# Calculating Stress (AISC)

1. Dec 16, 2008

### skaboy607

I have been carrying out a lab experiment on the buckling of axially loaded struts and for given set of data and various rod lengths, have to calculate the failure stress, Euler Stress, Rankine Failure Stress, Perry Robertson equation stress. I then plot these against the slenderness ratio. I have done this however the last section says:

'The AISC rules state for L/K values less that (L/K)c, where (L/K)c = sqrt (2pi^2E/sigma Y), plot the parabolic curve: sigma=sigmaY(1-((L/K)^2/2(L/K)c^2). Smoothly join this parabola into the euler curve for L/K values greater than (L/K)c.

I have taken E as 210Mpa, sigmaY as 600Mpa.

The question doesnt even make sense to me and then when I try and work out (L/K)c, I get an asnwer of 2.6 which is less than the lowest slenderness ration I have of wo which is for 20mm rod.

Any help is much appreciated.

Will

2. Dec 17, 2008

### mathmate

3. Dec 19, 2008

### skaboy607

I have got a value for (L/K)c and have 4 struts whose slenderness ration is less than it. I have calculated their stress using the parabolic equation: sigma=sigmaY(1-((L/K)^2/2(L/K)c^2).

How do I create a parabola from this?

Will

4. Dec 19, 2008

### mathmate

The formula
$$\sigma = \sigma_{c}(1-(L/K)^{2}/(L/K)_{c}^{2}$$)
is a parabola!

Also, you must have been misled to the value of E.
The elastic modulus of steel is of the order of 207000MPa, while the yield strength could be 600Mpa for high tensile steel.
If you are working with H.S. steel, then
Cc=$$\sqrt{(2\pi^{2}E/\sigma_{y})}$$
=82.52

Last edited: Dec 19, 2008
5. Dec 19, 2008

### skaboy607

Ok, forgive me but I am not to sure what to do with it now?

I have researched parabola's and they all seem to be in equation formats that don't look like this one?

I need to fit the parabola curve to the curve that I have for Euler Stress on the graph. I have actually been using my value for E as 210 Gpa taken from here and also my lecturer;

Is this not correct?

I get my value of (L/k)c = 83

Thanks for you help.

Will

6. Dec 19, 2008

### mathmate

If you plot $$\sigma$$ against l/r, you will be plotting the inelastic buckling load, since buckling occurs when the column reaches its yield stress, $$\sigma_{y}$$. This is true up to the critical l/r ratio, (l/r)c.
At this point, the column is considered long, and buckling will occur before the yield stress is reached. From this point on, Euler's buckling formula takes over, and increasing the yield strength of the column will not affect the buckling load.
Thus the complete column buckling load / stress is depicted by two functions, sigma(x) and Euler(x), where x represents the l/r ratio. The former applies for l/r from 0 to 83, and the latter applies for x(=l/r) = 83 and higher. The transition between the two curves is smooth, since by substituting the critical l/r into the parabola (sigma(x)), we get exactly $$\sigma_{y}$$/2 for both formulas.
Here is a plot of both curves.

Note that the transition is at l/r (=x) of 83, and 600 on the y-axis represent the yield stress of 600 Mpa.

7. Dec 20, 2008

### nvn

skaboy607: In your above posts, K = radius of gyration, sometimes called r. I will use the nomenclature sr = slenderness ratio = Le/K, and src = critical slenderness ratio = pi*(2*E/sigmaY)^0.5, where Le = column effective length, and K = radius of gyration. Therefore, just plot sigmac(sr) = sigmaY*[1 - 0.5(sr/src)^2], from sr = 0 to src, where sigmac(sr) is the critical (buckling) stress as a function of sr. In your case, src = pi*[2(210 000 MPa)/(600 MPa)]^0.5 = 83.12.

For sr > src, plot sigmac(sr) = E*(pi/sr)^2, the Euler critical (buckling) stress.

For a solid round rod of diameter d = 20 mm, K = (I/A)^0.5 = 5.000 mm.

8. Dec 22, 2008

### bortonj88

Alright Couchy. i sorted it out. ill send you it if you want it. turns out like a treat!! :D:D:D:D

And thanks to everyone in this thread, all replies really helpful. :)

Also mathmate, what program were using to draw those graphs, i think it may come in handy?

