ch07-distributed-forces-centroids-and-centers-of-gravity

ighth
dition
Vector Mechanics for Engineers: Statics
CE 102 Statics
Chapter 7
Distributed Forces:
Centroids and Centers of Gravity
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 1

ighth
dition
Vector Mechanics for Engineers: Statics
Contents
Introduction
Center of Gravity of a 2D Body
Centroids and First Moments of Areas
and Lines
Centroids of Common Shapes of Areas
Centroids of Common Shapes of Lines
Composite Plates and Areas
Sample Problem 7.1
Determination of Centroids by
Integration
Sample Problem 7.2
Theorems of Pappus-Guldinus
Sample Problem 7.3
Distributed Loads on Beams
Sample Problem 7.4
Center of Gravity of a 3D Body:
Centroid of a Volume
Centroids of Common 3D Shapes
Composite 3D Bodies
Sample Problem 7.5
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 2

ighth
dition
Vector Mechanics for Engineers: Statics
Introduction
• 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.
• 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.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 3

ighth
dition
Vector Mechanics for Engineers: Statics
Center of Gravity of a 2D Body
• Center of gravity of a plate
• Center of gravity of a wire
M
y
xW
=
xW
M
y
yW
=
=
x dW
yW
=
y dW
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 4

ighth
dition
Vector Mechanics for Engineers: Statics
Centroids and First Moments of Areas and Lines
• Centroid of an area
• Centroid of a line
x
xW = x dW
( At) = x ( t)
xA = x dA
dA
= Q
y
= first moment with respect to
yA = y dA = Qx
= first moment with respect to
y
x
x
xW = x dW
( La) = x ( a)
xL = x dL
yL = y dL
dL
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 5

ighth
dition
Vector Mechanics for Engineers: Statics
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
area dA’ of equal area at (-x,-y).
• The centroid of the area coincides with the
center of symmetry.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 6

ighth
dition
Vector Mechanics for Engineers: Statics
Centroids of Common Shapes of Areas
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 7

ighth
dition
Vector Mechanics for Engineers: Statics
Centroids of Common Shapes of Lines
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 8

ighth
dition
Vector Mechanics for Engineers: Statics
Composite Plates and Areas
• Composite plates
X
Y
W
W
=
=
xW
yW
• Composite area
X
Y
A =
A =
x A
yA
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 9

ighth
dition
Vector Mechanics for Engineers: Statics
Sample Problem 7.1
SOLUTION:
• Divide the area into a triangle, rectangle,
and semicircle with a circular cutout.
• Calculate the first moments of each area
with respect to the axes.
For the plane area shown, determine
the first moments with respect to the
x and y axes and the location of the
centroid.
• Find the total area and first moments of
the triangle, rectangle, and semicircle.
Subtract the area and first moment of the
circular cutout.
• Compute the coordinates of the area
centroid by dividing the first moments by
the total area.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 10

ighth
dition
Vector Mechanics for Engineers: Statics
Sample Problem 7.1
• Find the total area and first moments of the
triangle, rectangle, and semicircle. Subtract the
area and first moment of the circular cutout.
Q
Q
x
y
= + 506.210
= + 757.7 10
3
3
mm
mm
3
3
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 11

ighth
dition
Vector Mechanics for Engineers: Statics
Sample Problem 7.1
• Compute the coordinates of the area
centroid by dividing the first moments by
the total area.
X
=
x A
A
=
+ 757.7 10
13.82810
3
3
mm
mm
3
2
X
= 54.8 mm
Y
=
yA
A
=
+ 506.210
13.82810
3
3
mm
mm
3
2
Y
= 36.6 mm
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 12

ighth
dition
Vector Mechanics for Engineers: Statics
Determination of Centroids by Integration
xA =
yA =
xdA =
ydA =
x dxdy
y dxdy
=
=
x
y
el
el
dA
dA
• Double integration to find the first moment
may be avoided by defining dA as a thin
rectangle or strip.
xA
yA
=
=
=
=
x
x
y
y
2
el
dA
( ydx)
el
dA
( ydx)
xA
yA
=
=
=
=
x
a + x
2
y dA
y
el
el
dA
( a − x)
( a − x)
dx
dx
xA
yA
=
=
=
=
x
y
el
el
dA
2r
1
cos
3 2
dA
2r
1
sin
3 2
r
r
2
2
d
d
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 13

