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Welcome to the 5th lesson of Strength of Materials.
We will be discussing certain aspects
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of analysis of stress in this particular lesson.
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..
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Now, it is expected that once this particular
lesson is completed, one should be able to
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evaluate stresses on any plane through stress
transformation equations using Mohr’s
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circle. We will be looking into some more
aspects of it then evaluate principal stresses
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and locate principal planes at a particular
point on the stress body. Also, one should
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be
able to evaluate octahedral stresses which
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we will look into in this lesson.
.
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The scope of this particular lesson includes
recapitulation of transformation of equations
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which we have derived earlier, recapitulation
of evaluation of stresses using Mohr’s
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circle. We will be looking into some aspects
of octahedral stresses; we will define
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.octahedral plane and stresses acting on such
planes and then we will be looking into some
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examples of how to evaluate the stresses at
a particular point at a stress body.
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.
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We have discussed how to evaluate the stress
on a particular plane which is inclined at
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angle of θ with respect to x-plane and theses
we have termed as transformation
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equations. If we have a body which is acted
on by the rectangular stress components,
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which are σ x σ y in the y-direction and
the shearing stresses. We can compute stresses
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on any plane, the normal to which it is making
an angle θ with the x-plane. This is what
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we have defined as the normal stress on this
plane as, σ x prime and the shearing stress
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as τ x ' y ' .
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..
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The σ x prime is given in terms of σ x , σ
y and τ xy as ( σ x plus σ y ) by 2 plus
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( σ x minus σ y )
by 2 cos 2 θ plus τ xy sin2 θ . The normal
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stress on that x prime plane and τ x ' y
' - the
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shearing stresses in that particular plane
is equals to minus σ x minus σ y by 2 sin2
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θ plus
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τ xy cos2 θ . And also the normal stress
to the perpendicular plane in the x prime
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can be
written as ( σ x plus σ y ) by 2 and (minus
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σ x minus σ y ) by 2 cos 2 θ minus τ xy
sin2 θ . And
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eventually we have seen that the σ x prime
plus σ y prime is equal to σ x plus σ y
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which
gives us the first stress invariant.
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.
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.Thereby we had calculated the principal stresses
which are the maximum normal stress,
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the maximum and minimum in two different planes
it acts, and the magnitude of those
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principal stresses are, σ 1 is equal to ( σ
x plus σ y ) by 2 plus square root of ((( σ
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x minus σ y )
by 2)) whole square plus τ xy square). The
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σ 2 is the minimum principal stress is equal
to
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( σ x plus σ y ) by 2 minus square root
of ((( σ x minus σ y ) by 2)) whole square
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plus
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τ xy square). And the plane which gives us
the maximum normal stress or the principal
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stresses can be evaluated through this equation,
where tan2 θ P is equal to 2 τ xy by σ
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x
minus σ y . And in the physical plane this
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is denoted by θ P .
.
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Also, we have evaluated the maximum shear
stresses on planes and we had observed that
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the maximum shear stress occurs in planes
which are at an angle of 45 degrees with the
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difference to the principal planes. And the
magnitude of the maximum and minimum
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shear stresses are plus square root of ((( σ
x minus σ y ) by 2) whole square plus τ
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xy square)
and the minimum one is minus square root of
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((( σ x minus σ y ) by 2) whole square plus
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τ xy square). This we had elaborated last
time, τ
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max
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is equal to square root of
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(( σ x minus σ y ) by 2) whole square plus
τ xy square and τ
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min
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equals to minus of that
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quantity and the angle which can be evaluated
from the normal stresses and the shear
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stresses and this gives the plane shear stresses
are maximum and minimum.
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Also, we had looked in to how to compute the
stresses at any plane, using the concept of
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Mohr’s circle apart from the equations of
equilibrium. We had used the equations of
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equilibrium to evaluate stress at a particular
point; now we will look in to what we had
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looked into - how to evaluate any point using
Mohr’s circle of stress.
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.Let us assume that, at a particular point
in the stress body, the normal components
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of
stress which are acting, the rectangular stresses
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are: σ x , σ y the normal stress, τ xy
the
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shearing stress. Now if we like to represent
them in the Mohr’s circle, last time we
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had
seen that this is the sigma axis, this is
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the τ axis. And we can denote the centre
of the
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Mohr’s circle at a distance from the origin
which is at ( σ x plus σ y ) by 2, which
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was A,
and we have denoted this particular quantity
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as A. This particular point on the Mohr’s
circle denotes this plane in the physical
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space which is A, and there by this particular
point represents the normal stress σ x and
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the shearing stress τ xy . This particular
point
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represents this particular plane B, where
the normal stress is the σ y , and this is
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τ xy .
Now if we join these two lines which eventually
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pass through the center and in this
Mohr’s circle, the angle between these two
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planes is 180 degrees, which is 2 θ and in
the
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physical space between these two plates is
90 degrees; half of this particular angle.
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There
by this particular point is the maximum normal
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stress, this is the plane where maximum
normal stress acts; and this is the plane
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where the minimum normal stress acts.
From this particular diagram, we can observe
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that the maximum normal stress which we
generally denotes as σ 1 , this is nothing
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but equals to the distance from the origin
to the
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centre plus the radius. So this is ( σ x
plus σ y ) by 2 and this is the radius which
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is equals to
this hypotenuse which is this square plus
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this square (Refer Slide Time:). And this
distance is ( σ x plus σ x minus σ y ) by
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2.
