we will discuss about the stress.
The concept of
stresses. In this particular lesson we are going to look into some more aspects of analysis of STRESS.
Before we start this know must know what is the unit of stress and force.
The unit of Force is Newton (N) and unit of stress is (N by m square) or Pascal (Pa). This can be
represented in terms of also Mega Pascal (Mpa) or just Giga Pascal (GPa).
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Here is the answer of the first question, what is normal stress then after what is FBD?
- The force acting normal to the plane is known as normal stress. **'FBD' means free body diagram.
If you have a body which is a free body from a major body and is acted on by forces there will be
resulting forces into it which will keep the body in equilibrium. At a particular small element if
we take that this is a resulting stress then we can decompose this stress into two components, one
is along the normal direction of this particular cross section which is this and another component
along the plane of this particular section. The component which is acting normal to this particular
cross section is normally known as the normal stress. Normal stress is the normal component
perpendicular to the particular section.
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Note:- One should be able to understand the concept of stress in a body. Understand relevant stress
components, and then one should be able to understand why we need to go for the equations of
the equilibrium, for a given problem. One should be able to draw the free-body diagram and
evaluate the stress resultants from these diagrams and thereby compute the stresses.
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Question :-What are the axioms on which behavior of deformable member subjected to forces depend?
Answer:- We have the first thing which is equilibrium of forces. Fundamental laws of Newtonian
mechanics are used for the equilibrium of forces and for the body should be such that it must be
having the forces in x direction. Summation of forces in x direction should be equal to 0,
summation of forces in y direction should be equal to 0, and summation of forces in z direction
should be equal to 0. These equations must be satisfied to fit for the body which is in space. In a
two dimensional form, these equilibrium equations reduces to summation of forces in x direction
equal to 0, summation of forces in y direction equal to 0 and summation of movement of z is
equal to 0. Also the forces must satisfy the parallelogram of forces.
Now lets suppose that, we have two forces in the plane which is normal and plane forces in the direction of
the plane this must satisfy the law like the resultant should pass through the diagonal of the
parallelogram. Or, if we are talking about forces or stresses in three dimension, if we look into
this parallelepiped the forces acting in the x direction or the stress acting in the x direction the y
direction and the z direction the resultant of this must be acting along the diagonal of this
parallelepiped. This is the resulting stress of all these stress components.
Therefore either in two dimensional plane it should be in this configuration or in a three
dimensional plane it should be in this configuration which is the parallelogram of forces that
must be satisfied. So these are the two basic axioms based on which the forces act on the
deformable body are guided.
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- Stress multiplied by area on which they act produces force.
- At any section , vector sum of the forces keeps the body in equilibrium.
Now this stress when it is multiplied is acted on by external forces. If we take a small element in
which a stress is acting this stress multiplied by the area gives the force which we call as the
stress resultant and the total stress resultant is the stress multiplied by the elemental area
integrated over the whole of the area is the resulting force which we call as stress resultant. And
thereby, we assume that at any point it has the same behavior.
° So at any section the vector sum of the forces keeps the body in equilibrium and that is how the
stress resultant will be obtained for that particular section. So our job is to evaluate this stress
resultant. And once we know the test resultant we can compute stresses at that particular section.
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If we cut this particular body to a plane which is perpendicular or rather parallel to the y, z plane
then the normal drawn on this particular plane will be parallel to x-axis. For any plane when you
draw the normal to that particular plane and if that particular normal coincides with any of the
axis we designate that plane with the name of that particular axis.
For example, here we have cut this body through a plane which is parallel to y, z plane. So, if we
draw a normal on to this section this normal is going to be the parallel to x-axis and thereby this
particular plane we designate this plane as x-plane. Now on this x-plane we have at a particular
point the resulting stress which we call as R. If we take the component of this stress R in three
perpendicular axis direction then we have the stress acting in the direction of x which is normal
to this particular section and as per the definition of normal stress this is the normal stress which
is acting in the direction of x. If we take the component of R along y-axis or parallel to y-axis
then the plane we get as stress is acting in y direction.
