Ch.09 Center of Gravity and Centroid

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09. Center of Gravity and Centroid

HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Engineering Mechanics – Statics 9.01 Center of Gravity and Centroid

Chapter Objectives

• To discuss the concept of the center of gravity, center of mass,

and the centroid

• To show how to determine the location of the center of gravity

and centroid for a system of discrete particles and a body of

arbitrary shape

• To use the theorems of Pappus and Guldinus for finding the

surface area and volume for a body having axial symmetry

• To present a method for finding the resultant of a general

distributed loading and show how it applies to finding the

resultant force of a pressure loading caused by a fluid



§1. Center of Gravity, Center of Mass, and the Centroid of a Body

- How can we determine these weights and their locations?



- To design the structure

for supporting a water

tank, we will need to

know the weights of

the tank and water as

well as the locations

where the resultant

forces representing these

distributed loads are

acting


Center of Gravity

- Center of gravity: the point at which the entire weight of a body

may be considered as concentrated so that if supported at this

point the body would remain in equilibrium in any position



- A body is composed of an infinite

number of particles of differential size

and each of these particles have a

weight 𝑑𝑊

- These weights will form a parallel

force system, and the resultant of this

system is the total weight of the body,

which passes through a single point

called the center of gravity, 𝐺


- The location of the center of gravity 𝐺 with respect to the 𝑥,𝑦,𝑧

𝑥 = 𝑥 𝑑𝑊

𝑑𝑊, 𝑦 =

𝑦 𝑑𝑊

𝑑𝑊, 𝑧 =

𝑧 𝑑𝑊

𝑑𝑊



- The weight of the body: the sum of

the weights of all of its particles

𝑊 = 𝑑𝑊

- The location of the center of gravity

𝐺(𝑥 , 𝑦 , 𝑧 ) 𝑥 𝑊 = 𝑥 𝑑𝑊

𝑦 𝑊 = 𝑦 𝑑𝑊

𝑧 𝑊 = 𝑧 𝑑𝑊


Center of Mass of a Body



- The location of the center of mass 𝐶𝑚

with respect to the 𝑥,𝑦,𝑧 axes

Substitute 𝑑𝑊 = 𝑔𝑑𝑚 to

𝑥 = 𝑥 𝑑𝑊

𝑑𝑊,𝑦 =

𝑦 𝑑𝑊

𝑑𝑊, 𝑧 =

𝑧 𝑑𝑊

𝑑𝑊

⟹ 𝑥 = 𝑥 𝑑𝑚

𝑑𝑚, 𝑦 =

𝑦 𝑑𝑚

𝑑𝑚, 𝑧 =

𝑧 𝑑𝑚

𝑑𝑚

http://www.google.com.vn/url?sa=i&source=images&cd=&cad=rja&docid=nwyNTpoMmHYROM&tbnid=zhxsLSye4BpvtM:&ved=0CAgQjRwwAA&url=http://jamestownengineering.com/Municipal.html&ei=OyJHUcLiDuiQiAec5YHQBg&psig=AFQjCNGzWvtbeexq70rC4y6tGCxTDxvO1w&ust=1363702715281084

