# Homework Help: Euler's Buckling Load Formula

1. Jul 3, 2009

### Aerandirel

For an Intro engineering class we've working on a project to make bridges out of spaghetti. To determine if the different spaghetti components of my bridge are strong enough to hold the force that's being applied on them, Ive been trying to calculate the spaghetti's critical buckling load for the components in the bridge under compression. Im getting real whacky numbers when Ive been trying to calculate these forces. They arent matching up any where close to the compression strengths Ive measured in the lab.

Alright so the equation I need to solve this is F = (pi^2)*(E)*(I)/(L^2), or Euler's buckling Load formula. E = young's modulus of elasticity, I = cross section moment of inertia, which Ive been told is = pi*(R^4)4 (R is radius). E for spaghetti is somewhere around 5 gigaPascals.

Anyways, picking out one example from the data Ive collected I have a 10 cm long piece of past with a radius of .1 cm. The amount of force recorded when the spaghetti started to buckle during testing was 2.25 newtons, so I should get around this force when I use Euler's buckling formula right?

Attempt
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Ok so converting all measurement to meters I plug in F = [ (pi^2)*5*(pi*(.001m^4)/4) ] / .1 cm ^2. I work this out and get F = 3.87*10^-9 newtons....what am I doing wrong?

2. Jul 3, 2009

### PhanthomJay

Check your math and units. A GigaPaschal is (10)^9 Paschals.

3. Jul 3, 2009

### Aerandirel

Yes Ive checked a million times. Are you saying I should plug in the 5 GPA Youngs Modulus value into the equation as 5*10^9 instead of just 5? Ive done that too, and I still dont get the right Force values. Can anyone verify that im using the correct equation for this? Ive seen slightly different ones all over the internet.

Again Im trying to calculate Critical Buckling Load for a piece of dry spaghetti in compression. And btw, this is more of a self pursued experiment then one I was assigned, so this isnt something straight of my homework or anything, im jsut trying to get a better understanding of how my bridge will hold up.

Edit: Ok during testing I recorded a force of .9 Newtons in order to make the piece of spaghetti slightly buckle. By the Critical buckling formula I get a force of 3.5 Newtons. Is this reasonable for a max compression force for a 10 cm piece of dry spaghetti?

Last edited: Jul 3, 2009
4. Jul 3, 2009

### PhanthomJay

The buckling formula you stated is for an ideal column pinned at both ends, that is, free to rotate at the ends mut not free to translate. How was your test set up? You seem to have got two completely different values in your test. If in the test one end was fixed and the other free to rotate and translate.,the critical buckling load decreases by a factor of 4. (pi squared EI/(KL)^2 is the theoretical formula, where K=1 fior pinned ends and K=2 for the latter cantilever case). How good is your value for E??

5. Jul 3, 2009

### nvn

Aerandirel: Your computed value is still slightly incorrect, but close. Why did you say the measured buckling load was 2.25 N in post 1, but 0.9 N in post 3? Which is correct? Ensure you are measuring the spaghetti radius very accurately.

Most importantly, measure the straightness of your spaghetti. I.e., prior to loading, measure exactly the gap from the spaghetti centerline at midspan, to a straight line running from one end to the other end of the spaghetti; and let us know this value. This distance is called the eccentricity.

6. Jul 3, 2009

### Aerandirel

Sorry the 2.25 Newton value is correct, the .9 one is not. Im have a spreadsheet of data with forces for 25 different lengths and 4 different types of spghetti so it gets a little confusing.

Basically we went into the lab and tested the buckling load of 4 different types of spaghetti (4 different radii), and for each one recorded the buckling load and how it would change as you went from a 24 cm length to a 2 cm length (so students could see how the force exponentially increased as length went down.)

The test involved taking a piece of spaghetti, pressing one end down on a digital scale (where there was a piece of cloth taped down on the scale so the friction between the cloth and spaghetti would keep it from moving), and you would press down on the other end with your finger, and record the force on the scale at the point that the spaghetti would start to bend or buckle. So I think this means it was free to rotate but not free to translate, it was free to pivot, (the spaghetti end would not move from the point it was on), giving me a k=1 value, which is why I left it out of the equation, leaving L^2 in the denominator.

