 Welcome to lecture series on advanced geotechnical engineering. We introduced ourselves to module 4 and which is on stress strain relationship and shear strength of soils and this is lecture 2 in module 4. So we have discussed about the contents and then introduction to the Mohr's circle analysis. So in this lecture we will try to do in the detailed discussion on the Mohr's circle analysis and its applications and determination of the pole in the Mohr's 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 the stress paths in PQ space. So as we have discussed the Mohr's circle is a geometric representation of the two-dimensional stress state and is very useful to perform quick and efficient estimations. And it is also popularly used in geotechnical fields such as soil shear strength that anyhow we are going to discuss and earth pressures wherein you know 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 and analyze the complex geotechnical problems and predict the soil behavior. And in this the pole point on the Mohr's circle is a point so special that it can help to really 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 Mohr's circle is a point so special that it can help to readily find stresses on any specified plane by using diagram instead of complicated computation. Now this coming to the pole points a pole is a unique point located on the periphery of the Mohr's circle, the point of intersection of Mohr's stress circle and line drawn to the pole parallel to a given plane gives the stresses on that plane. So the point of intersection of Mohr's stress circle and the line drawn to the pole parallel to the given plane gives the stresses on that plane. So this pole point is also called as the origin of planes. So there are two ways the pole points can be established one is by relating to the direction of action of stresses the other one is the relating to the direction of the planes on which the stresses are acting. Suppose if there is a horizontal plane if there is a normal stress acting on that then we are referring an 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 actually referring as the direction of the plane on which the stresses are acting. Usually the pole point relating to the direction of the planes are in use, usually the pole point relating to the direction of the planes are in use, the direction of the planes whether it is horizontal or vertical is in use. Now here consider the procedure you know to establish a pole point for stresses. So what we need to do is that we have to draw the Mohr circle on the tau sigma space by knowing you know the sigma z and tau zx and sigma x and tau xz. So these are actually sigma x is normal to the vertical plane you can see here and sigma z is you know horizontal is actually normal to the horizontal plane that is sigma z here and this is tau zx. So if you look into this about the center of the element if you take moments we get that tau xz is equal to tau zx because we have to maintain the equilibrium of the element from the moment equilibrium point up here. So project the line so with by knowing these two points draw the Mohr circle on the tau sigma space then from this point you know sigma z tau zx is acting you know in the direction of the vertical direction drop a line wherever it actually cuts the Mohr circle and that is actually pole point from the stresses. So in this direction if you are able to cut then this is actually pole point from the stresses. Similarly next if you look into this from here you know when you have the sigma x which is in the acting in the horizontal direction the direction is in the horizontal direction but it is acting on the vertical plane then you know you actually get a point here which is again same PS. So this is actually pole point from the stresses the production of the line from the point sigma x tau xz in the line of action of sigma x that is horizontal till it intersect the circumference of the circle or projection of the line from the point sigma z tau zx in the line of action of sigma z vertical the both actually gives the pole point with reference to the one method of establishment that is actually called pole point for stresses. The pole points for the planes that means that the projection of the line from the point sigma z tau zx so here you can see that by following the usual procedure we actually have drawn the Mohr circle and here the sigma z and tau zx and sigma x tau xz so from this point so this is sigma z is actually 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 you know the which is you know parallel to this horizontal plane and till it intersects the Mohr circle and that point is regarded as the pole point. So 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 the sigma x tau xz here so here this is you know sigma x tau xz is you know this plane is vertical here so in the direction of this vertical plane draw the line till it actually intersects the point here so it will be again when you draw in this horizontal direction or in this direction you actually have a point here and from this point when you put this this is you know again the pole point from the planes. Now when we join this point with you know the major principal stress here sigma 1 the intercept with sigma axis at tau is equal to 0 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 you know from here to here and on this plane there is you know minor principal plane is acting on this plane now the major principal plane is acting. Now if you are actually 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 actually get the stresses that is sigma and tau, tau 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 pore stress circle when we said that mole points pole points for the planes and which is actually usually adopted wherein what we have to draw how to you know establish a pole point is that you know project the line from the point sigma z tau zx in the direction