ME 108 - Statics
Chapter 5
Distributed Forces
Ahmet Erkliğ
Section A
Center of Mass and Centroids
Applications
 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 are acting
 How can we determine
these weights and their
locations?
Center of Gravity (CG) and
Center of Mass (CM)
 Consider a three dimensional body
of any size and shape, having a mass m
 If we suspend the body, as shown
in the figure, from any point such as
A, B, or C, the body will be in
equilibrium under the action of tension
in the cord and the resultant W of the
gravitational forces acting on all particles
of the body
 For each instances we mark the line of action of the resultant force
 For all practicle purposes these lines of action will be cuncerrunt at
a single point G, which is called the center of gravity of the body
Center of Gravity (CG) and
Center of Mass (CM)
 To determine the mathematically the location of the CG
of any body, we apply the principle of moments to the
parallel system of gravitational forces
 The moment of the resultant gravitational force W about
any axis equals the sum of the moments about the same
axis of the gravitaional forces dW acting on all particles
treated as infinitesimal elements of the body
 The resultant of the gravitational forces acting on all
elements is the weight of the body and is given by
the sum W = ∫dW
 For example, if we apply the moment principle about the y-axis, the
moment about this axis of the elemental weight is x dW, and the
sum of these moments for all elements of the body is ∫x dW
 This sum of moments must equal , the moment of the sum
 Thus, W x = ∫ xdW
Center of Gravity (CG) and
Center of Mass (CM)
Similarly, we can sum moments about the x- and
z-axes to find the coordinates of G
With the substitution of W = mg and dW = g dm,
the expressions for the coordinates of G become
Center of Gravity (CG) and
Center of Mass (CM)
 The coordinates of G may be expressed in vector form as
 In which
r = xi + yj + zk and r = xi + yj + zk
Center of Gravity (CG) and
Center of Mass (CM)
 It is obvious that the previous equations are independent
of gravitational effects since g no longer appears
 They therefore define a unique point in the body which is a
function solely of the distribution of mass
 This point is called the center of mass, and clearly it
coincides with the center of gravity as long as the gravity
field is treated as uniform and parallel
Concept of Centroid
 The centroid C is a point which defines
the geometric center of an object
 The centroid coincides with the center
of mass or the center ofgravity only if
the material of the body is
homogenous (density or specific
weight is constant throughout the
body)
 If an object has an axis of symmetry,
then the centroid of object lies on that
axis
Centroid of Lines, Area, and Volume
(1) Lines. For a slender rod or wire of length L, crosssectional area A, and desity ρ, the body approximates a
line segment and dm = ρ A dL. If ρ and A are constant over
the length of the rod, the coordinates of the center of mass
become the coordinates of the centroid C of the line
segment.
Centroid of Lines, Area, and Volume
Centroid of Lines, Area, and Volume
Choosing Element for Integration
• Centroid is simple in concept, but not always simple in
evaluation. When possible, choose an incremental element
that reduces the number of integrations required.
example: Select appropriate area increments to find the
centroid of the shape shown.
x

 x˜ dA
 dA
~ ~
y
 y˜ dA
y
 dA
x, y = distances to area element dA
x
possible area increments:
y
y
y˜
x
x˜
(a)
dA  dxdy
xx
yy
double integral
with messy
limits
w
y
dy
y˜
x
x˜
(b)
x
(c)
dA  wdy
x  xw/2
dA  ___ dx
(need two
yy
single integral
with nice
limits
expressions)
xx
y  .....
two single
integrals
Example
Show that the centroid of the area is
x = 4a/7 and y = 2b/5
x=
y
a
b
y
3
3
b
a
x
Solution 1
a 3
x= 3 y
b
b
y
x˜
y˜ y
dx
a
a
1/ 3
b 3 
x  x  dx

x˜dA 0  a 

x

1/ 3
a 3
dA

b 
  a x  dx

0 
4
a
7

b
dA  ydx  
a
3
y 1b
y  
2 2 a
3
x
1/3

x  dx

1/3

x

xx
a
1 b 3
 2  a
y
˜
dA
y
 0 
 dA
2
=b
5
1/ 3

x 

1/ 3
b 3 
 x 
 a 
 dA
dx
Solution 2
y
 a 
dA  (a  x)dy   a  3 y 3  dy
 b 
yy
ax xa
x  x

2
2
1 a

  3 y3  a 
2b

x
a-x
b
dy
x˜
a
x
b
x˜dA

x

 dA

a
4
7
1  a 3
a  a y 3  dy
y

a
 2 b 3


3


b

0
b
a 3 

a

  b 3 y dy
0
b
y˜ dA

y

 dA
=b
2
5
a  a y 3 dy
y
  b 3 
0
 dA
Example 1
Centroid of a circular arc. Locate the centroid of
a circular arc as shown in the figure.
Solution
Example 2
 Centroid of a triangular area. Determine the distance h
from the base of a triangle of altitude h to the centroid of its
area.
Solution
Example
Solution
Example
Solution
example: Find the centroid of the shape shown.
y
2
y =x
1m
y=x
1m
2
x
A: x  y  9 20 m
example: A
homogeneous
uniform wire
is bent to the
shape shown.
Determine the
location of the
centroid.
2r 2
A: x  
3

y
r
45° x
45°
Example
2m
Locate the center of gravity x of the
homogeneous rod bent in the form of a
semicircular arc. The rod has a weight
per unit length of 0.5 N/m. Also,
determine the horizantal reaction at the
smooth support B and the x and y
components of reaction at the pin A.
Solution
2m
Solution
Extra Problem
Extra Problem