Also, we have the component which is acting in z direction. The stress which is acting parallel to
y or in the y direction we designate this as the stress tau acting in the plane x in the direction y
which we call as tau x, y.
Or if we designate this particular stress which is acting in the plane x along z direction then we
call these as tauxz. Thereby in this particular x-plane we have three stress components where one
is the normal to the plane and the other two are in the plane in the direction of y and z. The
normal stress we call as σ x and the other two components which are in the plane are called as
shearing stresses which are xy τ and xy τ .
Likewise, if we cut this body into a plane which is parallel to this z plane we cut along this then
on this plane on a particular point we can get three components of the stresses and they areσ y , τ
on the y plane in the x direction as yx τ , the shearing stress tau in the y plane in the z direction
as yz τ . Also, if we cut this body with a plane which is perpendicular or parallel to x, y plane and
if we plot the three stress components the stress which is normal to the plane gives out the
normal stress sigmaz and two stresses which are in the plane are on the z plane in the x-direction
and stress in the z-plane in the y-direction.
These are the nine stress components that we are going to get at a particular point. Now if this
particular body is cut in such a way that you take another plane which is at an infinite small
distance away from here if we cut it off by two parallel planes then we can get small a cubical
element on which you can plot the stresses.
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Now let us look into another aspect of it. Out of the stress components which we have obtained,
again let us call these as x-axis as usual, we call as y-axis and this as z-axis.
Now, if we take the movement of all these forces, now let us assume that this distance which we
have taken at a particular point of the body is dx, the vertical height being dy and along the z
direction tau with dz. Now if we like to take the movement of all the forces about z-axis, now in
this particular figure only the forces which will have relevance while taking the movement about
z-axis has been taken into account. Now this particular plane being x plane and the force which
is acting in the y-direction as per our nomenclature we call this as tauxy.
Accordingly this particular component of the stress which is acting in the direction of x and
acting on y-plane we call this as yx. Likewise, this is also tauyx and this is tauxy. Along with this
we have the other stresses like normal stress sigmax, sigmay and sigmaz and shearing
components as well in the other plane. Since only these forces or these stress components are
going to cause the movement other forces have not been shown here.
If we take the movement of all the forces about z-axis then the movement expression can be
written as tauyx which is the stress acting on the area dx by dz, so tauyx into dx into dz is the
force. The movement about the z-axis is the distance dy so this multiplied by dy is the movement
about the z-axis as tauyx which is clockwise in nature minus tauxy which is acting on the area
dz(dy). So tauxy into dy into dz is the force.
If we take the movement of this force with respect to z- axis then this is multiplied by the
distance dx. Assumingly that there are body force components in the x and y-direction, this x is
the body force per unit volume then along with this we have plus, but for the time being we are
neglecting the body force components because that is also not going to cause any moment as
such with respect to the z-axis.
again let us call these as x-axis as usual, we call as y-axis and this as z-axis.
Now, if we take the movement of all these forces, now let us assume that this distance which we
have taken at a particular point of the body is dx, the vertical height being dy and along the z
direction tau with dz. Now if we like to take the movement of all the forces about z-axis, now in
this particular figure only the forces which will have relevance while taking the movement about
z-axis has been taken into account. Now this particular plane being x plane and the force which
is acting in the y-direction as per our nomenclature we call this as tauxy.
Accordingly this particular component of the stress which is acting in the direction of x and
acting on y-plane we call this as yx. Likewise, this is also tauyx and this is tauxy. Along with this
we have the other stresses like normal stress sigmax, sigmay and sigmaz and shearing
components as well in the other plane. Since only these forces or these stress components are
going to cause the movement other forces have not been shown here.
If we take the movement of all the forces about z-axis then the movement expression can be
written as tauyx which is the stress acting on the area dx by dz, so tauyx into dx into dz is the
force. The movement about the z-axis is the distance dy so this multiplied by dy is the movement
about the z-axis as tauyx which is clockwise in nature minus tauxy which is acting on the area
dz(dy). So tauxy into dy into dz is the force.
