 Welcome to lecture series in the course advanced geotechnical engineering, we are actually discussing module 3 on compressibility and consolidation and in the lecture 1 we introduced ourselves to stresses due to surface loads, so in this lecture 2 we will continue further in this particular topic that is in module 3 lecture 2 we are going to concentrate on stresses in soil from the surface loads. And we have discussed that when the soil can be subjected to different shapes, the loading can be of different shapes, it can be angular shape or it can be a circular shape or it can be a rectangular area or it can be an area of irregular shape or it can be of like in the loading intensity can vary from 0 to Q and remain a constant like embankment construction or levee construction or dam construction or in case of landfills we have certain you know very flat slopes and the heights can even range up to 200 meters or so. So it will be interesting you know to learn about this particular topic and to compute the stresses in soil from the surface loads. So we have actually introduced in the previous lecture that Bosnian theory and Westergaard's theory and from the Bosnian theory the number of you know the deductions can be made and here we actually have tried to discuss about in the previous lecture about the vertical stress due to the strip load that means that if you are having a strip foundation and if it is connected with let us say if a wall which is actually has a continuous foundation running over length L then the strip load is the load transmitted by the structure of the finite width and infinitely length infinitely long soil length along the soil surface. So the sigma ZB can be calculated by treating the strip loads as the line loads by treating strip loads as the line loads. So sigma Z is equal to P0 by pi theta plus sin 2 theta by 2 within theta 1 to theta 2. So theta 1 and theta 2 are nothing but theta 2 minus theta 1 is equal to theta where they are depending upon the location of the point of interest of this particular stress that is from the left side edge of let us say that if you are having a strip load and if it is at a certain distance from this point and certain distance from this point and the angle subtended from with the vertical and with the horizontal here is theta 1 and then with this vertical and then this point is theta 2. So theta 2 minus theta 1 that is the angle is actually over the breadth of the foundation. So we actually have said that sigma Z the P0 is nothing but Qs, Qs by Q pi and theta plus sin 2 theta by 2 this is what actually we have discussed in the previous lecture. And the contours of the equal vertical stress of uniformly load intensity which are actually given here and the pressure bulbs it can be seen here under the strip area the depth of influence can actually go up to 3B and in case of square area the depth is limited. So the zone lying inside the vertical stress contour of value 0.2Q is described as the so within this zone and this is called as the pressure bulb the spread of the pressure bulb for the strip area is large compared to the square area. So this is here for the a square area or a square foundation or a footing subjected to uniform load intensity in a strip footing or strip foundation subjected to a uniform intensity is actually shown here and the zone within this is actually called as the bulb of pressure or pressure bulb and here what this figure shows is that under the strip area the zone of influence the depth of influence will be large compared to the square area. Now let us consider a strip carrying uniformly vertical loading on an infinite strip on the surface of a semi-infinite mass that means that let us assume that we are having a certain strip of width 2A it can be capital B is equal to 2A and the loading intensity is 0 at one point and it is Q at this point that is X is equal to 0 so this is along the X axis and this is along the depth axis that is Z axis. Now at X is equal to 0 the load intensity is 0 Q is equal to 0 and X is equal to 2A the loading intensity is equal to Q so there is a linearly varying loading intensity so that is the increasing vertical loading on an infinite strip and then this perpendicular to the plane of this figure let us assume that the load is actually is spreading over infinite length and in order to calculate the vertical stress in the due to increase in loading the type of loading like this what we do is that let us assume that we wanted to calculate the stress at a particular point P and if this let P be the point below this soil surface that is this coordinates of this point are X and Z where this distance is X and this distance is vertical depth is Z. Now for an elementary strip let us consider a small strip here that is of the breadth is equal to some ds the small strip and let us assume that the load intensity per unit length can be calculated by this particular ordinate, vertical ordinate can be calculated by similar triangles as Q by 2A that means that here at X is equal to S let us assume that there is a small strip of elementary strip of ds is there and it is at a distance S so this are vertical ordinate divided by S and this from the similar triangles here and this triangle we can calculate this magnitude as Q by 2A into S and that is the load and then the load intensity at that particular point into ds will give the so called load per unit width that is called the what we do is that we treat it like a small line load. Now let us assume that this distance is S so when this is equal to S X then this will be equal to X minus S. Now approximating as a line load so this is approximated as a line load now so the line load of certain intensity running of the infinite length and with the intensity of this line load is nothing but Q by 2A S into ds so now what we need to do is that we need to substitute in the expression for the vertical stress sigma Z is equal to 2Q by pi into Z cube