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Chapter 5: Distributed Forces: Centroids and Centers of Gravity 최해진 [email protected]

School of Mechanical Engineering

Contents Introduction

Theorems of Pappus-Guldinus

Center of Gravity of a 2D Body

Sample Problem 5.7

Centroids and First Moments of Areas and Lines

Distributed Loads on Beams

Centroids of Common Shapes of Areas Centroids of Common Shapes of Lines

Center of Gravity of a 3D Body: Centroid of a Volume

Composite Plates and Areas

Centroids of Common 3D Shapes

Sample Problem 5.1

Composite 3D Bodies

Determination of Centroids by Integration

Sample Problem 5.12

Sample Problem 5.9

Sample Problem 5.4 School of Mechanical Engineering 5-2

Introduction • The earth exerts a gravitational force on each of the particles forming a body. These forces can be replace by a single equivalent force equal to the weight of the body and applied at the center of gravity for the body. • The centroid of an area is analogous to the center of gravity of a body. The concept of the first moment of an area is used to locate the centroid. • Determination of the area of a surface of revolution and the volume of a body of revolution are accomplished with the Theorems of Pappus-Guldinus. • Determination of the area of a surface of revolution simplifies the analysis of beams subjected to distributed loads

School of Mechanical Engineering 5-3

Center of Gravity of a 2D Body 집중력은 없다 : 어떤 물체에 기계적으로 가한 모든 외력은 아무리 작아도 유한한 접촉면적에 분포되기 때문에 분포력으로 볼 수 있다.

※ “concentrated” forces does not exist in the exact sense

R : Resultant of

R

distributed forces

※ internal forces

C

C

C

Stress : internal distributed forces in solids

C

◎ 물체의 중심(Center of gravity ) : 물체의 각 부분에 작용하는 중력의 합력인 작용점 School of Mechanical Engineering 5-4

Center of Gravity of a 2D Body • Center of gravity of a plate

å Fz

• Center of gravity of a wire

Û W = DW1 + DW2 + L + DWn

모멘트 원리 : 임의의 축에 대한 중력의 합력 W의 모멘트는 물체의 미소 요소로 생각한 각 질점에 작용한 중력 dW의 모멘트의 합과 같다.

åMy

Û x W = å xi DWi = x1 DW1 +x2 DW2 + L + xn DWn Û x W = ò x dW

åMx

( x , y ) : 물체의

무게 중심의 좌표

Û yW = å yi DWi = y1 DW1 + y2 DW2 + L + yn DWn Û yW = ò y dW

School of Mechanical Engineering 5-5

Centroids and First Moment of Areas and Lines • Centroid of a line

• Centroid of an area

3 3 specific weight(비중량) g = r g (lb/ft or N/m )

x W = ò x dW Û x (gAt ) = ò x (g t )dA x A = ò x dA = Qy

( r g ) = (kg/m 3 )(m/sec2 ) = N/m 3 x W = ò x dW Þ x (g La ) = ò x (g a )dL x L = ò x dL

Þ y축에 대한 면적 A의 1차 모멘트

yL = ò y dL

yA = ò y dA = Qx Þ x축에 대한 면적 A의 1차 모멘트

\x =

1 xdA , ò A

y=

1 ydA ò A

( x , y ) : 면적 A의 도심 C의 좌표

\x =

1 xdL , ò L

y=

1 ydL ò L

( x , y ) : 선 L의 도심 C의 좌표 School of Mechanical Engineering 5-6

First Moments of Areas and Lines • An area is symmetric with respect to an axis BB’ if for every point P there exists a point P’ such that PP’ is perpendicular to BB’ and is divided into two equal parts by BB’. • The first moment of an area with respect to a line of symmetry is zero. • If an area possesses a line of symmetry, its centroid lies on that axis. • If an area possesses two lines of symmetry, its centroid lies at their intersection. • An area is symmetric with respect to a center O if for every element dA at (x,y) there exists an area dA’ of equal area at (-x,-y). • The centroid of the area coincides with the center of symmetry. School of Mechanical Engineering

5-7

Centroids of Common Shapes of Areas

School of Mechanical Engineering 5-8

Centroids of Common Shapes of Areas

School of Mechanical Engineering 5-9

Centroids of Common Shapes of Lines

School of Mechanical Engineering 5 - 10

Composite Plates and Areas • Composite plates X åW = å x W Y åW = å y W

• Composite area X å A = å xA Y å A = å yA

School of Mechanical Engineering 5 - 11

Sample Problem 5.1 SOLUTION: • Divide the area into a triangle, rectangle, and semicircle with a circular cutout. • Calculate the first moments of each area with respect to the axes.

