So, good morning to all of you, what I will do now, what we have studied yesterday mostly as we see as we have seen we have mostly you know reviewed the vector mechanics. So, through that we are able to correlate the vector mechanics for the application of forces. So, through vector mechanics we have learnt we can now add the forces to get the resultant. So, concurrent system of forces to get the resultant that was the first part. Then you remember we talked about the vector product. Now application of vector product that is the cross product was that we are able to take the moment of a force about a point. So, r cross f that concept was studied and thirdly what we have done is look at the cross triple product or mixed triple product. That means and that was important to get the moment of a force about an arbitrary axis. So, we have learnt that we can get the r cross f and then we are going to take the dot product of the moment vector with the unit vector lambda of the line. So, we have seen that lambda dot r cross f will give me the moment about that axis, so it is going to be a scalar quantity. So, once we have studied all of this, then we have really gone to the equivalent system because now the idea is that I should be able to represent a system by its force resultant as well as moment resultant. So, we say that it is a force couple system equivalent force couple system that is what we have studied. Now we said that this equivalent force couple system that we have we can actually simplify it for several system such as if we have a parallel force system and if we have a co-planar force system, we can actually replace them by a single force and this single force can be obtained just by the concept of equivalent system. That means we are trying to do the moment balance as well as force balance. I want say balance, but equating the moment of the original system with that of the resultant system. So, same concept we have studied when we try to find the centroid of bodies. So, remember in the centroid of bodies our first assumption that we made, suppose I have a thin plates. So, if I have thin plate then we said that okay thickness is uniform then what can happen? I am really considering elemental area small area dA and that dA therefore we will have a force which is simply given by the area multiplied by density multiplied by thickness. Now remember therefore what is happening this elemental load is not varying from the axis they are actually proportional to the area. So, what we have done we have discretized again this elemental you know the entire plate into elemental areas and again we looked at okay get the first moment. So, we really studied the first moment and that first moment should be equals to the net weight multiplied by the centroidal distance. So, again you can see that we are really equating the moment. So, ultimately once we have studied that okay we can actually you know solve this type of typical type of problems by simply equating the moments then we can again study one more important aspect that we see in lot of different applications that will be distributed load. So, what we are going to consider today is basically distributed load and we are going to buy the concept of centroids that means we have already learnt how to calculate centroids of small areas. So, we will be able to take those results and try to apply in order to find out the resultant load that is coming from the distributed load. So, today we are predominantly going to look at again distributed load concept now distributed load can come from different effects you can simply think of a beam and assume let us say beam you know if the thickness and area remains uniform throughout then I can say that it is a uniformly distributed load. So, we are really focusing on here see now if I really try to convert the load here the main thing that we are attempted to do here is that plotting the load per unit length. So, if you look at the unit so unit is really Newton per meter. So, how do I get that suppose I give you a beam for the time being just assume that we have uniform cross sectional area in that way remember if I multiply the cross sectional area with the width that means the B which is out of the plane. So, in this way I can actually convert it to a line load. So, ultimately what we are having is that we want to convert the weight to a line load. So, assume that we do have a line load or what with the distributed load. So, basically now we are saying that this distributed load is actually W x. So, W is a function of x now we want to replace this by a single resultant load which is W. So, the concept remains again same therefore what we really want to find out we want to find out the area under this load curve that is one part and then we want to decide that we want to calculate the centroid of this area where the net load will pass through. So, again we resort to the concept of equivalent system now you are looking at that this is one system where I have chosen an elemental area. So, this is a elemental area you can again see this is a thin strip that thin strip has an elemental area equals to W multiplied by dx. So, that is the elemental area now this will produce a moment and we can integrate that moment over the span x and that moment must be equals to the net weight that is the W that is the net load multiplied by the x bar which is the distance of the resultant load a single load from the y axis. So, ultimately first we are finding the area this is a very standard process that we have studied earlier. So, our area should be equals to the resultant of the load curve that is giving me the single load and then I have the line of action that line of action is given by simply x bar W multiplied equals to integral of x d W. So, x d W you can see here that d W is the elemental load. So, the elemental load is W dx multiplied by x.  So, it is exactly going to be same as that of the centroid. So, as long as we can really get the centroid of this then I can simply put the resultant load at that centroid. So, we will consider several problems now. Just look at this beam and let us assume that beam is under this loading. So, one section that you see that one span that is half of the beam has a uniformly distributed load look at the unit again it is Newton per meter. So, now another span has again a trapezoidal distribution. So, we have two distributions here. So, and on top of that remember we are also adding some concentrated force 100 Newton here as well as we are also applying a couple moment. Remember this is a free vector in this case. So, this is going to be 300 Newton meter.  