Mod-04 Lec-31 Lecture – 2 on Stress -strain relationship and shear strength of soils

Welcome to the lecture series on Advanced Geotechnical
Engineering. We introduced ourselves to Module 4 which is on Stress-strain relationship and
shear strength of soils. And this is lecture 2 in Module 4. So we have
discussed the contents and then Introduction to the Mohrís Circle Analysis. So in this
lecture, we will try to do a detailed discussion on the Mohrís Circle Analysis and its applications
and determination of the pole in the Mohrís Circle and how this can be used to find out
the stress on any given plane within the element. And thereafter we will introduce ourselves
to the principal stress space and stress paths in p q space. So as we have discussed, the Mohrís Circle
is a geometric representation of the two-dimensional stress state and is very useful for performing
quick and efficient estimations, it is also popularly used in geotechnical fields
such as soil shear strength that anyhow we are going to discuss and earth pressures,
varying activation of active conditions and passive conditions are discussed and also
in the bearing capacity theories.

So it is often used to interpret the test
data analyze complex geotechnical problems and predict the soil behavior
in these, the pole point on the Mohrís circle is a point so special that it can help to
readily find the stresses on a specified plane by using the diagram instead of complicated
computations. So the topic which we are going to discuss
the pole point on Mohrís circle is a point so special that it can help to readily find
stresses on any specified plane by using a diagram instead of complicated computation.
Now this coming to the pole points, a pole is a unique point located on the periphery of
the Mohrís circle. The point of intersection of Mohrís stress
circle and a line drawn through the pole parallel to a given plane gives the stresses on that
plane. So, the point of intersection of Mohrís stress
circle and a line drawn through the pole parallel to a given plane gives the stresses on that
plane. So this pole point is also called the origin of planes. So there are two ways
the pole points can be established. One is by relating it to the direction of action
of stresses.

The other one is related to the direction of the planes on which the
stresses are acting. Suppose there is a horizontal plane, if
there is normal stress acting on that. Then we are referring to a horizontal plane and if
there is a vertical plane on the element and the normal stress is acting on that then that
is also what we are referring to as the direction of the plane on which the stresses
are acting. Usually, the pole point relating to the direction of the planes is in use.
Usually, the pole point relating to the direction of the planes is in use. The direction of
the planes whether it is horizontal or vertical is in use. Now here consider the procedure to establish
a pole point for stresses. So what we need to do is that we have to draw the Mohrís
circle on the two-sigma space. By knowing the sigma z and tow zx and sigma
x and tow xz. So these are sigma x is normal to the vertical plane, as you can
see here and the sigma z is normal to the horizontal plane is sigma z here and
this is tow zx.

So if you look into this, the centre of the
element if you take moments, we get that tow xz is equal to tow zx because we have to maintain
the equilibrium of the element from the moment equilibrium point of view. So project the
line by knowing these two points, draw the Mohrís circle on the tow sigma space then
from this point sigma z, tow zx is acting in the direction of the vertical direction,
drop a line wherever it cuts the Mohrís circle and that is the pole point
from the stresses. So in this direction if you can cut
then this is the pole point from the stresses.

Similarly, next, if you look into
this, from here when you have the sigma x which is acting in the horizontal direction,
the direction is in the horizontal direction, but it is acting on the vertical plane then
you get a point here which is again the same PS. So this is the pole point from
the stresses. The projection of the line from the point
sigma x tow xz in the line of action of sigma x, that is horizontal, till it intersects
the circumference of the circle, or projection of the line from the point sigma z tow zx,
in the line of action of Sigma Z vertical, they both give the pole point with
reference to the one method of establishment that is called pole point for stresses. The pole points for the planes means
that, the projection of the line from the point sigma z tow zx, so here you can see
that by following the usual procedure we have drawn the Mohrís circle and here the
sigma z and tow zx and sigma x tow xz.

So from this point, this is sigma z is
acting, sigma z is acting which is perpendicular but it is acting on the horizontal plane. So in the direction of this plane draw a line
that is parallel to this horizontal plane and till it intersects the Mohrís circle
and that point is regarded as the pole point. The intersection point gives the pole point
PP for planes. So what we can notice here is that this is the point pole PP and from
the definition of pole point from the planes, similarly when you look into this sigma x
tow xz here, so here this is sigma x tow xz is, this plane is vertical here.

