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Welcome to lesson 3 on the course Strength
of Materials. We are going to discuss about
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the
analysis of the stress. We have already looked
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into some aspects of stress analysis.
.
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.Now it is expected that once this particular
lesson is completed one will able to compute
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stress
resultant and stresses in members subjected
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to axial forces, evaluate stresses at a point
on a body
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at any arbitrary plane, evaluate principal
stresses and locate principal planes and also
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compute
stress invariants.
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.
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Hence the scope of this particular lesson
includes: review of normal stress, concept
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of shear and
bearing stress, computation of stress on any
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arbitrary plane, concept of principal stress
and
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principal plane and concept of stress invariants.
We discussed the types of stress and specifically
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about the normal stress.
.
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.And we had noticed that if we take a body
which is axially loaded by a force P and if
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we take a
section and draw the free body diagram, this
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body is under the action of external force
P so at the
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chord section say “aa” there will be resulting
stress component which we call as stress resultant
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or the force which is resisting P. At every
point there will be a stress component and
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the normal
stress multiplied by the area will give the
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force. So integral of stress multiplied by
the small area
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integrated over this surface will give the
stress resultant which is equal to P.
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While making this kind of assumption that
every where state of the stress exists if
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the force acts
through the centriod of the section we assume
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that the particle of the material at every
point
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contributes to the resistance of this external
force and thereby we assume the homogeneity
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of this
material.
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We assume that at every point the same state
of stress exists. When a body is subjected
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to
external forces which are trying to cause
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traction in the member or trying to pull the
member we
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call these kinds of forces and thereby stresses
as tensile stresses whereas when the external
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forces are acting on the member trying to
push the member we call this kind of external
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forces
and the stresses as compressive stresses.
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.
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Let us look into the aspects of shear stresses.
Let us assume that this particular body is
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subjected
to the action of external forces P and the
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resistive forces thereby will be P by 2 and
P by 2.
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Now if we take the free body diagram of this
particular member if we cut it over here then
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we
will have this body which is acted upon by
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external force P by 2 hence the resistive
internal force
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will also be P by 2. If we take the free body
of the other part of the cut the resistive
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internal force
is P by 2 and this force will be resisted
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at this inter phase and this also will be
P by 2. Though,
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this force is eccentric with respect to the
centre line of this body but this thickness
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being smaller
we neglect the other effects because of these
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forces. This force will try to cause stress
at the
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contact between these two elements and this
contact area is so near which is the plan.
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When we
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.look from the top the top view of this body
looks like this and this shaded part indicates
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the
contact area between these two pieces of material.
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Now if we consider that this particular length
as “a” and this as “b” then you can
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define the shear
stress which is designated as tau equals to
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the force which is acting at the inter phase
which is P
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by 2 by a(b). So a(b) is the contact area
over which the shearing stress exits.
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.
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Many a times we come across situations where
some blocks are resting over another block
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and
transferring the forces from external sources.
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For example, if we have a body like this in
which,
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this is a block which is resting on a bigger
block and the smaller block is subjected to
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external
load P. If this block is placed, that is,
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the centroid of the top block is placed on
the centroid of the
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bottom block then at the interface between
these two blocks and at this inter phase there
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will be
normal stresses generated and these normal
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stresses we designate as bearing stress. By
the term
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bearing we mean that the bottom block is bearing
the load of the top block. So, if the contact
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area
between the smaller block and the bigger block,
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this equals to Ac then the bearing stress
sigma
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bearing can be written as the external force
P by Ac which is the contact area. So this
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is called as
bearing stress.
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Here is another example which is the concept
of the bearing stress. We have a rigid bar
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resting
on two supports
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and this is subjected to external force P.
Now if we take the free body diagram
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of this particular bar, the external force
is P so the resistive force at the support
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point, assuming
this P is acting to the centroid of this body
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is P by 2 and P by 2. This reactive force
will be in
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turn transmitted to the support. This is the
support and resistive force which is getting
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transmitted
on and depending on the contact area we have
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if this is length “a” and the width of
this body is
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“b” then this force P by 2 has contact
area which is a(b). Hence the bearing stress
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that is sigma
bearing is equal to P by 2 as the reacting
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force which is getting transmitted on this
support
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divided by a(b).
