 In this session, we will discuss about the failure plane and principal stress relationships. The learning outcome will be at the end of this session, students will be able to find the inclination of failure plane and principal stress relationship. Now, before we start, we know that the Mohr-Colem equation basically graphically represented here on x axis the normal stress and on y axis it is a shear stress and the equation is shear stress is equal to C plus sigma tan phi. So, here you see the two components are present a cohesion component and a friction component and this is shear strength here is nothing but this total ordinate. So, this total ordinate is nothing but the contribution of cohesion and the friction angle. So, taken together we get the shear strength. So, if I consider a small soil element present in a soil mass, then there exists certain stress system on this soil element. So, the enlarged view on the right hand side you can see the soil element is subjected to two principal stresses along the horizontal axis we have the minor principal stress sigma 3 and along the vertical axis we have the major principal stress sigma 1 and on any generally inclined plane we can find the normal stress and shear stress. So, you can get this particular equation from this Mohr-Circle concept. So, if I plot the Mohr circle corresponding to the given value of sigma 1 and sigma 3. So, we will get this Mohr circle which is shown here sigma 3 and sigma 1 and then if I say if I am interested to know the normal stress and shear stress on a plane inclined at an angle of theta, then that plane is to be drawn here from the pole point. So, this is the pole point where the two planes originate and from this point the inclined line of angle theta is drawn till it cuts the Mohr circle. Then wherever it cuts the Mohr circle then the coordinate of that particular point gives me a normal stress and shear stress present on that plane. So, from the graph we can find out normal stress. So, normal stress is nothing but this distance and the shear stress is nothing but this distance. So, suppose for normal stress for getting normal stress you have to measure this particular distance total distance. So, which will be equal to this distance that is from origin up to the center of the Mohr circle the distance is sigma 1 plus sigma 3 by 2 plus the remaining portion here. So, this remaining portion we can find as a cos component of this triangle. So, if I consider a triangle here at the center angle is 2 theta. So, this angle will be 2 theta because it is a property of the circle. So, if this line is making an angle theta here then when that intersection point is joined to the center again then the angle made by that line at the center is always 2 theta it is a property of the circle. So, we have a right angle triangle here having an angle 2 theta. So, this diagonal of this triangle is the radius of the Mohr circle which is sigma 1 minus sigma 3 by 2. So, the cos component of radius is this horizontal distance and sine component of this triangle is a vertical distance. So, I need a horizontal component which is the cos component of this diagonal. So, sigma 1 minus sigma 3 by 2 into cos 2 theta. So, this is added here this distance is added in the distance sigma 1 plus sigma 3 by 2 already you determine. So, the total gives me normal stress. So, this is what is the equation. So, normal stress on that plane is sigma 1 plus sigma 3 by 2 plus sigma 1 minus sigma 3 by 2 cos 2 theta. In the same way the vertical side of the triangle gives me shear stress and vertical side is a sine component of the diagonal. So, sine component is sigma 1 minus sigma 3 by 2 sine 2 theta. So, in this way the normal stress and shear stress can be derived from this Mohr circle. So, here this is what is a typical example where a soil mass is subjected to a failure say embankment where the slip is taking place along one particular plane and the strength and hollow pop that soil mass is shown here on the right hand side. So, if I consider the soil element here as shown in Y. So, since this particular element is not present on the failure plane therefore, the failure does not take place. So, whatever normal principle stresses are present on element Y accordingly if I draw the Mohr circle it will look like this because this Mohr circle would not touch to the failure non-low since this soil as or the soil element at Y has not subjected to failure whereas, if I consider the element present on the slip surface certainly it will touch to the failure non-low because this soil has failed here. So, the corresponding Mohr circle at X is drawn here with red killer and now this Mohr circle is touching to the failure non-low. So, therefore, we can identify the stable condition and the failure condition. So, if suppose the Mohr circle is below the failure non-low and if it is not touching to the failure non-low then that condition we call it as a stable condition and if the Mohr circle is touching to the failure non-low then that condition is called to do a failure condition. So, now what happens in real life initially there exist certain amount of stresses in horizontal direction and vertical direction and later on when we built a structure on the soil mass then the additional vertical pressure is been exerted on the soil sample. So, the Mohr circle accordingly will look like this. So, in a vertical direction the total vertical stress or the principal stress is sigma C plus sigma delta sigma sorry and sigma 3 is sigma C confining pressure. So, this particular Mohr circle has not touched to the failure non-low it means that failure has not taken place. Whereas, if I continue with the additional load the Mohr circle diameter goes on increasing still the failure has not taken place. If I further increases now the corresponding Mohr circle will touch to the failure non-low which indicates that the soil sample has failed. So, at particular value of normal vertical stress the failure has taken place. Take a review question quickly write the equation for finding normal stress and shear stress on any general inclined plane theta with horizontal which we discuss. Take a pause get the answer. So, this is an answer the shear stress is given by this equation and the normal stress is given by this equation which we already discussed. Now let us discuss about the orientation of failure plane. So, again look at this particular stress system a Mohr circle is drawn here and the failure envelope is shown. Then if suppose one plane is considered here. So, this plane is nothing but the failure plane because it is touching to the tangent here. So, this plane which is inclined at an angle of alpha is called as a failure plane. Now let us find the inclination of that failure plane. See if I consider this triangle by a dotted line here. So, if the angle is alpha at the pole point then the angle made by that dotted line at the center of the circle will be 2 alpha. So, here if I consider the triangle a big triangle here say a, b and c here at the center. If I consider triangle a, b, c then there are 2 alpha is an exterior angle. So, this exterior angle will be equal to the summation of 2 interior angles opposite interior angle. So, interior angle is phi and this angle is 90 because it is perpendicular to the failure envelope. So, theta sorry phi plus 90 will be equal to 2 alpha. So, this is what is written here 90 plus phi equal to 2 alpha. But we want the angle alpha therefore, let us rewrite this equation. So, alpha becomes 45 plus phi by 2. So, this is what is the inclination of the failure plane which is given by 45 plus phi by 2. In the same way we can discuss about the principal stress relationship at the failure. So, when the failure is taking place at that time the Mohr circle is drawn here and the failure envelope is drawn which is tangent. And if I consider the line here this particular distance will be c quad phi ok. Consider a small triangle here the intercept is cohesion and inclination is phi. Therefore, the horizontal side of the triangle is c quad phi. And the distance up to the center from the origin is sigma 1 plus sigma 3 by 2 that is known to us. So, again it is the tangent or the perpendicular from the tangent is drawn this is a radius of the Mohr circle it is sigma 1 minus sigma 3 by 2 ok. So, now let us consider the big triangle here and from the trigonometry we can write c quad phi right. So, c quad phi plus sigma 1 plus sigma 3 by 2 this gives me this total horizontal distance ok this distance. And it should be equal to the sign of this horizontal side is equal to the radius. So, we can see this particular equation. Now, this equation we can rewrite and we can simplify this by rearranging the terms and ultimately we get this particular simplification. And from the trigonometry these two 1 minus sin phi upon 1 plus sin phi is 45 plus phi by 2 and cos phi upon 1 minus sin phi is 45 plus sorry tan 45 plus phi by 2. Therefore, at the end we got a very two important relationship inter inclination of failure plane is 45 plus phi by 2 and principal stress relationship at failure is given by this equation. These are the references and thank you. Thank you very much.