 Welcome to the lecture series on advanced geotechnical engineering, we are introducing ourselves to module 4 stress strain relationships and shear strength of soils. So this module 4 and this is the lecture 1 of module 4 on the stress strain relationship and shear strength of soils. And we are going to cover this contents of this module in the following way, first we will introduce ourselves to stress strain and then we will try to understand about the Mohr circles and Mohr circle analysis and how to identify a pole and the principal stress space and the stress paths in PQ space and then we will try to discuss about the Mohr Coulomb failure criteria and its limitations and correlations with PQ space. Then it will be followed by the stress strain behavior, isotropic compression and pressure dependency, confined compression and large stress compression and definition of failure, interlocking concept and its interpretations and drainage conditions. Then we will actually introduce ourselves to different types of the shear test which are actually available to determine the shear strength in the laboratory like we have direct shear test and the triaxial test. And the triaxial behavior we will try to discuss in depth and then the stress state and analysis of unconfined compression, unconsolidated undrained, consolidated undrained and consolidated drained triaxial stress with special conditions and the special test particularly the extension test for tension and you know we will be discuss and thereafter we will try to concentrate on the stress paths in triaxial and octahedral plane, elastic modulus interpretations from the triaxial test and how this can be used further in understanding you know analysis of the geotechnical structures. So we have broadly divided this into the following you know contents where in first we will introduce ourselves to Mohr circles and then we will try to do you know the identification of the pole and principal stress space and stress paths in PQ space and failure criteria we will introduce and then you know we will try to look into all these contents in a you know in a systematic manner. The concept of the stress if you look into it if you take a cylindrical sample and if that if you consider this is the so called plane then the stress is actually is you know nothing but defined as a force by area. So defined as the force so this area is you know is collectively from the grain to grain contacts as well as the void area it is also called as the engineering area we have discussed with this while discussing seepage and you know the permeability. So defined as the force per unit area that is force for the that is the internal resistance per unit area so cannot be measured directly and gives no indications how forces are transmitted to the stress material and the manner of transfer of forces in solid crystalline material is different from point to point contact transfer like materials like in material transfer in materials like soil. So the manner of transfer of forces in solid crystalline material is different from point to point grain contact transfer in materials like soil. Now consider here a simple axial stress which actually has been applied with you know the compressive force F on acting on plane PQ in another figure which is there on the right hand side where in a plane PR which is inclined at an angle theta. Now we can actually divide you know we have said that on plane PQ the stress is nothing but sigma is equal to you know F by A but on plane PR if you look into it we get you know the T is equal to F sin theta that means that you resolve F components into horizontal component as with the vertical component then we get T is equal to F sin theta and then with that we actually get the normal force sigma n theta is equal to n by A by cos theta because A by cos theta is nothing but the area of the plane PR because it is inclined at an angle theta so that if A is that area that plane means PR so A by cos theta. So with that we can write that sigma n theta is equal to F by A cos square theta and tau theta is equal to we can write T cos theta that is the shear force acting on the area A with that we can write you know F by 2A sin 2 theta. So with this we actually have got stresses on the plane PR sigma n theta and tau theta and where in the derivation is worked out like this you know the A theta is nothing but A by cos theta and n is equal to F cos theta and n is equal to F cos theta and T is equal to F sin theta that is the F is resolved with an angle theta in horizontal component and vertical component and therefore sigma n theta is equal to which is nothing but the normal force divided by area area is nothing but A by cos theta n by A by cos theta which is nothing but n is equal to F cos theta if you substitute that we get F cos square theta by A. Similarly and tau theta is equal to T by A by cos theta so T is equal to nothing but F sin theta. So we can write F sin theta cos theta by A and with that we can write F sin 2 theta by 2A because sin 2 theta is equal to 2 sin theta cos theta. So