00:01
This problem number 28 so here in point h in point h you can see a support that support is called pin support and because of this type of support there will be two reactions one vertical as well as a horizontal reaction so let's name the vertical reaction as r of hy and the horizontal as r of xx now in point g there is another type of support you can observe that that is a roller support so because of this type of support there will be under reaction that will be a vertical reaction and let's name that particular reaction as r let's see now let's draw the forces here itself.
01:22
So this will be r of hy and this will be r of hx and here and here this will be r of g.
01:50
Now to calculate the forces in each member, first we drawn the free body diagram of this entire trust and now let's apply the equilibrium equations.
02:04
So first we have to create support reactions.
02:08
So from symmetry, summation along h equal to 0 gives you the only thing along h, x axis is r of hx, so that will be equal to 0.
02:27
And summation along y equal to zero that is now let's take the moment at it's first so that gives us minus of rg and the distance for its meter plus the 48 into 4 plus 4 plus 4 plus 4 that is 16 16 into 16 into 4 of course 48 equal to 0.
03:08
So from here, we can write rg equal to 192 kiloons.
03:18
Now, along y -r -hs, you get under equation, like summation along y gives us r -hy plus the 192, that is rg, minus 0, that is minus r -x.
03:40
Will be equal to 0 will be equal to 50 so this versus we can write rh might will be equal to minus of 140 kilo that means the direction is opposite so here this so here the direction we have given creates tension but because there is an it sign so that creates compression that's denoted by letter c and the rg creates compression rg also creates compression now let's move to the next let's denote like the angles the angle this one press this one yeah so here this will be theta 1 and this angle let's call it as theta 1 and the anchor in b that is this one this one will be theta 2 and this one this one this one that will be theta 2 and this one this one the that will be theta 3 and yeah that's all now let's find angles theta 1 will be equal to the inverse of 4 .5 divided by 12 that is 20 .6 5 theta 2 will be equal to time inverse of so theta 2 will be equal to time inverse of so so that gives us 3 by 4.
06:48
So that is 36 .87 degree.
07:03
Now, theta 3 will be equal to time inverse of 4 .5 divided by 4.
07:18
That will be equal to 48 .37 degree.
07:24
Yeah 4 .4 divided by yeah and theta doing be equal to this 4 .4 divided by 8 and theta 1 will be equal to 4 .4 divided by 12 now we found the angrize now let's go to each joined and to serve it first for joined a let's draw the diagram first so this is f a b there will be f ac some slope with time to theta 1 and the 48 kilometer so for joint the a let's resolve along y first summation along y equal to zero gives you minus of f ac c sine of theta 1 minus 48 will be equal to 0 so from here the fac will be equal to 48 divided by sign of theta 1.
09:10
So you know the angle of theta 1 that is 0 .65.
09:18
So that gives us 136 .109 that is 136 .109 that is 136 .1 .1 .9 that is 136 .1 .1, you know, my you know and the fac that is actually going outward so as this is going outward and this here so let's check it yeah as fac is going outward and but here and here that in answer there will be a negative sign so that means so that means the actual direction of your fac is in that creates compression let's denoted by the letter c the sign of theta 1 sign of 20 .65 that is 48 divided by sign of 28 .65 equal to my 136 .109 and there is minus in here so this is minus of this one so as that is minus so that is minus so that are actual direction is opposite to this one that creates compression denoted by the letter c now summation along x equal to 0 gives you f a c course of theta 1 plus of theta 1 plus fab equal to 0 so from here we can write fab will be equal to 128 kiloan open and if you b is going outward from the point a that creates tension let's denoted by the letter t now let's note the next joint that is joined c.
11:37
So join c.
11:52
If this is joint c, there will be f c b and just opposite to that fac and fbc.
12:11
Now let's resolve for c summation along it's.
12:19
Equal to 0 gives you fce this is along a line so f c e.
12:28
So f c.
12:30
So that it will be 1306 .1 .109 kilo nootan and here that is going outward that creates tension.
12:44
Let's turn out by the rarity.
12:46
Summation along y equal to zero gives you.
12:49
There is only fbc so the value for fbc so the value for f is equal to zero.
12:56
Now let's go to the next joint that is joint b for join b if this is join b there will be f bd b f ab and the f b b and f b b and f b fb e and f .b.
13:47
The fpc is c so you don't have to do that...