Area moment of inertia of inverted triangle?

[LFS]

Homework Statement

A inverted isosceles triangle gate with height a=3ft and base b=2ft is 6ft under the water (top of the inverted triangle). Find the Force on the gate and hp (the depth of the) center of pressure.

Homework Equations

hc= depth to gate + depth to centroid= 6+(1/3)•3= 6+1 = 7
Force = γ hc A = 62.4 • 7 • (0.5•2•3) = 1310 lb
Area Moment of Inertia of Triangle = Ic =ba³/36
So hp = hc + Ic/(hc•A) = 7 + [2*3³/36/(7*0.5*2*3)] = 7.07 ft

The Attempt at a Solution

What I don't understand is why Ic does not depends on whether the triangle is inverted or not. I looked >10 sites and always the triangle is "base down" and ALL of the other figures are symmetric so I cannot compare. (Sorry cannot put in sample link from wikipedia.)

I am a mathematician trying to help my son (studying engineering). I cannot find an integral for this to test how the "area of moment of inertia" is calculated (and I don't really understand what it is as I do with the centroid).

Thanks for any help. Linda

 

[tiny-tim]

Hi Linda!

What I don't understand is why Ic does not depends on whether the triangle is inverted or not. I looked >10 sites and always the triangle is "base down" and ALL of the other figures are symmetric so I cannot compare. (Sorry cannot put in sample link from wikipedia.)

I don't really understand why you think it would depend on anything other than shape.

You can find information on the second moment of area (area moment of inertia) at http://en.wikipedia.org/wiki/Second_moment_of_area and http://en.wikipedia.org/wiki/List_of_area_moments_of_inertia

 

[LFS]

Hiya Tiny
Thank-you for your response. Those were exactly the links I was going to put in (but wasn't allowed).

The reason I ask is that the centroid of a triangle is 1/3 of the height measured FROM the base (i.e. from the weighted position of the triangle).
So the (depth of the) centroid is NOT the same for an inverted triangle.

Gate is inverted triangle => hc=6+(1/3)3=6+1=7
Gate is NOT inverted triangle => hc=6+(2/3)3=6+2=8

So why doesn't the second moment depend on the "weighted" position of the triangle.
I understand that I may be mixing apples and oranges, it just doesn't fit in my head.

Thanks again, Linda

 

[tiny-tim]

The reason I ask is that the centroid of a triangle is 1/3 of the height measured FROM the base (i.e. from the weighted position of the triangle).
So the (depth of the) centroid is NOT the same for an inverted triangle.

No, the base is any of the three sides of the triangle, and then the height is as measured perpendicularly from that side.

As I said, moment of area only depends on the shape.

 

[LFS]

Hiya TinyTim,
I am sure you are absolutely correct that the area moment of inertia Ic depends only on shape. However, "area moment of inertia" is just 4 words to me (no physical meaning). Meanwhile, I did find the integral formula for computing the center of pressure (Fox) and calculated it using both a flat bottom and inverted isosceles triangle and then using the "area moment of inertia". Of course you get different depths from the centroid, but in both cases the formula Ic=WL³/36 wrt the centroid is indeed valid. I am sure it is of no interest to anyone, but I wrote it down and published it in scribd: doc/93403575. So I am happy :). Thanks again!