Chapter 9:
Center of Gravity and Centroid
Engineering Mechanics: Statics
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 area and volume for a surface of
revolution.
To present a method for finding the resultant of a
general distributed loading and show how it applies
to finding the resultant of a fluid.
Chapter Outline
Center of Gravity and Center of Mass for a
System of Particles
 Center of Gravity and Center of Mass and
Centroid for a Body
 Composite Bodies
 Theorems of Pappus and Guldinus
 Resultant of a General Distributed Loading
 Fluid Pressure

9.1 Center of Gravity and Center of
Mass for a System of Particles
Center of Gravity
 Locates the resultant weight of a system of
particles
 Consider system of n particles fixed within a
region of space
 The weights of the particles
comprise a system of parallel
forces which can be replaced
by a single (equivalent) resultant
weight having defined point G
of application
9.1 Center of Gravity and Center of
Mass for a System of Particles
Center of Gravity
 Resultant weight = total weight of n particles
WR  W


Sum of moments of weights of all the
particles about x, y, z axes = moment of
resultant weight about these axes
Summing moments about the x axis,
x WR  ~
x1W1  ~
x2W2  ...  ~
xnWn

Summing moments about y axis,
yWR  ~
y1W1  ~
y2W2  ...  ~
ynWn
9.1 Center of Gravity and Center of
Mass for a System of Particles
Center of Gravity
 Although the weights do not
produce a moment about z axis, by
rotating the coordinate system 90°
about x or y axis with the particles
fixed in it and summing moments
about the x axis,
z WR  ~
z1W1  ~
z2W2  ...  ~
znWn

Generally,
~
xm
~
ym
~
zm
x
;y 
,z 
m
m
m
9.1 Center of Gravity and Center of
Mass for a System of Particles
Center of Gravity
 Where x , y , z represent the coordinates of the center
of gravity G of the system of particles, ~x , ~y , ~z
represent the coordinates of each particle in the
system and W represent the resultant sum of the
weights of all the particles in the system.
 These equations represent a balance between the sum
of the moments of the weights of each particle and
the moment of resultant weight for the system.
9.1 Center of Gravity and Center of
Mass for a System of Particles
Center Mass
 Provided acceleration due to gravity g for every
particle is constant, then W = mg
 ~x m
 ~y m
 ~z m
x
;y 
,z 
m
m
m


By comparison, the location of the center of gravity
coincides with that of center of mass
Particles have weight only when under the influence
of gravitational attraction, whereas center of mass is
independent of gravity
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Center Mass
 A rigid body is composed of an infinite
number of particles
 Consider arbitrary particle
having a weight of dW
x
~x dW

 dW
;y 
~ydW

 dW
;z 
~z dW

 dW
9.2 Center of Gravity and Center
of Mass and Centroid for a Body
Center Mass
 γ represents the specific weight and dW = γdV
~xdV

x V
 dV
V
~ydV

;y V
 dV
V
~z dV

;z  V
 dV
V
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Center of Mass
 Density ρ, or mass per unit volume, is related
to γ by γ = ρg, where g = acceleration due to
gravity
 Substitute this relationship into this equation
to determine the body’s center of mass
 ~xdV
x V
 dV
V
 ~ydV
;y V
 dV
V
 ~zdV
;z  V
 dV
V
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Centroid
 Defines the geometric center of object
 Its location can be determined from
equations used to determine the body’s
center of gravity or center of mass
 If the material composing a body is uniform
or homogenous, the density or specific
weight will be constant throughout the body
 The following formulas are independent of
the body’s weight and depend on the body’s
geometry
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Centroid
Volume
 Consider an object subdivided into volume
elements dV, for location of the centroid,
~x dV

x V
 dV
V
~ydV

;y V
 dV
V
~z dV

;z  V
 dV
V
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Centroid
Area
 For centroid for surface area of an object,
such as plate and shell, subdivide the area
into differential elements dA
~x dA