And eventually this distance comes as ( σ
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x minus σ y ) by 2 and this is τ xy and
hence σ 1 is
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equal to ( σ x plus σ y ) by 2 plus square
root of (( σ x minus σ y ) by 2) whole square
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plus τ xy
square. And this gives you the minimum principal
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stress which is this distance minus the
radius, which will be ( σ x plus σ y ) by
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2 minus square root of (( σ x minus σ y
) by 2) whole
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square plus τ xy square .
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..
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Now let us look into the given state of stress
at a particular point as to how you evaluate
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this stress at a particular plane by constructing
the Mohr’s circle. Let us assume that we
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have a stress body where the stresses at a
particular point are: σ x is the normal stresses
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in
the x-plane and σ y is the normal stress
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in the normal plane, τ xy gives the shear
stress,
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which has positive shear this is on the positive
y-axis in this plane; and this is on the
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positive x-direction in this particular plane
and these are these components σ x , σ y
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and
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τ xy . As we have defined last time that,
this axis represents σ the normal stress
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axis, and
the y-axis represents the shearing stress
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axis.
We take the direction of the shear stress
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downwards, to have the compatibility of the
rotational angle which is in an anticlockwise
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direction, which will be represented in the
anticlockwise direction in Mohr’s circle
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as well. Thereby the normal positive stress
and
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the normal positive shear stress on this plane,
σ x and τ xy , if we represent on the Mohr’s
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plane, let us assume that, σ x > σ y , then
we have the plane which is represented as
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σ x
and τ xy as this particular point where normal
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stress is σ x
and shear stress is τ xy . We can
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represent this particular plane on this Mohr’s
plane wherein the stresses are σ x and τ
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xy .
Now please note here that the shear which
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is acting in the positive y-direction on x-plane
we have considered this as the positive shear.
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This particular shear along this
complimentary stress is causing the rotation
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in the anticlockwise direction which we
denoted as positive shear in the Mohr’s
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plane. Whereas the shear which is acting in
the yplane, this along the complimentary shear
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on the other side is causing rotation in the
clockwise direction. Based on this we are
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calling this as a negative shear on this Mohr’s
plane. If we try to represent the normal and
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shear stress in the Mohr’s plane, we have
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.positive σ y and the negative τ xy , which
represents point B, which is for this particular
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plane.
Now if we join these two lines as we have
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seen earlier, eventually this will pass through
the centre of the Mohr’s circle. And we
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have seen that the distance of the centre
of the
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Mohr’s circle from the τ axis is equals
to ( σ x plus σ y ) by 2, which is the average
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stress.
Now taking this as centre and OB or OA as
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radius if we plot the circle, eventually we
are
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going to get the Mohr’s circle.
Please note that the Mohr’s circle need
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not be constructed geometrically, if we can
represent them in this particular form we
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can compute the other stresses directly from
this
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diagram itself.
In this particular diagram, the maximum normal
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stress occurs at this particular plane and
the minimum normal stress occurs at this particular
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plane. And this normal stress we
denote as σ 1 . This particular normal stress
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we denote as σ 2 . From plane A if we move
by
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angle 2 θ P we come to the plane where the
maximum principle stress is located. So, in
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the physical plane if we rotate by angle θ
P in the anticlockwise direction this gives
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the
normal and perpendicular to the plane where
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the σ 1 x acts. That is how we decide about
the planes for the maximum principal stresses
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and correspondingly the maximum shear
stress.
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Likewise the maximum shear stresses, will
occur at this particular point which is the
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highest point in this circle. And this is
τ max which is eventually equal to the radius
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of
the circle. This is the radius so this is
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the positive maximum shear and this is the
minimum shear which is negative of this radius.
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This gives us the values of maximum
shear stresses.
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Please note that, at these points where the
maximum shear stresses occur there we will
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have some values of normal stress which is
in contrast to the one where we have the
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maximum normal stresses where shear stresses
are 0.
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Now if we like to evaluate the stress at a
particular plane from this Mohr’s diagram.
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This
is the plane, the normal to which is oriented
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at angle θ with x-axis. Since this is the
reference plane, from this particular plane
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we move in the anticlockwise direction as
that
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of the physical plane by an angle of 2 θ
; this is 2 θ then the point which we get
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on the
periphery of the circle represents this particular
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plane. The corresponding stresses
represent the normal stress and this represents
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the shearing stress. That is how we
compute stresses at any plane through Mohr’s
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circle.
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..
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Here is a question:
What is meant by state of pure shear? Now
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at a point in stress body as we have seen
that
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it is acted on by the normal stress and the
shearing stresses components, now in this
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particular body or the stress body if we do
not have any normal stress but we have only
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the shear stresses then we say that this particular
point of the stress body is subjected to
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pure state of shear.
Now if we look into Mohr’s diagram corresponding
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to this particular state of stress, we
have a stress body in which the state of stress
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is in the form shear alone, and this if we
try
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to represent in terms of the Mohr’s circle
then this is our reference axis σ and τ
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on this
particular plane where normal stress is 0,
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we have only τ positive so normal stresses
is 0
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τ is positive here, this is the point which
represents this particular plane.
