First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
9.1 Moments of Inertia
The moment of inertia is geometric property of an area that is used to determine the strength
of a structural member.it is sometimes referred to as the second moment of the area about an
axis.
By definition, the moments of inertia of a differential area about the x and y axis are:dIx = y 2 dA
dIY = x 2 dA
For the entire area A the moments of inertia are determined by integration;
Ix = ∫ y 2 dA
Iy = ∫ x 2 dA
A
A
9.2 Polar Moment of Inertia
The moment of inertia of dA about the pole O or z axis is referred as the polar moment.it is
defined as:dJO = r 2 dA
,
where r is the perpendicular distance from the pole(z axis) to the element dA
For the entire area the polar moment of inertia is:JO = ∫ r 2 dA = Ix + Iy
A
131
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
9.3 Parallel Axis Theorem for an Area
The parallel axis theorem can be used to find the moment of inertia of an area about any axis
that is parallel to an axis passing through the centroid and about which the moment of inertia
is known.
To develop this theorem, we will consider finding the moment of inertia of the shaded area
shown above about the x axis, the moment of inertia of dA about the x axis is
dIx = (ý + dy)2 dA
For the entire area
𝐼𝑥 = ∫ (𝑦́ + 𝑑𝑦)2 𝑑𝐴
𝐴
𝐼𝑥 = ∫ 𝑦́ 2 𝑑𝐴 + 2𝑑𝑦 ∫ 𝑦́ 𝑑𝐴 + 𝑑𝑦 2 ∫ 𝑑𝐴
𝐴
𝐴
𝐴
The first integral represents the moment of inertia about the centroidal axis 𝐼𝑥́
The second integral is zero
The third integral represent the total area.
Therefore;
𝐼𝑥 = 𝐼𝑥́ + 𝐴𝑑𝑦 2
132
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
A similar expressions can be written
𝐼𝑦 = 𝐼𝑦́ + 𝐴𝑑𝑥 2
𝐽𝑂 = 𝐽𝐶 + 𝐴𝑑 2
The form of each of these equations states that the moment of inertia for an area about an
axis is equal to its moment of inertia about a parallel axis passing through the area centroid
plus the product of the area and the square of the perpendicular distance between the axes.
9.4 Radius of Gyration of an Area
Provided the areas and moments of inertia are known, the radii of gyration are determined
from the formulas;
Ix
kx = √
A
k𝑦 = √
Iy
A
JO
k𝑂 = √
A
133
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Examples
Example(1):Determine the moment of inertia of the shaded area shown in figure below.
Solution(1):200
2
𝑦 2 (100 − 𝑥)𝑑𝑦 =
𝐼𝑥 = ∫ 𝑦 𝑑𝐴 = ∫
200
∫
0
𝐴
0
𝑦2
2
𝑦 (100 −
)
400
200
𝑦4
= ∫ (100𝑦 −
) 𝑑𝑦
400
0
= 107(106 )𝑚𝑚4
2
Solution(2):𝑑𝑥 𝑦 3
𝑦 2 1 3
𝑑𝐼𝑥 = 𝑑𝐼𝑥́ + 𝑑𝐴𝑦̃ =
+ 𝑦 𝑑𝑥 ( ) = 𝑦 𝑑𝑥
12
2
3
2
100
1 3
𝑦 𝑑𝑥
3
0
100
3
1
(400𝑥)2 𝑑𝑥 = 107(106 )𝑚𝑚4
=∫
3
0
𝐼𝑥 = ∫ 𝑑𝐼𝑥 = ∫
134
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Example(2):Determine the moment of inertia for the rectangular area shown below with respect to
A.The centroidal axis.
B. The axis through the base of the rectangle.
C.The polar moment of inertia through the centroid C.
Solution:𝐀. Ix́ = ∫ ý 2 dA
A
h/2
=∫
ý 2 (b dý )
−h/2
h/2
bh3
= b ∫ ý dý =
12
−h/2
2
B. Ixb = Ix́ + Ady 2
bh3
h 2 bh3
+ bh( ) =
12
2
3
C.JC = Ix́ + Iý
hb3
Iý =
12
1
JC =
bh(h2 + b2 )
12
135
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Example(3):Determine the moments of inertia of the triangular area about the x axis and y axis.
