1. No category

Chapter Six
A.Lecturer Saddam K. Kwais
Centriods and Centers of Gravity
6/1 Introduction
I.
The earth exerts a gravitational force on each of the particles forming a body.
These forces can be replace by a single equivalent force equal to the weight of
the body and applied at the center of gravity for the body.
II.
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.
III.
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.
6/2 Center of Gravity of a 2D Body
Center of gravity of a plate
∑My
xW = ∑ x ∆W
= ∫ x dW
∑Mx
yW = ∑ y ∆W
= ∫ y dW
6/3 Centroids and First Moments of Areas and Lines
I. Centroid of an area
68
Chapter Six
A.Lecturer Saddam K. Kwais
xW = ∫ x dW
Centriods and Centers of Gravity
Where:
γ = specific weight (weight per unit
volume) of the material
t = thickness of the plate
dA = area of the element
x (γAt ) = ∫ x (γt )dA
xA = ∫ x dA = ∑ x ∆A = Q y
= first moment with respect to y
yA = ∫ y dA = ∑ y ∆A = Q x
= first moment with respect to x
II. Centroid of a line
xW = ∫ x dW
x (γ La ) = ∫ x (γ a )dL
xL = ∫ x dL = ∑ x ∆L
yL = ∫ y dL = ∑ y ∆L
Where:
γ = specific weight of the material.
a = Cross sectional area of the wire.
dL = Length of the element
Notes:
1) 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.
.
69
Chapter Six
A.Lecturer Saddam K. Kwais
Centriods and Centers of Gravity
2) If an area possesses a line of symmetry, its centroid lies on that axis
3) If an area possesses two lines of symmetry, its centroid lies at their
intersection.
4) An area is symmetric with respect to a center O if for every element dA at (x,y)
there exists an area dA’ of equal area at (-x,-y). Thus the centroid of the area
coincides with the center of symmetry.
70
Chapter Six
A.Lecturer Saddam K. Kwais
6/4 Centroids of Common Shapes of Areas
71
Centriods and Centers of Gravity
Chapter Six
A.Lecturer Saddam K. Kwais
Centriods and Centers of Gravity
6/5 Composite Plates and Areas
1. Composite plates
X ∑W = ∑ x W
Y ∑W = ∑ y W
2. Composite area
X∑A= ∑xA
Y ∑ A = ∑ yA
EXAMPLE 1. For the plane area shown, determine
the first moments with respect to the x and y axes and
the location of the centroid.
SOLUTION:
Divide the area into a triangle, rectangle, and semicircle with a circular cutout.
72
Chapter Six
A.Lecturer Saddam K. Kwais
Centriods and Centers of Gravity
Q x = ∑ y A = +506.2 × 103 mm 3
Q y = ∑ x A = +757.7 × 103 mm 3
x A + 757.7 × 103 mm 3
∑
X=
=
⇒ X = 54.8 mm
∑ A 13.828 × 103 mm 2
Y
y A + 506.2 × 103 mm 3
∑
=
=
⇒Y
∑ A 13.828 × 103 mm 2
= 36.6 mm Ans.
EXAMPLE 2. The figure shown is made from a
piece of thin, homogeneous wire. Determine the
location of its center of gravity.
SOLUTION:
x=
y=
Ans.
∑ xL 600
=
=10in.Ans.
∑L
60
∑ yL 180
=
=3in.Ans.
∑L
60
73
Chapter Six
A.Lecturer Saddam K. Kwais
Centriods and Centers of Gravity
A
EXAMPLE 3. A uniform semicircular rod of
r
weight W and radius r is attached to a pin at A
O
and rests against a frictionless surface at B.
B
Determine the reactions at A and B.
Ay
SOLUTION:
↺ = 0
2 =!
Ax
2
=0
2r
⇒ = → $%&.
W
$( ! = 0 ⇒ $( = ⇒ $( = ← $%&.
↑ ', = 0
$, = 0 ⇒ $, = ↑
$ = - . ! tan 3 =
.
⇒ $ = -1 ! = ⇒ 3 = 72.34°
4
$ = 1.049
G
B
→ '( = 0
∴ $( =
2
1
.
= 1.049$%&.
= 0.328
74
Ans.
Chapter Six
A.Lecturer Saddam K. Kwais
Centriods and Centers of Gravity
6/6 Determination of Centroids by Integration
xA = ∫ x dA = ∫∫ x dxdy = ∫ xel dA
yA = ∫ y dA = ∫∫ y dxdy = ∫ yel dA
Double integration to find the first moment
may be avoided by defining dA as a thin
rectangle or strip.
xA = ∫ xel dA
x A = ∫ x el dA
= ∫ x ( ydx )
a+x
[ (a − x)dx]
2
yA = ∫ yel dA
=∫
y A = ∫ y el dA
y
= ∫ ( ydx )
2
= ∫ y [(a − x)dx]
EXAMPLE 4. Determine by direct integration the
location of the centroid of a parabolic spandrel.
SOLUTION:
• Determine the constant k.
y = k x2
b = k a2 ⇒ k =
y=
b
a2
x2
or
b
a2
x=
a
b1 2
y1 2
75
xA = ∫ xel dA
=∫
2r
1

