate system and all the off-diagonal values are zero
now we have learned the physical interpretation of principal stress that we can find three surfaces
where only normal stress exists however to find the principal stress for stress tensor
what we do is not trial and error rotating the faces each time and repeating here let's have
a brief introduction of how principle stress is calculated mathematically basically the process to
find principle stress is the process to calculate the eigenvalues of a three by three matches
either value problems occurs in many areas of science and engineering and in
analysing numerical methods in fact we will see eigenvalue problem multiple times in this course
so what is an eigenvalue problem it's a problem with given n by n matrix a find a scalar lambda
and vector x such that ax equals to lambda x and lambda will be the eigenvalues and vector x will
be the eigenvectors in the problem of finding principal stress we'll find three lambdas which
will be the three principal stress components correspondingly we'll find three eigenvectors
which form the principal coordinate system the reason that engineers are interested in finding
principal stress of a body is for some materials the state of them are determined by principle
stress just like what we discussed for torsion bar problem if the bar is made of brittle material
like chalk or concrete we need the principle stress to evaluate the material condition
because brittle material typically fails in tension in normal direction you