# Mechanics of Materials & Principal of superposition

Gold Member

## Homework Statement

A steel tube with a 32mm outer diameter and a 4mm thickness is placed in a vice and is adjusted so that the jaws tough the ends of the tube without exerting any pressure on them. The two forces shown are then applied to the tube. After these forces are applied, the vice is adjusted to decrease the distance between the jaws by 0.2mm. Determine the forces exerted by the vice on the tube at A and D.

http://img193.imageshack.us/img193/2700/13357119.th.jpg [Broken]

## The Attempt at a Solution

http://img51.imageshack.us/img51/2055/76147557.th.jpg [Broken]

Ok I did the question and got the answer after the professor solved it in class. However what I do not understand is why does he take the reaction forces as radially outward (in tension) and when he did the solution, he treated left as positive and right as negative.

This seems to go against convention. Even the solution manual has the same technique as he does. His explanation was that $$R_a$$ was in tension. Clearly from convention its in compression and so is $$R_d$$. Their answer also 'miraculously' show that it is in compression.

Last edited by a moderator:

Related Engineering and Comp Sci Homework Help News on Phys.org
Mapes
Homework Helper
Gold Member
In statics, you can draw any arrow in any direction. (If it's the "wrong" direction to produce a positive answer, you'll get a negative answer, of course.) The key is to sum the forces correctly. Here we'd want $-R_A-42+30+R_D=0$ or $R_A+42-30-R_D=0$.

He may have been saying that $R_A$ corresponds to a hypothetical tensile force, which it does. In the end, $R_A$ will turn out to be a negative number, indicating that the force was a positive compressive force all along. Does this answer your question?

Gold Member
Hmm, well that's only half of the answer. What I failed to mention was that when I used the right as positive and left as negative convention I got a rubbish answer. The work here isn't exactly statics, its mechanics of materials. This problem as you should see is statically indeterminate.

The way the problem is solved is by considering reduction in length $$\delta$$ created by the forces. Where $$\delta = \frac{FL}{AE}$$ (Where F is the force(s) in question, L is the length of the tube, A is its cross sectional area and $$E=\sigma \epsilon$$). We find the $$\delta$$ for each section (AB BC CD) and realize that total $$\delta_T= \delta_{AB} \delta_{BC} \delta_{CD} = 0$$

I.e. $$\Sigma_A^D \frac{FL}{AE}$$

When I used the regular convention For F (right as positive and left as negative) my answer came out totally wrong.

Mapes
Homework Helper
Gold Member
In mechanics the convention is different; a tensile force is positive and a compressive force is negative, regardless of the direction the force points in the global coordinate system. (These signs correspond to arrows pointing out of the beam.) Nevertheless, there should be no problem in assuming an unknown force to be tensile or compressive; the answer will reveal whether the guess is right. It takes some getting used to.

nvn