2- Handouts Lecture-42 43

COURSE:COURSE: CE 201 (STATICS)CE 201 (STATICS)

LECTURE NO.:LECTURE NO.: 42 & 4342 & 43

FACULTY:FACULTY: DR. SHAMSHAD AHMADDR. SHAMSHAD AHMAD

DEPARTMENT:DEPARTMENT: CIVIL ENGINEERINGCIVIL ENGINEERING

UNIVERSITY:UNIVERSITY: KING FAHD UNIVERSITY OF PETROLEUM KING FAHD UNIVERSITY OF PETROLEUM & MINERALS, DHAHRAN, SAUDI ARABIA& MINERALS, DHAHRAN, SAUDI ARABIA

TEXT BOOK:TEXT BOOK: ENGINEERING MECHANICSENGINEERING MECHANICS--STATICS STATICS by R.C. HIBBELER, PRENTICE HALLby R.C. HIBBELER, PRENTICE HALL

LECTURE NO. 42 & 43LECTURE NO. 42 & 43CENTROID OF COMPOSITE BODIESCENTROID OF COMPOSITE BODIES

Objectives:Objectives:►► To show how to determine centroid and center of To show how to determine centroid and center of

mass of the composite bodiesmass of the composite bodies

CENTROID AND CENTER OF MASS OF CENTROID AND CENTER OF MASS OF THE COMPOSITE BODIESTHE COMPOSITE BODIES

Procedure for locating center of gravity of a body or the centroidof a composite geometrical object

What is composite body?

A composite body consists of a series of connected simpler-shaped bodies, which may be rectangular, triangular, semicircular, etc.

• Using a sketch, divide the body or object into a finite number ofcomposite parts that have simpler shapes.

• If a composite part has a hole, or a geometric region having nomaterial, then consider the composite part without the hole and consider the hole as an additional composite part having negative.

• Establish the coordinate axes on the sketch and determine thecoordinates , ,x y z of the center of gravity or centroid of each part.

• Determine , ,x y z by applying the center of gravity equations

• If an object is symmetrical about an axis, the centroid of the objectlies on this axis.


Example of Composite Line Centroid

The given composite line can be divided intofollowing three parts having simpler shapes:

The coordinates , ,x y z of each part and the products of coordinates and length of each part are presented in the following table:

( )x mm ( )z mm 2(mm )xL( )y mm 2( )yL mm 2( )zL mmSegment L (mm)

1 π(60) = 188.5 60 –38.2 0 11310 –7200 02 40 0 20 0 0 800 03 20 0 40 –10 0 800 –200

ΣL = 248.5 mm; 11310xLΣ = mm2; 5600yLΣ = − mm2; 200zLΣ = − mm2


Example of Composite Line Centroid

ΣL = 248.5 mm; 11310xLΣ = mm2; 5600yLΣ = − mm2; 200zLΣ = − mm2

11310 45.5 mm Ans.248.55600 22.5 mm Ans.

248.5200 0.805 mm Ans.

248.5

xLxLyLyLzLzL

Σ= = =ΣΣ −

= = = −ΣΣ −

= = = −Σ


Example of Composite Area Centroid

The given composite area can be divided intofollowing three parts having simpler shapes:

(ft)x (ft)y 3(ft )xA 3(ft )yA12 (3)(3) 4.5=

Segment A (ft2)

1 1 1 4.5 4.5

2 (3)(3) = 9 –1.5 1.5 –13.5 13.5

3 –(2)(1) = –2 –2.5 2 5 –4

ΣA = 11.5 ft2; 4xAΣ = − ft3; 14yAΣ = ft3

411.5

0.348 ft Ans.14

11.5 1.22 ft Ans.

xAxA

yAyA

Σ −= =Σ

= −Σ

= =Σ

=


Example of Center of Mass of a Composite Body The given composite body can be divided into followingfour parts having simpler shapes: 0x y= =

45.815 14.6 mm Ans.3.141

zmzm

Σ= = =Σ

Σm = 3.141 kg; Σ zm = 45.815 kg.mm

For frustum of cone portion, γ = 8×10-6 kg/mm3 and for hemi-sphere portion, γ = 4×10-6 kg/mm3

PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1

Locate the center of mass ( , ,x y z ) of the four particlesshown above.

PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1Let the coordinates of the center ofmass of the particles are , , and x y z

Values of coordinates , , and x y z can be calculated using the followingexpressions:

, xm ym zmx y zm m m

Σ Σ Σ= = =Σ Σ Σ

Σm = = 10 kg; xmΣ = 13 m.kg; ymΣ = 23 m.kg; zmΣ = 0 13 23 01.3 m Ans.; 2.3 m Ans.; 0 m Ans.10 10 10

xm ym zmx y zm m m

Σ Σ Σ= = = = = = = = =Σ Σ Σ


Locate the centroid ( ,x y ) of the uniform wire bent in theshape shown above.

PROBLEM SOLVING: PROBLEM SOLVING: Example # 2Example # 2The given composite wire can be divided intofollowing five parts having simpler shapes:

y

x

2

3

1

4

5

(50, 150)mm

(75, 130)mm

(50, 65)mm

(0, 75)mm

(25, 0)mm

y

x

2

3

1

4

5

(50, 150)mm

(75, 130)mm

(50, 65)mm

(0, 75)mm

(25, 0)mm

2 2= 480 mm; 16500 mm ; 41200 mmL xL yLΣ Σ = Σ =

16500480

34.37 mm Ans.41200

480 85.83 mm Ans.

xLxL

yLyL

Σ= =Σ

=Σ

= =Σ

=


The gravity wall as shown above is made of concrete.Determine the location ( ,x y ) of the center of gravity G ofthe wall.


The given composite area can be divided into followingfour parts having simpler shapes:

y

x1

3.6m

0.4m

y

x1

3.6m

0.4m

y

x

2

0.6m

3.0m

3.0m

0.4m

y

x

2

0.6m

3.0m

3.0m

0.4m

+

y

x

30.

6m 3.0m

0.4m

1.8my

x

30.

6m 3.0m

0.4m

1.8m y

x

4

3.0m 0.6m

0.4m

3.0m

y

x

4

3.0m 0.6m

0.4m

3.0m− −

PROBLEM SOLVING: PROBLEM SOLVING: Example # 3Example # 3 y

x1

3.6m

0.4m

y

x1

3.6m

0.4m

y

x

2

0.6m

3.0m

3.0m

0.4m

y

x

2

0.6m

3.0m

3.0m

0.4m

+

y

x

3

0.6m 3.0m

0.4m

1.8my

x

3

0.6m 3.0m

0.4m

1.8m y

x

4

3.0m 0.6m

0.4m

3.0m

y

x

4

3.0m 0.6m

0.4m

3.0m− −

2 3 36.84 m ; 15.192 m ; 9.648 mA Ax AyΣ = = =∑ ∑15.192 9.6482.22 m Ans.; 1.41 m Ans.

6.84 6.84Ax Ayx yA A

Σ Σ= = = = = =Σ Σ


Locate the distance y to the centroid of the member’scross-sectional area.


0x =

1

4

32

7.5 in

1.5 in

1 in

1 in

2.5 in 2.5 in

6.0 in

1

4

32

7.5 in

1.5 in

1 in

1 in

2.5 in 2.5 in

6.0 in

2 317.25 in ; 44.25 inA AyΣ = Σ =44.25 2.565 in Ans.17.25

AyyA

Σ= = =

Σ

Multiple Choice Problems1. y-coordinate ( y ) of the centroid of the composite area, as shown below, is

(a) 55.0 mm (b) 97.5 mm (c) 85.9 mm (d) None of these

Ans: (c) Feedback:

2 2

2 2

(50) 25 110 15 105 (35) 177.5 393112.394 4 85.9 mm4575.6(50) 110 15 (35)

4 4

AyyA

π π

π π

× + × × + ×Σ= = = =Σ + × +

Multiple Choice Problems2. coordinates ( ,x y ) of the centroid of the composite area, as shown below, is

(a) (3, 2) in (b) (2, 3) in (c) (4, 3) in (d) (3, 4) in

Ans.: (a) Feedback:

6 2 5 2 6 1 72 3 in 6 2 2 6 24

6 2 1 2 6 3 48 2 in6 2 2 6 24

AxxAAyyA

Σ × × + × ×= = = =Σ × + ×Σ × × + × ×

= = = =Σ × + ×

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2- Handouts Lecture-42 43