John

Last edited: Dec 22, 2008
9. Dec 22, 2008

### skaboy607

Yo there Jonny boy-of course I would like to see it! Crack it on the old email!

Yea thanks to all your help guys.

Will

10. Dec 22, 2008

### mathmate

11. Dec 23, 2008

### bortonj88

thanks mathmate.

I was wondering if anyone could shine some light of a problem myself.....
why use all the different methods of stress analysis, standard failure stress, Euler Stress, Rankine Failure Stress, Perry Robertson equation stress, and the AISC rules??

Thanks
John

12. Dec 23, 2008

### mathmate

This is indeed a question that may have different answers from different point of view.
From mine, the major reasons why we do stress analysis at all are for safety and economy.

Safety is important if we would like to be sure that a bridge or a building does not collapse under loads it is expected to sustain during its lifetime. Assuming the engineer/architect knows the rudimentary rules of stress analysis, most of the time he can ensure safety by simply over-designing.

However, overdesigning leads to the second aspect, economy is compromised. Competing engineers may start to boast how his design is more economical, and hence more attractive to the investor.

Indeed, the two aspects are in conflict with each other, and society has reacted by enacting codes which dictate the minimum requirements for safety, such as AISC for structural steel, ACI for concrete, etc. By following the code, an engineer can now optimize his everyday design with respect to cost without compromising safety.

Who decides what's in the code, what are the rules? After all , the code is based on researches on the subject in question, and the subject-matter experts submit papers, discussions, and sometimes debates to decide on the content of the code, which is a distillation of the state-of-the-art knowledge, accumulated over the years by the prominent scientists, researchers, engineers, etc. However, the code covers only the more general cases.

So why does the ordinary engineer need to know about all the other complicated formulae, equations, and so on? One of the engineer's major tasks is to ensure that the product works as designed. He has to spot the exceptions not covered by the code, identify unexpected circumstances, and understand how the product behaves in different situations, and predict performances. All these other stress analysis methods enable him to do this part of the job.

So, all these different stress analysis methods, equations, procedures, are actually the tools in an engineer's toolbox. An engineer's expertise is often measured by how well he masters these tools to the advantage of his clients, in ensuring product performance, in achieving extra economy without compromising quality.

13. Jan 6, 2009

### skaboy607

If your still about to help mathmate, I have drawn the 2 curves on a graph in excel but my euler curve isnt as smooth as my AISC curve so there is no 1 point where they cross, have I done something wrong? My graph is attached.

Thanks

Will

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14. Jan 6, 2009

### mathmate

You could send the graph to some image server and post the link. It would be faster than to wait for the approval of the attachment as it requires human intervention to assure the quality of this site.
Does it look pretty much like the one shown in post #6 above?
If they don't cross, there is probably a little problem, as they are supposed to be tangent to each other.

Last edited: Jan 6, 2009
15. Jan 7, 2009

### skaboy607

16. Jan 7, 2009

### mathmate

It doesn't look right. The transition point should be tangential to each of the two curves, and should not cross. Could you check the formula, especially that of the blue curve?
Do you come up with the same formula as mine, which is indicated on the top right of the graph? The constants may be different, but the formula should be similar.

17. Jan 8, 2009

### skaboy607

18. Jan 8, 2009

### mathmate

The graph looks right, but I'd like you to calculate, if possible, the tangent point to make sure that they don't cross. I seem to see that they cross some how, but it could be an optical illusion.
Excellent work, I am glad that you cracked it!

19. Jan 13, 2009

### skaboy607

Hi mathmate

I haven't had a chance to calculate the tangents yet but was wondering if you had an answer to another question of mine. In this lab, we have values for stress and strain, but when using these values to calculate youngs modulus, it is different to the value I have used of 210Gpa? was wondering why this is?

Thanks

Will

20. Jan 13, 2009

### mathmate

The value of 210Gpa is dependent on the material you use, namely the type of steel. If it is in the range between 200 and 210, it should be normal. If it is very different, you may want to check your data, perhaps you have loaded the sample beyond the yield point, where it will be no longer elastic. Perhaps also you want to check the grade of steel you used.

What is the value that you have obtained, and how did you calculate it?