xA
 dA
A
~ydA

;y  A
 dA
A
~z dA

;z  A
 dA
A
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Centroid
Line
 If the geometry of the object such as a thin rod
or wire, takes the form of a line, the balance of
moments of differential elements dL about each
of the coordinate system yields
~x dL

xL
 dL
L
~ydL

;y  L
 dL
L
~z dL

;z  L
 dL
L
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Line
 Choose a coordinate system that simplifies as
much as possible the equation used to
describe the object’s boundary
Example
 Polar coordinates are appropriate for area
with circular boundaries
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Symmetry
 The centroids of some shapes may be partially or
completely specified by using conditions of
symmetry
 In cases where the shape has an axis of
symmetry, the centroid of the shape must lie
along the line
Example
 Centroid C must lie along the
y axis since for every element
length dL, it lies in the middle
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Symmetry
 For total moment of all the elements about
the axis of symmetry will therefore be
cancelled
~
 x dL  0 x  0

In cases where a shape has 2 or 3 axes of
symmetry, the centroid lies at the
intersection of these axes
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Procedure for Analysis
Differential Element
 Select an appropriate coordinate system, specify
the coordinate axes, and choose an differential
element for integration
 For lines, the element dL is represented as a
differential line segment
 For areas, the element dA is generally a
rectangular having a finite length and differential
width
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Procedure for Analysis
Differential Element
 For volumes, the element dV is either a
circular disk having a finite radius and
differential thickness, or a shell having a
finite length and radius and a differential
thickness
 Locate the element at an arbitrary point (x, y,
z) on the curve that defines the shape
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Procedure for Analysis
Size and Moment Arms
 Express the length dL, area dA or volume dV
of the element in terms of the curve used to
define the geometric shape
 Determine the coordinates or moment arms
for the centroid of the center of gravity of
the element
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Procedure for Analysis
Integrations
 Substitute the formations and dL, dA and dV into
the appropriate equations and perform
integrations
 Express the function in the integrand and in
terms of the same variable as the differential
thickness of the element
 The limits of integrals are defined from the two
extreme locations of the element’s differential
thickness so that entire area is covered during
integration
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Example 9.1
Locate the centroid of
the rod bent into the
shape of a parabolic arc.
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Differential element
 Located on the curve at the arbitrary point (x, y)
Area and Moment Arms
 For differential length of the element dL
dL 

Since x = y2 and then dx/dy = 2y
dL 

dx2  dy 2
2
 dx 
    1 dy
 dy 
2 y 2  1 dy
The centroid is located at
~
x  x, ~
yy
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
x
~
 x dL
L
 dL
L
1

0
1
0
x 4 y  1 dy
2
4 y 2  1 dy
0.6063
 0.410m
1.479
~
 ydL 1 y 4 y 2  1 dy
yL
 01
 dL  4 y 2  1 dy

L

0
0.8484
 0.574m
1.479

1 2
y
0
1

0
4 y 2  1 dy
4 y 2  1 dy
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Example 9.2
Locate the centroid of
the circular wire
segment.
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Differential element
 A differential circular arc is selected
 This element intersects the curve at (R, θ)
Length and Moment Arms
 For differential length of the element
dL  Rd

For centroid,
~
x  R cos 
~
y  R sin 
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
 ~x dL  / 2 R cos  R d
x
L
 dL
L


0
 /2
0
R d

R2 
 /2
0
R
cos  d
 /2
0
d
2R

 ~ydL  / 2 R sin  R d R 2  / 2 sin  d
0
yL
 0  /2

 /2
dL
R
d

R

0
0 d
L

2R

9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Example 9.3
Determine the distance
from the x axis to the
centroid of the area
of the triangle
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Differential element
 Consider a rectangular element having
thickness dy which intersects the boundary
at (x, y)
Length and Moment Arms
 For area of the element
dA  xdy 

b
h  y dy
h
Centroid is located y distance from the x
~
axis
yy
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
~
 ydA
b
h  y  dy