we have written tau theta is equal to F sin 2 theta by 2A. Now you know in order to locate the maximum shear stress and we can actually find out you know the angle of the plane over which this mass you know shear stress is maximum. So that we obtain by differentiating this tau theta with theta d tau theta divided by d theta is equal to 0 for maximum value of tau theta we get d tau theta by d theta is equal to F by A cos theta. Now for equal to 0 F by A cos theta is equal to 0 then cos theta is equal to 0 for theta is equal to 45 degrees or 135 degrees and tau theta maximum is nothing but you know F by 2A. So tau theta will be maximum either at 45 degrees angle or you know homogeneous for homogeneous material there will be another failure plane that is orthogonal to that and that is 45 and then plus 90 is 135. So tau theta max occurs on a plane with 45 degrees inclination with sigma and theta plane. So that maximum shear stress occurs you know on a plane which is inclined at 45 degrees you know that is we obtain by differentiating tau theta and equating it to 0 and then for conditions of with cos theta satisfying the conditions of cos theta is equal to 0 we have got theta is equal to 45 degrees or 135 degrees with that tau theta max is nothing but F by 2A. So this variation of the normal stress sigma n theta and a shear stress tau theta with angle of plane theta in cylindrical test specimen is actually plotted here and this is the normal stress normalized here. So you can see that sigma n theta is maximum at theta is equal to 0, sigma n theta is equal to 1 for this is the normalized value which is equal to 1 and 0 at 90 degrees and again subsequently it is maximum at 180 degrees whereas so this is actually variation for how the sigma n theta varies. So this is obtained based on the equation sigma n theta is equal to F by A cos square theta and this is obtained by using the equation tau theta is equal to F by 2A sin 2 theta wherein tau theta is maximum it occurs when theta is equal to 45 degrees. So you can see that this theta is equal to 45 degrees means the F by 2A that is actually is maximum here this is F by A. So the tau theta is maximum which occurs on a plane with theta is equal to 45 degrees and tau sigma n theta maximum on a plane with theta is equal to 0 degrees. Now let us do a small example for the simple axial stress a cylindrical specimen of rock say having 75 mm in diameter and 150 mm in height is subjected to an axial compressive force of 10 kilo Newton. So we need to find out find the normal stress sigma n theta and the shear stress tau theta on the on a plane inclined at 30 degrees to the radial direction and the maximum value of the shear stress and inclination of the planes on which the shear stress tau theta is equal to one of the tau theta max. So the area is nothing but 75 mm diameter so with that we can actually get the pi r square which is nothing but 4.42 into 10 to the power of minus 3 meter square. So sigma n theta which is given by F cos square theta by A so with this you know we can write F into F means 10 cos square and we have been asked to determine and plane 30 degrees to the radial direction. So F cos square 30 divided by area that is 4.40 into 10 to the power of minus 3 so we get 1696 kilo Pascal's and tau theta is nothing but F by 2A sin 2 theta that is nothing but 10 by you know 2 into 4.42 into 10 to the power of minus 3 into sin 60. So with that 979 kilo Pascal's is the tau theta. But if you look into this the tau theta maximum, tau theta maximum which is nothing but which occurs at 45 degree with if it is horizontal which is 45 degrees or tau theta maximum is equal to F by 2A. So F is nothing but here 10 kilo Newtons divided by 2 into 4.14 to the power of minus 3 which that we have got 1131 kilo Pascal's and then we have been asked to find out the inclination of the plane on which the shear stress tau theta is equal to one half of the tau theta max. So this is you know the inclination of the plane at which actually the shear stress is actually the tau theta max is half of one tau theta maximum. So tau theta maximum which is you know tau theta, so tau theta max by 2 is equal to tau theta max sin 2 theta. So with that we actually get sin 2 theta is equal to 1 by 2 so which is nothing but theta is equal to 15 degrees or 75 degrees, theta is equal to 15 degrees or 75 degrees. This is the you know the 15, this is the inclination 15 degrees or 75 degrees the inclination of the planes on which the shear stress tau theta is equal to one half of the tau theta max. Now let us again consider in order to formulate the Mohr's circle you know consider a body which is subjected to number of forces like which is shown here a body which is having in axis passing through O inclined at an angle alpha and subjected to forces F1, F2 and so on to F6 forces acting on a body and in a 2D plane. So the point of the application of force within a soil mass could be on a particle or a void. So the point of application of the force within a soil mass could be on a particle or in a void. As the void cannot