Solution:dA = x dy
b
x=b− y
h
𝑏
𝑑𝐴 = (𝑏 − 𝑦) 𝑑𝑦
ℎ
𝐼𝑥 = ∫ 𝑦 2 𝑑𝐴
𝐴
ℎ
𝑏
= ∫ 𝑦 2 (𝑏 − 𝑦) 𝑑𝑦
ℎ
0
ℎ
𝑏
= ∫ (𝑏𝑦 2 − 𝑦 3 ) 𝑑𝑦
ℎ
0
𝑏 3
𝑏 4 ℎ
=[ 𝑦 −
𝑦 ]
3
4ℎ
0
𝑏ℎ3
=
12
To find Iy
dA = 𝑦 dx
h
y=h− x
b
ℎ
𝑑𝐴 = (ℎ − 𝑥) 𝑑𝑥
𝑏
𝐼𝑦 = ∫ 𝑥 2 𝑑𝐴
𝐴
𝑏
ℎ
= ∫ 𝑥 2 (ℎ − 𝑥) 𝑑𝑥
𝑏
0
𝑏
ℎ
= ∫ (ℎ𝑥 2 − 𝑥 3 ) 𝑑𝑥
𝑏
0
ℎ 3 ℎ 4 𝑏 ℎ𝑏 3
=[ 𝑥 −
𝑦 ] =
3
4𝑏
12
0
136
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Example(4):Determine the radii of gyration for the shaded area shown in figure with respect to the x and
y axis
Solution:𝑦 2 = 3𝑥
𝑦 = √3 𝑥 1/2
dA = √3 x1/2 dx
y3
1
3
dIx = dx = (√3 x1/2 ) dx = √3 x 3/2 dx
3
3
5
Ix = ∫ dIX = ∫ √3 x 3/2 dx
A
2
5
2√3 5/2
=[
x ] = 34.81cm4
5
2
5
2
Iy = ∫ x dA = ∫ x 2 (√3x1/2 )dx
A
2
5
5
2√3 7/2
= ∫ √3x 5/2 dx = [
x ]
7
2
2
= 132.72cm4
5
5
2√3 3/2
1/2
A=∫2 √3 x dx = [ x ] = 9.644cm4
3
2
𝐼𝑥
34.81 1/2
𝑘𝑥 = √ = [
] = 1.899cm
𝐴
9.644
𝐼𝑦
132.72 1/2
𝑘𝑦 = √ = [
] = 3.7097cm
𝐴
9.644
137
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
9.5 Moment of Inertia for Composite Areas.
A composite area consists of a series of simpler parts or shapes such as rectangle, triangle
and circles. Provided the moment of inertia of each of these parts is known or can be
determined about a common axis, then the moment of inertia for the composite area about
this axis equals the algebraic sum of the moments of inertia for all its parts.
Procedure for Analysis
The moment of inertia for a composite area about a reference axis can be determined using
the following steps:1. Using a sketch, divided the area into its composite parts and indicates the perpendicular
distance from the centroid of each part to the reference axis.
2.If the centroidal axis for each part does not coincode with the reference axis, the parallel
axis theorem, should be used to determine the moment of inertia about the reference axis
3. The moment of inertia of entire area about the reference axis is determined by the
summing the results of its composite parts about this axis.
4. If a composite part has an empty region(hole),its moment of inertia is found by subtracting
the moment of inertia of this region from the moment of inertia of the entire part including
the region.
138
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
139
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Examples
Example(6):Determine the moment of inertia of the beam cross sectional area about the x axis.
Solution:Since the x axis pass through the centroid of both
rectangular segments then;
𝐼𝑥 = (𝐼𝑋 )1 + (𝐼𝑋 )2
100(2603 ) 92.5(2303 )
=
−
= 27.7(106 )𝑚𝑚4
12
12
140
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Example(2):Determine the moment of inertia for the cross sectional area of the member shown in figure
about the x and y centroidal axes.
Solution:Rectangle A and D
100(300)3
𝐼𝑥 = 𝐼𝑥́ + 𝐴𝑑 2 =
+ 100(300)(200)2
12
=1.425(109 )𝑚𝑚4
300(100)3
𝐼𝑦 = 𝐼𝑦́ + 𝐴𝑑 =
+ 100(300)(250)2
12
2
= 1.9(109 )𝑚𝑚4
Rectangle B
600(100)3
Ix =
= 0.05(109 )mm4
12
100(600)3
I𝑦 =
= 1.8(109 )mm4
12
The moment of inertia for the entire cross section
Ix = 2(1.425(109 )) + 0.05(109 ) = 2.9(109 )mm4
Iy = 2(1.9(109 )) + 1.8(109 ) = 5.6(109 )mm4
141
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Example(3):Determine the moment of inertia of the composite area about the y axis.
Solution:Iy = Iý + Ad2
(200)(300)3 1
=[
+ (200)(300)(200)2 ] +
36
2
(200)(300)3
[
+ (200)(300)(450)2 ] +
12
𝜋
[− (75)4 + (−𝜋(75)2 )(450)2 ] =
4
10.3 × 109 𝑚𝑚4
142
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Example(4):The polar moment of inertia for the area is Jc=642(106) mm4, about the z axis passing
through the centroid C.The moment of inertia about the 𝑦́ axis is 264(109) mm4, and the
moment of inertia about the x axis is 938(106) mm4.determine the area A.