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Center of a Volume



- The center of a volume or centroid 𝐶

or geometric center of the body with

respect to the 𝑥,𝑦,𝑧 axes

Substitute 𝑑𝑚 = 𝜌𝑑𝑉 to

𝑥 = 𝑥 𝑑𝑚

𝑑𝑚, 𝑦 =

𝑦 𝑑𝑚

𝑑𝑚, 𝑧 =

𝑧 𝑑𝑚

𝑑𝑚

⟹ 𝑥 = 𝑥 𝑑𝑉

𝑉

𝑑𝑉

𝑉

, 𝑦 = 𝑦 𝑑𝑉

𝑉

𝑑𝑉

𝑉

, 𝑧 = 𝑧 𝑑𝑉

𝑉

𝑑𝑉

𝑉


Center of an Area

- If an area lies in the 𝑥 − 𝑦 plane and is bounded by the curve

𝑦 = 𝑓(𝑥), then its Centroid 𝐶 𝑥 , 𝑦

𝑥 = 𝑥 𝑑𝐴

𝐴

𝑑𝐴

𝐴

, 𝑦 = 𝑦 𝑑𝐴

𝐴

𝑑𝐴

𝐴

- These integrals can be evaluated by performing a single integration

if we use a rectangular strip for the differential area element




Center of a Line

- If a line segment lies within the 𝑥 − 𝑦 plane and it can be

described by a thin curve 𝑦 = 𝑓(𝑥), then its Centroid 𝐶 𝑥 , 𝑦

𝑥 = 𝑥 𝑑𝐿

𝐿

𝑑𝐿

𝐿

, 𝑦 = 𝑦 𝑑𝐿

𝐿

𝑑𝐿

𝐿




- Example 9.1 Locate the centroid of the rod bent into the

shape of a parabolic arc

Solution

Differential Element: arbitrary point (𝑥,𝑦)

Area: 𝑥 = 𝑦2 → 𝑑𝑥/𝑑𝑦 = 2𝑦

𝑑𝐿 = (𝑑𝑥)2+(𝑑𝑦)2= 2𝑦 2 + 1𝑑𝑦

Moment Arms: 𝑥 = 𝑥, 𝑦 = 𝑦

Integrations


𝐿

𝑑𝐿

𝐿

= 𝑥 4𝑦2 +1𝑑𝑦

1

0

4𝑦2 +1𝑑𝑦1

0

= 𝑦2 4𝑦2 +1𝑑𝑦

1

0

4𝑦2 +1𝑑𝑦1

0

=0.6063

1.479= 0.41𝑚

𝑦 = 𝑦 𝑑𝐿

𝐿

𝑑𝐿

𝐿

= 𝑦 4𝑦2 +1𝑑𝑦

1

0

4𝑦2 +1𝑑𝑦1

0

=0.8484

1.479= 0.574𝑚




- Example 9.2 Locate the centroid of the circular wire segment

Solution

Polar coordinates will be used to solve this

problem since the arc is circular

Differential Element: arbitrary point (𝑅,𝜃)

Area: 𝑑𝐿 = 𝑅𝑑𝜃

Moment Arms: 𝑥 = 𝑅𝑐𝑜𝑠𝜃, 𝑦 = 𝑅𝑠𝑖𝑛𝜃

Integrations


𝐿

𝑑𝐿

𝐿

= 𝑅𝑐𝑜𝑠𝜃𝑅𝑑𝜃

𝜋/2

0

𝑅𝑑𝜃𝜋/2

0

= 𝑐𝑜𝑠𝜃𝑑𝜃

𝜋/2

0

𝑅 𝑑𝜃𝜋/2

0

=2𝑅

𝜋

𝑦 = 𝑦 𝑑𝐿

𝐿

𝑑𝐿

𝐿

= 𝑅𝑠𝑖𝑛𝜃𝑅𝑑𝜃

𝜋/2

0

𝑅𝑑𝜃𝜋/2

0

= 𝑐𝑜𝑠𝜃𝑑𝜃

𝜋/2

0

𝑅 𝑑𝜃𝜋/2

0

=2𝑅

𝜋




- Example 9.3 Determine the distance 𝑦 measured from the 𝑥

axis to the centroid of the area of the triangle

Solution

Differential Element

Arbitrary element (𝑥,𝑑𝑦) at 𝑦

Area: 𝑑𝐴 = 𝑥𝑑𝑦

=𝑏

ℎ(ℎ − 𝑦)𝑑𝑦

Moment Arms: 𝑦 = 𝑦

Integrations

𝑦 = 𝑦 𝑑𝐴

𝐴

𝑑𝐴

𝐴

= 𝑦

𝑏ℎ

ℎ − 𝑦 𝑑𝑦ℎ

0

𝑏ℎ

ℎ − 𝑦 𝑑𝑦ℎ

0

=

16𝑏ℎ2

12𝑏ℎ

=1

3ℎ



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- Example 9.4 Locate the centroid for the area of a quarter circle

Solution

Polar coordinates will be used

Differential Element (𝑂,𝑅,𝑑𝜃)