support any stress the stress is nothing but F by A where A is the gross cross sectional area which includes both grain to grain contact as well as voids. So which includes both grain to grain contact as well as voids. So as void cannot actually support any stress the stress is F by A where A is the gross cross sectional area both grain to grain contact as well as voids. So what we do is that these resolution of the forces if you do F1, F2, F3 into normal and shear components of acting on a plane passing through point O at an angle. So the expanded view of an element at O is actually shown here, the point O is here and if you consider you know the expanded view of an element and it actually shows like the element ABC where and AB on plane where the vertical force V is acting and which is nothing but you know if this area is say unit area and this is inclined at alpha then the area of this one is nothing but area is equal to 1 cos alpha, this area is nothing but 1 sin alpha. So 1 sin alpha, 1 cos alpha is the areas of AB and areas of BC and areas of AC is 1. So with that what will happen is that the vertical force is nothing but stress into area and horizontal force is nothing but horizontal stress into area. So the horizontal forces are written here and the stresses are also equivalent stresses are also written here, this is the normal force or normal stress and this is the shear force or shear stress. Now here we need to observe here particularly here in this the sign conventions need to be followed consistently. So the compressive stresses are actually positive because in the majority of the geotechnical engineering the stresses are in compression nature. So because of that the stresses are compressive stresses are positive in nature and the tensile stresses are negative and the positive shear stresses produce counter clockwise couple on the element or clockwise moments about a point outside the element. That means that if you are having a shear force acting like this, this is said to be positive if it is producing a counter clockwise moment about a point outside the element. That means that if it is producing a counter clockwise moment about a point outside the element. So if it is a clockwise moment about a point outside the element then the shear stress is said to be positive. So let us consider that if the shear stress is actually acting in this direction downward direction then in that case it will be negative. So the positive shear stresses produce counter clockwise couple on element and or clockwise moments about a point outside the element. So with this what we do is that we try to resolve the forces in horizontal direction and vertical forces and then we try to obtain the equilibrium conditions and then we try to simplify and see that what is the deduction which we are going to get. So at equilibrium the sum of forces in any direction must be 0. So we resolve in horizontal direction and vertical direction with that sigma fh is equal to 0 where h minus t cos alpha minus n sin alpha. That means that the components have been taken h is this horizontal direction is positive because this direction is positive and minus t cos alpha. So that is this portion has been taken t cos alpha that is this force and then we have n sin alpha that is due to the normal force here on the plane Ac. Now similarly sigma fv is equal to 0 where v minus t sin alpha minus n cos alpha is equal to 0. Now by dividing by the respective areas the stresses on the alpha plane that is the plane Ac are the normal stresses sigma alpha and tau alpha. That means that the stresses are here sigma alpha that is the normal stress on the plane inclined at an angle alpha and tau alpha that is the plane inclined at an angle alpha with the horizontal that is AB. Now the stresses on this alpha plane are the normal stresses sigma alpha and tau alpha. So sigma x sin alpha because that is the horizontal force minus tau alpha cos alpha minus sigma alpha sin alpha is equal to 0 similarly sigma y cos alpha plus tau alpha sin alpha minus sigma alpha cos alpha is equal to 0. So this is regarded as a by solving a so then we get sigma alpha is equal to sigma x plus sigma y by 2 plus sigma x minus sigma y by 2 cos 2 alpha and tau alpha is equal to sigma x minus sigma y by 2 sin 2 alpha. So we have got sigma alpha tau alpha stresses with the by knowing the application of the stresses the element is experiencing and on a plane which is inclined at alpha. So now these sigma alpha is equal to sigma x plus sigma y by 2 plus sigma x minus sigma y by 2 cos 2 alpha tau alpha is equal to sigma x minus sigma y by 2 sin 2 alpha. So by squaring and adding we get the equation of a circle. So by squaring the sigma alpha and tau alpha sigma alpha square plus tau alpha square and we get the equation of a circle with a radius sigma n x minus sigma y by 2 and at its center at sigma x plus sigma y by 2, 0. So when a circle is plotted in tau sigma space it is known as the so if we are this is actually the definition where when the circle is plotted in a tau sigma space it is known