00:01
Hi everybody.
00:03
So here, de and fgr zero first member.
00:09
De and fgr zero force member.
00:32
And here in point h, you can see a support.
00:38
That support is called roller support.
00:41
In point h, there is roller support.
00:43
You know the specialty of roller support the roller support because of this support there will be one force and that will be a vertical force because of the support there will be vertical force and let's name that force as h -way and in point a in point a there is another support and that support is known as pinned support and you know that special japan support because of this support there will be two forces one vertical as well as a horizontal force so let's name the vertical force as a y and there will be horizontal force also let's name the horizontal force as a x so let's draw the force forces here itself so here there will be a vertical force we named it as h -y and here here also there will be a vertical force we named it as a y and there will be a horizontal force also and we named it as a h now let's go to the problem so to use the method of we start with an analysis of the free body diagram of this ender trust then we look for joins connecting management two members with unknown internal forces to solve with false equilibrium equations so so the summation of moment at a equal to zero uses minus nine feet that is is in feature 9 feet 4 keeps multiplied by 4 kicks minus the distance to the next force is 18 feet 9 plus 9 18 feet the force is 4 kids itself and plus the net's distance to h way is 36 feet multiplied by h way equal to zero so from here we can directly write the value of h -way so we will get therefore h -way will be equal to three kids now let's find the summation along h's equal to zero that gives us h equal to zero as h is the only one along that's at senses now let's when summation along y equal to zero that gives us hy plus a y minus four keeps again there is a four keeps minus four keeps equal to zero from here we can substitute the value of h y we found here and we will get the value for a way directly the a way will be five to six so we got the value for h .y and ai and now let's move to that next job yeah let's move to each join and so this and let's just denote like h -y which way is going inward so that creates a compression let's denote this as c that denotes compression and a way also going inward so that creates compression so let's denote by the letter c now let's move to the first joint so first we are going to solve for joint a so if this is the joint a then there will be fab fac fac yes and there will be a y it's is already zero so we don't have to consider that so and we know the distances 9 this 9 feet this is 12 feet so the so the hypotenuse will be 15 feet now let's resolve this summation along y equal to 0 gives us minus 12 by 15 multiplied by fab plus a y equal to 0 therefore we know the value of a way substitute the value of a here that is five kids and then we will get that if a b will be equal to 6 .25 kids and as fab is going in one to the point a that creates compression so let's take noted by letter c now summation of h equal to zero gives us minus 9 by 15 9 by 15 f a b plus f a c equal to zero so we got the value for fab substitute the value of fab here and then we will get the value of f ac c and fac will be equal to 3 .75 keeps and fac is going outward from the point a so that creates a tension denoted as t.
08:34
You can use cost - theta and sine theta instead of this opposite by adjutant, and adjacent by hypotenuse also so here the distance are given so for use of calculation i used the distances now let's move to the next joint that is joint c let's use green color now let's move to nets join c so the diagram of join c will look like there will be if sorry there will be a forwards like this particularly upward fbc and there will be fce and there will be four kids vertically downward and facc so no so for joint c summation along h x equal to zero uses there is fce minus fac will be equal to zero.
10:17
We know the value of fac so just substitute the value of fac here and then we will get the value for faca that is fce will be equal to 3 .75 kicks and fc is going outward from the point c so that creates a tension let's denote it by letter c d.
10:47
Summation along y equals 0 gives us the fbc minus 4kis will be equal to 0 so that directly gives us the value for fbc that is 4 kicks and you can see here fbc is also going outward from the point c so that creates tension let's denote it by the letter t so we now solve for joint c now let's move to the next join that is joint b for joint b so let's draw the diagram for join b so if this is the point b there will be fpd and fbe and fbc particularly downward and the f -a -b so let's note the distances here itself this will be nine feet this is 12 feet so the hypotenuse will be 15 feet just like that here also this will be nine feet this will be 12 feet so the hypotenuse will be 15 feet so we got the values now let's go for the calculation just like the previous join summation all by equal to zero gives us 12 by 15 fab there will be 12 by 15 fab and again the opposite by hypotenuse 12 by 15 fbe minus fbc equal to zero we know the value of fbc and we know the value of f a b just substitute the values here in this equation then we can find the value for f b directly and we will get f b equal to 1 .25 keeps and you can see f b is actually going inward so fb creates a compression so let's denote it by the letter c now summation alone it's equal to zero uses fbd plus adjacent by opposite that is 9 by 15 fab minus again adjacent by hypotenuse sorry this the 15 is 9 by 15 fb e equal to 0 so here we just go to the value of f p and we know the value of fab we just found the value of fab from previous the f a b equal to 6 .25 so just substitute the values in this equation and then that uses the value for fbd therefore fbd will be equal to three kids and here if bd is actually going outward that creates a tension and let's denote it by the letter d so now we solve for joint b now let's move to the next join that is joint d join so the diagram for joint d will look like this there will be fdf there will be fdd and fde so here also summation that of h equal to zero gives us minus fdd plus fd fdf equal to zero that is we just the value of fbd just substitute the value here then we will get the value for fdf that will be equal to minus 3kx this native sign just indicates whatever the direction we given here and the actual direction is just opposite to the direction we given here that is the fds will actually be going inward okay so according to this one so if this is going inward that creates a compression actually so this is minus three keeps so the fdf is going inward so that creates compression denotransc so just like this here this is minus three keeps so according to that logic fbd then the actual fbd is going inward not going outward so that creates actually compression so this is not tension this is also compression now the summation over y equal to zero there is only fd along y so fd is just zero so you don't have to bother about that thing so let's move to the next join now that is joint e joint e so for joined e the diagram is kind of complex if this is joined e there will be f ef and there will be f be f be e and f be e and f eg and f eg and f e c c and the 4 kids given in the question.
19:28
So here let's indicate the distances also.
19:37
This will be 12 feet.
19:42
This will be 9 feet.
19:46
This will then be 15 as that is the hypotenuse here also this will be 12 this will be 9 then the hypotenuse will be 15.
20:00
So that's fine for joint e summation along y equal to zero gives us opposite by hypotenuse 12 by 15 f be plus again opposite by hypotenuse 12 by 15 f cf minus 4 x0 equal to 0 0 0 so therefore of fbe and the 4 kids is given in the question so there is all the unknown value is only f ef so therefore substituting the f b value uses the value of f f f that is 3 .75 kids and f f f f f is actually going outward from the point e so that creates a tension denoted by the letter t...