MOMENT OF INERTIA
Moment of Inertia - the summation of the product of each area by the square of its moment arm or
sometimes called the second moment of area. It is denoted by letter I.
Ix =‫ 𝟐𝒚 ׬‬dA
Iy = ‫ 𝟐𝒙 ׬‬dA
Polar Moment of Inertia - also known as second polar moment of area, is a quantity used to describe
resistance to torsional deformation (deflection), in cylindrical objects (or segments of cylindrical object)
with an invariant cross-section and no significant warping or out-of-plane deformation. (Denoted by J)
I = Iy + I x
Radius of Gyration - the radial distance to a point which would have a moment of inertia the same
as the body's actual distribution of mass, if the total mass of the body were concentrated there.
(Denoted by k)
k=
𝑰
𝑨
Transfer Moment of Inertia - movement of moment of inertia from one axis to another parallel axis.
Ix = Ixo + Ad2
where:
Ix - transfer moment of inertia at the parallel axis
Ixo - moment of inertia at the centroid of the area
A - centroidal axis of the area
d- distance of centroid of area from any parallel axis
Transfer Polar Moment of Inertia
J = Jo + Ad2
Moment of Inertia for Geometric Shapes
Shape
Rectangle
Moment of Inertia
Io =
bh3
Ixo =
Semi-Circle
Ix =
πr4
πr4
Ix =
16
Ixo =
A
k=
Ix
A
kz =
=
=
=
=
𝜋𝑟 2
2
3
ℎ
18
ℎ
6
2
r
2
r
𝜋𝑟 2
4
kx =
kx =
4
4
h
kx =
2
kxo = 0.264r
πab3
πba4
12
r
r
2
b
2
πab
Iyo =
h
kxo = 0.264r
πr4
Ixo = 0.055r4
Ellipse
πr2
2
πr4
8
I
A
ko =
4
Ixo = 0.11r4
Quarter Circle
bh
2
A
Ixo
ko =
36
12
Io
k=
𝑏ℎ3
bh3
Ixo =
J=
bh
3
Ix =
Circle
Radius of Gyration
ko =
12
Ix =
Triangle
Area
bh3
kx =
b
2
Properties of Structural Sections
Size
Areas of
Section
Ix
Iy
ẋ
ȳ
in2
in4
in4
in
in
Angle
5 x 3 x 1/2
6 x 4 x 1/2
8x4x1
8x6x1
3.75
4.75
11.00
13.00
2.60
6.30
11.6
38.8
9.50
17.4
69.6
80.8
1.75
1.99
3.05
2.65
0.75
0.99
1.05
1.65
Channel
10 in.-15.3 lb
12 in.-20.7 lb
4.47
6.03
66.90
128.1
2.3
3.9
0.64
0.70
0
0
Moment of Inertia for Composite Areas
When a composite area can be divided into geometric elements for which the moment of inertia are
known, the moment of inertia for the composite area is the sum of the moments of inertia for the
separate elements. Before the moment of inertia of the elements can be added, they must be found with
respect to the same axis.
Sample problems:
1. A rectangle is 3 in by 6 in. Determine the polar moment of inertia and the radius of gyration with
respect to the polar axis through the corner.
Solution
(3)(6)3
Ix =
= 216
3
(6)(3)3
Iy =
= 54
3
J = 216 + 54 = 270 in4
k=
270
= 3.87
3(6)
2. A hollow square cross section consists of an 8 in by 8 in square from which is subtracted a
concentrically placed square 4 in by 4 in. Find the polar moment of inertia and the polar radius of gyration
with respect to a z-axis passing through one of the outside corners.
Solution
(8)(8)3
(4)(4)3
Ix =
-[
+ (4)(4)(4)2] = 1088 in4
3
12
Iy = 1088 in4
J = Ix + Iy
J = 1088 + 1088 = 2176
k=
2176
= 6.73
82 − 4 2
3. Determine the moment of inertia of the T-section with respect to its centroidal xo axis.
Solution
A1 = 2(8) = 16
A2 = (2)(8) = 16
A = 16 + 16 = 32
Aȳ = A1y1 + A2y2
32ȳ = 16(1) + 16(6)
ȳ = 3.5
(2)(8)3
(8)(2)3
2
Ixo =
+ 2(8)(2.5) +
+ 8(2)(2.5)2
12
12
Ixo = 290.67 in4
4. Determine the moment of inertia of the area shown with respect to the centroidal area.
Solution
A1 = 12(1) = 12
A2 = 12(1) = 12
A3 = 6(1) = 6
A = 30
Aȳ = A1y1 + A2y2 + A3y3
30ȳ = 12(0.5) + 12(7) + 6(13.5)
ȳ = 5.7
12(1)3
1(12)3
6(1)3
2
2
Ixo =
+ 12(1)(5.2) +
+ (1)(12)(1.3) +
+ 6(1)(7.8)2
12
12
12
Ixo = 855.30 in4
1(6)3 12(1)3 1(12)3
Iyo =
+
+
12
12
12
Iyo = 164 in4
5. Find the moment of inertia about the indicated axis x for the figure shown
Solution
8(10)3
𝜋
Ix =
-[0.11(4)4 + (4)2(8.3)2] = 907.1 in4
3
2
6. A trapezoid is shown in the figure
a. Compute the area of the trapezoid
b. Compute the centroid of the trapezoid
c. Compute the centroidal moment of inertia of the trapezoid
Solution
1.
A1 = 6(6) = 36
(6)(6)
A2 =
(2) = 36
2
A = 36 + 36 = 72
or
(6+18)
A=
(6) = 72
2
2.
3.
Aȳ = A1y1 + A2y2
72ȳ = 36(3) + 36(2)
ȳ = 2.5
(6)(6)3
6(6)3 6(6)
2
Ixo =
+ 6(6)(0.5) + [
+
(0.5)2]2
12
36
2
Ixo = 198 in4