# How do I calculate the forces in a Truss?

1. Aug 18, 2016

### tandennis0703

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
I have spent the last 2 hours trying to figure out which part of the truss is compression or tension. Hopefully the second screenshot I took is correct. I tried figuring out the first few parts, just want to make sure that at least im on the right track.

2. Aug 18, 2016

### haruspex

Although it is usually easy to figure out in simple cases, it can be quite tricky. Fortunately, you rarely need to. Just make a stab at it in each case and write the equations accordingly. If the answer comes out negative then you guessed wrong, but so what? You still have the answer.

In fact, in the annotated diagram you put under "relevant equations" you seem to have guessed most wrongly.
And that is not what the "relevant equations" section is for. It's for standard equations that can be applied to a whole class of questions.

In the image of your working, how do you get Cx+Dx=0?
(Please do not post images of your algebra. Take the trouble to type it in. That makes it easier to read and easier to comment on. Images should be for diagrams and textbook extracts.)

3. Aug 19, 2016

### tandennis0703

Sorry for the mistakes. I never took statics in my course. I redid everything and used the method of summing the resultant forces in both the x and y direction. I am still a bit confused as to how to figure out whether or not it is a compression/ tension force. Hopefully my calculations are correct.

4. Aug 19, 2016

### haruspex

I did ask you to type in your algebra. I am having trouble reading some of the image, and if I want to comment on an equation I have no easy way to refer to it.
Check the signs in your equations for point B.

5. Aug 19, 2016

### tandennis0703

Joint A:

+^ΣFy=0 ; -2P+(2/3.61)FAB=0
FAB=3.61P

>+ ΣFx=0; -FAE+(3/3.61)(3.61)P=0
FAE=3P

Joint E:

+^ΣFy=0; FEB-P=0
FEB=P

>+ ΣFx=0; 3P - FED=0
FED=3P

Joint B:

+^ΣFy=0; (2/3.61)FBD-P-(2/3.61)(3.61)(P)=0
(2/3.61)FBD=3P
FBD=5.415P

>+ΣFx=0; FBC -(5.415)(3/3.61)(P)-(3.61)(P)(3/3.61)=0
FBC=7.5P

Hence:

stress BC= 7.5P/340 = 250

P=11333N

6. Aug 19, 2016

### haruspex

According to what you calculated at joint A, is AB under compression or
Now I can read it, it all looks right.

7. Aug 19, 2016

### tandennis0703

Thanks for the reply. However I am still a bit confused as to how I know whether or not a bar is undergoing compression or tension. Do I need my calculations in order to figure that out?

8. Aug 19, 2016

### haruspex

In general, yes. As I posted, it is not always possible to figure it out by simple inspection.