00:01
Hi, hi everybody.
00:03
So here this is the diagram given in the touch book.
00:08
So we can see here in point a there is a pin support and point g there is a roller support.
00:22
So let's just right here.
00:25
In point a there is a wind support and we know that we know that there is a wind support and we know that what pil support will do.
00:40
So do because of this support there will be two forces, one vertical and one horizontal force.
00:52
So one vertical force, say let's name the vertical force as say, rh, so it's a right? so it's point a.
01:09
So it's point a.
01:09
So it's let's name the vertical force as va that should be better and there will be one horizontal force also let's name the horizontal force h a horizontal h a and in point g in point g there is a roller support so due to because of this support there will be one force one vertical force so let's name that vertical force as vg the point is g so let's name according to the point it's vg and let's denote the forces here like this will be vg the vertical force vg due to this role as opposed and here there will be a va due to point a and also a ha due to point a.
02:39
Now there is one thing.
02:43
Now let's name the angles like let's say this is alpha and this is alpha and this this angle here that is beta and this angle here that is let's say gamma.
03:09
So this is alpha, this is beta, this is gamma.
03:18
Now in point g, you know we have a role support.
03:24
So as i said, because of this support, we have one external unknown reaction, we have vertical force vg and just remember in point a as i said because of this support because of this pinned support we have two external unknown force, vertical force and a horizontal force.
03:51
So now to calculate these unknown forces we need to write equations of equilibrium.
04:01
Let's write the equation of equilibrium.
04:06
Equation equilibrium.
04:23
Let's take the summation along at sizes.
04:28
Summation over h equal to zero gives us h a minus zero equal to zero, that is h .a.
04:39
And summation, let's consider the point g.
04:46
So moment at g equal to zero uses v a into nine that is three plus three plus three so it's nine this b a into this distance nine we are considering all the distance in meter itself minus the force one point five so 1 .5 into the same distance will be there.
05:31
9 minus the force 3, 3 kilo tonal.
05:39
3 into attribute better because otherwise this decimal and the multiplication will be confused.
05:51
Okay, so 3 into 3 into the my and minus 2 .25.
06:06
2 .25 like here this distance from a to b this distance will be here it's mentioned it is 2 .25 meter so the whole distance 9 and subtracted this 2 .25 and this one so minus the 3 into 4 .5 minus 3 into 2 .25 equal to 0 that is this 4 .5 actually this is kind of this middle point so if the whole distance is 9 so the half is 4 .5 so from g to this point the middle that is 4 .5 that's way 3 into 4 .5 and this 2 .25 is the distance from here to here this is given as 2 .25 so as this is symmetrical so this distance is also 2 .25 so this 3 into this 2 .25 so from here we will get v a equal to 6 km from this calculation we will get va equal to 6 kilo motor and is going in 1 so let's try that creates a compression did not play c now summation or y equal to 0 mission over by 2 uses v t plus 6 minus 1 .5 minus 3 minus 3 minus 3 minus 3 minus minus one more 1 .5 equal to zero that is this 1 .5 3 3 and this 1 .5 so that's how it is things and yeah that's pretty much and this 6 is this thing va so yeah this gives us bg equal to 6 km.
08:50
So bg is also 6 km so and bg is also going inward that creates compression so you know it does see now let's move on to the next path now we will write equations of equilibrium for 70 a.
09:23
So let's consider each and every joint and let's write the equilibrium equation of equilibrium for every joints.
09:35
So first we are considering joint a for joint a, for joint a, yeah.
09:49
We know the diagram for joint a will be like if this there will be a force s -a -b like this and s -a -c and the b the b a upper and we found the value that is 6 kilo newton so this is so it's six kilo not only just found value and yeah, this angle we denoted as alpha.
10:38
So let's say tan alpha.
10:44
Then anchor alpha will be 1 by 2 .25.
10:53
That is some way opposite.
10:58
So from here we will get alpha equal to 23.
11:02
Point minus six two degree so let's do a ruby summation over by equal to zero kinses yes a b sab sine alpha minus six plus 1 .5 minus this six plus 1 .5 minus this is plus 1 .25 so from here we will get the value of scb so the value of sab equal to 11 .0.
12:02
08 kilo newton and yeah this is c and as scb is going inward to the point a that creates a compression it's denoted by letter c.
12:23
Now let's take the summation of our along it.
12:30
This gives us sac minus 11 .08 this value into that will be cos alpha.
12:48
So we'll go to zero.
12:53
So from here, from here we will get the value of sac.
13:01
That will be 10 .1 2 5 kilo in a time so we got the values here as sac is going outward from the point a that creates a tension and let's denote it now let's move to the next joint let's consider let's take a question let's consider the next that is join b so for joint b first let's draw the diagram for join b if this is b there will be false right sbd s b c c and s a b and s a b that we're already calculated we got the value as 11 .08.
14:34
So it's terminated.
14:41
And we know this anchor, this anchor we named it as theta, and this anchor.
15:02
This anchor we named it as alpha.
15:07
Now let's go for calculating the equilibrium.
15:18
So let's first find time beta.
15:25
Pan of angle beta will be 1 minus 3 minus 2 .25.
15:33
That distance will be 3 minus 2 .25...