Finding the Centroid

8/4/2019 Finding the Centroid

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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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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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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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Each shape has a

centroid in the x

and y directions.


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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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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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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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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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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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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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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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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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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




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CentroidsExample for a

Composite Body

Find the centroid of the given body


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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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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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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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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




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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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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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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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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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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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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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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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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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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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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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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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ExampleThe pressure distribution on the wall is


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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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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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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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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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How does on find the forces on a submerge

surface at an angle?

Draw the the free-

body diagram.


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The free-body diagram would have

The pressure and

the weight of the

fluid.


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The resulting force distribution without the weight of the

water would look like,


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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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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

Finding the Centroid

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