ighth
dition
Vector Mechanics for Engineers: Statics
Sample Problem 7.2
SOLUTION:
• Determine the constant k.
• Evaluate the total area.
Determine by direct integration the
location of the centroid of a parabolic
spandrel.
• Using either vertical or horizontal
strips, perform a single integration to
find the first moments.
• Evaluate the centroid coordinates.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 14

ighth
dition
Vector Mechanics for Engineers: Statics
Sample Problem 7.2
SOLUTION:
• Determine the constant k.
y = k x
b = k a
b
y =
2
a
2
2
x
k =
2
a
b
2
or x =
b
a
1 2
y
1
2
• Evaluate the total area.
A = dA
=
ab
=
3
y dx =
a
0
a
b
2
x
2
dx
b
=
a
2
x
3
a
3 0
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 15

ighth
dition
Vector Mechanics for Engineers: Statics
Sample Problem 7.2
• Using vertical strips, perform a single integration
to find the first moments.
Q
Q
y
x
=
b
=
a
=
x
y
el
2
el
2
b
=
2a
dA =
x
4
4
4
5
x
5
a
0
dA =
a
0
xydx
2
=
a b
=
4
2
ab
=
10
a
0
y
ydx =
2
b
x a
a
0
1
2
2
a
b
x
2
2
x
dx
2
2
dx
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 16

ighth
dition
Vector Mechanics for Engineers: Statics
Sample Problem 7.2
• Or, using horizontal strips, perform a single
integration to find the first moments.
Q
Q
y
x
=
=
=
=
1
2
b
0
x
b
0
y
el
dA =
a
el
2
dA =
a
ay
−
b
1 2
a + x
2
2
a
−
b
y
( )
y
ydy
a − x
3
2
( )
a − x
dy
dy
2
dy
a b
=
4
=
2
ab
=
10
=
b
0
a
2
− x
2
a
ya
−
b
1 2
2
y
dy
1 2
dy
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 17

ighth
dition
Vector Mechanics for Engineers: Statics
Sample Problem 7.2
• Evaluate the centroid coordinates.
xA
=
Q
y
x
ab
3
=
a 2 b
4
x
=
3
4
a
yA
ab
y
3
=
=
Q
x
2
ab
10
3
y =
10
b
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 18

ighth
dition
Vector Mechanics for Engineers: Statics
Theorems of Pappus-Guldinus
• Surface of revolution is generated by rotating a
plane curve about a fixed axis.
• 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 = 2
yL
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 19

ighth
dition
Vector Mechanics for Engineers: Statics
Theorems of Pappus-Guldinus
• 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.
V = 2
yA
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 20

ighth
dition
Vector Mechanics for Engineers: Statics
Sample Problem 7.3
SOLUTION:
• Apply the theorem of Pappus-Guldinus
to evaluate the volumes or revolution
for the rectangular rim section and the
inner cutout section.
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.
= 7.8510
3
kg
m
3
• Multiply by density and acceleration
to get the mass and acceleration.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 21

ighth
dition
Vector Mechanics for Engineers: Statics
Sample Problem 7.3
SOLUTION:
• 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.
m = V =
W = mg
=
(
3 3)( 6 3) −9
3
7.8510
kg m 7.6510
mm 10
m
3 mm
( )(
2
)
60.0 kg 9.81m s
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
m = 60.0 kg
W = 589 N
5 - 22

ighth
dition
Vector Mechanics for Engineers: Statics
Distributed Loads on Beams
W
L
wdx = dA =
= A
0
• 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 (dW = wdx).
( )
OP W
=
L
xdW
( OP) A = xdA = xA
0
• A distributed load can be replace 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.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 23

ighth
dition
Vector Mechanics for Engineers: Statics
Sample Problem 7.4
SOLUTION:
A beam supports a distributed load as
shown. Determine the equivalent
concentrated load and the reactions at
the supports.
• The magnitude of the concentrated load
is equal to the total load or the area under
the curve.
• The line of action of the concentrated
load passes through the centroid of the
area under the curve.
• Determine the support reactions by
summing moments about the beam
ends.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 24