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..
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Here are the different stress components that
act on any
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arbitrary plane. Let us consider that A,
B, C are any arbitrary plane and o, x, y,
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and z is the reference axis system. As we
have noticed
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earlier the plane normal to which it coincides
with the axis we designate that in the name
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of the
particular axis. Likewise this particular
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plane is the x plane on which the normal stress
sigma x
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acts.
Likewise the plane oBC is normal to that coincides
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with y direction which is y-plane and the
normal stress sigma y on this plane. And the
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normal stress on the plane Aoc is the z-plane
is
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sigma z . Also, in those planes there are
shearing stresses x-plane in the y-direction
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will have the
stress which we call as tau xy , which we
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call as tau to the power XZ; on the x-plane,
in the ydirection, so tauxy. The component
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which is in the z-direction is designated
as tau to the power
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XZ. Likewise shear stress component in the
y-plane acting in the x-direction we call
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that as tau yx
and tau yz . Similarly this is tau zx and
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tau zy .
Let us assume that this arbitrary plane has
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a normal which is outward drawn normal is
n. This
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unit vector can be designated with reference
to xyz-plane. Let us assume that this is the
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reference
axis xyz. The unit normal is drawn here. If
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this vector makes an angle of α with x-axis,
beta with
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y-axis, and gamma with z-axis then we define
the cosine components in the x-direction as
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n x
which is cosine α; n y is the cosine beta;
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and n z as cosine gamma. Thereby the unit
vector this
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distance here can be represented by (n x ) whole
square plus (n z ) whole square; (n z ) whole
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square
plus (n y ) whole square will give this unit.
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So we have in effect (n x ) whole square plus
(n y ) whole
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square plus (n z ) whole square is equal to
1.
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Let us assume that on this arbitrary plane
we have the resulting stress vector as R and
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the
component of this resultant stress on this
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plane in the x, y, and z-direction be R x
, R y , and R z .
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Also let us assume that
the area of the arbitrary plane is dA which
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is the area of the plane ABC.
Now, if we take the projection of this area
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on x-plane which is AOB; so area of AOB is
dA into
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n x which is cosine of this area ABC on AOB.
Area BOC is the projection on the area ABC
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on y
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.plane which is dA into n y ; and area AOC
is the projection of ABC on the z-plane which
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is dA
into n z .
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.
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Now if
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we take the summation of all forces in the
x-direction where in the stress components
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involved will be sigmatau y in the x-direction,
and tauz x in the x-direction and R x then
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we can
write down the equilibrium equation in x-the
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direction. So equilibrium equation in the
xdirection will be R x into area dA; which
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is a force, stress resultant multiplied by
area minus
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sigma x dA n x the area of x-plane minus tau
yx which is acting in the y-plane times the
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area of yplane dA n y minus tau zx which is
the z-plane acting in x-direction multiplied
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by the area dA n z is
equal to 0. So the equation is as follows:
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R x dA minus sigma x dA n x minus tau yx dA
n y minus
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tau zx dA n z is equal to 0.
If we divide the whole equation by dA or in
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a limiting situation we get R x is equal to
sigma x n x
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plus tau yx n y plus tau zx n z . Similarly,
if we take the equations in the y and z-directions
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and write
down the equilibrium we will get R y is equal
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to tau xy n x plus sigma y n y plus tau zy
n z R y. And R z
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the resulting stress in the z-direction is
equal to tau xz (plane in the z-direction
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as) n x plus tau yz
(the y-plane in the z-direction) n y plus
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sigma z n z . That is R z is equal to tau
xz n x plus tau yz n y plus
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sigma z n z. These are the three resulting
stress components. So the stress components
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on the
arbitrary plane which are acting in the x,
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y and z-directions are represented in terms
of the stress
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components in the rectangular co-ordinate
system. This set of equations is normally
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designated
as Cauchy’s stress formula.
