finding the centroid
TRANSCRIPT
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Centroids and Distributed
LoadsENGR 221
February 10, 2003
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Lecture Goals
5.4 Centroids of Composite Bodies
5.6 Distributed Loads on Beams
5.7 Forces on Submerged Surfaces
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CentroidsComposite Bodies
The calculation of centroid uses the following
equation;
where AT is the total area and x and y bar are the
centroid of the body.
T
TArea Area
T
TArea Area
1
1
A x xdA x xdAA
A y ydA y ydA
A
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CentroidsComposite Bodies
The equation can be broken into integrals of
smaller areas.
i i
i i
T i i i i
TA A
T i i i i
TA A
1
1
A x x dA x x dAA
A y y dA y y dA
A
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CentroidsComposite Bodies
If each integral is replaced with its centroid and
area, the centroid of the entire body can be
computed using
i i
i i i i i i i i
A A
T i i i i
T
T i i i i
T
1
1
x dA x A y dA y A
A x x A x x AA
A y y A y y AA
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Centroids for Volume -
Composite BodiesThe same technique can be applied to finding thecentroid of the volume of a body using components
i i i
i i i i i i i i i i i i
V V V
T i i i i
T
T i i i i
T
T i i i i
T
1
1
1
x dV x V y dV y V z dV z V
V x x V x x V V
V y y V y y V V
V z z V z z V V
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CentroidsComposite Bodies
Each shape has a
centroid in the x
and y directions.
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CentroidsComposite Bodies
Each shape has acentroid in the x and
y directions. This
figure 5.1 out of your
text for 2-D figures.
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CentroidsComposite Bodies
Each volume has acentroid in the x, y,
and z directions.
This figure 5.2 out
of your text for 3-
D figures.
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CentroidsSimple Example for
a Composite Body
Find the centroid of the given body
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CentroidsSimple Example for
a Composite Body
To find the centroid,
i i
T
i i
T
1
1
x x AA
y y AA
Determine the area of the components
2
1
2
2
1120 mm 60 mm 3600 mm
2
120 mm 100 mm 12000 mm
A
A
1A
2A
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CentroidsSimple Example for
a Composite Body
The total area is
1
1
1 1
2
2
2 1
120 mm40 mm
3 3
60 mm60 mm 40 mm
3 3
120 mm
60 mm2 2
100 mm60 mm 110 mm
3 2
bx
hy h
bx
hy h
1A
2A
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CentroidsSimple Example for
a Composite Body
To centroid of each component
Compute the x centroid
1A
2A
2 2
T 1 2
2
3600 mm 12000 mm
15600 mm
A A A
i i
T
2 2
2
1
140 mm 3600 mm 60 mm 12000 mm
15600 mm
55.38 mm
x x A
A
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CentroidsSimple Example for
a Composite Body
To centroid of each component
Compute the y centroid
1A
2A
2 2
T 1 2
2
3600 mm 12000 mm
15600 mm
A A A
i i
T
2 2
2
1
140 mm 3600 mm 110 mm 12000 mm
15600 mm
93.85 mm
y y A
A
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CentroidsSimple Example for
a Composite Body
The problem can be done using a table to represent the
composite body.
BodyArea(mm2)
x (mm) y(mm)x*Area (mm3) y*Area (mm3)
Triangle 3600 40 40 144000 144000
Square 12000 60 110 720000 1320000
Sum 15600 864000 1464000
centroid (x) 55.38 mm
centroid (y) 93.85 mm
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CentroidsSimple Example for
a Composite Body
An alternative method of computing the centroid is
to subtract areas from a total area.
Assume that area isAssume that area is a large
square and subtract the small
triangular area.
1A
2A
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CentroidsSimple Example for
a Composite Body
The problem can be done using a table to represent the
composite body.