Because remember this is a planar force system now therefore, if we try to resolve the resultant force and resultant moment at O they will be perpendicular to each other and therefore, the entire system can be replaced by a single load as we have studied before. So, ultimately as I said that we are going to first calculate the centroid we are trying to get the equivalent concentrated load of this distributed load and then we are going to do the moment balance of all external forces and moments about O. So, just look at it systematically. So, what we are essentially doing here we are now first breaking this load into two parts one would be uniform distributed load over the entire span the other will be simply the triangular load. So, if we do that now the distributed load if we look at it firstly remember that we want to get the resultant load of this uniformly distributed load that is simply equals to 10 multiplied by 12. So, that is given by the area of this load and this one the one we have here triangular load again there will be a net load that is going to be there is half times 6 times 20. So, we have total load equals to this 180 Newton. Now question is this 180 Newton we want to find out where it is applied on the beam such that this distributed load is replaced by a single load. So, now what we are going to do we are simply remember this 10 times 12 this load has to be located at its own centroid that distance is 6 meter. So, what we have done area that is 10 times 12 multiplied by 6 plus we have done the triangular load is there right. So, we are taking the moment of this load and then moment of this. So, moment of the single load we have replaced this already that single load was half time 6 time point as you see multiplied by its distance that is 10. So, now that divide by the total load so total load was 180 therefore we have the solution is 7.33 meter. So, the ultimate idea is that we now have a single load that single load is located at 7.33 meter. Now remember we want to further reduce it by a single load equivalent load. So, therefore we are going back to yesterday's you know lectures. So, what we are doing basically now this entire system will has to be replaced by a single load. Now what is the single load which is coming 100 plus 180. So, that is the resultant right now remember that resultant will be applied here. So, 280 will be here and then you are going to get the moment resultant about the point O. So, about the point O what is the moment resultant moment resultant is simply going to be 180 multiplied by 7.33 plus 100 multiplied by 3 and this one remember is going to be subtracted. So, this is going to be negative 300. So, I am going to get a moment resultant about this point now how the direction of this applied that direction will be clockwise remember that direction of the moment will be ultimately clockwise. Because this is taken clockwise this is taken clockwise and this was taken anticlockwise. So, final moment here will be clockwise. Now what we have to do we have a resultant force which is applied downward here and we have a moment resultant here. So, I can simply shift that resultant load that is 280 Newton to some distance I am trying to find that distance. So, therefore, I am simply equating the moment. So, 280 multiplied by x bar which I want to find out. So, I want to find out R you know x bar. So, R is already known that R is the total R that is equals to 280 and therefore, we have x bar equals to the moment divide by the total load. So, that is 4.714 meter. So, therefore, as you can see now the entire system is being replaced by only a single resultant load that is acting at a distance 4.7 meter. But we have adopted here the concept of centroids also. So, what we have done basically we have discretized the entire load into small parts and we have looked at what are the individual centroids of these small you know loaded areas. So, we are simply taking the resultant load of this individual areas at those you know at the centroids of those areas. Now, we are again going to look at another problem that is very important that will be again hydrostatic pressure and you have seen this in several civil engineering and mechanical problems that in civil engineering problems specially related to dam we have the water in one side of the dam. So, that is going to you know apply the load on the dam. So, again we are interested to find out if I have a water pressure distribution then what is going to be the resultant load on that dam. You can also find from the resultant load what is the movement at a given point in the dam. Similarly, if you consider the mechanical system they also use different types of valves. Now, these valves are actually you know could be one way or two way but again you may have water in one side of the valve. So, therefore, this valve will take the you know water pressure. So, we again want to look at the resultant load that is applied on those valve. So, first thing that we do in this type of studies is that all of us know that what is the pressure at a point from a distance let us say or at a depth from the water level. So, we have the water level here let us say depth here is d minus x. So, here I have chosen the x from the bottom of the dam. So, let us say this is the x. So, therefore, the depth is d minus x. So, pressure is rho g d minus x remember the pressure will vary in a triangular fashion as we go from the top to bottom of the dam. Now, how I am going to convert this load to a distributed line load. So, my objective will be always to convert this pressure distribution to a line load and it can be easily done. Basically, we have to multiply it by the width the width of the dam. Now, dam can be very long in that case the width can be really long. So, in that case we can just look at maybe you know 1 meter distance or so ok. So, we will just take a section on the other direction in case of a valve there may be finite width. So, we can say that finite width is b. So, ultimately the step wise