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..
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So Cauchy’s stress formula is R x is equal
to sigma x n x plus tau yx n y plus tau zx
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n z ; R y is equal to
tau xy n x plus sigma y n y plus tau zy n
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z and Rz is equal to tau xz n x plus tau yz
n y plus sigma z n z .
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.
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Now let us look into, if we consider a plane
which has normal n and the direction cosines
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for the
normal are n x , n y and n z . Also, we
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assume that on this particular plane only
the normal stress
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acts in the direction of the normal to the
plane. Hence if we take the components of
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this in the x,
y, and z direction then as we have designated
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before R x as the resulting stress in the
x-direction;
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R y as the resulting stress in the y-direction
and R z as the resulting stress in the z-direction
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they
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.can be written in the direction cosines as:
R x as sigma n n x ; Ry as sigma n n y ; and
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R z as sigma n
nz.
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.
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Exactly in the same form the way we have evaluated
Cauchy’s stress formula taking the
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equilibrium equations in the x, y and z-direction
we can compute the resulting forces in x,
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y, and
z-direction in terms of sigma n .
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In the previous case we had forces in the
x direction as R x dA is equal to sigma x
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dA n x plus tau yx
dA n y plus tau zx dA n z and by dividing
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the whole equation by dA we had R x is equal
to sigma x
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n x plus tau yx n y plus tau zx n z . Exactly
in the same form in place of R x now we have
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sigma n n x
the resulting stress in the x-direction and
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this multiplied by area gives the force in
the xdirection, this equals to sigma x dA
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n x plus tau yx dA n y plus tau zx dA n z
. Hence from this we can
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00:24:26,730 --> 00:24:43,400
write sigma n n x is equal to sigma x n x
plus tau yx n y plus tau zx n z .
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..
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This is what is represented here the equations
of equilibrium in the three directions x,
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y, and z.
Now these equations can be rearranged and
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can be written as (sigma x minus sigma n)
n x plus
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tau yx n y plus tau zx n z is equal to 0;
tau xy n x plus (sigma y minus sigma n ) n
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y plus tau zy n z is equal
to 0; tau xz n x plus tau yz n y plus (sigma
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00:25:54,770 --> 00:26:05,500
z minus sigma n) n z is equal to 0.
We have already seen that tau yx is equal
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to tau xy; tau zx is equal to tau xz al; tau
zy is equal to tau yz .
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Now these three equations can be thought of
as simultaneous equations containing n x , n
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00:26:21,309 --> 00:26:26,150
y and n z
and we can evaluate the values of n x , n
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y and n z . Now, if we expand this particular
equation we
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will get a cubical equation in sigma n .
.
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00:26:38,820 --> 00:26:57,850
.This is the cubical equation in sigma n : (sigma
n ) whole cube minus l 1 (sigma n ) whole
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square plus
l 2 sigma n minus l 3 is equal to 0. And once
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we solve this cubical equation we are expected
to get
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three roots which eventually will turn out
be real and we designate those roots as sigma
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00:27:18,600 --> 00:27:23,471
1 , sigma 2
and sigma 3 . And corresponding to each values
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of sigma we will get the values of n x , n
y and n z .
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00:27:51,290 --> 00:27:52,290
.
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So here we have the equations on the arbitrary
plane on which the stress in absolutely normal
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00:28:00,400 --> 00:28:12,289
and we have observed that we can get simultaneous
equations n x , n y and n z in terms of sigma
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00:28:12,289 --> 00:28:16,880
x
tau xy , tau yx and tau yz; and sigma x , sigma
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00:28:16,880 --> 00:28:20,130
y and sigma z .
.
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00:28:20,130 --> 00:28:38,919
.This is the cubical equation in the terms
of I 1 , I 2 , and I 3 and for each of these
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sigma values if we
are interested to compute the values of n
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x , n y , and n z we can do that using these
equations along
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with the expression we have n x square plus
n y square plus n z square is equal to 1.