0
y A
 h h
A dA 0 b h  y  dy
h
1 2
bh
h
6


1
bh 3
2
h
y
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Example 9.4
Locate the centroid for
the area of a quarter
circle.
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Method 1
Differential element
 Use polar coordinates for circular boundary
 Triangular element intersects at point (R,θ)
Length and Moment Arms
 For area of the element

1
R2
dA  ( R )( R cos  ) 
d
2
2
Centroid is located y distance from the x axis
2
~
x  R cos 
3
2
~
y  R sin 
3
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
 / 2 2
~
R2

x
dA

0  R cos   d
x
A
 dA
3

 / 2 R2
0
A

2
 /2
0
d
d
4R
3
y
 ~ydA
A
 dA
A

 2
 2 R   / 2 cos  d

 0
3



4R
3
 / 2 2

0
2
 R d  2 R   / 2 sin  d
R
sin




 0
3
3
 2



 /2
 / 2 R2
0 d
0 2 d
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Method 2
Differential element
 Circular arc element
having thickness of dr
 Element intersects the
axes at point (r,0) and
(r, π/2)
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Area and Moment Arms
 For area of the element

dA  2r / 4dr
Centroid is located y distance from the x axis
~x  2r /  ~y  2r / 
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
 ~x dA 0R 2r  24r  dr  R r 2 dr

  0
xA

R 2r
 R
dr
r dr
 dA
0 4
2 0
A

4R
3
 ~ydA 0R 2r  24r  dr  R r 2 dr

  0
yA

R 2r
 R
dA
dr
r dr

0 4

0
2
A

4R
3
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Example 9.5
Locate the centroid of
the area.
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Method 1
Differential element
 Differential element of thickness dx
 Element intersects curve at point (x, y), height y
Area and Moment Arms
 For area of the element
dA  ydx

For centroid
~
xx
~
y  y/2
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
x
~
 x dA
A
 dA
A
1

1 3
x dx
0
1 2
x dx
0
0


1
0 y dx 
xy dx
0.250
 0.75m
0.333
 ~ydA 1( y / 2) y dx 1( x2 / 2) x2 dx
yA
 0 1
 0 1
2
y
dx
x
 dA

 dx

A

0.100
 0.3m
0.333
0
0
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Method 2
Differential element
 Differential element
of thickness dy
 Element intersects
curve at point (x, y)
 Length = (1 – x)
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Area and Moment Arms
 For area of the element
dA  1  x  dy

Centroid is located y distance from the x axis
~x  x   1  x   1  x
 2  2
~y  y
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
~
 x dA 11  x  / 21  x  dx
x
A
 dA

0
0 1  y dy
1
1
0 1  x  dx
A
1 1
1  y  dy

0
2
0.250
 0.75m
0.333
 ~ydA 1 y1  x  dx 1 y  y3 / 2 dy
yA
 01
 01
 dA  1  x  dx  1  y dy


A

0
0.100
 0.3m
0.333
0

9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Example 9.6
Locate the centroid of
the shaded are bounded
by the two curves
y=x
and
y = x2.
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Method 1
Differential element
 Differential element of thickness dx
 Intersects curve at point (x1, y1) and (x2, y2), height y
Area and Moment Arms
 For area of the element
dA  ( y2  y1) dx

For centroid
~
xx
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
x
~x dA

 


0
0
 1
 1
 dA 0  y2  y1  dx 0 x  x2 dx
A
A
1
 12  0.5m
1
6
1
x y2  y1  dx
1
x x  x 2 dx
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Method 2
Differential element
 Differential element
of thickness dy
 Element intersects
curve at point (x1, y1)
and (x2, y2)
 Length = (x1 – x2)
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Area and Moment Arms
 For area of the element

dA   x1  x2  dy
Centroid is located y distance from the x axis
~x  x   x1  x2   x1  x2
2
2
 2 
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
x
~
 x dA
A
 dA
x


