Centre of gravity and centroid
• Centre of gravity
– Known as centre of mass
• The centre of mass of an object is the point where the whole
of the object is assumed to be concentrated
– Refers to masses or weights
Center of gravity - center of mass in a uniform gravitational field R 
 mi ri
 mi
The centre of gravity of
the leaning tower of Pisa
lies above its base of
support , so the tower is in
stable equilibrium
Locating the centre of mass of a
symmetrical object
• Find the centre of mass of the system of particle shown
• Centroid
– Ignoring weight and considering only volume
– Different densities of material= the centre of gravity and centroid
do not always coincide
– Homogeneous = the centre of gravity and centroid would be
coincide.
– Type of centroid
• Centroid of lines (rope.wire)
• Centroid of areas (x,y)
– Simple areas
– Composite areas
• Centroid of volume (x,y,z)
Centroid of line
• for one-dimension object such as rope, wire and cable.
• measured in length only.
• the curve object, the centre of gravity is not located in the object
and can be determined using formula of derivation.
• the centroid can be determined by:
The centroid of line whether straight, curve or composite lines can only
be determined if the elements are connected and made up from a
homogenous material. For the straight line, the centroid lies at a distance
L/2 from reference axis. Normally, the centroid of line is always coincides
with the center of gravity
Example
Determine the centroid of a thin homogeneous wire
Centroid of simple areas
Shape
y
Area A
b
3
h
3
1
bh
2
0
4r
3
r 2
2
4r
3
4r
3
r 2
b
2
h
2
y
1. Triangle
x
h
y
x
b
2. Semicircle
y
r
3. Quarter circle
x
y
x
y
r
y
x
4. Rectangle
x
yx
h
b
y
x
4
bh
Centroid of composite area
• Composite areas are the combination of simple areas.
• The determination of centroid is more easier if calculate in the table.
1st Step:Divide the areas into several parts.
2nd Step: State the reference axis in the figures.
3rd Step: Find x1, x2 and y1, y2 from the reference axis
that you stated.
4th Step: Find the centroid.
Ax = A1x1 + A2x2
Ay = A1y1 + A2y2
Total area, ΣA = A1 + A2
x
 Ax ,
A
y
 Ay
A
Example 1 : centroid of simple area
y
x
b 6
  3m
2 2
y
h 4 .5

 2.25 m
2
2
4.5 m
y
x
6m
x
Centroids
y
x 0
y
4.5 m
y
x
6m
x
h 4 .5

 2.25 m
2
2
Centroids
y
x 0
y 0
y
4.5 m
x
6m
x
Example 2 : centroid of composite area
Calculation steps:
y
1. Divide the areas into simple
shapes
150 mm
2. State the reference axes
3. Find A1, A2, x1, x2, y1 and y2
based on the reference axes
20 mm
2
4. Find the centroid
1
x
120 mm
x
 Ai xi
 Ai
x  75 mm
y
 Ai y i
 Ai
y  91.82 mm
30 mm
Section
Area A (mm2)
x (mm)
y (mm)
Ax (mm3)
Ay (mm3)
1
3600
75
60
270000
216000
2
3000
75
130
225000
390000
 Ai  6600
 Ai xi  495000  Ai y i  606000
Example 3
y
150 mm
20 mm
2
120 mm
3
20 mm
diameter
hole
1
Ai xi

x
 Ai
Ai y i

y
 Ai
x  75 mm
y  93.41 mm
x
30 mm
Section
Area A (mm2)
x (mm)
y (mm)
Ax (mm3)
Ay (mm3)
1
3600
75
60
270000
216000
2
3000
75
130
225000
390000
3
314.16
75
60
 23561.9
 18849.6
 Ai  6285.84
 Ai xi  471438 .1  Ai y i  587150.4
Example 4
y
150 mm
20 mm
x
2
20 mm
diameter
hole
120 mm
3
Ai xi

x
 Ai
Ai y i

y
 Ai
x  75 mm
y  26.59 mm
1
30 mm
Section
Area A (mm2)
x (mm)
1
3600
75
60
270000
 216000
2
3000
75
10
225000
30000
3
314.16
75
60
 23561.9
 Ai  6285.84
y (mm)
Ax (mm3)
Ay (mm3)
18849.6
 Ai xi  471438 .1  Ai y i  167150
Finding a centroid for 3D body diagram
Example
Determine centre of mass or centre of gravity
The difference between calculating the centroid using volume, mass, or weight
is simply a scale factor, since mass and weight are proportional to volume as
indicated below.
 (Greek letter rho) = mass density (typically expressed in kg/m3 or
slug/ft3)
m = V
 (Greek letter gamma) = weight density or specific weight (typically in N/m3
or lb/ft3)
W = mg = gV = V
the centroids for mass and weight can be found as
follows:
x 
x 
x m
m
i
y m
m
i
z m
m
x W
W
y W
W
z W
W
i
i
i
i
i
y 
y 
i
i
i
i
i
z 
i
i
z 
i
i
i
i
Centroid of common shape in 3D
Centroid of common shape in 3D
(continue)
PROBLEM SOLVING
Given:
Two blocks of
different materials are
assembled as shown. The
weight densities of the materials
are
A = 86.8 kN / m3 and
B = 260.4 kN / m3.
Find: The center of gravity of
this assembly.
Plan: Follow the steps for
analysis
Solution
1. In this problem, the blocks A and B can be considered
as two segments.
PROBLEM SOLVING
Weight = w =
 (Volume in cm3)
wA = 86.8 (0.5) (6) (6) (2) / (10)3 = 3.125 N
wB =
260.4 (6) (6) (2) / (10)3 = 18.75 N
Segment
w (N)
x (cm)
y (cm)

z (cm)
w x
(N.cm)
w y
w z
(N.cm) (N.cm)
A
B
3.125
18.75
4
1
1
3
2
3
12.5
18.75
3.125
56.25
6.25
56.25

21.88
31.25
59.38
62.5
PROBLEM SOLVING (continued)
~
x = ( x w) / ( w ) = 31.25/21.88 = 1.47 cm
~
y = ( y w) / ( w ) = 59.38/21.88 = 2.68 cm
~
z = ( z w) / ( w ) = 62.5 /21.88 = 2.82 cm
Application : Distributed load on Beam
Illustration: The lumber on the rack shown below distributes the
weight evenly across the supporting beam. This uniform loading
is represented by a load curve with equal length lines. The total
weight (resultant) equals the area under the load curve and it acts
at the centroid of the load curve.
Application: Water tank
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 act.
How can we determine these weights and their locations?
Application : Tilt-slab construction
In tilt-slab construction, we
have a concrete wall (with
doors and windows cut out)
which we need to raise into
position. We don't want the
wall to crack as we raise it,
so we need to know
the center of mass of the
wall. How do we find the
center of mass for such an
uneven shape?
Solution:
1.
find the centroid of an area with straight sides
2.
concept to areas with curved sides where we'll use integration.