 We will just go through the tutorial, because I think I want to give you some time to kind of look at the tutorial problems carefully. The first problem is now displayed on the hydrostatic force that we have discussed. Just read the problem, I will give a brief hint here to solve this problem. So you have this gate AB, now gate AB is hinged here. So there is a hinge at B. So the gate can only open clockwise and it can try to rotate about the point B, just remember that. So due to the resultant force that will come, the resultant force will be coming from, I am just giving hints, the resultant force will come from the pressure distribution of this, pressure distribution of that side as well as weight of the body. There will be resultant force. Now you decide on the fact that what should be the resultant force, rather what should be the direction of the resultant force such that the gate will not open or gate is just about to open. So here the again question is when should gate is just about to open. So there is a resultant force, just think of that way it is very conceptual problem, think in terms of resultant force and what has to happen to the resultant force such that the gate is just about to open. And again in this problem only difference is you have water on both sides. If you do not have water on this side, then you have weight coming down and you have also resultant pressure distribution that will be perpendicular to this gate. So think this way and one more important thing I would like to state here, there will be a pressure distribution from the right hand side which will act perpendicular to the gate. From this you will have a pressure distribution. So line load resultant that will also be perpendicular to the gate. There is a difference of these two. So ultimately you could also look at what is the differential height of the water. This differential height of the water will actually produce the distributed load on one side of this gate. The differential height will only produce a distributed load on one side of this gate. See if you calculate P from both sides and take the net P, that net P will be on from this side. So I am just giving in hint. Now you give me the solution. Is the hint is clear? There is a net pressure difference if you choose any point and that net pressure is always going to act from the bottom of this gate and it should be perpendicular to the plane. Notice that 6 meter wide is the gate. On the perpendicular to this you have 6 meters. So that gate is basically you know 6 meter wide. So it is a rectangular shaped gate. 6 meter wide means the width of the gate is 6 meter which is perpendicular to the slide. So just give me the answer for the height. You can post your result on the module. It is a very conceptual problem. You can do it in one line by using the triangle rule. I started now some of you are reporting the results. It appears that yes some of you are absolutely correct. So this is 4.5 meter would be the answer. So answer is 4.5 meter of the H. Just look at this concept carefully. See as I said now you have what are on both sides. So if you choose any point at any given depth, you have pressure calculated from this side, pressure calculated from this side. The difference will always be equals to H minus 4. So what is being shown here? Let us say this is the gate AB. On the gate AB you are always going to get the pressure that is equals to rho g H minus 4. Is that clear? That pressure needs to be multiplied by the width. So that width is equals to 6. So you get the uniformly distributed load. So that will be rho g H minus 4 multiplied by 6. Please excuse me. Here the 6 is missing. Please add that. There will be a 6 multiplication factor 6. So the resultant force due to the hydrostatic pressure distribution that is coming from the both side is simply going to act. You can take it as if it is acting on one side of the gate and it is uniformly distributed. So we know what is the resultant. So the value of Fp can be obtained. So this is the Fp that is the force due to resultant force, net force due to the difference in the hydrostatic pressure. So that Fp value can be easily obtained. So that will be rho g H minus 4 multiplied by 6 multiplied by 10. And again you have a weight mg that weight is already given. That weight is equals to 50,000 kg. Now if you think of it, now this weight mg and this load Fp that will create a resultant force. Now that resultant force remember that resultant force has to pass through the B. Or in other words that line of action of the resultant force has to be parallel to AB. Because if it is so then that resultant force R will not create any moment about B. So from this we can clearly get, so ultimately what is my unknown that H. So ultimately what you have this theta is known to you. The theta being known that is the main point about it. And ultimately what you can do, you can simply get it by the triangle load as you said the R should be decomposed into two parts. One part will come from the Fp that is from the pressure and it is going and also you have a weight here. So component you know once you take the resultant of these two R that has to be parallel to the gate. So if you do this way now I have the R here that R is obtained and remember when I am obtaining R, so there are two values here coming into play. So you look at this carefully. So ultimately the answer will be 4.5 meter because that theta is being known, this theta is known. The theta can be computed. Let us move on to the next problem. So it is a hollow steel cone. So we have a hollow steel cone and there is a pin hole, very small hole at the top and it is filled with water. The question is what is the minimum weight of the cone which will prevent the water from uplifting the cone and flowing out. So we are interested in finding out what is the minimum weight of the cone. Remember it is filled