 Today, I will start my fourth lecture in this series and that is related to soil exploration and this will be probably the last lecture on soil exploration. Now, in this lecture I will be discussing about a technique one is geophysical exploration. So, how to use this technique to determine the soil properties. So, what are the different types of geophysical exploration methods are available. So, those things I will discuss in this lecture. Now, first there is the these are the different types of geophysical exploration methods. So, one is seismic refraction survey, one is seismic reflection survey, then cross hole seismic survey and then resistivity survey. So, I will discuss one by one about the different type of this surveying methods and how to use this method to determine the soil properties. Now, first I will start about this seismic refraction survey. Now, seismic refraction survey survey is used to determine the wave propagation velocity through various soil layers in the field and to obtain the thickness of each layer. So, by the using of this surveying method, we can determine the velocity of the wave that is passing through that medium and the thickness of the each layers by which this velocity is passing. Now, next is the seismic refraction survey are conducted by impacting the surface either by hammer blow or by a small explosive charge. So, that means we have to create the waves and with the help of these waves we have to determine the soil properties and the velocity of that wave and the respective thickness of the layers. Now, to how we will generate these waves? So, that we can generate either by the hammer blow or by small explosive charge. So, if I go this diagram, so this is the diagram of this seismic refraction survey method and these are the in three layer systems. So, this is first layer whose velocity of the wave is V 1, this is V 2 is the velocity of the wave and V 3 is the velocity of the wave in the three layers. And suppose in the A, we have to create one, we have to create one wave here. So, by using the hammer blow or by small explosive charge, if we do the hammer blow here or by the small explosive charge, we can create that wave. So, here after this hammer blow, that wave will be generated and it will pass through this different direction. So, now it will pass through this layer, it will come here. So, it will either reflect or it will different. So, now with the help of these methodology, we can determine the seismic properties or the velocity of that particular wave which is generated and then the thickness of this each layer. Now, this impact of this ground creates two types of waves. Now, we are talking about the waves and we are talking about the velocity of these waves. Now, after this impact, only two types of waves are generated. One is called this P waves that is plane waves or another is called S wave that is shear wave. So, these two types of stress wave will be generated after this impact and these two types of wave. So, from these two types of waves, this P waves travels faster than S wave. So, that means, if we want to record, that means, somewhere we have to detect the or we have to if we place some geophones which will receive those waves after this refraction. So, if we go for the first arrival of the receiving of this wave. So, that means, as P wave travels faster than the S wave. So, as the first the wave that will receive in the geophone that is P wave. So, that means, in the plane wave that the velocity we are talking about the velocity of the wave. So, that means, here the velocity of the wave will be the velocity of the P wave. So, we will generate two types of waves by using this hammer blow or this small explosive charge and by using this hammer blow, we will generate two waves P wave and S wave, but as these P waves travel faster than this S wave. So, first wave that we will receive by the geophone or by the receiver which are placed at different distance from the A, this is the point of source and this B, C, D are the point of receivers or geophones where we will receive first the P wave. So, the velocity that we will determine that is basically the velocity of this P waves that is generated by this hammer blow or the explosive charge. Now, how we will determine this velocity and how we will determine the thickness of each layer? Basically, by using this method, we will determine the velocity of this wave P wave that is passing through the different soil layers and the thickness of each soil layer by which this wave is passing. So, now, and this thickness we will get, so that means, by we will get an idea that where the different types of soils are there, the thickness of each soil layer and where the bed rocks are present. So, this idea we will get by this serving method. So, now, this velocity and we know that different soil layers have its own velocity range of this P wave velocity. So, if we can determine the velocity of this wave, so by using this range, we can get an idea which type of soil is it in this layer, what are the different types of soils are present. So, that idea also we will get. So, now, first if I