So in the direction of this vertical plane,
draw the line till it intersects the point here. Then so if you again
draw in this horizontal direction or this direction, you have a point
here and from this point, when you put this, this is again the pole point from the planes. Now when we join this point with the major
principal stress here, sigma 1 by intercept with sigma x is at tow is equal to is the
major principal stress and here minor principal stress that is sigma 3 which is here. So when
you draw a line like this, when you draw a line from here to here and on this plane there
is the minor principal plane acting. On this plane now the major principal plane is acting.

Now if you are trying to find out,
wanted to find a stress on a particular plane within the element, let us say an angle alpha
within the element, that can be found out by just drawing or dropping a line such that
you can get these stresses that is sigma and tow,
tow and sigma at this particular any element state here. So with that, we will be able to
find out the stresses in the elements, the stress states in the elements at any given
plane within the element. So in the Mohrís stress circle, when we said
that pole points for the planes and which is usually adopted wherein what we
have to draw, how to establish a pole point is that, project the line from the point sigma
z tow zx in the direction of the plane on which these stresses are acting till it intersects
the circumference of the circle and intersection point gives the pole point PP for planes.

Now what we have further discussed is that
when we join this with the major principal stress magnitude here this plane is nothing
but the major principal plane on which the major principal stress is actually acting
and here on this, this is a minor principal plane. So these principal planes are the planes
on which the shear stresses are zero. Shearless planes are called principal planes and
we have a major principal plane and a minor principal plane which a two-dimensional
a case whereby ignoring the intermediate principal stress, what we say is that major principal
stress and minor principal stress on principal planes can be identified.

Now the pole of the Mohrís circle is defined
as thus now if a line is drawn from the pole to a point on
the circle where the stresses are tow I and sigma I, then in that XZ plane the line is
parallel to the plane on which the tow I and Sigma I act. Now use the pole point if you look into
is, basically to find out the stresses of Sigma C, Sigma A, TOW AC, and TOW CA on the plane
which is inclined at theta, to the plane on which sigma z acts.

So now we can see that
this is sigma z is acting on this plane and sigma x is acting on this
vertical plane and there is tow zx is acting here and tow xz is acting here. So in step 1, locate the pole point for
the planes extending point E horizontally. So what we do is that we locate the pole
point that is we know that the stress state here is that sigma z tow zx and sigma x tow
xz. Construct the Mohrís circle with this as the diameter and from this the sigma z
is acting on a horizontal plane. To extend this line, project this line, till
it intersects the periphery of this circle, that is the circumference of the circle, then
that point is regarded as the pole point PP. Now what we do is, locate the pole point
for the planes extending point E horizontally in the direction of the plane on which the
sigma z is acting, that is horizontal, so we have drawn the horizontal line.

Then, what we have to do in step 2, is to,
draw a line parallel to the plane in which Sigma c acts. So now draw a line parallel to the
plane on which sigma c acts, so this sigma c is acting at an angle, let us say
that certain inclination. So draw a line parallel to this line and here we draw a line. To extend
a line from point D through the center of this circle till it intersects the circle. So from here draw the circle so that you get
the sigma a and tow ac that is on these stress states and then we get sigma c and tow ca.
So what we have drawn is that we have drawn a line parallel to this plane on which the
sigma c is acting.

Now what we have established is that we established elements at the D and
Dí here and these are the stresses at this state on the sigma c and tow ca which are
nothing but here and here it indicates that sigma a and tow ac. So when we join, let us say again here to
the major principal stress minor principal stress, then this will become the major principal
plane and this will become the minor principal plane. Now this is shown diagrammatically,
here once again, use of the pole point to locate the stresses sigma and tow at an angle
theta to the reference stress direction.

So here what we have done is we have got.
Mohr circle on the tow sigma space and we with sigma z tow zx which are acting
here and sigma x and tow xz by following the sign conventions which are defined yesterday
we defined in the previous lecture, we can draw, indicate this in tow
sigma space and draw a Mohrís circle with this as the diameter. Then this sigma, wherever it intersects the
sigma axis, this will become the major principal stress that is measured from here origin to
this point and this is the magnitude that is measured from here to here that is Sigma
3.

