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Welcome to the course on the strength of materials.
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In the last lesson I introduced the concept
of
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stresses.
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In this particular lesson we are going to
look into some more aspects of analysis of
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stresses.
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..
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Here are some of the questions posed in the
last session.
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The first question was; what is the unit
of force and stress?
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The unit of Force is Newton (N) and unit of
stress is (N by m square) or Pascal (Pa).
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This can be
represented in terms of also Mega Pascal (Mpa)
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or just Giga Pascal (GPa).
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.
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What is the definition of normal stress?
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It can be defined as stress acting normal
to the plane.
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..
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If you have a body which is a free body from
a major body and is acted on by forces there
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will be
resulting forces into it which will keep the
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body in equilibrium.
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At a particular small element if
we take that this is a resulting stress then
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we can decompose this stress into two components,
one
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is along the normal direction of this particular
cross section which is this and another component
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along the plane of this particular section.
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The component which is acting normal to this
particular
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cross section is normally known as the normal
stress.
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Normal stress is the normal component
perpendicular to the particular section.
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.
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What is meant by free body diagram?
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.Free body diagram is a diagram of a body
as a whole acted on by external resistive
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forces or a
part of it.
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It can be the whole body acted on by active
forces and resistive forces or if we cut any
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section from the body and if they are also
acted on by the same forces and the resistive
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forces
which will keep the body in equilibrium then
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we call that as the free body diagram of that
particular configuration.
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Let us say we have a body which is supported
at these two points and acted on by forces.
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If we
have to draw free body diagram of this, we
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will draw a diagram subjected to the external
forces.
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This may not be the right diagram, the right
diagram will be, if you take a body acted
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on by
active force and also at the support point
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we have the resistive forces which will keep
the body in
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equilibrium.
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Then this is the true free body diagram of
the body.
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Now, supposing if we want to cut the free
body clearly and separate it into two halves
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then we
have this half which is acted by active and
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resistive forces and this section will have
resistive
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force related which will keep this part of
the body in equilibrium.
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Then this particular part also is
a part of the free body of whole structural
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form.
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So, free body diagram is essentially, when
they
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are acted on by the active forces and the
resistive forces of the whole body or part
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of a body
subjected to the external active and resistive
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forces and the internal resistive forces.
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.
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What are the axioms on which behavior of deformable
member subjected to forces depend?
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We have the first thing which is equilibrium
of forces.
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Fundamental laws of Newtonian
mechanics are used for the equilibrium of
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forces and for the body should be such that
it must be
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having the forces in x direction.
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Summation of forces in x direction should
be equal to 0,
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summation of forces in y direction should
be equal to 0, and summation of forces
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in z direction
should be equal
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to 0.
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These equations must be satisfied to fit for
the body which is in space.
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In a
two dimensional form, these equilibrium equations
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reduces to summation of forces in x direction
equal to 0, summation of forces in y direction
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equal to 0 and summation of movement of z
is
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equal to 0.
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Also the forces must satisfy the parallelogram
of forces.
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.Supposing we have two forces in the plane
which is normal and plane forces in the direction
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of
the plane this must satisfy the law like the
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resultant should pass through the diagonal
of the
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parallelogram.
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Or, if we are talking about forces or stresses
in three dimension, if we look into
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this parallelepiped the forces acting in the
x direction or the stress acting in the x
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direction the y
direction and the z direction the resultant
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of this must be acting along the diagonal
of this
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parallelepiped.
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This is the resulting stress of all these
stress components.
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Therefore either in two dimensional plane
it should be in this configuration or in a
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three
dimensional plane it should be in this configuration
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which is the parallelogram of forces that
must be satisfied.
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So these are the two basic axioms based on
which the forces act on the
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deformable body are guided.
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.
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Also, the mechanical properties of the material
are essential to be satisfied.
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These properties of
materials are to be established through laboratory
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experiments.
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Hence the equilibrium of forces
and mechanical properties are the main aspects
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of behavior of the members subjected to forces.
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..
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.
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Here are some of the aspects.
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Now
in this lesson we are going through the aspect
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of stress.
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One should be able to understand the concept
of stress in a body.
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Understand relevant stress
components, and then one should be able to
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understand why we need to go for the equations
of
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the equilibrium, for a given problem.
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One should be able to draw the free-body diagram
and
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evaluate the stress resultants from these
diagrams and thereby compute the stresses.
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..
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The components are subjected to stresses due
to external forces.
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Now, we will be evaluating
stresses in structural components in a systematic
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manner.
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How do you go?