as the more circle of stress it represents the state of stress at a point in at equilibrium and it applies to any material not just soil and note that the scales are for the tau and sigma have to be same to obtain a circle. So whatever from the reduced derivation we actually have obtained sigma alpha and tau alpha in terms of sigma x and sigma y and alpha and by squaring and adding we actually have got a form of equation which is representing the circle and that circle is having a radius of sigma x minus sigma y by 2, sigma x is the stress applied in the x direction, sigma y is the direction applied in this stress applied in the y direction and with a center at sigma x plus sigma y by 2, 0. So when this circle is plotted on tau sigma space it is known as the more circle of stress. So it represents the state of stress at a point at equilibrium and it applies to any material not just soil. So the most stress circle is also called as two dimensional this is in two dimensional this is graphical representation of stress relationship at equilibrium and they are discovered by Cullman in 1866 and developed in detail by Moore in 1882. So this is the development actually goes back into the solid mechanics and then it is used in soil mechanics. So the stresses are represented in the form of a circle so considering any point P x y on the circle equation of the circle can be written as x minus s whole square plus y square is equal to r square where r is the radius of the circle and x and y are the coordinates of point on the circle and s is the horizontal distance of the center of the circle from the origin. So by getting this you can see that this is the circle where this stress is actually sigma 1 so this magnitude from here to here measured is sigma 1 and this is on the tau 0 space and sigma 2 this is this ordinary sigma 2 and this is the radius of the circle so this is called as the Moore stress circle on tau sigma space. Now this deduction can be obtained like this you know the graphical derivation is that once again we can say that normal stress acting on any plane at an angle theta which is nothing but you know in terms of sigma n theta is equal to we can write like sigma x cos square theta plus tau xy sin 2 theta plus sigma y square sin square theta so and using sin square theta is equal to 1 minus cos 2 theta by 2 cos square theta is equal to 1 plus cos 2 theta by 2 and with that we actually get sigma n theta is equal to sigma x plus sigma y by 2 sigma n theta minus sigma x plus sigma y by 2 is equal to 1 by sigma x minus sigma by cos 2 theta plus tau xy sin 2 theta. So shear stress acting on any plane at an angle theta is given by tau theta is equal to half sigma x minus sigma z sin 2 theta plus tau xy sin 2 theta. So tau theta is equal to half sigma y minus sigma x sin 2 theta plus tau xy cos 2 theta. So squaring and adding the equation so again this is actually one form of you know deriving we already reduce this but this also says that you know the equation which is actually reduced to a form of equation of a circle where sigma n theta minus s whole square plus tau theta square is equal to r square and this represents equation of a circle. So sigma n theta minus s that is you know this one plus tau theta square is equal to r square this is nothing but the radius term. So this is represented in the graphical form like this the graphical form when you interpret you will get the number of you know the unknown parameters can be reduced. So this is the circle with radius r at which with a center on tau sigma plot at point sigma is equal to s and tau is equal to 0 that is s is equal to sigma x plus sigma y by 2 which is also nothing but sigma 1 plus sigma 2 by 2. So because this is sigma 1 and this is sigma 2 so this is also given as sigma 1 plus sigma 2 by 2. Now sigma 1 sigma 2 are the principal stresses as tau is equal to 0 on the x axis. So shear less planes are called principal planes so sigma 1 and sigma 2 are the principal stresses as tau is equal to 0 on the x axis. So sigma y and the sigma z and tau z y tau z are the boundary stresses which helps to plot the circle and sigma n theta and tau theta are the normal and shear stresses on a plane at an angle theta to the sigma z plane and sigma n theta and tau theta can be found on more circle by traveling clockwise around the circle from the stress point to a distance 2 theta at the center of the circle and sigma 1 is at an angle alpha to the plane of sigma z. So these are the angle 2 alpha and this is 2 theta which is actually shown here. Now let us consider for some examples of the more circles for different conditions of the stresses that is biaxial compression, biaxial compression and tension and biaxial pure shear. So here when you consider this one with sigma 1 and sigma 2 both are actually compressive stresses acting on an element and the two dimensional conditions have been selected and the more circle represents in a towards the positive side you can see that they center at this particular point and in case here we have a compressive stress and elongation