# Homework Help: Forces in rods due to temperature

1. Sep 28, 2010

### sailsinthesun

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

2. Relevant equations
Sum Fx=0
Sum Fy=0
displacement=alpha*(DeltaT)*L
displacement=sum of (P*L)/(A*E)

3. The attempt at a solution
I'm really not even sure where to start. I have a FBD drawn with Fb, Fc, and Fd and these equations written down, but not really sure where to go from there.

2. Sep 28, 2010

### Staff: Mentor

Your attachment is not posting for some reason...

3. Sep 28, 2010

### Staff: Mentor

I'll try to post it:

Oh well, at least the link is posted now. I'm not going to be of much help on the problem, but can you say a few things about what a change in temperature will do do tension or compression of a rod? What does your FBD look like?

Last edited by a moderator: Apr 25, 2017
4. Sep 28, 2010

### sailsinthesun

Well since temperature is increasing it will increase the forces in all rods due to expansion. The FBD is point A with Fb to the left, Fc to the right and 60* above horizontal, and Fd to the right 60* below horizontal.

One source of confusion is that my equation involves deltaT or change in temp. but the problem just says "increases to 50*C" with no initial temperature.

Last edited by a moderator: Apr 25, 2017
5. Sep 29, 2010

### nvn

sailsinthesun: Attach your FBD as a PF attached file by clicking the "Manage attachments" link. Or perhaps even better, post your FBD at imageshack.us. But if you post an in-line image from imageshack, do not let your in-line image be wider than 640 pixels. If your image file is wider than 640 pixels, just post a plain text link (url) to the image, instead of embedding the image within your post.

Actually, it is generally better to just post a text link to the image, regardless of the width, instead of imbedding an image in a post.

Last edited: Sep 29, 2010
6. Sep 29, 2010

### HyperSniper

Here's the image resized so it will show up on the forum:

[PLAIN]http://img692.imageshack.us/img692/5014/problemm.jpg [Broken]

I've been working on this problem too. I feel like I understand the case where it is a rod that is attached to a wall and then is heated, but what is confusing me is the fact that in this case there are three rods all expanding and two are attached at an angle. How do I figure out the forces that each rod feels from the temperature effects in this case?

Last edited by a moderator: May 4, 2017
7. Sep 29, 2010

### HyperSniper

Okay, so the only answer I can come up with that makes sense is where all the rods are subject to the same amount of force. Here's my work:

(Sorry, but I can't get the math formatting to work properly).

Area of cross-section of rods:

d=25 mm

A=((pi)(d^2))/4= 490 mm^2

The rods are in equilibrium:

$$\Sigma$$F_y=0

Compatibility:

delta=PL/AE

delta_A/B=0=delta_T-delta_F

0=(alpha)(DELTA*T)L-FL/AE

Solving for force:

F=(alpha)(DELTA*T)AE

F=(10^-6)(12)(50 C)(490 m^2*10^-6)(200 Pa*10^9)

F=58 800 N

Since, all the rods are the same dimensions, they all are subject to the same thermal stress, and therefore the same amount of force (58 800 N).

Since they are also in equilbrium:

F_Ay=58 800*sin(60 deg)-58 800*sin(60 deg)=0

F_Ax=-58 800*cos(60 deg)*2+58 800=0

No rod feels any more force than the others.

So therefore, each rod develops a force of 58 kN.

Am I doing something wrong?

Last edited: Sep 29, 2010
8. Sep 29, 2010

### nvn

HyperSniper: I currently think you made a good assumption that the word "to" in the problem statement is probably a typographic mistake, and should instead be "by."

Regarding your current solution, I am not quite sure yet, but I think your solution assumes rod AB goes back to its original location, right? I am currently thinking we cannot necessarily make that assumption (although it might turn out to be true in this particular problem). Is there any way you could write your equations such that the final horizontal location, x, is an unknown, then use compatibility to say the x values for all three rods are equal?

Last edited: Sep 29, 2010
9. Sep 29, 2010

### HyperSniper

Yeah, I was thinking something similar too, because rod AB is held in place by a hinge at A, it may not be in the same location that it is in initially. I'll take a look at it again later.

10. Sep 29, 2010

### sailsinthesun

Thanks for getting the picture posted hypersniper. That does look like you're on the right track but I'm not sure about the placement of A due to forces. It does seem that if rods are all expanding, the point A would move to the right slightly and cause both 60* angles to be slightly larger.

11. Sep 29, 2010

### nvn

HyperSniper: Nice work. Just a couple of comments. Generally always maintain four or five significant digits throughout all your intermediate calculations. Then round only the final answer to three (or four) significant digits.

Also, always leave a space between a numeric value and its following unit symbol. E.g., 25 mm, not 25mm. See the international standard for writing units (ISO 31-0).

12. Sep 29, 2010

### HyperSniper

Opps, sorry about that. I was just getting frustrated with LaTeX and wanted to save time in getting my work on the forum, however I did actually carry out all the significant digits in my intermediate calculations, it just so happened that it came out to exactly 58 800 N.

I checked with a friend who worked this problem independently and he got the same answer, so wikiality leads me to believe my answer is correct.

13. Sep 29, 2010

### nvn

HyperSniper: Try not rounding area A, and you will get a slightly different answer. Nonetheless, nice work.