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How do you solve
given a problem and show the particular free
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body diagram of the problem considering that
the
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external forces and the resistive forces acting
on the member and finally once we know the
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stress
resultant from the equations of the equilibrium.
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We should be able to calculate the stresses
at
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different sections as we desire.
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.
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.The approach for analysis of external forces
should be through the development of free
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body
diagram for evaluation of reactive and internal
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forces and thereby evaluation of developed
stresses due to external forces.
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.
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The concept of stress has been already discussed
which is at a particular smaller element and
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is
the intensity of the forces acting on the
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area.
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We had defined that force for smaller unit
area as
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the stress.
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.
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.Now this stress when it is multiplied
is acted on by external forces.
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If we take a small element in
which a stress is acting this stress multiplied
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by the area gives the force which we call
as the
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stress resultant and the total stress resultant
is the stress multiplied by the elemental
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area
integrated over the whole of the area is the
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resulting force which we call as stress resultant.
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And
thereby, we assume that at any point it has
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the same behavior.
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.
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So at any section the vector sum of the forces
keeps the body in equilibrium and that is
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how the
stress resultant will be obtained for that
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particular section.
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So our job is to evaluate this stress
resultant.
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And once we know the test resultant we can
compute stresses at that particular section.
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.
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.We are interested in knowing the resultant
stress vector in a particular section.
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Let us look into a
body.
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.
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This is x-axis, this is y-axis and this is
z-axis.
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This is a body which is acted on by external
forces.
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If we cut this particular body to a plane
which is perpendicular or rather parallel
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to the y, z plane
then the normal drawn on this particular plane
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will be parallel to x-axis.
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For any plane when you
draw the normal to that particular plane and
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if that particular normal coincides with any
of the
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axis we designate that plane with the name
of that particular axis.
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For example, here we have cut this body through
a plane which is parallel to y, z plane.
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So, if we
draw a normal on to this section this normal
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is going to be the parallel to x-axis and
thereby this
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particular plane we designate this plane as
x-plane.
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Now on this x-plane we have at a particular
point the resulting stress which we call as
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R. If we take the component of this stress
R in three
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perpendicular axis direction then we have
the stress acting in the direction of x which
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is normal
to this particular section and as per the
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definition of normal stress this is the normal
stress which
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is acting in the direction of x.
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If we take the component of R along y-axis
or parallel to y-axis
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then the plane we get as stress is acting
in y direction.
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Also, we have the component which is acting
in z direction.
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The stress which is acting parallel to
y or in the y direction we designate this
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as the stress tau acting in the plane x in
the direction y
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which we call as tau x, y.
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Or if we designate this particular stress
which is acting in the plane x along z direction
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then we
call these as tau xz . Thereby in this particular
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x-plane we have three stress components where
one
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is the normal to the plane and the other two
are in the plane in the direction of y and
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z.
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The
normal stress we call as σ x and the other
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two components which are in the plane are
called as
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shearing stresses which are τ xy and τ xy
.
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.Likewise, if we cut this body into a plane
which is parallel to this z plane we cut along
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this then
on this plane on a particular point we can
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get three components of the stresses and they
are σ y , τ
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on the y plane in the x direction as τ yx
, the shearing stress tau in the y plane in
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the z direction
as τ yz . Also, if we cut this body with
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a plane which is perpendicular or parallel
to x, y plane and
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if we plot the three stress components the
stress which is normal to the plane gives
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out the
normal stress sigma z and two stresses which
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are in the plane are on the z plane in the
x-direction
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and stress in the z-plane in the y-direction.
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These are the nine stress components that
we are going to get at a particular point.
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Now if this
particular body is cut in such a way that
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you take another plane which is at an infinite
small
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distance away from here if we cut it off by
two parallel planes
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then we can get small a cubical
element on which you can plot the stresses.
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.
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Thereby the three faces which are visible
to us are in front of the board, if we look
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into that this
particular plane let us say this is x-axis,
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this is y-axis, and this z-axis.
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The stress on this particular
plane the direction of x is the normal stress
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sigma x is in the x-plane, on the x-plane
the stress
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acting in the y-direction is tau xy and on
the x-plane the stress acting in the z-direction
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is
designated as tau xz .
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Likewise, on this plane we will have stress
acting in the y-direction the normal stress
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which is
sigma y and for the positive direction of
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x on the y-plane we have the shearing stress
which is
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designated as tau y-plane in the x-direction
and on the y-plane in z-direction we have
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the stress
which we designate as tau yz . This particular
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plane the normal to z we call it as the z-plane
and on
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this we have the stresses acting the normal
to the z as sigma z .
