Sheridan Institute of Technology and Advanced Learning

Faculty of Applied Science and Technology

School of Mechanical and Electrical Engineering & Technology

_____________________________________________

MATH22558 Integral Calculus - 1155_56353

Project Report

Instructor: Patrick Keenan

Class #: 56353

Paritosh Chahal (991345820)

Date of Submission: 11th August 2015

1

Table of Contents

Objective: ...................................................................................................................................................... 2

Introduction: ................................................................................................................................................. 2

Techniques of finding moment of inertia: .................................................................................................... 3

Formula of Moment of inertia of an object with constant Mass “m”: ..................................................... 3

1. Moment of inertia for a piece of mass “dm” about center of mass: ................................................ 5

2. Moment of inertia of an arbitrary plane area about the x-axis: ....................................................... 6

3. Moment of inertia of an arbitrary plane area about the y-axis: ..................................................... 11

4. Moment of inertia of a volume about the x-axis: ........................................................................... 11

5. Moment of inertia of a volume about the y-axis: ........................................................................... 11

Proof of Moment of Inertia of shapes: ...................................................................................................... 11

1. Cylinder: .......................................................................................................................................... 11

2. Rectangles ....................................................................................................................................... 13

3. Moment of Inertia of rectangle about x-axis: ................................................................................. 14

4. Cylinder about some axis: ............................................................................................................... 15

5. Moment of inertia of area bounded by 2 curves: about y-axis: ..................................................... 16

6. Moment of inertia of area bounded by 2 curves: about x-axis: ..................................................... 17

Appendix: .................................................................................................................................................... 18

References: ................................................................................................................................................. 21

2

Objective: The purpose of this report is to prove the moment of inertia formula of any given object through

integration.

Introduction: As learned in Applied Mechanics 1, moment of inertia is defined as the measure of a body to

resist rotation [in other words, moment of inertia can be described as the “mass” in rotational

dynamics formulas (fig1)].

Fig 1: Inertia of an object has the same job as “mass” of an object has in linear dynamics

Moment of Inertia is an important subject in many different fields of mechanical engineering.

For instance, moment of inertia is widely used in measuring the strength of different type of

beams, rods, and building structures. To elaborate, as learned in the course Mechanics of

Materials, “Area Moment of Inertia” (I) is used to find the maximum bending stress1, maximum

horizontal shearing stress2, shear flow

3 along a shear plane, and maximum deflection

5 of

cantilever beams (fig 2) and simply supported beam (fig 3).

3

Fig 2: Cantilever Beam

Fig 3: Simply supported Beam

In addition, other than “Area Moment of Inertia” (I), the “Polar Moment of Inertia” (J) is used to

measure find the maximum torsion5 a rod (solid of hollow) can resist.

Other than the “structural (beam is technically considered as a structure) mechanics field”, the

moment of inertia also has many uses in the dynamics field. TO elaborate, moment of inertia can

be used to find how fast an object can accelerate while rotating with reference to its center, how

much time does a crane (or pulley system) takes to lift an object with mass “x”, and energy

transfer at different instances when a wheel is rolled down a slope (which help in understanding

the designs of transportation systems such as cars, buses, trucks etc.).

Techniques of finding moment of inertia: Even though moment of inertia has same definition for every rigid object in the universe;

different techniques are used to find moment of inertia depending on the shape and type of

object.

Formula of Moment of inertia of an object with constant Mass “m”:

The basic formula of Moment of Inertia is as given:

𝐼 = 𝑀 ∗ 𝑅2

Where:

M= mass of the object

R= Distance from the center

4

Looking at the formula above, one would say that the above formula is a quadratic function (as

learned in MATH 17688- Mathematics 1) where inertia is doubled as the length of the constant

mass “M” from the center of rotation is increased by 1 unit.