for the element in the horizontal direction that is the tension in the horizontal direction and here there are no normal stresses only the element is actually subjected to biaxial pure shear the element is actually subjected to shear like as shown here. So in this case the more circle is actually symmetrical about this tau sigma axis. So the biaxial compression these biaxial stresses are represented by a circle which plots in positive sigma space and passing through stress points sigma 1 and sigma 2 and on the tau 0 axis. So the biaxial stresses are represented by a circle which plots in positive sigma space that is the circle is actually plotted in a positive sigma space and with stress point sigma 1 and sigma 2 on the tau 0 axis that is these are the points which are actually acting on the tau 0 that this is the this plane on this tau is equal to 0 is tau 0 axis it is called where the stress here at this point where this meets the you know the sigma axis that is called sigma 1 here and where the circle intercepts this point is sigma 2. The center of the circle is located on the tau 0 axis at the stress point that is sigma 1 plus sigma 2 by 2 and the radius of the circle has a magnitude of sigma 1 minus sigma 2 by 2. So the radius of the circle has a magnitude of nothing but sigma 1 minus sigma 2 by 2 and the maximum stress also is sigma 1 minus sigma 2 by 2. So this is the more circle which is so biaxial compression the more circle will be in the on the tau sigma space it will be on the positive sigma space that is the positive sigma space that is this is actually called as positive sigma space. Then the center is located at the stress point that is sigma 1 plus sigma 2 by 2 so the center is actually located at a stress point sigma 1 plus sigma 3 sigma 2 by 2 and the radius of the circle has the magnitude of sigma 1 minus sigma 2 by 2 which is nothing but so the radius of the circle is nothing but the maximum shear stress ordinate which is nothing but the sigma 1 minus sigma 1 minus sigma 2 by 2 that is this is the maximum shear stress located here in the positive side and negative side. And biaxial compression or tension that means the element is actually subjected to both biaxial compression and tension in this case the stress circle extends into both positive and negative sigma space. So the sigma space which is towards the positive side is called as a positive sigma space and sigma space which is actually towards the negative side is called the negative sigma space similarly tau that is the tau space which is above is called as the positive space and below is called the negative space. The center of the circle is located on the tau 0 axis at stress point sigma 1 plus sigma 2 by 2 and has the radius sigma 1 minus sigma 2 by 2 and this is also the maximum value of the shear stress which occurs at the direction at 45 degrees to the sigma 1 direction and the normal stress is 0 in directions plus or minus theta to the directions of sigma 1 where cos theta cos 2 theta is equal to minus sigma 1 plus sigma 2 divided by sigma 1 minus sigma 2 by sigma 1 minus sigma 2. So cos 2 theta is equal to minus of sigma 1 plus sigma 2 divided by sigma 1 minus sigma 2. So for the biaxial compression and shear you can see that this is the compression and this is the tension no shear forces then in that case the Mohr's circle is actually extends to positive sigma space and then negative sigma space also and the center is sigma 1 plus sigma 2 by 2 and the shear force is sigma, exo shear force is sigma 1 minus sigma 2 by 2 and the radius is nothing but sigma 1 minus sigma 2 by 2 and this also the maximum value of the sigma 1 minus sigma 2 by 2 is also maximum value of the shear stress which occurs at the direction 45 degrees to the sigma 1 direction and the normal stress is 0 in direction plus or minus theta to the direction of sigma 1 where cos theta is equal to minus of sigma 1 plus sigma 2 divided by sigma 1 by sigma 2. And now a case of biaxial pure shear in this case the circle has a radius equal to tau xy which is equal in magnitude to tau yz but opposite in sign the center of the circle is at tau 0 and sigma 0 the principal stresses are sigma 1 sigma 2 are equal in magnitude but opposite in sign and they are equal in magnitude to tau zy, so the directions of sigma 1 sigma 2 are at 45 degrees to the directions of tau zy and tau yz. So the circle is represented like this wherein with that you know you have got sigma 1 the major the sigma 1 here ordinate here and sigma 2 here and this is the shear stress that is tau zy and tau z, tau zy and tau yz is represented like this tau yz and tau zy. So this is in case of the biaxial pure shear. Now consider an example where the stress on a circle and a soil mass shown in the following figure we need to determine the principal stresses using Mohr's circle and the element is actually subjected, so this axis is y axis and this axis is z axis and this plane is AB and this plane is AC, so this plane is horizontal plane and this plane is vertical plane and this is an inclined plane inclined at 