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• # How to determine the strength in aluminum?

Discussion in 'Calculations' started by FL_STEVE, Nov 11, 2014.

1. ### FL_STEVEMember

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Hello my name is Steve and I'd like to say thanks for letting me join.

First please forgive me if this is a stupid question, unfortunately I am not an engineer nor a metallurgist and would really appreciate the input of more intelligent individuals.

I have had a design in mind for quite a while and designed it using CAD software, I had a prototype made and it looks and functions great, my dilemma is the strength in 1 spot, picture attached.

I have looked up on Wiki 6061 T6 & 7075 T6 strengths and understand the concept of PSI, but my math is very bad and I do not know to calculate it's strength described below.

Material is aluminum 6061 T6 & picture attached, this is a 2.5" length round rod with a .194" in diameter, .350 from the end of the rod I have a hole .094 (changed to .069 1/16" diameter) in diameter which leaves .050 (now .063) on either side of the hole. So the Q? is it strong enough? It will have approximately a 5-10 lbs pull straight back (10 lbs is a lot), there will be a ring on the end which you will pull this gadget, then let go and that is it, so pull, let go, repeat.

How many times will you pull this?
(Answer) in one day +- 30-50 times.

How many times a year:?
(Answer) depending on the individual everyday.

From what I have read on Wiki aluminum will stretch with time, Will this piece with such a small pull weight? Should I consider using 7075 T6 which will make it stronger?

I can not thank you guys enough for any input....

Steve

2.
3. ### K.I.S.S.Well-Known Member

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Hi Steve,

When is America going to realise that Metric is just so much better than Imperial...?
Anyway, there's a couple of things that require clarification before a properly useful answer can be attempted - firstly, what CAD software are you using? Many programs have simulation functions that allow you to define loads and constraints and assign material properties - which will then very simply do the job for you.

You mention that it will be a ring that will exert the load on the walls of the hole - what diameter is the ring, and is it operated by a finger?
There is a lot more to consider with this than if it was a circular pin - in this case, the load will not be uniform, and the force will not be linear if a human being is pulling back the ring.
Obviously, the smaller diameter the ring, the greater the pressure that will be applied to the edges of the hole.
This will cause the side walls of the hole to deform, and this is probably a far greater risk than longitudinal stretch of the thinnest cross section of the component walls.
Also, the acceleration of the component has to be taken into consideration - if the load is quite heavy, a persons natural reaction is to exert an initially powerful acceleration to induce movement, and if this component is restricted in its travel by some form of hard stop, then this force is directly applied to the two interacting components involved in the initial movement.
All metals are subject to some degree of elastic creep when a load that exceeds their elastic limit is applied, but all metals will not exhibit this tenancy if their elastic limit is not reached. Also, you have to discriminate between dynamic loading (as in your component) and static loading (which is far more likely to induce longitudinal deformation).

So, I suppose that what I'm saying is that the problem you think you have, is not actually the problem you will face in reality.

4. ### FL_STEVEMember

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Thank you so much for the info & your response, you are correct it is not a ring that will be used, please forgive me for not giving all info, I do not have a patent on this idea as of now and am also trying to do.

The CAD sw was a free copy from emachineshop, design your product and send the design via the SW to be made by the same company. I believe it is a basic 3D modeling CAD sw, not near the quality of the others out their, it works that is for sure...

So more explanation, there is a secondary part that encases the rod then a spring pin (roll pin) 1/16" (.069 give or take) in diameter that holds both parts together.... As far as acceleration there is really non, as fast as a human can pull using a 10 lbs pull, it only comes back so far about 1.5" max.. The fatigue that I was referring too was the walls where the hole is not the entire rod. I may have misread your post.

The pic below shows the opening of the second part...

Thank you, Steve

Last edited: Nov 14, 2014
5. ### FL_STEVEMember

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Thanks again

6. ### K.I.S.S.Well-Known Member

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Hi Steve,

Thanks for the clarification, but as of today I'm taking my first holiday in three years, so I'll be off the radar for a couple of weeks.
If some other kind forum member would like to assist Steve...?
If not, I'll assist when I'm back, but in my humble opinion the intent of this forum isn't really to spew out answers to specific engineering problems, but rather to guide someone seeking an answer in the right direction.

With that in mind, try the following - calculate the cross section of the smallest area of your part. This is very easy to do.http://www.engineeringcalculator.net/cross_section_properties.html will provide a good start. As your part has a hole in it, you will need to separately calculate the extent of material that your hole removes from the solid body and subtract this from the first answer. This is also easy - http://www.wikihow.com/Calculate-Volume Admittedly, this will not provide you with a precise answer as the cylinder you are calculating will have linear faces, whereas your component will have a circular element at the extremes of the length limits, but it will still provide you with a sufficiently good result for you to base some first pass calculations on.