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So, for a trivial
solution n x , n y , n z is equal to 0 which
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00:29:01,480 --> 00:29:04,289
is not really going to give us the solution
because n x
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00:29:04,289 --> 00:29:10,350
square plus n y square plus n z square is
equal to 1 now for a trivial solution of the
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00:29:10,350 --> 00:29:16,090
simultaneous
solution we said that the determinant of the
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coefficients of n x , n y , and n z is equal
to 0. And if we
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do that, we have (sigma x
minus sigma n ) tau xy tau xz tau xy (sigma
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00:29:47,240 --> 00:30:01,670
y minus sigma n ) tau yz tau xz tau yz
(sigma z - sigma n ) is equal to 0.
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00:30:01,670 --> 00:30:10,130
So if we say that the determinant of this
equals to zero for trivial solution we get
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00:30:10,130 --> 00:30:18,289
the cubical
equation in sigma n cube minus I 1 sigma n
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square plus I 2 sigma n minus I 3 where the
values of I 1 is
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00:30:30,419 --> 00:30:46,210
equal to sigmax plus sigma y plus sigma z
; I 2 is equal to determinant sigmax sigma
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00:30:46,210 --> 00:31:02,549
y tau xy plus
sigma y sigma z tau yz plus sigma z sigma
211
00:31:02,549 --> 00:31:19,010
x tau xz ; I 3 is equal to sigma x sigma y
sigma z plus tau xy tau xz
212
00:31:19,010 --> 00:31:32,800
tau yz and this is symmetrical tau xy , tau
xz , tau yz these are the three coefficients
213
00:31:32,800 --> 00:31:38,010
I 1 , I 2 , and I 3 .
.
214
00:31:38,010 --> 00:31:45,500
.Now the plane on which this stress vector
is fully normal is known as principal plane.
215
00:31:45,500 --> 00:31:50,350
This is
what we considered in the previous situation
216
00:31:50,350 --> 00:31:53,039
where we had taken the arbitrary plane and
we
217
00:31:53,039 --> 00:32:00,510
considered the stress vector which is along
the normal and thereby we obtained the cubical
218
00:32:00,510 --> 00:32:07,390
equations in sigma n and from which we obtained
the three roots sigma sigma 1 , sigma 2 and
219
00:32:07,390 --> 00:32:10,090
sigma 3 .
In this particular plane the normal stress
220
00:32:10,090 --> 00:32:13,610
is acting along the normal of the plane. So
the plane on
221
00:32:13,610 --> 00:32:25,380
which this stress vector is wholly normal
is called as principal plane. The stress on
222
00:32:25,380 --> 00:32:28,840
this principal
plane which is absolutely normal is called
223
00:32:28,840 --> 00:32:34,110
as principal stress. And since the stress
acting is in the
224
00:32:34,110 --> 00:32:43,929
normal direction there are no tangential stresses,
the principal stress is the resultant normal
225
00:32:43,929 --> 00:32:48,740
stress
so on a principal plane there are no tangential
226
00:32:48,740 --> 00:32:59,880
or shearing stresses, they are 0.
227
00:32:59,880 --> 00:33:00,880
.
228
00:33:00,880 --> 00:33:06,690
.Now we have seen the cubical equation in
sigma n from which we have obtained three
229
00:33:06,690 --> 00:33:11,890
roots
sigma 1 , sigma 2 and sigma 3 . And we have
230
00:33:11,890 --> 00:33:16,360
looked into three coefficients I 1 , I 2 and
I 3 I 1 is equal to
231
00:33:16,360 --> 00:33:22,330
sigma x plus sigma y plus sigma z ; I 2 if
we expand determinants I 2 is equal to sigma
232
00:33:22,330 --> 00:33:24,049
x sigma y
(tau xy ) whole square plus sigma y sigma
233
00:33:24,049 --> 00:33:25,049
z (tau yz ) whole square plus sigma z sigma
x (tau zx ) whole
234
00:33:25,049 --> 00:33:27,850
square; and I 3 is equal to sigma x sigma
y sigma z plus 2 tau xy tau yz tau zx minus
235
00:33:27,850 --> 00:33:37,309
sigma x tau yz square
minus sigma y tau xz square minus sigma z
236
00:33:37,309 --> 00:33:43,799
taux y square. We call these as invariants.
