Chapter 05 - Distributed Forces-Centroids and Centers of Gravity.pdf

ME 1222
APPLIED MECHANICS
DISTRIBUTED FORCES: CENTROIDS
AND CENTERS OF GRAVITY
Department of Mechanical Engineering, General Sir John Kotelawala Defence University
CHAPTER 05

COURSE OUTLINE
Department of Mechanical Engineering, General Sir John Kotelawala Defence University
1. Introduction
2. Statics of Objects
3. Rigid Bodies: Equivalent System of Forces
4. Equilibrium of Rigid Bodies
5. Distributed Forces: Centre of Gravity
6. Friction
7. Introduction to Vibration

3
CONTENT
3
CHAPTER 5 – Distributed Forces: Centroids and Centers of Gravity
1. Introduction
2. Center of Gravity of a 2D Body
3. Centroids and First Moments of Areas and Lines
4. Centroids of Common Shapes of Areas
5. Centroids of Common Shapes of Lines
6. Composite Plates and Areas
7. Determination of Centroids by Integration
8. Theorems of Pappus-Guldinus
9. Distributed Loads on Beams
10. Center of Gravity of a 3D Body: Centroid of a Volume
11. Centroids of Common 3D Shapes
12. Composite 3D Bodies

Department of Mechanical Engineering, General Sir John Kotelawala Defence University 5
1. INTRODUCTION
•The earth exerts a gravitational force on each of the particles forming a
body.
•These forces can be replaced by a single equivalent force equal to the
weight of the body and applied at the center of gravity for the body.
•The centroid of an area is analogous to the center of gravity of a body.
The concept of the first moment of an area is used to locate the centroid.
•Determination of the area of a surface of revolution and the volume of a
body of revolution are accomplished with the Theorems of Pappus-
Guldinus.

2. CENTER OF GRAVITY OF A 2D BODY
• Center of Gravity of a Plate
σ 𝑀𝑥 : ത
𝑦𝑊= 𝑦1∆𝑊1 + 𝑦2∆𝑊2 + … . . +𝑦𝑛∆𝑊
𝑛
σ 𝑀𝑦 : ҧ
𝑥𝑊= 𝑥1∆𝑊1 + 𝑥2∆𝑊2 + … . . +𝑥𝑛∆𝑊
𝑛
Taking moments of W about the y and x axes:
ҧ
𝑥 =
𝑥1∆𝑊1 + 𝑥2∆𝑊2+ . . . +𝑥𝑛∆𝑊
𝑛
𝑊
ത
𝑦 =
𝑦1∆𝑊1 + 𝑦2∆𝑊2+ . . . +𝑦𝑛∆𝑊
𝑛
𝑊
Solving equations:
𝑾 = න 𝒅𝑾 ഥ
𝒙 =
‫׬‬ 𝒙 𝒅𝑾
𝑾
ഥ
𝒚 =
‫׬‬ 𝒚 𝒅𝑾
𝑾

2. CENTER OF GRAVITY OF A 2D BODY
• Center of Gravity of a Wire
Contd..
σ 𝑀𝑥 : ത
𝑦𝑊= 𝑦1∆𝑊1 + 𝑦2∆𝑊2 + … . . +𝑦𝑛∆𝑊
𝑛
σ 𝑀𝑦 : ҧ
𝑥𝑊= 𝑥1∆𝑊1 + 𝑥2∆𝑊2 + … . . +𝑥𝑛∆𝑊
𝑛
Taking moments of W about the y and x axes:
ҧ
𝑥 =
𝑥1∆𝑊1 + 𝑥2∆𝑊2+ . . . +𝑥𝑛∆𝑊
𝑛
𝑊
ത
𝑦 =
𝑦1∆𝑊1 + 𝑦2∆𝑊2+ . . . +𝑦𝑛∆𝑊
𝑛
𝑊
Solving equations:
𝑾 = න 𝒅𝑾 ഥ
𝒙 =
‫׬‬ 𝒙 𝒅𝑾
𝑾
ഥ
𝒚 =
‫׬‬ 𝒚 𝒅𝑾
𝑾