45 degrees to plane AB and these are the shear stresses and these are the horizontal stresses. So here it is given as 50 kilo Pascal's in horizontal direction compression and this is also compression 100 kilo Pascal's and this shear is 25 kilo per meter square. So the magnitude of the normal and shear stresses on plane AC is shown in the figure, plane AC we need to determine what are the magnitude of the normal and shear stresses on plane AC which is actually shown in the figure here and the magnitude of the principal stresses using Mohr's circle. So the available information is that sigma y that is the stress acting on the horizontal direction that is the 50 kilo per meter square in the y direction and sigma z 100 kilo per meter square that is the stress in the z direction and then tau yz that is tau yz is 25 kilo per meter square the tau yz which is acting the shear acting on this plane is 25 kilo per meter square. So the step one in constructing the Mohr's circle is that mark a point sigma y tau yz and point q that is sigma z, tau zy on the tau sigma coordinate system. So we have a tau sigma coordinate system and we marked actually in the tau sigma space we marked point p and q. Then in the next step is that join p and q with a straight line then draw the circle. So the draw the circle considering intersecting the point of s axis and line pq that is the point o at the center and the distance op as the radius. So with this what we have got is that we have got a circle now that is the Mohr's circle and this is called as Mohr's circle. Now the principal stresses are nothing but the point where this plane where the shear stress is 0 principal stresses where the plane where is the shear stress is 0 that is where the Mohr's circle cross a sigma axis. So by measuring from the graph we can actually get the sigma 1 as 110.36 kilo Pascal's here and sigma 2 here as the 39.64 kilo Pascal's that is from the measurements of the graph. Now we can actually further find out angle 2 alpha that is 2 times the angle between sigma z plane and major principal plane and major principal plane is inclined at 22.5 degrees to the sigma z plane and minor principal plane is inclined at 1, 1, 2.5 degrees to the sigma z plane. So sigma z plane is nothing but the plane on which sigma z acts. So after this then plane inclined at 35 degrees to the sigma z plane and the stresses on the plane at 35 degrees to sigma z plane is obtained by the point so at this point so we need to get if you look into the problem magnitude of the normal and shear stress on plane AC which is shown in the figure. So this plane is inclined at 35 degrees so the 35 degrees is actually represented here 35 degrees to the sigma z plane and this is actually is 70 degrees. So 35 degrees to sigma z plane when you put that this is the point where we get the sigma n theta that this horizontal ordinate is sigma n theta and vertical ordinate is tau theta. So from the measurements of the graphically we can get sigma n theta is equal to 42.96 kilo Pascal's and tau theta is equal to 14.94 kilo Pascal's. So this how you know what we have got is that from the graphical interpretation we actually have got sigma n theta and tau theta for the type of the stresses which are actually given by using the Mohr's circle concept. Now the three dimensional stresses on a cubical element the elements are actually subjected suppose if you consider an element within the soil it is subjected to the following stresses in x and y and z directions and on each plane there will be one vertical stress and two shear stresses. So you can see that on this xy plane we have vertical stress and you know the shear stress tau xy and the tau zx which is actually acting here and similarly on this plane sigma y and then these are the shear stresses which are actually acting and similarly on this plane and on this plane or the stresses are shown on the visible planes clearly. So these are actually represented in the matrix form like this the three dimensional stress at a point can be represented as t alpha t sigma is equal to sigma x tau xy tau zx tau yz sigma y tau yx and tau xz tau xy and sigma x sigma terms are the normal stresses and tau terms are the shear stresses. So please note that sigma terms are the normal stresses the tau terms are the shear stresses and the total six terms are independent and then we have sigma x sigma y and sigma z tau xy tau yz and tau zx and then if the references axis are in the directions of 1, 2, 3 and which is nothing but the direction of the principal stresses then in the geotechnics we are actually suppose if you are having a cylindrical sample or a let's say cubical sample we have if it is in the major vertical stress direction sigma 1 then it is called sigma 1 as the major principal stress and sigma 2 as the intermediate principal stress and sigma 3 as you know minor principal stress. For a sample which is cylindrical in nature the sample will be having a vertical stress that is sigma 1 major principal stress and 2 the stresses parallel to the acting on the plane that is nothing