CE 102 Statics

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Vector Mechanics for Engineers: Statics

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CE 102 Statics

Chapter 7

Distributed Forces: Centroids and Centers of Gravity



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Contents

Introduction

Center of Gravity of a 2D Body

Centroids and First Moments of Areas

and Lines

Centroids of Common Shapes of Areas

Centroids of Common Shapes of Lines

Composite Plates and Areas

Sample Problem 7.1

Determination of Centroids by

Integration

Sample Problem 7.2

Theorems of Pappus-Guldinus

Sample Problem 7.3

Distributed Loads on Beams

Sample Problem 7.4

Center of Gravity of a 3D Body:

Centroid of a Volume

Centroids of Common 3D Shapes

Composite 3D Bodies

Sample Problem 7.5



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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.



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Center of Gravity of a 2D Body

• Center of gravity of a plate

dWy

WyWyM

dWx

WxWxM

y

y

• Center of gravity of a wire



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Centroids and First Moments of Areas and Lines

x

QdAyAy

y

QdAxAx

dAtxAtx

dWxWx

x

y

respect toh moment witfirst

respect toh moment witfirst

• Centroid of an area

dLyLy

dLxLx

dLaxLax

dWxWx

• Centroid of a line



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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.



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Centroids of Common Shapes of Areas



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Centroids of Common Shapes of Lines



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Composite Plates and Areas

• Composite plates

WyWY

WxWX

• Composite area

AyAY

AxAX



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Sample Problem 7.1

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

SOLUTION:

• Divide the area into a triangle, rectangle, and semicircle with a circular cutout.

• Compute the coordinates of the area centroid by dividing the first moments by the total area.

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

• Calculate the first moments of each area with respect to the axes.



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Sample Problem 7.1

33

33

mm107.757

mm102.506

y

x

Q

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



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Sample Problem 7.1

23

33

mm1013.828

mm107.757

A

AxX

mm 8.54X

23

33

mm1013.828

mm102.506

A

AyY

mm 6.36Y

• Compute the coordinates of the area centroid by dividing the first moments by the total area.



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Determination of Centroids by Integration

ydxy

dAyAy

ydxx

dAxAx

el

el

2

dxxay

dAyAy

dxxaxa

dAxAx

el

el

2

drr

dAyAy

drr

dAxAx

el

el

2

2

2

1sin

3

2

2

1cos

3

2

dAydydxydAyAy

dAxdydxxdAxAx

el

el• Double integration to find the first moment

may be avoided by defining dA as a thin rectangle or strip.



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Sample Problem 7.2

Determine by direct integration the location of the centroid of a parabolic spandrel.

SOLUTION:

• Determine the constant k.

• Evaluate the total area.

• Using either vertical or horizontal strips, perform a single integration to find the first moments.

• Evaluate the centroid coordinates.



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Sample Problem 7.2

SOLUTION:

• Determine the constant k.

2121

22

22

2

yb

axorx

a

by

a

bkakb

xky

• Evaluate the total area.

3

30

3

20

22

ab

x

a

bdxx

a

bdxy

dAA

aa



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Sample Problem 7.2

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

1052

2

1

2

44

2

0

5

4

2

0

22

2

2

0

4

2

0

22

abx

a

b

dxxa

bdxy

ydAyQ

bax

a

b

dxxa

bxdxxydAxQ

a

a

elx

a

a

ely



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Sample Problem 7.2

• Or, using horizontal strips, perform a single integration to find the first moments.

10

42

1

22

2

0

2321

2121

2

0

22

0

22

abdyy

b

aay

dyyb

aaydyxaydAyQ

badyy

b

aa

dyxa

dyxaxa

dAxQ

b

elx

b

b

ely



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Sample Problem 7.2

• Evaluate the centroid coordinates.

43

2baabx

QAx y

ax4

3

103

2ababy

QAy x

by10

3



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• 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.

LyA 2



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• 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.

AyV 2



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Sample Problem 7.3

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 determine the mass and weight of the rim.

