Moments of Inertia 1.9 Polar Moment of Inertia .29

Highway and Transport. Engineering Department First Class-Second Course College of Engineering Engineering Mechanics

Mustansiriyah University 2020-2021 Lec.Rana Hashim

م.رنا هاشم 131

قسم الطرق والنقل

كلية الهندسة .الجامعة المستنصرية

Moments of Inertia 1.9

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 = y2dA dIY = x2dA

For the entire area A the moments of inertia are determined by integration;

Ix = ∫ y2

A

dA Iy = ∫ x2

A

dA

Polar Moment of Inertia .29

The moment of inertia of dA about the pole O or z axis is referred as the polar moment.it is

defined as:-

dJO = r2dA ,

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 = ∫ r2

A

dA = Ix + Iy






Parallel Axis Theorem for an Area .39

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 = (y + dy)2dA

For the entire area

𝐼𝑥 = ∫ (�� + 𝑑𝑦)2 𝑑𝐴𝐴

𝐼𝑥 = ∫ ��2

𝐴

𝑑𝐴 + 2𝑑𝑦 ∫ ��𝑑𝐴 + 𝑑𝑦2 ∫ 𝑑𝐴𝐴𝐴

The first integral represents the moment of inertia about the centroidal axis 𝐼��

The second integral is zero

The third integral represent the total area.

Therefore;

𝐼𝑥 = 𝐼�� + 𝐴𝑑𝑦2






A similar expressions can be written

𝐼𝑦 = 𝐼�� + 𝐴𝑑𝑥2

𝐽𝑂 = 𝐽𝐶 + 𝐴𝑑2

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.

Radius of Gyration of an Area .49

Provided the areas and moments of inertia are known, the radii of gyration are determined

from the formulas;

kx = √Ix

A

k𝑦 = √Iy

A

k𝑂 = √JO

A






Examples

-:Example(1)

Determine the moment of inertia of the shaded area shown in figure below.

-:(1)Solution

𝐼𝑥 = ∫ 𝑦2𝑑𝐴𝐴

= ∫ 𝑦2(100 − 𝑥)𝑑𝑦200

0

=

∫ 𝑦2(100 −𝑦2

400)

200

0

= ∫ (100𝑦2 −𝑦4

400) 𝑑𝑦

200

0

= 107(106)𝑚𝑚4

-:Solution(2)

𝑑𝐼𝑥 = 𝑑𝐼�� + 𝑑𝐴��2 =𝑑𝑥 𝑦3

12+ 𝑦 𝑑𝑥 (

𝑦

2)

2

=1

3𝑦3𝑑𝑥

𝐼𝑥 = ∫ 𝑑𝐼𝑥 = ∫1

3𝑦3𝑑𝑥

100

0

= ∫1

3(400𝑥)

32𝑑𝑥 = 107(106)𝑚𝑚4

100

0






-: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. Ix = ∫ y2

A

dA

= ∫ y2(b dy)h/2

−h/2

= b ∫ y2h/2

−h/2

dy =bh3

12

B. Ixb = Ix + Ady2

bh3

12+ bh(

h

2)2 =

bh3

3

C.JC = Ix + Iy

Iy =hb3

12

JC =1

12bh(h2 + b2)






-:Example(3)

Determine the moments of inertia of the triangular area about the x axis and y axis.

-:Solution

dA = x dy x = b −b

hy

𝑑𝐴 = (𝑏 −𝑏

ℎ𝑦) 𝑑𝑦

𝐼𝑥 = ∫ 𝑦2𝑑𝐴𝐴

= ∫ 𝑦2 (𝑏 −𝑏

ℎ𝑦) 𝑑𝑦

ℎ

0

= ∫ (𝑏𝑦2 −𝑏

ℎ𝑦3) 𝑑𝑦

ℎ

0

= [𝑏

3𝑦3 −

𝑏

4ℎ𝑦4]

0

ℎ

=𝑏ℎ3

12

To find Iy

dA = 𝑦 dx y = h −h

bx

𝑑𝐴 = (ℎ −ℎ

𝑏𝑥) 𝑑𝑥

𝐼𝑦 = ∫ 𝑥2𝑑𝐴𝐴

= ∫ 𝑥2 (ℎ −ℎ

𝑏𝑥) 𝑑𝑥

𝑏

0

= ∫ (ℎ𝑥2 −ℎ

𝑏𝑥3) 𝑑𝑥

𝑏

0

= [ℎ

3𝑥3 −

ℎ

4𝑏𝑦4]

0

𝑏

=ℎ𝑏3

12






-:Example(4)

Determine the radii of gyration for the shaded area shown in figure with respect to the x and

y axis

-:Solution

𝑦2 = 3𝑥

𝑦 = √3 𝑥1/2

dA = √3 x1/2dx

dIx =y3

3dx =

1

3(√3 x1/2)

3dx = √3 x3/2 dx

Ix = ∫ dIXA

= ∫ √3 x3/2dx5

2

= [2√3

5x5/2]

2

5

= 34.81cm4

Iy = ∫ x2dAA

= ∫ x2(√3x1/2)dx5

2

= ∫ √3x5/2dx = [2√3

7x7/2]

2

55

2

= 132.72cm4

A=∫ √3 x1/25

2dx = [

2√3

3x3/2]

2

5

= 9.644cm4

𝑘𝑥 = √𝐼𝑥

𝐴= [

34.81

9.644]

1/2

= 1.899cm

𝑘𝑦 = √𝐼𝑦

𝐴= [

132.72

9.644]

1/2

= 3.7097cm






.Moment of Inertia for Composite Areas .59

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.