with water. So ultimately I will just give a hint. We will be interested in finding out total uplift force that is being created by the pressure distribution inside the cone. So we are interested in finding out the total uplift force that is being created by the pressure distribution that is inside the cone due to the water. So this will definitely involve integration approach because you will have to now take a strip. That strip will be circular in nature on the periphery of this cone. So in the perimeter of the cone you just have to take a strip and try to figure out what is the pressure on this strip. Therefore what is the small force on that strip and the component, the vertical component of that small force on the strip you have to find out and then integrate it from you know the height from 0 to h whatever way you want to choose. So it is a bit complex problem but ultimate concept is we have to find out there is a pressure distribution how to find out the net uplift force. That uplift force must be balanced by the weight of the cone. That is the minimum weight of the cone that is required so that you know the water is just about to flow out. You can maybe try to work out this problem if you have not done it yet but try to do in terms of algebraic quantities involved. In other words you can have rho g h you know like that or so you can consider rho g h like that you can solve this problem. So far I did not have the I did I do have the correct answers coming up. So the correct answer is 41.8 kilo Newton lot of remote centers are posting their answer so 41.8 kilo Newton that is the final answer. So solution is being displayed now it requires some effort in terms of the integration what we have done here if you think of it. So basically we have taken a strip here. So you can see that strip but that strip if I have to draw it in this you know plane basically you are looking at something like this. So in 3D it should look like this. That is the strip I am talking about. Now remember so at a distance z first we know the pressure that pressure is actually given here. So that is rho g z. Now we are interested to find out what is the small elemental force that is being acting on that strip. So that can be found out by saying that we have pressure multiplied by 2 pi r. So 2 pi r is the total perimeter of this multiplied by d s. So you can see that d s. So d s is really here. So that is your d s. Now the main part here is that you have to really look at that there is a pressure that pressure is acting really here like this. So that is the you know type of pressure being acted. So we do have to take the vertical component of this. So this d r if you look at it this d r we are simply taking the vertical component of that d r. So right here. So d r cosine alpha so we are only interested on this one d r cosine alpha. So let us look at this way therefore now it is basically from the geometry I have to get few relationship because I want to ultimately integrate it from 0 to the height. So my you know I have to convert my integration or integral in terms of d z. So now look at so we got here in terms of d s. So that first let us take care of that. So d s has relationship with the d z. d z by d s should be equals to sin alpha that one part. Second part was what is r. So r is at a depth z what is the radius. So that is r z. So r z is displayed here. So this is r z. r is a function of z now. So we have to get that r z. So now you see the r z can be found in terms of z by the similar triangle relationship here. So this can be taken as r. So we can clearly see. So here I have also marked r in this way also. So that is r. So r by z is now equals to 1 over tan alpha. So r equals to z by tan alpha. So ultimately we have converted d s in terms of d z and also we have converted r. We have found r in terms of z. So everything is now on z. Now you can go ahead and try to do the integration carefully. So substitute everything here and ultimately we see clearly that finally once you integrate it out it will be 4 pi rho g by 3. So that is the answer. And therefore I have 41.8 kilo Newton. The concept is clear. So it was really done through direct integration and remember again we have exercised here how to do the integration on a slanted height as well. So you have a slanted height we had converted it to in terms of z. Now it was a bit complex problem but as long as we understand the strip carefully. So that strip has to be understood. And remember this is important again. We are going to discuss many problem on moment of inertia and so on and so forth in the next section. These kind of strips are very very important. No matter you know whether we want to calculate the centroids or whether we want to get the moment of inertia. The concept of strip how to take it will be very very important. So any discussions for maybe 2, 3 minutes or maybe 5 minutes. Any discussions on the distributed forces that is distributed force on beam as well as distributed in a hydrostatic pressure force. Hydrostatic force coming from the pressure. 