It is to be
237
00:33:43,799 --> 00:33:49,710
noted that the principal stress which we have
calculated at a particular point remains the
238
00:33:49,710 --> 00:33:53,429
same
irrespective of the reference axis system
239
00:33:53,429 --> 00:33:56,539
we take.
In this particular case we have taken x, y
240
00:33:56,539 --> 00:34:00,159
and z as the rectangular axis system. Supposing
at that
241
00:34:00,159 --> 00:34:06,210
particular point if we take different axis
system which is represented as x prime, y
242
00:34:06,210 --> 00:34:12,190
prime and z
prime if we write down corresponding normal
243
00:34:12,190 --> 00:34:15,470
stresses and shearing stresses as sigma x
prime,
244
00:34:15,470 --> 00:34:25,720
sigma y prime and sigma z prime then we can
observe that the values of I 1 , I 2 and I
245
00:34:25,720 --> 00:34:28,970
3 which are
represented in terms of the normal stresses
246
00:34:28,970 --> 00:34:33,059
and shearing stresses they remain the same
because
247
00:34:33,059 --> 00:34:38,440
the principal stresses at that particular
point remains unchanged.
248
00:34:38,440 --> 00:34:39,440
.
249
00:34:39,440 --> 00:34:57,129
.That is why these coefficients I 1 , I 2
and I 3 are known as stress invariants. Let
250
00:34:57,129 --> 00:35:10,880
us see how to solve
these stresses in some physical problems?
251
00:35:10,880 --> 00:35:17,280
In this particular example here there are
two plates
252
00:35:17,280 --> 00:35:33,150
which are connected together by a bolt and
it is subjected to a pull - external force
253
00:35:33,150 --> 00:35:45,569
P. This is the
plan of the two plates and if we take a section
254
00:35:45,569 --> 00:35:49,700
we cut the plate here and cut the plate here
and
255
00:35:49,700 --> 00:35:53,030
view from this side then the section looks
like this.
256
00:35:53,030 --> 00:36:00,420
Therefore the width of the plate is 200 mm
and the thickness of the plate is given as
257
00:36:00,420 --> 00:36:06,839
10 mm and
the tensile pull that the plate is subjected
258
00:36:06,839 --> 00:36:15,289
to is 50kN. Now we will have to compute average
normal stress at a section where there are
259
00:36:15,289 --> 00:36:25,089
no holes for the bolts. So, if we cut the
section here and
260
00:36:25,089 --> 00:36:37,950
draw the free body diagram.
Here we have the plate which is put in a three
261
00:36:37,950 --> 00:36:51,119
dimensional form and
this is acted on by a load 50
262
00:36:51,119 --> 00:37:02,259
kN, the width of the plate is given as 200
mm and the thickness of the plate is given
263
00:37:02,259 --> 00:37:08,459
as 10 mm.
So, at this particular section the resistive
264
00:37:08,459 --> 00:37:12,339
force for this external force 50kN which is
acting
265
00:37:12,339 --> 00:37:22,660
through the centroid of this plate is R and
hence from the equilibrium of these forces
266
00:37:22,660 --> 00:37:26,380
R is equal
to 50 kN.
267
00:37:26,380 --> 00:37:36,079
Hence the normal stress sigma at this particular
cross section where there are no holes is
268
00:37:36,079 --> 00:37:40,890
equals
to R divided by the cross sectional area sigma
269
00:37:40,890 --> 00:37:47,661
is equal to R by A is equal to 50 into 10
cube N by
270
00:37:47,661 --> 00:38:06,650
200 into 10. This gives us the normal stress
as 25 N by mm square and 1000 mm is equal
271
00:38:06,650 --> 00:38:11,680
to 1
meter so this is equal to 25 into (10) whole
272
00:38:11,680 --> 00:38:19,430
power 6 N by m square. Since we designate
N by m
273
00:38:19,430 --> 00:38:32,319
square as P so it is 25 into (10) whole power
6 P and since (10) whole power 6 P gives the
274
00:38:32,319 --> 00:38:41,819
mPa,
hence we are going to write this as 25 mPa.