if you want to go ahead first what is the hydrostatic pressure at a distance x from the bottom that is rho g d minus x. Second how do I convert it to a line load at a distance x that will be w x which is nothing but p x multiplied by b. Remember, therefore, the small force that is acting on this rectangular area that is the strip we have chosen. So, on that strip that strip is infinitesimally small with a length here dx and width b. So, the area is b dx. So, therefore, the small load that is acting we call this dr that is simply given by rho g d minus x b multiplied by dx ok. So, this is the w x multiplied by the dx we can treat like that. So, we can simply say w x multiplied by the dx that is going to be the small load dr. So, therefore, what is the total thrust? So, again total thrust is simply going to be now you are really looking at this. So, we have to simply integrate this small elemental load from let us say 0 to the depth. So, 0 to d that is the height 0 to d of the dam. So, we are going to do that and we are going to get the net load. Now, this is really the integration approach we are again following, but we need not to do that. Remember, we already know that the resultant load will simply be given by the area under this load curve which can be calculated. What is the area under this load curve? Remember here the value is rho g d multiplied by b. So, rho g d b is the value of the distributed load. So, that is the base and height is d and then we can get the area. So, rho g d b multiplied by d divide by 2. So, half times base times height will give me the same result. And we also studied that what is the centroid of this? Everyone will know that centroid is simply going to be located at d over 3, but if you want to prove that fine. So, then how to prove that? Then we are going to get the moment of this elemental load. So, the moment of this elemental load will be simply x multiplied by d r. So, that is the integration approach. We get that and then we say the moment produced by the elemental load. So, this moment at the base should be equals to r times x bar. So, finally we are going to get x bar equals to d 3. Now, here the point I am trying to make this is again you know we are basically reviewing how we have calibrated of a body. So, I have really followed the same approach as we have done. So, this is really taking the first moment and so on so forth to get that centroid of that area. So, now we have already learnt that centroid will be at a distance d over 3 and the resultant load that will be simply given by the total area of this distributed load. So, we will just try to take a quick problem. So, just understand the problem carefully here. So, there is a automatic valve that consist of a square plate of uniform thickness. So, you can see the valve here and you see the cross section of this entire valve. So, that is the cross section it is a square cross section. Now remember the valve is pivoted at A. So, it is hinged at A that means this valve can rotate about A. Now remember it cannot rotate counter clockwise because some arrangement is made to prevent the counter clockwise rotation. So, it is only allowed to rotate clockwise. So, what we want to find out is that determine the depth of water d for which the valve will open. So, think of it like this way I keep filling the water keep filling the water. So, ultimately due to this water level I am going to always get some pressure distribution. Therefore, some distributed load on this valve and I can always find out what is the resultant of this distributed valve where it is active on that valve. So, as I keep filling the water there will be a situation when the resultant load will be passing through the hinge. So, remember when should the valve will open the valve will open when it is going to see a moment that is applied at point O which is clockwise right. That means at the limit we can say as soon as the resultant load if the resultant load is really just little bit above point A the valve is going to open. So, in the borderline what we are saying that the resultant load must pass through A. So, the ultimately if we want to look at the solution key to this problem I want to determine the depth of water and that depth of water can be found when the resultant load is passing through the point A or passing through the hinge. That means in other words the centroid of this result you know of the distributed load that we are going to see here the centroid of that should be at 100 millimeter. So, the ultimately what I have just described now is in this slide right here. So, ultimately is the center of pressure that is the resultant load must pass through A, but there are steps involved that we have to carry out because there are several situations can now happen. Just think of this problem what are the typical situations can happen. Let us say the depth of the water is below the valve or let us say the depth of the water is up to the valve. Let us say depth of the water is up to the valve now the height of the valve is 225 millimeter right. If the depth of the water is up to the valve then 225 millimeter in this case what would be my centroid of the resultant load because I am always going to get a triangular load distribution right. So, due to that the centroid will be of that one will be 225 by 3. So, 225 by 3 is always going to be less than 100 therefore the gate cannot open clear. So, therefore as we can see that the depth of the water has to be on top of the valve. So, finally what we have concluded that if the depth of the water is up to the valve then the gate is not going to open because my resultant load is not going to pass through the hinge, but it is going to be below the hinge the depth will be 225 by 3 that is the distance of the resultant load from the bottom. So, 75 is less than 100 therefore the gate cannot open the valve cannot open. So, we have to really look at now when the depth of the water is above the valve ok. So, now we consider this depth D. So, remember what kind of line load distribution now I am having on the valve. If the depth is greater than the 225 millimeter which is the valve height then the pressure distribution on the valve if you see if you convert it into line load it is simply going to by a trapezoidal right. So, it is going to look like a