cosθ  r 2 dθ 
3
2

yA = ∫ yel dA
=∫
2r
1

sinθ  r 2 dθ 
3
2

Chapter Six
A.Lecturer Saddam K. Kwais
Centriods and Centers of Gravity
• Evaluate the total area.
A = ∫ dA
a
 b x3 
2
x dx = 
= ∫ y dx = ∫

2
2 3
a
a

 0

0
ab
=
3
a
b
• Using either vertical or horizontal strips, perform a single integration to find the
first moments.
a
 b 2
Q y = ∫ xel dA = ∫ xydx = ∫ x 
x dx
2
a


0
a
 b x4 
a 2b
=
 =
2
4
 a 4  0
2
a
y
1 b 2
Q x = ∫ yel dA = ∫ y dx = ∫ 
x  dx
2
2
2

0 a
a
 b2 x5 
ab 2
=
 =
4 5
2
a

 0 10
• Evaluate the centroid coordinates.
xA = Q y
ab a 2 b
x =
3
4
3
x = a Ans.
4
yA = Q x
ab ab 2
y =
3
10
y=
3
b Ans.
10
76
Chapter Six
A.Lecturer Saddam K. Kwais
Centriods and Centers of Gravity
EXAMPLE 5. Determine the location of the
r
centroid of the arc of circle shown.
α
0
α
SOLUTION:
y
Since the arc is symmetrical with respect to the x axis,
7
8 = 0. A differential element is chosen as shown, and
θ=α
r
dθ
the length of the arc is determined by integration.
=
=
0
9 = : ;9 = : ;< = : ;< = 23
>=
dL = r dθ
θ
x = r cosθ
>=
The first moment of the arc with respect to the y axis is
=
=
θ=α
?, = : @;9 = : cos <;< = : cos < ;<
.
>=
?, = . Csin <D=>= = 2 . sin 3
Since ?, = @̅ 9, we write
@̅ 23 = 2 . sin 3 ⇒ @̅ =
>=
sin 3
$%&.
3
6/7 Theorems of Pappus-Guldinus
• Surface of revolution is generated by rotating a plane curve about a fixed axis.
77
x
Chapter Six
A.Lecturer Saddam K. Kwais
Centriods and Centers of Gravity
• 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.
A = 2π yL
• Body of revolution is generated by rotating a plane area about a fixed axis.
• Volume of a body of revolution is equal to the generating area times the distance
traveled by the centroid through the rotation.
78
Chapter Six
A.Lecturer Saddam K. Kwais
Centriods and Centers of Gravity
EXAMPLE 6. 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 ρ = 7.85 × 103 kg m 3
determine the mass and weight of the rim.
SOLUTION:
1. Apply the theorem of Pappus-
Guldinus to evaluate the volumes or
revolution for the rectangular rim
section and the inner cutout section.
2. Multiply by density and acceleration
to get the mass and acceleration.
(
)(
)
3

m = ρV = 7.85 × 103 kg m 3 7.65 × 106 mm 3 10 − 9 m 3 mm  ⇒ m = 60.0 kg


(
)
W = mg = (60.0 kg ) 9.81 m s 2 ⇒ W = 589 N
79