For the plane area shown, determine the first moments with respect to the x and y axes and the location of the centroid.

• Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout. • Compute the coordinates of the area centroid by dividing the first moments by the total area.

School of Mechanical Engineering 5 - 12

Sample Problem 5.1

• Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout.

Q x = +506.2 ´ 103 mm 3 Q y = +757.7 ´ 103 mm 3 School of Mechanical Engineering 5 - 13

Sample Problem 5.1 • Compute the coordinates of the area centroid by dividing the first moments by the total area. x A + 757.7 ´ 103 mm 3 å X = = å A 13.828 ´ 103 mm 2 X = 54.8 mm

y A + 506.2 ´ 103 mm 3 å Y = = å A 13.828 ´ 103 mm 2 Y = 36.6 mm

연습문제: 5.3, 5.13, 5.20 School of Mechanical Engineering 5 - 14

Determination of Centroids by Integration x A = ò xdA = òò x dxdy = ò xel dA yA = ò ydA = òò y dxdy = ò yel dA

x A = ò xel dA = ò x ( ydx ) yA = ò yel dA y = ò ( ydx ) 2

• Double integration to find the first moment may be avoided by defining dA as a thin rectangle or strip.

x A = ò xel dA a+x [ (a - x )dx] =ò 2 yA = ò yel dA = ò y [(a - x )dx ]

x A = ò xel dA =ò

2r æ1 ö cosq ç r 2 dq ÷ 3 è2 ø

yA = ò yel dA =ò

2r æ1 ö sin q ç r 2 dq ÷ 3 School èof2Mechanical ø Engineering

5 - 15

Sample Problem 5.4 SOLUTION: • Determine the constant k. • Evaluate the total area. • Using either vertical or horizontal strips, perform a single integration to find the first moments. Determine by direct integration the location of the centroid of a parabolic spandrel.

• Evaluate the centroid coordinates.

School of Mechanical Engineering 5 - 16

Sample Problem 5.4 SOLUTION: • Determine the constant k. y = k x2 b = k a2 Þ k = y=

b a2

x2

or

b a2 x=

a b1 2

y1 2

• Evaluate the total area. A = ò dA a

é b x3 ù b 2 = ò y dx = ò 2 x dx = ê 2 ú êë a 3 úû 0 0a ab = 3 a

School of Mechanical Engineering 5 - 17

Sample Problem 5.4 • Using vertical strips, perform a single integration to find the first moments. a

æ b ö Q y = ò xel dA = ò xydx = ò xç 2 x 2 ÷dx ø 0 èa a

é b x4 ù a 2b =ê 2 ú = 4 êë a 4 úû 0 2

a y 1æ b ö Q x = ò yel dA = ò ydx = ò ç 2 x 2 ÷ dx 2 ø 02èa a

é b2 x5 ù ab 2 =ê 4 ú = êë 2a 5 úû 0 10

School of Mechanical Engineering 5 - 18

Sample Problem 5.4 • Or, using horizontal strips, perform a single integration to find the first moments. b 2 a+x a - x2 (a - x )dy = ò Q y = ò xel dA = ò dy 2 2 0

1 b æç 2 a 2 = ò a 2 0 çè b

2 ö a b y ÷dy = ÷ 4 ø

a æ ö Q x = ò yel dA = ò y (a - x )dy = ò yç a - 1 2 y1 2 ÷dy è ø b a 3 2ö ab 2 æ = ò ç ay - 1 2 y ÷dy = 10 ø b 0è b

School of Mechanical Engineering 5 - 19

Sample Problem 5.4 • Evaluate the centroid coordinates. xA = Q y ab a 2b x = 3 4

3 x= a 4

yA = Q x ab ab 2 y = 3 10

y=

3 b 10

School of Mechanical Engineering 5 - 20

Theorems of Pappus-Guldinus

• Surface of revolution is generated by rotating a plane curve about a fixed axis.