275
00:38:41,819 --> 00:38:50,250
Also, we will have to compute the average
shear
276
00:38:50,250 --> 00:38:59,609
stresses in the bolts.
The force which is acting here gets transmitted
277
00:38:59,609 --> 00:39:03,969
from this particular plate to this plate through
this
278
00:39:03,969 --> 00:39:12,210
connection where these two plates are connected
by the bolts. And as a result if we draw free
279
00:39:12,210 --> 00:39:21,450
body at this interface the force P will be
get transmitted at this interface and the
280
00:39:21,450 --> 00:39:25,089
two bolts which
are connecting these two plates together will
281
00:39:25,089 --> 00:39:36,349
be subjected to this force P. So we have one
bolt
282
00:39:36,349 --> 00:39:48,599
.here and another bolt here of diameter 20
mm and these two bolts will be resisting the
283
00:39:48,599 --> 00:40:02,289
force P by
2 and P by 2.
284
00:40:02,289 --> 00:40:09,729
The plate is subjected to load P and this
P is transmitted through the top plate through
285
00:40:09,729 --> 00:40:16,490
the
interconnecting bolt. So the force which will
286
00:40:16,490 --> 00:40:22,789
be resisted by these bolts is half of P, hence
the
287
00:40:22,789 --> 00:40:30,499
stress which is acting in the bolt which is
the force on this particular area is called
288
00:40:30,499 --> 00:40:37,039
as the shearing
stress on the plane of the bolt. So the average
289
00:40:37,039 --> 00:40:48,249
shear stress in each bolt which is tau average,
tau is
290
00:40:48,249 --> 00:41:10,719
equal to P by 2 by pi by 4 (d) whole square
(divided by the area of the bolt) and P
291
00:41:10,719 --> 00:41:24,969
is equal to 50;
25 into 10 cube N by pi by 4 into 400 is equal
292
00:41:24,969 --> 00:41:32,519
to 250 by pi N by m square which is so much
of
293
00:41:32,519 --> 00:41:38,420
mPa. So this is the average stress in the
bolt.
294
00:41:38,420 --> 00:41:53,140
Thirdly, we will have to compute the bearing
stress between the bolts and the plates. When
295
00:41:53,140 --> 00:41:56,519
the
force is getting transmitted from one plate
296
00:41:56,519 --> 00:41:59,559
to another it is getting transmitted through
these
297
00:41:59,559 --> 00:42:04,089
interconnecting bolts.
.
298
00:42:04,089 --> 00:42:12,430
And if we look into the transmission of the
force in little greater detail we will see
299
00:42:12,430 --> 00:42:21,400
that this is the
plate, here two bolts are present and this
300
00:42:21,400 --> 00:42:26,420
plate is being pulled by force P. Now the
transfer of
301
00:42:26,420 --> 00:42:37,999
force from the plate to the bolts which are
here when plate is being pulled, this part
302
00:42:37,999 --> 00:42:47,940
of the plate,
this is the bolt, the hub part of this particular
303
00:42:47,940 --> 00:42:52,930
plate comes in contact with this particular
bolt and
304
00:42:52,930 --> 00:43:01,719
there may be release in the contact between
the plate and the bolt surface.
305
00:43:01,719 --> 00:43:10,680
Basically the plate is resting on this surface
of the bolt. On an average sense we take the
306
00:43:10,680 --> 00:43:19,420
projection of this surface which is the diameter
of this bolt and the contact area is here
307
00:43:19,420 --> 00:43:23,130
which is
diameter times the thickness of the plate.
308
00:43:23,130 --> 00:43:29,789
So the contact surface which we get is half
the
309
00:43:29,789 --> 00:43:39,619
perimeter the projection of which is d and
the thickness of the plate at that particular
310
00:43:39,619 --> 00:43:46,559
interface.
This is the area which is in contact with
311
00:43:46,559 --> 00:43:52,380
the plate and the bolt.