Area: 𝑑𝐴 =1

2𝑅 𝑅𝑑𝜃 =

1

2𝑅2𝑑𝜃

Moment Arms: 𝑥 =2

3𝑅𝑐𝑜𝑠𝜃, 𝑦 =

2

3𝑅𝑠𝑖𝑛𝜃


𝐴

𝑑𝐴

𝐴

=

23𝑅𝑐𝑜𝑠𝜃

𝑅2

2𝑑𝜃

𝜋/2

0

𝑅2

2𝑑𝜃

𝜋/2

0

=

23𝑅 𝑐𝑜𝑠𝜃𝑑𝜃

𝜋/2

0

𝑑𝜃𝜋/2

0

=4𝑅

3𝜋


𝐴

𝑑𝐴

𝐴

=

23𝑅𝑠𝑖𝑛𝜃

𝑅2

2𝑑𝜃

𝜋/2

0

𝑅2

2𝑑𝜃

𝜋/2

0

=

23𝑅 𝑠𝑖𝑛𝜃𝑑𝜃

𝜋/2

0

𝑑𝜃𝜋/2

0

=4𝑅

3𝜋



Integrations


- Example 9.5 Locate the centroid for the area

Solution 1


Arbitrary element (𝑦,𝑑𝑥) at 𝑥

Area: 𝑑𝐴 = 𝑦𝑑𝑥

Moment Arms: 𝑥 = 𝑥, 𝑦 = 𝑦/2

Integration


𝐴

𝑑𝐴

𝐴

= 𝑥𝑦𝑑𝑥

1

0

𝑦𝑑𝑥1

0

= 𝑥3𝑑𝑥

1

0

𝑥2𝑑𝑥1

0

=0.250

0.333= 0.75𝑚


𝐴

𝑑𝐴

𝐴

= (𝑦/2)𝑦𝑑𝑥

1

0

𝑦𝑑𝑥1

0

= (𝑥2/2)𝑥2𝑑𝑥

1

0

𝑥2𝑑𝑥1

0

=0.100

0.333= 0.3𝑚




- Example 9.5 Locate the centroid for the area

Solution 2


Arbitrary element (1 − 𝑥,𝑑𝑦) at (𝑥, 𝑦)

Area: 𝑑𝐴 = (1 − 𝑥)𝑑𝑦

Moment Arms: 𝑥 = 𝑥 +1−𝑥

2=

1+𝑥

2 𝑦 = 𝑦

Integration


𝐴

𝑑𝐴

𝐴

= [ 1 + 𝑥 /2](1 − 𝑥)𝑑𝑦

1

0

(1 − 𝑥)𝑑𝑦1

0

=0.250

0.333= 0.75𝑚


𝐴

𝑑𝐴

𝐴

= 𝑦(1− 𝑥)𝑑𝑦

1

0

(1− 𝑥)𝑑𝑦1

0

= 𝑦 − 𝑦

32 𝑑𝑦

1

0

1 − 𝑦 𝑑𝑦1

0

=0.100

0.333= 0.3𝑚




- Example 9.6 Locate the centroid of the semi-elliptical area

Solution


Arbitrary element (𝑑𝑥,𝑦)