• Area of a surface of revolution is equal to the length of the generating curve times the distance traveled by the centroid through the rotation. A = 2p yL School of Mechanical Engineering 5 - 21

Theorems of Pappus-Guldinus

• Body of revolution is generated by rotating a plane area about a fixed axis. • Volume of a body of revolution is equal to the generating area times the distance traveled by the centroid through the rotation. V = 2p y A School of Mechanical Engineering 5 - 22

Sample Problem 5.7 SOLUTION: • Apply the theorem of Pappus-Guldinus to evaluate the volumes or revolution for the rectangular rim section and the inner cutout section. • Multiply by density and acceleration to get the mass and acceleration. The outside diameter of a pulley is 0.8 m, and the cross section of its rim is as shown. Knowing that the pulley is made of steel and that the density of steel is r = 7.85 ´ 103 kg m 3 determine the mass and weight of the rim. School of Mechanical Engineering 5 - 23

Sample Problem 5.7 SOLUTION: • Apply the theorem of Pappus-Guldinus to evaluate the volumes or revolution for the rectangular rim section and the inner cutout section. • Multiply by density and acceleration to get the mass and acceleration.

(

)(

)

3ö æ m = rV = 7.85 ´ 103 kg m 3 7.65 ´ 10 6 mm 3 ç10 -9 m 3 mm ÷ è ø 2 W = mg = (60.0 kg ) 9.81 m s

(

연습문제: 5.35, 5.42, 5.59

)

m = 60.0 kg W = 589 N School of Mechanical Engineering 5 - 24

Distributed Loads on Beams

L

W = ò wdx = ò dA = A 0

(OP )W = ò xdW L

(OP ) A = ò xdA = x A 0

• A distributed load is represented by plotting the load per unit length, w (N/m) . The total load is equal to the area under the load curve.

• A distributed load can be replace by a concentrated load with a magnitude equal to the area under the load curve and a line of action passing through the area centroid. School of Mechanical Engineering 5 - 25

Sample Problem 5.9 SOLUTION: • The magnitude of the concentrated load is equal to the total load or the area under the curve. • The line of action of the concentrated load passes through the centroid of the area under the curve. A beam supports a distributed load as shown. Determine the equivalent concentrated load and the reactions at the supports.

• Determine the support reactions by summing moments about the beam ends.

School of Mechanical Engineering 5 - 26

Sample Problem 5.9 SOLUTION: • The magnitude of the concentrated load is equal to the total load or the area under the curve. F = 18.0 kN • The line of action of the concentrated load passes through the centroid of the area under the curve. X =

63 kN × m 18 kN

X = 3.5 m

School of Mechanical Engineering 5 - 27

Sample Problem 5.9 • Determine the support reactions by summing moments about the beam ends.

å M A = 0 : B y (6 m ) - (18 kN )(3.5 m ) = 0 B y = 10.5 kN

å M B = 0 : - Ay (6 m ) + (18 kN )(6 m - 3.5 m ) = 0 Ay = 7.5 kN

School of Mechanical Engineering 5 - 28

Center of Gravity of a 3D Body: Centroid of a Volume

• Center of gravity G r r - W j = å (- DW j ) r r r r rG ´ (- W j ) = å [r ´ (- DW j )] r r r r rGW ´ (- j ) = (å r DW ) ´ (- j )

W = ò dW

r r rGW = ò r dW

• Results are independent of body orientation, x W = ò xdW

yW = ò ydW

z W = ò zdW

• For homogeneous bodies, W = g V and dW = g dV

x V = ò xdV

yV = ò ydV

z V = ò zdV School of Mechanical Engineering 5 - 29

Centroids of Common 3D Shapes

연습문제: 5.71, 5.77,5.113

School of Mechanical Engineering 5 - 30

Composite 3D Bodies • Moment of the total weight concentrated at the center of gravity G is equal to the sum of the moments of the weights of the component parts.

X åW = å xW

Y å W = å yW

Z åW = å zW

• For homogeneous bodies, X åV = å xV

Y å V = å yV

Z åV = å zV

School of Mechanical Engineering 5 - 31

Sample Problem 5.12 SOLUTION: • Form the machine element from a rectangular parallelepiped and a quarter cylinder and then subtracting two 0.03-m- diameter cylinders.

Locate the center of gravity of the steel machine element. The diameter of each hole is 1 cm.