3. CENTROIDS AND FIRST MOMENTS OF AREAS AND LINES
• Centroid of an Area
Magnitude ΔW of the weight of an element of the plate
∆W= 𝛾𝑡∆𝐴
Similarly,
W= 𝛾𝑡𝐴
σ 𝑀𝑥 : ത
𝑦𝐴= 𝑦1∆𝐴1 + 𝑦2∆𝐴2 + … . . +𝑦𝑛∆𝐴𝑛
σ 𝑀𝑦 : ҧ
𝑥𝐴= 𝑥1∆𝐴1 + 𝑥2∆𝐴2 + … . . +𝑥𝑛∆𝐴𝑛
Substituting for ∆𝑊 and W in the moment equations and dividing
throughout by 𝛾𝑡,
ഥ
𝒙 =
‫׬‬ 𝒙 𝒅𝑨
𝑨
ഥ
𝒚 =
‫׬‬ 𝒚 𝒅𝑨
𝑨

3. CENTROIDS AND FIRST MOMENTS OF AREAS AND LINES
• Centroid of a Line
Magnitude ΔW of the weight of an element of wire
∆𝑊 = 𝛾𝑎∆𝐿
ഥ
𝒙 =
‫׬‬ 𝒙 𝒅𝑳
𝑳
ഥ
𝒚 =
‫׬‬ 𝒚 𝒅𝑳
𝑳
Contd..

4. FIRST MOMENTS OF AREAS AND LINES
• An area is symmetric with respect to an axis BB’ if for every
point P there exists a point P’ such that PP’ is perpendicular
to BB’ and is divided into two equal parts by BB’.
• The first moment of an area with respect to a line of
symmetry is zero.
• If an area possesses a line of symmetry, its centroid lies on
that axis.
• If an area possesses two lines of symmetry, its centroid lies
at their intersection.
• An area is symmetric with respect to a center O if for every
element dA at (x,y) there exists an equal area of dA’ at (-x,-y).
• The centroid of the area coincides with the center of
symmetry.

5. CENTROIDS OF COMMON SHAPES OF AREAS

5. CENTROIDS OF COMMON SHAPES OF AREAS Contd..

6. CENTROIDS OF COMMON SHAPES OF LINES
Contd..

7. COMPOSITE PLATES AND AREAS
• Composite Plates




=
=
W
y
W
Y
W
x
W
X
• Composite Area




=
=
A
y
A
Y
A
x
A
X

SAMPLE PROBLEM 5.1
For the plane area shown,
(a) Determine the first moments with respect to the x and y axes and the
location of the centroid.

SAMPLE PROBLEM 5.1
STRATEGY:
Break up the given area into simple components, find the centroid of each component, and
then find the overall first moments and centroid
MODELING:
Solution :
Contd..

SAMPLE PROBLEM 5.1
ANALYSIS:
a. First Moments of the Area.
Find the total area and first moments of the triangle, rectangle, and
semicircle. Subtract the area and first moment of the circular cutout
3
3
3
3
mm
10
7
.
757
mm
10
2
.
506

+
=

+
=
y
x
Q
Q
b. Location of Centroid.
•Compute the coordinates of the area centroid by dividing the first
moments by the total area.
2
3
3
3
mm
10
13.828
mm
10
7
.
757


+
=
=


A
A
x
X
mm
8
.
54
=
X
2
3
3
3
mm
10
13.828
mm
10
2
.
506


+
=
=


A
A
y
Y
mm
6
.
36
=
Y
Contd..

( )
 
( )
 
dy
x
a
y
dA
y
A
y
dy
x
a
x
a
dA
x
A
x
el
el
−
=
=
−
+
=
=




2
8. DETERMINATION OF CENTROIDS BY INTEGRATION






=
=
=
=
=
=
dA
y
dy
dx
y
dA
y
A
y
dA
x
dy
dx
x
dA
x
A
x
el
el • Double integration to find the first moment may
be avoided by defining dA as a thin rectangle or
strip.
( )
( )
ydx
y
dA
y
A
y
ydx
x
dA
x
A
x
el
el




=
=
=
=
2






=
=






=
=








d
r
r
dA
y
A
y
d
r
r
dA
x
A
x
el
el
2
2
2
1
sin
3
2
2
1
cos
3
2

SAMPLE PROBLEM 5.4
Determine by direct integration the location of the centroid of a
parabolic spandrel.