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.On the z-plane in the x-direction we have
tau zx and then this is the z-plane with the
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y-direction
which is in this particular direction the
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positive direction of y so this is tau zy
. Likewise on the
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00:22:16,010 --> 00:22:23,510
other three faces which are hidden from this
side this face, this face and this face also
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00:22:23,510 --> 00:22:29,510
have the
three components of the stresses which are
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00:22:29,510 --> 00:22:33,720
sigma tau and in the two planes we have tau.
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00:22:33,720 --> 00:22:40,020
In this particular plane normal to this which
is acting in the negative x-direction will
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00:22:40,020 --> 00:22:44,820
have the
stress sigma x in this particular direction.
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The
face will have the shearing stress component
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00:22:52,421 --> 00:22:56,490
this
being the x-plane and in the opposite direction
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of y we will have the shearing stress that
is tau xy ,
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in the x-plane but in the z-direction opposite
to the positive z-axis we will have x, z so
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this is
tau xy and this is tau x, z.
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Likewise on the y-plane we will have the stress
acting perpendicular to it which is sigma
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y . The
y-plane in the x-direction will have the shearing
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stress but it will be in the opposite direction
of x
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which is being the other end.
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On this y-plane in the z-direction we will
have stress tau yz . This is
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tau yx and this is tau yz . Likewise on the
other side of the z-plane we will have the
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normal stress
acting which is sigma z and on the z-plane
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we have stresses in the x-direction which
is tau zx and
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in the z-plane we have tau zy . These are
the stress components that will be acting
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on this
particular body at a point.
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.
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Let us look into the state of stress at a
point.
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As we have seen, the state of stress at a
point is
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represented by those nine stress components.
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Now three stage state are represented by the
three stress components like sigma x , tau
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xy and tau xz
and likewise in other planes.
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So all these three stress components are written
on three mutually
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perpendicular axis and they are represented
by a term which we generally written as tau
233
00:25:24,270 --> 00:25:28,450
ij which
is called as a stress tension.
234
00:25:28,450 --> 00:25:29,450
..
235
00:25:29,450 --> 00:25:45,130
So stress tensor is tau ij where i represents
the (x, y, z) directions or the (x, y, z)
236
00:25:45,130 --> 00:25:50,050
plane and j
represents the three stress directions.
237
00:25:50,050 --> 00:25:56,270
So if we expand these on the x-plane ad when
i is x we
238
00:25:56,270 --> 00:26:12,130
have j as xyz, so the stress components are
tau xx, tau xy and tau xz . When i is the
239
00:26:12,130 --> 00:26:17,430
y plane we will
have three components in x, y and z direction.
240
00:26:17,430 --> 00:26:28,530
So we will have tau yx , tau yy and tau yz
. Likewise
241
00:26:28,530 --> 00:26:34,120
when i is in the z-plane we will have three
components in the x, y and z-direction.
242
00:26:34,120 --> 00:26:45,620
We have tau zx ,
tau zy and tau zz . Now the components which
243
00:26:45,620 --> 00:26:51,059
are tau xx , tau yy and tau zz are normal
to the plane.
244
00:26:51,059 --> 00:27:01,929
Let
us call them as sigma x , this as sigma y
245
00:27:01,929 --> 00:27:08,660
and this as sigma z .
So, we have the stress components sigma x
246
00:27:08,660 --> 00:27:13,460
, tau xy and tau xz which are acting in the
x-plane, we
247
00:27:13,460 --> 00:27:23,370
have tau yx , sigma y , tau yz in the y-plane,
we have tau zx , tau zy and sigma z which
248
00:27:23,370 --> 00:27:27,390
are acting on the
z-plane or in the z-direction.
249
00:27:27,390 --> 00:27:37,900
These are the nine stress components we have
at a point.
250
00:27:37,900 --> 00:27:38,900
..
251
00:27:38,900 --> 00:27:41,330
Now let us look into another aspect of it.
252
00:27:41,330 --> 00:27:45,220
Out of the stress components which we have
obtained,
253
00:27:45,220 --> 00:27:58,560
again let us call these as x-axis as usual,
we call as y-axis and this as z-axis.
254
00:27:58,560 --> 00:28:09,919
Now, if we take the movement of all these
forces, now let us assume that this distance
255
00:28:09,919 --> 00:28:13,280
which we
have taken at a particular point of the body
256
00:28:13,280 --> 00:28:21,091
is dx, the vertical height being dy and along
the z
257
00:28:21,091 --> 00:28:25,370
direction tau with dz.