Body Area(mm2) x (mm) y(mm) x*Area (mm
3) y*Area (mm3)
Square 19200 60 80 1152000 1536000Triangle -3600 80 20 -288000 -72000
Sum 15600 864000 1464000
centroid (x) 55.38 mm
centroid (y) 93.85 mm
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CentroidsExample for a
Composite Body
Find the centroid of the given body
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CentroidsExample for a
Composite Body
Determine the area of the components
1
2
2
2
2
3
2
190 mm 60 mm
2
2700 mm
120 mm 90 mm
10800 mm
40 mm2
2513.3 mm
A
A
A
1A
2A 3A
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CentroidsExample for a
Composite Body
The total area is
1
1
1 1
2
2
2 1
3
3
90 mm30 mm
3 3
60 mm
60 mm 40 mm3 3
90 mm45 mm
2 2
120 mm60 mm 120 mm
3 2
4 40 mm490 mm 73.02 mm
3 3
60 mm 20 mm 40 mm 120 mm
bx
h
y h
bx
hy h
rx b
y
1A
2A3
A
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CentroidsExample for a
Composite BodyBody Area(mm2) x (mm) y(mm) x*Area (mm3) y*Area (mm3)
Triangle 2700 30 40 81000 108000
Square 10800 45 120 486000 1296000
Hemisphere -2513.27 73.02 120 -183528.00 -301592.89
Sum 10986.73 383472.00 1102407.11
centroid (x) 34.90 mmcentroid (y) 100.34 mm
3
i i 2
T
1 1102407.11 mm
10986.73 mm
100.34 mm
y y AA
3
i i 2
T
1 383472.00 mm
10986.73 mm
34.90 mm
x x AA
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CentroidsExample for a
Composite Body
An Alternative Method would be to
subtract to areas
Body Area(mm2) x (mm) y(mm) x*Area (mm
3) y*Area (mm3)
Triangle -2700 60 20 -162000 -54000Square 16200 45 90 729000 1458000
Hemisphere -2513.27 73.02 120 -183528.00 -301592.89
Sum 10986.73 383472.00 1102407.11
centroid (x) 34.90 mm
centroid (y) 100.34 mm
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CentroidsClass Problem
Find the centroid of the body
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CentroidsClass Problem
Find the centroid of the body
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Distributed Loads
How do you determine the equivalent load
(magnitude) and its location?
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Distributed Loads
Treat the load as a body and determine its area
(magnitude) and its centroid (location of the
resultant)
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Distributed Loads
So that
L
L L
1
W f x dx
xW x f x dx x x f x dxW
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Distributed LoadsExample
A beam supports a distributed load, determinethe equivalent concentrated load.
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Distributed LoadsExample
The load can be broken up into two triangular loads
where the magnitude of the load can be determined
1
2
11500 N/m 6.0 m 4500 N
21
4500 N/m 6.0 m 13500 N2
W
W
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Distributed LoadsExample
The center of the loads are
Total load is
1
2
6.0 m2.0 m
3 3
6.0 m6.0 m 4.0 m
3 3
Lx
L
x L
T 1 24500 N 13500 N
14000 N
W W W
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Distributed LoadsExample
The location of the resultant load is
i i
T
1
12 m 4500 N 4 m 13500 N
18000 N
3.5 m
x x WW
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Distributed LoadsExample
One can use the table method to find theloading acting on the beam.
Body Area(N) x (m) x*Area (N-m)
Triangle 1 4500 2 9000
Triangle 2 13500 4 54000
Sum 18000.00 63000.00
centroid (x) 3.50 m
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Distributed LoadsExample
An alternative loading would be to use a distributedload of (1.5 kN/m) and ramp load of 3 kN/m /m.
A
B
A
B
1500 N/m
4500 N/m 1500 N/m 3000 N/m
x 3 m
2x 6 m 4 m
3
w
w
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Distributed LoadsExample
An alternative loading would be to use a distributedload of (1.5 kN/m) and ramp load of 3 kN/m /m.