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)

12−

92.5(2303)

12= 27.7(106)𝑚𝑚4






-:)2Example(

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

𝐼𝑥 = 𝐼�� + 𝐴𝑑2 =100(300)3

12+ 100(300)(200)2

=1.425(109)𝑚𝑚4

𝐼𝑦 = 𝐼�� + 𝐴𝑑2 =300(100)3

12+ 100(300)(250)2

= 1.9(109)𝑚𝑚4

Rectangle B

Ix =600(100)3

12= 0.05(109)mm4

I𝑦 =100(600)3

12= 1.8(109)mm4

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






-:)3Example(

Determine the moment of inertia of the composite area about the y axis.

-:Solution

Iy = Iy + Ad2

= [(200)(300)3

36+

1

2(200)(300)(200)2] +

[(200)(300)3

12+ (200)(300)(450)2] +

[−𝜋

4(75)4 + (−𝜋(75)2)(450)2] =

10.3 × 109𝑚𝑚4






-:)4Example(

) mm4, about the z axis passing 6The polar moment of inertia for the area is Jc=642(10

) mm4, and the 9axis is 264(10 �� through the centroid C.The moment of inertia about the

. ) mm4.determine the area A6is 938(10moment of inertia about the x axis

-:Solution

JC = Ix + Iy

642 × 106 = Ix + 264 × 106

Ix = 378(106)mm4

Ix = Ix + Ad2

938 × 106 = 378 × 106 + Ad2

560 × 106 = 𝐴(200)2

A = 14 × 103mm2






-:Example(5)

Determine the moment of inertia of the shaded area with respect to y axis

-:Solution

For area 1

𝐼𝑦 = ∫ 𝑥2𝑑𝐴

dA = (10 − y)dx

Iy = ∫ x2(10 − y)dx10

0

Iy = ∫ x2 (10 −𝑥2

10) dx

10

0

= ∫ (10𝑥2 −𝑥4

10) 𝑑𝑥 = 1333𝑐𝑚4

10

0

For area 2

𝐼𝑦 =𝑏ℎ3

3=

10(63)

3= 720𝑐𝑚4

For area 3

𝐼𝑦 =10(6)3

36+

10(6)(42)

2= 540𝑐𝑚4

For the total area

𝐼𝑦 = 1333 + 720 + 540 = 2593𝑐𝑚4






-:)6Example(

Determine the distance �� to the centroid of the beam cross sectional area, then determine the

moment of inertia about the �� axis.

-:Solution

To find the centroid 𝑦��(mm) ��(mm) A(𝑚𝑚2) segment

) 375(103 75 50(100) 1

101.56(103) 12.5 325(25) 2

-125(103) -50 25(100) 3

351.5625(103) 15.625(103) ∑

�� =∑ ��𝐴

∑ 𝐴=

351.56(103)

15.625(103)= 22.5𝑚𝑚






To find the moment of inertia about the x axis,

use the parallel axis theorem.

𝐼𝑥 𝐴 𝑑𝑦2 𝐼��(𝑚𝑚4) 𝑑𝑦

(𝑚𝑚)

𝐴(𝑚𝑚2) segmen

t

17.9(106) 13.78(106) 50(100)3

12

52.5 50(100) 1

1.236(106) 0.812(106) 325(25)3

12

10 325(25) 2

15.22(106) 13.14(106) 25(100)3

12

72.5 25(100) 3

Ix = ∑ Ix = 34.41 × 106mm4






Home Works

)3H.W(

Locate the centroid �� for the cross sectional area for the angle, then finds the moment of

inertia about the �� centroidal axis.

4Ans:64cm

H.W(4)

Determine the moment of inertia about the y axis.

4Ans:307cm






(additional)Example

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.y12 = 2bx1 y2 = 2x2 − 2b

w = x2 − x1 =y

2+ b −

y2

2b

𝑑𝐴 = [𝑦

2+ 𝑏 −

𝑦2

2𝑏] 𝑑𝑦

Ix = ∫ y2

A

dA = ∫ y2 (y

2+ b −

y2

2b) dy

2b

0

= ∫ (y3

2+ by2 −

y4

2b) dy

2b

0

= [𝑦4

8+

𝑏𝑦3

3−

𝑦5

10𝑏]

0

2𝑏

=22𝑏4

15

B. 0 ≤ x ≤ b h = √2bx

b ≤ x ≤ 2b h = √2bx − (2x − 2b)

𝐼𝑦 = ∫ 𝑥2𝑑𝐴 = ∫ 𝑥2√2𝑏𝑥 𝑑𝑥𝑏

0𝐴

+ ∫ 𝑥2(√2𝑏𝑥 − 2𝑥 + 2𝑏)𝑑𝑥2𝑏

𝑏

= ∫ √2𝑏 𝑥5/2𝑑𝑥𝑏

0

+ ∫ (√2𝑏 𝑥5/2 − 2𝑥3 + 2𝑏𝑥2)𝑑𝑥2𝑏

𝑏

= √2𝑏 [2

7𝑥7/2]

0

𝑏

+ [2√2𝑏

7𝑥7/2 −

2𝑥4

4+

2𝑏𝑥3

3]

𝑏

2𝑏

=73𝑏4

42

Moments of Inertia 1.9 Polar Moment of Inertia .29

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