 Hello everyone, myself, Ms. Shailaja Diavarkonda, assistant professor in civil engineering department, Wolchen Institute of Technology, Sulakuru. In today's lecture, we are going to study analysis of pin-jointed play frames, learning outcomes. At the end of this session, students will be able to explain assumptions made in the analysis of truss. They can classify the different types of frames and the students will able to analyze the cantilevered truss with the method of section assumptions. In the analysis of frame, it is assumed as the two members of the frames are connected with the pin-jointed or it is called as hinged. The loads are acting at the joints only. It will not act in between the member. The self-hat of the members are negligible, means in analysis we will consider the only external forces were neglecting the self-hat of all the members. Then members are having uniform cross sections. The centroid is located along the longitudinal axis. In the frame, the members which are having uniform cross section, the centroid of those sections are located along the same longitudinal section. So, these are the common assumptions which are considered in the analysis of frames. Classification of frames. The frames are classified as perfect frame, deficient frame and redundant frame. When you are having the number of joints are sufficient to resist, the applied load is known as perfect frame. If you observe this figure, here you are having three members and three joints. If you put these number of members and joints in this equation, m is equal to 2j minus 3, where m is equal to the number of members and j is equal to number of joint. If you put these numbers in this equation, you will get 3 is equal to 3. So, whenever you are satisfying the equation m is equal to 2j minus 3, it is known as the perfect frame. Deficient frame. When the number of members are less than the required for perfect frame, it is known as deficient frame. If you observe this figure, here number of members are 4 and number of joints are 4. If you put these numbers in equation m is equal to 2j minus 3, it will give 4 fine numbers, means members are 4 and this 2j minus 3 will get the 5, means number of members are less this called as deficient frame. Redundant frame. When the number of members are more than that of required for perfect frame is called as redundant frame. If you see this figure, here the number of members are 11 and the number of joints are 6. If you solve this equation, you will get 11 and 9 numbers, means number of members are 11 and number of joint and 2j minus 3 is equal to 9. So, number of members are more than required for perfect frame, so it is called as redundant frame. Whenever you are analyzing the frame, we are considering the various forces acting on the frame. Normally, the nature of force is of two types, one is compression and one is tensile. Compression means when the force is acting towards the joint, again whenever the force we will consider it has direction, so the force may be in horizontal direction, vertical direction and if you consider the movement it may be clockwise or anticlockwise. Whenever the force is acting upward and towards the right, it is considered as a positive and the clockwise movement is also considered as the positive. Whenever the force is acting downward towards the left and the movement of the force is anticlockwise, it is considered as the negative. Then there are various methods to analyze the pin-jointed rest that is methods of joints, method of section and graphical section. In this lecture, we are going to study how to analyze the frame by using method of section. Method of section is adopted when you have to analyze a few members of the trust, means you are having a large trust and you have to find out the forces in a very few number of forces in the overall trust. In that case, you have to go for the method of section and if the method of joint fails to start with analysis for getting a joint more than two unknowns. Normally, we will start the analysis of any joint where you are getting only two unknowns because our equilibrium equations are only two, that is summation f of x and summation f of y, another one is summation of movements. We are having only two equilibrium equations, so whenever at which joint you are getting only two unknowns, from that joint only we have to start the analysis. In this method, after determining the reaction, a section line is drawn which should not pass through the more than three members. Let us see further how we will implement all these three points into the analysis. And the magnitude and nature of forces in the members BC, CF and EF of the loaded trust as shown in figure, means you have to find out the forces in member BC, CF and EF, okay. So now, we will see further how to find out. Here they have not given any inclination of these members and again it is a cantilever type of trust. One end of trust is fixed and another end is free. You can observe at this joint D, there is no support. It is only the force of 20 kilo Newton is acting at the point D and at point A, from point A to G, it is a fixed wall, so it is called as cantilever type of trust. So first of all, you have to find out the inclination of these members. So theta 1 is the inclination at the point D. So it is calculated by tan theta 1 is equal to Y by X. Y means it is the total height of the trust and again X means it is the total length of the trust. So it is tan theta 1 is equal to 4 by 9, so you will get 23.96 degree. So then, as I told, first you have to start from the joint where you are having only two unknowns. If you observe this total trust at joint D only you are getting two unknowns. Means the force of 20 kilo Newton is acting at the joint D and the force in DC and the force in member DE are the unknown. So by considering the equilibrium conditions that is summation F of Y and F of X, we have to find out the force in member DC and the force in member DE. So considering the summation F of Y means it is summation of vertical forces. So F of V is equal to 0, it is F DC sin 23.96 because this member DC is inclined DC. So it has having two components vertical and horizontal. So here it is F DC sin 23.96 minus 20 because this 20 kilo Newton is acting downward and the vertical component of this force DC is acting upward. So for this it is positive and this 20 kilo Newton is considered as negative. So you are getting the force in member DC is 49.24 kilo Newton. This T means it is tensile force, the nature of force. Already at starting we have assumed that the nature of force in member DC is tensile. So you are getting the positive answer. So means whatever you have considered the nature of force it is correct. Then summation of horizontal forces is equal to 0. You have to consider all the horizontal components. So here horizontal components are force in member DE and the horizontal component of force in member DC. So both are acting left. The sign is negative for the both. If you solve this further you will get the force in member DE 45 kilo Newton and again it is negative. Initially we have considered the nature of force in member DE is tensile but you are getting negative answer means the nature of force is opposite what you have assumed. If you are assuming the compression and you are getting the negative value means it is the actual nature of force in the member is tensile. To pass the section line from the three members that is BC, CF and CE after passing the section line you will get this free body diagram. Now consider a joint where you will get the maximum forces are 0. If you consider the moment at joint C you will get the forces in all these members 0. Means the moment of the forces which are passing through the moment center are 0. So you will get the force in EF. If here it is the force in member you will get 45.11 kilo Newton compression. Compression means you can observe in this figure I have initially considered it as a tensile but the value is in negative means whatever nature of force you have considered it is opposite to the its actual nature. So this will be compressive. Then again the considering the joint equilibrium of joint C consider the summation of vertical forces is equal to 0 and summation of horizontal forces is equal to 0. Now you will get two equations by solving these two simultaneous equations you will get the force in CF and the force in member CB. So likewise you have to find out the forces in few members from the overall trust by using the method of section. You pause this video and try to answer this questions. These are the answers and these are the references considered for the study. Thank you.