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The perpendicular plane which is at an angle
of 90 degrees in the Mohr’s circle will
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be
180 where normal stresses is 0 and we have
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negative shear of equal amount of magnitude
so these are the two points on the Mohr’s
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circle. Since σ x and σ y both are 0 so
eventually this will be the centre of Mohr’s
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circle. If we draw a circle with centre as
this
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particular point and this radius from O to
A, we get the Mohr’s circle for this kind
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of
stress which is acting. Thereby this gives
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us the maximum value of normal stress which
is
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σ 1 and this is the minimum value of normal
stress which is σ 2 .
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Please note that here the values of σ 1 and
σ 2 , which is in the opposite direction
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is equals
to the shearing stress τ . So the maximum
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normal stress is τ and the minimum normal
stress also is τ , but it is the negative
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τ .
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.And if we look into the plane, now this is
the plane which is representing this particular
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face which is face A, so this is the plane
which is representing the face A and from
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here
the plane on which the maximum normal stress
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acts is at angle of 90 degrees, which is 2
θ P . So θ P in the physical plane, it will
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be 45 degrees.
With respect to this if we draw a normal which
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is at angle of 45 degrees, the plane on
which maximum normal stress acts is this.
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Maximum normal stress is equal to σ 1 , which
is equals to τ . The other normal stress
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σ 2 will be acting on the plane which is
perpendicular to this but in the opposite
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direction which is σ 2 , which is negative
τ .
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So, if we have a stress representation which
is like you have the stress τ here, the normal
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stress which is acting in this plane, which
is τ here. This also represents the state
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of pure
shear. Mind that this particular plane is
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at an angle of 45 degrees with reference to
the xplane. So either we represent that at
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a point in stress body, the stresses are purely
in the
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form of shear or in the form of normal stresses,
having the magnitude as that of a shear
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which is normal tensile in the maximum normal
stress direction and the compressive τ in
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the other normal direction. So this is state
of pure shear.
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.
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Now what is the co-ordinate of the centre
of Mohr’s circle and what is its radius?
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By this
time you have observed that the values of
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the radius and the position of the centre
in the
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Mohr’s circle which we have represented
now, this is σ and this is τ . The centre
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is
located at the distance of ( σ x plus σ
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y ) by 2. Based on these we have drawn the
Mohr’s
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circle. The plane which is representing is
having σ x and τ xy . There by this particular
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distance is ( σ x minus σ y ) by. So the
radius R is equals square root of ( σ x minus
211
00:24:54,040 --> 00:24:56,750
σ y ) by 2
212
00:24:56,750 --> 00:25:04,100
.square plus τ xy square. The centre is at
the distance of ( σ x plus σ y ) by 2 and
213
00:25:04,100 --> 00:25:06,179
the radius
of the circle is square root of ( σ x minus
214
00:25:06,179 --> 00:25:11,549
σ y ) by 2 square plus τ xy square.
.
215
00:25:11,549 --> 00:25:18,129
Now what happens if σ x plus σ y is equal
to 0 and σ 1 is equal to σ 2 ? This also
216
00:25:18,129 --> 00:25:27,790
we
observed that, the maximum, the Mohr’s plane
217
00:25:27,790 --> 00:25:33,970
again, this is σ and this is τ , and if
we
218
00:25:33,970 --> 00:25:49,980
draw the Mohr’s circle, the centre is ( σ
x plus σ y ) by 2 and this is σ 1 , this
219
00:25:49,980 --> 00:25:54,190
is σ 2 . If σ x plus
220
00:25:54,190 --> 00:26:01,190
σ y is equal to 0, this particular point
moves to this for the state of pure shear.
221
00:26:01,190 --> 00:26:07,140
If σ 1 is
equal to σ 2 , this reduces, means the circle
222
00:26:07,140 --> 00:26:10,320
reduces to point and there by there are no
shear
223
00:26:10,320 --> 00:26:17,649
in the particular plane.
224
00:26:17,649 --> 00:26:18,649
..
225
00:26:18,649 --> 00:26:29,679
Having looked into the aspects and how to
evaluate the stresses through equations of
226
00:26:29,679 --> 00:26:35,950
transformations and through the Mohr’s circle
let us look into some more aspects of
227
00:26:35,950 --> 00:26:46,019
stresses which is known as Octahedral stress.
To find the octahedral stress, in fact the
228
00:26:46,019 --> 00:26:54,279
octahedral stresses are the stresses which
act on the octahedral planes. We will have
229
00:26:54,279 --> 00:26:59,840
to
know what we really mean by octahedral planes.
230
00:26:59,840 --> 00:27:04,740
The octahedral plane is the plane which
is equally inclined to all the three principal
231
00:27:04,740 --> 00:27:08,370
axes of reference. This is generally termed
as
232
00:27:08,370 --> 00:27:17,470
octahedral plane.
Now, if we look in to the reference plane,
233
00:27:17,470 --> 00:27:23,210
it is said that the, normally we represent
the
234
00:27:23,210 --> 00:27:29,490
stresses in terms of rectangular stress component
which are σ x , σ y and τ xy with respect
235
00:27:29,490 --> 00:27:36,470
to
x, y and z axis.
