Continuing with our discussion of structural analysis statics review, in this segment well look at trusses. So, trusses was a topic we discussed in some detail in the statics module section IIIa. And here, we're dealing predominantly with plain, simple trusses which have rigid members, in other words, no defamation, and they are pin jointed. In other words, there's no moment at the joints. So, simple trusses look something like this, and they are combined of rigid members, such as these, and all of the members are two force members, which means that they are under the action of only two forces which are applied at the ends, and the forces are called linear. In other words, the line of action of the forces lies along the line of the member. So, they look either like this. The members are either only in tension or compression, and this gives ways in turn to a sign convention that a member under tension, we denote as being under a positive force and a member under compression we say is under a negative force. We use two methods of solving trusses, the method of joints and the method of sections. So for the method of joints, we imagine that all of the members are detached from their pins, or joints, nodes, And that each member is in equilibrium regardless of the value of the force in that member. Therefore, the equilibrium of the truss is reduced to simply equilibrium of the pins. So in this case here we have a simple triangular truss, ABF, so the approach is to isolate the joints and apply the equilibrium equations to each joint. And we only have two equations, sum of the forces in the x direction is zero, and sum in the y direction is equal to zero. There's no moment equation because there is no moment about the joints. If we detach the members from the joint. So, for example, at joint or node A here, we have two members connected, AF and AB, and if we imagine them detached from that point, then the directions of the forces are as shown. If the member is in compression, the force arrow is pointing towards the node, and if the member is in tension, the force arrow is pointing away from the node. So we will always start with the joint with the fewest arms, so that we can solve it. Let me give an example on that. We have a weight with a mass of 100 kilograms which is supported as shown and the dimensions are shown. So this is a simple truss. It had a pin support here at A and a roller support at C. So generally, there would be three, unknowns in this case, therefore this is a statically determinate structure that we can solve. So the questions are, first of all we want to solve for the force in the member AB. That is this member. It's most nearly which of these alternatives? So to do this, I will isolate the node B here. And the node B then looks like this. We have the weight, which is of course acting straight downwards. And the forces in the members AB and BC where I have assumed that they are in tension. In other words, the force arrows are pointing away from the node. And I recommend you always do this. Don't try to guess the direction ahead of time. Just assume that they're intention and let the sign tell you whether they're actually intention or compression. So now I can apply equilibrium to this node point. So first of all, summing the forces in the vertical, I'm sorry, in the horizontal In the x direction here we have sum Fx is equal to zero and the forces we have are the horizontal component of AB which is negative because it's pointing to the left in the negative direction plus the component of the force BC which is positive and it's in the rightwards direction. And the cosines of the angles here follow because this is a three, four, five triangle here and this is a one two square root of five triangle here. Now we can't solve it yet, because we have two unknowns there and only one equation. So, now we have to apply the equation for equilibrium in the vertical, the y direction, so if Fy is equal to zero. And, here, the equation is this. Again, we have the false components of AB, components of those forces in the vertical direction, and now we also have the additional gravity force, the weight of the object, which is acting straight downwards, therefore, which is negative W. It's in the minus y direction. Now we have two equations and two unknowns and we can solve them and solving those, you find that the force in AB is equal to W, the weight. The weight is equal to the mass times acceleration due to gravity. 100 kilograms times 9.81 meters per second squared Which gives 981 Newtons. So the answers is D. Next we're going to solve for the force in the member BC. This member right here and it is most nearly which of these alternatives? So in this case we've already solved it from the first part, we already know what AB is and substitution that back into this equation we can solve for BC. And the answer is BC is 2 over square root of 5 times W or 2 over square root of 5 times 100 times 9.81. Mass times gravity is equal to 877 N, and the answer is C. Finally, we're going to find the force in the remaining member AC, the top horizontal one here. It is which of these alternatives? So, here is our free body diagram of the node C. And here we recognize that this is a roller support here. It's assumed friction-less, therefore, there's no component of force parallel to the slope, therefore the force reaction here must be perpendicular in that case. So the reaction here is Nc which is, I'll call it Nc, which is in this direction. Now, to solve in this case, we have two unknowns here. The force in AC and the reaction at Nc. So again, we could solve this by summing the forces in the horizontal and the vertical