Structure mechanics basic through problem
Structural Mechanics
it looks like a funny figure but it will serve as a path breaking problem as we proceed through the presentation so don't run away and cope up with me one thing for sure if you are able to bear with me to last you will develop a new skill of handling problem this exercise is intended to introduce the essential concepts and principles of the engineering mechanics of solid it is meant as an overview so do not be disturbed by the variety of concepts or the range of vocabulary we will try to grabs the essential workings of the device and begin to see the relevance of the concepts and principles of engineering mechanics to an understanding of how it functions and how it might be made to work better so this is a problem of two frame mounted on door joint and attached to Springs at end so our aim is to determine the behavior of the structure.
as the applied load increases from zero to any finite value so let's just pause for a minute and try to understand what this problem speaks so this will be the deflected position right so what will happen now when you will apply it load at point B well this our frame will deflect like this so what you can infer from that what you can infer from the problem
well as the load
due to this deflection there rollers will move in the horizontal direction right when the load will when the load will increase hair over rollers will move and lead this direction this roller
will move in this direction and the stored water will move like this right then as you can see that the length of these two frames is equal it is same and also the boundary conditions for both the frames for both the linkages is same so what you can say that this problem is a mirror image about a so we can say that the problem is symmetric right due to this movement there will be compression in the strings so let me summarize all the points the vertical displacement is related to u1 and u2 well vertical displacement here is this there is the vertical displacement and this u1 and u2 is horizontal displacement but they are related right then the second point is the structure is symmetric about a right yeah because length is same here the boundary conditions are same same load is applied for the structure is symmetric about a so both Delta and you are measured with respect to under formed configuration oh
I'm sorry for the spelling mistake it's an under formed configuration that SP increases that means as you are increasing the load which got it will cause to increase the U then I was including simultaneously you will increase an increase in you is compression in Springs so I think I have explained the problem in detail right now and I think now we should start solving of a problem using some equations okay yeah proceeding further so what we'll do first is first we'll
isolate at the point P
is applied it will cause a frame to deflect and it is very easy for you to tell all the physical changes because as you are as you have already visualize the problem in the beginning right and
and as we know from the concept of principle
equilibrium that we have already covered in our first lecture that the resultant force needs to be zero at any isolated point for a equilibrium right so let me mention here the node B this is not be
here okay this this is not me yeah also the angle B little linkage is theta from the horizontal so let's first do an horizontal force balance and see what symmetry does to the poor's balance so when you will equate the venule to the horizontal force balance what you will get is F 1 cos theta is equals to F 2 cos theta right here let me show you the things this is F 1 this is f 2 so F 1 cos theta acting in this direction and F 2 cos theta acting in this direction so you can say for equilibrium F 1 cos theta should equal to F 2 cos theta by canceling cos theta from the both sides what you will get is F 1 is equals to F 2 right so this is the conclusion that because of symmetry f1 is equal to f2 now doing the same force balance for the vertical direction what we'll get is P is equals to F plus F sine theta plus F
sine theta how you're getting this is right e for your full speed is acting this is the force P which is acting in downward direction and you will resolve this force f1 it will get resolved in this and then what tickly in this direction similarly this force f2 is horizontally here and vertically here the first equilibrium this force is upward whose should equals the downward force so f sine theta plus F sine theta is equals to P and already we know that F F 1 is equals to F 2 that's why I'm not writing F 1 sine theta plus F 2 sine theta directly you can say that P is equals to 2 F sine theta ok so these are the two findings that we got from node B now we'll take another node that is node D now doing the static equilibrium at node D what we'll what we'll get from when we'll do the force balance at node B we are taking this section here at node D let me show you the force balance well we will get these equations now how we'll get these equations I'll take you now when you build up when you will do for node deal with which is here right now the this this F 2 you are going to resolve ok so this F 2 first horizontal
force will be in this direction and vertical force will be in this downward
direction right similarly this joint this joint is going to provide you one reaction are so that reaction will act upward my next video
lecture is on these constraints only so you will get a very good very clear idea after listening to the third lecture but till now just know that this reaction will act at work now when our this whenever spring will compress to this side okay tower force will generate in this way so this will be like our effect so this is our FS that I can say yes so now if you'll do the post balance what you will get is this F 2 cos F 2 or F F cos theta is equals to F s obviously