The bearing stress is the function of the
312
00:43:52,380 --> 00:44:00,390
contact area so the sigma bg bearing is equal
to the force
313
00:44:00,390 --> 00:44:11,829
P and since we have two bolts each bolt will
get half the forces so P by 2 divided by the
314
00:44:11,829 --> 00:44:12,829
contact
315
00:44:12,829 --> 00:44:29,849
.area which is dt; sigma bg is equal to P
by 2 by dt is equal to 25 into 10 cube N by
316
00:44:29,849 --> 00:44:35,009
20 into 10 is
equal to 125 MPa.
317
00:44:35,009 --> 00:44:36,009
.
318
00:44:36,009 --> 00:44:47,680
Here is
319
00:44:47,680 --> 00:44:58,099
another example where we are interested to
evaluate the principal stresses and the stress
320
00:44:58,099 --> 00:45:08,410
invariants. We have learnt how to compute
the principal stresses from the cubical equations
321
00:45:08,410 --> 00:45:13,999
and
the stress invariant components. Now the state
322
00:45:13,999 --> 00:45:19,469
of stress at a point is given by this, this
is the
323
00:45:19,469 --> 00:45:30,240
stress tensor where sigma x is equal to minus
5 units; sigma y is equal to 1 unit; and sigma
324
00:45:30,240 --> 00:45:36,369
z is
equal to 1 unit and tau xy is equal to minus
325
00:45:36,369 --> 00:45:49,369
1; tau xz is equal to 0; and tau yz is equal
to 0. So, if we
326
00:45:49,369 --> 00:46:01,449
compute the stress invariants, as we have
seen, I 1 is equal to sigma x plus sigma y
327
00:46:01,449 --> 00:46:11,359
plus sigma z ;
sigma x is equal to minus 5 plus 1 plus 1,
328
00:46:11,359 --> 00:46:24,519
is equal to minus 3. I 2 is equal to sigmax
sigma y (tau xy )
329
00:46:24,519 --> 00:46:57,480
whole square plus sigma y sigma z (tau yz
) whole square plus sigma z sigma x (tau zx
330
00:46:57,480 --> 00:47:58,479
) whole square is
equal to (minus 5 minus 1) plus (1 minus 0)
331
00:47:58,479 --> 00:48:07,669
plus(minus 5 minus 0) is equal to minus 10.
332
00:48:07,669 --> 00:48:08,669
.
333
00:48:08,669 --> 00:48:16,349
.Likewise if you compute I 3 is equal to minus
6. Hence the cubical equation which you get
334
00:48:16,349 --> 00:48:22,431
is,
sigma n cube minus I 1 into (sigma n ) whole
335
00:48:22,431 --> 00:48:26,660
square here I 1 is equal to minus 3 plus 3
(sigma n )
336
00:48:26,660 --> 00:48:41,960
whole square, I 2 is equal to minus 10 hence
minus 10sigma n plus 6 is equal to 0.
337
00:48:41,960 --> 00:48:50,259
Now sigma n is equal to 1 we will make this
function as 0, since sigma n in 1 of the root
338
00:48:50,259 --> 00:48:55,729
of the
equation, sigma n minus 1 if we take, sigma
339
00:48:55,729 --> 00:49:02,869
n square so this gives sigma n cube minus
sigma n
340
00:49:02,869 --> 00:49:20,519
square plus 4sigma n square minus 4sigma n
minus 6sigma n plus 6; so
341
00:49:20,519 --> 00:49:36,849
minus 6 into (sigma n minus
1) is equal to 0. So we get (sigma n minus
342
00:49:36,849 --> 00:49:45,359
1) into (sigma n square plus 4sigma n minus
6) is equal
343
00:49:45,359 --> 00:49:56,170
to 0. From this we will get sigma n is equal
to 1, minus 2 plus or minus square root of
344
00:49:56,170 --> 00:50:02,469
10. These
are the roots of this equation. These are
345
00:50:02,469 --> 00:50:06,640
the three principal stresses. These are the
three invariants
346
00:50:06,640 --> 00:50:21,140
I 1 , I 2 , and I 3 . That is how we can compute
principal stresses and stress invariants.