School of Mechanical Engineering 5 - 32

Sample Problem 5.12

School of Mechanical Engineering 5 - 33

Sample Problem 5.12

X = å xV

Y = å yV

Z = å zV

åV = (5.024 ´ 10 ) (63.347 ´ 10 ) -7

-3

åV = (- 13.52 ´ 10 ) (63.347 ´ 10 ) -7

(

-7 V = 25 . 22 ´ 10 å

-3

) (63.347 ´ 10 ) -3

School of Mechanical Engineering 5 - 34

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School of Mechanical Engineering

Contents Introduction

Theorems of Pappus-Guldinus

Center of Gravity of a 2D Body

Sample Problem 5.7

Centroids and First Moments of Areas and Lines

Distributed Loads on Beams

Centroids of Common Shapes of Areas Centroids of Common Shapes of Lines

Center of Gravity of a 3D Body: Centroid of a Volume

Composite Plates and Areas

Centroids of Common 3D Shapes

Sample Problem 5.1

Composite 3D Bodies

Determination of Centroids by Integration

Sample Problem 5.12

Sample Problem 5.9

Sample Problem 5.4 School of Mechanical Engineering 5-2

Introduction • The earth exerts a gravitational force on each of the particles forming a body. These forces can be replace by a single equivalent force equal to the weight of the body and applied at the center of gravity for the body. • The centroid of an area is analogous to the center of gravity of a body. The concept of the first moment of an area is used to locate the centroid. • Determination of the area of a surface of revolution and the volume of a body of revolution are accomplished with the Theorems of Pappus-Guldinus. • Determination of the area of a surface of revolution simplifies the analysis of beams subjected to distributed loads

School of Mechanical Engineering 5-3

Center of Gravity of a 2D Body 집중력은 없다 : 어떤 물체에 기계적으로 가한 모든 외력은 아무리 작아도 유한한 접촉면적에 분포되기 때문에 분포력으로 볼 수 있다.

※ “concentrated” forces does not exist in the exact sense

R : Resultant of

R

distributed forces

※ internal forces

C

C

C

Stress : internal distributed forces in solids

C

◎ 물체의 중심(Center of gravity ) : 물체의 각 부분에 작용하는 중력의 합력인 작용점 School of Mechanical Engineering 5-4

Center of Gravity of a 2D Body • Center of gravity of a plate

å Fz

• Center of gravity of a wire

Û W = DW1 + DW2 + L + DWn

모멘트 원리 : 임의의 축에 대한 중력의 합력 W의 모멘트는 물체의 미소 요소로 생각한 각 질점에 작용한 중력 dW의 모멘트의 합과 같다.

åMy

Û x W = å xi DWi = x1 DW1 +x2 DW2 + L + xn DWn Û x W = ò x dW

åMx

( x , y ) : 물체의

무게 중심의 좌표

Û yW = å yi DWi = y1 DW1 + y2 DW2 + L + yn DWn Û yW = ò y dW

School of Mechanical Engineering 5-5

Centroids and First Moment of Areas and Lines • Centroid of a line

• Centroid of an area

3 3 specific weight(비중량) g = r g (lb/ft or N/m )

x W = ò x dW Û x (gAt ) = ò x (g t )dA x A = ò x dA = Qy

( r g ) = (kg/m 3 )(m/sec2 ) = N/m 3 x W = ò x dW Þ x (g La ) = ò x (g a )dL x L = ò x dL

Þ y축에 대한 면적 A의 1차 모멘트

yL = ò y dL

yA = ò y dA = Qx Þ x축에 대한 면적 A의 1차 모멘트

\x =

1 xdA , ò A

y=

1 ydA ò A

( x , y ) : 면적 A의 도심 C의 좌표

\x =

1 xdL , ò L

y=

1 ydL ò L

( x , y ) : 선 L의 도심 C의 좌표 School of Mechanical Engineering 5-6

First Moments of Areas and Lines • An area is symmetric with respect to an axis BB’ if for every point P there exists a point P’ such that PP’ is perpendicular to BB’ and is divided into two equal parts by BB’. • The first moment of an area with respect to a line of symmetry is zero. • If an area possesses a line of symmetry, its centroid lies on that axis. • If an area possesses two lines of symmetry, its centroid lies at their intersection. • An area is symmetric with respect to a center O if for every element dA at (x,y) there exists an area dA’ of equal area at (-x,-y). • The centroid of the area coincides with the center of symmetry. School of Mechanical Engineering

5-7

Centroids of Common Shapes of Areas

School of Mechanical Engineering 5-8

Centroids of Common Shapes of Areas

School of Mechanical Engineering 5-9

Centroids of Common Shapes of Lines

School of Mechanical Engineering 5 - 10

Composite Plates and Areas • Composite plates X åW = å x W Y åW = å y W

• Composite area X å A = å xA Y å A = å yA

School of Mechanical Engineering 5 - 11

Sample Problem 5.1 SOLUTION: • Divide the area into a triangle, rectangle, and semicircle with a circular cutout. • Calculate the first moments of each area with respect to the axes.

