IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Distributed Forces:
Centroids and Centers of Gravity
22.3.2016
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.
22.3.2016
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
M
• Center of gravity of a wire
y
x W   x W
  x dW
x
y W   y W
  y dW
22.3.2016
Dr. Engin Aktaş
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IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Centroids and First Moments of Areas and Lines
• Centroid of an area
• Centroid of a line
x W   x dW
x W   x dW
x At    x t dA
x  La    x  a dL
x A   x dA  Q y
 first moment wit h respect to y
yA   y dA  Qx
 first moment wit h respect to x
22.3.2016
Dr. Engin Aktaş
x L   x dL
yL   y dL
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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.
22.3.2016
Dr. Engin Aktaş
5
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Centroids of Common Shapes of Areas
22.3.2016
Dr. Engin Aktaş
6
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Centroids of Common Shapes of
Lines
22.3.2016
Dr. Engin Aktaş
7
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Composite Plates and Areas
• Composite plates
X W   x W
Y W   y W
• Composite area
XAxA
Y  A  yA
22.3.2016
Dr. Engin Aktaş
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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.
22.3.2016
• 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.
22.3.2016
Dr. Engin Aktaş
Qx  506  103 mm 3
Q y  758  103 mm 3
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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 A  757.7 10 mm

X

 A 13.828 10 mm
3
3
3
2
X  54.8 mm
 yA
 506.2 103 mm 3
Y

3
2
A
13.828

10
mm

Y  36.6 mm
22.3.2016
Dr. Engin Aktaş
11
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Determination of Centroids by Integration
x A   xdA   x dxdy   xel dA
yA   ydA   y dxdy   yel dA
x A   xel dA
  x  ydx 
yA   yel dA

22.3.2016
y
 ydx 
2
• Double integration to find the first moment
may be avoided by defining dA as a thin
rectangle or strip.
x A   xel dA
ax
 a  x dy 

2
yA   yel dA
  y a  x dy 
Dr. Engin Aktaş
x A   xel dA

2r
1

cos   r 2 d 
3
2

yA   yel dA

2r
1

sin   r 2 d 
3
2

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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.
22.3.2016
• 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  k x2
b  k a2  k 
y
b
a2
x2
b
a2
x
or
a
b1 2
y1 2
• Evaluate the total area.
A   dA
a
 b x3 
b 2
  y dx   2 x dx   2 
a
 a 3 0
0
ab

3
a
22.3.2016
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
 b

Q y   xel dA   xydx   x 2 x 2 dx

0 a
a
 b x4 
a 2b
 2  
4
 a 4  0
2
a
y
1 b

Qx   yel dA   ydx    2 x 2  dx
2

02a
a
 b2 x5 
ab 2
 4  
 2a 5  0 10
22.3.2016
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.
b 2
ax
a  x2
a  x dy  
Q y   xel dA  
dy
2
2
0
1 b  2 a 2
  a 
2 0 
b
2

a
b
y dy 

4

a


Qx   yel dA   y a  x dy   y a  1 2 y1 2 dy


b
a 3 2
ab 2

   ay  1 2 y dy 
10

b
0
b
22.3.2016
Dr. Engin Aktaş
16
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem 5.4
• Evaluate the centroid coordinates.
xA  Q y
ab a 2b
x

3
4
3
x a
4
yA  Q x
ab ab 2
y

3
10
22.3.2016
Dr. Engin Aktaş
y
3
b
10
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IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Distributed Loads on Beams
L
W   wdx   dA  A
0
OP W   x dW
L
OP A   x dA  xA
• A distributed load is represented by plotting the load
per unit length, w (N/m) . The total load is equal to
the area under the load curve.
• 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
22.3.2016
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.
22.3.2016
• 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.
63 kN  m
X
18 kN
22.3.2016
Dr. Engin Aktaş
X  3.5 m
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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 :  Ay 6 m  18 kN 6 m  3.5 m  0
Ay  7.5 kN
22.3.2016
Dr. Engin Aktaş
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