but sigma 2 intermediate principal stress and minor principal stress and the cylindrical sample as sigma 2 is equal to sigma 3 we generally refer it as sigma major principal stress and minor principal stress that is sigma 1 as the major principal stress and the sigma 3 as the minor principal stress and the sigma 2 as the intermediate principal stress. So the Mohr stress circle particularly in the three dimensional stress conditions is a no simple method exist to draw a Mohr circle to represent the general case all normal and shear stresses acting on all the six phases of a cube. So two simple cases can be represented by using the Mohr circles as given below and one is that a cubical element having only normal stresses on its faces and a cubical element which has only normal stresses acting on pair of opposite faces and both normal and shear stresses on remaining two pairs of faces. So with this slide what we wanted to convey is that no simple method exist to draw a Mohr circle to represent a general case and all normal and shear stresses acting on all the six faces of a cube that means that all normal and shear stresses acting on all six faces of cube cannot be represented and two simple cases can be represented by using three Mohr circles, A which is the case A is the cubical element having only normal stresses and another case which is cubical element which has only normal stresses acting on pair of opposite faces and both normal and shear stresses on the remaining two pairs of faces. So it can be proved that the stress conditions on any plane within the element must fall within the shaded area but it is usually sufficient to be able to determine the stresses on planes which are perpendicular to at least one opposite pair of element boundary faces. So stresses on these planes lie on the circle bounding the shaded area so these stresses for the case A where block which is actually having only normal stresses you can say the sigma 1, sigma 2 and sigma 3 are shown here. So in this case the Mohr circle is actually represented as this is the you know circle and with sigma 1 and sigma 3 and sigma 2 is the intermediate stress that is the intermediate principle stress. So here in this case A it can be proved that the stress conditions on any plane with within the element must fall within the shelled area that is this shelled area and it is usually sufficient to be able to determine the stresses on planes which are perpendicular to at least one opposite pair of element boundary faces. So in case B it is represented where we are having normal stresses and at least two faces are actually having the shear stresses then in that case we draw three Mohr circles here this is the Mohr circle 1 and second and then third one. So now here first you are drawing this Mohr circle and then after knowing this as the major principle stress and this as the minor principle stress sigma 2 and sigma 3 then this is the third Mohr circle and the second Mohr circle as once we have this one and that is the minimum stress that is the third Mohr circle. So in case of case B which depicts a cubical element with compressive normal stresses acting on all six faces and shear stresses on two pairs of opposite faces. So again in this case the stresses on all planes within the element lie within the shelled area and with stresses on all planes which are perpendicular to at least one pair of element faces lying on one of the boundary circles. The sequence of drawing the circles consists of first drawing of locating the stress point sigma z tau z y and sigma y and tau z then drawing the circle that means that first we have to locate this sigma z and tau z y and sigma y and tau y z and drawing the circle one that is the first one is the drawing this circle one locating these two points and then dropping here perpendicular and these are the sigma z axis that is measured from here and this is sigma y. So this drawing this points then drawing the circle through these with the center on the tau 0 axis tau 0 axis so this is the first Mohr circle which is drawn the second one locates this locates the principal stresses sigma 1 sigma 2 so once we draw the circle one we have got the opportunity to identify sigma 1 and sigma 2. Then as the third principal stress is known now the circles 2 and 3 can now be circles 2 and 3 2 and 3 can be 2 and 3 can be that is 2 and then finally the third circle that is the with intermediate principal stress sigma 2 can be drawn. So this locates the principal stresses sigma 1 sigma 2 and as the third principal stress is known the circles 2 and 3 are drawn subsequently. So in this case sigma 1 is greater than sigma 2 and then sigma 3 so as it has been told in geotechnics it is actually convenience to use sigma 1 as the major principal stress sigma 2 as the intermediate principal stress sigma 3 as the minor principal stress. So let us look into a problem where in a piece of sandstone is cut into the shape of a cube with 100 mm long edges and the forces of 5 kilo Newton 10 kilo Newton and 20 kilo Newton are applied respectively uniformly