Now the stress point on the Mohrís circle is from here we draw a line parallel
to this, so you actually can locate this pole that is PP, or you can draw a line parallel
to this plane, so you this vertical plane you will again get from either from
this or this, establish the origin of the planes. Then from here what we need to do is that,
we wanted to determine the stresses of sigma c and tow ca at an angle theta, so draw a
line parallel to the plane on which sigma c and tow ca are acting. So with that what
we get is that we get this stress states like sigma c after locating D by drawing
an angle theta, below with PPE.

Then drawing D and Dí passing through the
center of the Mohrís circle, we get the stress state sigma a tow ac, that
is sigma a tow ac and then we get sigma c and tow ca. So by using this procedure, like
use of pole point P to locate the stresses sigma c and tow ca at an angle theta to the
reference section is demonstrated in this particular slide. Now the fundamental relationships by the Mohrís
the circle can be discussed. So here are two sigmas in space in Mohrís circle which is actually
shown here where you can see that the stresses are acting sigma 1 and sigma 3 are
acting on this, sigma 1 is acting on plane OA, and sigma 3 is acting
on plane OB which is inclined at an angle theta. So what we need to do is that again
by doing sigma 1 and sigma 3, we can draw the Mohrís circle and from here locate
PP or OP, that is the pole, from here draw the line parallel to this plane on which this
is acting.

So this is inclined at theta. So draw a line.
So from here what we can draw is that, if we look into this the shear stress
will be maximum here that is tow max is equal to sigma 1 minus sigma 3 by 2 which is nothing
but the radius of this Mohrís circle and this point is sigma 1, measured from here
to here and this point is sigma 3, the minor principal stress. So what we get is that
this is the major principal plane and this is the minor principal plane. The shear
stress, the principal shearing stress occurs at the line joining this point to
the point where the maximum shear stress occurs, that is 45 degrees. So we have two max which
is on the upper portion of the Mohrís circle and the bottom portion of Mohrís circle.
So the maximum shear stress often called the principal shearing stress has a magnitude
of sigma 1 minus sigma 3 by 2 which equals the radius of the Mohrís circle.

So the maximum
shearing stress often called the principal shearing stress has a magnitude of Sigma 1
minus sigma 3 by 2 which equals the radius of the Mohrís circle. So the principal shearing stresses occur on
planes inclined at 45 degrees. The principal shearing stresses occur on planes
inclined at 45 degrees. The principal or the maximum shearing stresses
occur on planes inclined at 45 degrees. Now another aspect is that, the conjugate shearing
stresses. That is the shearing stresses on planes at right angles to each other are
numerically equal but are of opposite sign. So numerically equal, because the element
has to be maintained equilibrium by t.

When you take a moment about the center of the
element we get tow zx is equal to tow xz or tow xy is equal to tow yx or tow yz is equal
to tow zy. So these stresses are called conjugated shearing stresses. So the conjugate shearing
stresses are represented here. So we can see that if you are having these shearing stresses
at this and this point, we call these are, which is on the upper, they are equal
in magnitude but the only thing is that they are opposite in the sign. So one is positive,
other one is negative here.

So this is indicated here. These two are called conjugating
shearing stresses. So these are conjugating shearing stresses. Now another aspect is that what is the
obliquity and the resultant stress? So the resultant stress on any plane has a magnitude
expressed by, we have an element, let us say sigma and tow, so the resultant is
nothing but root over sigma square plus tow square at any particular point on the Mohrís
circle and has an obliquity which is equal to the tan inverse of tow by sigma.

So if we look into, let us say that, this
is the point C on the Mohrís circle to which we are referring, now what we do is that from
the origin that is where, from here to this point, so this is called the maximum
obliquity angle, which is nothing but alpha is equal to tan inverse tow by sigma, this
is tow and this ordinate is sigma and this magnitude is nothing but root over sigma square
plus two squares which is the resultant.

So the angle of obliquity is nothing but alpha
is equal to tan inverse tow by sigma. Now further if you look into it, in this particular
slide where it has been shown that we have got in the same situation that Mohrís circle,
but here if you look into this here, when indicating this sigma 1 and sigma 3 act at
an inclination theta. So what we have drawn is that we have located the pole and from
here what we have drawn is that we try to determine the stresses on any plane at an inclination
theta, so that is this point, the point where it touches the, the point where it
is tangential to this line. When you draw from the origin, and if the
the point where it gets tangential and that point and this point when you join and
when you look into that this is also, so this angle is called, this angle joining
from this point to this point is called as the maximum angle of obliquity.
So the maximum of all the possible obliquity angles on various planes is called the maximum
angle of obliquity, which is alpha m.