1
0
 x2  / 2 x1  x2  dx
 x
1
0
A

1

1
 x2  dx
1 1
1
2
y

y
dy
0
 21
 12  0.5m
1
0 y  y dx 6





1
0
 
y  y /2

1
0


y  y dy
y  y dx
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Example 9.7
Locate the centroid for the
paraboloid of revolution,
which is generated by
revolving the shaded area
about the y axis.
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Method 1
Differential element
 Element in the shape of a thin disk, thickness dy,
radius z
 dA is always perpendicular to the axis of revolution
 Intersects at point (0, y, z)
Area and Moment Arms
 For volume of the element
 
dV  z 2 dy
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
 For centroid
~
yy
Integrations
~
 ydV
yV
 dV
V
 66.7mm
 


 z  dy
100
0
y z dy
100
0
100
2
2

100 
0
y 2 dy
100
100 
0
y dy
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Method 2
Differential element
 Volume element in the form of thin
cylindrical shell, thickness of dz
 dA is taken parallel to the axis of
revolution
 Element intersects the
axes at point (0, y, z) and
radius r = z
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Area and Moment Arms
 For area of the element
2rdA  2z 100  y dz

Centroid is located y distance from the x axis
~
y  y  100  y  / 2  100  y  / 2
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
y
~
 ydV
100  y  / 22z 100  y  dz


 2z 100  y  dz
100
V
0
 dV
100
0
V
  z 10 4  10  4 z 4  dz
100

0
100
2 
0
 66.7 m


z 100  10  2 z 2 dz
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Example 9.8
Determine the location of
the center of mass of the
cylinder if its density
varies directly with its
distance from the base
ρ = 200z kg/m3.
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
View Free Body Diagram
Solution
For reasons of material symmetry
x  y0
Differential element
 Disk element of radius 0.5m and thickness dz since
density is constant for given value of z
 Located along z axis at point (0, 0, z)
Area and Moment Arms
 For volume of the element
dV   0.5 dz
2
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
 For centroid
~
z z
Integrations
~
 z dV
~
z V
 dV
V

1 2
z dz
0
1

0 zdz
1
z (200 z ) 0.52 dz

 01
2


dz
5
.
0

)
z
200
(
0
 0.667m
9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
 Not possible to use a shell
element for integration
since the density of the
material composing the
shell would vary along the
shell’s height and hence the
location of the element
cannot be specified
9.3 Composite Bodies



Consists of a series of connected “simpler”
shaped bodies, which may be rectangular,
triangular or semicircular
A body can be sectioned or divided into its
composite parts
Provided the weight and location of the center of
gravity of each of these parts are known, the
need for integration to determine the center of
gravity for the entire body can be neglected
9.3 Composite Bodies

Accounting for finite number of weights
~
xW
x
W

~
yW
y
W
~
zW
z
W
Where
represent the coordinates of the center
of gravity G of the composite body
~
represent the coordinates of the center
x, ~
y , ~z
of gravity at each composite part of the body
represent the sum of the weights of all
W
the composite parts of the body or total weight
x , y, z
9.3 Composite Bodies


When the body has a constant density or
specified weight, the center of gravity coincides
with the centroid of the body
The centroid for composite lines, areas, and
volumes can be found using the equation
 ~x W
x
W
 ~y W
 ~z W
y
z
W
W
However, the W’s are replaced by L’s, A’s and V’s
respectively
9.3 Composite Bodies
Procedure for Analysis
Composite Parts
 Using a sketch, divide the body or object into
a finite number of composite parts that have
simpler shapes
 If a composite part has a hole, or a
geometric region having no material,
consider it without the hole and treat the
hole as an additional composite part having
negative weight or size
9.3 Composite Bodies
Procedure for Analysis
Moment Arms
 Establish the coordinate axes on the
sketch and determine the coordinates
of the center of gravity or centroid of
each part
9.3 Composite Bodies
Procedure for Analysis
Summations
 Determine the coordinates of the
center of gravity by applying the center
of gravity equations
 If an object is symmetrical about an
axis, the centroid of the objects lies on
the axis
9.3 Composite Bodies
Example 9.9
Locate the centroid of the wire.
9.3 Composite Bodies
Solution
Composite Parts
Moment Arms
 Location of the centroid for each piece is
determined and indicated in the diagram
9.3 Composite Bodies
Solution
Summations
Segment
L
(mm)
x (mm)
y (mm)
z (mm)
xL
(mm2)
yL
(mm2)
zL
(mm2)
1
188.5
60
-38.2
0
11 310
-7200
0
2
40
0
20
0
0
800
0
3
20
0
40
-10
0
800
-200
11 310
-5600
-200
Sum
248.5
9.3 Composite Bodies
Solution
Summations
~
x L 11310
x