236
00:27:36,470 --> 00:27:44,660
Instead of representing in x, y and z form
if we represent reference axis system which
237
00:27:44,660 --> 00:27:52,000
is
denoted by principal stresses σ 1 , σ 2
238
00:27:52,000 --> 00:28:04,809
and σ 3 then if we have plane which is equally
inclined to these three reference axis then
239
00:28:04,809 --> 00:28:11,960
we call this particular plane as octahedral
plane. And since this particular plane is
240
00:28:11,960 --> 00:28:16,090
equally inclined to these reference planes,
the
241
00:28:16,090 --> 00:28:23,480
direction cosines of this particular plane
on these planes, n x , n y , n z they are
242
00:28:23,480 --> 00:28:28,580
going to be
same.
243
00:28:28,580 --> 00:28:29,580
..
244
00:28:29,580 --> 00:28:37,639
So this is what is indicated here: For octahedral
planes
245
00:28:37,639 --> 00:28:46,750
which are equally inclined to the
reference principal axes, they have the direction
246
00:28:46,750 --> 00:28:54,580
cosine of n x , n y and n z all equal. We
have seen earlier that the (n x ) whole square
247
00:28:54,580 --> 00:28:56,420
plus (n y ) whole square plus (n z ) whole
square
248
00:28:56,420 --> 00:29:03,390
the direction cosines of three planes is equal
to 1. So there by it leads us to the value
249
00:29:03,390 --> 00:29:06,220
of n x
is equal to n y is equal to n z as plus or
250
00:29:06,220 --> 00:29:15,870
minus square root of 3 of the octahedral planes.
.
251
00:29:15,870 --> 00:29:21,190
So this is the representation of the octahedral
planes, in fact the planes which are equally
252
00:29:21,190 --> 00:29:30,320
inclined with reference to the principal axes
system, we get eight such planes. This
253
00:29:30,320 --> 00:29:42,490
represents σ 1 , this is represents σ 2
and this is represents σ 3 . Now we will
254
00:29:42,490 --> 00:29:43,490
get eight such
255
00:29:43,490 --> 00:29:50,779
.planes which are equally inclined to these
three reference axis and three reference planes.
256
00:29:50,779 --> 00:29:58,390
That is why these particular planes are called
octahedral planes. So the stresses which act,
257
00:29:58,390 --> 00:30:08,070
the normal stress, the shearing stresses which
act on these octahedral planes, we call
258
00:30:08,070 --> 00:30:12,100
those stresses as octahedral stresses.
.
259
00:30:12,100 --> 00:30:22,409
Now for the computation of the octahedral
stresses on the octahedral plane we use
260
00:30:22,409 --> 00:30:32,480
Cauchy’s stress formula which we have derived
already. Where in at any plane which is
261
00:30:32,480 --> 00:30:44,429
having outward normal n the resulting stress
in the x-direction, y-direction and in the
262
00:30:44,429 --> 00:30:52,200
zdirection, can be represented in terms of
the normal rectangular stress components:
263
00:30:52,200 --> 00:30:56,409
σ x ,
σ y , σ z and the corresponding shear stresses:
264
00:30:56,409 --> 00:31:09,179
τ xy , τ yx, and τ zx along with the direction
cosines n x , n y , and n z .
265
00:31:09,179 --> 00:31:13,480
Now we have already seen that the octahedral
planes are the planes which can equally
266
00:31:13,480 --> 00:31:23,600
inclined to the three principal reference
axes σ 1 , σ 2 ,and σ 3 . Thereby if we
267
00:31:23,600 --> 00:31:27,370
take the
stress in the σ 1 direction, we will have
268
00:31:27,370 --> 00:31:33,499
only σ 1 and shearing stresses will be absent;
because principal stress, principal plane
269
00:31:33,499 --> 00:31:37,110
does not have the shearing stresses.
270
00:31:37,110 --> 00:31:38,110
..
271
00:31:38,110 --> 00:31:55,320
So if we take those components as, σ x is
equal to σ 1 , σ y is σ 2 and σz as σ
272
00:31:55,320 --> 00:31:59,350
3 . The
shearing stress components on this are zero
273
00:31:59,350 --> 00:32:08,529
because those are the principal planes.
Now if we choose any plane having outward
274
00:32:08,529 --> 00:32:18,670
normal n on which the rectangular stress
components are σ 1 , σ 2 , and σ 3 then
275
00:32:18,670 --> 00:32:26,321
correspondingly the components of the resultant
stress could be R x , R y , and R z is equal
276
00:32:26,321 --> 00:32:31,950
to σ 1 n x from Cauchy‘s stress formula
σ 2 n y and
277
00:32:31,950 --> 00:32:43,440
σ 3 n z . So in the Cauchy‘s stress formula
we are replacing σ x by σ 1 , σ y by σ
278
00:32:43,440 --> 00:32:45,240
2 , and σz by
279
00:32:45,240 --> 00:32:51,671
σ 3 the shearing stress components as 0.
Hence R x is equal to σ 1 n x , R y is equal
280
00:32:51,671 --> 00:32:58,289
to σ 2 ny
and R z is equal to σ 3 nz.
281
00:32:58,289 --> 00:32:59,289
..