ighth
dition
Vector Mechanics for Engineers: Statics
Sample Problem 7.4
SOLUTION:
• The magnitude of the concentrated load is equal to
the total load or the area under the curve.
F
= 18.0 kN
• The line of action of the concentrated load passes
through the centroid of the area under the curve.
X
63 kN
m
=
18 kN
X = 3.5 m
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 25

ighth
dition
Vector Mechanics for Engineers: Statics
Sample Problem 7.4
• Determine the support reactions by summing
moments about the beam ends.
( 6 m) − ( 18 kN)( 3.5 m) 0
M A = 0 : B
=
y
B y
= 10.5 kN
( 6 m) + ( 18 kN)( 6 m − 3.5 m) 0
M B = 0 : − A
=
y
A y
= 7.5 kN
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 26

ighth
dition
Vector Mechanics for Engineers: Statics
Center of Gravity of a 3D Body: Centroid of a Volume
• Center of gravity G
r
r
G
G
W
W
−W j
=
( − W
j )
( −W j ) = r
( − W j )
( − j ) = ( rW
) ( − j )
= rdW
dW rGW
=
• Results are independent of body orientation,
xW
xV
xdW yW = ydW zW =
= zdW
• For homogeneous bodies,
W = V and dW = dV
xdV yV = ydV zV =
= zdV
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 27

ighth
dition
Vector Mechanics for Engineers: Statics
Centroids of Common 3D Shapes
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 28

ighth
dition
Vector Mechanics for Engineers: Statics
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.
X W = xW Y W = yW Z W = zW
• For homogeneous bodies,
X V = xV Y V = yV Z V = zV
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 29

ighth
dition
Vector Mechanics for Engineers: Statics
Sample Problem 7.5
SOLUTION:
• Form the machine element from a
rectangular parallelepiped and a
quarter cylinder and then subtracting
two 1-in. diameter cylinders.
Locate the center of gravity of the
steel machine element. The diameter
of each hole is 1 in.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 30

ighth
dition
Vector Mechanics for Engineers: Statics
Sample Problem 7.5
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 31

ighth
dition
Vector Mechanics for Engineers: Statics
Sample Problem 7.5
X
= xV V
=
( ) ( )
3.08in
4
5.286 in
3
X = 0.577 in.
Y
= yV V
=
( ) ( )
− 5.047 in
4
5.286 in
3
Y = 0.577 in.
Z
= zV V
=
( ) ( )
1.618in
4
5.286 in
3
Z = 0.577 in.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
5 - 32

Problem 7.6
y
20 mm 30 mm
Locate the centroid of the plane
area shown.
36 mm
24 mm
x
33

y
20 mm 30 mm
36 mm
24 mm
Problem 7.6
Solving Problems on Your Own
Locate the centroid of the plane area
shown.
Several points should be emphasized
when solving these types of problems.
x
1. Decide how to construct the given area from common shapes.
2. It is strongly recommended that you construct a table
containing areas or length and the respective coordinates of
the centroids.
3. When possible, use symmetry to help locate the centroid.
34

C 1
C 2
y
20 + 10
Problem 7.6 Solution
Decide how to construct the given
area from common shapes.
30
24 + 12
10
Dimensions in mm
x
35

y
20 + 10
C 1
Problem 7.6 Solution
Construct a table containing areas and
respective coordinates of the
centroids.
C 2
Dimensions in mm
30
24 + 12
10
x
A, mm 2 x, mm y, mm xA, mm 3 yA, mm 3
1 20 x 60 =1200 10 30 12,000 36,000
2 (1/2) x 30 x 36 =540 30 36 16,200 19,440
S 1740 28,200 55,440
36

y
20 + 10
Then
Problem 7.6 Solution
XS A = S xA
X (1740) = 28,200
C 1
or
X = 16.21 mm
C 2
Dimensions in mm
30
24 + 12
and
YS A = S yA
10
x
Y (1740) = 55,440
or Y = 31.9 mm
A, mm 2 x, mm y, mm xA, mm 3 yA, mm 3
1 20 x 60 =1200 10 30 12,000 36,000
2 (1/2) x 30 x 36 =540 30 36 16,200 19,440
S 1740 28,200 55,440
37

Problem 7.7
A
w A
a
24 kN 30 kN
1.8 m
0.3 m
B
w B
The beam AB supports two
concentrated loads and
rests on soil which exerts a
linearly distributed upward
load as shown. Determine
(a) the distance a for which
w A = 20 kN/m, (b) the
corresponding value w B .
38