If we take the movement of this force with respect to z- axis then this is multiplied by the
distance dx. Assumingly that there are body force components in the x and y-direction, this x is
the body force per unit volume then along with this we have plus, but for the time being we are
neglecting the body force components because that is also not going to cause any moment as
such with respect to the z-axis.
Therefore here it is z is equal to 0 for equilibrium. This produces tauyx is equal to tauxy. In effect
this means that the cross term tauyx and tauxy are equal. Likewise if we take the movement of
forces about x and y-axis and take the relevant forces we can see that tauzx is equal to tauxz and
tauyz is equal to tauzy. This gives us that the cross hearing terms are equal. So if we look into
stress and strain which we had tauij the tauij is equal to sigmax, tauxy, tauxz, tauyx, sigmay and
tauyz, tauzx, tauzy and sigmaz if we write in the matrix form. This is the stress tensor.
Now for the equality of the shear we have obtained tauxy is equal to tauyx, tauxz is equal to tauzx,
tauyz is equal to tauzy. Thereby stress tensor can be written as tauij in the matrix notation as
sigmax, tauxy and tauxz. Now tauxy is equal to tauyx we will write this as tauxy, sigmay and tauyz
and zx and xz being the same we write this as tauxz, tauyz, tauyz and sigmaz and thereby it
reduces to the six stress components sigmax, sigmay and sigmaz, tauxy, tauxz and tauyz and this
we find is symmetrical in nature so the stress tensor has a symmetric form.
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Having known that the stress at particular point is acting which are combinations of normal
stresses and shearing stresses let us look into that if we have a body and if we are interested to
find out the change in stress from one point to another then the change of the stresses is from
point to point, we need certain equations to be solved and those equations are called as equations
of equilibrium.
Coming back to the body here, for example we have a body which is stress and we like to find
out the change in stress from this point to this point. So we need these changes to be evaluated
through these equations of equilibrium. Now as usual we call this as x-axis, this as y-axis and
this as z-axis. Now on this particular plane which is normal to the x-plane we have normal
stresses known as sigmax.
We have two shearing stress components in the x-plane acting in y-direction called as tauxy. We
have stress in the x-plane in the z-direction which we call as tauxz. When it comes to this
particular plane which is at a distance of dx from this plane and likewise let us assume that this
length is dy and this is dz so the stress which will be acting in this which is the normal stress will
have a component as sigmax plus del sigmax del x which is acting over the length dx. Likewise
we will have the shearing stress component tauxy which is varying from this end to this end we
will have tauxy plus del tauxy del x by dx.
We will have x in the z-direction that is tauxz and tauxz in on this particular plane so when it is
coming to this plane there is a change over the length dx so tauxz plus del tauxz del x by dx.
Likewise the stress in this particular plane normal to this which is the y-plane will have sigmay,
the stress acting normal to this is sigmay plus del sigmay del y by dy the length. The shearing
stress component on the y-plane acting in the direction of x will have tauyx, plus del tauyx del y
by dy the length. On this plane we will have sigmaz and the normal stress on the front z-plane is
sigmaz plus del sigmaz del z over the length dz and so on.
Now if we take the forces which are acting in the x-direction and sum them up as for the
equations of equilibrium the summation of all the forces in the x-direction must be equal to 0. If
we write down the forces in the x-direction we have sigmax plus del sigmax del x by dx. So in
the equation we have sigmax plus del sigmax del x by dx and acting over the area dy and dz
minus sigmax acting over the area dy and dz.
Also, in this particular direction we have plus (tauyx plus del tauyx del x(dx)) and delta yx by del
y by dy acting over the area dx and dz minus tauyx, dx dz plus we have a term in the z plane
acting on the x direction which is tauzx plus del tauzx del z by dz into dx and dy the area minus
tauzx(dx and dy) plus if we assume that X is the body force per unit volume then this multiplied
by dx, dy and dz is equal to 0. So from this we will get del sigmax del x, if we cancel out these
terms and divide the whole equation by dx, dy and dz we have del sigmax del x plus del tauyx by
del y plus del tauxz by del z plus x is equal to 0 where x is the component of the body force.