by X square plus Z square to the raise 2 in this for Q substitute Q by 2A S into ds so that what we do is that because of this small strip load we get what is the small d sigma Z then when we integrate for the entire length of the 0 to 2A what we get is that the increase in vertical stress due to the particular load intensity. So this further we can actually continue like sigma Z is equal to the integral of d sigma Z and with by substituting you know the 1 by 2A and 2Q by pi into when S limits are nothing but this S is ranging from 0 that is X is equal to 0 and X is equal to 2A that is S is equal to 0 to S is equal to 2A and Z cube into S ds by X minus S whole square plus Z square to the raise 2. Now here what we did is that we substituted for substituted Q by 2A into S into ds for Q and X minus S for X and Z is kept it as Z so by integrating and then simplifying what we get is that sigma Z is equal to Q by 2 pi into X by A alpha minus sin 2 alpha sin 2 delta. So this alpha and delta which are nothing but this inclination that is this delta and then this is lambda this is alpha and this is delta. So let us assume that we are actually have interesting have a point at the this particular point here let us say that in that case what will happen is that the delta will be equal to 0 when this load intensity when we are actually interested in calculating this stress at this particular point then the load intensity at this particular point you know then in that case this delta will be equal to 0. Now we are actually having a case where alpha that is covering the breadth of this so called strip of width dA and delta. Now this with delta is equal to 0 and sigma Z is equal to Q by 2 pi into X by A alpha where after once we get the based on the you know the different depth requirements and once we compute this alpha then alpha need to be expressed in the radiance. So with sigma Z is equal to Q by 2 pi into X by A alpha minus sin 2 delta and when delta is equal to 0 that is right below the you know where the load intensity is actually high and then we can actually calculate that the sigma Z as Q by 2 pi into X by A into alpha. Now let us consider further we by using the same concept let us see that how you know we are we can actually construct the you know the stresses below the embankment loading or let us assume that we are having certain area loading over certain area because once we you know we learn about the consolidation and in order to calculate the settlements you need to calculate what is the increase in stress due to loading let us say embankment loading or due to landfill loading or due to certain you know dam or levy loading. So in this case let us consider a simple embankment is you know simplified with having an horizontal distance A and horizontal distance of the slope portion is A and this is actually B. So this vertical the load intensity is simplified by you know having Q is the load intensity and as we go down here the load intensity falls here. So here what we can do is that we can actually use the same concept now at this point here this point is right lies right below the right below this point Q and alpha 2 is the synchronization and alpha 1 is the synchronization and now what we do is that we will use the method of superposition and what we do is that in order to compute the vertical stress at a semi infinite mass due to embankment loading what we need to do is simply extend this line further and calculate by using again similar triangles between this triangle and this particular triangle and this particular you know triangle. So what we get is that we can actually calculate this width is equivalent to B so from the similar triangles again we can actually get what is the load intensity that is nothing but Q into B by A. So this particular loading is assumed as a fixious loading and we converted this embankment loading into a triangular strip having load intensity varying from Q0 to Q into 1 plus B by A and A plus B is the total horizontal distance of the strip and the point is actually lies right below the that is at the center of the embankment. So this angle is nothing but the alpha 1 plus alpha 2 and this width is nothing but A plus B now because as this loading is not there so what we need to do is that this particular stress additional loading due to this fixious portion what we assume need to be deducted so that point the stress due to that one is say sigma z 1 and stress due to this portion is say sigma z 2 so sigma z at point A is equal to sigma z 1 minus sigma z 2 so here this particular triangle which is nothing but having intensity Q into B by A and breadth B so this triangle is having B and then this inclination because what this breadth B the angle is alpha 2 so this is the alpha 2 angle. So once we determine simplify further by using we have said that if the loading intensity if the load under the reference is actually right below the you know the wherever the load intensity is maximum the point below the soil surface then using sigma z is equal to Q by 2 pi into x by A alpha now what we do is that we substitute A plus B for x and A plus B by 2 for A alpha 1 plus alpha 2 for alpha. So for sigma z 1 that is the stress sigma z 1 that is the stress due to you know this particular point due to the entire this triangle so the load intensity is nothing but Q plus B by A into Q that is why we have written for Q we have written Q plus B by A into Q 2 pi into for x now we are writing because the breadth the strip width is nothing but A plus B so we are writing A plus B and then A plus B by 2 that is for A that is nothing but this portion now into alpha 1 plus alpha 2 so and similarly you know for sigma z 2 which is nothing but Q into B by A into 2 pi into B by B by 2 into alpha 2 so when we you know take the difference that sigma z is equal to sigma z 1 minus sigma z 2 is nothing but Q by pi Q is nothing but the magnitude of the embankment so how to get that is by