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: Itrz(L,a) is the moment about the z axis of the equilateral triangle with side L, clipped by a at each corner (with a≤L/2). We will also use the same strategy, writing the moments of the complete shape as a combination of its four fragments:
Iz(L)=Itrz(L,a)+3[Iz(a)+m(a)d2],
where d=(L−a)/√3. Using the results obtained for the full triangle immediately yields:
Itrz(L,a)=√348μ[L4−3a4−12a2(L−a)2]
Similarly, from:
Iy(L)=Itry(L,a)+Iy(a)+2[Iy(a)+m(a)(L−a)2/4]
we get:
Itry(L,a)=√396μ[L4−3a4−12a2(L−a)2]=Itrz(L,a)/2
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, Iz=2Iy 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 Itry,z(2L,L)=Iy,z(L). In this case, one retrieves the situation shown in the illustration to the previous post.
No comments:
Post a Comment