 Hello everyone, myself, Nagaraj Kuppa, working as an assistant professor in the Department of Civil Engineering, Palchand Institute of Technology, Solapur. So in the present video lecture, I am going to explain about derivation of bearing capacity equation. So these are the learning outcomes from the video lecture. At the end of this session, students will be able to derive Terzaghi's bearing capacity equation. So these are the contents present in the video lecture. So this is a picture which is related to derivation point. In the previous video lecture, already I have explained assumptions of Terzaghi's bearing capacity theory. So based on the assumptions bearing capacity theory, so it is a drunk picture which is related to bearing capacity equation. So as usual, it is a footing which is carrying the load, quview, ground condition is horizontal, gamma, once again unit weight of soil sample, gamma into df is nothing but overburden pressure, therefore p equal to gamma into df at the base of footing. So here we are dividing entire this zones into three components. One is active zone, second one, radial shear zone, third one, passive zone. So active zone, in the active zone, soil is in limit equilibrium state. So in the due to the overburden pressure, there is inducing passive zone. So due to the presence of passive zone as well as active zone, there may be chances of inducing radial shear zone. We are calling it as zone number 2. So angle at A equal to 5, angle at B is also 5. Due to the condition of limit equilibrium of soil sample, angle at CAB equal to angle at CBA. This quview is vertical and symmetrical. Therefore here due to the presence of quview, the PP, passive pressure is acting in upward direction at an angle of 5 and C which is acting in inclined direction. So where C is nothing but cohesion of soil sample. So here we have to calculate, first we have to calculate weight of wedge. So weight of wedge, we have to calculate weight of wedge ABC. Now weight of wedge, angle at A and angle at B are equal. It is angle at A, it is 5, angle at B is also 5. By using this picture, we have to calculate weight of wedge. Weight of wedge ABC equal to, now I will apply tan rule for this figure, tan phi equal to opposite divided by adjacent. Opposite is x adjacent is B by 2, therefore tan phi equal to x divided by B by 2. Therefore now x equal to B by 2 into tan phi. Basically we know that gamma equal to W by V, where gamma is nothing but unit weight of soil, weight of soil sample divided by volume of soil sample. Therefore what we want to calculate now? We want to calculate weight of soil sample, therefore gamma into V. Therefore W equal to gamma into V, W equal to once again substitute all the values above values in the equation gamma into 2 into 2 times of this area, 2 times of triangle, it means first triangle as usual. So it is a right side triangle and left side triangle, therefore 2 triangles, half into base into height. So B by 2 it is base, x is height, therefore x equal to B by 2 into tan phi. So substitute all the values in the equation, here we got weight of wedge equal to 1 by 4 into gamma into B square into tan phi. Now downward force equal to upward force. So for this figure let us make this equation downward force equal to upward force. What are the downward forces now? So Q U into B it is a downward force, weight of wedge is also acting in downward force, P P it is acting in upward direction, C is also acting in upward direction. Therefore now Q U into B plus 1 by 4 into gamma into B square into tan phi equal to 2 P P, here P P for right side, P P for left side, therefore P P plus P P 2 P P 2 P P plus 2 C Li sin phi. So cohesion which is acting in inclined direction, therefore Li into sin phi. Now we have to calculate Li for CB surface, Li for CA surface. Let us I will talk about CB surface, CB equal to once again apply cos rule for the for this triangle. CB equal to B by 2 divided by cos phi. Let us substitute this B by 2 divided by cos phi in the equation number 1, therefore equation number 1 becomes Q U into B plus 1 by 4 into gamma into B square into tan phi equal to 2 terms of passive pressures plus 2 C. So Li is nothing but B by 2 into 1 divided by cos phi into sin phi, sin phi divided by cos phi is nothing but once again tan phi. Therefore Q U into B equal to 2 P P plus B into C tan phi minus 1 by 4 gamma into B square into tan phi. So take this term 1 by 4 gamma into B square into tan phi towards right side it becomes once again negative. So final equation, equation number 2 it becomes Q U into B equal to 2 P P plus B into C tan phi minus 1 by 4 gamma into B square into tan phi. Let us consider passive pressure which is resulting into 3 components. One is due to unit weight of soil, due to cohesion of soil, due to surcharge load Q. If we are considering cohesion term, it means unit weight of soil sample and surcharge load both becomes ignorable quantity. If we are considering gamma term, gamma is nothing but unit weight of soil sample, therefore while calculating surcharge weight overburden pressure Q equal to once again gamma into D f therefore we must consider once again unit weight of soil sample. So C becomes ignorable quantity. So Q U into B equal to 2 times of P P, gamma plus P P cohesion plus P P related to surcharge so plus B into C tan phi minus 1 by 4 gamma into B square into tan phi. So let us consider 2 times of P P with respect to cohesion equal to C into N C, 2 P P with respect to surcharge weight Q into N Q, 2 P P with respect to unit weight of soil sample half into B into gamma into N gamma. So N C N gamma and N Q are related to bearing capacity factors. So when we are considering bearing capacity factors N C, N Q and N gamma so 1 by 2 it becomes ignorable quantity. So B into C it is once again ignorable quantity therefore Q U into B equal to so substitute all the values in the equation B C into N C plus B C B into Q into N Q plus half into B square into gamma into N gamma. So common term is B, so B B get cancels at last finally equation Q U equal to C into N C plus Q into N Q plus half into B into gamma into N gamma where Q is nothing but ultimate bearing capacity of soil, cohesion of soil. So over burden pressure, base of footing, unit weight of soil sample N C, N Q and N gamma are Terzaghi's bearing capacity factors. We are going to determine Terzaghi's bearing capacity factors by this chart. So general shear failure and local shear failure for corresponding friction angle. So these are the N C value, N Q value and N gamma value. Now you may pause the video and the correct answers are the parameters N C, N Q, N gamma in the equation of bearing capacity failure are known as bearing capacity factors. So what is the value of N gamma for 35 degree so it is 42.4 from this chart for corresponding 45 degree the value of N gamma so it is 42.4. So these are the references for the preparation of this video lecture. Thank you.