if you know the embankment loading having let us say you know today if is embankment is constructed with a field material having a unit weight of 20 kilo Newton per meter cube then 20 into let us say embankment height is say 5 meters then it is about 120 kilo Pascal so 120 kilo Newton per meter square so 120 R Q by pi into A plus B by A into alpha 1 plus alpha 2 minus B by A into alpha 2. Now so this is indicated as sigma z is equal to I influence factor that is the IE that is IE is nothing but the influence factor for embankment loading where IE is nothing but this entire multiplication that is this 1 by pi into A plus B by A into alpha 1 plus alpha 2 minus B by A into alpha 2 which is actually nothing but you know the function of 1 by pi into function of A by z into B by z where A by z is nothing but that the distance A or that horizontal portion of the embankment and the distance B which is horizontal portion of the slope of the embankment slope portion of the embankment and the horizontal portion of the from the crest of the embankment to the center line of the embankment where we can write that A by z and B by z so sigma z is equal to IE into Q so this for this actually the Osterberg charts are available and where we can actually use this Osterberg charts and then try to get the stress values say for example here the IE embankment that is IE influence factors are given and here the values of A by z ranging from 0.01 to 10 are given here and this can be used for even for embankment with B is equal to 0 that means that if you are having only triangular strip loading then the entire B by z is equal to 0 so then this curve we have to use and then we actually have depending upon the A by z value we can actually calculate the influence factor sigma z into this influence factor into the load intensity sigma z is equal to IE into Q we will be able to get what is the increase in vertical stress due to you know the embankment loading. So what we need to notice that this point will give the embankment stress at this point or in this case if you are talking about you know the stress suppose B is equal to 0 then that means that the stress actually occurs only at the center point. Let us say that we are having you know the loading which is symmetrical about you know let us say about the x axis where you know at the center it is actually having a load intensity Q and then on the both the sides actually is reducing to 0. So in this case what we need to do is that we need to calculate the influence factor and multiply with 2 so that you know we get the load intensity due to this and then load intensity due to this one. For example here when we have this is the symmetry of the embankment and if you wanted to take the stress due to embankment of the remaining portion then this multiplied into 2 we will be able to get the stress due to the complete embankment that is the total embankment distance. So this Osterberg charts are actually used for calculating the vertical stresses for the embankment loadings where you know the theory which is actually is deduced from the triangular strip having you know linearly varying increasing load and from there what we have deduced is that we have deduced the sigma z is equal to sigma z1 minus sigma z2 and from there we have actually got the expression for IE as a function of A by z and A is nothing but the horizontal distance of the embankment in the slope portion and B is nothing but the horizontal distance of the embankment from the crest of the embankment to the center line of the embankment. So by using this the increase in vertical stress at different points can be calculated. Let us look into an example having you know example problem having a 5 meter high embankment basically it is to be constructed as shown in the figure and if the unit weight of the compacted soil is 18.5 kilo Newton per meter cube and calculate the vertical stress due to solely to the embankment loading. So there can be initial effective stresses but however what we are interested is that by treating this mass as weightless medium we calculate what is the increase in the vertical loading due to solely due to the embankment loading only at point A at point B and at point C. So if you notice that you know we do not have the symmetry here but is at the center but we need actually the embankment loading. So if you look into this portion at point A this portion of the embankment causes the loading and this portion of the embankment causes the loading. So if you call this portion of the embankment as portion A and portion B and the sigma z at A is equal to sigma z A1 and sigma z A2. If this is portion A this portion is called as A1 this portion called as A2 then the sigma z A is nothing but sigma z A1 and sigma z A2 so both need to be added. Now here when it comes to this point we are having an embankment of you know this particular shape so the stress due to this portion and stress due to this portion and stress this to this portion need to be deducted. So this portion this stress need to be deducted. Now when it comes to this point here we construct a fictitious embankment and then afterwards we again remove the loading due to the load due to the stress increase in stress due to this much portion. So with that what will happen is that we get the increase in stress at a point C away from the embankment also. So here what we need to do is that by using the same notations and by using the method of superposition we can actually calculate the vertical stresses at point A, point B and point C. The method is that we first have to you know see how the embankments can be divided and if any fictitious portion need to be added it can be added and then you know then again the stress due to that particular features portion need to be deducted. Once that is done we will actually able to get the stresses at point A, point B, you know point C and if required if you are having any point here at the toe