of the plane on which the stresses are acting till it intersects the circumference of the circle and the intersection point gives the pole point pp for planes. Now what we have further discussed is that when we have when we join this with the major principal stress magnitude here and this plane is nothing but the major principal plane on which the principal major principal stress is actually acting and here on this is the minor principal plane. So these principal planes are the planes on which these shear stresses are 0, principal planes are shear less planes are called principal planes and we have major principal plane and minor principal plane and in which actually in the two-dimensional case where by ignoring the intermediate principal stress what we say is that major principal stress and minor principal stress principal planes can be identified. Now from the pole of the Mohr's circle is defined as thus now if a line is drawn from the pole to a point on the circle where the stresses are tau i and sigma i then in that exact plane the line is parallel to the plane on which the tau i and sigma i act. So the definition whatever we have discussed is that if a line is drawn from the pole to a point on the circle where the stresses are tau i and sigma i then in the exact plane the line is parallel to the plane on which the tau i and sigma i act. Now use of the pole point if you look into is to basically to find out the stresses sigma c, sigma a, tau ac and tau ca on the plane on which is inclined at theta to the plane on which sigma z acts. So now we can see that this is you know the sigma z is actually acting on this plane and sigma x is actually acting on this vertical plane and there is tau zx is acting here and tau xz is acting here. So in the step one locate the pole point for the plane's extending point e horizontally. So what we do is that you locate the pole point that is we know that the stress state here is the sigma z tau zx and sigma x tau xz construct the Mohr circle with this as the diameter and from this you know the sigma z is actually acting on the horizontal plane. So extend this you know line project this line till it intersect the periphery of the circle and that is the circumference of the circle then that point is regarded as the pole point pp. Now what we do is that locate the pole point for the planes extending point e horizontally in the direction of the plane on which the sigma z is actually acting that is as horizontal so we have drawn the horizontal line. Then what we have to do in the step two is that draw a line parallel to the plane which sigma c acts. So now draw a line parallel to the plane on which sigma c acts. So this sigma c is actually acting at an angle let us say that certain inclination. So draw a line parallel to this line and here you draw a line. So extend a line from point d through the center of the circle till it intersects the circle. So from here draw the circle so that we get the sigma a and tau ac that is you know on this stress states and then we get sigma c and tau c a. So this is what we have written what 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 element at the d and let us say d dash here and these are the stresses at this state of the sigma c and tau c a which are nothing but here and here it indicates that sigma e, sigma a and you know tau ac. So when we join let us say again here to the major principle plane major principle stress here minor principle stress then this will become the major principle plane and this will become the minor principle plane. Now this is actually shown diagrammatically here once again use of the pole point to locate the stresses sigma and tau at an angle theta to the reference stress direction. So here what we have done is that we have got more circle on the tau sigma space and we with sigma z tau zx which are actually acting here and sigma x and tau xz by following the sign conventions which are defined yesterday we actually can defined in the previous lecture. We can actually draw a indicate this in tau sigma space and draw a more circle with this as the diameter. Then the sigma wherever it intersects the sigma axis this will become the major principle 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 you know the stress point on the more circle is actually now 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 write this vertical plane you will actually 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 sigma c and tau ca at an angle theta. So draw a line parallel to the plane on which sigma c and tau ca are acting. So with that what we get is that we get the stress states like sigma c from after locating d by drawing an angle theta below with the pp e then you know drawing you know d and d dash passing through the center of the more circle we actually get the stress state sigma a tau ac that is sigma a tau ac and then we get sigma c and tau ca. So by using this procedure like use of pole point p to locate the stresses sigma and sigma c and tau ca at an angle theta to the reference section is demonstrated in this particular slide. Now the fundamental relationships by the more stress circle can be discussed. So here a tau sigma you know in a space a more circle which is actually shown here where you know you can see that the stresses are actually acting sigma 1 and sigma 3 are acting on this sigma 1 is actually acting on OA plane OA and sigma 3 is actually acting on plane OB and which is inclined at an angle you know you can say that inclined at an angle theta. So what we need to do is that again by knowing sigma 1 and sigma 3 we can draw the more circle and from here to locate pp or op that is the pole draw the you know draw the from here draw the line parallel to this plane on which this is actually acting. So this is inclined at theta so draw a line and from here what we can actually draw is that if you look into this and the shear stress will be maximum here