For the plane area shown, determine the first moments with respect to the x and y axes and the location of the centroid.

• Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout. • Compute the coordinates of the area centroid by dividing the first moments by the total area.

School of Mechanical Engineering 5 - 12

Sample Problem 5.1

• Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout.

Q x = +506.2 ´ 103 mm 3 Q y = +757.7 ´ 103 mm 3 School of Mechanical Engineering 5 - 13

Sample Problem 5.1 • Compute the coordinates of the area centroid by dividing the first moments by the total area. x A + 757.7 ´ 103 mm 3 å X = = å A 13.828 ´ 103 mm 2 X = 54.8 mm

y A + 506.2 ´ 103 mm 3 å Y = = å A 13.828 ´ 103 mm 2 Y = 36.6 mm

연습문제: 5.3, 5.13, 5.20 School of Mechanical Engineering 5 - 14

Determination of Centroids by Integration x A = ò xdA = òò x dxdy = ò xel dA yA = ò ydA = òò y dxdy = ò yel dA

x A = ò xel dA = ò x ( ydx ) yA = ò yel dA y = ò ( ydx ) 2

• Double integration to find the first moment may be avoided by defining dA as a thin rectangle or strip.

x A = ò xel dA a+x [ (a - x )dx] =ò 2 yA = ò yel dA = ò y [(a - x )dx ]

x A = ò xel dA =ò

2r æ1 ö cosq ç r 2 dq ÷ 3 è2 ø

yA = ò yel dA =ò

2r æ1 ö sin q ç r 2 dq ÷ 3 School èof2Mechanical ø Engineering

5 - 15

Sample Problem 5.4 SOLUTION: • Determine the constant k. • Evaluate the total area. • Using either vertical or horizontal strips, perform a single integration to find the first moments. Determine by direct integration the location of the centroid of a parabolic spandrel.

• Evaluate the centroid coordinates.

School of Mechanical Engineering 5 - 16

Sample Problem 5.4 SOLUTION: • Determine the constant k. y = k x2 b = k a2 Þ k = y=

b a2

x2

or

b a2 x=

a b1 2

y1 2

• Evaluate the total area. A = ò dA a

é b x3 ù b 2 = ò y dx = ò 2 x dx = ê 2 ú êë a 3 úû 0 0a ab = 3 a

School of Mechanical Engineering 5 - 17

Sample Problem 5.4 • Using vertical strips, perform a single integration to find the first moments. a

æ b ö Q y = ò xel dA = ò xydx = ò xç 2 x 2 ÷dx ø 0 èa a

é b x4 ù a 2b =ê 2 ú = 4 êë a 4 úû 0 2

a y 1æ b ö Q x = ò yel dA = ò ydx = ò ç 2 x 2 ÷ dx 2 ø 02èa a

é b2 x5 ù ab 2 =ê 4 ú = êë 2a 5 úû 0 10

School of Mechanical Engineering 5 - 18

Sample Problem 5.4 • Or, using horizontal strips, perform a single integration to find the first moments. b 2 a+x a - x2 (a - x )dy = ò Q y = ò xel dA = ò dy 2 2 0

1 b æç 2 a 2 = ò a 2 0 çè b

2 ö a b y ÷dy = ÷ 4 ø

a æ ö Q x = ò yel dA = ò y (a - x )dy = ò yç a - 1 2 y1 2 ÷dy è ø b a 3 2ö ab 2 æ = ò ç ay - 1 2 y ÷dy = 10 ø b 0è b

School of Mechanical Engineering 5 - 19

Sample Problem 5.4 • Evaluate the centroid coordinates. xA = Q y ab a 2b x = 3 4

3 x= a 4

yA = Q x ab ab 2 y = 3 10

y=

3 b 10

School of Mechanical Engineering 5 - 20

Theorems of Pappus-Guldinus

• Surface of revolution is generated by rotating a plane curve about a fixed axis.