282
00:32:59,289 --> 00:33:06,059
So the resultant stress R square in terms
of σ 1 , σ 2 and σ 3 is equal to R x square
283
00:33:06,059 --> 00:33:12,429
plus R y
square plus R z square is equal to σ 1 square
284
00:33:12,429 --> 00:33:18,460
n x square plus σ 2 square n y square plus
σ 3 square n z square.
285
00:33:18,460 --> 00:33:25,720
Now if σ is the normal stress and τ is the
shearing stress on this particular plane,
286
00:33:25,720 --> 00:33:29,269
which
we are defining as octahedral plane, then
287
00:33:29,269 --> 00:33:33,049
the normal stress sigma can be represented
in
288
00:33:33,049 --> 00:33:39,919
terms of σ 1 , σ 2 and σ 3 as σ 1 n x
square, σ 2 n y square, and σ 3 square n
289
00:33:39,919 --> 00:33:45,080
z . These we
compute by writing the equations of equilibrium
290
00:33:45,080 --> 00:33:50,600
in the normal direction as we have done
in the previous cases.
291
00:33:50,600 --> 00:34:01,230
The τ square shearing stress can be represented
as R square minus σ square R square
292
00:34:01,230 --> 00:34:08,970
being this value substituted as σ square
if we take and substitute here and if we simplify
293
00:34:08,970 --> 00:34:14,890
we get an expression which is like this: n
x square n y square ( σ 1 minus σ 2 ) whole
294
00:34:14,890 --> 00:34:19,710
square
plus n y square n z square ( σ 2 minus σ
295
00:34:19,710 --> 00:34:28,190
3 ) whole square plus n z square n x square
( σ 3 minus σ 1 ) whole square. These two
296
00:34:28,190 --> 00:34:36,610
values σ and τ are the stress on any plane,
which
297
00:34:36,610 --> 00:34:43,460
has a normal n. Now for octahedral planes,
we have noted earlier that, n x , n y , n
298
00:34:43,460 --> 00:34:47,860
z are of
the equal magnitude, n x is equal to n y is
299
00:34:47,860 --> 00:34:53,170
equal to n z is equal to plus or minus 1 by
square
300
00:34:53,170 --> 00:34:57,440
root of 3.
Now if we substitute the values of n x , n
301
00:34:57,440 --> 00:35:03,340
y , and n z in this expression we can get
the normal
302
00:35:03,340 --> 00:35:11,910
stress on the octahedral plane which is equal
to ( σ 1 plus σ 2 plus σ 3 ) by 3. Likewise
303
00:35:11,910 --> 00:35:16,580
the
τ square is equal to 1 by 9 ( σ 1 minus
304
00:35:16,580 --> 00:35:19,770
σ 2 ) whole square plus 1 by 9 ( σ 2 minus
σ 3 ) whole
305
00:35:19,770 --> 00:35:28,270
square plus 1 by 9 ( σ 3 minus σ 1 ) whole
square.
306
00:35:28,270 --> 00:35:29,270
..
307
00:35:29,270 --> 00:35:36,320
The values of normal stress, σ octahedral
on the σ plane having n x , n y and n z as
308
00:35:36,320 --> 00:35:39,100
1 by
square root of 3, is equals to ( σ 1 plus
309
00:35:39,100 --> 00:35:45,890
σ 2 plus σ 3 ) by 3. If you remember we
have defined
310
00:35:45,890 --> 00:35:50,600
the values of σ 1 plus σ 2 plus σ 3 as
the sum of the normal stress component as
311
00:35:50,600 --> 00:35:54,570
equals to
the first invariant. So σ 1 plus σ 2 plus
312
00:35:54,570 --> 00:35:59,700
σ 3 is equal to I 1 the first invariant,
( τ octahedral)
313
00:35:59,700 --> 00:36:07,430
whole square is equal to [( σ 1 minus σ
2 ) whole square plus ( σ 2 minus σ 3 ) whole
314
00:36:07,430 --> 00:36:10,610
square
plus ( σ 3 minus σ 1 ) whole square by 9]
315
00:36:10,610 --> 00:36:17,590
or 9 into ( τ octahedral) whole square if
we expand
316
00:36:17,590 --> 00:36:23,130
these and simplify we get in this particular
form which is 2( σ 1 plus σ 2 plus σ 3
317
00:36:23,130 --> 00:36:28,740
) whole
square minus 6 ( σ 1 σ 2 plus σ 2 σ 3
318
00:36:28,740 --> 00:36:35,010
plus σ 3 , σ 1 ).
Now σ 1 plus σ 2 plus σ 3 we have noted
319
00:36:35,010 --> 00:36:45,640
as the first invariant I 1 . This particular
expression σ 1 σ 2 plus σ 2 σ 3 plus σ
320
00:36:45,640 --> 00:36:52,650
3 σ 1 is known as the second invariant I
2 . So we can
321
00:36:52,650 --> 00:36:58,300
write down the expression for τ octahedral
in terms of these two invariants I 1 and I
322
00:36:58,300 --> 00:37:03,430
2
which is square root of 2[I 1 square minus
323
00:37:03,430 --> 00:37:11,650
3I 2 ) to the power 1/2] by 3.
