(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

this is really getting confusing, i need to find the second moment of inertia Ixx and Iyy of the upside down beam, please see attachment

i've used and studied the other threads relating to the same topic and used wikipedia for the parallel axis theorem but am still confused, an explanation about the approach would be appreciated.

https://www.physicsforums.com/showthread.php?t=129591

http://en.wikipedia.org/wiki/Parallel_Axis_Theorem

http://en.wikipedia.org/wiki/Second_moment_of_area

https://www.physicsforums.com/showthread.php?t=88567

diagram dimensions:

a = 150/2 mm = 75mm

b = 150mm

c = d = 50mm

assume x axis is horizontal and y axis is vertical

assume origin is at the a distance a (150/2 = 75mm) from rhs and a distance c + b (150 + 50 = 200 mm) from top

2. Relevant equations

I used a textbook example, but the t beam was the opposite to the one i need

y_c neutral axis =

[ (b/2)(b2a) + (b + c/2)(2dc) ] / [ (b2a) + (c2d) ]

Ixx =

(1/12)(2a)(b^3) + (2ab)(y_c - b/2)^2 + (1/12)(2d)(c^3) + (c2d)(b + c/2 -y_c)^2

Iyy =

(1/12)(b)(2a)^3 + (1/12)(c)(2d)^3

3. The attempt at a solution

a = 150/2 mm = 75mm

b = 150mm

c = d = 50mm

using the above dimensions i got these values

y_c = 93.18 mm from bottom

Ixx = 841.5*10^5 mm^4

Iyy = 463.5*10^5 mm^4

i know they are wrong, but don't know why.

is the calculation for the second moment of inertia for a t beam with the same dimensions different to that of the identical beam that is upside down? why? would it be the same if i switched the base dimensions?

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update

i recaculated by y_c and got 50mm from bottom, and Ixx = 5.18125*10^-4 m^4

i still can't get Iyy

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# Homework Help: Second moment of inertia of upside down t beam

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