that is like say B1 that also can be calculated. Now by let us say we have actually discussed about you know the linearly varying load and an embankment stresses due to embankment loading and for certain type of loading in the loading areas we can also have you know the circular areas carrying the uniform intensity that means that it can be a ring foundation or it can be a foundation for chimney with a raft of having 2R diameter let us say and in that case you have a ring a raft of having 2R diameter so that means it be subjected to or let us say that you we have say oil storage tank having you know diameter of about some 46 meters or so then if it is subjected to oil of certain unit weight or certain height and it is subjected to certain increase in stress. So in order to calculate the increase in stress due to you know any oil storage tank having certain diameter or with certain load intensity we can actually then in practical applications we may have this type of circular areas carrying uniform pressures. Sometimes we also can have a you know the ring foundations that means that if you are actually having R1 and R2 where R2 is greater than R1 so R2 minus R1 portion that much is the breadth of the ring and over which the loading is actually happens with uniform intensity Q. So using Bosnik's point load solution so this can be reduced for example in this particular slide and we are actually having a circular shape loaded area having at the center and we are actually calculating the increase in vertical stress at the center point of the below the center of the you know this so called circular area and subjected to uniform load intensity. If R is the radius that is 0 here and extending here now consider a you know a small sector where you know with an angle d alpha then you know this portion this length of the arc which is nothing but L into R into d alpha L is equal to R into d alpha and assume that the small strip is actually having you know this so called this distance radial distance is dr the strip width here is dr and it is at a distance smaller from the center. Now what we do is that we calculate this particular you know the this length is nothing but rd alpha rd alpha d alpha is the angle so dr into rd alpha is the you know this area and assume that the point load in this area is nothing but q into dr into rd alpha so what we are doing is that we have taken a small strip having the dr into rd alpha dimensions and we are multiplying by the uniform load intensity q then point load on the elementary area is given by q into dr into rd alpha so we are actually using the Bosnik's point load solution. Now due to that so called you know the small element load is subjected to increase in intensity q into dr into rd alpha we can calculate d sigma z is equal to 3q dr into rd alpha so is nothing but 3q by 2 pi into zq by r square plus z square to the raise pi by 2 this is what actually we have discussed for q capital q what we substituted is that the small dq is nothing but q into dr into rd alpha by 2 pi into zq by r square plus z square over the raise pi by 2. Now increase in vertical stress at a due to the entire loaded area so in order to get this what we need to do is that the entire area is 0 to 2 pi alpha is actually ranging from here alpha is equal to 0 here and then alpha is equal to 2 pi and the radius that is the this circle and then this this bit is actually given by r is equal to 0 to r so we need to do the double integration sigma z is equal to d sigma z is equal to integral of d sigma z is equal to integral of alpha is equal to 0 to alpha is equal to 2 pi integral of r is equal to 0 to r is equal to 2 pi and for the the small d sigma z we can substitute here where 3q by 2 pi into z cube into r so this particular r into r square plus z square to the raise pi by 2 into dr d alpha. So once we substitute and you know we integrate the do the double integration and substitute what we get is that sigma z is equal to q into within square bracket brackets 1 minus 1 by 1 plus r by z whole square to the raise 3 by 2. So this particular you know portion this particular component of this particular equation is calculated as influence factor for the circular loaded area. So for different r by z values so this this this particular increase in stress due to at the center of the loaded area. Suppose if you are actually wanted due to load due to the if you if the increase in stress due to this loaded area away from the center then we need to have we need to use some other influence factor values but this is actually valid for the increase in vertical stress due to the load subjected to uniform load intensity having q and that too at the center of the loaded area only. So sigma z is equal to q into 1 minus 1 by r 1 by r by z whole square to the raise 3 by 2 that is a q into i c where i c is nothing but the influence factor for the circle loaded area. So this we can actually get by using charts also. So here the circle area carrying uniform pressure so the chart which is actually given here in terms of 2 r by z where r d is equal to capital 2 r and i c that is the influence factor. So what we can get is that by knowing the d by z for a let us say that we have a d by z value of 1 and the influence factor is 0.3. So that means that 0.3 into q that is the increase in stress due to the particular circle load area and this vertical stress is at the center of the circular loaded area. So consider an example here where a rectangular concrete tower is provided on a raft on a ring foundation and mostly these chimney foundations particularly in the coal base of Dharma power stations in steel plants you will find these chimneys and they are resting on ring foundations and in this particular problem the inner diameter of the ring foundation is given as 6 meters and outer diameter is given as 12 meters that means that here the inner diameter so the breadth of this ring foundation is nothing but 12 minus 