that is tau max is equal to sigma 1 minus sigma 3 by 2 which is nothing but the radius of this more 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 you know the major principal plane and this is the minor principal plane and you know the shear stress principal shearing stress is actually occurring at a the line joining this point to the point where the maximum shear stress occurs that is 45 degrees. So we have tau max which is on the you know the upper portion of the more circle and then bottom portion of the more circle. So the maximum shear stress often called the principal shearing stresses as a magnitude of sigma 1 minus sigma 3 by 2 and which equals the radius of the more 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 more circle. So the principal shearing stresses occurs on planes inclined at 45 degrees, the principal shearing stress occurs on planes inclined at 45 degrees, the maximum the principal shearing stress or the maximum shear stress occurs on planes inclined at 45 degrees. Now the another aspect is that the conjugate shearing stresses that means that you know the shearing stresses on planes at right angles to each other are numerically equal but are opposite sign. So numerically equal because the element has to be maintained equilibrium when you take a moment about the center of the element we get tau zx is equal to tau xz or tau xy is equal to tau yx or tau yz is equal to tau zy. So these stresses are called conjugate shearing stresses, so here the conjugate shearing stresses are represented here so we can see that if you are having these shearing stresses at this point and at this point so we call these are you know which is on the upper they are equal in magnitude but only thing is that they are opposite in the sign. So one is positive other one is negative here, so this is indicated here you know this is actually called these two are called as conjugating shearing stresses, so these are conjugating shearing stresses. Now the another aspect is that what is the obliquity and the resultant stress, so the resultant stress on any plane has a magnitude expressed by you know we have an element let us say sigma and tau, so the resultant is actually nothing but root over sigma square plus tau square at any particular point on the Mohr's circle and has an obliquity which is equal to tan inverse of tau by sigma. So if you look into let us say that this is the point C on the Mohr's circle which we are referring. Now what we do is that from the origin that is where you know the from here to this point, so this is actually called as the you know the maximum is the obliquity angle that is which is nothing but alpha is equal to tan inverse tau by sigma, this is tau and then this is this ordinate is sigma and this magnitude is nothing but root over sigma square plus tau square which is the resultant. So the obliquity angle of obliquity is nothing but alpha is equal to tan inverse tau by sigma. Now further if you look into it in this particular slide where it has been shown that you know we have got in the same situation that Mohr's circle but here if you look into this here when indicate the sigma 1 and sigma 3 acting at an inclination theta, so what we have drawn is that we located the pole and from here you know what we have drawn is that we try to determine the stresses on any plane at inclination theta, so that is this point. Point where it actually touches the you know the point where it is tangential to this line when you draw from the origin and if the point where it actually gets tangential and that point and this point when you join and you know when you look into that this is also in the beginning. So this angle is actually called as the this angle joining from this point to this point is actually called as the maximum angle of obliquity. So the maximum of all the possible obliquity angles on various planes is called maximum angle of obliquity which is alpha m the maximum of all possible obliquity angles on the various planes is called as the maximum angle of obliquity and the coordinates of the point of tangency are the stresses on the plane of maximum obliquity and is less than the plane of principle shear. So if you look into this the principle shear is nothing but the maximum shear stress. So you know here the maximum shear that means that the failure is most likely to happen but you know when you say that you know the limiting obliquity is that is a basically criteria which is used to you know initiation of the slip indicate the initiation of the slip. So the maximum obliquity is you know less than the plane of the you know you can see that the is less than the plane of the principle shear because since the limiting obliquity is the criterion of the slip and whereas in the plane of principle shear is actually liable to happen. So the coordinates of point of tangency are the stress on the plane of the maximum obliquity and this which is actually you know the angle which is joining to the maximum shear stress is actually less than this you know maximum angle of obliquity. So if you look into this you know 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 actually most likely to happen but here this point which actually you know indicates the maximum limiting obliquity angle is basically is the criterion which is actually specified for initiation of the slip and whereas in the plane of principle shear is actually is liable to happen. So now consider an example where you know we need to draw the most stress circle at failure on cylindrical specimen of stiff clay with a shear strength of 100 kilo Pascal's and if the radial stress is obtained constant at a maintain constant at 80 kilo Pascal's by using the pole point method find the inclination theta to the radial direction of the planes on which the shear stresses is one of the maximum shear stress and the determine the normal stresses acting on this plane. So we need to determine the