Problem 7.7
A
w A
a
24 kN 30 kN
1.8 m
0.3 m
B
w B
Solving Problems on Your Own
The beam AB supports two
concentrated loads and
rests on soil which exerts a
linearly distributed upward
load as shown. Determine
(a) the distance a for which
w A = 20 kN/m, (b) the
corresponding value w B .
1. Replace the distributed load by a single equivalent force.
The magnitude of this force is equal to the area under the
distributed load curve and its line of action passes through
the centroid of the area.
2. When possible, complex distributed loads should be
divided into common shape areas.
39

A
20 kN/m
a
24 kN 30 kN
C
0.3 m
B
w B
Problem 7.7 Solution
Replace the distributed
load by a pair of
equivalent forces.
0.6 m 0.6 m
R I
R II
1
We have R I = (1.8 m)(20 kN/m) = 18 kN
R II =
2
1
2
(1.8 m)(w B kN/m) = 0.9 w B kN
40

a
24 kN 30 kN
0.3 m
Problem 7.7 Solution
A
C
B
w B
0.6 m 0.6 m
R I = 18 kN
R II = 0.9 w B kN
(a)
+
SM C = 0: (1.2 - a)m x 24 kN - 0.6 m x 18 kN
- 0.3m x 30 kN = 0
or a = 0.375 m
(b) + SF y = 0: -24 kN + 18 kN + (0.9 w B ) kN - 30 kN= 0
or w B = 40 kN/m
41

Problem 7.8
3 in
2 in
2 in
y
1 in
r = 1.25 in
x
For the machine element
shown, locate the z coordinate
of the center of gravity.
z
0.75 in
2 in
2 in
r = 1.25 in
42

3 in
2 in
2 in
y
1 in
r = 1.25 in
x
Problem 7.8
Solving Problems on Your Own
For the machine element
shown, locate the z coordinate
of the center of gravity.
z
0.75 in Determine the center of
2 in
gravity of composite body.
r = 1.25 in
2 in
For a homogeneous body
the center of gravity coincides
with the centroid of its volume. For this case the center of gravity
can be determined by
X S V = S x V Y S V = S y V Z S V = S z V
where X, Y, Z and x, y, z are the coordinates of the centroid of the
body and the components, respectively.
43

3 in
2 in
2 in
y
1 in
r = 1.25 in
Problem 7.8 Solution
Determine the center of gravity
of composite body.
z
2 in
2 in
r = 1.25 in
Divide the body into
five common shapes.
0.75 in
x
II
First assume that the machine
element is homogeneous so
that its center of gravity will
coincide with the centroid of
the corresponding volume.
III
V
y
IV
I
x
z
44

z
II
III
V
y
IV
I
x
3 in
2 in
2 in
r = 1.25 in
x
z
0.75 in
2 in
2 in
y
1 in
r = 1.25 in
V, in 3 z, in. z V, in 4
I (4)(0.75)(7) = 21 3.5 73.5
II (/2)(2) 2 (0.75) = 4.7124 7+ [(4)(2)/(3)] = 7.8488 36.987
III -(11.25) 2 (0.75)= -3.6816 7 -25.771
IV (1)(2)(4) = 8 2 16
V -(/2)(1.25) 2 (1) = -2.4533 2 -4.9088
S 27.576 95.807
Z S V = S z V : Z (27.576 in 3 ) = 95.807 in 4
Z = 3.47 in
45

Problem 7.9
y
y = kx 1/3
a
Locate the centroid of the volume
obtained by rotating the shaded area
about the x axis.
h
x
46

y
y = kx 1/3
h
a
x
Solving Problems on Your Own
Locate the centroid of the volume
obtained by rotating the shaded area
about the x axis.
The procedure for locating the
centroids of volumes by direct
integration can be simplified:
Problem 7.9
1. When possible, use symmetry to help locate the centroid.
2. If possible, identify an element of volume dV which produces
a single or double integral, which are easier to compute.
3. After setting up an expression for dV, integrate and determine
the centroid.
47