6 meters that is 6 meters and on this you know annually if you say that we are having a annual ring having you know the breadth of 6 meters. So on this area due to you know the loading the different sorts of loading due to the shelf of the shell of the you know this element then you know the subjected to a 150 kilo Pascal's of loading. So we need to calculate what is the vertical stress at the center at 6 meter depth below the foundation. So for this you know we have if you notice here the central portion is not subjected to any loading. So what we need to do is that you calculate the in order to get the sigma ZA you know calculate the vertical stress due to the entire area that is entire diameter that is 24 meters diameter and minus the stress instant intensity due to you know this particular portion because this is particular portion is not loaded only this portion is subjected to load then you know what we can get is that we will get that net increase in stress due to the loading on the ring foundation. So we need to use the same equation that is in case of R first we need to use here 12 meters then we get the sigma Z then afterwards we need to use the R is equal to 6 meters then we get the vertical stress that is 1 q into 1 minus 1 by 1 plus 12 by 6 whole square to the raise 3 by 2 minus q into 1 minus 1 by 6 by 6 whole square to the raise 3 by 2. So the net difference sigma z 1 minus sigma z 2 we get the so called the increase in vertical stress at a depth 6 meters below the so called reinforced concrete tower having a ring foundation which is subjected to load intensity of 150 kilo pastels. So after having you know looking to you know the so called you know the circled area and the embankment loadings let us also look into the vertical stresses below a rectangular loaded area where in this particular figure where we it can be seen that this is the x axis and this is y axis and this is the uniform vertical load q per unit area and we are having a rectangular loaded area L is the length along the x axis and B is the breadth along the y axis and this is the jet. So we are actually calculating at one of its corners of the rectangle so let us say that if you are having a rectangular area of 2L by 2B so that means that we actually have to multiply with the four times of that particular you know load intensity by this is nothing but the method of superposition. So again you know in order to calculate the increase in load intensity due to a rectangular loaded area. So here if you want to calculate let us say for square loaded area when L is equal to B so with that also you are able to get for the square loaded area. So here let us take for L and B and at one of the corners like you know this corner is under consideration now so we are actually calculating the this x is equal to 0 y is equal to 0 and the along the z axis that is you know the depth z. So you know we can actually calculate what is the increase in stress due to so again what we do is that we actually use the both sticks point load solution as we have done in the in the circular load. The stress at point P due to you know so the small increase in the stress vertical stress due to point point load acting on the small step which is actually is darkened portion here having dx along the x axis and dy along the y axis. So the area is dx into dy and the loading intensity is q per unit area. So q into dx dy is the small point load here acting at this particular point. So using d sigma z is equal to 3 q z q by 2 pi r to the power of 5 and with r square is equal to r square plus z square is equal to x square plus y square plus z square. So here what we can do is that 3 into for q what we do is that small q into dx dy. So small q into dx dy into z cube divided by 2 pi into for r what r for r what we are writing is that x square plus y square plus z square that is root over to the raise pi. So we get x square plus y square plus z square is equal to the raise pi by 2. So d sigma z is equal to we have got something like 3 q dx dy z cube by 2 pi x square plus y square plus z square to the raise pi by 2. So here by using this vertical stress below a rectangular loaded area on the surface. So for this total increase in vertical stress at point p due to the entire loaded area can be determined by integrating with the limits like x is equal to 0 to x is equal to l and y is equal to 0 to y is equal to b. If you integrate in these limits like with l and b then we actually get one of the at one of the corners what is the increase in stress due to the loaded area. So this is simplified with sigma z is equal to q into i sigma. So here this i sigma or i r it can be like a rectangular loaded area where see i sigma is equal to 1 by 4 pi into 2 m1 divided by m square plus n square plus 1 to the raise 1 by 2 into divided by m square plus n square plus m square n square plus 1 into m square plus n square plus 2 divided by m square plus n square plus 1 plus tan inverse 2 mn m square plus n square plus 1 to the raise 1 by 2 divided by m square plus n square minus m square and square plus 1 to the bracket process. So where here m is nothing but b by z n is nothing but l by z. So let us look once again m is nothing but m is nothing but b by z, n is nothing but l by z, l is along the length axis and b is along the breadth axis. So here in this m is nothing but b by z, n is nothing but l by z. So for this we have the Fordham charts, vertical stress below the rectangular loaded area on the surface can be obtained by using Fordham charts. The values for i sigma is equal to i can be obtained by using this Fordham chart which is given by Fordham in 1948 and where m is equal to b by z and i and these are the different curves for n is equal to 0.1, n is equal to 0.2, 0.3, 0.4 like this up to 2. So here by knowing these values so we get the influence factor and let us assume that we are actually calculating for a square area having b then we need to take say b