IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Distributed Forces:
Centroids and Centers of Gravity
13.12.2015
Dr. Engin Aktaş
1
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Introduction
• The earth exerts a gravitational force on each of the particles
forming a body. All these forces can be represented 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 similar to the center of gravity
of a body. The concept of the first moment of an area is
used to locate the center of gravity.
13.12.2015
Dr. Engin Aktaş
2
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Center of Gravity of a 2D Body
• Center of gravity of a plate
M
• Center of gravity of a wire
y
xW 
 xW
 x dW
  y W

M
x
yW

13.12.2015
 y dW
Dr. Engin Aktaş
3
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Centroids and First Moments of Areas and Lines
• Centroid of an area
xW 
x  At  
xA 
 x dW
xW 
 x  t dA
 x dA
 y dA
h respect to
y
 Qx
 first moment wit
13.12.2015
x  La  
 Qy
 first moment wit
yA 
• Centroid of a line
h respect to
 x dW
 x  a dL
xL 
 x dL
yL 
 y dL
x
Dr. Engin Aktaş
4
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
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.
13.12.2015
Dr. Engin Aktaş
5
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Centroids of Common Shapes of Areas
13.12.2015
Dr. Engin Aktaş
6
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Centroids of Common Shapes of
Lines
13.12.2015
Dr. Engin Aktaş
7
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Composite Plates and Areas
• Composite plates
W   xW
Y W   yW
X
• Composite area
 A xA
Y  A   yA
X
13.12.2015
Dr. Engin Aktaş
8
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 5.1
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.
13.12.2015
• 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.
Dr. Engin Aktaş
9
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 5.1
• Find the total area and first moments of the
triangle, rectangle, and semicircle. Subtract the
area and first moment of the circular cutout.
13.12.2015
Dr. Engin Aktaş
Q x   506  10 mm
3
Q y   758  10 mm
3
3
3
10
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 5.1
• Compute the coordinates of the area centroid by dividing the first
moments by the total area.
X 
 xA
A
 757 . 7  10 mm
3

13.828  10 mm
3
3
2
X  54 . 8 mm
Y 


yA
A
 506 . 2  10 mm
3

13.828  10 mm
3
3
2
Y  36 . 6 mm
13.12.2015
Dr. Engin Aktaş
11
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Determination of Centroids by Integration
• Double integration to find the first moment
may be avoided by defining dA as a thin
rectangle or strip.
x A   x dA   x dx dy   x el dA
y A   y dA   y dx dy   y el dA
xA 

yA 

13.12.2015

x el dA

x  ydx
y
el
y
xA 

dA
 2  ydx 

yA 

x

el
ax
y
2
el
xA 
dA
  a  x dy 
dA
 y a  x dy 
Dr. Engin Aktaş

yA 

x

dA
1 2

cos   r d  
3
2

2r
y

el
el
dA
1 2

sin   r d  
3
2

2r
12
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 5.4
SOLUTION:
• Determine the constant k.
• Evaluate the total area.
• Using either vertical or horizontal
strips, perform a single integration to
find the first moments.
Determine by direct integration the
location of the centroid of a parabolic
spandrel.
13.12.2015
• Evaluate the centroid coordinates.
Dr. Engin Aktaş
13
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem
5.4
SOLUTION:
• Determine the constant k.
y  kx
2
b  ka
2
b
 k 
a
y 
b
a
2
x
2
2
a
x 
or
b
1 2
y
1 2
• Evaluate the total area.
A
 dA
a

 y dx  
0

a
 b x3 
2
x dx   2

2
a
 a 3 0
b
ab
3
13.12.2015
Dr. Engin Aktaş
14
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 5.4
• Using vertical strips, perform a single integration
to find the first moments.
a
Qy 

x el dA 

xy dx 
 b 2
x
  2 x  dx

0 a
a
2
 b x4 
a b
 

2 4 
4
 a
 0
Qx 
 y el dA  
a
2
1 b 2
y dx   
x  dx
2
2
2

0 a
y
a
2
 b2 x5 
ab
 

4 5 
10
 2 a
 0
13.12.2015
Dr. Engin Aktaş
15
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 5.4
• Or, using horizontal strips, perform a single
integration to find the first moments.
Q y   x el dA  
ax
2
b
 a  x dy  
0
a
2
x
2
dy
2
2
2
b

1  2 a
a b
 a 
y dy 

2 0
b 
4
a

1
Q x   y el dA   y  a  x dy   y  a 
y
1 2

b
b
a

3
   ay 
y
1 2

b
0
13.12.2015
Dr. Engin Aktaş
2
ab

 dy 
10

2

 dy

2
16
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 5.4
• Evaluate the centroid coordinates.
xA  Q y
x
ab
2

3
a b
x 
4
3
a
4
yA  Q x
y
ab

3
13.12.2015
Dr. Engin Aktaş
ab
10
2
y 
3
b
10
17
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Distributed Loads on Beams
• A distributed load is represented by plotting the load
per unit length, w (N/m) . The total load is equal to
W   w dx   dA  A
the area under the load curve.
L
0
OP W

 x dW
L
OP  A   x dA
 xA
• A distributed load can be replaced by a concentrated
load with a magnitude equal to the area under the
load curve and a line of action passing through the
centroid of the area.
0
13.12.2015
Dr. Engin Aktaş
18
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 5.9
SOLUTION:
• 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.
A beam supports a distributed load as
shown. Determine the equivalent
concentrated load and the reactions at
the supports.
13.12.2015
• Determine the support reactions by
summing moments about the beam
ends.
Dr. Engin Aktaş
19
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 5.9
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
13.12.2015
Dr. Engin Aktaş
X  3 .5 m
20
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 5.9
• Determine the support reactions by summing
moments about the beam ends.
M
A
 0:
B y  6 m   18 kN
3 .5 m   0
B y  10 . 5 kN
 M B  0 :  A y  6 m   18 kN  6 m  3 .5 m   0
A y  7 . 5 kN
13.12.2015
Dr. Engin Aktaş
21
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Centroids and Centers of Gravity