Centers and Centroids
Center of gravity
Center of mass
Centroid of volume
Centroid of area
Centroid of line
Department of Mechanical Engineering
Center of Gravity
A point where all of the
weight could be concentrated
without changing the external
effects of the body
To determine the location of
the center, we may consider
the weight system as a 3D
parallel force system
Department of Mechanical Engineering
Center of Gravity –
discrete bodies
The total weight is W Wi
Thelocation of the center can be
found using the total moments
1
M yz WxG Wi xi xG Wi xi
W
1
M zx WyG Wi yi yG Wi yi
W
1
M xy WzG Wi zi zG Wi zi
W
Department of Mechanical Engineering
Center of Gravity –
continuous bodies
The total weight is W dW
Thelocation of the center can be
found using the total moments
1
M yz WxG xdW xG xdW
W
1
M zx WyG ydW yG ydW
W
1
M xy WzG zdW zG zdW
W
Department of Mechanical Engineering
Center of Mass
A point where all of the mass could be concentrated
It is the same as the center of gravity when the body is assumed
to have uniform gravitational force
Mass of particles
1 n 1 n 1 n n
xC xi mi yC yi mi zC zi mi m mi
m i m i m i i
Continuous mass
1 1 1
xG x dm yG y dm zG z dm m dm
m m m
Department of Mechanical Engineering
Example: Center of discrete mass
List the masses and the
coordinates of their centroids in a
table
Compute the first moment of
each mass (relative to the planes
of the point of interest)
Compute the total mass and total
first moment
Compute the center
Department of Mechanical Engineering
Center of mass – list of mass and the coordinates
Labels Mass xi (m) yi (m) zi (m)
(kg)
A 1 0.3 .24 0.0
B 2 0.15 0.4 0.0
C 1 0.3 0.4 0.27
D 2 0.3 0.0 0.27
E 1 0.0 0.2 0.27
Department of Mechanical Engineering
Center of discrete mass –
calculation of the center
1st moment of mass
Mass # Mass xi (m) yi (m) zi (m) mixi miyi mizi
(kg)
A 1 0.3 0.24 0.0 0.3 0.24 0.0
B 2 0.15 0.4 0.0 0.3 0.8 0.0
C 1 0.3 0.4 0.27 0.3 0.4 0.27
D 2 0.3 0.0 0.27 0.6 0.0 0.54
E 1 0.0 0.2 0.27 0.0 0.2 0.27
total 7 1.5 1.64 1.08
The center 1.5/7 1.64/7 1.08/7
xc yc zc
This method applies to discrete weights, lines, areas etc
Department of Mechanical Engineering
Centroids of Volumes
Volumes made of sub vols
n n n n
1 1 1
xC
V
xV
i
i i yC
V
yV
i
i i zC
V
zV
i
i i V Vi
i
xi, yi, zi = centroids of the sub volumes
Vi = volumes of the segments
Continuous volumes
1 1 1
xC x dV yC y dV zC z dV V dV
V V V
Department of Mechanical Engineering
Centroids of Areas
Areas made of segments
1 n 1 n 1 n n
xC xi Ai yC yi Ai zC zi Ai A Ai
A i A i A i i
xi, yi, zi = centroids of the area segments
Ai = Areas of the segments
Continuous areas
1 1 1
xC x dA yC y dA zC z dA A dA
A A A
Department of Mechanical Engineering
Centroids of Lines (xc, yc, zc)
Lines made of segments
1 n 1 n 1 n n
xC xi Li yC yi Li zC zi Li L Li
L i L i L i i
xi, yi, zi = centroids of the line segments
Li = length of the segments
Continuous lines
1 1 1
xC
L x dL yC
L y dL zC
L z dL L dL
Department of Mechanical Engineering
Tables of special volumetric bodies, areas, and lines
These tables are helpful when the centroid of a composite
body (composed of volumes, areas, or lines) is in question
In the following table, the centroids of the body are relative
to the given origin O
Department of Mechanical Engineering
Department of Mechanical Engineering
Department of Mechanical Engineering
Department of Mechanical Engineering
Department of Mechanical Engineering
Continuous bodies – crucial tasks
Choosing the coordinate system
Determining the differential element for the integration
Determining the lower and upper limits of the integral
Carefully perform the integration (may require integration table)
Department of Mechanical Engineering
Example:
dA = differential element = b dy
This is not the only choice of the differential element !!
Department of Mechanical Engineering
Example:
Many possibilities of differential elements and coordinate system
Department of Mechanical Engineering
Please read example problems 5-17 and 5-18
5-17 Centroid of line segments
5-18 Centroid of a cone
Department of Mechanical Engineering
Problem 5-80: Centroid?
