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

1. Dec 2, 2008

### gnarkil

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.

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

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?

**************************
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

#### Attached Files:

• ###### tbeam.JPG
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Last edited: Dec 3, 2008
2. Dec 3, 2008

### PhanthomJay

Hah, Wiki will confuse the best of us!

It shouldn't make any difference on your moment of inertia calcs as to whether the beam is upside down or right side up, as long as you are calculating same about the neutral axes of the section.

I am not sure where your getting your numbers for the neutal axes calculations, about which I assume you desire to calculate the moments of inertia.

To calculate the neutral axis, you needn't do any calculation for the yy axis; since the beam is symetrical, it must lie along what you have shown as a dashed line. For the neutral xx axis determination, just sum moment areas about the base. The section consists of 2 rectangles; Find the area of each; then the area of the first times the distance from its centrioid to the base, plus the area of the second times its distance from its centroid to the base, is equal to the total area of both rectangles times its unknown distance to the base. Solve for this unknown distance. Once you correctly note the position of the xx neutral axis, then you can use the parallel axis theorem to calculate Ixx . You do not seem to be doing this properly. For the Iyy calculcation, for example, you don't need the parallel axis theorem; Iyy is just c(2a)^3/12 + (b)d^3/12

3. Dec 3, 2008

### gnarkil

thanks for that, i figured out the y_c to be 75mm.

if for instance the t beam was reflected about the x axis so that it was right side up (looks like a T), would the Ixx be calculated using the Iyy calculations you explained above? And then I would use the parallel axis theorem to find Ixx?

i'm still a little confused with solving Ixx using the parallel axis theorem

convert mm to m

Ixx = bh^3/12 + 0.015(0.075^2) = (0.15)((0.15 + 0.05)^3)/12 + 0.015(0.075^2) = 0.0013m^4

where b is base, h is height, A is area, d is = y_c

doesn't seem correct...

4. Dec 3, 2008

### PhanthomJay

good, that's 75mm up from the bottom
no, everything will be the same, except the neutral xx axis will be located 75 mm from the top. Just turn your paper or your computer upside down.
Yeah, your getting confused on the parallel axis theorem. You've got to break up your t-section into 2 rectangles, the same way you did for calculating the neurtal axis. Then calculate Ixx for the first rectangle about its centroid , add Ad^2 to it, where A is its area and d is the distance from its centroid to the neutral axis, (d=50mm), and add the Ixx for the lower rectangle about its centroid, and add A d^2 to it, where A is the area of the lower rectangle, and d, the distance from its centroid to the neutral axis, works out coincidentally also to be 50mm.