33 mkg 1085.7

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.



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Sample Problem 7.3SOLUTION:

• Apply the theorem of Pappus-Guldinus to evaluate the volumes or revolution for the rectangular rim section and the inner cutout section.

3393633 mmm10mm1065.7mkg1085.7Vm kg 0.60m

2sm 81.9kg 0.60mgW N 589W

• Multiply by density and acceleration to get the mass and acceleration.



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Distributed Loads on Beams

• 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 (dW = wdx).

AdAdxwWL

0

AxdAxAOP

dWxWOP

L

0

• 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.



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Sample Problem 7.4

A beam supports a distributed load as shown. Determine the equivalent concentrated load and the reactions at the supports.

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.

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



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Sample Problem 7.4

SOLUTION:

• The magnitude of the concentrated load is equal to the total load or the area under the curve.

kN 0.18F

• The line of action of the concentrated load passes through the centroid of the area under the curve.

kN 18

mkN 63 X m5.3X



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Sample Problem 7.4

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

0m .53kN 18m 6:0 yA BM

kN 5.10yB

0m .53m 6kN 18m 6:0 yB AM

kN 5.7yA



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

• Center of gravity G

jWjW

jWrjWr

jWrjWr

G

G

dWrWrdWW G

• Results are independent of body orientation,

zdWWzydWWyxdWWx

zdVVzydVVyxdVVx

dVdWVW and

• For homogeneous bodies,



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Centroids of Common 3D Shapes



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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.

WzWZWyWYWxWX

• For homogeneous bodies,

VzVZVyVYVxVX



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Sample Problem 7.5

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

SOLUTION:

• Form the machine element from a rectangular parallelepiped and a quarter cylinder and then subtracting two 1-in. diameter cylinders.



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Sample Problem 7.5



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Sample Problem 7.5

34 in .2865in 08.3 VVxX

34 in .2865in 5.047 VVyY

34 in .2865in .6181 VVzZ

in. 577.0X

in. 577.0Y

in. 577.0Z

33

y

x

20 mm 30 mm

Problem 7.6

36 mm

24 mm

Locate the centroid of the planearea shown.

34

Solving Problems on Your Own

Several points should be emphasized when solving these types of problems.

Locate the centroid of the plane area shown.

y

x

20 mm 30 mm

36 mm

24 mm

1. Decide how to construct the given area from common shapes.

2. It is strongly recommended that you construct a table containing areas or length and the respective coordinates of the centroids.

3. When possible, use symmetry to help locate the centroid.

Problem 7.6

35

Problem 7.6 Solutiony

x

24 + 12

20 + 10

10

30

Decide how to construct the given area from common shapes.

C1

C2

Dimensions in mm

36


x

24 + 12

20 + 10

10

30

C1

C2

Dimensions in mm

Construct a table containing areas and respective coordinates of the centroids.

A, mm2 x, mm y, mm xA, mm3 yA, mm3

1 20 x 60 =1200 10 30 12,000 36,0002 (1/2) x 30 x 36 =540 30 36 16,200 19,440 1740 28,200 55,440

37


x

24 + 12

20 + 10

10

30

C1

C2

Dimensions in mm

XA = xAX (1740) = 28,200

Then

or X = 16.21 mm

and YA = yA

Y (1740) = 55,440

or Y = 31.9 mm

A, mm2 x, mm y, mm xA, mm3 yA, mm3

1 20 x 60 =1200 10 30 12,000 36,0002 (1/2) x 30 x 36 =540 30 36 16,200 19,440 1740 28,200 55,440

38

Problem 7.7

The beam AB supports two concentrated loads and rests on soil which exerts a linearly distributed upward load as shown. Determine (a) the distance a for which wA = 20 kN/m, (b) the corresponding value wB.

wA wB

A B

a 0.3 m24 kN 30 kN

1.8 m

39

The beam AB supports two concentrated loads and rests on soil which exerts a linearly distributed upward load as shown. Determine (a) the distance a for which wA = 20 kN/m, (b) the corresponding value wB.

wA wB

A B

a 0.3 m24 kN 30 kN

1.8 m


1. Replace the distributed load by a single equivalent force. The magnitude of this force is equal to the area under the distributed load curve and its line of action passes through the centroid of the area.