by 2 and b by 2 and then we get the you know at the center to get this stress at the center of an area which is having loaded b by 2, b by b dimensions then in that case you have to take for b by 2 and in that case m is equal to 1 and depending upon the depth you can actually get based on that what you can get the influence factor and that influence factor into 4 times you have to multiply to get the load intensity to get the sigma z. So vertical stress below rectangular area on the surface can be given by sigma z is equal to q into i sigma. So i sigma is obtained let us say that at z is equal to 2 meters and b is equal to 2 meters then m is equal to 1 and n is equal to let us say that 0.3. So we will actually get the influence factor i is equal to about 0.07. So 0.07 multiplied by q into you know the configuration whatever we consider we have to use that one. So this is the Fodum chart which is given by Fodum in 1948 and these are used for the in order to calculate the vertical stress below a rectangular load area and it is possible for so if you are actually calculating to calculate the settlements in a soil layer particularly at the mid depth of the soil layer due to a rectangular load area then we have to use this particularly in place you know we have to add the increase in vertical stress due to this to the initial effective stress so that we get sigma 0 dash plus delta sigma. This delta sigma is the thing but at that particular point in the center of the clay layer how much the loaded area imposes the vertical stress can be calculated. So this is used even in calculating vertical stresses and in settlement estimations at all. So this is another view of the Fodum chart and where in a similar way it is actually expressed where m and i r or i sigma and where you know this is for the different n values which is given. So this is again to get the vertical stress at the one of the corners. Now let us try to take one example where we are actually having a load of 1500 kilo Newton is required to be carried on a foundation having a 2 meter square at a shallow depth of in a shallow depth in a soil bus. And we need to determine the vertical stress at a point below 5 meter below the surface below the center of the foundation assuming that the load is uniformly distributed over the foundation and assuming that load acts as a point load at the center of the foundation. So here we are actually you know assuming that the load is uniformly distributed over the foundation and then we are having a square foundation so 2 meter by 2 meter area it is not rectangular area it is a square area. And another assumption what we have so the prime of AC the load intensity is nothing but uniform pressure is nothing but 1500 kilo Newton divided by 2 square that is 1500 divided by 2 square that is 375 kilo Newton per meter square. And if you look the 375 kPa into area we get the 1500 kilo Newton the load. So first is that we are assuming as a load intensity the increase in load the intensity is 375 kilo Pascal's. And so by taking we need to calculate at a depth of point 5 meter below the center of the foundation that means that we are having a square 2 meter by 2 meter. So here what we have is that you know we have 1 meter by 1 meter 4 squares are there and then at each corner that is at each one of the corner let us say that left corner we are actually getting for the increase in vertical stress at a certain depth of 5 meters and after getting the influence factor from the curve then 4q into IR. So here the 4 is multiplied because of we are actually divided that area into 4 and 4 into 375 into 0.018 so these are about 27 kilo Pascal's. And now the second portion of the problem as asked is that assuming that the load acts as a point load at the center of the foundation. So assuming that when r by z is equal to 0 we know that IP that influence factor for the point load is 0.4775 and it also can be given as 0.478 and by for the sigma z is nothing but q by z square into IP. So 1500 divided by z is nothing but 5 meters so 1500 by 5 square into influence factor for the assumption of point load is 0.478. So if you look into this the increase in vertical stress due to the point load is actually comes to be point load assumption for the in the given problem comes to us 29 kilo Pascal's which is actually more than what we have assumed for the uniform pressure. So we need to note here the point load assumption should not be used if the depth to the point x depth to the you know if the depth to the this particular point x under the reference is less than the 3 times the larger dimensions. For example in this larger dimension of the foundation is say you know 2 meters 2 into 3 is 6 meters and 6 into the depth is say 5 meters so it is if it is less than you know the so called 3 times the larger dimensions then this particular point load assumption is not valid that means z 1 as 1 e to calculate the assuming as the uniform pressure. Now we have discussed about rectangular areas having square areas or sometimes we may have combination of the areas so vertical stress due to a regular shaped area. Suppose if you are having irregular shaped area loaded with uniform loading density so it can be a combination of circular area or it can be of any shape let us say that it can be of any shape so if we need to calculate the vertical stress of any irregular shaped area and for this the new mark 1942 has developed you know the influence charts basically to compute the vertical stress and these influence charts are also there for horizontal and the shear stresses also within the soil mass due to the load area of any shape and irregular shape or it can be any shape in the sense it can be square area or it can be rectangular area and the point of interest if it is known then we can actually you know place the area drawn to the scale as per the procedure outline when you mark then you know we can calculate the at below any point on either