258
00:28:25,370 --> 00:28:34,450
Now if we like to take the movement of all
the forces about z-axis, now in
259
00:28:34,450 --> 00:28:40,500
this particular figure only the forces which
will have relevance while taking the movement
260
00:28:40,500 --> 00:28:45,600
about
z-axis has been taken into account.
261
00:28:45,600 --> 00:28:50,310
Now this particular plane being x plane and
the force which
262
00:28:50,310 --> 00:28:59,429
is acting in the y-direction as per our nomenclature
we call this as tau xy .
263
00:28:59,429 --> 00:29:06,059
Accordingly this particular component of the
stress which is acting in the direction of
264
00:29:06,059 --> 00:29:14,070
x and
acting on y-plane we call this as yx.
265
00:29:14,070 --> 00:29:27,049
Likewise, this is also tau yx and this is
tau xy . Along with this
266
00:29:27,049 --> 00:29:34,059
we have the other stresses like normal stress
sigma x , sigma y and sigma z and shearing
267
00:29:34,059 --> 00:29:36,370
components as well in the other plane.
268
00:29:36,370 --> 00:29:42,419
Since only these forces or these stress components
are
269
00:29:42,419 --> 00:29:48,850
going to cause the movement other forces have
not been shown here.
270
00:29:48,850 --> 00:29:56,620
If we take the movement of all the forces
about z-axis then the movement expression
271
00:29:56,620 --> 00:30:05,160
can be
written as tau yx which is the stress acting
272
00:30:05,160 --> 00:30:12,020
on the area dx by dz, so tau yx into dx into
dz is the
273
00:30:12,020 --> 00:30:14,800
force.
274
00:30:14,800 --> 00:30:23,450
The movement about the z-axis is the distance
dy so this multiplied by dy is the movement
275
00:30:23,450 --> 00:30:37,090
about the z-axis as tau yx which is clockwise
in nature minus tau xy which is acting on
276
00:30:37,090 --> 00:30:39,650
the area
dz(dy).
277
00:30:39,650 --> 00:30:47,049
So tau xy into dy into dz is the force.
278
00:30:47,049 --> 00:30:51,309
If we take the movement of this force with
respect to z- axis then this is multiplied
279
00:30:51,309 --> 00:30:53,900
by the
distance dx.
280
00:30:53,900 --> 00:31:10,180
Assumingly that there are body force components
in the x and y-direction, this x is
281
00:31:10,180 --> 00:31:23,220
the body force per unit volume then along
with this we have plus, but for the time being
282
00:31:23,220 --> 00:31:26,950
we are
neglecting the body force components because
283
00:31:26,950 --> 00:31:30,540
that is also not going to cause any moment
as
284
00:31:30,540 --> 00:31:33,460
such with respect to the z-axis.
285
00:31:33,460 --> 00:31:43,660
.Therefore here it is z is equal to 0 for
equilibrium.
286
00:31:43,660 --> 00:31:53,010
This produces tau yx is equal to tau xy . In
effect
287
00:31:53,010 --> 00:32:02,450
this means that the cross term tau yx and
tau xy are equal.
288
00:32:02,450 --> 00:32:09,799
Likewise if we take the movement of
forces about x and y-axis and take the relevant
289
00:32:09,799 --> 00:32:20,521
forces we can see that tau zx is equal to
tau xz and
290
00:32:20,521 --> 00:32:34,090
tau yz is equal to tau zy . This gives us
that the cross hearing terms are equal.
291
00:32:34,090 --> 00:32:43,890
So if we look into
stress and strain which we had tau ij the
292
00:32:43,890 --> 00:32:59,950
tau ij is equal to sigma x , tau xy , tau
xz , tau yx , sigma y and
293
00:32:59,950 --> 00:33:12,750
tau yz , tau zx , tau zy and sigma z if we
write in the matrix form.
294
00:33:12,750 --> 00:33:15,919
This is the stress tensor.