trapezoid. So, ultimate idea is that if I know the centroid of the trapezoid if I can calculate the centroid of the trapezoid that centroidal distance is where the resultant load is applied is in that and the problem demands that the resultant must pass through the hinge. So, therefore that centroidal distance is equals to 100 millimeter already, but the unknown quantity here is D right. So, we have to find the D such that the centroid of this is equals to 100 millimeter that is one approach. The other way we can look at is just the regular integration approach. So, we are saying now we want to say the what is the pressure at a distance x. So, pressure is rho G D minus x what is the distributed line load. So, W x that is acting at a distance x should be pressure P x multiplied by the width B clear. And therefore, the elemental force that is acting is equals to W x multiplied by D x. So, if you look at this D x in this way it is D x is here. So, W x D x is this elemental force. So, now what the total force on the plate will be R that R has to be calculated. So, in this case what is the R 0 to 225 puts be my integration limit right because this D x D x is varying between 0 and 225 millimeter. So, I am going to get this. So, I calculated the R already. Now, I have to find out the moment due to this elemental force. So, ultimately the moment produced by the elemental force at the base that I am going to calculate that is x D R right. So, I again do the simple integration in this case. So, x D R is going to give me this. Remember everything is coming out in terms of D, D is my unknown. So, ultimate idea is now R multiplied by x bar that should be equals to the moment that is produced by this distributed load. So, I am just equating the moment due to the resultant load and due to the distributed load. So, ultimately you see that the center of pressure let us say this is x bar instead of alpha is equals to simply m by R. Remember that alpha is equals to nothing but 100 that is given in the problem already ok. And therefore, what we see that m equals to h multiplied by R. So, from this h is already equals to 100 in order for get to open right and we can find the D. D equals to 400 and 50 millimeter ok. So, in this problem what we have understood is that as we keep filling the water whenever the resultant is passing through the hinge it is just above the hinge it is going to give it a small clockwise moment and due to this the valve is going to rotate clockwise and it is going to open. Now, as I said we have again you know looked at the concept of centroid. I am not saying that we always have to follow this integration approach. We could have just simply looked at the centroid of this area trapezoidal area. We can get the area of this trapezoid. We know what is the centroidal distance of the trapezoid and this centroidal distance should have just been 100 mm to solve for the depth D clear. So, now we are going to look at a similar problem. However, in this problem what we have done instead of taking a square gate we have seen we have taken it in terms of an isosceles triangle. So, it has 225 is the base and height is also equals to 225 millimeter. So, base is 225 and height is 225. Now, remember what is happening in this problem your distributed line load is going to completely change from that of the square plate that is the only difference. So, the ultimate idea again would be how do I get the distributed load on this triangular area. So, let us go again step by step. So, remember pressure distribution will always be triangular that is no problem. So, pressure distribution if I look at it it is always going to be triangular. For the time being just assume that the water level in this case is 225 millimeter just to begin the problem let us see what is happening. So, let us assume the water level is 225 millimeter and try to find out where the resultant of the distributed load is going to be. So, let us start. So, therefore, now we see the pressure at a distance x if the height is 225 millimeter of the water then the pressure at a distance x. Now, x is always measured from the bottom of the dam. So, therefore, the pressure is rho g 225 minus x. Now, this has to be multiplied by the width in order to convert it to distributed load. Now, this width of this strip varies as you go from 0 to the height of the valve. The width is going to vary that is the only trick in this problem. Now, what is that width I am talking now? That width is now dependent on x. So, we can use the similar triangle to find the width in terms of x since the width is varying because the pressure needs to be multiplied by the width if I want to get the line load distributed line load on this triangular plate. So, I have b by 225 from the similar triangle equals to x by 225. In this case since it is an isosceles triangle therefore, we have b equals to x is that clear.  Now, that is wonderful. So, what we get? We get a quadratic line load. So, variation of line load is really going to be like this. Remember at x equals to 0 it is 0, at x equals to 225 which is the height of the valve it is again 0 and it is going to attain a maximum peak which is given by simply 225 divided by 2. So, that means the line load in this case is going to form this shape of a parabola let us say and we know the centroid of this is exactly going to be at the middle of it. So, if we consider the depth to be 225 millimeter the height in the depth of the water 225 divided by 225 millimeter then the resultant load should be at 225 divided by 2 that means 112.5 which is greater than 100. So, the gate will open the valve will open is that clear. So, if you think reverse way you know intuitively just think it out therefore, in order for the resultant load to be at 100 millimeter that means, if the centroid is at 100 millimeter then what is the depth of the water? Any answer in the module? What is the depth of the water? I think some of you are writing yes that is a great answer. So, the answer will be 200 millimeter. See let us see when we are telling the students you know we do not have to in this case we need not to do any math once we get the distributed line load correctly we should be able to simply you know get the solution to this particular problem. So, ultimately the next two slides I am not going to show you what is happening