So we get the normal stress and the shearing
324
00:37:11,650 --> 00:37:17,990
stress on the octahedral plane. Now, if σ
1
325
00:37:17,990 --> 00:37:26,840
plus σ 2 plus σ 3 is equal to 0 for a particular
stress condition then on octahedral plane
326
00:37:26,840 --> 00:37:29,760
we
will not get any normal stress and thereby
327
00:37:29,760 --> 00:37:34,720
octahedral plane will be subjected to only
shearing stress and that is what is represented
328
00:37:34,720 --> 00:37:42,580
through this expression.
329
00:37:42,580 --> 00:37:43,580
..
330
00:37:43,580 --> 00:37:50,140
Having known these transformation equations,
the concept of the Mohr’s circle, and the
331
00:37:50,140 --> 00:37:57,570
concept of the octahedral stresses acting
on the octahedral planes let us look into
332
00:37:57,570 --> 00:38:02,810
some
examples on how we compute stresses on any
333
00:38:02,810 --> 00:38:10,080
plane if we know the stress at a particular
point on a body through the rectangular stress
334
00:38:10,080 --> 00:38:18,060
components like σ x , σ y and τ xy in two
dimensional plane.
335
00:38:18,060 --> 00:38:34,160
The problem is, the state of stress at a point
in a body is represented here where we have
336
00:38:34,160 --> 00:38:47,820
σ x as 20MPa, σ y as 30 MPa and the shearing
stress as 20MPa. Now what we will have
337
00:38:47,820 --> 00:38:55,530
to do is to find the normal and shear stresses
acting on plane aa using transformation
338
00:38:55,530 --> 00:39:02,190
equations. So we have to find out the stress
on this particular plane. This particular
339
00:39:02,190 --> 00:39:07,790
plane
is inclined to the vertical at an angle of
340
00:39:07,790 --> 00:39:19,690
30 degrees. So, if we take the normal to this
particular plane the normal makes an angle
341
00:39:19,690 --> 00:39:32,750
of 30 degrees with x-axis.
So, if we write down the transformation equations
342
00:39:32,750 --> 00:39:53,250
for the evaluation of the stresses the
2 θ is equal to 60 degrees. We have the stresses
343
00:39:53,250 --> 00:39:56,680
acting on a particular body at a point; σ
x
344
00:39:56,680 --> 00:40:22,130
as 20, σ y as 30, and τ xy as 20MPa. We
are interested to find stress at a particular
345
00:40:22,130 --> 00:40:34,810
plane
which is at an angle of 30 degrees with x-axis.
346
00:40:34,810 --> 00:40:54,140
We have σ x is equal to plus 20, σ y is
equal to plus 30, τ xy is equal to plus 20
347
00:40:54,140 --> 00:41:01,730
and 2 θ is equal to 60 degrees. The
transformation equation as we have observed
348
00:41:01,730 --> 00:41:12,010
is equal to σ x prime which is angle θ is
equal to 30 degrees is equal to ( σ x plus
349
00:41:12,010 --> 00:41:24,110
σ y ) by 2 plus ( σ x minus σ y ) by 2
cos 2 θ plus
350
00:41:24,110 --> 00:41:35,680
τ xy sin2 θ .
Now if we substitute the values of rectangular
351
00:41:35,680 --> 00:41:39,070
stress components then this is equal to ( σ
x
352
00:41:39,070 --> 00:41:48,080
plus σ y ) by 2 is equal to (20 plus 30)
by 2 that is 25; plus ( σ x minus σ y ) by
353
00:41:48,080 --> 00:41:50,000
2 is equal to
354
00:41:50,000 --> 00:42:06,830
.20 minus 30 by 2 is equal to (minus 10 by
2) cos 60 degrees; τ xy is equal to 20 sin
355
00:42:06,830 --> 00:42:20,030
60
degrees. So this is equals to 25 minus 2.5
356
00:42:20,030 --> 00:42:34,300
20 into square root of 3 by 2 so this gives
us the
357
00:42:34,300 --> 00:42:51,420
value of 22.5 plus 17.32 is equal to 39.82
MPa. This is the value of normal stress on
358
00:42:51,420 --> 00:42:54,390
this
particular plane which is inclined at an angle
359
00:42:54,390 --> 00:43:00,920
of 30 degrees with the vertical.
.
360
00:43:00,920 --> 00:43:11,920
Similarly, we can compute the value of shear
stress on that plane which is τ x prime y
361
00:43:11,920 --> 00:43:28,070
prime is equal to σ x minus σ y by 2sin
2 θ plus τ xy cos2 θ . This is equal to
362
00:43:28,070 --> 00:43:42,230
minus σ x is
equal
363
00:43:42,230 --> 00:43:56,920
to 20, σ y is equal to minus 30 by 2 sin
60 degrees plus τ xy is equal to 20 cos 60
364
00:43:56,920 --> 00:43:57,920
degrees.
365
00:43:57,920 --> 00:43:58,920
..
366
00:43:58,920 --> 00:44:07,590
So this equals to 5 into square root of 3
by 2 plus 20 into 1 by 2 is equal to 4.33
367
00:44:07,590 --> 00:44:14,670
plus 10
plus 14.33 MPa so the value of the normal
368
00:44:14,670 --> 00:44:23,430
stress, σ x prime is equal to 39.82 MPa and
the
369
00:44:23,430 --> 00:44:34,670
value of shear stress is equals to 14.33 MPa.