 45.5mm
L
248.5
~
y L  5600
y

 22.5mm
L
248.5
~
z L  200
z

 0.805mm
 L 248.5
9.3 Composite Bodies
Example 9.10
Locate the centroid of the plate area.
9.3 Composite Bodies
Solution
Composite Parts
 Plate divided into 3 segments
 Area of small rectangle considered “negative”
9.3 Composite Bodies
Solution
Moment Arm
 Location of the centroid for each piece is
determined and indicated in the diagram
9.3 Composite Bodies
Solution
Summations
Segment
A (mm2)
x (mm)
y (mm)
xA (mm3)
yA (mm3)
1
4.5
1
1
4.5
4.5
2
9
-1.5
1.5
-13.5
13.5
3
-2
-2.5
2
5
-4
Sum
11.5
-4
14
9.3 Composite Bodies
Solution
Summations
~
xA  4
x

 0.348mm
 A 11.5
~
y A 14
y

 1.22mm
 A 11.5
9.3 Composite Bodies
Example 9.11
Locate the center of mass of the
composite assembly. The conical
frustum has a density of
ρc = 8Mg/m3 and the hemisphere
has a density of ρh = 4Mg/m3.
There is a 25mm radius
cylindrical hole in the center.
9.3 Composite Bodies
View Free Body Diagram
Solution
Composite Parts
 Assembly divided into 4 segments
 Area of 3 and 4 considered “negative”
9.3 Composite Bodies
Solution
Moment Arm
 Location of the centroid for each piece is
determined and indicated in the diagram
Summations
 Because of symmetry,
x  y0
9.3 Composite Bodies
Solution
Summations
Segment
m (kg)
z (mm)
zm (kg.mm)
1
4.189
50
209.440
2
1.047
-18.75
-19.635
3
-0.524
125
-65.450
4
-1.571
50
-78.540
Sum
3.141
45.815
9.3 Composite Bodies
Solution
Summations
~
z m 45.815
z

 14.6mm
m
3.141
9.4 Theorems of Pappus and
Guldinus
A surface area of revolution is generated by
revolving a plane curve about a non-intersecting
fixed axis in the plane of the curve
 A volume of revolution is generated by revolving
a plane area bout a nonintersecting fixed axis in
the plane of area
Example
 Line AB is rotated about
fixed axis, it generates
the surface area of a
cone (less area of base)

9.4 Theorems of Pappus and
Guldinus
Example
 Triangular area ABC rotated
about the axis would
generate the volume of
the cone

The theorems of Pappus and Guldinus are used
to find the surfaces area and volume of any
object of revolution provided the generating
curves and areas do not cross the axis they are
rotated
9.4 Theorems of Pappus and
Guldinus
Surface Area
 Area of a surface of revolution = product of
length of the curve and distance traveled by the
centroid in generating the surface area
A  r L
9.4 Theorems of Pappus and
Guldinus
Volume
 Volume of a body of revolution = product of
generating area and distance traveled by the
centroid in generating the volume
V  r A
9.4 Theorems of Pappus and
Guldinus
Composite Shapes
 The above two mentioned theorems can be
applied to lines or areas that may be
composed of a series of composite parts
 Total surface area or volume generated is the
addition of the surface areas or volumes
generated by each of the composite parts
A   ~
r L
V   ~
r A
9.4 Theorems of Pappus and
Guldinus
Example 9.12
Show that the surface area of a sphere is
A = 4πR2
and its volume
V = 4/3 πR3
9.4 Theorems of Pappus and
Guldinus
View Free Body Diagram
Solution
Surface Area
 Generated by rotating semi-arc about the x
axis
 For centroid,
r  2R / 

