 Hello everyone, myself, Ms. Shailaja Dhavarkonda, Assistant Professor in Civil Engineering Department, Wolchen Institute of Technology, Sulapur. In today's lecture, we are going to study Analysis of Pin-Jointed Plain Frames by the method of section. Learning outcomes. At the end of this session, the students will be able to analyze the symmetric type of truss with the method of section. For the analysis of truss, we are having these three methods that is method of joint, method of section and graphical method. These two are the analytical methods and second one is graphical method. Now the method of section is adopted when you have to find out the forces in few members of the truss from the entire truss and whenever you are not able to start with the method of joint means normally we are starting the analysis of frame from the joint where you are getting only two unknowns because we are having the equilibrium equations or two that is summation f of x and summation f of y, ok. So in which truss you are not getting these conditions, in that case you have to go for the method of section and again in method of section you have to pass the section line from only three members which forces are unknown. These are the sign conventions which are considered for the forces. Now we will see how to find out the forces in members by method of section. Here in this problem, determine the forces in members of f h, h g and g i in the truss as shown in figure 1. Each load of 10 kilo Newton and all triangles are equilateral with sides of equal to 4 centimeter means from this entire truss which is shown in figure you have to find out the forces in only the members f h, h g and g i means you have to find out the forces in only three members. Many times in case of large trusses it will be more time consumable for the analysis means if you are considering the method of joint you have to go for the one by one joint means first joint A then joint B, C, D so likewise you have to calculate. So for this type of trusses which are symmetrical in geometry as well as the load is acting at all joints are equal. So they are mentioned in problem whatever loads are acting at joints those all are 10 kilo Newton and these trusses are like the equilateral triangles of sides 4 meter. So the inclination of each member is 60 degree means it is trusses symmetric about the y axis. So you can consider either any part of the truss either. So in this we have to find out the forces in only three members so we will go for directly the method of section. First you have to calculate the reactions at the support because whenever you are starting the analysis you have to find out the joint where you will get only two unknowns. If you observe this entire truss at no joint you will get only two unknowns. So first you have to find out the reactions at the support. If you observe this truss at joint A it is fixed support and at joint O it is roller support. Fixed support means you are getting two reactions that is one is horizontal reaction and second one is vertical reaction and at roller support you are getting only one vertical reaction. So if you observe this entire truss there is no horizontal force acting on the truss. So the horizontal reaction at point A is 0. Now you have to calculate the only vertical reactions that is RA and RO at both the ends. Due to symmetry of the truss as I told in this truss the loads are acting at joints are same and the geometry of truss is also same. So you can directly divide the total load as the reactions of the support means total load is 70 kilo Newton. So this 70 kilo Newton is directly divided in two support, two reactions that is reaction RA and reaction RO means it is possible in the case only where you are having the loads acting are equal and the truss is having equal geometry. Then as we know we have to pass the section line only from the three members where you have to find out the forces. So in this problem we have to find out the force in member FH, force in member HG and force is member GI. So I have passed the section line AA from only these three members. Please keep in mind you have to pass the section line only three members itself. Then only the method of section is applicable. Then after passing the section line either you can consider the any part of the truss. Here I have considered the first part. So after passing section line I have considered the first part of the truss and we have already calculated the reaction at support A also. Now the inclination of this force HG is 60 degree because they have mentioned in problem all triangles are equilateral triangles. So the angles of each or the inclination of each member is 60 degree. Now consider the joint where the moment of the maximum forces will be 0. If you observe this figure in this if you consider the moment about joint G so you will get the moment of all these forces that is force in GI, force in HG, force in GE and force in GFR 0. Now after passing the section line I have considered the first part of the truss and we have already calculated the reaction also. Now consider the point where you will get the maximum forces 0. If you observe this truss at point G here the number of forces are passing 4 from the point G means force in member GI, force in member GH, force in member GF and force in member GE. If you consider the moment at point G the moment of all these forces are 0 because these forces are passing through the moment center then you will get the force in member FH. So consider the moment at point G so sum of moments at G is equal to 0. So you will get this RA into distance is 12 because this distance is having 4 meter so 4 plus 4 plus 4 so it is RA into 12 then this reaction is acting clockwise direction about the point G because reaction is in upward direction so it is acting clockwise direction so we have taken it as a positive. Then these all 10 kilo Newton external loads are acting in anticlockwise direction with respect to the point G so we have taken it as a negative. So from this equation we will get the force in member FH because the force in the moment of this force FH is about G is force in FH into 3.46, 3.46 is the vertical distance of the force FH from the point G so the total force you will get 69.36 kilo Newton. So here it is compression because you will get the negative initially what we have assumed this force is tensile force but you got the negative answer means the nature of this force is opposite what you have assumed so here it is compressive force. Then by considering the equilibrium of the joint further we have to go for the summation of a vertical forces is equal to 0 and summation of the horizontal forces is equal to 0. By considering the summation of vertical forces is equal to 0 you have to calculate the force in member GH. So what are the vertical forces one is upward reaction at point A and another are the external loads that is the 3, 10 kilo Newtons which are acting downwards so this reaction is considered as positive and remaining forces are considered as negative. So likewise you have to find out the force in member GH and force in member GI. So instead of going for the method of joint in a large trusses we have to we are adopting the method of section to find out the forces in few members. So it is the similar type of problem means this is also a symmetric geometry of type of truss. So you have to find out the force in member CE, CD and BD ok so we will see further. So as I told if the loads are acting are same so the reactions are directly half of the total load. So here you will get the vertical reaction at A and vertical reaction at H is equal to 9 kilo Newton because no horizontal forces are acting on the truss. So horizontal reaction at H sorry horizontal reaction at support A is 0 ok so you will get only vertical reactions R A and R H is equal to 9 kilo Newton. Then pass the section line from the members where you have to find out the forces. So this is the section line A is passed through the CE, CD and BD ok after passing the section line the same procedure will be proceed you have to find out the inclination by considering the geometry of the truss ok. Now by considering the geometry of the figures you have to find out the inclination of this member. So here if you consider the triangle ABC so this theta 1 is equal to tan theta is equal to y by x y is the distance from the B to C ok. So you will get the theta 1 is equal to 33.69 degree and the theta 2 it is the inclination of force CE because this inclination will be again different. So this theta 2 will be 18.43 degree ok. Now by considering the moment about the joint C you have to first find out the force in member BD ok. So here you will get only the moment of reaction about the point C which is clockwise. So have considered it as a positive moment and the moment of the force F BD. So which is anticlockwise so we have considered here it is a negative. So you will get force in BD is equal to 13.5 kilo Newton. And again considering the equilibrium of the joint you have to consider summation of vertical forces is equal to 0 and summation of horizontal forces is equal to 0. So by considering summation of vertical forces you have to calculate the force in member CE and FD. You will get two equations by solving the simultaneous equations you will get the force in member CE and force in member CD. So likewise you have to consider the method of section and you have to find out the forces ok. So now you pause this video and try to answer these questions. These are the answers. These are the references considered for the study. Thank you.