and on normal to the three pairs of the faces of the cube. So a piece of sandstone is cut into the shape of 10 centimeter in size you know having the long edges of the size of the edges 10 centimeters and forces of 5 kilo Newton 10 kilo Newton and 20 kilo Newton are applied uniformly and normal to the three pairs of faces of the cube and evaluate the major intermediate and minor principal stresses in the rock and subsequently draw the more circles of the stress and we need to also find out what is the maximum shear stress in the rock. So by knowing the forces the element is actually subjected then you know we can actually calculate what are the you know what is the maximum shear stress within the rock and what are the major and intermediate and minor principal stresses the element which is the sandstone element or a piece of rock is subjected. Now for this you know what we take is that the area of the each face is A is equal to 0.01 meter square wherein there is nothing but 10 centimeter into 10 centimeter 100 centimeter square in terms of meters it is 0.01 meter square. This three principal stresses major principal stress sigma 1 is equal to 20 into 10 to the power of minus 3 divided by 0.01 that is 2 mega Pascal's and intermediate principal stress which is nothing but sigma 2 which is 10 into 10 to the power of minus 3 divided by 0.01 is 1 mega Pascal and minor principal stress sigma 3 is equal to 5 into 10 to the power of minus 3 divided by 0.01 that is 0.5 mega Pascal's. Now we have got sigma 1, sigma 2 and sigma 3 the stresses this we have calculated based on the sandstone which is with nothing but the force which has been subjected divided by that area of the face of that particular sandstone rock piece and with that we have got these major principal stress as 2 mega Pascal's and intermediate principal stress as 1 mega Pascal's and minor principal stress sigma 3 as 0.5 mega Pascal's and with this what we have got is that now on the tau sigma space where the tau the scale which is actually here is you know on the equal scale with the equal scale when you represent and we can write that on the sigma which is here. So the major principal stress being the sigma with this as sigma 1 and sigma 2 that is intermediate principal stress as 1 mega Pascal's so with 1 by 1 mega Pascal's by 2 that is 0.5 mega Pascal's as radius can draw a circle so with that we get this first circle. Then you know once we get this one then with this as sigma 2 that is 1 mega Pascal's and sigma 3 as 0.5 mega Pascal's so 1 minus 0.5 that is 0.25 mega Pascal's as radius we can draw another circle that is the this one and then one more circle is actually possible that is sigma 1 and that is the major principal stress and minimum principal stress that is 0.5 that means that 1.5 you know mega Pascal's as radius so we have drawn circle 1, circle 2 and then circle 3. So in this we can actually see that the maximum shear stress in the element which is actually subjected which is the radius of the largest Mohr circle so the maximum shear stress is nothing but the radius of the largest Mohr circle. So if you look into this the maximum shear stress which is actually given by this circle is nothing but 0.25 mega Pascal's and here this one is nothing but 0.5 mega Pascal's but however if you look into this circle which is having a 0.75 mega Pascal's as the radius so the maximum shear stress is yielded the element is actually subjected is about 0.75 mega Pascal's so what we have done in this particular problem is that we actually by knowing the forces subjected to the by the this the cubical shape sandstone piece we actually have calculated the what are the stresses acting on the major and intermediate and minor principal stresses and then we have drawn the Mohr circle of stress and then we also find out what is the maximum shear stress in the rock piece. So for this what we have done is that we actually have calculated the stresses with force by area and then we plotted of these stresses on the tau sigma space with drawing 3 Mohr circles where in order to deduce the maximum shear stress here the tau max is actually here indicated and this is the this circle actually yields the you know the maximum shear stress so the sum the for the type of forces which are has been subjected the rock piece actually has been subjected to a shear stress of about 0.75 mega Pascal's. So if we are having say this much shear stress and in order to ever to the failure along that particular plane then you know the material should have adequate shear strength. So the shear strength which is actually you know counters the shear stress which is actually resulted due to the external loading system. So in our further lectures what we try to look into is that how further we discuss about on this concept of this Mohr circles and the how to locate a pole and then how this different element conditions can be used in you know understanding about the stress states in a you know for the elements and then we deduce you know we connect ourselves to you know the stress parts in the PQ space.