Question

How to calculate the second moment of area for an equilateral triangle ؟؟

Answer 1

about centroid? and in what direction?
Do you want to derive from first principles (by integration), or you need just the value?
See if the following can help if you just need the values:
http://www.geocities.ws/xpf51/MATHREF/TC_FORMULAS.html
Otherwise post again with your requirements.

Answer 2

Yes about centroid ,I need the derivation from first principles in the y axis direction ...

Answer 3

Thanks for the link but I need the derivation too..

Answer 4

As shown in the diagram, we have an equilateral triangle with
s=side length
h=triangle height=(sqrt(3)/2)s
y0=centroid distance from base = h/3 = (sqrt(3)/6)s

By symmetry, we will calculate (left) half of the triangle and double the results.
Using left extremity as the origin, the left slant side has equation
y=(2h/s)x=sqrt(3)x

By definition,
Ix=\(\large \int\int_A (y-y_0)^2dA\)

As shown in diagram, we will integrate with respect to y over a vertical strip, and then with respect to x.

Ix=\(\large 2\int_0^{s/2}\int_0^{sqrt(3)x}(y-y_0)^2~dy~dx\)
if you evaluate the double integral, you should get \(\frac{\sqrt{3}}{96}s^4\), exactly the same as what you would get using the formula \(\frac{bh^3}{36}\).
I will show a few intermediate steps to help you check your work.

= \(\large 2\int_0^{s/2} \int_0^{\sqrt 3 x} (y^2-2y_0y+y_0^2)~dy~dx\)
....
=\(\large 2\int(_0^{s/2}(\sqrt 3 x^3-3y_0x^2+\sqrt 3 y_0^2x)~dx\)
....
=\( \large [\frac{1}{32}-\frac{1}{24}+\frac{1}{48}]\sqrt 3 s^4\)

=\(\Large \frac{\sqrt 3}{96}s^4\)

Answer 5

Thanks a lot @mathmate ^_^

Answer 6

You're welcome! :)