1, 2, 3, 5. Is there any question? If it is an irregular body how can we calculate the center of gravity and moment by using an integration method? So your question is for a very irregular shaped body how can I calculate the center of gravity right or let us say centroid even think of do not go to the center of gravity let us say centroid ok. So let us try to discuss this. So what I will do I will try to go to this and try to tell you. One more query sir. Can you give some idea about non uniformly varying load? There are 2 questions. One question is for irregular shape very irregular shape how do I calculate centroid or let us say centroid of gravity. The second question is non uniformly varying load ok. So how do I do that? Now let me answer like this. Do I understand how to calculate first because the first question answer to the first question will second the suffix. Now we will do for the next question answer to the next question. In the first part suppose I have a very irregular shape I do not know what that shape is let us say I have a very irregular shape ok. First of all direct integration method is going to be extremely difficult unless otherwise I do not know the function right. So I have to really get if I look at the x and y let us say from here if this is my P x y I have to first know any point here should be defined in terms of x and y. So I have to know the function ok. Now that function if it is a very irregular shape it is difficult to do that way. The usual point here is that still if you can do little bit more effort in computer ok. Let us say you do little bit more effort in computer I can always discretize the shapes in terms of some finite shapes those are known ok. Remember these finite shapes the way I will solve this problem I will know some finite shapes in those finite shapes I already know the centroids ok right. Now ultimately what is there? I have n number of elements there I have n number of element which can go even in thousands right. My ultimate goal is to find out what the net centroid of this right. So I am interested in finding out what is my ultimate centroid that I do not know let us say it is here. So my force is downward in this case right. If I consider let us say uniform plate if I think of uniform plate weight is acting downwards right. So ultimately what we have to do first of all I need to get the total area. Total area should be just sum of ai but remember this n this n is very very important here it can go to you know order of thousands or anything because I am discretizing it in many shapes. Next thing is that each and every small shape that I have chosen will produce a moment. So I have to now calculate the first moment of each shape right about the y axis. So I will now calculate the q y. So q y is nothing but i equals to 1 to n ai xi right. So you go to n and you go to n and you go to any shape here this shape has a predefined centroid. So what is my xi? xi of this is simply this right and ai is also clear. So once my q y is known then ultimately what I am doing that x bar multiplied by the net area right should be equals to q y. So I am equating the first moment remember area multiplied by rho t. See rho t is very silent here okay and q y you know this here also you can take the rho t which is very silent. Rho time t that is density time thickness multiplied by the area that will give me the weight of the small small shapes right okay. So ultimately this rho t, rho t does not come into play. So in principle I should be able to calculate x bar. So see the computerization is a big thing okay you can take any geometry but remember once we go to more advanced studies we have to do this finite elements. There is another technique. So finite element you know always takes care of that how to mesh the entire area and from those elements we can calculate those individual areas are known and therefore once the individual areas are known I can do everything because those are defined shape. It could be triangle, it could be you know four-noded element, eight-noded element or whatever okay. It could be a different shape for which my know the centroids okay. So the second problem was the distributed load. Let us say non-uniformly distributed load does not matter. How does it matter to me now? I have a beam let us say okay I have a very non-uniformly varying load right okay. But that non-uniformly load let us say it is a parabola but it is defined by some function clear. So once it is defined by some function I have no problem whatsoever to calculate the centroid of this right. If it is let us say you know if it has some zig-zag shape let us say some zig-zag shape then again we are going back to the previous concept. See ultimately this concept is same as that concept as long as I understand this I should be able to do this. Now if my shape is smooth like this it is non-uniformly varying but still I know the shape then I should be able to simply do the integration. I take this rectangular strip right okay. The rectangular strip I will take here and let us say I am interested in finding out the X bar. So ultimately remember this has a X this is my DX okay. So what is my DA now? So this is Y. So DA is nothing but Y DX is that clear okay. So total area should be simply integral of Y DX going from 0 to X remember Y equals to K X square you have to find what is K right. So I can integrate the area therefore now you take the first moment of this element with respect to the Y axis right. So what is the first moment of this element? This is simply given by XYDA that is the first moment of sorry so that is XDA right. So that is XYDA that is the first moment of that area right. So I have to equate that with the first moment of the entire area with a single load. So now equivalent system right. So I am really trying to see where is my resultant load of this is. So let us say that resultant load is that you know given by R. So in this case R equals to A. So this is really equals to A okay right and this X bar. So now R times X bar or rather A times X bar should be equals to integral of XDA. This is the basic and that is the fundamental you give me any problem as long as I understand the first moment how to take it right. Now this system and that system is purely equivalent. This is and this one is equivalent system that is what we are trying to say nothing else because ultimately what is happening we I am breaking this system into small, small rectangular elements right and I am taking the moment of those elements and that is equated with the moment of the net force which is coming from the net area right that is it okay. So it is in general this concept is very important. Now I think we can take a break maybe for 20 minutes.