MOMENT OF INERTIA OF
COMPOSITE BODIES
MOMENT OF INERTIA
Measure of resistance to angular acceleration;
 The summation of the product of each area by the
square of its moment arm or sometimes called the
second moment of area.

I X = I X O + Ad 2
I X O = moment of inertia at the centroid of the area
d = y = distance of centroid of area from any parallel axis
I X = transfer moment of inertia at the parallel axis
x’ or 𝑥𝑜 = centroidal axis
x = any axis parallel to the centroidal axis.
MOMENT OF INERTIA OF SIMPLE GEOMETRIC AREAS
𝑥′
I𝑥′ =
b h3
3
Ixo = 0.11 r 4
xo
Problem #1: Determine the moment of inertia of the T-section
with respect to its centroidal xo axis.
Problem #1: Determine the moment of inertia of the T-section with respect
to its centroidal xo axis.
AT =
A1 + A2 = 16 in2 + 16 in2
𝐀𝐓 = 𝟑𝟐 𝐢𝐧𝟐
2 in
y1 =
= 𝟏 𝐢𝐧
2
8 in
y2 = 2 in +
= 𝟔 𝐢𝐧
2
𝐀𝟐
2.5"
2.5"
𝐀𝟏
AT ȳ = A1 y1 + A2 y2
6"
= 𝟑. 𝟓"
1"
32 in2 (ȳ) = 16 in2 (1 in) + 16 in2 (6 in)
ȳ = 3.5 in
𝐈 𝐱−𝐀𝟏
I X o = I X + Ad 2
IXo =
8 in 2 in 3
12
A2 = 2 in x 8 in = 𝟏𝟔 𝐢𝐧𝟐
𝐝𝟏 𝟐
+ 16 in2 (3.5 in − 1 in)2
𝐈 𝐱−𝐀𝟐
+
A1 = 2 in x 8 in = 𝟏𝟔 𝐢𝐧𝟐
𝐀𝟏
2 in 8 in 3
12
𝐀𝟐
𝐝𝟐 𝟐
+ 16 in2 (6 in − 3.5 in)2
𝐈Xo = 𝟐𝟗𝟎. 𝟔𝟕 𝐢𝐧𝟒
Problem #2: Find the moment of inertia about the indicated X
axis for the shaded area shown.


Problem #2: Find the moment of inertia about the indicated X axis for
the shaded area shown.
𝐈𝐗 𝐨 −𝐀𝟏
4r
= 1.70"
3π
Ix =
+ 80 in2 ( 5 in)2
− 0.11 4 in
8.30"
𝐀𝟏
8 in 10 in 3
12
𝐈𝐗 𝐨 −𝐀𝟐
𝐀𝟐
𝐝𝟏 𝟐
𝐀𝟏
4
𝐝𝟐 𝟐
𝐀𝟐
− 25.133 in2 (10 in −
𝐈𝐱 = 𝟗𝟎𝟔. 𝟏𝟑 𝐢𝐧𝟒
5"
𝐀𝐧𝐨𝐭𝐡𝐞𝐫 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
𝐈𝐱
I X = I X O + Ad 2
𝐈𝐗 𝐨 −𝐀𝟐
Ix = 8 in 10 in
3𝐀
3
𝟐
𝐝𝟐 𝟐
− 25.133 in2 (10 in −
A1 = 10 in x 8 in = 𝟖𝟎 𝐢𝐧𝟐
π(4 in)2
πr 2
=
A2 =
= 𝟐𝟓. 𝟏𝟑𝟑 𝐢𝐧𝟐
2
2
− 0.11 4 in
𝐈𝐱 = 𝟗𝟎𝟔. 𝟏𝟑 𝐢𝐧𝟒
4
4 4 in
3π
)2
4 4 in
3π
)2