Matrices and Tensors - Continuum Mechanics of Anisotropic Materials - page 377

In the special case of the area moment of inertia this formula becomes

I 0 ¼ I centroid þfð d d Þ 1 ð d d Þg A ; (A.140)

where the I are now the area moments of inertia and the mass of the object M O has
been replaced by the area of the thin plate A .

Example A.9.6

Consider again the rectangular prism of Example A.9.1. Determine the mass
moment of inertia tensor of that prism about its centroid or center of mass.

Solution: The desired result, the mass moment of inertia about the centroidal axes is
the I cm in (A.139) and the moment of inertia about the corner, I 0 , is the result
calculated in Example A.9.1,

2 3

b 2 c 2 4 ð þ Þ 3 ab 3 ac

I 0 ¼ r

abc
12

4 5 :

a 2 c 2 3 ab 4 ð þ Þ 3 bc

a 2 b 2

3 ac 3 bc 4 ð þ Þ

The formula (A.139) is then written in the form

I 0 fð I cm ¼ d d Þ 1 ð d d Þg M O ;

where M o ¼ r abs . The vector d is a vector from the centroid to the corner,

ð

1
2

d ¼ ae 1 þ be 2 þ ce 3 Þ:

Substituting I 0 and the formula for d into the equation for I above, it follows that the
mass moment of inertia of the rectangular prism relative to its centroid is given by

2 3

b 2 c 2 ð þ Þ 0 0

I cm ¼ r

abc
12

4 5 :

ð a 2 þ c 2 Þ

0 0 Æ

ð a 2 þ b 2 Þ

0 0

Example A.9.7

Consider again the thin right-triangular plate of Example A.9.4. Determine the area
moment of inertia tensor of that right-triangular plate about its centroid.

Solution: The desired result, the area moment of inertia about the centroidal axes is
the I Area

centroid in (A.140) and the moment of inertia, I 0 Area , about the corner is the result
calculated in Example A.9.4,

bh
24

2 h 2

bh

I 0 Area ¼

:

2 b 2

hb