We derived a value for E out of own testing by finding the slope of the stress/strain in another test. Although hard to find, most sources on the internet quote the E for spaghetti being around 4-5 GPa. The value we derived from our tests was almost exactly 4 GPa, (this was an average, as the slope seemed to change based on the length of spaghetti - which I know was probably due to human error, as the E should be a constant in a material correct?)

Last edited: Jul 3, 2009
7. Jul 4, 2009

### nvn

Aerandirel: OK, good. Here are some things you should check to try to make your computed buckling load more accurate.

(1) Are you sure the spaghetti was exactly 100 mm in length for the 2.25 newton buckling load? Being off on the length by only 1 mm would cause a 2 % error in your computed buckling load.

(2) Can you quantify the eccentricity (mentioned in post 5) for the typical 100 mm spaghetti, which started bending at 2.25 N?

(3) Are you sure the spaghetti diameter is exactly 2.00 mm? And are you sure the cross section is a perfect circle? Or is it an ellipse? Let us know the exact dimensions. You need to measure the diameter extremely accurately, to at least the nearest 0.05 mm. Being off on the diameter by only 0.1 mm would cause a 21 % error in your computed results.

(4) Use the average modulus of elasticity. Why are you using 5 gigapascals, instead of 4? The average value (or less) will generally get you closer to the observed buckling load.

(5) It would be helpful if you determine the spaghetti compressive yield strength. Look at your stress-strain plots and determine the stress at which your stress-strain plots become nonlinear; and let us know this stress value.

8. Jul 5, 2009

### Aerandirel

1. It was very close to 100mm but seeing as we were using a standard ruler for measurememnt it could have been off by 1 mm, but this probably isnt what is causing the huge difference between what I recorded and what I calculated.

2. At the time of testing we didnt measure this. It was the standard straight spaghetti if that means anything but as far as the measure of its eccentricity goes I cant tell you.

3. 10 spaghetti sticks were measured side by side with a standard ruler, and that length was divided by 10 to find the diameter of the spaghetti. Again a standard ruler was used, so while we were trying to be very accurate in our measurememnts we could have been off a little, but at the most it would have been .1 mm, no more than that.

4. Alright I used the 4 gigapascals value but I still get a force of around 3.25 N...

5. Yea I was kind of confused on exactly how we found the average modulus of elasticity...Individually we never really got to the conclusion ourselves, I think Ill have to schedule time with my instructor to make sure I get this all understood correctly, thanks for everyones help though, atleast im getting a value thats in the single GPa now instead of in the thousands, so Ive gotten a lot closer to wear I can estimate the rough force it should hold, ill review with my instructor so I can better understand the stress vs strain plots and then go back to the possible sources of error NVN raised. Thanks all.

9. Jul 6, 2009

### nvn

Aerandirel: (4) Why do you say you got a force of 3.25 N? E*0.25(pi^3)(r^4)/L^2 = (4000 MPa)*0.25(pi^3)[(1.00 mm)^4]/(100 mm)^2 = 3.10 N, not 3.25. Double-check all your calculations.

(3) I think it would be very easy to overestimate the spaghetti diameter by at least 0.04 mm per strand, using your measurement method. And unless you are extremely careful, I think you could be overestimating the diameter by 0.1 mm per strand. Therefore, assuming the diameter is 1.90 mm, this would get you down to a computed buckling load of 4000*0.25(pi^3)(0.95^4)/100^2 = 2.53 N.

Use a variety of measurement methods and instruments to check and double-check your diameter measurements. Also, test your force measuring device, to test its accuracy and to calibrate it. E.g., weigh a known volume of water (density, rho = 1000 kg/m^3 = 1 g/cm^3 = 1 g/mL). Measure the volume of water with a very accurate graduated cylinder that measures milliliters. If you do not have a highly accurate graduated cylinder, you can measure the inside dimensions (mm) of a straight, cylindrical or rectangular container, and convert mm^3 to mL (by multiplying mm^3 by 0.001 mL/mm^3). Unless you independently test your testing equipment, then you have no idea if it is even close to correct.

(6) Doesn't your spaghetti immediately start bending even when you apply only a very small force? Therefore, how or when do you define it as "buckled"? I.e., what is your criterion during the experiment to say, "This specimen has now buckled"?