 Hello everyone, myself, Ms. Shailaja Dehavarkonda, Assistant Professor in Civil Engineering Department, Vulture Institute of Technology, Solapur. In today's lecture, we are going to study analysis of pin-jointed plane frames by method of joint. At the end of this video, the viewers will able to determine the forces interest member by method of joint. Now, to analyze the pin-jointed frames, these are the methods used. Method of joints, method of section and graphical method. From these three, method of joint and method of section are the analytical methods. Method of joint. At any joint, the members meeting and the loads acting constitute a system of concurrent forces. Means, whenever at a joint, the forces are acting is called as concurrent force system. Hence, two equations of equilibrium can be formed at each joint. Means, if you consider a single joint, it will have two equilibrium equations. In this method, after determining the reactions, the joint is selected where there are only two unknown forces. As we know, there are only two equilibrium conditions. So, you have to select the joint which will have only two unknown forces. Then, using equilibrium equations for the system of force at joint, we have to calculate the forces in members. Now, we will see how to find out the forces in all the members of this truss. Figure shows a pin-jointed truss having all the members of 3 meter long and these are the external forces. Before going to further calculations, we have to consider assumptions we have made in the analysis that is the internal forces of these members are negligible. So, we are considering the only external forces and we will go for the analysis. Now, if you observe the entire truss, you have to find the joint such that which will have only two unknowns. If you observe this entire truss at joint G, you will get only two unknowns that is external force and the force in member FG and force in member GA. So, at joint G, we will get two unknowns that is forces member GF and force in member GA and one external load is 20 kilo Newton. So, as we know, the two equilibrium conditions are there. So, we can easily find out these two forces. Then, in analysis, whatever downward forces are there, those we will consider negative and upward forces we will consider positive. And the left side of the joint we will consider positive and right side of the joint we will consider negative means it is the direction of horizontal force. Now, by using two equilibrium conditions, we have to analyze the joint G. Here, the summation of vertical forces is 0. What are the vertical forces? One is external force 20 kilo Newton and another one is force in member FG. This member FG is inclined. So, here you will get two components that is horizontal component as well as vertical component. So, vertical component of force FG is FG sin 60. And here, initially, we have assumed the nature of force in member GF is tensile and the nature of force in member G is also tensile. Tensile means the force is acting away from the joint. So, by using the equilibrium equation summation F of Y, you can find out the force in member GF. So, here the force in member GF you will get 23.1 kilo Newton tensile. Then, by using second equation that is summation F of horizontal, summation of horizontal forces is equal to 0. You will get force in member GE. So, here what we have considered? Force in member GE is tensile force. Then, but after calculation, you are getting it. The force in member GE is compression means here as we consider right side of the joint it is negative. So, here you will get minus FG minus GF cos 60 means horizontal component of this member GF. So, you will get here GE is equal to 11.55 kilo Newton compression. So, exact nature of force in member GE is compression force. Then, whenever you are going to next joint, you have to consider the nature of force what you are getting. Now, after analyzing joint G, you have to go for joint F because here you have finding, you have calculated the force in member GF and force in member GE. If you consider the joint E, here how many unknowns are there? One is vertical reaction, force in member EF, force in member ED and force in member EC. So, we cannot analyze the joint E. So, we will go for joint F whereas one external force is there that is 10 kilo Newton and already we have calculating force in member GF. So, we will analyze the joint F. We got the nature of force GF is tensile force and it is inclined. So, we will get here horizontal and vertical component of force GF and as well as it is same for force in EF. So by considering the two equilibrium equations, you have to find out, you have to find out the forces in these members EG and sorry EF, FE and FD. So, the sum of vertical force is equal to 0, you will get F force in member FE sin 60 minus FGE sin 60 is equal to 0. So, you will get force in member FE and force in member GF is equal to 23.1 kilo Newton compression. It is not compression, it is tensile force. Next, by considering some of horizontal forces, you will get summation F of H is equal to 0. By considering some of horizontal forces, you will get force in FD that is 31.1 kilo Newton. The nature of force is tensile force. So, by considering the two equilibrium equations, you have to analyze the each joint. After finding or analyzing joint G and joint F, you have to go for the joint E, means you have to find out the reactions at support A and support E. At support A, there are two reactions because it is in support. So, you will get one horizontal and one vertical reaction. But at support E, it is roller support. So, you will get only vertical reaction. Now, vertical reactions we are finding by taking the moment of all these external forces. Here, we are taking the moments of only external forces because we have assumed that the internal forces in the members are negligible. So, by taking the moment of all the external forces, you have to calculate the reactions that is vertical reaction at support E is 58.17 kilo Newton and vertical reaction at support A is 31.83 kilo Newton. In entire truss, only one horizontal external load is there. So, you will get the horizontal reaction at support A is 10 kilo Newton. After calculating the reactions, if you consider joint A, there are only two unknowns because horizontal reactions already we have calculated and vertical reaction also we have calculated. So, by using two equilibrium equations, you can find easily force in member AC and force in member AB. So, here force in member AB we have considered compression. So, it is minus FAB sin 60 vertical component of the force AB and plus 31.83. It is vertical reaction. It is acting upwards. So, it is positive. So, here you will get force in member AB is equal to 36.75 kilo Newton compression. Then next, sum of horizontal forces is equal to 0. So, by considering sum of horizontal forces that is force in member AC minus force in member AB cos 60 plus horizontal reaction H is equal to 0. So, here you will get the force in member AC is equal to 8.38 kilo Newton tension. So, likewise you have to consider the single joint and how to analyze. The same procedure is carried for the joint B. After calculating force in AB, you can go for joint B because at joint B one external force is there and force in AB also you have calculated. So, you can find out by using two equilibrium equations force in member BD and force in member BC. The same procedure is carried for the joint C. If you observe joint C force AC or force in member AC is also you have calculated and force in member BC also you have calculated. So, you can find easily force in member CE and force in member CD by using two equilibrium equations. So, likewise you have to analyze each and every joint. Now, these are the answers for the above question and nature of the forces. Now, you pause the video and try to solve this problem. These are the answers of the problem which show nature of forces and the magnitude of force. These are the references considered for the study. Thank you.