Body Area(N) x (m) x*Area (N-m)
Uniform 9000 3 27000
Triangle 9000 4 36000
Sum 18000.00 63000.00
centroid (x) 3.50 m
Di t ib t d L d E l
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Distributed LoadsExample
Problem
Determine the resultant R of the system of distributed
loads and locate its line of action with respect to the
left of the support for
Di t ib t d L d E l
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Distributed LoadsExample
Problem
Break the problem into three parts
1
2 3
1 1
2 2
3 3
150 lb/ft 15 ft
2250 lb1 1
300 lb/ft 150 lb/ft 7 ft2 2
525 lb
1 1400 lb/ft 150 lb/ft 8 ft
2 2
1000 lb
W w L
W w L
W w L
Di t ib t d L d E l
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Distributed LoadsExample
Problem
Break the problem into three parts
1
2 3
1
1
2
3
3
15 ft
2 2
7.5 ft
7 ft
3 3
2.33 ft
8 ft15 ft
3 3
12.33 ft
L
x
LW
Lx L
Di t ib t d L d E l
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Distributed LoadsExample
Problem
Put it in a table format
1
2 3
Body Area(lb) x (ft) x*Area (lb-ft)
Area 1 2250 7.5 16875
Area 2 525 2.333333 1225
Area 3 1000 12.33333 12333.33333
Sum 3775.00 30433.33
centroid (x) 8.06 ft
Di t ib t d L d Cl
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Distributed LoadsClass
Problem
Determine the resultant R of the system of distributed
loads and locate its line of action with respect to the
left of the support for
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Force on Submerged Surfaces
In a fluid at rest, the weight of the liquid will
create a pressure on the surface of a body. This
pressure is defined as the hydrostatic pressure.
where PA is pressure absolute, P0 is the initial
pressure and g is the specific weight of the fluid in
F/L3 and d is the depth.
A 0P P dg
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Force on Submerged Surfaces
The density of fluid, r is multiplied by the g to get
the specific weight of the fluid and PG (gauge
pressure) is defined as.
G A 0
g
P P P
d gd
g r
g r
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Force on a Submerged Surface
The pressure acts as a function of depth.
*width
*width
F y P
yg
gd
R
1
* *width2
R d dg
The resultant force, R
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Force on a Submerged Surface-
ExampleA 3- by 3 ft gate is placed in a
wall below water level as shown.
Determine the magnitude andlocation of the resultant of the
forces exerted by the water on
the gate. (g =62.4 lb/ft3)
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Force on a Submerged Surface-
ExampleThe pressure distribution on the wall is
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Force on a Submerged Surface-
ExampleThe pressure distribution on the wall is
3
2 ft
2
3
5 ft
2
62.4 lb/ft 5 ft 3 ft
124.8 lb/ft
62.4 lb/ft 5 ft
312 lb/ft
P d
P d
g
g
How does one obtain the distribution force?
MULTIPLY by the width (3 ft)
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Force on a Submerged Surface-
ExampleThe equivalent load on wall is
2
2 ft
2
5 ft
* 124.8 lb/ft 3 ft
374.4 lb/ft
* 312 lb/ft 3 ft
936 lb/ft
F P w
F P w
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Force on a Submerged Surface-
ExampleThe equivalent force is
1
2
374.4 lb/ft 3 ft 1123.2 lb
1936 lb/ft 374.4 lb/ft 3 ft
2
842.4 lb
F
F
1 21
2
3 ft5 ft 3.5 ft
2
3 ft5 ft 4.0 ft
3
x
x
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Force on a Submerged Surface-
ExampleUse a table to find the location
1
2
Total force is 1965.6 lb at 3.71 ft
from the surface.
Force Area(lb) x (ft) x*Area (lb-ft)Uniform 1123.2 3.5 3931.2
Triangular 842.4 4 3369.6
Sum 1965.60 7300.80
centroid (x) 3.71 ft
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Force on a Submerged Surface
How does on find the forces on a submerge
surface at an angle?
Draw the the free-
body diagram.
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Force on a Submerged Surface
The free-body diagram would have
The pressure and
the weight of the
fluid.
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Force on a Submerged Surface
The resulting force distribution without the weight of the
water would look like,
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Force on a Submerged Surface
If we were take a look at the distribution on a non-linear
surface the results would
The force can
be represented
as:
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Force on a Submerged Surface
Class Problem
The quick action gate AB is 1.75 ft
wide and is held in it closed
position by a vertical cable and by
hinges located along its top edge B.
For a depth of water d = 6-ft
determine the force acting on the
gate and location of the force.
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Homework (Due 2/17/03)
Problems:
5-2, 5-4, 5-6, 5-12, 5-14
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An automatic value consists of asquare plate 225 by 225 mm,
which pivoted about a horizontal
axis through A located at a distance
h=100 mm above the lower edge.Determine the depth of the water d
for which the valve will open.
Bonus Slides
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The bent plate ABCD is 2 m wide and is hinged at A.
Determine the reactions on A and D for the water level.
Bonus Slides