The maximum of all the possible obliquity
angles on various planes is called as maximum angle of obliquity. The coordinates of the
point of tangency or the stresses on the plane of maximum obliquity are less than the
plane of principal shear. So if you look into this, the principal shear
is nothing but the maximum shear stress. So here is the maximum shear which means that the
failure is most likely to happen, but when you say that, limiting obliquity is that it
is a criterion that is used to initiation of the slip, indicate the initiation
of the slip.

So the maximum obliquity is less than the plane of the principal shear because
since the limiting obliquity is the criterion of the slip whereas in the plane of the principal
shear is liable to happen. So the coordinates of the point of tangency
are the stress on the plane of the maximum obliquity, which is the angle at which
is joining to the maximum shear stress is less than this maximum angle of obliquity.
So if you look into this, this is not the maximum shear stress, the maximum shear stress
is here. So here at the point of maximum shear stress with radius R is equal to sigma 1 minus
sigma 3 by 2 the failure is most likely to happen. But here, this point indicates
the maximum limiting obliquity angle is the criterion that is specified
for initiation of the slip whereas in the plane of principal shear, it is
liable to happen. So now consider an example where you need
to draw the Mohrís stress circle at failure on a cylindrical specimen of stiff clay with
a shear strength of 100 kPa, and if the radial stress is maintained constant at 80 kPa.

By
using the pole point method find the inclination theta to the radial direction of the planes
on which the shear stress is one-half of the maximum shear stress, and determine the normal
stresses acting on these planes. So we need to determine the magnitude of the
normal stresses acting on these planes and the condition which is given here is that
find the inclination theta to the radial direction of the planes on which the shear stress is
one-half of the maximum shear stress. Now what has been given is that, the maximum
shear strength which is nothing but 100 kPa, so which means that the maximum ordinate is
sigma 1 minus sigma 3 by 2 is two max is equal to 100, so with this as radius,
sigma 1 minus sigma 3 by 2 as the radius, draw the Mohrís circle on two sigma space,
so then we have got the maximum shear stress with 100 kPa here, minus 100 kPa here and
the circle when it intersects with the sigma axis at sigma 1 is 280 kPa and the sigma
3 is 80 kPa.

So tow max is the radius of the circle, so
sigma 1 is equal to sigma 3 plus 2 tow max, so tow max is given as 100, so with
that what we can find out is that sigma 1 is nothing but 280. So the
diameter of the Mohrís circle is nothing but 280 minus 80 which is 200 kPa, and the radius
is nothing but two max is equal to sigma 1 minus sigma 3 by 2 which is 100 kPa.
So the plot, Mohrís circle based on the above information i.e. radius and the two points
on the circle.

Once we do that, now the condition that has been given is that 50% of maximum shear
stress, draws a line. That means that when you are drawing a line that intersects the Mohrís circle, drawing a line at two is equal to
50 kPa which is half of the max. Now as the principal stresses are acting on edges on
the sigma 3, this plane is horizontal because of the cylindrical sample,
so sigma 1 is acting on the horizontal plane, and sigma 3 is acting on the minor principal plane.

So this itself will become like a, PP is nothing
but the pole here. So from here draw the line which intersects at the point where sigma 3 is
identified and that is pole point PP. Now from here what has been asked is that the
inclination of the plane on which tow is equal to tow max by 2 by drawing a line from
here, this point is indicated by the condition that we have given, and with that we can
get on this 15 degrees and at this particular point, we can get sigma n 15 as 267
kPa and another ordinate is nothing but 50 kPa, then you can see that when we draw a
line at 75 degrees, now not possible because this point intersects at these two
points, so when you draw the line here, from these 75 degrees with horizontally, it is
at 75 degrees with horizontal and this magnitude at this stress sigma n 75 degrees indicates
that 93 kPa. So draw the line from the pole to intersect the intersection of 50 kPa, a
line on the Mohrís circle. So what we have based on the given
data, we try to establish and determine the normal stresses acting on these planes.

Now here with the given information two planes
are possible so we could get for this same shear stress two different normal
stresses can be seen. One is that plane inclined at 75 degrees, the other one is actually at
15 degrees. Now consider an example problem where the
Mohrís circle of the total stress and element having a size of 40mm by 40mm, it is a cube.
So a two-dimensional state has been represented here and the x-axis and z-axis
are shown within the element and for estimating the shear forces and all, with
mustard color-filled rectangles are shown basically to show, you can see that
this is conjugate shearing stress shear force F4, and this is the conjugate shear force
F3.