For surface area,
A  ~
r L;
2R 

A  2   R  4R 2
 
9.4 Theorems of Pappus and
Guldinus
Solution
Volume
 Generated by rotating semicircular area about
the x axis
 For centroid,
r  4 R / 3

For volume,
V  ~
r A;
4R   1 2  4 3

V  2    R   R
 3   2
 3
9.5 Resultant of a General
Distributed Loading
Pressure Distribution over a Surface
 Consider the flat plate subjected to the loading
function ρ = ρ(x, y) Pa
 Determine the force dF acting on the differential area
dA m2 of the plate, located at the differential point
(x, y)
dF = [ρ(x, y) N/m2](d A m2)
= [ρ(x, y) d A]N
 Entire loading represented as
infinite parallel forces acting on
separate differential area dA
9.5 Resultant of a General
Distributed Loading
Pressure Distribution over a Surface
 This system will be simplified to a
single resultant force FR acting through
a unique point on the plate
9.5 Resultant of a General
Distributed Loading
Pressure Distribution over a Surface
Magnitude of Resultant Force
 To determine magnitude of FR, sum the
differential forces dF acting over the plate’s
entire surface area dA
FR   F ;   ( x, y )dA   dV
A

V
Magnitude of resultant
force = total volume under
the distributed loading
diagram
9.5 Resultant of a General
Distributed Loading
Pressure Distribution over a Surface
Location of Resultant Force
x ( x, y )dA  xdV

x

  ( x, y)dA  dV
y ( x, y )dA  ydV

y

  ( x, y)dA  dV
A
A
A
A

V
V
V
V
Line of action of action of the resultant force
passes through the geometric center or centroid
of the volume under the distributed loading
diagram
9.6 Fluid Pressure




According to Pascal’s law, a fluid at rest creates a
pressure ρ at a point that is the same in all
directions
Magnitude of ρ measured as a force per unit
area, depends on the specific weight γ or mass
density ρ of the fluid and the depth z of the point
from the fluid surface
ρ = γz = ρgz
Valid for incompressible fluids
Gas are compressible fluids and thus the above
equation cannot be used
9.6 Fluid Pressure
Consider the submerged plate
 3 points have been specified

9.6 Fluid Pressure




Since point B is at depth z1 from the liquid
surface, the pressure at this point has a
magnitude of ρ1= γz1
Likewise, points C and D are both at depth z2
and hence ρ2 = γz2
In all cases, pressure acts normal to the
surface area dA located at specified point
Possible to determine the resultant force
caused by a fluid distribution and specify its
location on the surface of a submerged plate
9.6 Fluid Pressure
Flat Plate of Constant Width
 Consider flat rectangular plate of constant
width submerged in a liquid having a specific
weight γ
 Plane of the plate makes an angle with the
horizontal as shown
9.6 Fluid Pressure
Flat Plate of Constant Width
 Since pressure varies linearly with depth,
the distribution of pressure over the plate’s
surface is represented by a trapezoidal
volume having an
intensity of ρ1= γz1
at depth z1 and
ρ2 = γz2 at depth z2
9.6 Fluid Pressure



