00:01
We'll be looking at structural analysis.
00:02
Consider the problem where we need to determine the magnitude of the forces in the members, the members of the structure as well as their nature.
00:15
To solve this problem, we apply the method of joints.
00:18
Before there, let us look at the support reaction at a as well as at c.
00:28
So we take submission of for a moment about c to be equal to zero.
00:37
We're going to have at 30 times four, two times four, plus arrow, the vertical component of the reactor at a times six, plus 15, times three, equal to zero.
01:06
So we're going to have that component equal to 12 .5 kilo newton.
01:25
Therefore we can have a rc to be 27 using upward and downward forces okay so we should not be upward and downward forces arrive into 27 .5 kilo newton now if we look at joint f it's a good place to start from jf you can have uh we have something that look like this sorry we have uh this force okay these are joints, just coming in.
02:10
Then we have this force coming out, f, force f, so to f, e, going from f to e.
02:27
And from f to joint, and we have at this joint, the forces should be coming out of the joint.
02:33
Except those that are naturally going in like this particular mutine.
02:39
Okay, it's given in the question.
02:42
Okay this is f such with f a going from f to a then f to then okay so we can take solution of forces in the horizontal direction i'm going to have a katty kati plus f with f e equal to zero so you solve this as i'm going to have f such with f e is equal to kati so that will be equal to kati so that you give you negative party to newton so it tells you that this force or this member is under compression of the negative indicating a member under compressive force so we can take summation of forces if you take some function of forces in the assumption of forces in the wide vertical direction.
04:05
We are going to have minus f .a minus f sosp equal to 0.
04:19
So that tells you that that force f is equal to 0.
04:39
Then we can look at join d.
04:44
We take join d, block this one out, let's say we take join d, call it jd.
04:50
D.
04:52
Okay, we can have a promise, okay, we have this to look like this.
05:00
From inspection, we are going to have ve, fdc, d .c.
05:15
Now if you look at this force from inspection, they are actually zero, when you solve it out, and they are actually two forces forming two two members forming a joint at d and there's no external force applied or reaction coming applied to the joint.
05:36
So if you won't solve it at the summation along the horizontal direction and total of the particular direction, you have to arrive at this but this force is to be equal to 0.
05:52
Dc.
05:55
Okay? so you are zero.
05:57
So we can look at joint.
06:02
You can look at joint another joint let's say joint c we are going to arrive at this value sorry this diagram okay this one coming out you said it forces should be coming after the joint this i don't see okay this one is f source rate c b now this angle is 53 .1 degree using information given in the diagram 3 .1 degree use the information given in the question the diagram solve the problem to solve the to determine the angles rather okay we're already that this is f so through cd and that is equal to zero if we determine already if this force is coming down so we take summation of forces in the horizontal direction you are going to have minus f so see b minus f, subscript, c .e.
07:42
Costs, 3 .1 degree equal to zero.
07:50
Okay, then you can also take cb...