draw the method, so suppose this is the surface or the ground surface or ground layer. So, here this is the A is the source where we will do this hammer blow. So, now, after this and these are three layers, one is say this layer, this is two layers. So, this is our layer number one, this is layer number two and this is layer number three. So, probably first we will go for this three layer system. So, now, this technique can be used for any number of layers. Now, as we are talking about that, we will calculate the velocity of this layer that is the velocity of this wave that is P wave. So, we will denote this is the velocity P V P 1 is the velocity of the first layer, then V P 2 is the velocity of P wave in the second layer and V P 3 is the velocity of the wave in the third layer. So, what will happen? Once this after this hammer blow, this wave will generate and it will follow different directions. So, now, one wave that will travel say suppose this is one way that is travelling. So, this is this is the interface of the first and the second layer. So, first it may reflect in this portion. So, it will travel this and some wave that will refract also this wave will refract. So, we are now first talking about the seismic refraction survey. So, now, we will consider only this refraction part. So, now, this is the angle say alpha 1 by which this layer is making angle with this perpendicular line that is alpha this ray is making that is angle alpha 1 with this perpendicular line. And say another angle which is say alpha 2 that is the angle by which this layer is passing or layer is refracting by this angle which is the angle which is making by this horizontal layer. And this is another angle say alpha 3 which is also the angle this layer this ray refraction ray is making with this vertical line. So, this is the first angle alpha 1 this is the alpha 3. Now, what will happen that now as this survey one limitation is this survey is only applicable if this v p 2 is greater than v p 1. So, under this situation we can able to this survey that mean v p or we can write another thing the v p 3 is greater than v p 2 is greater than v p 1. So, that means as we will go further in further in the lower direction. So, that means to undertake this survey we should that the velocity of the wave should be greater than the is velocity of the wave in the previous layer. So, velocity of this wave that will increase as we will go in the deeper layer. So, as this v p 2 is greater than v p 1. So, this ray will go towards travel away from this vertical line. So, now we can by this law we can write that sin alpha 1 divided by v p 1 is equal to sin alpha 3 divided by v p 2. Now, another thing is that depending on this alpha 1 angle this alpha 3 angle is the angle that will change. So, once particular situation if that we can see this alpha 3 is equal to 90 degree. So, that will happen when if the alpha 1 angle such that that after this refraction this alpha 3 angle is 90 degree. So, it means that after the refraction this alpha 3 is it is 90 degree that means this ray after the refraction will follow this path. So, that means as this angle is 90 degree alpha 3. So, it will follow this path. So, under this situation if alpha 3 is 90 degree then the corresponding alpha 1 is called the critical angle. So, if alpha is 90 degree then alpha 1 is called the critical angle. So, this is called critical angle of incidence. So, now if we put alpha 3 is equal to 90 degree then we can write that sin alpha 1 divided by v p 1 will be equal to sin 90 degree is 1 by v p 2. So, alpha 1 will be equal to sin inverse v p 1 divided by v p 2. So, this alpha 1 angle is the critical angle under this situation. So, what will happen that if we put receivers at different locations. Suppose this is the position of one receiver say r 1 and this is the position of another receiver r 2 and this is the position of another receiver r 3. So, depending upon this receiver that. So, every time this ray now after this thing it will pass from a to say b and then it will follow this path and then it will again. So, as this here the velocity change means this v p 1 is less than v p 2. So, this ray will go towards the vertical line. So, it will follow this path. So, this it will receive one ray. So, here also if we put the receiver or geophone. So, it may follow this path and it may also follow this path. So, similar thing will happen for this. This is for the two layer system. So, I will go for the three layer systems further. So, then in three layer system what will happen? So, it will follow this path then it will this fashion it will go and then it will follow this root here. Suppose this angle is now the critical angle for this second layer and third layer. So, this alpha 1 is the critical angle with respect to first layer and second layer and this angle say alpha 1 dash is the second angle with respect to second layer and third layer. So, it will follow this path then again it will follow this fashion. So, we can also receive if we put the another geophone say r 4 and we will receive this ray also. So, this is passing through first second and this