347
00:50:21,140 --> 00:50:22,140
.
348
00:50:22,140 --> 00:50:31,140
.Now in this we have another example problem
in which mass is hung by two wires AB and
349
00:50:31,140 --> 00:50:35,630
BC
and the cross sectional areas of these two
350
00:50:35,630 --> 00:50:42,089
wires are given as 200 mm square and 400 mm
square.
351
00:50:42,089 --> 00:50:52,039
And if the allowable tensile stress of these
wire material is limited to 100 MPa then you
352
00:50:52,039 --> 00:51:00,880
will
have to find out the mass M that can be safely
353
00:51:00,880 --> 00:51:05,719
supported by these wires. This particular
problem
354
00:51:05,719 --> 00:51:17,220
can be solved by taking the free body diagram
of this.
355
00:51:17,220 --> 00:51:18,220
.
356
00:51:18,220 --> 00:51:42,329
To summarize what we have learnt in this particular
lesson; first we recapitulated on different
357
00:51:42,329 --> 00:51:55,119
kinds of stresses and those stresses are the
normal stress, and the normal stress on an
358
00:51:55,119 --> 00:51:58,989
axially
loaded bar. What is the maximum normal stress
359
00:51:58,989 --> 00:52:05,709
that acts on an axially loaded bar is the
axial pull
360
00:52:05,709 --> 00:52:13,839
P or compressive force push by cross sectional
area and cross sectional area which is minimum
361
00:52:13,839 --> 00:52:19,730
is
normal to the axis of the bar.
362
00:52:19,730 --> 00:52:24,440
.Thereby we have seen the relationship between
the normal stress corresponding shearing
363
00:52:24,440 --> 00:52:35,749
stresses. We have evaluated the stresses on
any arbitrary plane which we generally designate
364
00:52:35,749 --> 00:52:40,449
as
Cauchy’s stress formula. And from this Cauchy’s
365
00:52:40,449 --> 00:52:50,260
formula we arrived at concepts of principal
plane and principal stresses and we have said
366
00:52:50,260 --> 00:52:52,849
that the principal plane is the one on which
the
367
00:52:52,849 --> 00:52:59,140
stress is fully normal and thereby tangential
stresses are at 0 and the shearing stresses
368
00:52:59,140 --> 00:53:03,709
in the
plane is 0. While computing the principal
369
00:53:03,709 --> 00:53:08,969
stresses we have seen different coefficients
which we
370
00:53:08,969 --> 00:53:14,650
designated as stress invariants I 1 , I 2
and I 3 .
371
00:53:14,650 --> 00:53:24,920
We have noticed that I 1 , I 2 and I 3 are
the functions of normal stress at a point
372
00:53:24,920 --> 00:53:28,829
which are sigma x ,
sigma y and sigma z and the corresponding
373
00:53:28,829 --> 00:53:45,130
shearing stresses at
the particular point. We have seen
374
00:53:45,130 --> 00:53:52,549
some examples to demonstrate how to evaluate
stress at different points. We have tried
375
00:53:52,549 --> 00:53:55,299
to give
you the concept of normal stress, we have
376
00:53:55,299 --> 00:53:58,910
computed the normal stress at a particular
cross
377
00:53:58,910 --> 00:54:08,299
section, we computed the shearing stresses
in the bolt, where the normal force which
378
00:54:08,299 --> 00:54:12,730
is acting on
the plate is transferred into the bolt cross
379
00:54:12,730 --> 00:54:21,799
section and then we have computed the bearing
stresses, the bearing stresses are acting
380
00:54:21,799 --> 00:54:27,449
at the contact area between the plate and
the bolt and
381
00:54:27,449 --> 00:58:00,410
finally we have computed
382
00:58:00,410 --> 00:58:18,619
the
principal stresses and
383
00:58:18,619 --> 00:59:35,430
stress invariants.
.
384
00:59:35,430 --> 00:59:35,430
.