So here the element is subjected
to shear like this and these are the normal stresses and these are the normal forces F2,
and this is normal force F1.So in the given figure, the normal loads are applied
to the faces of the soil cube having the 40mm by 40mm by 40mm dimensions and F1 is 0.45
kN, F2 is 0.3 kN and the shear loads are F3 is equal to F4 is equal to 0.1 kN. The sides
of the soil cube are each one having 40 mm?

To construct a Mohrís circle
of the total stress and find the magnitude of the principal total stresses and the direction
of the principal planes in the soil. So what we have been asked is that we have
been given forces, so each area of the face is nothing but 40mm by 40mm, in meters it
is when you convert millimeters into meters an area of each phase is 0.016 meters
square, so by putting let us say, F1 divided by that area, we will get the stress sigma
Z and sigma X is nothing but F2 by this area then we can get this shear stress F4 on the
plane divided F4 divided by that area and similarly, F3 divided by this area you will
get the shear stress on the sample. So now we do that as we discussed we determined
sigma Z which is 281 kilo Newton per meter square and sigma X is 187. 5 kilo Newton per
meter square. Now tow xz is equal to tow zx is equal to 0.1 divided is due to F3
and F4. So with that, we have got the data, which is given based on the given data we
establish sigma Z and sigma X and the tow xz and tow zx which are acting on
the element.

Now what we have done is that by knowing the stress rates here that is the
magnitudes are a little bit different, but they can be ignored and so here with this
stress stated like 281.25 kilo Newton per meter square and 62.5 kilo Newton per meter
square at this point, as well as this point by the appropriate using the sign convention. We can actually draw the Mohr circle and on
the tow sigma space which is drawn to the selected scale, then what we can do is that
from this because it is acting on a horizontal plane, draw a line to locate
the pole, so this is the point P, which is the pole. Now this is the major principal
stress, the ordinate is with the given dimensions which are given in the figure.

The sigma 1 magnitude is 306 kPa and sigma3
is 154 kPa, which is measured from here to here and which is measured from here to here,
so the line joining this when you drop a line to this one, this is the major principal plane,
which is inclined at 26 degrees with the horizontal line, which is drawn within the Mohr circle
to locate the plane and this is inclined at 90 degrees plus 26 which is about 116 degrees,
90 plus 26 and this is nothing but the minor principal plane.

So in the given problem, we have been asked
to calculate the magnitude of the major principal plane and minor principal planes, and also
the direction of the planes are indicated in this particular problem, which is the solution,
that is given for the problem. Now, another example here is if you look into
it, there is an element that is subjected to, a cylindrical element, subjected
to the minor principal stress is 12 kPa and major principal stress is 52 kPa. You can
see that these elements do not have any shear stresses, so hence they can be called
shear-less planes, this is a major principal stress and minor principal stress.

The major
principal stress is acting on the horizontal plane and minor principal stress
is acting on the vertical plane. So as can be noted here, the sigma 1, with
the sigma 1 52 kPa and sigma 3, so with 52 minus 12 that is about 40 kPa as the diameter
draws the Mohr circle. Once you draw the Mohr circle, this is point A which is on
the sigma axis 52 kPa is indicated here, this is the minor principal
stress, so this is acting on a horizontal plane, so from here draw a line to locate
the pole P, so this is the pole P. Now what we need to do is that, we need to,
let us say that the angle alpha is 35 degrees, so on this angle 35 degrees, when you draw
from this point to this point, so point intersects at C, so at point C wherever it intersects
the Mohr circle, we can interpret the sigma A and tow A, so that is nothing
but 39 kPa and this is 18.6 kPa.

So what we have got from this problem
is that we have for a given plane, which is inclined at alpha is equal
to 35 degrees within the element, so first we drew the Mohr circle based on the
information, which is given that sigma 1 is given 52 kPa, sigma 3 is given 12 kPa. So with 40 kPa as a diameter, we have drawn
a Mohr circle and then we have identified the pole P and from the pole P, we can actually
draw a line at 35 degrees, suppose it is let us say 60 degrees then that line will
be located at this point.