remember the main logic that we have looked at the parabolic line loading once we have the center of gravity should be at L over 2. So, 225 by 2 if the depth of water is 225 then that simply my you know line load CG will be at 225 by 2 which will allow the gate to open. See in order to avoid that now what we are saying the depth has to be less than 225 millimeter so therefore in order for resultant to pass through this hinge so depth should be 100 is already given that is the centroidal distance. So, that multiplied by 2 so 200 millimeter will be the depth now as I said that you can actually verify all of this through the integration direct integration approach. Let us say if you really you know if the students cannot get the concept at all they can simply adopt an integration procedure they will land up getting if the height is 225 millimeter if they assume then you will see that x bar is coming to be 112.5 for this problem this will be greater than 100 millimeter. So, the gate will open from that we can say the height of the water should be then less than 225 millimeter. So, the next slide if you consider the height of the water to be less than 200 millimeter 225 millimeter then just use d here so everywhere instead of 225 millimeter height of the water is d you start the problem you go through the integration approach try to locate the centroid and you will ultimately get an answer that centroid is equals to if you look at m by r centroid will be d over 2. So, d over 2 is already being said to be 100 mm right d over 2 is 100 mm. So, ultimately you know you get the d what is arcs that is 200. So, centroid here will be ultimately 100 mm that is your height clear. So, point that I am trying to make you we have looked at this problem conceptually then you know if one has to follow the integration approach he or she can actually adopt the integration approach also, but ultimate outcome will be same only thing is that it is lengthy process and you are spending unnecessary time because we have not looked at the pattern of the distributed load carefully. Once the pattern of the distributed load is clear then we can solve this problems very easily by taking the concept of centroids. So, we are simply calculating the area under the distributed load that is my resultant force and the centroid of it which is also known to us for simple steps. So, we have covered you know these two problems. So, these two topics you know the topics that we have really covered how to get the distributed load and typically these are what we do at IIT also. So, one is the distributed load on a beam and another one is the hydrostatic pressure. So, how to look at the problems involving hydrostatic pressure? Now briefly what I will do to discuss you know into little bit of depth I am going to go to and try to discuss this hydrostatic pressure on inclined as well as curved surfaces. So, now we are going to look at the inclined surface. Let us say you have a submerged body again it could be just a plate let us say or let us say it is a surface of the dam which is inclined. Now if it is inclined then we can still work around it with whatever we have learnt because remember what is happening the pressure that is going to act on each and every element that is going to be the perpendicular to the plane. So, the pressure is always going to be perpendicular to plane therefore if we convert the pressure to the distributed load right. So, again in this case the pattern is going to be trapezoid because here you can see this depth is d1 let us say depth is d1 then this value that you see here that will be rho g multiplied by d1 right that is the pressure and we have to multiply by the width. So, width is something that is perpendicular to the slide. Similarly here if say if the height is d2 then it is simply rho g d2 this height is going to be rho g d2 this value right here distributed load rho g d2 multiplied by the width perpendicular to the plane of the paper. So, as such the inclined section will not have any difference than a vertical section just remember that it is not going to have any difference except for that load is now going to applied perpendicular to its own plane. So we can basically solve this using the same technique that we have followed because I know where is the centroid of this. So, therefore the resultant load is going to act through the centroid and we know what is this resultant load that is simply going to be the area of this load curve. Problem appears when we are going to deal with curved surface. So, if we have a curved surface it is now not easy to solve this class of problem using integration approach. So, we will try to look at some mechanistic approach instead of integration approach in order to solve this kind of problem. So, what kind of loading I am going to see on top of this curved surface. So, what we have to do in this case remember we are interested in finding out the resultant load. So, what we will do essentially instead of looking at this resultant load that is applied by the water onto the curved surface I am now going to study a problem whereby the resultant load is being applied by the curved surface onto the volume that is occupied by the curved surface. So, we are simply going to look at Newton's third law. So, what is done here we are going to have a small body where the volume occupied by the liquid or the water is being detached. So, think of this volume now that is A D B. So, remember the volume you have a third direction also that third direction depends on the width of the dam or width of the valve as we have studied. So, we are talking about now this one. So, this volume is detached on that volume what are the loads going to come into play. Now, this fluid volume is determined by two surfaces one is the A D one is the D B and remember I also have this surface on this surface the idea is that I am trying to find the resultant load. So, this resultant load that is applied by the fluid on the curved surface is simply reversed here. So, ultimately what we have to find out we have to find out this load that is R 1, R 2 and the weight of the liquid itself. So, ultimately this free body diagram has four different forces one force is coming from the water that is on top of this surface if there is no water on top of this surface then this force