Same formula (stated above) can be used for finding the moment of inertia of 2D with constant

area:

𝐼 = ∆𝐴 ∗ 𝑅2

Where:

A= Area of the shape (constant)

R= Distance from the center of rotation

Using the above two formulas, we will derive formulas for the following in the next few

sections:

1. Moment of inertia for a strip about center of mass

2. Moment of inertia of an arbitrary plane area about the x-axis

3. Moment of inertia of an arbitrary plane area about the y-axis

4. Moment of inertia of a volume of revolution about the x-axis.

5. Moment of inertia of a volume of revolution about the y-axis.

5

1. Moment of inertia for a piece of mass “dm” about center of mass: Suppose we have a cylinder with radius “R”, mass “M” and height “h” (fig 4):

Fig 4: Cylinder with height “h” and radius “R”

Suppose we want to find the moment of Inertia of a thin strip of material with mass dm (fig 5

shows the circular cross section of the cylinder) with reference to the center axis (red)

NOTE: As learned in Math

Fig 5: The cross-section of cylinder in figure 4

Moment of inertia for the single strip is:

𝐼 = 𝑥2(𝑑𝑚) [1]

Above equation can be to any arbitrary shape.

R

h

x

6

In terms of second moment of inertia:

Let the “dm” piece in figure 5 has area “dA”:

𝐼 = 𝑥2(𝑑𝐴) [2]

2. Moment of inertia of an arbitrary plane area about the x-axis:

Fig 6: Random arbitrary shape

To find the moment of inertia of the arbitrary shape (with area “A”) above, one has to

take the sum of all small “dA” from equation [2]. So the moment of inertia of an arbitrary

shape (with reference to center of mass of the shape is:

(𝑟2)∑𝑑𝐴 = ∫ 𝑟2 ∗ 𝑑𝐴𝐴

0 [3]

Now to find the moment of inertia of the same arbitrary shape above with respect to the

x-axis (fig 6); one may need to use the parallel axis theorem.

Fig 7: Parallel axis theorem

7

The moment of inertia is the least when the reference (rotation) axis is at the center of gravity of

any arbitrary shape; as the reference axis increased further away from the center of mass, the

moment of inertia of the arbitrary shape increases:

𝐼 = 𝐼𝑐𝑚 +𝑀𝑑2(Parallel axis theorem formula) [4]

𝐼 = 𝐼𝑐𝑚 + 𝐴𝑑2 (Second Moment of Inertia) [5]

1. Say we have an arbitrary shape (of area “A”), and we have to find the moment of inertia

of this shape with respect to origin “O” (fig 7)

Fig 7

NOTE: The rotation axis is “coming out of the paper”

i. Moment of inertia of the shape with respect to center of gravity:

𝐼 = ∑𝑇2(𝑑𝐴) [6]

ii. Changing the reference point from “CG” to “O” (fig 8), the moment of inertia will

be:

𝐼 = (𝑇 + 𝑅)2𝑋𝑑𝐴 [7] (where R is given distance between “O” and “CG”)

NOTE: (𝑇 + 𝑅)2 = (𝑅𝑥 + 𝑇𝑥)2 + (𝑅𝑦 + 𝑇𝑦)

2

y

x

O

T

8

Fig 8

So the equation 7 changes to 𝐼 = ∫ [(𝑅𝑥 + 𝑇𝑥)2 + (𝑅𝑦 + 𝑇𝑦)

2] 𝑋 𝑑𝐴

iii. Using the binomial expansion on “updated” equation 8:

𝐼 = ∫ [((𝑅𝑥2) + 2(𝑅𝑥 )(𝑇𝑥) + (𝑇𝑥

2)) + ((𝑅𝑦)2 + 2(𝑅𝑦)(𝑇𝑦) + (𝑇𝑦)

2) ]𝑋 𝑑𝐴

𝐼 = ∫ [(𝑅𝑥2) + (𝑅𝑦)

2 + (𝑇𝑦)2+ (𝑇𝑥

2) + 2(𝑅𝑥 )(𝑇𝑥) + 2(𝑅𝑦)(𝑇𝑦)]𝑋 𝑑𝐴 [9]

iv. We can divide equation 9 into 4 different notation to make the proof a bit more

simpler:

𝐼 = [∫(𝑅𝑥2) + (𝑅𝑦)