This is the point where you can get the stress in
the element at the onset of failure, in particular a plane of inclination, and another important
point is that in this the major principal plane is this and the minor principal plane is
indicated by a line perpendicular to this major principal plane, so the major principal
the plane is horizontal here and the minor principal plane is here, which is perpendicular to the
major principal plane. Now the same procedure is explained
here, plot the Mohr circle to some convenient scale and establish the pole point or the
origin of the planes. A line through the pole inclined at an angle alpha is equal to 35
degrees from the horizontal plane would be parallel to the plane of the element and that
is of interest and the intersection is at point C from there we can calculate the
sigma alpha, which is 39 kPa, and tow alpha is 18.6 kPa which are acting along
the plane, which is inclined at 35 degrees with the major principal plane.

Now consider a similar example, The only thing is
that in this particular case, the element in space is inclined at 20 degrees which means
that the element is considered within the space with 20 degrees inclination with
the horizontal, so the element under the same stresses have been considered. But the only thing is that this element in space
is actually subjected to an inclination of 20 degrees and again we are interested
along this 35 degrees plane what are the state of the stresses? So with the given information
now with 52 kPa and 12 kPa, now draw the Mohr circle so what we get is that we draw the
Mohr circle. Now what will happen is that because the inclination of the plane on which
the major principal stress is acting has changed now, so because of that so parallel
to this plane, the location of the pole point changes.

Previously, the pole point was here when it
was horizontal now it got lifted so you can see that draw a line from AP and where
it intersects, so the pole is located here, so from here if you put this
35 degrees then again it will intersect at sigma n is equal to 39 kPa and two energy
is equal to 18.6 kPa so this sigma 1, is the major principal plane now, and this
is the minor principal plane with magnitudes that are given like 12 kPa and 52
kPa, so this is the major principal plane and this is a minor principal plane. So the resulting sigma alpha is equal to 39
kPa and tow alpha are equal to 18.6 kPa are the same because nothing has changed except
the orientation of the space of the element, so when you have the orientation of the element
is changed within the space then the internal stresses will not change but
the pole point locations and the inclinations of the planes depend upon the orientation
of the element is under consideration.

So now we will take another example wherein
the stresses are shown in the figure below. Wherein you can see that this is an element
and which is subjected to, this is the horizontal plane where sigma V is equal to 6 MPa which
is subjected here and we have got tow is equal to 2 MPa and this is minus 2
MPa and this element are under tension and this is the shear stress, which is
minus 2 MPa here and minus 2 MPa here.

So we have the, in the vertical direction
it is compression and in the lateral direction it is, we have the element under tension.
So what we need to determine is to evaluate the stresses of sigma alpha and tow alpha when
alpha is equal to 30 degrees within the element and as shown in the element and evaluate sigma
1 and sigma 3 when alpha is equal to say 30 degrees, so we need to calculate what is the
location of the maximum principles stress and minor principal stress. So the procedure is that plot the state of
the stress on the horizontal plane that is (six, two) at point A and note that the shear
stress makes a clockwise movement about A, so the shear stress is making clockwise,
shear stress is making a clockwise movement about A, so from the consideration
which we have defined in the previous lecture, shear stress makes a clockwise movement about
A, so therefore when the shear stress is making a clockwise movement then it is actually
regarded as positive.

And if you look into this here, the A is a
point outside the element, so this if it makes a clockwise movement it
is regarded as positive, so in this case, if you consider element A it is making
clockwise movement, but when you take this particular one, the shear stress is actually
counterclockwise movement about this point outside the element, so this is actually
regarded as negative. So this point is negative and
again here this is making a clockwise movement so this is positive, so these two
shear stresses are positive and these two shear stresses are negative. So plot the point B (minus four, minus two),
the shear stress on the vertical plane is negative since it makes counterclockwise
movement, so clockwise shear stresses are positive, and counterclockwise shear stresses
are negative, so points A and B are two points on the circle, so construct the Mohr circle
with center at (one, zero) that is 1 MPa and zero ordinate. Find the pole by drawing the horizontal line
through A or vertical line through B and find the state of the stress on the planes inclined
at alpha that is 30 degrees and draw the line PC.