y
x
dx
Problem 7.9 Solution
Use symmetry to help locate the
centroid. Symmetry implies
y = 0 z = 0
z
y = kx 1/3
r
x
Identify an element of volume dV
which produces a single or
double integral.
Choose as the element of volume a disk or radius r and
thickness dx. Then
dV = r 2 dx
x el = x
48

y
x
dx
Problem 7.9 Solution
Identify an element of volume dV
which produces a single or
double integral.
z
r
x
dV = r 2 dx
x el = x
Now r = kx 1/3 so that
y = kx 1/3
dV = k 2 x 2/3 dx
At x = h, y = a : a = kh 1/3 or k = a/h 1/3
Then dV = x 2/3 dx
a 2
h 2/3
49

y
x
dx
Problem 7.9 Solution
Integrate and determine the centroid.
dV =
a 2
h 2/3
x 2/3 dx
z
y = kx 1/3
r
x
V = x 2/3 dx
0
h
a 2
a 2
h 2/3
= x 5/3
3
h 2/3 5
3
=
5
a 2 h
[ ]
h
0
Also
x el dV = x ( x 2/3 dx) = [ x 8/3 ]
0
h
a 2
h 2/3 a 2
3
h 2/3 8
= a 2 h 2 3
8
50

y
x
dx
Problem 7.9 Solution
Integrate and determine the centroid.
V =
3
5
a 2 h
z
r
x
3
x el dV = a 2 h
8
2
y = kx 1/3
Now
xV = xdV:
3
5
3
8
x ( a 2 h) = a 2 h 2
x =
5
8
h
y = 0 z = 0
51

Problem 7.10
d
1.8 ft
30 o
A
B
The square gate AB is held in the
position shown by hinges along its
top edge A and by a shear pin at B.
For a depth of water d = 3.5 ft,
determine the force exerted on the
gate by the shear pin.
52

Problem 7.10
Solving Problems on Your Own
d
1.8 ft
30 o
A
B
The square gate AB is held in the
position shown by hinges along its
top edge A and by a shear pin at B.
For a depth of water d = 3.5 ft,
determine the force exerted on the
gate by the shear pin.
Assuming the submerged body has a width b, the load per unit
length is w = bgh, where h is the distance below the surface of
the fluid.
1. First, determine the pressure distribution acting perpendicular
the surface of the submerged body. The pressure distribution
will be either triangular or trapezoidal.
53

Problem 7.10
Solving Problems on Your Own
d
1.8 ft
30 o
A
B
The square gate AB is held in the
position shown by hinges along its
top edge A and by a shear pin at B.
For a depth of water d = 3.5 ft,
determine the force exerted on the
gate by the shear pin.
2. Replace the pressure distribution with a resultant force, and
construct the free-body diagram.
3. Write the equations of static equilibrium for the problem, and
solve them.
54

Problem 7.10 Solution
A
1.7 ft
Determine the pressure distribution
acting perpendicular the surface of the
submerged body.
P A
P B
(1.8 ft) cos 30 o
P A = 1.7 g
P B = (1.7 + 1.8 cos 30 o )g
B
55

L AB /3
P 1 =
P 2 =
L AB /3
L AB /3
1
2
1
2
1.7 g
A
P 1
P 2
A y
(1.7 + 1.8 cos 30 o )g
Problem 7.10 Solution
A x
Replace the pressure
distribution with a
resultant force, and
(1.8 ft) cos 30 o construct the free-body
diagram.
F B
The force of the water
B
on the gate is
1 1
P =
2
Ap = A(gh)
2
(1.8 ft) 2 (62.4 lb/ft 3 )(1.7 ft) = 171.85 lb
(1.8 ft) 2 (62.4 lb/ft 3 )(1.7 + 1.8 cos 30 o )ft = 329.43 lb
56

L AB /3
1.7 g
A
A y
A x
F B
(1.8 ft) cos 30 o
Problem 7.10 Solution
L AB /3
L AB /3
P 1
P 2
Write the equations of
static equilibrium for the
problem, and solve them.
B
(1.7 + 1.8 cos 30 o )g
P 1 = 171.85 lb P 2 = 329.43 lb
+
S M A = 0:
1
3
2
3
( L AB )P 1 + ( L AB )P 2
- L AB F B = 0
1
3
2
(171.85 lb) + 3 (329.43 lb) - F B = 0 F B = 276.90 lb
F B = 277 lb
30 o
57