magnitude of the normal stresses and acting on these planes and the condition which is given is that find the inclination theta to the radial direction of the planes on which the shear stress is one of the maximum shear stress. Now what has been given is that the maximum shear strength which is nothing but 100 kilo Pascal's so that means that the maximum ordinate is actually sigma 1 minus sigma 3 by 2 that is tau max is equal to 100 so with this as radius sigma 1 minus sigma 3 by 2 as the radius draw the Mohr circle on tau sigma space. So then we have got the maximum shear stress with 100 kilo Pascal's here minus 100 kilo Pascal's here and the circle when it actually intersects with sigma axis at sigma 1 it is 280 kilo Pascal's and sigma 3 is 80 kilo Pascal's so tau max is the radius of the circle so sigma 1 is equal to sigma 3 plus tau 2 tau max so tau max is actually is given as 100 so with that what we can actually find out is that sigma 1 so which is nothing but 280 so the diameter of the Mohr circle is nothing but 280 minus 80 that is 200 kilo Pascal's the radius is nothing but tau max is equal to that is sigma 1 minus sigma 3 by 2 that is 100 kilo Pascal's so the plot Mohr circle based on the above information and radius and two points on the circle. Once we do that now the condition has been given is that 50% of max shear stress a draw a line that means that when you are actually doing the draw a line which is intersecting the Mohr circle so draw a line at tau is equal to 50 kilo Pascal's which is half of tau max. Now as the principle stresses are acting on edges of sigma 3 so here this plane is actually horizontal because the cylindrical sample so sigma 1 is acting on the horizontal plane sigma 3 is acting on the minor principle plane so this itself will become like a PP is nothing but the pole here so from here draw the line it intersects at point where sigma 3 is identified and that is point pole point PP. Now from here what has been asked is that the inclination of the plane on which tau is equal to tau max by 2 which we by drawing a line from here this point is actually indicated by the condition which we have given and with that we can actually get on this 15 degrees and at this particular point we can actually get sigma n 15 as 267 kilo Pascal's and another ordinate is nothing but 50 kilo Pascal's then when we can see that when we draw a line at 75 degrees now possible that because this point actually intersects at these two points so when you draw the line here from the 75 degrees with horizontal it is at 75 degrees with horizontal and this magnitude of the stress sigma 1 75 degrees indicates that 93 kilo Pascal's so draw the line from the pole to intersect the intersection 50 kilo Pascal's a line on the and the Mohr circle so what we have actually based on the given data we try to establish determine the normal stresses acting on these planes. Now here with the given information two planes are possible so we could actually get for this same shear stress two different normal stresses can be seen one is at plane inclined at 75 degrees the other one is actually at the 15 degrees. Now consider an example problem where the Mohr circle of the total stress and element having a size of 40 mm by 40 mm is a cube so a two dimensional state has been actually represented here and the x axis and z axis are actually shown within the element and for elucidating the shear forces and all the with a mustard color filled rectangles are actually shown basically to show we can see that this is conjugate shearing stress shear force F4 and this is the conjugate shear force F3 and so the shear the element is actually subjected to shear like this and these are the normal stresses and these are the normal forces F2 and this is actually normal force F1. So in the given figure the normal loads are applied to the phases of the soil cube having 40 mm by 40 mm and 40 mm dimensions and F1 is 0.45 kilo Newton F2 is 0.3 kilo Newton and the shear loads are F3 is equal to F4 is equal to 0.1 kilo Newton the sides of the soil cube are each one is actually having 40 mm the construct the Mohr 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 phase is nothing but 40 mm by 40 mm in meters it is when we convert into millimeters into meters it is area of the each phase is 0.016 meter 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 this plane divided by F4 divided by that area and similarly F3 divided by this area you will get the shear stresses of the sample. So now what we do is that we 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 tau xz is equal to tau zx is equal to 0.1 divided by that is due to F3 and F4. So with that we have got the data which is given based on the given data we establish a sigma z and sigma x and tau xz and tau zx which are actually acting on the element. Now what we have done is that by knowing the stress states here that is the magnitudes are you know little bit different but however they can be ignored and so here with this stress state like 281.25 kilo Newton per meter square and 622.5 kilo Newton per meter square and this point as well as this point by the appropriate using the sign convention we can actually draw the Mohr circle and on the tau sigma space which is drawn to the selected scale. Then what we can do is that from this because it is actually acting on horizontal plane so draw a line to locate the pole. So this is the point P which is the pole. Now this is the major principle stress the ordinate is with the given dimension which are given in the figure the sigma 1 magnitude is 306 kilo Pascal and sigma 3 is 154 kilo Pascal 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 principle plane 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 that is about 116 degrees you know 90 plus 26 and this is nothing but the minor principle plane. So in the given problem we have been asked to you know calculate the magnitude of the major