2 + ∫(𝑇𝑦)2+ (𝑇𝑥

2) + ∫2(𝑅𝑥 )(𝑇𝑥) + ∫ 2(𝑅𝑦)(𝑇𝑦)] ∗ 𝑑𝐴 [10]

v. In equation 10 the first notation has a constant value which sums up to “R”:

∫[(𝑅𝑥2) + (𝑅𝑦)

2] 𝑑𝐴 = 𝑅2∫𝑑𝐴 = 𝑅2(𝐴)

vi. Second notation is quite similar to equation 6:

∫[(𝑇𝑦)2+ (𝑇𝑥

2)] 𝑑𝐴 = 𝑇2(𝐴) = 𝐼𝐶𝐺

R

T

Tx

Ty

Ry

Rx

9

vii. For final two notation, one may observe that they have two constants respectively

that can be taken out:

∫[2(𝑅𝑥 )(𝑇𝑥)] 𝑑𝐴 = 2𝑅𝑥 ∫(𝑇𝑥)𝑑𝐴 [10]

∫[2(𝑅𝑦 )(𝑇𝑦)] 𝑑𝐴 = 2𝑅𝑦 ∫(𝑇𝑦)𝑑𝐴[11]

The above two equation is very similar to the center of mass equation:

𝐶𝐺𝑥 =∫𝑥′ ∗ 𝑑𝐴

𝐴

𝐶𝐺𝑦 =∫𝑦′ ∗ 𝑑𝐴

𝐴

Where “x’” and “y’” is the effective distance between the center of mass (i.e.

gravity) point and origin of the x-y axes (fig 9).

Fig 9

Since the “x” and “y” axis (which helped us find the distance dA) from fig 8 is within the center

of gravity (fig 10), the sum of all the “Ty” and “Tx” values will sum up to zero*.

*NOTE: According to vector rules, the values of “Ty” and “Tx” on one side are positive; and

negative on other side.

10

Fig 10

The above statement proves the equation 10 and 11 will sum up to 0.

∫[2(𝑅𝑥 )(𝑇𝑥)] 𝑑𝐴 = 2𝑅𝑥 ∫(𝑇𝑥)𝑑𝐴 = 0

∫[2(𝑅𝑦 )(𝑇𝑦)] 𝑑𝐴 = 2𝑅𝑦 ∫(𝑇𝑦)𝑑𝐴 = 0

viii. Summing them all up gives us the parallel axis theorem formula:

𝐼𝑥 = 𝐼𝐶𝐺 + (𝑅2(𝐴))

2. Therefore, to find the moment of inertia about the “x-axis” of the arbitrary shape from

figure 8:

𝐼𝑦 = 𝐼𝐶𝐺 + (𝑅𝑦2(𝐴))[12] (where Ry is a given distance between the axis and point “CG”

of body).

Y

X

11

3. Moment of inertia of an arbitrary plane area about the y-axis: To find the moment of inertia about the y-axis; the parallel axis theorem is used:

𝐼𝑦 = 𝐼𝐶𝐺 + (𝑅𝑥2(𝐴))[13] (Where Rx is a given distance between the axis and point “CG”

of body)

4. Moment of inertia of a volume about the x-axis: The moment of inertia of a solid 3D model (with a definite volume) [fig 4] have similar

techniques as finding second moment of inertia of 2D arbitrary shapes.

The moment of inertia of 3D model with reference from center of axis (fig 4) is:

𝐼 = ∫ 𝑥2 ∗ 𝑑𝑚

NOTE: 𝑑𝑚 = 𝑑𝑣 𝑋 𝜌 [Volume X density]

𝐼𝑐𝑔 = 𝜌 ∫ 𝑥2 ∗ 𝑑𝑣 [14]

To find the moment of inertia of a volume with respect to x-axis; we use parallel axis theorem:

𝐼𝑥 = 𝐼𝑐𝑔 + (𝑅𝑦2(𝑀))[15] (where Ry is a given distance)

5. Moment of inertia of a volume about the y-axis: To find the moment of inertia of a volume with respect to y-axis; we use parallel axis theorem:

𝐼𝑦 = 𝐼𝑐𝑔 + (𝑅𝑥2(𝑀))[16] (Where Rx is a given distance)

Proof of Moment of Inertia of shapes:

1. Cylinder: Let’s use the above information gathered to make the equations above to find the moment of

inertia of a cylinder about its center axis from fig 4:

12

From equation 14 we know:

𝐼𝑐𝑔 = 𝜌 ∫ 𝑥2 ∗ 𝑑𝑣

Fig 11

Sum of all the small all the thin cylindrical shell with thickness “dr” (fig11) can help us to find

the moment of inertia of the solid cylinder

𝑑𝑣 = (𝑅𝑙𝑎𝑟𝑔𝑒𝑟2 − 𝑅𝑠𝑚𝑎𝑙𝑙𝑒𝑟

2 )𝜋(ℎ)

𝑑𝑣 = (𝑅𝑙𝑎𝑟𝑔𝑒𝑟 − 𝑅𝑠𝑚𝑎𝑙𝑙𝑒𝑟)(𝑅𝑙𝑎𝑟𝑔𝑒𝑟 + 𝑅𝑠𝑚𝑎𝑙𝑙𝑒𝑟)𝜋(ℎ)

𝑑𝑣 = (𝑑𝑟)(𝑅𝑙𝑎𝑟𝑔𝑒𝑟 + 𝑅𝑠𝑚𝑎𝑙𝑙𝑒𝑟)𝜋(ℎ)

𝑑𝑣 = (2)(𝑅𝑙𝑎𝑟𝑔𝑒𝑟 + 𝑅𝑠𝑚𝑎𝑙𝑙𝑒𝑟)

2𝜋(ℎ)(𝑑𝑟)

NOTE: (𝑅𝑙𝑎𝑟𝑔𝑒𝑟+𝑅𝑠𝑚𝑎𝑙𝑙𝑒𝑟)

2 is the average distance “R” from center of the circle

𝑑𝑣 = (2)𝑅𝜋(ℎ)(𝑑𝑟)

𝐼𝑐𝑔 = 𝜌∫𝑅2(2𝑅)𝜋(𝑑𝑟)(ℎ)

𝐼𝑐𝑔 = 2 𝜌(𝜋)(ℎ)∫𝑅3(𝑑𝑟)

dr

13

𝐼𝑐𝑔 = 2 𝜌(𝜋)(ℎ)∫𝑅3(𝑑𝑟)

𝐼𝑐𝑔 = 2 𝜌(𝜋)(ℎ) (𝑅4

4)

𝐼𝑐𝑔 = (𝑅2) 𝜌(𝜋)(ℎ) (

𝑅2

2)

𝐼𝑐𝑔 =1

2(𝑚)(𝑅2)[17]

2. Rectangles

Fig 12

Now say we have to find the moment of inertia of the rectangle above with fixed dimensions of

“L * W” with respect to the axis passing through the center of gravity of the shape.

Fig 13

𝐼𝑐𝑔 = ∫𝑅2(𝑑𝐴)

𝐼𝑐𝑔 = ∫ 𝑅2(𝑊 ∗ 𝑑𝑦)𝐿/2

−𝐿/2 [NOTE: Let R = some distance “y”)

L

W

14

𝐼𝑐𝑔 = 𝑊 ∫ 𝑦2(𝑑𝑦)

𝐿/2

−𝐿/2

𝐼𝑐𝑔 = 𝑊 [1

3(𝐿

2)3

− {−1

3(𝐿

2)3

}]

𝐼𝑐𝑔 = 𝑊 [2

24(𝐿3)]

𝐼𝑐𝑔 = [𝐿3𝑊

12] [18]

3. Moment of Inertia of rectangle about x-axis:

Fig 14

𝐼𝑋 = ∫𝑅2(𝑑𝐴)

𝐼𝑥 = ∫ 𝑅2(𝑊 ∗ 𝑑𝑦)𝐿/2

−𝐿/2 [NOTE: Let R = some distance “y”)

𝐼𝑥 = 𝑊∫𝑦2(𝑑𝑦)