Magnitude of the resultant force FR = volume of
this loading diagram and FR has a line of action
that passes through the volume’s centroid, C
FR does not act at the centroid of the plate but at
point P called the center of
pressure
Since plate has a constant
width, the loading diagram
can be viewed in 2D
9.6 Fluid Pressure
Flat Plate of Constant Width
 Loading intensity is measured as force/length
and varies linearly from
w1 = bρ1= bγz1 to
w 2 = bρ2 = bγz2
 Magnitude of FR = trapezoidal area
 FR has a line of action that passes through
the area’s centroid C
9.6 Fluid Pressure
Curved Plate of Constant Width
 When the submerged plate is curved, the pressure acting
normal to the plate continuously changes direction
 For 2D and 3D view of the loading distribution,

Integration can be used to determine FR and location of
center of centroid C or pressure P
9.6 Fluid Pressure
Curved Plate of Constant Width
Example
 Consider distributed loading acting on the
curved plate DB
9.6 Fluid Pressure
Curved Plate of Constant Width
Example
 For equivalent loading
9.6 Fluid Pressure
Curved Plate of Constant Width
 The plate supports the weight of the liquid Wf
contained within the block BDA
 This force has a magnitude of
Wf = (γb)(areaBDA)
and acts through the centroid of BDA
 Pressure distributions caused by the liquid acting
along the vertical and horizontal sides of the
block
 Along vertical side AD, force FAD’s magnitude =
area under trapezoid and acts through centroid
CAD of this area
9.6 Fluid Pressure
Curved Plate of Constant Width
 The distributed loading along horizontal side AB
is constant since all points lying on this plane are
at the same depth from the surface of the liquid
 Magnitude of FAB is simply the area of the
rectangle
 This force acts through the area centroid CAB or
the midpoint of AB
 Summing three forces,
FR = ∑F = FAB + FAD + Wf
9.6 Fluid Pressure
Curved Plate of Constant Width
 Location of the center of pressure on the plate is
determined by applying
MRo = ∑MO
which states that the moment of the resultant
force about a convenient reference point O, such
as D or B = sum of the moments of the 3 forces
about the same point
9.6 Fluid Pressure
Flat Plate of Variable Width
 Consider the pressure distribution acting on
the surface of a submerged plate having a
variable width
9.6 Fluid Pressure
Flat Plate of Variable Width
 Resultant force of this loading = volume
described by the plate area as its base and
linearly varying pressure distribution as its
altitude
 The shaded element may be used if integration is
chosen to determine the volume
 Element consists of a rectangular strip of area dA
= x dy’ located at depth z below the liquid
surface
 Since uniform pressure ρ = γz (force/area) acts
on dA, the magnitude of the differential force dF
dF = dV = ρ dA = γz(xdy’)
9.6 Fluid Pressure
Flat Plate of Variable Width
FR   dA   dV  V
A


Centroid V defines the point which FR acts
The center of pressure which lies on the surface
of the plate just below C has the coordinates P
defined by the equations
~
~
x dV
y ' dV


x
y' 
 dV
 dV
V
V

V
V
V
This point should not be mistaken for centroid of
the plate’s area
9.6 Fluid Pressure
Example 9.13
Determine the magnitude and location of the
resultant hydrostatic force acting on the submerged
rectangular plate AB. The
plate has a width of 1.5m;
ρw = 1000kg/m3.
9.6 Fluid Pressure
Solution
 The water pressures at depth A and B are
 A   w gz A  (1000kg / m3 )(9.81m / s 2 )( 2m)  19.62kPa


 B   w gz B  (1000kg / m3 )(9.81m / s 2 )(5m)  49.05kPa
Since the plate has constant
width, distributed loading
can be viewed in 2D
For intensities of the load at
A and B,
wA  b A  (1.5m)(19.62kPa)  29.43kN / m
wB  b B  (1.5m)( 49.05kPa)  73.58kN / m
9.6 Fluid Pressure
Solution
 For magnitude of the resultant force FR created by the
distributed load
FR  area of trapezoid


