Mechanics of Materials & Principal of superposition

[djeitnstine]

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

The Attempt at a Solution

http://img51.imageshack.us/img51/2055/76147557.th.jpg

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.

Anyone can clarify this please?

 

[Mapes]

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?

 

[djeitnstine]

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]

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]

djeitnstine: As Mapes mentioned, you can establish any free-body diagram (global) coordinate system you wish. Then you can draw any particular unknown vector in any direction you wish (the positive or negative direction). If you draw an unknown vector in the negative direction, and its magnitude turns out positive, then the force is in the global negative direction. After you obtain a force result from this free-body diagram, you then transform the force to the element local coordinate system for each element. A universal convention for the element coordinate system is, tension positive, compression negative. You therefore finally see, from the element local, not global, coordinate system perspective, what force or stress is actually on the material.

 

[djeitnstine]

Ok thanks, I also discussed it with my professor and found out that made a mistake when I worked the problem out using convention.