Then you should find out the specific elastic (Youngs) modulus of the material you want to make the component from - Google it, the Wikipedia page is quite good.

You then need to calculate the load placed on the component, and I don't think you have calculated that yet. Even a simple strain gauge like a fishermans weighing scale will provide you with useful info.

With these three basic pieces of information you can work out whether your part can withstand the forces that your design subjects it to. Not completely of course, but enough to know if it is possible, or it simply won't work or break within a month.

You really, really shouldn't bother with anything like a Patent (which is very expensive) until you have at least done some preliminary research like the type I've suggested above.

I wrote an article for this site about a month ago that included a paragraph regarding Patents - if you look at it, you'll see my viewpoint on them. In your case, I think your money can be spent far more profitably.

So, Steve, here's the challenge - take a week or two to work out the things I've said will be useful. It's very easy to say that your maths is bad, but that's probably only because you had a bad maths teacher - if you really want to do this, then get serious about it - there is a large void between an inventor and a designer.
If you're willing to try and attempt a explanation of the forces and stresses involved in your part, then I will be willing to help where and when I can, and I can promise you that if you put your explanation on this site, the only people that will attempt to ridicule you will be self important idiots who didn't know it themselves until a few months ago.

Everyone else will encourage you.

All the best,

K.I.S.S.

7. ### Auto EngineerActive Member

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Hi Steve,

Just a thought that might help you. A very long time ago when I was at University they had a machine that loaded wires etc and produced graphs of the force etc, if you take your prototype and ask a lab technician to load test it for you, that might save you a lot of headaches if the math is a problem?

8. ### K.I.S.S.Well-Known Member

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Hi Steve,

OK, I'm almost on holiday...
As Auto Engineer suggested, a automated strain gauge is also a very easy way of determining what you need to know with empirical testing (always the most reliable way of testing...).
I would add that in order to save costs, be sure that you inform the tester that you want to determine the point of nonelastic deformation, as opposed to the failure load. Most strain gauge machines can determine this.
But please don't give up on the idea of also trying to work it out for yourself - it's a very powerful tool that can help you in many ways.
And it's usually bad teachers that produce bad students, but these days the internet can provide you with all sorts of resources - it's really just a question of knowing where to look.

OK, I'm off to the airport. Good luck

All the best,

K.I.S.S.

Last edited: Nov 16, 2014
9. ### FL_STEVEMember

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Nov 2014
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KISS I hope your vacation was a good one, I cannot thank you all for your help and responses. I am so sorry for the delay in my response, but things have changed since I have posted. KISS I read through the articles you posted and as I expected they are way over my head and unfortunately my mathematics is not that good. What I was hopping for was a statement such as:

"If you know that T6 60XX has a strength of 45,000 PSI per square inch, then the smallest point which is 1/16" (.062) thick should be 2,813 PSI squared, take 45,000 divided by 16 which ='s +- 2,813 PSI on either side." something like that lol....

As far as AUTO's suggestion, first I have to check to see if I am allowed on the campuses, so I did something similar, I put 1 together, took a hold of 1 end with 1 hand and the same with the other and pulled as hard as I could, released, dismantled it then measured the rod, no changed in the length, I repeated approximately 8 times and it did not change the length. Well this is a good thing if it did not change from this I an confident I have a product that will do what will hold up, + I have already decided to construct it out of 7075T.

Unfortunately all is on hold no more \$\$\$ after prototype.... Long story short, my investor is on too many black label prescription drugs, I have known this guy since I was 19 and he was 18, I am over 50 today....

I guess it was not meant to be for me to make something out of my life.... Thank you all again and best wishes to all.....

Last edited: Jan 18, 2015
10. ### K.I.S.S.Well-Known Member

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FL_STEVE,
With all due respect, bollocks.
!.) If you have any serious belief behind your idea, find another investor. This doesn't appear to be a highly complex (and therefore inexpensive) design.
2.) The articles I've posted make no reference to mathematics or calculus - the last one was drafted as nothing more than an encouragement and a reality check for people such as yourself.
3.) Either you cut and pasted the 2nd paragraph, or you actually took the time to explore such avenues - so you're closer than you think...
4.) You actually did some basic empirical experimentation regarding elongation testing - feel free to ask more in this forum about general methods that will improve the accuracy of this...
5.) It's not only Universities that have the technology you require - it's also cheaply available on a commercial basis - again, Google will help you.
6.) Snap out of it.

All the best,

K.I.S.S.

Joined:
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