295
00:33:15,919 --> 00:33:34,890
Now for the equality of the shear we have
obtained tau xy is equal to tau yx , tau xz
296
00:33:34,890 --> 00:33:48,610
is equal to tau zx ,
tau yz is equal to tau zy . Thereby stress
297
00:33:48,610 --> 00:33:56,169
tensor can be written as tau ij in the matrix
notation as
298
00:33:56,169 --> 00:34:09,399
sigma x , tau xy and tau xz . Now tau xy is
equal to tau yx we will write this as tau
299
00:34:09,399 --> 00:34:20,770
xy , sigma y and tau yz
and zx and xz being the same we write this
300
00:34:20,770 --> 00:34:29,110
as tau xz , tau yz , tau yz and sigma z and
thereby it
301
00:34:29,110 --> 00:34:36,450
reduces to the six stress components sigma
x , sigma y and sigma z , tau xy , tau xz
302
00:34:36,450 --> 00:34:42,960
and tau yz and this
we find is symmetrical in nature so the stress
303
00:34:42,960 --> 00:34:54,859
tensor has a symmetric form.
304
00:34:54,859 --> 00:34:55,859
.
305
00:34:55,859 --> 00:35:04,039
If we write down the stress components in
a two dimensional plane, then we call those
306
00:35:04,039 --> 00:35:09,180
stresses as
the elements in the plane stress.
307
00:35:09,180 --> 00:35:20,529
The stress
components which we will have here are sigma
308
00:35:20,529 --> 00:35:27,479
x the normal stress this being the xdirection,
this being the y-direction we have the stress
309
00:35:27,479 --> 00:35:38,539
sigma y so we are concentrating on the xyplane
and this is also sigma y and this is sigma
310
00:35:38,539 --> 00:35:53,569
x . This being the x-plane the stresses acting
in ydirection is tau xy . This being the y-plane
311
00:35:53,569 --> 00:35:57,440
and this is acting in the x-direction according
to our
312
00:35:57,440 --> 00:36:01,719
designation nomenclature we call this as tau
yx .
313
00:36:01,719 --> 00:36:11,489
But since tau yx is equal to tau xy we call
this as tau xy . And so are these stresses
314
00:36:11,489 --> 00:36:19,200
which are tau xy
and tau xy . Since all the shearing stress
315
00:36:19,200 --> 00:36:26,170
components are xy we can call this simply
as tau.
316
00:36:26,170 --> 00:36:38,010
Therefore we have normal stress components
sigma x , sigma y and shearing stress tau.
317
00:36:38,010 --> 00:36:39,010
..
318
00:36:39,010 --> 00:36:52,130
Having known that the stress at particular
point is acting which are combinations of
319
00:36:52,130 --> 00:36:56,210
normal
stresses and shearing stresses let us look
320
00:36:56,210 --> 00:37:01,069
into that if we have a body and if we are
interested to
321
00:37:01,069 --> 00:37:11,239
find out the change in stress from one point
to another then the change of the stresses
322
00:37:11,239 --> 00:37:15,900
is from
point to point, we need certain equations
323
00:37:15,900 --> 00:37:19,150
to be solved and those equations are called
as equations
324
00:37:19,150 --> 00:37:23,799
of equilibrium.
325
00:37:23,799 --> 00:37:36,539
Coming back to the body here, for example
we have a body which is stress and we like
326
00:37:36,539 --> 00:37:40,690
to find
out the change in stress from this point to
327
00:37:40,690 --> 00:37:42,119
this point.
328
00:37:42,119 --> 00:37:49,440
So we need these changes to be evaluated
through these equations of equilibrium.
329
00:37:49,440 --> 00:37:57,489
Now as usual we call this as x-axis, this
as y-axis and
330
00:37:57,489 --> 00:38:02,719
this as z-axis.
331
00:38:02,719 --> 00:38:12,089
Now on this particular plane which is normal
to the x-plane we have normal
332
00:38:12,089 --> 00:38:21,940
stresses known as sigma x .
We have two shearing stress components in
333
00:38:21,940 --> 00:38:38,910
the x-plane acting in y-direction called as
tau xy . We
334
00:38:38,910 --> 00:38:48,019
have stress in the x-plane in the z-direction
which we call as tau xz . When it comes to
335
00:38:48,019 --> 00:38:53,440
this
particular plane which is at a distance of
336
00:38:53,440 --> 00:38:57,479
dx from this plane and likewise let us assume
that this
337
00:38:57,479 --> 00:39:08,859
length is dy and this is dz so the stress
which will be acting in this which is the
338
00:39:08,859 --> 00:39:25,910
normal stress will
have
339
00:39:25,910 --> 00:39:40,029
a component as sigma x plus del sigma x del
x which is acting over the length dx.
340
00:39:40,029 --> 00:39:42,710
Likewise
we will have the shearing stress component
341
00:39:42,710 --> 00:39:47,921
tau xy which is varying from this end to this
end we
342
00:39:47,921 --> 00:39:56,990
will have tau xy plus del tau xy del x by
dx.
