# Trying to break a rod

1. Nov 13, 2009

### signorvahid

Hi guys,

Sorry if I’m posting In the wrong place, I’m a newiee:D

I have a problem (seriously, it is not homework), no equations, no numbers, just a mechanical question. I’m trying to break a rod for instance (The rod has the same diameter althrough except for one point where it becomes thinner). I know where it will break since it is a little thinner or weaker in a part of it. Now, I put it on a holding device put force on exactly adjacent to the weak line (where the thin line is). Is the amount of force required to fracture the same rod going to change, by changing where the holding device is going to grip the rod? (I made a picture of the situation)

Please guys, i've been checking this post 10 times every hour!!

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Last edited: Nov 13, 2009
2. Nov 13, 2009

### signorvahid

can someone at least move the post to somewhere else if i've posted it in the wrong place, please?

3. Nov 13, 2009

### sylas

By putting the force directly on the weak part, you make it most unlikely to break at that point. It is most likely to break near to the holding device; and it becomes more likely to break as you increase the distance from the holding point to the point at which the force is applied.

Think "torque".

Cheers -- sylas

4. Nov 13, 2009

Killer question.The way I see it at the moment the further away the holding device the more the rod will bend before breaking.This bending will increase the separation of the molecules along the upper part of the rod and this increase will extend to a certain amount within the defect thereby weakening it even further and causing it to break more easily but,I am not sure if this is correct and I will give it some further thought.

5. Nov 13, 2009

### semc

IMHO lets assume static equilibrium when the same force is applied at the same point and assume the rod is on the verge of breaking but not yet hence equilibrium. By taking torque about the other end of the rod we can clearly see that a greater torque is generated when the support is near the torque so it should depend on the distance?

6. Nov 13, 2009

### jambaugh

It sound like you are trying to shear the rod at that point. If so then the amount of force needed will be independent of where you clamp it. You'd want to clamp it as close to the cut point as possible to prevent bending and to get a cleaner break. This is why scissors a.k.a. "shears" with a loose screw tear paper (or metal or whatever) instead of cutting cleanly.

Now if you want to break it open by bending (like a hinge) the force needed will be a function of the distance of the applied force from the break point but again independent of the clamp point. If you then clamp too far from the intended point you may not get a break there but rather a bend or break closer to the clamp even though the bar is thinner at the intended point. You'll also need to take this into consideration if you are shearing as above since the bar doesn't know whether you want to shear it or bend it.

7. Nov 13, 2009

### signorvahid

Thanks a lot to all you guys, that was a lot of relief :D

8. Nov 13, 2009

### signorvahid

You see guys, I’ve actually done this in an experiment where I separated the two parts of the rod, stuck them back together with sth, and then I measured the SHEAR STRESS required to detach it (the distance between the force applied and the glueing line was minimal in all my work). Now my master says I have to redo the whole thing because I have an unequal LEVER ARM on the other side of the rod (from the detachment to the clamp) because I didn't control this part exactly(even though the differences were about millimeters). now the question was that does this affect the final force calculated?(which you've said it doesn't so far) and is there any way I can prove this to him in a physical/mechanical way?

Thanks again for all your replies

Last edited: Nov 13, 2009
9. Nov 13, 2009

### jambaugh

Theory can tell you what happens in idealistic situations but when it comes to making precision measurements you may want to at the very least run a few trials to verify there is no significant effect. Given the rod will deflect some (more the farther from the clamp you apply the shear force) and so you're applying a force at a slight angle then there may be an issue of shear strength vs tensile strength if the joint is hinging open slightly instead of purely shearing.

This can impart error in your measurements. Only systematic testing can say whether it is significant.

Mostly though he is correct because he is in charge so I recommend you do it again with great care to clamp it uniformly close to the shear site.

10. Nov 15, 2009

### signorvahid

Thank you jambaugh, but the thing is i've tested hundreds of these, i've tested it in both ways and the difference is negligible (compared o the final enormity of the force). + the results I have now are really logical and well undestood with the method I used. The only problem is convincing him so I can graduate. do u think there would even be a theoretical explanation I could convince him with it?

11. Nov 15, 2009

### jambaugh

The clincher would be to show that the deformation of the bar within the range of clamping distances yields variations of effect below or at the error tolerances of your data, especially by showing the angle of deflection at the point where the force is applied is on the same order as the angle tolerances at which you are applying the force.

I take it you are testing shear strength of a joint much weaker than the material of the actual bar? (Otherwise it would have bent dramatically before breaking unless you kept the clamp very close.)

I think it is a simple matter to give a ball-park determination of the angle tolerances of the clamping and then show that it is much larger than any angle due to deformation of the bar as far as that goes. You can then argue that this means either the university needs to provide you with much more expensive test equipment so you can clamp at super high tolerances w.r.t. angle... or he needs to acknowledge that this has no significant effect and so neither does the slight bending of the bar under the applied force.

At worst you could do a small set of trials with nearly identical shear strengths but different clamp distances and show empirically that the clamp distance is not correlated to your data. (A simple bit of statistical analysis.) I suggest you pick 4 or 5 distances and 3 or 4 trials per distance. Select the trials in random order so there is no time bias possible.

The actual measurement needn't be significant w.r.t. your work (i.e. you needn't use the same joint material as your earlier data, just something on the same order of magnitude w.r.t. shear strength.)

12. Nov 16, 2009