Department of Mechanical Engineering
Centroid of an area – area integration
Key components:
– The differential element and its definition
– The limits of the integration
– The moment arms
Department of Mechanical Engineering
Centroid of an area – vertical differential element
The area of the differential element
x
dA hdx h y2 y1 b b
x
b1
a a
dx
dA
h
Department of Mechanical Engineering
Centroid of an area – vertical differential element
The limits of the integration
– Lower limit x = 0
– Upper limit x = a
x=a
x=0
Department of Mechanical Engineering
Centroid of an area – vertical differential element
Performing the integral to obtain the area
x
a
1 2 3/ 2
a a
A dA b1
dx b x x
0 0
a a 3 0
1 2 3/ 2 2 ab
b a a ab 1
a 3 3 3
dA
Department of Mechanical Engineering
Centroid of an area – Getting the 1st moment of area about y
axis - My
My needs a moment arm parallel to x-axis
The arm is from the y axis to the centroid of the
element, here for the element it is x
M y s x dA xdA
dA q a
x
x
xhdx bx1
dx
0 0 a
Department of Mechanical Engineering
Centroid of an area – Getting the 1st moment of area
about y axis (My) and the x coordinate of the centroid
Performing the integration for the 1st moment of area
x
a a
M y xdA xhdx bx1
dx
0 0 a
a
1 1 2 5/ 2 1 1 2 5/ 2 1
b x2 x b a 2 a ba 2
2 a5 0 2 a 5 10
Calculating the x coordinate of the centroid
xC
xdA
a 2b / 10
0.3a
dA ab / 3 Department of Mechanical Engineering
Centroid of an area – Getting the 1st moment of
area about x axis - Mx
Mx needs a moment arm parallel to y
The arm is from the x axis to the centroid of the element
M x y1 y2 dA
1
2
2
The centroid of the
rectangular element dA
is [ x, (y1+y2)/2]
1
(y1+y2)/2 y1
x
Department of Mechanical Engineering
Centroid of an area – Getting the 1st moment of area
about x axis (Mx) and the y coordinate of the centroid
Performing the integration for moment area
Mx
1
y1 y2 dA
2
1 x 1 x x
b b dA b b
b b
dx
2 a 2 a a
a
b x
2 a
b 2
1x 2
ab 2
1 dx x
2 0 a 2 2 a 0 4
Calculating the y coordinate of the centroid
Mx ab 2 / 4
yC 0.75b
dA ab / 3
Department of Mechanical Engineering
Problem 5-79: Centroid?
Department of Mechanical Engineering
Problem 5-79: Solution
dAv
xc, yc=x, y/2
Department of Mechanical Engineering
Problem 5-79: Solution
dAv
xc, yc=x, y/2
Department of Mechanical Engineering
Problem 5-79: Solution
dAv
xc, yc=x, y/2
Department of Mechanical Engineering
Centroids of composite bodies
Possible elemental bodies:
– Basic areas
– Basic volumes
– Line segments
Similar method to centroid of discrete mass
Pay attention to the centroid of the elemental
bodies
Department of Mechanical Engineering
Centroid of a composite area
The composite = A square - a full circle - a quarter circle
Department of Mechanical Engineering
Centroids of the elemental areas
Area 1
120mm
120mm 160/
Area 3 4r/3
Area 2
4r/3
60mm
See Table 5-1
60mm Department of Mechanical Engineering
Calculation of the centroid relative to O
Label Area xi (mm) yi (mm) Aixi (1000 Aiyi (1000
mm3) mm3)
1 57600 120 120 6912 6912
2 -11309 100 80 -1130.9 -904.72
3 -11309 240- 240- -2138.2 -2138.2
160/ 160/
Total 34982 3642.8 3869.1
The centroid 104.1 110.6
Department of Mechanical Engineering
Centroid of composite volume and line
Similar method to composite area can be applied (use volume
and length instead of area)
Use Table 5-1 and 5-2 to determine the centroid of the
elemental bodies
Department of Mechanical Engineering
Decomposition of the line body
Straight line segments Semicircular arc
Department of Mechanical Engineering
How about this?
Department of Mechanical Engineering
Distributed loads on structural members
Tasks:
– Find the resultant
– Find the location of the resultant
Distributed loads:
– Continuous distribution xR
R
involves some area integral
– Composite of simple distribution
– A combination of the two
Department of Mechanical Engineering
Distributed load
The magnitude
L
R dR w( x)dx
0
The location
L
xdR xw( x)dx
d 0
R R
Department of Mechanical Engineering
Composite of simple distributed load
R1 R3
R2
1
R1 2.300 300 N
2 1.33 300 5 1800 9.33 600
R2 6.300 1800 N
d
300 1800 600
R3
1
4.300 600 N 5.56m
2 Department of Mechanical Engineering
Continuous distributed load
x
y wmax sin
2L
w=y
1 1
L
x
L L
x d xw( x)dx wmax x sin dx
R w( x)dx wmax sin dx R R0 2L
0 0
2L L
wmax 4 L2 x 2 L x
2 Lwmax x
L
2 Lwmax 2 sin x cos
cos 0.637 wmax L R 2L 2L 0
2 L 0
0.637
Department
L of Mechanical Engineering
Summary
Moment about a point O is given by a vector product;
~ ~ ~
Mo r F
The magnitude of the moment is
~
M o M o M o M ox M oy M oz
2 2 2
Moment analysis:
– Scalar approach
– Vector approach
Moment about a line OB
Mo = r x F Moment about point O
MOB = [(r x F) . e] e Moment about line OB
e is the unit vector along OB
O is any point on the line OB
Department of Mechanical Engineering
Summary
1
Couples M yz WxG Wi xi xG Wi xi
W
Equivalent force-couple system 1
M zx WyG Wi yi yG Wi yi
Finding resultant of general force W
system 1
M xy WzG Wi zi zG Wi zi
W
Center of weights and masses
1
M yz WxG xdW
W
Centroids of areas, lines and xG
xdW
volumes
1
Distributed load M zx WyG ydW yG ydW
W
1
M xy WzG zdW zG zdW
W
Department of Mechanical Engineering