principle plane and minor principle planes and also the direction of the planes which are indicated in this particular problem which is the solution which is actually given for the problem. Now another example here if you look into it there is an element which is actually subjected to a cylindrical element subjected to you know the minor principle stress is 12 kilo Pascal's and major principle stress is you know 52 kilo Pascal's you can see that these elements are not having any shear stresses so hence they can be called as you know shear less planes. So this is major principle stress and minor principle stress the major principle stress is actually acting on the horizontal plane and minor principle stress is actually acting on the vertical plane. So as can be noted here the sigma 1 with the sigma 1 52 kilo Pascal's and sigma 3 so with 52 – 12 that is about 40 kilo Pascal's as the diameter draw the you know Mohr circle once you draw the Mohr circle so this is the point A which is on the sigma axis which is 52 kilo Pascal's is actually indicated here this is the minor principle stress. So this is actually acting on the horizontal plane so from here draw a line to locate the pole P so this is the you know pole P now what we need to do is that we need to you know let us say that the angle alpha is 35 degrees. So on this you know this angle 35 degrees when you draw from this point to this point so point intersects at C so at point C whenever it actually intersects the Mohr circle we can actually interpret you know the sigma A and tau A so that is nothing but 39 kilo Pascal's and this is 18.6 kilo Pascal's. So what we have actually got from this problem is that we actually have for a given plane which is actually inclined at alpha is equal to 35 degrees within the element. So first we have drawn the Mohr circle based on the information which is given that is sigma 1 is given 52 kilo Pascal's sigma 3 is given 12 kilo Pascal's so with 40 kilo Pascal's as a diameter we have drawn a Mohr circle and then you know we have indicated identified the pole P and from the pole P you know we can actually draw a line at 35 degrees suppose if it is let us say 60 degrees then that will be line will be located at such is that point this point then you know this is the point where you know you can actually get the stresses in the element at the onset of failure at a particular plane of inclination and another important point is that you know in this the major principle plane is this and minor principle plane is actually indicated by a line perpendicular to this major principle plane. So major principle plane is in horizontal here and minor principle plane is here which is perpendicular to the major principle plane. Now the same procedure is actually 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 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 and from there we can actually calculate the sigma alpha which is 39 kilo Pascal's and tau alpha is equal to 18.6 kilo alpha Pascal's which are actually acting at the along the plane which is inclined at 35 degrees with the major principle plane. Now consider similar example only thing is that in this particular case the element in space is inclined at 20 degrees that means that the element is actually considered within the you know space with 20 degrees inclination with the horizontal. So the element under same stresses have been considered but only thing is that this element in space is actually subjected to an inclination of 20 degrees and we again we are interested in you know this along 35 degrees plane what are the state of the stresses. So with the given information now with 52 kilo Pascal's and 12 kilo Pascal's now you know 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 principle stress is actually acting is changed now. So because of that so parallel to this plane you know the location of the pole point changes previously the pole point was here when it was horizontal now it got lifted up so you can see that draw a line from AP and where it actually interacts so the pole is actually located here. So from here if you put this 35 degrees then again you know it will intersect at sigma n is equal to 39 kilo Pascal's and tau n is equal to 18.6 kilo Pascal's so this sigma one here is the major principle plane now and this is the minor principle plane with the magnitudes which are actually given like 12 kilo Pascal's and 52 kilo Pascal's so this is the major pencil plane and this is mine principle plane. So the resulting sigma alpha is equal to 39 kilo Pascal's and tau alpha is equal to 18.6 kilo Pascal's are same because nothing has changed except the orientation of the space of the element. So when you have the orientation of the element is actually 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 take another example wherein the stresses are actually shown in the figure below, wherein you can see that this is element and which is subjected to this is the horizontal plane where sigma v is equal to 6 mega Pascal's which is actually subjected here and we have got tau is equal to 2 mega Pascal's here 2 mega Pascal's and this is minus 2 mega Pascal's here this is the you know this element is under tension and this element is this is actually the shear stress which is minus 2 mega Pascal's here and minus 2 mega Pascal's here. So we have the in the vertical direction it is compression and the lateral direction it is you know we have the element is under tension. So what we need to determine is that evaluate the stresses sigma alpha tau 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 30 degrees. So we need to calculate what is the location the maximum principal stress and minor principal stress. So the procedure is that plot the state of the stress on the horizontal plane that is 6 to at point A and note that the shear stress makes a clockwise moment about A. So the shear stress is actually making clockwise shear stress is