side or outside the so here if you notice here we can calculate the vertical stress either the point can lie inside or outside as a whole we did not require to divide into you know the method of superposition will not be used. So here we can actually calculate and point on either side or outside the load area so for this you know the basis of this you know deducing this new mark chart is the circular loaded area subjected to uniform loaded density so we actually have deduced an expression for increase in vertical stores at the center of the circle for a circular loaded area okay. So based on this you know particularly by using this concept the new mark developed these influence charts now let us see that how this actually you know conceptualize and how this can be used in calculating the vertical stresses due to area of any shape irregular shape or any regular shape where for the point lying either inside or outside the loaded area. Now consider a circular low area of R1 loaded with a uniform load intensity q so that means that you are having a circular loaded area of having dimension R1 and O at the center and to be divided that is arbitrarily it has been divided into 20 sectors so like this OBC like that it is been divided into 20 sectors that means that each the center angle is about 18 degrees and then BC is that arc length like that this entire area is divided into 20 sectors. So this entire area is loaded to with a load intensity of q now as I said that this is actually from the you know the circular load principle vertical stress at point O and a depth z below its base for one sector area is nothing but q by 20 because that is we are actually talking about the one sector area into 1 minus 1 by 1 plus R1 where R is equal to R1 we are substituted now R1 by z whole square to the rise 3 by 2. Now if that left side of the circular area is actually equivalent to load intensity of q by so if you are having now when we say that this total number of numeric circles are divided into 10 so there will be total number of numeric circles are 10 and these are fixed arbitrarily by q by 10 into 20 so the influence factor for you know is nothing but 0.005q so this is the influence factor which was considered. So for the calculating the radius of the first numeric circle so if this is assumed as if the left hand side portion of this equation is assumed as the equal to 0.005q is equal to q by 20 into 1 minus 1 by 1 plus R1 by z whole square to the rise 3 by 2 by solving we get R1 by z is equal to 0.27 that means that the first numeric circle will have 0.27 times the diameter z so this is for the increase in verticals just due to an irregular shaped area loaded with an uniform load intensity and in this particular slide we have got the radius of the first numeric circle. Now if the circle is drawn with radius R1 equal to 0.27 z and the area divided into 20 area units each area unit will be produced in a vertical stress equal to 0.05q at a depth z below the center that is the meaning. Now what we do is that we take another circle where radius is R2 and R2 greater than R1 and the second concentric circle of radius be drawn and divide into 20 area units and the total stress due to area units O, B, C and B, B dash, C, C dash that is that this portion O, B dash, C dash this portion is let us assume that this portion is 0.005q we have set and this portion also we assume that is as a 0.005q. So in order to get the radius R2 now what we do is that 2 into 0.005q is equal to q by 20 into 1 minus 1 by 1 plus R2 by z whole square to the raise 3 by 2. So solving we get R2 by z is equal to 0.4 in the same way by doing for third circle 3 into 0.005q and fourth circle 4 into 0.005q similarly 10th circle 10 into 0.005q what we get is that we get the radius of the different radii of the different numeric circles. So R1, R2, R3, R4, R8, R9 so this all concentric circles radius can be calculated. Now the equation for the radius of the 10th circle is given by as I said 10 into 0.005q. So this q by 20, q by 20 is equal to 1 minus 1 by 1 plus R10 by z whole square to the raise 3 by 2 by solving we get R10 by z as infinity. So the 10th circle of the numeric chart is actually for the vertical stress is infinity. So with this we can actually get the numeric circles and the vertical stress due to irregular shaped area can be found out and here in this particular table the circle number and the radius R by z1, R by z is equal to 1, this is for first circle, second circle and the 10th circle the radius is infinity. So by knowing these values we can actually calculate construct the circles and once the more numeric chart is ready then let us say that we wanted to calculate the stress increase in stress due to circular loaded area and the point is outside the loaded area and that point has to be placed in the center of that numeric chart and what we need to calculate is that the number of sector points covered within that foundation area drawn to the scale on the generally here a transparent sheet is used or tracing paper sheet is used to draw the loaded area so that the number of sectors covering the loaded area can be calculated and with that we can actually calculate the influence that number into that influence factor into the load intensity will be able to get the increase in stress. So here in this particular slide a numeric chart is actually given where in actually we have these one first circle, second circle, third circle and this is the third the last circle which is at infinity so only the nine circles are shown here. So numeric influence chart for vertical stress and influence values per unit pressure that is 0.005. So here sigma z is equal to 0.005 qn and n is the number of influence areas covered by the suppose example we have the rectangular area so in this area what we do is that we calculate how many number of this rectangular areas are there and some approximations can be made because we cannot actually