343
00:39:56,990 --> 00:40:13,739
We will have x in the z-direction that is
tau xz and tau xz in on this particular plane
344
00:40:13,739 --> 00:40:16,349
so when it is
coming to this plane there is a change over
345
00:40:16,349 --> 00:40:29,839
the length dx so tau xz plus del tau xz del
x by dx.
346
00:40:29,839 --> 00:40:38,170
Likewise the stress in this particular plane
normal to this which is the y-plane will have
347
00:40:38,170 --> 00:40:45,509
sigma y ,
the stress acting normal to this is sigma
348
00:40:45,509 --> 00:40:56,369
y plus del sigma y del y by dy the length.
349
00:40:56,369 --> 00:40:59,650
The shearing
stress component on the y-plane acting in
350
00:40:59,650 --> 00:41:16,760
the direction of x will have
tau yx , plus del tau yx del y
351
00:41:16,760 --> 00:41:23,249
by dy the length.
352
00:41:23,249 --> 00:41:36,279
On this plane we will have sigma z and the
normal stress on the front z-plane is
353
00:41:36,279 --> 00:41:49,519
sigma z plus del sigma z del z over the length
dz and so on.
354
00:41:49,519 --> 00:41:59,599
.Now if we take the forces which are acting
in the x-direction and sum them up as for
355
00:41:59,599 --> 00:42:02,509
the
equations of equilibrium the summation of
356
00:42:02,509 --> 00:42:08,829
all the forces in the x-direction must be
equal to 0.
357
00:42:08,829 --> 00:42:17,119
If
we write down the forces in the x-direction
358
00:42:17,119 --> 00:42:53,670
we have sigma x plus del sigma x del x by
dx.
359
00:42:53,670 --> 00:42:58,930
So in
the equation we have sigma x plus del sigma
360
00:42:58,930 --> 00:43:16,160
x del x by dx and acting over the area dy
and dz
361
00:43:16,160 --> 00:43:31,700
minus sigma x acting over the area dy and
dz.
362
00:43:31,700 --> 00:43:52,510
Also, in this particular direction we have
plus (tau yx plus del tau yx del x(dx)) and
363
00:43:52,510 --> 00:44:07,650
delta yx by del
y by dy acting over the area dx and dz minus
364
00:44:07,650 --> 00:44:18,079
tau yx , dx dz plus we have a term in the
z plane
365
00:44:18,079 --> 00:44:38,690
acting on the x direction which is tau zx
plus del tau zx del z by dz into dx and dy
366
00:44:38,690 --> 00:44:54,059
the area minus
tau zx (dx and dy) plus if we assume that
367
00:44:54,059 --> 00:44:59,019
X is the body force per unit volume then this
multiplied
368
00:44:59,019 --> 00:45:07,400
by dx, dy and dz is equal to 0.
369
00:45:07,400 --> 00:45:22,609
So from this we will get del sigma x del x,
if we cancel out these
370
00:45:22,609 --> 00:45:37,910
terms and divide the whole equation by dx,
dy and dz we have del sigma x del x plus del
371
00:45:37,910 --> 00:45:54,380
tau yx by
del y plus del tau xz by del z plus x is equal
372
00:45:54,380 --> 00:46:02,479
to 0 where x is the component of the body
force.
373
00:46:02,479 --> 00:46:12,219
Likewise if we take the equilibrium of the
forces which are acting in the y and z-direction
374
00:46:12,219 --> 00:46:20,849
we get
two other sets of equations and they are,
375
00:46:20,849 --> 00:46:37,339
del tau xy by delx plus del sigma y by del
y plus del tau yz
376
00:46:37,339 --> 00:46:53,079
by del z plus y the body component force is
equal to 0 del tau xz by del z plus del tau
377
00:46:53,079 --> 00:47:05,009
yz del y plus
del sigma z del z plus z is equal to 0.
378
00:47:05,009 --> 00:47:17,529
These are called the equations of equilibrium.
379
00:47:17,529 --> 00:47:18,529
.
380
00:47:18,529 --> 00:47:25,150
These equations of equilibrium can be written
down in a two dimensional form as well.
381
00:47:25,150 --> 00:47:33,380
You can
designate these in xy-plane, here we have
382
00:47:33,380 --> 00:47:42,749
x and here we have y, this is sigma x and
the variation
383
00:47:42,749 --> 00:47:56,059
along the x is sigma x plus del sigma x del
x by dx the length where this is the length
384
00:47:56,059 --> 00:48:12,119
dx and this is
dy, this is tau xy plus del tau xy del x by
385
00:48:12,119 --> 00:48:15,680
dx the length.
