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The second moment of area of a rectangle is $I_{yy} = \dfrac{bh^2}{12}$

b : base h : height

What is the second moment of area of a T beam like this one?

beam section

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  • $\begingroup$ Hint, find the centroid of the 'T' beam, use the relationship between the centroid and second moment of inertia. $\endgroup$ – Sam Farjamirad Nov 21 '18 at 18:55
  • $\begingroup$ This looks like a homework question. In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. What have you tried to solve this yourself? Please edit your question to include this information. $\endgroup$ – Wasabi Nov 21 '18 at 19:42
  • $\begingroup$ I think the question with the attached graphic is clear enough and has potentially educational value. $\endgroup$ – kamran Nov 21 '18 at 20:04
  • $\begingroup$ First you need to find the first moment of area to discover Iyy before you can apply it to second moment of area. $\endgroup$ – Rhodie Nov 23 '18 at 1:16
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First let's find the centroid of the shape:

Assuming each shading has its own different modulus of elasticity E1,E2,.. and just for the sake of this example let's say the ratio of E's are as follows.

E1 for the top flange = 1

E2 stem =1.2

E3 trapezoid bases = 1.3

We set the base as Y axis and find the centroid.

the trapezoid moment = $ 1.3 ((b_{ ta} - b_a )(H_2/2 )+((b_{ ta} -b_a)(h_1/2)(h_1/3 + h_2/2)$

The stem moment + $1.2( b_a\times h_p/2) $

the flange moment =$ b_0 h_d (h_p-h_d/2) $

And the centroid is sum of these area moments divided by sum of weighted areas.

$\overline{Y} =\frac {\sum\ area\ moments} {\sum\ weighted \ areas} $

And I of the beam is the I of individual part plus parallel axis moment of inertia.

$ I= ( \sum A_n\times E_n\times Y_n^2 ) +\sum I_n $

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  • $\begingroup$ First you need to find the first moment of area to discover Iyy before you can apply it to second moment of area. $\endgroup$ – Rhodie Nov 23 '18 at 1:15

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