they should cancel each other for the equal
they should be equal and opposite they are equal and opposite so riding in the yellow manner we are getting is f cos theta minus FS is equals to 0 similarly if you take the horizontal horizontal vertical forces it will be F sine theta this F sine theta and the acting which is acting in downward direction is equals to R or again in writing in the another mat form we will get R minus F sine theta is equals to 0 see we are not taking force P here because force P is acting only on node B right right now here what we are considering these forces that are acting on node D ok so I think it's clear now if it is if it is not clear pause the video resolve the forces and see if you're getting the same equations you take any direction you take any positive or negative you will end up with these equations only okay so this is all about the force balance at not ignored B and E but not due to the force deflection equation you know positive post post deflection equation that F is equals to K into u where F is the force K is the spring constant and use the deflection this equation you must might have come across in your engineering till now so now what we are going to do is we have concluded each and everything we have derived everything now we have to use these
things okay whatever equations we have we have just derived now we have to use these equations to get one result ok so from
the problem statement only
that we have assumed that the links are continuous right links are continuous so which means that they they are not going to elongate or they are not going to compare that means length of that link is constant right and this also means that the angle see initially it was making theta naught and after the deflection is it is making angle theta that means that they are independent of each other because length anyways length
is is going to be same okay so if and if if I want to know like what is the deflection horizontal deflection you so how you can refer that how you can find the deflection from the geometry itself see this anyways this is the same this is going to be the same this is an initial position okay and this one is a final after deflection so u1 is this so he feels subtract this length from this you'll get your you so first angle is theta naught so this length will be L
into cos theta and then other will be L cos theta naught so if you subtract these two lengths you will get your deflection that means horizontal deflection you so similarly you can use
the same geometry so you can use the same trigonometry method for finding ur hope for finding this horizontal deflection right so you will be L we have taken n common so L cos theta minus cos theta naught and Delta is equals to L sine theta naught minus sine theta clear yeah so now we have to use this equations now
we are done with the for
ce balance now we need to get these things arranged so the first equation that was
obtained you from the node B was P is equal to 2 F into sine sine theta right and the other equation was f into cos theta minus FS is equals to 0 now if I'm going to take first equation okay just a
little bit let's take this only what you'll get from here is f is equals to P by 2 sine theta and from this you get this FS is equals to F cos theta right here yeah so this is the equation F this this equation you are getting from here ok FS is equals to F cos theta and from
this you are extracting air that F is equals to P by 2 okay it's divided by sine theta ok so now I just simply substitute this
there's the this. f here okay so FS is equals to P by 2 into cos theta by sine theta right now you have we have done
this equation FS is equals to K into u so if you have the value of affairs you have key you can find the value of U so u will be P by 2 K into cos theta into sine theta right but from the trigonometry what you have derived here is L cos theta minus cos theta not unit Delta is equals to L sine theta naught minus L sine theta so simply using these geometrical relations what you have finally got is these equations right clear so using all the equations and simply by substituting the values and using one equation into another P are
ended up with these two equations so what you can say by looking at the equations hmm because these are the final ones that we have got this
definitely that means that they are telling you something equation speaks right so think if you closely look at the equations they are in parametric form remember if you remember like for a circle you know that X is equals to R cos theta and Y is equals to R sine theta these are the simple parametric circle equations that you know so it simply means that if if you are going to plot these equations what you will get is a variation of result how your
parameters are pairing okay so if you plot the both equations and see the variation of results one can also eliminate the theta to point the analytical expression but this is something you need to find right now we are plotting and yeah this is the graph
that you will get when you will plot the result now can you relate so here what what does erection y-axis whatever is your constants B is equal to 2 K into L and Delta by L so you can see that how it is varying and what are the results that we have got right the ranges between theta naught to theta and theta to theta naught yeah even if you take any fear not you will get your result so thank you for your attention I hope your concepts are clear now and if not and we'll cover some more problems in there were subsequent lectures so subscribe my channel and and keep exploring
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