interface of the third layer and these rays are passing through the first and the interface of first and second layer. So, now the question is that these rays are passing after the reflection. Now, in this from A to this different receiver this ray can travel directly from A to different receiver. So, there is a two rays one is passing through this different layers after the reflection of this total system another is passing directly from A to different receivers. Now, we are talking about the first arrival time of the ray. So, this initially it in the initial portion if this distance is small then this the direct ray that is travelling directly from A to this receiver that will arrive first faster compared to this refracted rays. But after a certain distance this refracted ray or this refracted ray that will travel or that will arrive in the receiver faster than the direct ray. So, that distance is called critical distance. So, after the critical distance the refracted ray will receive or refracted ray will be received by this geophone. So, that means faster than this direct ray. So, say suppose this distance is critical say x c. Now, if x or any distance say x is any distance from the source say x is any distance from the source if x c x is less than x c then the direct ray that will receive by the receiver before the refracted ray. And if x is greater than equal to x c or greater than x c then the refracted ray that will receive by the receiver before the direct ray. And at if x is equal to x c then both the rays will be received at a same distance same time. So, these are the three conditions. So, that means the time travel time for this direct ray and the travel time by this refracted ray will be same if x is equal to x c. So, this is the total system. Now, by using this total system we will determine the velocity of this different velocity of the wave within different layers say layer 1 layer 2 layer 3 and the thickness of each layer. Now, so again if I draw this layer. So, suppose this is the source. So, this ray will travel in this fashion. So, this is say a and then this will travel in this fashion and then it will receive by the receiver at. So, this angle is alpha 1 equal to alpha c. This angle will be also alpha c and this thickness of the layer say h 1 and this is the velocity of the layer 1 v p 1 and this is the velocity of the layer 2 v p 2. Now, if I consider another point say b. So, at a distance of say x 1 and c distance is at a distance of say x n. So, now if x n is say the critical distance. So, that means at the c point both the ray that means the direct rays from a to c and this refracted rays from a b then d to c. So, a b d c and from a to c that both the rays will travel in the same fashion. So, we receive by the geophone or receiver at the c at the same time. But then at the b point the direct ray will receive first or before the refracted ray. So, now we can write that for this c condition the time required for the direct ray to travel from a to c that means travel from a to c. So, that distance required will be x n divided by v p 1. So, this will be that means direct ray travel from a to c will be x n v 2 1. Now, another time that is t r which is travel from a to c via b and. So, this is another ray of the refracted ray that is travel from a to c also, but via this a b p d and d c. So, this time required. So, if this angle is alpha c and this distance is h 1 or the thickness of this layer is h 1 or you can write this thickness of this layer is h 1 and this is the v p 1 is the velocity of the first layer v p 2 is the velocity of the second layer and this h 1 is the thickness of the first layer. So, this a b distance a b distance will be equal to h divided by cos alpha c. Now, if this is h this angle is alpha. So, we can determine a b that is equal to h divided by cos alpha c. Similarly, c d is also equal to h is equal to h is equal divided by cos alpha c and this distance b d that is equal to x n because x n is the total distance minus this. So, x n minus c e plus a f. So, x n is the total distance minus c e plus a f that will be equal to b d or e f b d or. So, b d and e f will be x n minus e c minus a f. So, we can write that this will be x n minus c e plus a f. So, this c e will be equal to h or h 1 this is also h 1 h 1 alpha c. So, ultimately we can write x n minus 2 h 1 tan alpha. So, now, we know a b distance a b distance a b distance we know c d distance we know p d distance. So, if I want to write the what will be time required t r that will be equal to b d a a b that is a b divided by v p 1 then plus b d divided by v p 2 and plus d c divided by v p 1. So, these are the three distance passing from a b then b d and d 2 c. So, here for this a b distance a b distance a b distance it is passing through first layer for c d it is also passing through second layer a first layer, but for b d it is passing through the second layer. So, we have to consider the second layer velocity when you calculate the time for this b d portion. So, now if I put this value of a b b d and d c. So, this expression will be h 1 divided by v p 1 cos alpha c plus x n minus 2 h 