23 August 2014

Moment of inertia of a clipped triangle

After calculating the moment of inertia for an equilateral triangle, let us consider the same shape, but with clipped corners, as in the drawing below:

We will preserve the notations of the previous post, adding the superscript "tr" for the truncated shape: \(I_{z}^{\text{tr}} (L,a)\) is the moment about the \(z\) axis of the equilateral triangle with side \(L\), clipped by \(a\) at each corner (with \(a \leq L/2\)). We will also use the same strategy, writing the moments of the complete shape as a combination of its four fragments:
\[I_{z}(L) = I_{z}^{\text{tr}} (L,a) + 3[I_{z}(a)+m(a)d^2] ,\]
where \(d=(L-a)/\sqrt{3} .\) Using the results obtained for the full triangle immediately yields:
\[ I_{z}^{\text{tr}} (L,a) = \frac{\sqrt{3}}{48} \mu \left [ L^4 - 3a^4 - 12 a^2 (L-a)^2\right ] \tag{3}\]
Similarly, from:
\[I_{y}(L) = I_{y}^{\text{tr}} (L,a)+I_{y}(a) + 2[I_{y}(a)+m(a)(L-a)^2/4]\]
we get:
 \[ I_{y}^{\text{tr}} (L,a) = \frac{\sqrt{3}}{96} \mu \left [ L^4 - 3a^4 - 12 a^2 (L-a)^2\right ] = I_{z}^{\text{tr}} (L,a)/2 \tag{4}\]
The clipped shape preserves the threefold symmetry of the original one, so the same conclusion as to the in-plane isotropy of the inertia tensor holds. Also, \( I_{z} = 2  I_{y}\) in both cases; I'm sure there is some elegant way to explain this, but I can't find it.
A quick check of results (3) and (4) is that \( I_{y,z}^{\text{tr}} (2L,L) = I_{y,z}(L) .\) In this case, one retrieves the situation shown in the illustration to the previous post.

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