actually making clockwise moment about A. So from the consideration which we have defined in the previous lecture shear stress makes a clockwise moment about A. So therefore when the clockwise moment when the shear stress is actually making clockwise moment then it is actually regarded as positive and if you look into this here for the A is the element outside the A is the point outside the element. So about this if it actually makes a clockwise moment it is regarded as the positive. So in this case if you consider element A it is actually making clockwise moment but when you take this particular one this shear stress is actually making counter clockwise moment about this a point within outside the element. So this is actually regarded as negative. So this point is actually is negative and this is negative and again here this is making clockwise moment so this is actually positive. So these two shear stresses are positive and these two stress are negative. So plot the point B minus 4 minus 2 the shear stress on the vertical plane is negative since it makes counter clockwise moment. So clockwise shear stresses are positive so and counter clockwise shear stress are negative. So point A and B are two points on the circle so construct the Mohr's circle with center at 1 0 that is 1 mega Pascal's and 0 ordinate. Find the pole by drawing the horizontal line through A and R vertical line through B and find the state of the stress on the plane inclined at alpha that is 30 degrees and the draw line PCA. So this is actually explained here so what we have done is that we know the stress states from the defined sign conventions we have located the A that is 6 2 which is written here plot the state of the stress on the horizontal plane 6 2 at point A. Note that the shear stress makes a clockwise moment about A and therefore it is positive so that we have located and now to locate point B what we have said is that plot point B minus 4 minus 2 the shear stresses on the vertical plane is negative since it makes a counter clockwise moment so we have located here. So with this as the diameter and with center as 1 0 we draw a Mohr's circle on the tau sigma space. So now we can see that the origin is here then the Mohr's circle actually I have attended to the negative portion also. Now what next step we need to do is that after having drawn the Mohr's circle find the pole by drawing the horizontal line through A so draw the from this known stress state here draw the line horizontal because this is on the known stress state is here draw the line parallel to this plane. So it intersects the Mohr's circle at point B or you can draw a vertical line through this parallel to this plane on which the stresses are acting. So it intersects at the P so this is the pole P. So from here what we have been asked is that calculate the stresses at an inclination 30 degrees within the element. So from P with PA draw a line which actually intersects the PC so the point C is nothing but we get this stress state that is sigma n is 1.8 mega Pascal's and the tau A is actually 5.3 mega Pascal's. 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 mega Pascal's and which is m and m dash we actually have got these points here this is the minimum response stress this is the maximum shear and this is also a point of maximum shear. So this inclination of PM is actually 34 degrees and this magnitude of this is actually about 5.4 mega Pascal's. Now the next issue is that to draw you know line passing through this point this is nothing but the sigma 1 and this is sigma 3. So with this we can actually find out what is the magnitude of the sigma 1. So this sigma 1 magnitude is 6.4 mega Pascal's and this is nothing but in tension that is sigma 2 is equal to minus 4.4 mega Pascal's. So this is the you know minor principal plane and this is the maximum principal plane the inclinations with the planes are actually indicated here. So in the figure what have been asked is that we actually need to find out the state of the stress when the element is actually having a plane inclined at alpha is equal to 30 degrees and lines drawn from P to sigma 1 to sigma 3 establish the orientation of the major and major minor principal planes that we have actually done through here in this particular exercise in the problem where we have actually identified and calculated what is the maximum principal stress and minor principal stress and also you know when the given element is actually subjected to different state of stresses and we have actually tried to understand what is the you know how we can actually determine and interpret the several parameters very efficiently. Now as we know that the total stress is nothing but effective stress plus pore water pressure. So at a given condition you know we can actually draw a different Mohr circles for total stresses and effective stresses. So here you know in this particular slide wherein we can actually see that Mohr circle for total and effective stresses. So here on a tau sigma plot if you look into 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 we have sigma 1 as the major principal stress sigma 3 as minor principal stress with this sigma 1 minus sigma 3 as the diameter draw the Mohr circle. Now you know this is the again the pole point here. Now you know from this point at an inclination theta you can actually draw the element where sigma 0 and tau 0 tau theta are the you know the stresses on that particular point and that particular plane which are actually acting sigma 3 is acting perpendicular to this plane and tau theta is actually acting on this plane along this plane. So but when we use sigma 1 dash is equal to sigma 1 minus u where u is the pore water pressure and sigma dash 3 is equal to sigma 3 minus u when we do that the circle actually shifts to the left hand side. So in case of effective stress circle which is on the left hand side of the total stress circle and with the