So this is the solution of this particular
370
00:44:34,670 --> 00:44:53,550
problem where we are computing the normal
stress and the shearing stresses.
371
00:44:53,550 --> 00:44:54,550
.
372
00:44:54,550 --> 00:45:00,800
Let us look into this example 2 where we will
have to solve this particular problem or
373
00:45:00,800 --> 00:45:08,920
rather we have to solve this which is represented
in 1 using Mohr’s circle of stress. In the
374
00:45:08,920 --> 00:45:15,960
previous problem we used the transformation
equations for the solution of the stresses
375
00:45:15,960 --> 00:45:16,960
at
376
00:45:16,960 --> 00:45:22,280
.any plane. Now we are going to evaluate the
stresses on any plane using the Mohr’s
377
00:45:22,280 --> 00:45:34,040
circle of the stress.
If we look into the state of stress which
378
00:45:34,040 --> 00:45:40,500
we had we represent that as σ x is equal
to 20
379
00:45:40,500 --> 00:45:55,600
MPa, σ y is equal to 30 MPa and τ xy is
equal to 20MPa. As we had represented in a
380
00:45:55,600 --> 00:46:05,130
Mohr’s plane, here this is σ axis, this
is τ axis.
381
00:46:05,130 --> 00:46:14,390
First of all if we locate the plane, where
the normal stress is 20 and the shearing stress
382
00:46:14,390 --> 00:46:22,440
is
20 then we go in this direction as 20 and
383
00:46:22,440 --> 00:46:29,600
the positive shear stress as 20. Now this
particular shear stress is positive because
384
00:46:29,600 --> 00:46:36,590
the shear stress on this plane and the
complimentary shear they are making a rotation
385
00:46:36,590 --> 00:46:49,830
which is anticlockwise in nature which
we call as positive, so this is 20 and this
386
00:46:49,830 --> 00:46:58,360
is 20.
Let us represent this plane which is 90 degrees
387
00:46:58,360 --> 00:47:05,520
with this x plane. Eventually in Mohr’s
circle it is 180 degrees so we represent σ
388
00:47:05,520 --> 00:47:13,790
y and the τ as negative, τ is negative because
the shearing stress on this plane is causing
389
00:47:13,790 --> 00:47:19,830
a rotation which is clockwise in nature so
this
390
00:47:19,830 --> 00:47:41,770
is negative τ , this is σ y and this is
τ xy . If we join these two points, this
391
00:47:41,770 --> 00:47:46,300
particular point is
eventually the centre of the Mohr’s circle
392
00:47:46,300 --> 00:48:14,930
which is equal to ( σ x plus σ y ) by 2.
Now if we plot Mohr’s circle, this is the
393
00:48:14,930 --> 00:48:18,150
point representing x-plane and this is the
point
394
00:48:18,150 --> 00:48:26,480
which is representing y-plane. This is the
maximum normal stress which is σ 1 and this
395
00:48:26,480 --> 00:48:30,970
is
the minimum normal stress which is σ 2 . Now
396
00:48:30,970 --> 00:48:35,430
we are interested to evaluate the stress at
a
397
00:48:35,430 --> 00:48:41,700
plane, which is making an angle of 30 degrees
with x-plane.
398
00:48:41,700 --> 00:48:46,680
Eventually that will make 60 degrees in the
Mohr’s circle with respect to this. So with
399
00:48:46,680 --> 00:48:53,600
reference to this line, say line OA if we
represent that plane which is represented
400
00:48:53,600 --> 00:49:00,560
by this
point say C, this particular angle is 60 degrees.
401
00:49:00,560 --> 00:49:12,110
Here we know this magnitude, we can
compute this magnitude. This is σ x OO prime
402
00:49:12,110 --> 00:49:16,400
is equal to ( σ x plus σ y ) by 2, which
is
403
00:49:16,400 --> 00:49:20,810
equals to (20 plus 30) by 2, which is equal
to 25 MPa.
404
00:49:20,810 --> 00:49:32,860
Let us say this is A prime, which is A. OA
prime eventually is equal to ( σ x plus σ
405
00:49:32,860 --> 00:49:40,760
y ) by 2
minus σ x , which is σ x is 20 here. Eventually
406
00:49:40,760 --> 00:49:54,640
OA prime is equal to 5. And AA prime is
equal to 20. So the radius OA is equal to
407
00:49:54,640 --> 00:50:02,040
square root of 20 square plus 5 square is
equal
408
00:50:02,040 --> 00:50:21,320
to 20.62 MPa. Now we are interested to find
out stress at this particular plane, if we
409
00:50:21,320 --> 00:50:26,820
drop
a perpendicular to the σ axis, this will
410
00:50:26,820 --> 00:50:35,850
gives us the value of σ x on this plane,
that is the
411
00:50:35,850 --> 00:50:42,500
value of normal stress, let us call as σ
x prime, if we call this as x prime plane,
412
00:50:42,500 --> 00:50:47,350
and this
magnitude will gives us the magnitude τ,x
413
00:50:47,350 --> 00:50:56,330
prime y prime.