direction. Which would give us two equations, which we could solve for the two unknowns. However, a simpler approach is to rotate our coordinate system to new coordinates, x prime and y prime, where y prime is perpendicular to the surface or parallel to MC. And now if I do my force balance in the x prime direction, I don't need to worry about NC because it has no component in that direction and I can solve for AC directly. So I sum the forces in the x prime direction, sum Fx prime is equal to zero which leads to this equation minus BC minus AC times 2 over square root of 5 is equal to zero. And, we have already solved for BC from here so, substituting it we find that AC is minus BC squared of 5 over 2. Or AC is equal to minus W which is minus 981 Newtons and the closest answer is D. So in this case for these three members we see that the force in AB is positive, in other words this is in tension. The force in BC was also positive, so this is also intention, but the force in AC was negative. In other words, this member is in compression. And, again, for other examples, see the statics module, section IIIa. The other method of solution was the method of sections, which is discussed in static section IIIb. And in this method, we divide the truss into two distinct sections. We cut it in two and we replace the cut bars by their internal forces which then become external forces, and the basic principal here is that if the structure or the truss whole in z, is in equilibrium, then each of the individual cut sections must also be in equilibrium. For example, this was the case that we looked at in the static section. We have a simply supported truss here it's statically determinate with a load L as show. Let's suppose that we wanted to find the force in the member BE. To do that by the Method of Joints would be quite lengthy. We would have to start at node A then move to B or F, etc., and work our way through until we got to the unknown. However, we can do it more quickly by the method of sections by simply taking a cut through here. So, I'll take a cut through there which cuts through the unknown member like this and the free body diagram of the entire structure looks like that, where the reactions at A and D are denoted by R1 and R2. And now I divide this into 2 separate sections which look like this. So focusing our attention just on the left hand section the free body diagram of the left cut section looks like this where the forces in the cut members are replaced by external forces labeled as EF, BE and BC. And in this case we are trying to solve for the force BE, the force in the member BE. So to do this, I can sum the forces for example in the vertical direction. So summing the forces in the vertical direction we have R1, L, which is actually downwards, which is negative and the vertical component of the force BE which is, BE sine Theta. So, rearranging I see that the force in BE is equal to L minus R1 divided by sine Theta which we can solve given that we know L and the reaction R1. So we can get the solution very quickly and easily by this method. Let me do an example. The truss is loaded as shown, where the magnitude of the load L is five kilo Newtons. And the question is, we want to find the force in the member CG, this member here. Is it most nearly which of these alternatives? So, firstly, we note that by symmetry, the vertical components of the reaction forces at AY, at A and F, Ay and Fy are both equal to 3L over 2 and we can find that by simple statics of the entire diagram. So now I'm going to make a cut here so I make my cut through here Which cuts through the member that I'm interested in and re-draw the left hand section as a free body diagram, which looks like this. Where I've assumed that the forces in CD, CG, and GH are all intention, in other words their forces are pointing in the directions as shown. And here is the vertical reaction force which I've added in there 3L over 2 at A. So to find CG, I can simply sum the forces in the vertical direction. Sum Fy is equal to zero. And here then we have the forces are 3L over 2 acting upwards, L over 2 acting downwards, therefore it's negative, L downwards, also negative, and the component of CG is CG times 3 over 5, from the geometry of that three, four, five triangle, is equal to zero. So, solving that equation, we find rather surprisingly, that the force of CG is equal to zero. And, a member which has no force in it is called a zero force member, and therefore, the closest answer is C. The next question, the same problem. The force in the member GH. So, now we want to calculate the force in this bottom member here, GH, or HG, is it most nearly which of these alternatives? So to solve this, I'm going to take moment about the point C here is the simplest way to do it, and I'll assume that moments in the clockwise direction are positive for my sign convention. It does not matter which direction you assume here you will get the same answer ultimately. So taking moments about the point C we have the moment of the reaction forces here L over two and three L over two, three L over two times its moment arm times three is a positive moment because that's rotating in the clockwise direction about the point C. Minus the moment due to L over two because it's acting in the counter clockwise direction minus GH times it's moment arm which is three meters. And again, it's negative because it's rotating in the counterclockwise direction. So solving that equation, we find that GH is equal to L, which is given as 5 kiloNewtons. Therefore the answer is D. So, for other similar examples for solving trusses by the method of sections see Statics Section IIIb.