00:01
We'll be looking at structural analysis, consider the problem at hand who want to determine the maximum force p, that the structure can be loaded with, okay? and to solve this problem, we'll apply the method of joints, okay, to solve it.
00:19
So, but before they let us look at the support reaction.
00:22
For the support reactions, you can see that the structure is symmetrical.
00:27
Therefore the reaction at a is going to be p over 2 and the reaction at c is also going to be p over 2 okay so you can look at the joint a so we'll apply the method of joint okay so let us look at joint a if you look at joint a you will discover the forces there's this force here okay and force now this is our joint a now you are going to determine the angles using the information given in the question to do that okay you can call this because it's coming out of the joint a and going to another joint b call it sub a b now this angle is inclined to 45 degrees okay so use the information given in the question the structure there to do it to do that i will have another force like this okay this one we can call it this is a s suffraith a ad and the angle there is actually 14 point you do define it out you are going to arrive at 14 .0 4 degrees okay now you can apply of course this is what we got not quite long p over 2 for arrow or so 3a okay so we can look take summation of forces in the uh vertical direction to be zero so when we when we do this we are going to arrive at p over two plus f such with a b sign 45 degrees okay plus f a d sub sub sub a d okay sign 14 14 0 .04 degrees to be equal to 0.
03:06
You can call this equation 1.
03:09
Then if we take some function of forces in the horizontal direction, they are going to have, the forces are going to be, just look, use the diagram.
03:26
That's why the diagram is there.
03:27
You have that force s with ab, okay? when you resolve it, you'll give you cost 45 ,000.
03:37
Degrees okay in the horizontal direction you said it f substitute a ad cost 14 .04 degree is equal to zero you can call these equation two okay okay now if we solve these two equations and we'll consider write equation two making a f a b disrupt of the formula okay you are going to arrive at fab to be equal to plugging in the values of your cosine wherever all the values you need to put in 317 okay i'm going to arrive at this value f ad okay now you can plug this back into into equation one okay when you do that you are going to arrive at your f -a -a -b to be equal to minus 0 .9443.
05:23
Okay? 0 .943p.
05:33
You can see that this member has a negative sign attached.
05:37
So it's a member under compression.
05:39
Okay? that's in nature.
05:42
Now we can look at another joint.
05:45
Let's take join b, let's call it just for sure jb.
05:50
Okay, we have this force coming like this.
05:54
We have this force coming straight down.
05:57
We have this force.
05:58
Okay.
05:59
Now, if you check the forces, check the diagram.
06:05
Okay, use the information, the figures, dimensions given in the diagram.
06:08
Okay.
06:09
You will have this angle to be actually 45.
06:20
This angle is going to give you 45 degrees.
06:25
Okay? these are joint b.
06:28
You said the forces should be coming out of the joint, except those that are naturally going in.
06:36
So we are going to have this one to be also 45 degrees.
06:45
Okay? so we just use the information given in the question.
06:51
This is going out.
06:53
This is fb or s of b .c...