 Hello friends, in today's session I am going to deliver lecture on equilibrium, basically we focus on connected body, muscle, between party, assistant professor, department of civil engineering, Vulture Institute of Technology, Solaapur. So my dear friends, these are the learning outcomes for today's session. At the end of this particular session, student will be able to, first, explain the concept of equilibrium, and second, analyze and calculate the forces present in connected body, first, that is static equilibrium. So my dear friends, have you heard about this particular term, equilibrium? The best example is Tegafor, you just apply equal forces. Equal forces means having equal magnitude of that particular forces, and the direction is opposite one. So see here, what do you mean by static equilibrium? The system of forces that keeps the body at a rest, it's said to be in equilibrium. Means here, system of forces, it try to keep the body, it is in equilibrium condition. Means you just apply equal magnitude forces, but the direction is different, and the body it is in a rest condition. So we call it as your static equilibrium. And similarly, system of forces that keeps the body at a movable condition. Means here also, you just apply the forces in opposite direction, having same magnitude with the body it is in a movable condition, we called as dynamic equilibrium. See here, the state of body is not affected by the action of force system in equilibrium. Generally, equilibrium is applicable to those system of forces whose resultant action is zero. Second term, what do you mean by equilibrium? My dear friends, we just focus on this particular figure, when body is present here, number of forces are present on this particular body, force p, q, r is nothing but your resultant, is nothing but your equilibrium. Force that brings the system of forces, the force that bring the system of forces that is p and q. In equilibrium is known as equilibrium, means what it indicates here, p and q, both forces are present here by considering these two particular forces, pair of forces, find out the resultant. Now the equilibrium is exactly in opposite direction of the resultant, means see it is always equal, means the magnitude, opposite in a direction, so resultant is present here, the direction is opposite and collinear with the resultant of the system. And it is collinear with this particular system, that is nothing but equilibrium is there. So body showing here resultant and equilibrium. Third point, when a system of force is in equilibrium, means when we apply the forces on this particular body, the resultant and equilibrium, both are in opposite direction, same magnitude. So when system of forces is in equilibrium, its equilibrium and resultant both are zero. So we need to analyze this particular problem connected to the body, we are identifying, we are adopting some steps, determination of forces. First step, select such a joint where one known force should be there, means in your problem statement what is the geometry they are given, by considering that particular geometry we just select such a joint where one force should be known, two forces should be unknown there, if more than two forces are present we are applying different equilibrium equations, if two forces are unknown there we are applying here different theorem. So that part we will discuss. Step two, draw the free body diagram by considering any one joint. So what it indicates here, consider any one joint and try to draw free body diagram for that particular joint. In that particular joint you just keep in your mind external forces, reaction you just mention as it is and remaining the unknown members, that particular unknown members you just mention tensile nature. Step three, we have just bifurcated into two parts, first one apply the static equilibrium equations, that is summation f of x, summation f of y and summation m is equal to 0. So here we are just assuming some direction, all right hand side forces, upward forces and clockwise rotary effect, positive sign, downward forces, left hand side forces and anti clockwise rotary effect, negative sign. Similarly we are applying Lamm's theorem and last one step find the unknown forces. Next we solve one problem, we just focus on this particular figure, a system of connected flexible cables shown in a figure is supporting two vertical forces, 200 Newton, 250 Newton at point B and D. So this is point B, point D is there. Determine the forces in the segments of cable, so A, B, B, C, P, D, then D these are nothing but the segments of this particular cables. So first here the angles they are mentioned, forces they are given, we just identify the forces present in this particular cables. So first one you just consider any one joint, now here I am just consider joint D, see here I am just consider joint D because two unknown forces are present and one known force is there, here three unknown forces are present, that is the force I select joint D. In that particular joint D angles they are mentioned here, so by considering these previous angles easily can prepare a geometry and find out the remaining angles with respect to x and y axis. So joint D is present here, downward direction one force, this is joint D, so DE member is present here, DB member is present and these are nothing but your angles, you just calculate by considering this particular geometry. So one 35 degree, so you just observe here x and y axis is present, this is 90 degree, this is 45 degree is there, so according to that easily you can calculate this particular angle, one 35, similarly calculate second angle, 360 degree minus these two it gives you third angle. Now first step apply Lame's theorem, in that particular Lame's theorem we are considering this particular force and you just consider opposite angle that is sign of opposite angle between remaining two forces. So see here 250 this particular force in numerator and denominator you just consider sign 105 degree is there, is equal to second one DE, sign 120, DB, sign 135. Now by considering this particular equation you just refer pairs because 250 already they were given, this particular two equation you just consider 250, sign 105 is equal to DE, sign 120 is there. So easily you can calculate the member force present in segment D and it comes 22414 Newton, give the equation number one. Similarly consider 250, sign 105 and this last one DB, sign 135, so by calculating the answer is present here, the force pavarence in member DB is 183.01 Newton that is the equation number two. Similarly, consider joint B, in that particular joint B two inclined members are present show the x and y component. So by considering this particular component with respect to x axis you just consider the angles, so apply static equilibrium equation that is summation h is equal to 0. So first one BC cos 60 plus BD cos 30, so BD is present here cos 30 minus BA, right hand side force is positive, left hand side force is negative sign is there. So by considering equation number one and two put the value here, so one equation is present that is equation number three. Similarly applies second static equilibrium equation summation v is equal to 0, all vertical upward positive sign all vertical downward negative sign. So BC sin 60, so BC sin 60 is present here, similarly you just consider BD sin 30 downward negative minus 200 downward negative. So put the value of BD here already we have calculated give the equation number four. So the value of BC is 334, 6.60 Newton. Now by considering equation number three and four calculate the value of BA, it comes 326.7 just like you just calculate the member forces present in this particular joint. Now my dear friends you just pause this particular video and try to give the answer of this particular question. Please verify this answer to prepare this particular session I refer this particular references. Thank you.