1
(3)( 29.4  73.6)  154.5 N
2
This force acts through the
centroid of the area
1  2(29.43)  73.58 
h 
(3)  1.29m
3  29.43  73.58 
measured upwards from B
9.6 Fluid Pressure
Solution
 Same results can be obtained by considering
two components of FR defined by the triangle
and rectangle
 Each force acts through its associated
centroid and has a magnitude of
FRe  (29.43kN / m)(3m)  88.3kN
Ft  (44.15kN / m)(3m)  66.2kN

Hence
FR  FRe  FR  88.3kN  66.2kN  154.5kN
9.6 Fluid Pressure
Solution
 Location of FR is determined by summing
moments about B
M R B   M B ;
(154.5)h  88.3(1.5)  66.2(1)
h  1.29m
9.6 Fluid Pressure
Example 9.14
Determine the magnitude of the resultant
hydrostatic force acting on the surface of a seawall
shaped in the form of a parabola. The wall is 5m
long and ρw = 1020kg/m2.
9.6 Fluid Pressure
Solution
 The horizontal and vertical components of the
resultant force will be calculated since
 B   w gz B  (1020kg / m 2 )(9.81m / s 2 )(3m)  30.02kPa

Then
wB  b B  5m(30.02kPa)  150.1kN / m

Thus
1
Fx  (3m)(150.1kN / m)  225.1kN
3
9.6 Fluid Pressure
Solution
 Area of the parabolic sector ABC can be
determined
 For weight of the wafer within this region
Fy  (  w gb)( area ABC )
1
 (1020kg / m )(9.81m / s )(5m)[ (1m)(3m)]  50.0kN
3
2

2
For resultant force
FR  Fx2  Fy2  (225.1) 2  (50.0) 2
 231kN
9.6 Fluid Pressure
Example 9.15
Determine the magnitude and location of
the resultant force acting on the triangular
end plates of the wafer of the water trough.
ρw = 1000 kg/m3
9.6 Fluid Pressure
View Free Body Diagram
Solution
 Magnitude of the resultant force F = volume of
the loading distribution
 Choosing the differential volume element,
dF  dV  dA   w gz (2 xdz)  19620zxdz

For equation of line AB
x  0.5(1  z )

Integrating
1
F  V   dV   (19620) z[0.5(1  z )]dz
V
1
0
 9810 ( z  z 2 )dz  1635N  1.64kN
0
9.6 Fluid Pressure
Solution
 Resultant passes through the centroid of the
volume
 Because of symmetry
x 0

For volume element
z
~
 z dV
V
VdV
1
z (19620) z[0.5(1  z )]dz

0

1635
1

9810 ( z 2  z 3 )dz
0
1635
 0.5m
Chapter Summary
Center of Gravity and Centroid
 Center of gravity represents a point where the
weight of the body can be considered
concentrated
 The distance to this point can be determined by
a balance of moments
 Moment of weight of all the particles of the body
about some point = moment of the entire body
about the point
 Centroid is the location of the geometric center
of the body
Chapter Summary
Center of Gravity and Centroid
 Centroid is determined by the moment balance
of geometric elements such as line, area and
volume segments
 For body having a continuous shape, moments
are summed using differential elements
 For composite of several shapes, each having a
known location for centroid, the location is
determined from discrete summation using its
composite parts
Chapter Summary
Theorems of Pappus and Guldinus
 Used to determine surface area and volume of a
body of revolution
 Surface area = product of length of the
generating curve and distance traveled by the
centroid of the curve to generate the area
 Volume = product of the generating area and the
distance traveled by the centroid to generate the
volume
Chapter Summary
Fluid Pressure
 Pressure developed by a fluid at a point on a
submerged surface depends on the depth of the
point and the density of the liquid according to
Pascal’s law
 Pressure will create a linear distribution of
loading on a flat vertical or inclined surface
 For horizontal surface, loading is uniform
 Resultants determined by volume or area under
the loading curve
Chapter Summary
Fluid Pressure
 Line of action of the resultant force passes
through the centroid of the loading
diagram
Chapter Review
Chapter Review
Chapter Review
Chapter Review
Chapter Review
Chapter Review