Axial Stress

[Guest]

Below is a problem I have, I need to work out the axial stress in each of the three differant parts, top bit is ok, When I get to the middle bit am I right in thinking that the value I use for force is simply 10kN because the forces are acting against each other and you take the smaller from the larger to find the overcoming force?

Then for the bottom one I am looking at 15kN as they are added up?

[Sonny Clarke]

[Haz]

Well that has to be a warning for spam.

2. Spam Minimum Penalty - 1 Warning Point

Please refrain from posting meaningless threads, one word (or short) non-sense posts. Repeat offenders will be warned, and possibly moved into New Members Chat.

[Sam Nichols]

Well that has to be a warning for spam.

Well that has to be a warning for being a tosser..

[PaRtZ]

Well I can certainly see your logic, but for the middle one, surely the lower one still has to come into effect? Looking at the bottom part, it has 10kn up, 5 kn down, so does that mean the bottom part has an overall resultant of 5kn up? THEN you apply your first logic to give result 15kn downwards

Then, as 15kn is downwards for the middle bit, you can use that but this is where Im getting confused, because you've already used the lower components tofind out the middle bit

[Guest]

argh.

Now I can see your logic!

Can anyone tell us who's right?

[Shaun H]

Can anyone tell us who's right?

I'm gunna be brave and say neither of you :$

Did some similar stuff for my mechanics module (which I passed ) except that involved resolving a billionty and one forces.

ANYWAY, here's my solution:

As the top is fixed, you can work from the bottom up, anything above a given point won't affect that point, as the fixing point is at the top. < That's quite badly explained sorry. Hope you can see where I'm coming from...

Edited July 14, 2008 by Shaun H

[Guest]

Cheers shaun, the tension and compression issues shouldn't effect my calculations should it?

Stress still = force /area yes?

So I just need to amend my calculations with respect to the forces involved yeh?

[Shaun H]

Cheers shaun, the tension and compression issues shouldn't effect my calculations should it?

Stress still = force /area yes?

So I just need to amend my calculations with respect to the forces involved yeh?

Since stress is a pressure measurement (F/a as you quite rightly said) I believe you take any tension to be a negative value as it effectively produces a negative pressure in the material.

[Guest]

OK I think i'm clear now, worked them out;

For the top bit I get an area of 0.0019634m^2 so;

-15000/0.0019634 = -7639808.5N/m^2

Just seems like a huge number is there something really obvious i'm missing out on?

[Shaun H]

OK I think i'm clear now, worked them out;

For the top bit I get an area of 0.0019634m^2 so;

-15000/0.0019634 = -7639808.5N/m^2

Just seems like a huge number is there something really obvious i'm missing out on?

I think it's simply due to the tiny area, it's only a few mm after all.

EDIT: Btw the SI unit for pressure is Pa (equal to N/m^2), could be one extra mark?

Edited July 14, 2008 by Shaun H

[Guest]

cheers shaun, Been intending in rounding up the figures to Pascals but didn't want to unless I was sure I had the fundamentals down.

Looking at it now I suppose it's not all that peculiar...

Thanks a million that's one less thing to worry about!