SAMPLE PROBLEM 5.4
2
1
2
1
2
2
2
2
2
y
b
a
x
or
x
a
b
y
a
b
k
a
k
b
x
k
y
=
=
=

=
=
Solution : STRATEGY:
First, express the parabolic curve using the parameters a and b. Then
choose a differential element of area and express its area in terms of a, b,
x, and y. We illustrate the solution first with a vertical element and then a
horizontal element.
MODELING:
a. Determination of the Constant k.
Contd..

SAMPLE PROBLEM 5.4
• Choosing the differential element, the total area of the
region is,
3
3
0
3
2
0
2
2
ab
x
a
b
dx
x
a
b
dx
y
dA
A
a
a
=








=
=
=
=



Solution : ANALYSIS :
a. Vertical Differential Element
Contd..

SAMPLE PROBLEM 5.4
• Using vertical strips, perform a single integration to find the first
moments.
10
5
2
2
1
2
4
4
2
0
5
4
2
0
2
2
2
2
0
4
2
0
2
2
ab
x
a
b
dx
x
a
b
dx
y
y
dA
y
Q
b
a
x
a
b
dx
x
a
b
x
dx
xy
dA
x
Q
a
a
el
x
a
a
el
y
=








=






=
=
=
=








=






=
=
=






Contd..

SAMPLE PROBLEM 5.4
• Using horizontal strips, perform a single integration to find
the first moments.
( )
( )
10
4
2
1
2
2
2
0
2
3
2
1
2
1
2
1
2
0
2
2
0
2
2
ab
dy
y
b
a
ay
dy
y
b
a
a
y
dy
x
a
y
dA
y
Q
b
a
dy
y
b
a
a
dy
x
a
dy
x
a
x
a
dA
x
Q
b
el
x
b
b
el
y
=






−
=






−
=
−
=
=
=








−
=
−
=
−
+
=
=








b. Horizontal Differential Element.
Contd..

SAMPLE PROBLEM 5.4
• Evaluate the centroid coordinates.
4
3
2
b
a
ab
x
Q
A
x y
=
=
a
x
4
3
=
10
3
2
ab
ab
y
Q
A
y x
=
=
b
y
10
3
=
Contd..

9. THEOREMS OF PAPPUS-GULDINUS
• Area of a surface of revolution is equal to the length
of the generating curve times the distance traveled
by the centroid through the rotation.
L
y
A 
2
=
• Surface of revolution is generated by rotating a plane curve about a fixed axis.
Theorem 01

9. THEOREMS OF PAPPUS-GULDINUS
• Volume of a body of revolution is equal to the
generating area times the distance traveled by
the centroid through the rotation.
A
y
V 
2
=
• Body of the revolution is generated by rotating a plane area about a fixed axis.
Theorem 02

SAMPLE PROBLEM 5.7
The outside diameter of a pulley is 0.8 m, and the cross-section of its rim is as shown.
Knowing that the pulley is made of steel and that the density of steel is
Determine the mass and weight of the rim.
3
3
m
kg
10
85
.
7 
=


SAMPLE PROBLEM 5.7
• Apply the theorem of Pappus-Guldinus to evaluate the volumes or
revolution for the rectangular rim section and the inner cutout
section.
• Multiply by density and acceleration to get the mass and
acceleration.
( )( ) 







=
= − 3
3
9
3
6
3
3
mm
m
10
mm
10
65
.
7
m
kg
10
85
.
7
V
m  kg
0
.
60
=
m
( )( )
2
s
m
81
.
9
kg
0
.
60
=
= mg
W N
589
=
W
Solution : STRATEGY :
MODELING AND ANALYSIS :
Contd..