get the accurate number of the rectangular these areas. So let us say this is 1, 2, 3 like that when you count you may get the 29.5 or it can be approximate to the whole number and that number is taken multiplied into q into 0 this influence factor then we can actually get the you know first of all any loaded area if it is there the loaded area is drawn on the tracing paper to a scale such that the length of the scale line on the chart represents the depth z at which the vertical stress is required. So if z is equal to 5 meters so that the length of the scale line is actually drawn to that so the loaded area is drawn on the tracing paper, to a scale such that the length of the scale line on the chart represents the depth z at which the vertical stress is required. So depth z is accounted by making the area drawn to that particular scale. So the rolled areas drawn on the tracing paper to a scale such that the length of the scale line on the chart represents the depth z at which the vertical stress is required. So let us take a problem where a rectangular foundation having 6 meter by 3 meter carries a uniform pressure of 300 glopascals and near the surface of the soil mass. So here what we have is that determine the vertical stress at a depth 3 meter below the point A that is at this point 3 meters below this point A. So we have a rectangular area here and this point A is here and this point is 1.5 meter away from the long edge of the foundation and using the influence factors and the new marks influence chart. So by using the rectangular loaded areas so because this area is not subjected load what we have to do is that we have to use the method of superposition. So we assume that entire area is loaded now and this portion which is actually not having any load so we have to take the apply the negative loading. So by calculating the vertical stress due to these two triangles so that we will get at this particular point and minus the stress due to these two triangles with the negative loading what we get is the so called vertical stress at this point. So this minus this what we get is the vertical stress at the point. Now by using the new marks chart what we can do is that we can actually draw we need to calculate at a depth of you know this stress depth 3 meter below the point A. So with scale draw to 3 meters you can draw the area and put this point A with depth scale and put this point A at the center of the new mark chart then calculate the number of squares and with that we can actually calculate what is the increase in the stress. So the solution is calculated as explained here 2 x 300 x 0.193 minus 2 x 300 x 0.120. So this is the due to the influence factors the rectangle load area and by using the Fodden chart we have got this and using the new mark chart we get as part of artifact low Pascal which is nothing but the load area is drawn on the tracing path to the scale such that the length scale line of the chart represents the depth z is equal to 3 meters at which vertical stress is required suppose if z is equal to 5 meters then the scale of the loaded area changes. The area is positioned such that point A is at the center line of the chart and number of the influence areas are calculated in this particular case we get 30. So 0.005 x 30 x 300 we get 45 kilo Pascal. So this how you know the you can actually calculate suppose if you are having a irregular shape and that is also can be applied and calculated vertical stresses, shear stresses and the harder can be calculated. So this is another problem where the determined vertical stress increase for a point at a depth of 6 meter below the center of the invert of a newly built spread footing where you know 3 meter by 4 meter in area so we need to calculate at this particular point. So this is the 2000 kilo Newton load is given so what we need to do is that again use the method of superposition and at the center so that we can actually get the vertical stress due to this portion of the triangle and this portion of the rectangular area, this portion of the area, this portion of the area and this portion of the area and by method of superposition we can actually calculate the what is the increase in stress at this particular point. So like this we actually have the number of application problems. So we take this particular problem having a strip footing of 2 meter wide area covers in uniform pressure of 250 kilo Pascal's on the surface of the deposit of sand. The water table is at the surface and the saturated unit weight of the sand is 20 kilo 20 kilometer cube and k0 is 0.4 so determine the effective vertical and horizontal stresses at a 0.3 meter below the center of the footing before and after the application of the pressure. So before the application of the pressure you calculate vertical stress that is effective stress and multiplied by k0 value we get the horizontal stress. After loading then by applying calculating for the strip loading equation for the sigma jet we can actually calculate what is the increase in stress. So with that we can actually get before loading and after loading what are the increase in the stresses. So in this particular lecture we try to understand about the stresses due to surface loads and where in when you have got the different shapes and how we can actually use the Bosnik's theory for calculating the stresses particularly primarily we have primarily we have covered about the increase in vertical stresses. Of course there is a possibility that horizontal stresses and the shear stresses also can be calculated. And so these stresses cause the settlements in the soil and also undergoes the volume changes can takes place. So further we actually look into the concepts of how this loaded areas cause the increase in the consolidation and or if the consolidation does not happen how we can actually accelerate the consolidation so that the settlements before the constructional structure can be anticipated.