1. Lecture 5 – Moment of Inertia of Non-symmetric Shapes
In general, most cross-sectional shapes of structural members are symmetric
(i.e., mirror image on both sides of both neutral axes). The determination of
section properties for these symmetric shapes involves plugging in numbers into
the formulas as discussed in Lecture 4.
Non-symmetric shapes are those that are NOT mirror image on both sides of one
or both of the neutral axes. Examples of non-symmetric shapes are as follows:
Angles
T-Sections
Channels
Sections w/ unsymmetric
holes
The procedure for determining the moment of inertia of these non-symmetric
shapes involves two steps:
1) Determine location of centroid of entire shape, y and x
y=
Σ( A piece y piece )
ΣA piece
where: Apiece = area of the individual piece
ypiece = dist. to centroid of the individual piece
x=
Σ( A piece x piece )
ΣA piece
2) Determine the transformed moment of inertia, It by using the “Parallel
Axis Theorem”
I t = ∑ ( I + Ad 2 ) piece
where: I = moment of inertia of the piece
A = area of the piece
d = y – ypiece
Lecture 5 - Page 1 of 8

2. Example 1
GIVEN: A “T” shaped beam using a nominal wood 2x8 web with a 2x4 top flange
REQUIRED: Determine
a) The location of the neutral axis y
b) The transformed moment of inertia about the strong axis
c) The moment of inertia about the weak axis
Piece “B”
3½”
1½”
7¼”
yB
y
Piece “A” (centered
under piece “B”)
yA
Datum
1½”
a) Location of neutral axis y:
Piece
Area
(1.5”)(7.25”) = 10.88 in2
A
B
(1.5”)(3.5”) = 5.25 in2
16.13 in2
Totals:
y=
y
½(7.25”) = 3.625”
7.25” + ½(1.5”) = 8”
Σ( A piece y piece )
ΣA piece
= 81.44 in3
16.13 in2
y = 5.05 in.
Lecture 5 - Page 2 of 8
(Area)y
(10.88)(3.625”) = 39.44 in3
(5.25 in2)(8”) = 42 in3
81.44 in3

3. b) Determine transformed moment of inertia about the strong “x” axis:
Area
Piece
A
10.88 in
3.625”
39.44 in
B
5.25 in2
8”
42 in3
(3.5" )(1.5" ) 3
= 0.98in 4
12
Totals:
16.13 in2
81.44 in3
Ad2
48.61 in4
2
y
(Area)y
3
I
(1.5" )(7.25" ) 3
= 47.63in 4
12
d = y-ypiece
3.625 - 5.05
= -1.425”
8 - 5.05 = 2.95”
(10.88)(-1.425)2
= 22.09 in4
(5.25)(2.95)2
= 45.69 in4
67.78 in4
I x = ∑ ( I + Ad 2 ) piece
= 48.61 in4 + 67.78 in4
Ix= 116.39 in4
c) Determine the moment of inertia about the weak “y” axis:
Piece “B”
3½”
1½”
Neutral axis of
piece “A” and “B”
7¼”
Piece “A” (centered
under piece “B”)
1½”
Since the neutral axis of both pieces line–up over each other, the total
moment of inertia is the sum of the moment of inertias of the pieces.
Iy = IA + IB
Iy =
(7.25)(1.5) 3 (1.5)(3.5) 3
+
= Iy = 7.4 in4
12
12
Lecture 5 - Page 3 of 8

4. Example 2
GIVEN: A steel W18x35 “I” beam reinforced with a ½” x 8” steel plate welded to
the bottom flange of the beam as shown below
REQUIRED: Determine:
a) The location of the neutral axis “y”
b) The transformed moment of inertia about the strong axis
c) The moment of inertia about the weak axis
d) The section modulus about the strong axis
e) The radius of gyration about the strong axis
Piece 2
W18x35 steel beam (Area = 10.3 in2)
(Ix = 510 in4)
(Iy =15.3 in4)
8.85”
From
textbook
17.70”
8.85”
y
Y2
Datum
0.5”
Piece 1
Y1 = 0.25”
½” x 8” steel plate welded to center
of bottom flange of beam
Make a Table as shown below:
Piece
1
Area
4 in2
0.25”
(Area)y
1 in3
2
10.3 in2
9.35”
96.3 in3
510 in4
Totals:
14.3 in2
97.3 in3
510.08 in4
y
I
(8" )(0.5" ) 3
= 0.08in 4
12
d = y-ypiece
6.80”-0.25” = 6.55”
(4)(6.55)2 =
171.6 in4
6.80”–9.35” = -2.55”
(10.3)(-2.55)2 =
66.98 in4
a) Determine location of neutral axis “y”:
y=
Σ( A piece y piece )
ΣA piece
=
97.3in 3
14.3in 2
y = 6.80”
Lecture 5 - Page 4 of 8
Ad2
238.58 in4

5. b) Determine Transformed Moment of Inertia about Strong Axis:
I x = ∑ ( I + Ad 2 ) piece
= 510.08 in4 + 238.58 in4
Ix= 748.66 in4
c) Determine Moment of Inertia about Weak Axis:
W18x35 steel beam (Area = 10.3 in2)
(Ix = 510 in4)
(Iy =15.3 in4)
4”
4”
8”
Since the neutral axis of both pieces line–up over each other, the total
moment of inertia is the sum of the moment of inertias of the pieces.
Iy = I1 + I2
(0.5" )(8" ) 3
Iy =
+ 15.3in 4
12
Iy = 36.63 in4
Lecture 5 - Page 5 of 8

6. d) Determine the Section Modulus about Strong Axis:
S strong =
=
I strong
y strong
748.66in 4
6.80"
Sstrong = 110.1 in3
e) Determine Radius of Gyration about Strong Axis:
rstrong =
rstrong =
I s trong
Atot
748.66in 4
14.3in 2
rstrong = 7.24”
Lecture 5 - Page 6 of 8

7. Example 3
GIVEN: Repeat Example 1, using a 2x4 top flange and a 2x8 web.
REQUIRED: Using AutoCAD, determine the following:
a) The location of the neutral axis y
b) The transformed moment of inertia about the strong axis
d) The moment of inertia about the weak axis
Piece “B”
3½”
1½”
7¼”
yB
y
Piece “A” (centered
under piece “B”)
yA
Datum
1½”
Lecture 5 - Page 7 of 8

8. Using AutoCAD, draw the shape shown and make a “REGION” out of it:
Results are exactly the same as in Example 1
Lecture 5 - Page 8 of 8