Structure mechanics Free body diagram
Hello friends welcome to the sixth lecture of the series so in this this is a third and a very easy problem of Freebody diagram,
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as we have already covered some problems on trusses and beams previously I hope you can solve this by yourself and then just mash the answer so the problem approach that you are using may be little bit different from what I am using here but anyways solution will come exactly the same in the end so we will start a verb approach by analyzing the structure here first so let's investigate the problem
so this is a trust math truss member which is subjected to a load at some location which is offset from the center so you can say that P is acting at a distance of l1 from one end and l2 from the other end but L 1 is not equals to L 2 so from intuition if you imagine that the two
persons are holding a block and if the block is not at the Centre if the block is at center then obviously the the both will carry the same load but if you shift the load to one side then one need to apply more load to hold the block and similarly we can say that the reactions and forces in the structure will not be same due to the UN symmetric of load here so in this example a truss structure has mounted on two supports and subjected to loading so how many constraints are there there are two constraints so one is the pin joint that can't move in both the directions so definitely he will not be happy to avoid that motion automatically so there will be a reaction force so this is the reaction forces of pin joint that is one in the horizontal direction and one in the vertical direction if you want to remember this just assume yourself as a pin joint and there is a ball in front of you the ball will not let you move as it applies force on you now if you're Superman and apply too much force to break the ball then definitely there will be no constraint and in terms of mechanical failure mechanic of material of mechanics will term it as a failure but in case of a roller joint it is allowed to move in one direction so there will be no reaction in that direction and that direction will be horizontal in that the in that case as you know that role in roller support only you will get one one reaction that is in the vertical direction so let's start our mathematical equations so as we remember first we need to isolate the structure and replace the support by its reaction and then we will find the reactions so starting from our equilibrium conditions so this is the structure
at Point a we are getting VA and H a and at point B we are getting only V we know a chain know H be on point B as you know it is a roller support so now I am applying the equilibrium conditions so first I am taking moment about point B you can also take moment about point a it's totally your wish but as we have discussed previously you have to take either at B or either at bay either at a or either at B to reduce the number of unknowns and to make the equations easy so when you'll take this moment about point B the reaction you will get is this thing now you as you can see VA is acting in the upward direction so any sign convention you can take what I am taking here is I'm taking clockwise as positive so VA multiplied the perpendicular distances length of a B which is equals to L 1 plus L 2
So, when VA is multiplied by L 1 and L 2 it will create a clockwise movement similarly P when P is multiplied by L 2 it will create a anti-clockwise moment so P will get a negative sign and VA we are assuming clockwise is positive so via term will getting a positive sign so when you solve this you will get VA value as P into L 2 by L 1 plus L 2 so this is the one equation that you're getting now then you will resolve then you'll equal decreed the forces in X direction you will get that H a is equals to zero because on VA because in this structure only P is acting which is a vertical force if P is inclined at any angle so that means you have to resolve the in to solve the force and you'll get some of the horizontal component and some vertical component but if this is a simple case so HG will come as u similarly when you will equate the force in the Y direction what you will get is VA plus VB minus B so what is the thing is we are considering upward forces as positive since VA and VB are acting upward here and P is acting downwards VA plus VB minus P is equal to zero I hope why this minus sign is coming is clear to you so,
What you will get VB is P minus V now when you solve this equation you are getting VA in terms of P and you are getting VB in terms of P so just substituting any any value of V and V being we are calculating vb here now you need to make one conclusion here at C since L 1 is greater than L 2 definitely VB will be greater than V similarly if you'll get if you'll shift the load to the side of 8 this to the to the point a then L 2 will become greater so at that point VA will become greater okay so let's take a force balance ad joint
So this is
called method of joint in which we impose the equilibrium conditions at the joint here we need to find the forces in BC and BD what we need to do first is we need to resolve the forces first and then we will look then we will apply the same our equilibrium conditions so when you'll do this you have a VB is acting in the vertical direction and this force in BC since it is inclined at angle theta you will resolve this and we will get oneness FBC sine theta in the vertical direction and fvc cos theta in the horizontal direction similarly for the member of BD since BD is not at any angle it is in horizontal direction only there will be no vertical component for force in BD okay so I think this should this force resolution is done so to find the forces in BC and BD let's draw Freebody diagram that we have already covered now equate the forces in y direction so you'll get vb + f vc sine theta is equals to 0 you can get the value of force in bc since since vb we have already calculated in the previous section so you will get force in f bc force in bc here then substituting the value of VB in terms of pc because P is the un be is a known value and l1 and l2 will be the known value when you start solving this equation VB and VA are we are calculating so we will substitute the value of V behave similarly equation of summation of forces in X direction will be Co so this equation will come and again when you substitute the value okay again when you substitute the value you will get the force in BD so this was a very easy numerical very easy problem and I think this is good so if you liked the Lecture please subscribe for more interesting Lecture
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Thank you.