1 tan alpha c divided by v p 2 plus h 1 v p 1 cos alpha c. So, now, again we know that that sin alpha c that is equal to sin alpha c that is equal to v p 1 v p 2 as previously I have derived that sin alpha 1 is v p 1 by v p 2. So, alpha 1 is equal to alpha c. So, sin alpha c is equal to v p 1 by v p 2. So, cos alpha c that we can write will be equal to root over 1 minus sin square alpha c. So, if we put this alpha c sin alpha c value here. So, cos alpha c will be finally root over v p 2 square minus v p 1 square divided by v p 2. So, that means this one will be v p 1 square divided by v p 2 square. So, finally we can write v p 2 square minus v p 1 square divided by v p 2. So, in the same process we can write that tan alpha c is equal to sin alpha c divided by cos alpha c. So, that is equal to sin alpha c is v p 1 divided by v p 2 then divided by root over v p 2 square minus v p 1 square divided by v p 2. So, finally the expression of tan will get v p 1 root over v p 2 square minus v p 1 square. So, this is the expression of tan of this tan. Now, if I put the expression of this sin alpha c cos alpha c tan alpha c in our main expression that is this expression. So, we can write that t r will be equal to so, t r will be equal to v p 2 into h 1 divided by v p 2 into h 1 divided by v p 2 into h 1 divided by v p 1 root over v p 2 square minus v p 1 square. So, we have to put cos value here cos is root over v p 2 square minus v p 1 square divided by v p 2. So, then plus x n divided by v p 2 minus 2 h 1 you have to write the tan alpha c tan alpha c tan value that is v p 1 then v p 2 root over v p 2 square minus v p 1 square. Then the next part again this v p 2 into h 1 divided by v p 1 root over v p 2 square minus v p 1 square. So, now, if we simplify this expression. So, we will get x n divided by v p 2 plus 2 h 1 divided by v p 2 square minus v p 1 square. We take the tan alpha c tan alpha c tan this thing common then we will get v p 2 square minus v p 1 square divided by v p 1 and v p 2. So, this is we are taking common this v p 1 then v p 2. Now, finally, if I further simplify this expression we will get finally, t r will be x n divided by v p 2 divided plus 2 h 1 into root over 1 by v p 1 square minus 1 by v p 2 square. So, this will be the final expression of the time required to travel by this refracted ray from a to c. Now, the next step that as we have assumed that. So, suppose this is our equation number 2 and the direct ray equation that we have solved this is t d 1 this is equation number 1. So, as we have mentioned this is the critical distance. So, at critical distance suppose the c point is the critical at the critical distance. So, this at critical distance or t d that will be equal to t r that means, time required to travel by the direct ray and the refracted ray both are same. So, now, if we put this expression then at the critical distance that we can write this is our x c. So, at x c x c distance. So, x c x c divided by v p 1 that will be equal to x c divided by v p 2 plus 2 h 1 root over 1 by v p 1 square minus root over 1 by v p 2 square. So, root over 1 by v p 1 square minus 1 by v p 2 square. So, if we further simplify this expression we can write the 2 h 1 plus root over 1 by v p 1 square minus 1 by v p 2 square that is equal to x c 1 by v p 1 minus 1 by v p 2. So, if we further simplify this expression we can finally get that h 1 will be equal to x c divided by 2 into root over v p 2 minus v p 1 divided by v p 2 plus v p 1. So, this will give you the expression of the layer here we have written h 1 because this is the first layer we have taken. In general we can write that h or the thickness of the layer will be equal to the critical distance divided by 2 into root over the velocity of the wave in the second layer minus velocity of the wave in the first layer by velocity of the wave in the second layer plus velocity of the wave in the first layer. So, now, we have this is say expression number 3. So, we have 3 expression that is for the direct way expression and then the second one is the refactored time required expression and this is the expression for the this one is the third one is the expression for the thickness of the layer. Now, the way using this expression. So, now, in the first figure where I have explained that we can put this receiver at different distance. So, that means, at the beyond the critical distance we can put the receivers where we will get first the refracted ray. So, beyond the critical distance and in the critical distance if we put the receiver that the where the wave that will receive first that is the refracted wave. So, if we put this receiver beyond this critical distance at different points and we can receive the time required for this refracted ray because this is the first arrival of the refracted rays and the time required then we can draw a draw a graph. So, in this distance this is the distance x and this is the time t this time is the first arrival time of the rays. So, if we put the receiver beyond the critical distance at different points then we will get the first arrival time that is the refracted ray and if we put the