# moments of inertia of a right-triangle

• posted

Hi,

When deriving Ix and Iy for a triangle, I know the equation to use is, say for Ix

Ix=int(y^2)dA

and that i can express y in terms of x so y=f(x),

but what do I need to change dA into to make it work?

Any help would be greatly appreciated.

• posted

dA ~= dx*y

David A. Smith

• posted

So if there was right triangle who's hypotenuse passes through the origin of a cartesian plot and the point (b,h):

y=(h/b)x

so dA=ydx=(h/b)x dx

Now:

Ix=int(y^2)dA = int(h^2 / b^2) (h / b) x dx = int (h^3 / b^3) x dx =(h^3 x^2) / (2 b^3)

where x is from 0 to b so:

Ix=(h^3b^2) / (2b^3) = h^3 / 2b ?

Oh man, Im wrong again!

could you please point out where I slipped up?

• posted

Whoops, I've got it i think:

y=(h/b)x

dA=ydx

Ix=int(y^2)dA

=> Ix=int(y^3)dx

Ix=int [ (h^3 / b^3) x^3 ] dx

Ix= h^3 x^4 / 4(b^3)

• posted

My lecture notes say:

for a triangle with base=b and height=h,

Ix = b h^3 / 12

Iy= b^3 h / 4

Ixy = b^2 h^2 /8

I really don't understand how Ix and Iy can be so different. Im really worried about this so is anyone can help i would be very grateful.

• posted

Also, if the x centroid location is 2b/3 and the y centroid location is h/3, the area is hb/2

why cant I just multiply the area by the square of the centroid's distance from the origin (I=md^2) ???

• posted

Adam: Here's a tip. Unless you are making additional provisions using parallel axis theorem or double integration, always keep your infinitesimal strip parallel to the axis about which you are computing moment of inertia. Try it again using this concept and see if you now get the correct answer.

Because you can't compute moment of inertia, all at once, for a large area (one that has significant height perpendicular to the axis about which you are calculating moment of inertia) unless you use parallel axis theorem.

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