becomes 0. If there is water then it is going to be simply uniformly distributed load is that clear on this surface as we all know it is a vertical plane now. So, the distribution will be a triangular because pressure will act perpendicular to this plane and that pressure is going to vary as we come down. So, I am going to get a triangular load here clear. So, ultimately we have uniformly distributed load due to the pressure because pressure is going to be same everywhere right and in this case I am going to have a varying load that is the triangular load distribution here W I can already find out that is the weight of the body that can be found very easily because I know the area of this and therefore, what we can simply study the resultant of this system is going to be the R that is given by this clear. So, the concept is very simple in this case as such what we have to find out what is my R 1, what is my R 2 and what is my W. Remember R 1 is going to be 0 if there is no water above this surface. So, only thing that I will be now interested in R 2 and W once I solve for R 2 and W then problem is actually solved. So, it is more complex problem, but if you think of from the mechanistic angles all I am saying the resultant of these three forces is going to be the resultant of the net pressure distribution that is going to act on this curved surface. So, if we buy that concept. So, ultimately just look at again this is just a demonstration purpose what we have studied is basically first we detach the volume of the liquid. So, weight is coming into play that is one force then the resultant R 1 of the forces exerted on the A D on this surface then the R 2 force that is exerted on this surface D B and we are interested in finding out the resultant negative R of the forces exerted by the curved surface on the liquid. Now, remember this negative R has the same line of action as that of positive R that is coming from the liquid to the curved surface. Now, as I said W R 1 and R 2 can be determined by standard methods you can simply look at you know areas of small distributed loads to get the R 1 and R 2 and W of course. These are just coming from the area under the load curve and we also know where they will act because I know the centroid of those small small areas clear. So, ultimately once we have those we should be get we should be able to get what is the negative R and from the negative R we can find what is the R which has the same line of action, but opposite in sense. So, we will study a very complex problem, but in a very simplified form. So, just try to study this problem. So, it is a concrete dam and remember all the units here in foot and ultimately when we are looking at the specific weights we need that then we are going to simply look at pound per foot cube. So, it is a FPS unit is being followed. So, let us forget about that for the time being. So, what are in questions draw the free body diagram of the dam by considering all the forces acting remember I have the water here and this shape is a parabola. This shape right here where the water is being held this shape is a parabola from C to B I have a parabola. Now, we have the shape of the dam. So, shape of the dam can again be discretized. So, you can take as a triangular area here you can take it as a rectangular area and then again a parabola here. So, dam can be discretized in three areas in order to get the total weight or different forces that is going to come. So, now represent all the forces by an equivalent force couple system at A just understand it very carefully. This is really the equivalent system in a force couple system represent all the forces by an equivalent force couple system at A at the base of the dam. Remember since it is a planner problem this equivalent force couple system that we are going to get at A resultant you know force is going to be perpendicular to the moment or in other words moment is going to act perpendicular to the force. So, the resultant force since it is a planner system I can replace it by a single resultant load we have already done that. So, the third question part C replace the equivalent force couple system by a single force acting on the dam. The resultant pressure exerted by the water on the face BC of the dam that is what we have discussed. How to get the resultant of the pressure forces exerted by the water on the face of the dam? One thing just note what is being asked consider a one foot thick section. So, that means on the other direction third direction I have one foot thick section. So, let us look at very conceptually not look at the numbers how I am going to solve this problem I have to draw the free body diagram part first. Remember the free body diagram is the heart of the solution to this problem if I cannot draw the free body diagram it is not going anywhere. So, if I just try to quickly look at the free body diagram how it should look like very very clearly ok. Just look at the dam part this is my concrete dam how I have taken all the loads coming from the dam. So, I have really discretized this dam into three different parts this is my triangular part this is my rectangular part and this is the parabolic part is that clear? Does anyone has any question how do I find out this loads if I know the shape of this dam I do not think so, because that is now crystal clear to us that how I am going to find out this load ok. So, let us not get into the numbers. So, I can find out the load here let us say I can find out the load here I can find out the load here I also know the line of action of these loads. So, these loads are coming from the area right and then we have to simply multiply by the width this one foot multiplied by the specific weight specific weight is already given is that clear to everyone we know the line of action of this load for example, where it is acting this is a triangle. So, it should be from two-third from this and one-third from this base right. So, that is there. So, it is six foot this is in the centre ok. So, here also I know where it is because that we have to go back and look at for this parabola where should be the line of action of this load you can show that it is three foot if this the total is the ten foot ok. Now, only thing that remains on this part now as I said this part what are the things coming into