386
00:48:15,680 --> 00:48:24,789
This is sigma y plus del sigma y del y by
dy
387
00:48:24,789 --> 00:48:37,480
the length, then we have tau this is tau yx
and this gives the variation of tau which
388
00:48:37,480 --> 00:48:51,789
is tau yx plus del
tau yx by del y(dy).
389
00:48:51,789 --> 00:49:00,239
These are the stresses in the two dimensional
plane and if we take the equilibrium of the
390
00:49:00,239 --> 00:49:08,710
forces in
the x-direction then we can obtain the equations
391
00:49:08,710 --> 00:49:10,549
of the equilibrium in two dimensional plane
392
00:49:10,549 --> 00:49:28,229
.which could be del sigma x del x plus del
tau and as yx and xy being the same we can
393
00:49:28,229 --> 00:49:34,460
write this
as tau xy del y plus the component of the
394
00:49:34,460 --> 00:49:38,130
body force x is equal to 0.
395
00:49:38,130 --> 00:49:52,479
The other equations will be del tau xy by
del x plus del sigma y del y plus y is equal
396
00:49:52,479 --> 00:49:54,869
to 0.
397
00:49:54,869 --> 00:49:58,199
So these
are the equations of equilibrium in two dimensional
398
00:49:58,199 --> 00:50:03,089
planes.
399
00:50:03,089 --> 00:50:17,989
Now, having known these stresses at
a point, equations of equilibrium and how
400
00:50:17,989 --> 00:50:20,900
to evaluate those stresses or how to write
down those
401
00:50:20,900 --> 00:50:28,799
stresses at different planes if we have to
evaluate the stresses in an axially loaded
402
00:50:28,799 --> 00:50:33,319
member, then
we have a member or we have a body in which
403
00:50:33,319 --> 00:50:40,329
we have a force acting in the axial direction,
so
404
00:50:40,329 --> 00:50:45,930
let us call this body acted on by force P.
.
405
00:50:45,930 --> 00:50:54,329
Now if we like to evaluate the stresses at
any inclined plane let us cut this body by
406
00:50:54,329 --> 00:50:57,950
an inclined
plane.
407
00:50:57,950 --> 00:51:18,609
And if we draw the free body diagram of this
then we have
408
00:51:18,609 --> 00:51:22,299
the body in this form.
409
00:51:22,299 --> 00:51:28,769
Here we
have the resistive force P which is acting.
410
00:51:28,769 --> 00:51:39,660
On this, the resulting force or the stress
resultant is
411
00:51:39,660 --> 00:51:48,930
acting in this particular direction to equilibrate
the body.
412
00:51:48,930 --> 00:51:58,959
Now we can take the components of
this force in the normal direction, normal
413
00:51:58,959 --> 00:52:07,950
to this plane and along the plane which will
give you
414
00:52:07,950 --> 00:52:16,839
the three forces of the stress component which
is normal which we call as the stress
415
00:52:16,839 --> 00:52:36,829
corresponding to normal and the two shearing
stress components tau.
416
00:52:36,829 --> 00:52:56,339
If we concentrate on the two dimensional plane,
if we take the axially loaded member
417
00:52:56,339 --> 00:53:02,089
the
member is subjected to the load in the axial
418
00:53:02,089 --> 00:53:04,460
direction.
419
00:53:04,460 --> 00:53:11,979
Let us take a plane which is cutting this
body in this form and let us assume that this
420
00:53:11,979 --> 00:53:18,859
plane is making an angle of theta with the
vertical.
421
00:53:18,859 --> 00:53:29,469
If we take the free body diagram of this particular
body
422
00:53:29,469 --> 00:53:42,249
this is angle theta, we have the force
acting here as P, the resistive force acting
423
00:53:42,249 --> 00:53:48,499
on this body to keep the equilibrium is P
so this will
424
00:53:48,499 --> 00:54:00,589
have two components one along the normal and
one along the plane of this particular section.
425
00:54:00,589 --> 00:54:08,589
Now this angle being theta, if we drop a perpendicular
here this angle will also be theta hence
426
00:54:08,589 --> 00:54:11,140
this particular angle is also theta.
427
00:54:11,140 --> 00:54:18,749
So the force component along this is P cos
theta and the force
428
00:54:18,749 --> 00:54:23,890
component along this direction is P sin theta.