Now if we represent this angle by θ , then
414
00:50:56,330 --> 00:51:03,960
θ is equals to tan power minus 1 AA prime
by
415
00:51:03,960 --> 00:51:15,140
OA prime and AA prime being 20 and OA prime
being 5, so this is tan power minus 1 4
416
00:51:15,140 --> 00:51:27,960
≈ 76 degrees. So
if we represent this angle by α , then α
417
00:51:27,960 --> 00:51:37,020
is equal to 180 minus 76 minus
418
00:51:37,020 --> 00:51:57,680
.60 is equal to 44 degrees. OC being the radius,
then normal stress σ x prime is equal to
419
00:51:57,680 --> 00:52:12,900
the distance OO prime plus radius cos α .
So the σ x prime is equal to the central
420
00:52:12,900 --> 00:52:19,530
distance; σ x is equal to ( σ x plus σ
y ) by 2 plus R
421
00:52:19,530 --> 00:52:36,750
cos α . R is equal to square root of [( σ
x minus σ y ) by 2] whole square plus τ
422
00:52:36,750 --> 00:52:52,300
xy square
20.62, as we have seen. So σ x prime is equal
423
00:52:52,300 --> 00:53:01,710
to ( σ x plus σ y ) by 2 is equal to 25
plus
424
00:53:01,710 --> 00:53:20,670
20.62 2 into cos α , cos 44 degrees is 0.72.
This is going to give us a value of 39.85
425
00:53:20,670 --> 00:53:27,270
MPa,
which is similar to which we have computed
426
00:53:27,270 --> 00:53:36,740
through the transformation equation. The
shearing stress on this plane, τ x prime
427
00:53:36,740 --> 00:53:42,480
y prime from the Mohr’s circle, if you look
into
428
00:53:42,480 --> 00:54:07,120
τ x prime y prime is equal to R sin α is
equal to 20.62 into 0.7 is equal to 14.43
429
00:54:07,120 --> 00:54:14,670
MPa.
This is similar to the one which we have computed
430
00:54:14,670 --> 00:54:18,310
using the transformation equation.
.
431
00:54:18,310 --> 00:54:25,460
From this you can compute the value of normal
stress represented through this point
432
00:54:25,460 --> 00:54:31,760
which is equal to ( σ x plus σ y ) by 2
is equal to R. Now ( σ x plus σ y ) by 2
433
00:54:31,760 --> 00:54:42,140
is equal to 25,
and R is 20.62, that will give the value as
434
00:54:42,140 --> 00:54:49,380
45.62. σ 1 is equal to 45.62. σ 2 is equal
to
435
00:54:49,380 --> 00:54:59,870
( σ x plus σ y ) by 2, this distance minus
the radius which is 25 minus 20.62 which is
436
00:54:59,870 --> 00:55:05,760
equal
to 4.38. We get the values of maximum and
437
00:55:05,760 --> 00:55:12,250
minimum normal stress or the maximum and
the minimum principal stresses. The value
438
00:55:12,250 --> 00:55:18,210
of the radius gives the maximum value of the
positive and negative shear, which is equals
439
00:55:18,210 --> 00:55:23,000
to 20.62 MPa.
So from the Mohr’s circle we can compute
440
00:55:23,000 --> 00:55:30,570
the stress at any plane and the maximum
principal stress and the minimum shear stress
441
00:55:30,570 --> 00:55:37,540
without even using the transformation
equations and without going for the geometrical
442
00:55:37,540 --> 00:55:52,710
construction of the Mohr’s circle.
443
00:55:52,710 --> 00:55:53,710
..
444
00:55:53,710 --> 00:55:56,960
The state of a stress at a particular point
is shown in the figure. Determine the normal
445
00:55:56,960 --> 00:55:58,950
and
the shear stresses acting on the vertical
446
00:55:58,950 --> 00:56:04,250
plane using transformation equations.
.
447
00:56:04,250 --> 00:56:09,320
This is the state of a point in a stress at
a body, you have to evaluate the principal
448
00:56:09,320 --> 00:56:12,700
stress
and the maximum shear stresses and the associated
449
00:56:12,700 --> 00:56:14,850
normal stress using the Mohr’s circle.
450
00:56:14,850 --> 00:56:15,850
..
451
00:56:15,850 --> 00:56:20,330
These are the values of the principal stresses
at a particular point; you have to evaluate
452
00:56:20,330 --> 00:56:23,840
the octahedral stresses.
.
453
00:56:23,840 --> 00:56:29,850
So in this particular lesson, what we have
done is we have recapitulated the aspects
454
00:56:29,850 --> 00:56:35,140
which
we have discussed in the particular last lesson,
455
00:56:35,140 --> 00:56:40,010
and we have evaluated the octahedral
stresses and also solved some examples to
456
00:56:40,010 --> 00:56:42,190
demonstrate the evaluation of stresses on
any
457
00:56:42,190 --> 00:56:43,260
plane.
458
00:56:43,260 --> 00:56:44,260
..
459
00:56:44,260 --> 00:56:49,900
Here are some more questions:
What happens to octahedral stresses when first
460
00:56:49,900 --> 00:56:53,830
invariant is zero?
What is the value of the shear stress where
461
00:56:53,830 --> 00:59:11,320
maximum normal stress occurs? What is the
value of normal stress where maximum shear
462
00:59:11,320 --> 00:59:34,569
stress occurs?
463
00:59:34,569 --> 00:59:34,569
.