magnitude of the you know effective major principal stress is nothing but sigma 1 dash is equal to sigma 3 sigma 1 minus u and as effective minor principal stress is nothing but sigma 3 dash is equal to sigma 3 minus u. So the difference is nothing but you know u again when we actually locate the pole point the pole will not change and from this inclination theta and this stress state sigma 0 dash theta and tau dash theta we can see that the difference between these two these two horizontally these two are separated by a you know distance u and you can see that both in effective stress circles and total stress states you know the more circles are actually having identical diameters. So we can actually 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 tau dash theta and sigma dash theta are effective stresses acting on the plane inclined at angle theta by examining the circles so we note that tau dash theta is equal to tau theta and the sigma dash theta is equal to sigma theta minus u. So what we can actually conclude is that for a given state of total stress changes in pore water pressure have no effect on the effective shear stresses they alter only the effective normal stresses. So one important conclusion which we have deduced here is that for a given state of total stress the change in pore pressure have no effect on the effective shear stresses they alter only the effective normal stresses. So this is actually represented here tau dash theta is equal to tau theta where you can see that the vertical ordinate you know this is identical to that of though the magnitudes are different the vertical ordinate is actually tau dash theta is equal to tau theta that is indicated here and also sigma dash theta is equal to sigma theta minus u. 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 tau 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 tau dash theta is equal to tau theta and the sigma dash theta is equal to sigma theta minus u. So in the for a given state of the total stress changes in the pore water pressures 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 you know how a you know the more 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 that how we can actually determine the major principle planes and the minor principle planes and we can actually say that the principle planes are the ones where the shear stresses are 0 that is shear less planes are called principle planes. So while determining the shear strength of a soil you know if you are able to create these conditions then they are eligible to be called as the shear less planes or principle planes and which is you know we will be discussing in the forthcoming lectures while discussing about the drag shear test and then further we have discussed that the more circles of the total and effective stresses the effective stress circle and total stress circles the diameters are identical and again you know there is no there is reason why as such there is no tau dash theta. So tau dash or tau is equal to tau because it is equal and only thing what actually gets you know different is that the sigma dash theta will actually change it as sigma dash theta is equal to sigma theta – u. So for a given state of the total stress the changes in pore water pressure have no effect on the effective shear stresses and they are only alter the effective normal stresses. So we have discussed you know about the more circles 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 have no effect on the effective shear stresses they alter only the effective normal stresses because from the demonstration here tau dash theta is equal to tau theta and sigma dash theta is equal to sigma theta – u. Further we can actually look that use the pole construction on the effective stress more circle to calculate the effective stresses on any plane is exactly same way as we use the pole construction to calculate the total stresses. So if you are having the effective stress conditions and effective stress more circle also the same way we can actually use the pole construction method and another thing is that the position of the pole in the more circle of effective stress is same as in the more circle of the total stress and the location of the principle 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 actually the procedure is same and the principle planes you know in the effective stress conditions and total stress condition they remain same, that means that the position of the pole in major principle major circle of effective stress is same as the more circle of total stress and the location of the principle planes of total and effective stresses in soil are identical. So the positions of the pole in more circle of effective stress is same as in the more circle of total stress and the locations of principle planes of total and effective stresses in the soil are identical. So that can be seen from the diagram here, when you look into this here the major principle plane is here and in the effective stress conditions and minor principle plane here it has not changed. Here you can see that this is the major principle plane even the total stress conditions and here this is the minor principle plane that is a perpendicular one in this case for the total stress conditions also and the pole is here for the total stress condition and here also the pole is here. So in a way what is actually changing is that sigma dash theta which is actually nothing but you know sigma theta minus u. So the u is that pore water pressure which is you know the separating for the normal stress only but whereas you can see that the tau dash theta is equal to tau theta that means that you know it will not affect the effective shear stresses. So this you know with this we actually have you know discussed in length about the Mohr circles for you know and then pole interpretations and we connected to you know Mohr circles of the total and effective stresses.