429
00:54:23,890 --> 00:54:29,420
Now, if we say that the cross sectional area
is A
430
00:54:29,420 --> 00:54:39,000
.and the cross sectional area of this as A
prime then the stress which is acting in this
431
00:54:39,000 --> 00:54:42,509
particular
inclined plane if we say that normal stress
432
00:54:42,509 --> 00:54:46,690
sigma theta and theta being designated by
this
433
00:54:46,690 --> 00:54:54,079
particular plane which has got an angle theta
in the vertical, sigma theta equals to the
434
00:54:54,079 --> 00:54:58,479
normal
force component which is P cos theta divided
435
00:54:58,479 --> 00:55:01,980
by this area A prime.
436
00:55:01,980 --> 00:55:12,989
And A prime from geometrical property we can
say A prime is
437
00:55:12,989 --> 00:55:14,999
equal to A by cos theta.
438
00:55:14,999 --> 00:55:19,980
Then P
cos theta by A by cos theta is equal to P
439
00:55:19,980 --> 00:55:26,810
by A cos square theta.
440
00:55:26,810 --> 00:55:34,279
And the stress which is acting in
the plane is P sin theta.
441
00:55:34,279 --> 00:55:45,999
The stress tau theta is equal to P sine theta
by Acos theta so this
442
00:55:45,999 --> 00:55:56,119
eventually is going to give us P by 2A by
sin 2 theta.
443
00:55:56,119 --> 00:55:59,019
So these are the two stress components on
this inclined plane.
444
00:55:59,019 --> 00:56:06,299
The normal plane is P by A cos square theta
and the stress which is parallel to
445
00:56:06,299 --> 00:56:13,420
the plane which is the shearing component
is P by 2A sin2 theta.
446
00:56:13,420 --> 00:56:20,259
So the maximum value of
sigma theta is when cos square theta is equal
447
00:56:20,259 --> 00:56:26,130
to 1 and is theta is equal to 0.
448
00:56:26,130 --> 00:56:31,339
And for this sin 2
theta as 90 degrees and 2 theta being 90 degrees
449
00:56:31,339 --> 00:56:36,779
so theta being 45 degrees is the maximum value
of the shearing stress.
450
00:56:36,779 --> 00:56:41,309
So the maximum value of normal is P by A and
the maximum value of
451
00:56:41,309 --> 00:56:43,299
shearing stress is P by 2A.
452
00:56:43,299 --> 00:56:48,400
Eventually the relationship between tau theta
and sigma theta is that
453
00:56:48,400 --> 00:57:02,819
tau theta is half of sigma theta.
454
00:57:02,819 --> 00:57:03,819
.
455
00:57:03,819 --> 00:57:08,480
We have seen the maximum normal stress which
is P by A and maximum shearing stress is P
456
00:57:08,480 --> 00:57:11,049
by
2A.
457
00:57:11,049 --> 00:57:28,549
Shear stresses are at this particular body
is acted on by the force P eventually the
458
00:57:28,549 --> 00:57:33,960
resistive
force will be P by 2 and P by 2.
459
00:57:33,960 --> 00:57:41,930
At this level if we take the free body at
this part if I cut here then
460
00:57:41,930 --> 00:57:51,449
this particular body will have P by 2 and
this is also going to be P by 2.
461
00:57:51,449 --> 00:57:57,470
So at this particular
section this is P by 2 and its resistive is
462
00:57:57,470 --> 00:57:59,900
going to be P by 2.
463
00:57:59,900 --> 00:58:06,689
Now here there is little amount of eccentricity
for this force to be transferred over here
464
00:58:06,689 --> 00:58:10,709
and this
we ignore as the thickness is being smaller.
465
00:58:10,709 --> 00:58:14,920
This P by 2 is called as the shearing force
and this
466
00:58:14,920 --> 00:58:22,199
shearing force divided by this area which
is acted on between these two plates is called
467
00:58:22,199 --> 00:58:23,689
as the
shearing stress.
468
00:58:23,689 --> 00:58:32,249
If we say this width is b and this width is
t then the shearing stress is (P by 2) by
469
00:58:32,249 --> 00:58:36,999
b to the power t is equal to tau.
470
00:58:36,999 --> 00:58:37,999
..
471
00:58:37,999 --> 00:58:53,819
Now, here we are given a problem where two
plates are connected by two bolts for which
472
00:58:53,819 --> 00:58:57,459
we
have to evaluate normal stress, shear stress
473
00:58:57,459 --> 00:59:45,069
and the wearing stress.
474
00:59:45,069 --> 00:59:45,069
.