Area: 𝑑𝐴 = 𝑦𝑑𝑥

Moment Arms: 𝑥 = 𝑥, 𝑦 = 𝑦/2

Integration

𝑥 = 0


𝐴

𝑑𝐴

𝐴

=

𝑦2(1− 𝑥)𝑑𝑦

2

−2

𝑦𝑑𝑥2

−2

= 1 −

𝑥2

4𝑑𝑥

2

−2

1 −𝑥2

4𝑑𝑥

2

−2

=4/3

𝜋= 0.424




- Example 9.7 Locate the 𝑦 centroid for

the paraboloid of revolution

Solution


Arbitrary element: thin disk

Volume: 𝑑𝑉 = 𝜋𝑧2 𝑑𝑦

Moment Arms: 𝑦 = 𝑦

Integration

𝑥 = 0

𝑦 = 𝑦 𝑑𝑉

𝑉

𝑑𝑉

𝑉

= 𝑦(𝜋𝑧2)𝑑𝑦

100

0

(𝜋𝑧2)𝑑𝑦100

0

=100𝜋 𝑦2𝑑𝑦

100

0

100𝜋 𝑦𝑑𝑦100

0

= 66.7𝑚𝑚




- Example 9.8 Determine the location of the center of mass of

the cylinder if its density varies directly with the distance from

its base, i.e., 𝜌 = 200𝑧𝑘𝑔/𝑚3

Solution


Arbitrary element: disk element thickness 𝑑𝑧

Volume: 𝑑𝑉 = 𝜋0.52𝑑𝑧

Moment Arms: 𝑧 = 𝑧

Integration

𝑥 = 0, 𝑦 = 0

𝑧 = 𝑧 𝑑𝑉

𝑉

𝜌𝑑𝑉

𝑉

= 𝑧 200𝑧 [𝜋 0.52 ]𝑑𝑧

1

0

200𝑧 [𝜋 0.52 ]𝑑𝑧1

0

= 𝑧2𝑑𝑧

1

0

𝑧𝑑𝑧1

0

= 0.667𝑚



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Fundamental Problems

- F9.1 Determine the centroid (𝑥 , 𝑦 ) of the shaded area




- F9.2 Determine the centroid (𝑥 , 𝑦 ) of the shaded area




- F9.3 Determine the centroid 𝑦 of the shaded area




- F9.4 Locate the center mass 𝑥 of the straight rod if its mass

per unit length is given by 𝑚 = 𝑚0(1 + 𝑥2/𝐿2)




- F9.5 Locate the centroid 𝑦 of the homogeneous solid formed

by revolving the shaded area about the 𝑦 axis




- F9.6 Locate the centroid 𝑧 of the homogeneous solid formed

by revolving the shaded area about the 𝑧 axis



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§2. Composite Body

- A composite body consists of a series of

connected “simpler” shaped bodies, which

may be rectangular, triangular,

semicircular, …

- Such a body can often be sectioned or

divided into its composite parts and,

provided the weight and location of the

center of gravity of each of these parts are

known

- We can then eliminate the need for

integration to determine the center of

gravity for the entire body



§2. Composite Body

- The center of gravity of 𝐺

𝑥 =∑𝑥 𝑊

∑𝑊, 𝑦 =

∑𝑦 𝑊

∑𝑊, 𝑧 =

∑𝑧 𝑊

∑𝑊

𝑥 , 𝑦 , 𝑧 : the coordinates of the center of gravity 𝐺 of the

composite body

𝑊: the weights of the composite parts of the body

𝑥 , 𝑦 , 𝑧 : the coordinates of the center of gravity of each

composite part of the body

- The Centroid for composite lines, areas and volumes can be

found using relations analogous to the above one



§2. Composite Body

- Example 9.9 Locate the centroid of the wire

Solution

𝑥 =∑𝑥 𝐿

∑𝐿=

11310

248.5= 45.5𝑚𝑚, 𝑦 =

∑𝑦 𝐿

∑𝐿=

−5600

248.5= −22.5𝑚𝑚,

𝑧 =∑𝑧 𝑊

∑𝑊=

−200

248.5= −0.805𝑚𝑚



§2. Composite Body

- Example 9.9 Locate the centroid of the plate area

Solution

Composite Parts: divide the wire into 3 segments

Moment Arms: the centroid are located as indicated

Summary

𝑥 =∑𝑥 𝐴

∑𝐴=

−4

11.5= −0.348𝑚

𝑦 =∑𝑦 𝐴

∑𝐴=

14

11.5= 1.22𝑚



§2. Composite Body

- Example 9.10 Locate the center of mass of the assembly. The

conical frustum has a density of and the

hemisphere has a density of There is a 25-

𝑚𝑚-radius cylindrical hole in the center of the

frustum

Solution

Composite Parts

(1) + (2) − (3) − (4)



§2. Composite Body

Composite Parts: (1) + (2) − (3) − (4)


Summary



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§2. Composite Body

Composite Parts: (1) + (2) − (3) − (4)


Summary

The center of mass

𝑥 = 0

𝑦 = 0

𝑧 =∑𝑧 𝑚

∑𝑚=

45.815

3.142= 14.6𝑚𝑚




- F9.7 Locate the centroid (𝑥 , 𝑦 , 𝑧 ) of the wire bent in the shape




- F9.8 Locate the centroid 𝑦 of the beam’s cross-sectional area




- F9.9 Locate the centroid 𝑦 of the beam’s cross sectional area




- F9.10 Locate the centroid (𝑥 , 𝑦 ) of the cross-sectional area




- F9.11 Locate the center of mass (𝑥 , 𝑦 , 𝑧 ) of the homogeneous

solid block



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- F9.12 Locate the center of mass (𝑥 , 𝑦 , 𝑧 ) of the homogeneous

solid block



Documents

Ch.09 Center of Gravity and Centroid