So this is explained here. So what we have done is that we know the stress
rates upon the defined sign conventions. We have located the A that is (six, two), which
is written here. Plot the state of the stress on the horizontal plane (six, two) at point
A. Note that the shear stress makes a clockwise movement about A and therefore it is positive,
so we have located and now to locate point B. What we have said is that the plot point
B (minus four, minus two). The shear stresses on the vertical plane are negative since
they make a counterclockwise movement, so we have located them here. So with this as the diameter and with the center
as (one, zero), we draw a Mohr circle on the two-sigma space. So now you can see that the
origin is here and then the Mohr circle I had extended to the native portion also.

Now what next step do we need to take after
having drawn the Mohr circle find the pole by drawing the horizontal line through A,
so draw from this known stress state here, draw the line horizontally because the known
stress state is here, draw the line parallel to this plane, so it intersects the Mohr circle
at point B or you can draw a vertical line parallel to this plane on which the stresses
are acting, so it intersects at P, so this is the pole P, so from here, what we have
been asked that calculate the stresses at an inclination of 30 degrees within the element,
so from P with PA, draw a line that intersects with PC, so the point C is nothing
but we get the stress state that is sigma N is 1.8 MPa and the tow A is 5.3
MPa. But if you look into this one, this particular
portion M, which is sigma 1 minus sigma 3 by 2, which is nothing but plus or minus 5.4
MPa and which is M and Mí we have got these points here, this is the maximum
shear and this is also a point of maximum shear.

So this inclination of PM is 34 degrees and
the magnitude of this is about 5.4 MPa. Now the next issue is that to draw a line
passing through this point, this is nothing but the sigma 1 and this is sigma 3, so with
this, we can find out what is the magnitude of the sigma 1, so this sigma 1
magnitude is at 6.4 MPa and this is nothing but in tension that is sigma 2 is equal to
minus 4.4 MPa. So this is the minor principal plane and this
is the maximum principal plane. The inclinations with the planes are indicated here.
So in the figure what has been asked is that we need to find out the state of
this stress when the element is having plane inclined at 30 degrees and lines drawn
from P to sigma 1 and sigma 3 established the orientation of the major and minor principal
planes.

That we have done herein
this particular exercise, in the problem where we identified and calculated what
is the maximum principal stress and minor principal stress and also when the given elementary
subject to the different states of stress and we have tried to understand
what is, how we can determine and interpret the several parameters very
efficiently. Now as we know total stress is nothing
but effective stress plus pore water pressure. So given this condition we can draw
a different Mohrís circle for total stresses and effective stresses. So here, in this particular
slide wherein we can see that more circles for total and effective stresses, so
here on the tow, sigma plot if you look at this on the right-hand side the circle in
white color indicates the total stress circle and this is nothing but the effective stress
circle.

So on the, when you have sigma one as the
major principal stress and the sigma three as the minor principal stress, with this sigma
one, sigma three as the diameter draw the Mohr circle. Now this is the again the pole
point here, now from this point at an inclination theta we can draw an element where
sigma not and tow theta are the stresses on that particular plane which are acting
purpose sigma 3 is acting perpendicular to this plane and tow theta acting along
this plane. So when you use Sigma one dash is equal
to sigma one minus U, where U is the pore water pressure and sigma 3 dash is equal to
sigma 3 minus U, when we do that the circle shifts to the left-hand side, so
in the case of an effective stress circle which is on the left-hand side of the total stress
circle and with the magnitude of the effective major principal stress is nothing but sigma
one dash is equal to sigma 1 minus U and this effective minus principal stress is nothing
but the sigma 3 dash is equal to sigma 3 minus U.

So the difference is nothing but U. Again
when you locate the pole point, The pole will not change, and from this inclination
theta and this stress state sigma dash theta and tow dash theta we can see that the difference
between these two horizontally, these two are separated by distance U. You can see that both ineffective stress
circles and total circles state the Mohr circles have identical diameters. So we can summarize the points like
the effective stress circle has the same diameter as the total stress circle and is separated
from it by the pore water pressure. The effective stress circle has the same diameter as the
total stress circle and is separated by the pore water pressure and the stresses to dash
theta and sigma dash theta are effective stresses acting on the plane inclined at angle theta. By examining the circles we note that tow
dash theta is equal to tow theta and sigma dash theta is equal to sigma theta minus U
so what we can conclude is that for a given state of total stress changes in pore
water pressure does not affect the effective stresses, they alter only the effective normal
stresses.