Solution:JC = Ix́ + Iý
642 × 106 = Ix́ + 264 × 106
Ix́ = 378(106 )mm4
Ix = Ix́ + Ad2
938 × 106 = 378 × 106 + Ad2
560 × 106 = 𝐴(200)2
A = 14 × 103 mm2
143
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Example(5):Determine the moment of inertia of the shaded area with respect to y axis
Solution:For area 1
𝐼𝑦 = ∫ 𝑥 2 𝑑𝐴
dA = (10 − y)dx
10
Iy = ∫ x 2 (10 − y)dx
0
10
𝑥2
) dx
10
Iy = ∫ x 2 (10 −
0
10
= ∫ (10𝑥 2 −
0
𝑥4
) 𝑑𝑥 = 1333𝑐𝑚4
10
For area 2
𝑏ℎ3 10(63 )
𝐼𝑦 =
=
= 720𝑐𝑚4
3
3
For area 3
𝐼𝑦 =
10(6)3 10(6)(42 )
+
= 540𝑐𝑚4
36
2
For the total area
𝐼𝑦 = 1333 + 720 + 540 = 2593𝑐𝑚4
144
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Example(6):Determine the distance 𝑦̅ to the centroid of the beam cross sectional area, then determine the
moment of inertia about the 𝑥́ axis.
Solution:To find the centroid
segment
1
2
3
∑
A(𝑚𝑚2)
50(100)
325(25)
25(100)
15.625(103 )
𝑦́ =
(mm)𝑦́
75
12.5
-50
́
(mm)𝑦𝐴
375(103 )
101.56(103 )
-125(103 )
351.5625(103 )
∑ 𝑦́ 𝐴 351.56(103 )
=
= 22.5𝑚𝑚
∑𝐴
15.625(103 )
145
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
To find the moment of inertia about the x axis,
use the parallel axis theorem.
segmen
𝐴(𝑚𝑚2 )
t
𝑑𝑦
𝐼𝑋́ (𝑚𝑚4 )
𝐴 𝑑𝑦 2
𝐼𝑥
17.9(106 )
(𝑚𝑚)
1
50(100)
52.5
50(100)3
12
13.78(106 )
2
325(25)
10
325(25)3
12
0.812(106 ) 1.236(106 )
3
25(100)
72.5
25(100)3
12
13.14(106 ) 15.22(106 )
Ix = ∑ Ix = 34.41 × 106 mm4
146
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Home Works
H.W(3)
Locate the centroid 𝑦̅ for the cross sectional area for the angle, then finds the moment of
inertia about the 𝑥́ centroidal axis.
Ans:64cm4
H.W(4)
Determine the moment of inertia about the y axis.
Ans:307cm4
147
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Example(additional)
Determine the second moment of area for the shaded region shown in figure with respect to
A.The x axis
B.The y axis
Solution:A.y1 2 = 2bx1
y2 = 2x2 − 2b
y
y2
w = x2 − x1 = + b −
2
2b
2
𝑦
𝑦
𝑑𝐴 = [ + 𝑏 − ] 𝑑𝑦
2
2𝑏
2b
Ix = ∫ y 2 dA = ∫
A
2b
=∫
0
0
3
y
y2
y 2 ( + b − ) dy
2
2b
y
y4
2
( + by − ) dy
2
2b
2𝑏
𝑦 4 𝑏𝑦 3
𝑦5
22𝑏 4
=[ +
−
] =
8
3
10𝑏 0
15
B. 0 ≤ x ≤ b
h = √2bx
h = √2bx − (2x − 2b)
b ≤ x ≤ 2b
𝑏
𝐼𝑦 = ∫ 𝑥 2 𝑑𝐴 = ∫ 𝑥 2 √2𝑏𝑥 𝑑𝑥
𝐴
0
2𝑏
+ ∫ 𝑥 2 (√2𝑏𝑥 − 2𝑥 + 2𝑏)𝑑𝑥
𝑏
𝑏
= ∫ √2𝑏 𝑥
2𝑏
5/2
𝑑𝑥 + ∫ (√2𝑏 𝑥 5/2 − 2𝑥 3 + 2𝑏𝑥 2 )𝑑𝑥
0
𝑏
𝑏
2𝑏
2
2√2𝑏 7/2 2𝑥 4 2𝑏𝑥 3
73𝑏 4
= √2𝑏 [ 𝑥 7/2 ] + [
𝑥 −
+
] =
7
7
4
3 𝑏
42
0
148
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
2020-2021
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
149
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