𝐿

0

𝐼𝑥 = 𝑊 [1

3(𝐿

1)3

− {−1

3(0)3}]

𝐼𝑥 = [𝐿3𝑊

3] [19]

15

Using Parallel Axis Theorem:

𝐼𝑥 = 𝐼𝐶𝐺 + (𝑅𝑦2(𝐴))

𝐼𝑥 = [𝐿3𝑊

12] + ((

𝐿

2)2(𝐿 ∗ 𝑊))[NOTE:𝑅𝑦

2 = (𝐿

2)2]

[𝐿3𝑊

12] +

(

(𝐿2

4) (𝐿 ∗𝑊)

)

[𝐿3𝑊

12] + (

3𝐿3𝑊

12)

4𝐿3𝑊

12

𝐼𝑥 = [𝐿3𝑊

3] [19]

4. Cylinder about some axis:

Fig 15

Using Parallel Axis Theorem:

𝐼𝑦 = 𝐼𝑐𝑔 + (𝑅2 (𝑀))[16] (Where R is a given distance between y-axis and CG point of

cylinder)

𝐼𝑦 = (1

2𝑀𝑅2) + (𝑅2 (𝑀))

𝐼𝑦 = (3

2𝑀𝑅2)[21]

R

Some axis

16

5. Moment of inertia of area bounded by 2 curves: about y-axis: Area bounded by curves y = x

2 and y = x [fig 16]:

Fig 16: Area bounded by the two functions

Since we are not given the centroid of the “area”, we can find it using integration (and

afterwards use parallel axis theorem), or derive an equation to find moment of inertia of areas

about x-axis.

I = r2

(A) = ∑(r2)(dA); to make the area if fig 16 follow this equation, the following should be

done:

Fig 17

y

x

x

dA

17

NOTE: In fig 17, dA = (dx)(x-x2), and x = R

𝐼 = ∫ (𝑥)2(𝑥 − 𝑥2)𝑑𝑥1

0

𝐼 = ∫ 𝑥3 − 𝑥4𝑑𝑥1

0

𝐼 = (1

4𝑥4) − (

1

5𝑥5) + 𝑐

Use definite integral in the next few steps to find the answer (do it yourself, for practice).

6. Moment of inertia of area bounded by 2 curves: about x-axis: Same function as before:

NOTE: dA (“x” in terms of y)= [(y)0.5

-(y)]dy, and R = “y”

𝐼 = ∫ (𝑦)2(𝑦0.5 − 𝑦)𝑑𝑦1

0

𝐼 = ∫ (𝑦1 − 𝑦2)𝑑𝑦1

0

𝐼 = (1

2𝑦2) − (

1

3𝑦3) + 𝐶

Use definite integral in the next few steps to find the answer (do it yourself, for practice).

y

dA

18

Appendix: 1. Maximum ending stress:

19

2. Horizontal shearing stress:

20

3. Shear Flow:

21

References: http://calculus.nipissingu.ca/tutorials/area_volume.html (Volumes by Cylindrical shells)

http://ummalqura-phy.com/HYPER1/icyl.html

http://www.intmath.com/applications-integration/4-volume-solid-revolution.php

https://www.youtube.com/watch?v=6I7iKz72HW4

http://www.intmath.com/applications-integration/6-moments-inertia.php

http://www.ce.siue.edu/examples/Worked_examples_Internet_text-only/Data_files-Worked_Exs-

Word_&_pdf/Moment_Inertia_Integration.pdf

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

http://www.strucalc.com/engineering-resources/normal-stress-bending-stress-shear-stress/

http://timbertoolbox.com/beamdesign.htm

http://www.slideshare.net/msheer/lesson-06-shearing-stresses (slide 8)

http://engineering-references.sbainvent.com/strength_of_materials/shear-flow.php

Morrow, H., & Kokernak, R. (2014). Statics and strength of materials (7th ed., pp. 305-336). Upper

Saddle River, NJ: Prentice Hall.

Walker, Keith M. "Equilibrium." Applied Mechanics for Engineering Technology. Upper Saddle River, NJ:

Pearson/Prentice Hall, 2004. 283-300. Print.