So an important conclusion that we have deduced
here is that for a given state of the stress the change in pore water pressure does not affect the effective shear stresses, the altered remedy is effective in normal stresses.
So this is represented here tow dash theta is equal to tow theta. Where you can see that the vertical ordinate,
is identical to that of, though the magnitudes are different, but the vertical ordinate that
tow dash theta is tow theta, which is indicated here. And also sigma dash
theta is equal to sigma theta ñU. So the effective stress circle has the same
diameter as the total stress circle and is separated from it by the pore water pressure
and the stresses are tow dash theta and sigma dash theta which are the effective stresses
acting on the plane inclined at theta.

So by examining the circles, we can see that tow
dash theta is equal to tow theta and that sigma dash theta is equal to sigma theta minus
U. So for a given state of the total stress, changes in the pore water pressure have no
effect on the effective shear stresses, they alter only in the effective normal stresses. So in this particular lecture we actually
have tried to understand how the Mohr circles can be drawn and then what are the applications
of this and then we try to establish the origin of the planes and the pole point P and then
we also have discussed how we can determine the major principal planes and the
minor principal planes, and we can say that the principal planes are the once
where the shear stresses are zero. Shearless planes are called principal planes.
So while determining the shear strength of soil if you can create this condition
then they are eligible to be called the shearless planes or principal planes and
which will be discussed in the forthcoming lectures discussing the triaxial
test.

And then further we have discussed that the
Mohrís circle of the total and effective stresses, the effective stress circle and
total stress circle the diameters are identical and again there is no reason as such there
is no tow dash theta, so tow dash is equal to tow because it is equal, and the only thing
what gets different is that the sigma dash theta will change it to sigma
dash theta is equal to sigma theta minus U. So for a given state of the total stress the
changes in pore water pressure do not affect the effective shear stresses and they only
alter the effective normal stresses. So we have discussed the Mohrís circle
for the total and effective stresses, so we can conclude that for a given state of the
total stress, the changes in pore water pressure do not affect the effective shear stresses.
They alter only the effective normal stresses because from the demonstration here two two-dash
theta is equal to tow theta and sigma dash theta is equal to sigma minus U.

Further, we can look at the uses of
pole construction on the effective Mohrís circle to calculate the effective stress on
any plane in the same way as we use pole construction to calculate the total
stresses. So if you have effective stress conditions are effective stresses Mohrís
circle also the same way we can use the pole construction method another thing
is that the position of the pole in the Mohrís circle of the effective stress is the same as
in the Mohrís circle of the total stress and the location of the principal planes of
the total and effective stresses in soil are identical. So when we have got the total stress conditions
and effective stress conditions then what we said is that the pole construction which
is the procedure is the same and the principal planes in the effective stress condition
and total stress condition remain the same. That means that the position of the pole in the major
circle of effective stress is the same as the Mohrís circle of total stress and the location
of the principal planes of the total and effective stresses in soil are identical.

So the positions of the pole in Mohrís circle
of effective stress are the same as in the Mohrís circle of the total stress and the locations
of the principal planes of total and effective stresses in the soil are identical. So that
can be seen from the diagram here. When you look at the major principal plane
here and in the effective stress conditions and the minor principal plane is here, it does
not change, here you can see that this is the major principal plane even in the total
stress conditions and here this is the minor principal plane that is the perpendicular
one in this case for the total stress condition also. And the pole is here from the total
stress condition and here also the pole is. So in a way what is changing is that
sigma dash theta which is actually nothing but sigma theta minus U.

So the U is that
pore water pressure which is there separating for the normal stress only, but whereas you
can see that the tow dash theta is equal to tow theta, which means that it will not affect
the effective shear stresses. So with this, we have discussed in
length about the Mohrís circles for and then pole interpretations and we connected two
Mohrís circles of the total and effective stresses. NPTEL
NATIONAL PROGRAMME ON TECHNOLOGY ENHANCED LEARNING NPTEL
Principal Investigator IIT Bombay
Prof. R. K. Shevgaonkar Prof. A. N. Chandorkar Head CDEEP
Prof. V.M. Gadre Producer
Arun Kalwankar Project Manager
Sangeeta Shrivastava Online Editor/Digital Video Editor
Tushar Deshpande Digital Video Cameraman
Amin Shaikh Jr. Technical Assistant
Vijay Kedare Project Attendant
Ravi Paswan Vinayak Raut Copyright
CDEEP IIT BOMBAY.

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