Likewise if we take the equilibrium of the forces which are acting in the y and z-direction we get
two other sets of equations and they are, del tauxy by delx plus del sigmay by del y plus del tauyz
by del z plus y the body component force is equal to 0 del tauxz by del z plus del tauyz del y plus
del sigmaz del z plus z is equal to 0. These are called the equations of equilibrium.
These equations of equilibrium can be written down in a two dimensional form as well. You can
designate these in xy-plane, here we have x and here we have y, this is sigmax and the variation
along the x is sigmax plus del sigmax del x by dx the length where this is the length dx and this is
dy, this is tauxy plus del tauxy del x by dx the length. This is sigmay plus del sigmay del y by dy
the length, then we have tau this is tauyx and this gives the variation of tau which is tauyx plus del
tauyx by del y(dy).
These are the stresses in the two dimensional plane and if we take the equilibrium of the forces in
the x-direction then we can obtain the equations of the equilibrium in two dimensional plane.
which could be del sigmax del x plus del tau and as yx and xy being the same we can write this
as tauxy del y plus the component of the body force x is equal to 0.
The other equations will be del tauxy by del x plus del sigmay del y plus y is equal to 0. So these
are the equations of equilibrium in two dimensional planes. Now, having known these stresses at
a point, equations of equilibrium and how to evaluate those stresses or how to write down those
stresses at different planes if we have to evaluate the stresses in an axially loaded member, then
we have a member or we have a body in which we have a force acting in the axial direction, so
let us call this body acted on by force P.
Now if we like to evaluate the stresses at any inclined plane let us cut this body by an inclined
plane. And if we draw the free body diagram of this then we have the body in this form. Here we
have the resistive force P which is acting. On this, the resulting force or the stress resultant is
acting in this particular direction to equilibrate the body. Now we can take the components of
this force in the normal direction, normal to this plane and along the plane which will give you
the three forces of the stress component which is normal which we call as the stress
corresponding to normal and the two shearing stress components tau.
If we concentrate on the two dimensional plane, if we take the axially loaded member the
member is subjected to the load in the axial direction. Let us take a plane which is cutting this
body in this form and let us assume that this plane is making an angle of theta with the vertical.
If we take the free body diagram of this particular body this is angle theta, we have the force
acting here as P, the resistive force acting on this body to keep the equilibrium is P so this will
have two components one along the normal and one along the plane of this particular section.
Now this angle being theta, if we drop a perpendicular here this angle will also be theta hence
this particular angle is also theta. So the force component along this is P cos theta and the force
component along this direction is P sin theta. Now, if we say that the cross sectional area is A.
and the cross sectional area of this as A prime then the stress which is acting in this particular
inclined plane if we say that normal stress sigma theta and theta being designated by this
particular plane which has got an angle theta in the vertical, sigma theta equals to the normal
force component which is P cos theta divided by this area A prime.
And A prime from geometrical property we can say A prime is equal to A by cos theta. Then P
cos theta by A by cos theta is equal to P by A cos square theta. And the stress which is acting in
the plane is P sin theta. The stress tau theta is equal to P sine theta by Acos theta so this
eventually is going to give us P by 2A by sin 2 theta. So these are the two stress components on
this inclined plane. The normal plane is P by A cos square theta and the stress which is parallel to
the plane which is the shearing component is P by 2A sin2 theta. So the maximum value of
sigma theta is when cos square theta is equal to 1 and is theta is equal to 0. And for this sin 2
theta as 90 degrees and 2 theta being 90 degrees so theta being 45 degrees is the maximum value
of the shearing stress. So the maximum value of normal is P by A and the maximum value of
shearing stress is P by 2A. Eventually the relationship between tau theta and sigma theta is that.
See below.
This is the end of this chapter.
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