ME 201 Engineering Mechanics: Staticsemp.byui.edu/MILLERG/ME 201/Supplemental Material... ·...

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ME 201Engineering Mechanics: Statics

Unit 9.1

Center of Gravity and Center of Mass

Composite Bodies

Centroid

Point which defines geometrical center of an

object

Determined by geometry for simple shapes

and by computing moments of an element

about an axis for more complex shapes

Also referred to as First Moment Area

Centroid of Simple Shapes

2

~ wx

2

~ hy

hwA

Rectangle

h

wx

y

c

Triangle

h

bx

y

c

3

~ bx

3

~ hy

hbA 2

1

Circle

x

y

c

r

0~ x 0~ y

4

22 d

rA

Centroid of Simple Shapes

0~ x3

4~ ry

2

2

1rA

Semi-Circle

x

y

cr

Quarter-Circle

3

4~ rx

3

4~ ry

2

4

1rA

x

y

cr

Centroids of Composite Areas

Centroids of more complex shapes can

be found by the following procedure

Divide into numbered, simple areas

Find centroid of each area

Sum moment areas relative to origin

Shape Area x y xA yA

1

2

3

1

2

3

A

AxX

~

A

AyY

~

x

y

Example Problem Solution

252/50~502/100~

5000100*50

1

1

1

y

x

A

50~2.121)*3/(50*4100~

39272/50*

3

3

2

3

y

pix

piA

7.663/5050~7.663/2*100~

25002/100*50

2

2

2

y

x

A

Given:

Dhole = 50 mm

Rarc = 50 mm

Find:

Centroid of composite

Solution:

Divide into simple areas

Find centroid of each area

50

100

+(70,40)1

23

4

40~70~

19634/50*

4

4

2

4

y

x

piA

Example Problem Solution

shape

1 5000 50.0 25.0 250000 125000

2 2500 66.7 66.7 166667 166667

3 3927 121.2 50.0 476032 196350

4 -1963 70.0 40.0 -137445 -78540

∑ 9463 755254 409476

mmA

AxX 8.79

9463

755254~

mmA

AyY 3.43

9463

409476~

x~ y~ Ax~ Ay~A

Solution:

Divide into simple areas

Find centroid of each area

Sum area & moment areas

Divide summations to find

composite centroid

Centroid

Point which defines geometrical center of an

object

Determined by geometry for simple shapes

and by computing moments of an element

about an axis for more complex shapes

Also referred to as First Moment Area

Centroid of Simple Shapes

2

~ wx

2

~ hy

hwA

Rectangle

The centroid of a rectangle or parallelogram lies

at the intersection of its diagonals.

h

wx

y

c

Centroid of Simple Shapes

3

~ bx

3

~ hy

hbA 2

1

Triangle

The centroid of a triangle lies at the intersection

of its median.

h

bx

y

c

Centroid of Simple Shapes

0~ x 0~ y

4

22 d

rA

Circle

The centroid of a circle lies at its center point.

x

y

c

r

Centroid of Simple Shapes

0~ x3

4~ ry

2

2

1rA

Semi-Circle

x

y

cr

Quarter-Circle

3

4~ rx

3

4~ ry

2

4

1rA

x

y

cr

Centroids of Composite Areas

Centroids of more complex shapes can

be found by the following procedure

Divide into numbered, simple areas

Find centroid of each area

Sum moment areas relative to origin

Shape Area x y xA yA

1

2

3

1

2

3

A

AxX

~

A

AyY

~

x

y

Example ProblemCentroid of Composite Area

50

100

Given:

Dhole = 50 mm

Rarc = 50 mm

Find:

Centroid of composite

+(70,40)

Example ProblemGiven:

Dhole = 50 mm

Rarc = 50 mm

Find:

Centroid of composite50

100

+(70,40)

Example Problem Solution

252/50~502/100~

5000100*50

1

1

1

y

x

A

50~2.121)*3/(50*4100~

39272/50*

3

3

2

3

y

pix

piA

7.663/5050~7.663/2*100~

25002/100*50

2

2

2

y

x

A

Given:

Dhole = 50 mm

Rarc = 50 mm

Find:

Centroid of composite

Solution:

Divide into simple areas

Find centroid of each area

50

100

+(70,40)1

23

4

40~70~

19634/50*

4

4

2

4

y

x

piA

Example Problem Solution

shape

1 5000 50.0 25.0 250000 125000

2 2500 66.7 66.7 166667 166667

3 3927 121.2 50.0 476032 196350

4 -1963 70.0 40.0 -137445 -78540

∑ 9463 755254 409476

mmA

AxX 8.79

9463

755254~

mmA

AyY 3.43

9463

409476~

x~ y~ Ax~ Ay~A

Solution:

Divide into simple areas

Find centroid of each area

Sum area & moment areas

Divide summations to find

composite centroid

Center of Gravity & Mass

Centroid (Center of Area)

Center of Gravity

Center of Mass

A

AxX

~

A

AyY

~

W

WxX

~

W

WyY

~

W

WzZ

~

m

mxX

~

m

myY

~

m

mzZ

~

In Class Exercise

Compute the y centroid of object if the radius

of the circle is 3”

3” 3”

6”

9”

y = 4.42 “

Solution

Shape Area ybar yA

rect 108 4.5 486

2-tri -27 3 -81

cir -14.14 7.73 -109.23

66.86 4.42 295.77

In Class Exercise

Solution

I-Clicker QuestionIn-Class Exercise

Indicate the number of In-Class problems

which you were able to complete.

A. 0

B. 1

C. 2