play in this part we have still not detached this volume. So, what we are showing here if I just consider this volume right here you have a load that is taken by the weight of this fluid. So, you have the weight of this fluid that is going to take this load right and then this phase is now open on this phase you are simply going to get the distributed triangular load right it is as similar to what we have shown before. So, if we go back here I am really interested in finding out this R 2 and as I said there will be a triangular pressure distribution on this phase. So, this triangular pressure distribution will create the R 2. So, if we look at the numbers now so it is clear how do we get the W 1 it is all here how do you get the W 2 W 3 and W 4 all of this can be done from the concept that we have studied. This one remember I have to get this P that is my R 2. So, that will be given by simply the area of this triangle right. So, if I know the area of this triangle I can get this force P clear now what is being asked find out the equivalent force couple system at A. So, to get the equivalent force couple system now we just simply look at that what is the force in this direction in the x direction let us say. So, that is going to be equals to H right. So, in that direction it will act. So, that H will be simply given by this value it is simply the horizontal force there is no other force that is acting horizontal force. Similarly, all of this vertical forces will also induce a net resultant force that net resultant force will be equals to V which is nothing but sum of all the vertical forces. So, the V will be 47840. Just remember it is going to be opposite of what is being shown here. So, H is on the left hand side V is also going to be on the down outside moment resultant you can also calculate I am not going to go through that just take the moment of all these forces at this point. So, now that is the equivalent force couple system at A all of us know now how to get that right equivalent force couple system here can be now represented which is composed of one horizontal force one vertical force. So, there is again these two will have one single resultant H and V if you look at it and then along with that I have a moment result. So, therefore again it is a planner system. So, I should be able to now replace you know take this V on the right hand side right such that V times distance that V times distance is D that V times distance should be equals to the moment resultant M. So, to get the distance D I have moment resultant M multiplied divide by the V which is going to give me the distance. So, now the single force that you are going to see here is basically again V and H this is going to give a single force which will be you know your R that is acting based on what the values of V and H are. So, there is going to be a single force acting in this dam at a distance D, but just tell me is there any part that looks complicated here if we understand the centroid of all of this individual area does not appear to be. So, the next part is how do I get the. So, let us go back to the problem one more time that last part was find the resultant of the pressure forces exerted by the water on the phase B c of the dam. We should have just taken this problem we should have just taken this part instead of going through all of this, but again it is a good exercise for the equivalent force couple system. Remember we are going to replace the effect of all the forces by a single resultant force that we have already done. Now, this part would have been done separately to get this part as I have illustrated before we will simply detach this volume. So, if I detach this volume. So, just look at this part right here if I detach this volume right here I have a weight right of this liquid and this p already being solved this p is coming from the tangular pressure distribution right that p was solved in the previous slide. If you look at this tangular pressure distribution you can see here that is the p right. So, once we get that then we are going to have now you have basically two forces here w 4 and p and the resultant that r that I am interested to find out can simply be you know used from the triangle rule. So, you can simply adopt the triangle rule to find out what is that r. So, in this case it is negative r that is being pointing this way finally, remember the liquid is going to apply the resultant force on the other direction, but that will be plus r this way. So, what is shown here is the plus r that plus r is going to be this much is that clear. So, we have now learnt the distributed forces and how to you know look at load on a beam right. So, distributed load how to convert to a single load. So, we have studied a simple problem on load on a beam and then we have also you know studied the distributed pressure that is the pressure due to the water how to resolve that into a single force. So, I will just keep maybe few minutes five minutes for just discussions then we can do a small tutorial on this just to make sure that we are on same page. So, any discussions please raise your hands or please flag in the moodle. Yeah, 1 2 0 3, go ahead we can hear you. Yes, we can hear you. Nyanamuni College of Engineering from 1 2 0 3, how do you calculate the centroid or centroid gravity for irregular geometry look like a human being because so far we have discussed it for all cases regular geometrical figure. If it is I want to calculate a centroid gravity for human being how do you calculate is there any standard procedure for that please sir. Yes, so I see even I am not aware of any standard procedure that is available. So, the your question is can I really try to find out the centre of gravity of any arbitrary shape and sizes right or let us think of not the you know centre of gravity at that point think of even a plate of arbitrary geometry just a plate ok right. So, that is a two dimensional body much easier to understand remember you can always discretize this body in a regular shape is in that I can always see there is a finite element method so, if we buy the concept of finite elements any shape right can be discretized any arbitrary shape can be discretized in small regular shapes at the edges what happens at the edges they will only differ. So, therefore remember depending