# Second moments of area of a simply supported regular I beam

1. May 28, 2012

### Steve Collins

I’m having trouble getting my head around this example and solution that was given as part of a revision pack for an upcoming exam. Any explanation would be gratefully received.

Question:

A simply supported regular I beam is 90mm wide and 120mm high. The top and bottom flanges are 10mm thick, as is the web. It is 6m long and carries a UDL of 6KN/m over the length of its top flange.
Determine the second moment of area Ixg about the horizontal axis which passes through the centroidal axis of the beam.

Equations:

Second moment of area of a rectangle about an axis through its centroidal axis, Igg= (bd^3)/12

Second moment of area of a rectangle about its base, Izz= (bd^3)/3

Parallel axes theorem, Iaa= Ig+A(y)^2 (I’m not sure how to put the – over the y using word)

Solution:

Total area A= 4x40x10+120x10= 2800mm^2 … easy enough

Centroid y = [(2x10x40x5)+(2x10x40x115)+(120x10x60)]/2800 = 60mm

Again this is easy enough and I can see the logic, but this is where I am struggling:

From second moment of area of a rectangle about its base, Izz= (bd^3)/3 and parallel axes theorem Iaa= Ig+A(y)^2.

Ixx= 1/3x40x10^3x2 + 1/3x10x120^3 + 2x [1/12 x 40x10^3 + 40x10x115^2] = 16.37x10^6 mm^4

An explanation of the different sections of the line above would help me a lot. I’m confused with the use of Igg= (bd^3)/12 and Izz= (bd^3)/3.

Ig= 16.37x10^6 – 2800 x 60^2 = 6.29x10^6 mm^4… so Ixx is in fact Iaa?

Thanks for taking the time to read this and I hope that question is clear.

Cheers Steve.

2. May 28, 2012

### PhanthomJay

When using the parallel axis theorem for determining the 2nd moment of area with respect to the axis passing horizontally through its cg, use the I of the rectangles about their own cg's. These are all of the form bh^3/12. To this number, you must add the products of the individual rectangle areas times their distance^2 from their own cg to the member cg

3. May 30, 2012