receivers within the critical distance at different distance from the source point from the source that ray will give you the first arrival time or the first arrival ray that is the direct ray. So, now if I plot the expression of different expression the time versus. So, first up to the critical distance that the time required that will follow this expression number 1. So, that will follow and this is will be the linear graph. So, it will follow this fashion. So, next it will follow this expression number 2 for the next one. So, this will follow this fashion this will follow this expression 2. So, this is follow this one. So, at this that means there is a 2 graphs where the slopes are different. So, we can point out the point of intersection of these 2 straight lines. So, this straight line. So, from initial point to this point this will give us the critical distance because at this critical distance it will follow the expression 1. So, that is the first arrival time for the direct ray and this is this graph represents the time versus distance of the refracted ray. Now, if we extend this line. So, this is the time t i where x is 0. Now, the slope of this. So, this is for the equation 1 this is for the equation 2. Now, slope of this first say suppose 0 to a and say b. So, slope of this 0 to a line will give us 1 by v p 1 because from this expression we can see the slope is 1 by v p 1. So, this is t versus x n graph. So, that the slope is 1 by v p 1. So, similarly the slope of this graph that will give us the 1 by v p 1. Similarly, the slope of this second graph a second that means the expression 2 that will this is the slope. So, that will give us the 1 by v p 2. So, that means the slope of this second graph that is equal to 1 by v p 2. So, slope of this first a 2 0 to a that will give us 1 by v p 1 and slope of this second graph second portion that will give us 1 by v p 2. Now, if there is 3 layer system then we will get the another straight line and slope of this curve will give us 1 by v p 3. So, and then we can extend this line also. So, similarly we will get the slope of say different straight portion and that slope will give you the 1 by v p or 1 by of the different layers. Now, from this if we know the slope then we will easily determine the velocity of the each layer. Here this will determine the velocity of the first layer by this graph will be determine the velocity of the second layer by this graph will determine the velocity of the third layer. If we can if we draw this graph and you can determine this slope. Now, from this graph we know the velocities of the each layer. Now, the next step how to determine the thickness of this each layer of this. So, if I go for this third expression that the h 1 is equal to x e divided by 2. Now, from this expression. So, from this graph we can determine this is the intersection point corresponding distance that will give us the h c. So, now, if we can determine this h c value. Now, we know this velocity of this different layer. So, we can put the h c we will determine the thickness of the first layer. And another way we can determine the thickness of the. So, this is for the thickness of the first layer. Now, what we will get the thickness of the others layers also in general time we can determine the thickness of this layer. So, now, if in expression 2 that is in this expression if x n is or x is 0. And if in this expression 2 x is 0 which is because this second state portion represent the expression 2. So, here if x is 0 if x c is 0 or x is or x is equal to 0. Then t r will be equal to 2 h 1 root over 1 by v p 1 square minus 1 by v p 2 square. Or here we can write t r in terms of t i. t i is the time where x is equal to 0. So, we will get this value. Now, from this graph we can determine what will be the value of t i. If we extend this graph in this direction we will get the t i value. Now, if we put this t i value here and we know the velocity of the each layer from this slope then we can easily determine the thickness of the first layer. Similarly, for the second layer also if we extend the third straight line and then we will get the time at 0 distance. Then by using the t i value general expression we can also determine the thickness of this second layer also. If we know the thickness of the first layer and the velocity of the first second and the third layer because here we will get the velocity of the first second and third layer and here we will get the thickness of the first layer. So, by using this extend in this line we will get the thickness of the second layer also. So, in this fashion we can determine the velocity of the each layer and the thickness of this each layer and this is for this horizontal slope. Now, this technique we can use for inclined layer also. Suppose, here the layer is horizontal in this way, here also we can use this technique. Now, if this any layer is inclined this is for the inclined layer, there also we can use this technique and determine the angle, this angle of inclination of the layer, velocity of this of the wave between these two layers and the thickness at any particular point from the