• Area of a surface of revolution is equal to the length of the generating curve times the distance traveled by the centroid through the rotation. A = 2p yL School of Mechanical Engineering 5 - 21

Theorems of Pappus-Guldinus

• Body of revolution is generated by rotating a plane area about a fixed axis. • Volume of a body of revolution is equal to the generating area times the distance traveled by the centroid through the rotation. V = 2p y A School of Mechanical Engineering 5 - 22

Sample Problem 5.7 SOLUTION: • Apply the theorem of Pappus-Guldinus to evaluate the volumes or revolution for the rectangular rim section and the inner cutout section. • Multiply by density and acceleration to get the mass and acceleration. The outside diameter of a pulley is 0.8 m, and the cross section of its rim is as shown. Knowing that the pulley is made of steel and that the density of steel is r = 7.85 ´ 103 kg m 3 determine the mass and weight of the rim. School of Mechanical Engineering 5 - 23

Sample Problem 5.7 SOLUTION: • Apply the theorem of Pappus-Guldinus to evaluate the volumes or revolution for the rectangular rim section and the inner cutout section. • Multiply by density and acceleration to get the mass and acceleration.

(

)(

)

3ö æ m = rV = 7.85 ´ 103 kg m 3 7.65 ´ 10 6 mm 3 ç10 -9 m 3 mm ÷ è ø 2 W = mg = (60.0 kg ) 9.81 m s

(

연습문제: 5.35, 5.42, 5.59

)

m = 60.0 kg W = 589 N School of Mechanical Engineering 5 - 24

Distributed Loads on Beams

L

W = ò wdx = ò dA = A 0

(OP )W = ò xdW L

(OP ) A = ò xdA = x A 0

• A distributed load is represented by plotting the load per unit length, w (N/m) . The total load is equal to the area under the load curve.

• A distributed load can be replace by a concentrated load with a magnitude equal to the area under the load curve and a line of action passing through the area centroid. School of Mechanical Engineering 5 - 25

Sample Problem 5.9 SOLUTION: • The magnitude of the concentrated load is equal to the total load or the area under the curve. • The line of action of the concentrated load passes through the centroid of the area under the curve. A beam supports a distributed load as shown. Determine the equivalent concentrated load and the reactions at the supports.

• Determine the support reactions by summing moments about the beam ends.

School of Mechanical Engineering 5 - 26

Sample Problem 5.9 SOLUTION: • The magnitude of the concentrated load is equal to the total load or the area under the curve. F = 18.0 kN • The line of action of the concentrated load passes through the centroid of the area under the curve. X =

63 kN × m 18 kN

X = 3.5 m

School of Mechanical Engineering 5 - 27

Sample Problem 5.9 • Determine the support reactions by summing moments about the beam ends.

å M A = 0 : B y (6 m ) - (18 kN )(3.5 m ) = 0 B y = 10.5 kN

å M B = 0 : - Ay (6 m ) + (18 kN )(6 m - 3.5 m ) = 0 Ay = 7.5 kN

School of Mechanical Engineering 5 - 28

Center of Gravity of a 3D Body: Centroid of a Volume

• Center of gravity G r r - W j = å (- DW j ) r r r r rG ´ (- W j ) = å [r ´ (- DW j )] r r r r rGW ´ (- j ) = (å r DW ) ´ (- j )

W = ò dW

r r rGW = ò r dW

• Results are independent of body orientation, x W = ò xdW

yW = ò ydW

z W = ò zdW

• For homogeneous bodies, W = g V and dW = g dV

x V = ò xdV

yV = ò ydV

z V = ò zdV School of Mechanical Engineering 5 - 29

Centroids of Common 3D Shapes

연습문제: 5.71, 5.77,5.113

School of Mechanical Engineering 5 - 30

Composite 3D Bodies • Moment of the total weight concentrated at the center of gravity G is equal to the sum of the moments of the weights of the component parts.

X åW = å xW

Y å W = å yW

Z åW = å zW

• For homogeneous bodies, X åV = å xV

Y å V = å yV

Z åV = å zV

School of Mechanical Engineering 5 - 31

Sample Problem 5.12 SOLUTION: • Form the machine element from a rectangular parallelepiped and a quarter cylinder and then subtracting two 0.03-m- diameter cylinders.

Locate the center of gravity of the steel machine element. The diameter of each hole is 1 cm.

School of Mechanical Engineering 5 - 32

Sample Problem 5.12

School of Mechanical Engineering 5 - 33

Sample Problem 5.12

X = å xV

Y = å yV

Z = å zV

åV = (5.024 ´ 10 ) (63.347 ´ 10 ) -7

-3

åV = (- 13.52 ´ 10 ) (63.347 ´ 10 ) -7

(

-7 V = 25 . 22 ´ 10 å

-3

) (63.347 ´ 10 ) -3

School of Mechanical Engineering 5 - 34

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