2- handouts lecture-42 43
TRANSCRIPT
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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
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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
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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.
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CENTROID AND CENTER OF MASS OF CENTROID AND CENTER OF MASS OF THE COMPOSITE BODIESTHE COMPOSITE BODIES
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
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CENTROID AND CENTER OF MASS OF CENTROID AND CENTER OF MASS OF THE COMPOSITE BODIESTHE COMPOSITE BODIES
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
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CENTROID AND CENTER OF MASS OF CENTROID AND CENTER OF MASS OF THE COMPOSITE BODIESTHE COMPOSITE BODIES
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
=
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CENTROID AND CENTER OF MASS OF CENTROID AND CENTER OF MASS OF THE COMPOSITE BODIESTHE COMPOSITE BODIES
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
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1
Locate the center of mass ( , ,x y z ) of the four particlesshown above.
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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
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 2Example # 2
Locate the centroid ( ,x y ) of the uniform wire bent in theshape shown above.
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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
=
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 3Example # 3
The gravity wall as shown above is made of concrete.Determine the location ( ,x y ) of the center of gravity G ofthe wall.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 3Example # 3
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− −
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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
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 4Example # 4
Locate the distance y to the centroid of the member’scross-sectional area.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 4Example # 4
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
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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 + × +
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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 × + ×