source that will also of this layer that will also and determine. So, this is seismic refraction serving. Now, next one that I will explain that is seismic refraction serving. So, first one is a refraction survey and this is refraction survey. Now, in this serving because this is this are the refraction the first one in the refraction. So, then in this survey also we will create one impact. Suppose, this is the source A, now it will create one impact and the ray will travel in this direction then it will reflect and then it will be received by this receiver at C and this B is the center section point between the layer. So, this is first layer this is second layer where velocity is v p 1 for the first layer and velocity of the second layer. Suppose, this angle is alpha this angle is also alpha. Now, here also this wave can travel from A to C directly and via this root A B C. Now, the time required for the direct travel from A to C again we can write if this distance is again x n we can write this is x n divided by B p 1. Another time required from travel from A to C then via B that we will get that A B plus B C divided by B p 1. So, now we can write that alpha tan alpha that will be equal to x n divided by 2 into h. Suppose, this thickness of this layer is h so this is for the second layer because here we will not be able to determine this velocity of the second layer we will only determine the velocity of the first layer and we can determine this is if there is any bedrock here then we can use this technique because in that case also but here we are not able to determine this velocity of the second layer. Now, this is alpha 1 now we can write A B will be equal to B C that is equal to h square plus x n divided by 2 whole square. Now, we can write this t that is the time required for this ray via B C from A to C that will be equal to 2 into root over h square plus x n divided by 2 whole square divided by V p 1. Now, further simplifying if we simplify this expression then you will get the expression of h in this form that is V p 1. This is root over V p 1 t square minus x n square. So, this will give us the expression of h in this form so this is one first equation and this is second equation. So, here also if we put the receiver at different distance from the source and then if we draw the graph so again this is distance x and this is the time. So, for the first expression for the first equation that is the this equation t d x n divided by V p i and if we put at different distance of the receiver then this x n will change similarly the t d will also change and it will follow a linear fashion and this the slope of that linear graph will give you the 1 by V p 1. So, we can draw that linear graph so this is for equation 1 and slope of this graph that will give us 1 by V p 1. Now, again if I draw the slope or the graph of this second equation so it will follow this path. So, this is in parabolic in nature of the second expression this t for this second expression if x n equal to 0 then we will get a time t. So, if I draw the second expression time versus x n so that will follow a parabolic nature graph and if the x n is 0 then we will get a time say t 0 so this time say t 0 is the time. So, it for this so we will get the if we draw this first equation curve the different distance though we will get the velocity of this layer now we have to determine the thickness of that first layer or that particular layer. So, at x equal to 0 t will be equal to t 0 that is equal to 2 h root over V p 1 because in this expression if I put x n equal to 0 then t will be equal to t 0 will be 2 h divided by V p 1. So, that is 2 h V p 1. So, now from this expression if I know this t 0 now using this expression also we can write h is equal to half t 0 V p 1. So, from this graph we have to use the 2 graphs. So, from the first graph the slope we will get the 1 by V p 1 from that we will get the velocity of this layer and from this second graph where it is intersecting the time axis there we will get the t 0. If we know this t 0 we will put this t 0 here and the velocity and then by using this expression we will get the thickness of this layer. So, in this 2 serving that method that I have discussed. So, for this serving we will get the thickness of this different layers and the velocity of the wave that is passing through this layer at different layer. So, by this first seismic refraction serving this seismic refraction serving if we use this seismic refraction serving that is where we will can determine the thickness of different layers and the velocity of the waves at different layers. But the conditions is that under this situation where this V p or the velocity of the wave will be greater or that will increase if I go in the further depth. So, that is one condition, but where we will get the thickness and velocity of the each layer. And for the seismic refraction serving where we will get the thickness of the wave the thickness and the velocity of one particular layer. Now in the next class I will discuss the other 2 geophysical serving method and then I will start this shallow foundation part. Thank you.