on what kind of edge you have you can you may adopt a you know smooth line there. So, instead of let us say you have little bit of curve I can always that little curve can be jointed based on a small line. Therefore, the point that I am trying to make any arbitrary geometry can be discretized into finite elements small small elements right. Now, you take moment of all those elements so you take the first moment of that element right that should be equals to the you know net weight that again you can calculate by summing the effect of small small weights right. So, that net moment times x bar equals to your first moment see ultimate idea is you should be able to discretize the body if I cannot discretize the body in known shapes I cannot get the answer is that clear clear. Hello sir we are having one more doubt sir regarding this hydrostatic pressure the problem that you saw no sir with an isosceles triangle cross section sir. I am sorry. The problem with isosceles. Yeah, yeah triangular cross section triangular isosceles triangle yes. Triangular cross section yeah. Here the pattern of distribution how we are coming to know that it is an parabolic. Yes. So, now again going back to the question came from 1 2 0 3 is that when we have considered the triangular section ok how we have decided on the parabolic line distribution of the load right. So, distributed load was parabolic now that came from the very basics what we have done first I have calculated the pressure at a depth x which is rho g multiplied by the depth. So, pressure is not changing now remember that was multiplied by the width. So, pressure has a unit Newton per meter square I have multiplied by the width right that width was b right that will give me the distributed load which is in line load Newton per meter because I am multiplying by width. Now in the triangular case the problem was the it is a since it is triangular my width was varying at different depths clear. Now I had to relate that width I had to relate that width in terms of the depth ultimately since it is isosceles triangle the width was equals to the depth think of it. So, width the b was equals to x now you have pressure equals to rho g x let us say you multiply again by x therefore it becomes a parabola clear. So, distributed load becomes parabolic and we know what is the center of that parabola is that clear now ok. So, that width plays a very important role when we are converting to the line load that width plays and very very important role and that is the whole point of discussion there. So, square plate that is not coming right because b is not a function of the depth ok. 2 1 5 go ahead we can hear you sir. Sir is there any problem on which we can find the shifting of center of gravity or center of mass as we then did in class 12. So, your question is is there any problem where you have you know shifting of center of mass see the center of mass will not shift center of mass of not shift will just come to that topic in the next section will not shift the center of mass, but what will try to get in problems involving will try to get the inertia moment of inertia about some other axis. So, let us say I know the solution on the center of mass right then how do I shift it using the parallel axis theorem to other points that is what we are going to discuss we are definitely going to discuss that in the next section, but again your question is relevant to the application what application we are talking about ok. Center of mass will not change ok. But sir is there a problem sir when a man is walking in a boat is still in the water then there is some shifting of center of gravity. No that is absolutely not the question center of gravity remember what is changing the result and load on the body. Center of mass of the let us say human being is working on a raft right the raft problem we have solved right the floating floating a floating boat they are the no we have not say center of mass of that floating boat is not changing what is changing due to the presence of those children's right the ultimate result and load that is shifting. So, that cannot be told to be center of mass say center of mass is always localized to a body what can change if you have other forces on to it it will simply shift the location of the net result and load. So, the body will start now try to rotate its center of mass because you have we are going to have the moment about the center of mass is that clear ok. So, the body will try to rotate just think no if you have a just floating something is floating you go and step why should it try to always rotate because ultimately the resultant force due to your presence on the boat that has been shifted. So, that resultant force has the weight of the floating device as well as weight of you that is giving you the net weight that is being shifted. So, net resultant force has been shifted and due to that you are going to get a moment about the actual c g of the body. Moment about that center of gravity. Yes, yes that center of gravity. Center of gravity will remain at the position. You have your center of gravity. The net resultant will create a moment about the center of gravity. That is why you are trying to tilt ok is that clear. Then it will rotate. Ok. Thank you. There is an imbalance in this case because you do not have a support.


Pretty much just spending most of my time in the cages, working on stuff, really trying to find two of my swing and figure out ways to get back to hitting the ball solid. Say find two in your swing. Did anything kind of click for you? I mean, I just really been working on trying to hit the ball up the middle more, hit line drives up the middle and, you know, trying to get the ball off the ground on the pole side, you know, because I'm just not having been getting those doubles, you know, down the right field line or to right field that normally do and, you know, just been smothering it a bit. Singles day walk, feel like this is something you can build off of? Yeah, definitely.  And then the, you know, had a good battle with the LFD, you know, almost had a double down the line on the slider.  Basically, do you feel okay?   I feel like all the mobility stuff that I've been trying to get has been working and, you know, obviously now it feels really good, you know, during the season.