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Machinery’s Handbook #1: Pushrod/Column Buckling

December 14, 2011
Ellipse Inertia

Being on Christmas Break from Engineering school I dug through some of the stuff I have stashed at other peoples houses and found my copy of the Machinery’s Handbook. I opened to a random page and quickly recognized the equation for column buckling and realised the book must be full of applications to race car technology. So this will be the first of many Machinery’s Handbook posts, where I take something from the book and apply it to race car design.

I will be referencing page numbers, so to specify, I am using this edition:

Flipping to page 285 brings you to the Columns section. Obviously the thickness of F1 pushrods are much less than the lengths, this ratio is the Slenderness Ratio.

The equation I’ll be using is on page 286:

P_{cr} = \frac{\pi^{2} I E}{l^{2}}

Example

Above image is from formula1techandart. The image links to the site.

Let’s make some assumptions:

  • Failure will be by buckling, not compression
  • Ignore for now Anistropic (directional) properties
  • Assume a hollowed elliptical cross section
  • Ellipse 20mm X 50mm
  • 875mm length
  • End conditions will be “round round” or “pinned pinned”
  • 125GPa Young’s Modulus for Carbon Fiber
  • Ellipse thickness of 5mm

First, lets find the value for I, Moment of Inertia. Flipping to page 242 a hollow ellipse is found. Were there no hollow ellipse in the tables, a solid ellipse could have been subtracted from a larger solid ellipse.

I = \frac{\pi}{4} (a^{3} b - c^{3} d)

However for column buckling, we want to consider the weakest Moment of Inertia. The weakest moment of inertia is found in this case by swapping a for b, and c for d. The dimensions in the way they are to be considered for this application are drawn this way and are correct to the equations.

I = \frac{\pi}{4} (.01^{3} .025 - .005^{3} .02) = 1.767*10^{-8} [m^{4}]

Compare this to 3E-8 for a hollow rectangular section of the same dimensions. It makes sense that the ellipse has about half the moment of inertia of the rectangle.

P_{cr} = \frac{\pi^{2} 1.767*10^{-8} 125*10^{9}}{.875^{2}} = 28475[N]

Based on these many simplifying assumptions the pushrod will fail at under 30,000 Newtons. In reality, materials can also fail due to compression, not just buckling. Also, carbon fiber is not isotropic but anistropic, which I will cover in a later example. I have no idea what the dimensions of an F1 car pushrod are, and I’d care to bet the section is close to an ellipse, but actually more streamlined. Due to the different shape, the equation for finding the moment of inertia of the area would change.

Just for a means of comparison, let’s imagine adding a rib down the center of the ellipse in order to stiffen it and increase the moment of inertia. Let the section be 5mm thick like before

I_{add} = \frac{1}{12} (.02^{3} .005) = 3.333*10^{-9} [m^{4}]

Adding this inertia to the one before gets a new critical load of 33,846 Newtons of force. However, if you look at what I just did you’ll see that I am adding too much inertia to the cross section. In reality, the rib can only be .01m long since the outher thickness of the ellipse is .02m and the thickness is .005m leaving only .01m of cross sectional height for the rib. After making what seems to be that small change, the new critical load is 29,147 Newtons.

This small discrepancy changed the increased load carrying ability from 672 Newtons to the incorrect 5371 Newtons with the impossibly long rib. This is a factor of 8, which just so happens to be 2^3. Double the length, to the third power, creates an error of 8 times.

Given the big difference in answers given quick assumptions it will be interesting to see what happens when different carbon fiber orientations and mass ratios will do to the results of this example.

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