10. DISTRIBUTED LOADS ON BEAMS
• A distributed load is represented by plotting the load per unit
length, w (N/m) . The total load is equal to the area under
the load curve.

 =
=
= A
dA
dx
w
W
L
0
( )
( ) A
x
dA
x
A
OP
dW
x
W
OP
L
=
=
=


0
• A distributed load can be replaced by a concentrated load
with a magnitude equal to the area under the load curve and
a line of action passing through the area centroid.

SAMPLE PROBLEM 5.9
A beam supports a distributed load as shown.
Determine the equivalent concentrated load and the reactions at
the supports.

SAMPLE PROBLEM 5.9 Contd..
Solution : STRATEGY:
The magnitude of the resultant of the load is equal to the area under
the load curve, and the line of action of the resultant passes through
the centroid of the same area. Break down the area into pieces for
easier calculation and determine the resultant load.
MODELING and ANALYSIS:
a. Equivalent Concentrated Load.
• The magnitude of the concentrated load is equal to the total load or
the area under the curve.
kN
0
.
18
=
F

Solution :
MODELING and ANALYSIS:
b. Line of Action
• The line of action of the concentrated load passes through the
centroid of the area under the curve.
kN
18
m
kN
63 
=
X m
5
.
3
=
X
c. Reactions
• Determine the support reactions by summing moments about
the beam ends.
( ) ( )( ) 0
m
.5
3
kN
18
m
6
:
0 =
−
=
 y
A B
M
kN
5
.
10
=
y
B
( ) ( )( ) 0
m
.5
3
m
6
kN
18
m
6
:
0 =
−
+
−
=
 y
B A
M
kN
5
.
7
=
y
A

11. CENTER OF GRAVITY OF A 3D BODY: CENTROID OF A VOLUME
• Center of gravity G • Results are independent of body orientation,


 =
=
= zdW
W
z
ydW
W
y
xdW
W
x


 =
=
= zdV
V
z
ydV
V
y
xdV
V
x
dV
dW
V
W 
 =
= and
• For homogeneous bodies,

12. CENTROIDS OF COMMON 3D SHAPES

13. COMPOSITE 3D BODIES
• Moment of the total weight concentrated at the center of
gravity G is equal to the sum of the moments of the
weights of the component parts.





 =
=
= W
z
W
Z
W
y
W
Y
W
x
W
X
• For homogeneous bodies,





 =
=
= V
z
V
Z
V
y
V
Y
V
x
V
X

SAMPLE PROBLEM 5.12
Locate the center of gravity of the steel machine element. The diameter
of each hole is 1 in.

STRATEGY:
This part can be broken down into the sum of two volumes minus two
smaller volumes (holes). Find the volume and centroid of each volume
and combine them using equations to find the overall centroid.
Solution :
The machine part can be obtained by adding a rectangular
parallelepiped (I) to a quarter cylinder (II) and then subtracting two 1-
in.-diameter cylinders (III and IV).
Determine the volume and the coordinates of the centroid of each
component and enter them in a table.
MODELING:

Solution : ANALYSIS:
Centroids of components.

SAMPLE PROBLEM 5.12
( ) ( )
3
4
in
.286
5
in
048
.
3
=
= 
 V
V
x
X
( ) ( )
3
4
in
.286
5
in
5.047
−
=
= 
 V
V
y
Y
( ) ( )
3
4
in
.286
5
in
.555
8
=
= 
 V
V
z
Z
in.
577
.
0
=
X
in.
955
.
0
−
=
Y
in.
618
.
1
=
Z
Contd..

SAMPLE PROBLEM 5.13
A body consisting of a cone and hemisphere of radius r fixed on the same
base rests on a table, the hemisphere being in contact with the table.
Find the greatest height of the cone, so that the combined body may stand
upright.

SAMPLE PROBLEM 5.13

REFERENCES
❖ Vector Mechanics for Engineers – Statics and Dynamics (11th Ed.),
Beer F.P., Johnston E.T. Jr., McGraw-Hill, INC

Chapter 05 - Distributed Forces-Centroids and Centers of Gravity.pdf

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