2. When possible, complex distributed loads should be divided into common shape areas.

Problem 7.7

40

Problem 7.7 Solution

wB

A B

a 0.3 m24 kN 30 kN

20 kN/m

C

0.6 m 0.6 m

RI RII

We have RI = (1.8 m)(20 kN/m) = 18 kN12

RII = (1.8 m)(wB kN/m) = 0.9 wB kN12

Replace the distributed load by a pair ofequivalent forces.

41


wB

A B

a 0.3 m24 kN 30 kN

C

0.6 m 0.6 mRI = 18 kN RII = 0.9 wB kN

(a) MC = 0: (1.2 - a)m x 24 kN - 0.6 m x 18 kN - 0.3m x 30 kN = 0 or a = 0.375 m

(b) Fy = 0: -24 kN + 18 kN + (0.9 wB) kN - 30 kN= 0 or wB = 40 kN/m

+

+

42

Problem 7.8

y

x

0.75 inz

1 in2 in2 in

3 in

2 in2 in

r = 1.25 in

r = 1.25 in

For the machine elementshown, locate the z coordinateof the center of gravity.

43


y

x

0.75 inz

1 in2 in2 in

3 in

2 in2 in

r = 1.25 in

r = 1.25 in

X V = x V Y V = y V Z V = z V

where X, Y, Z and x, y, z are the coordinates of the centroid of the body and the components, respectively.

For the machine elementshown, locate the z coordinateof the center of gravity.

Problem 7.8

Determine the center of gravity of composite body. For a homogeneous bodythe center of gravity coincides

with the centroid of its volume. For this case the center of gravity can be determined by

44


x

0.75 inz

1 in2 in2 in

3 in

2 in2 in

r = 1.25 in

r = 1.25 in

Determine the center of gravityof composite body.

First assume that the machineelement is homogeneous sothat its center of gravity willcoincide with the centroid ofthe corresponding volume.

y

x

z

I

IIIII

IVV

Divide the body into five common shapes.

45

y

x

z

I

IIIII

IVV

y

x

0.75 inz

1 in2 in2 in

3 in

2 in2 in

r = 1.25 in

r = 1.25 in

V, in3 z, in. z V, in4

I (4)(0.75)(7) = 21 3.5 73.5 II (/2)(2)2 (0.75) = 4.7124 7+ [(4)(2)/(3)] = 7.8488 36.987III -(11.25)2 (0.75)= -3.6816 7 -25.771IV (1)(2)(4) = 8 2 16 V -(/2)(1.25)2 (1) = -2.4533 2 -4.9088

27.576 95.807

Z V = z V : Z (27.576 in3 ) = 95.807 in4 Z = 3.47 in

46

Problem 7.9

Locate the centroid of the volume obtained by rotating the shaded area about the x axis.

h

a

x

yy = kx1/3

47


1. When possible, use symmetry to help locate the centroid.

Locate the centroid of the volume obtained by rotating the shaded area about the x axis.

h

a

y = kx1/3

x

y

The procedure for locating the centroids of volumes by direct integration can be simplified:

2. If possible, identify an element of volume dV which produces a single or double integral, which are easier to compute.

3. After setting up an expression for dV, integrate and determine the centroid.

Problem 7.9

48


x

y

z x

dx

y = kx1/3

r

Use symmetry to help locate the centroid. Symmetry implies

y = 0 z = 0

Identify an element of volume dV which produces a single or double integral.

Choose as the element of volume a disk or radius r and thickness dx. Then

dV = r2 dx xel = x

49


x

y

z x

dx

y = kx1/3

r

Identify an element of volume dV which produces a single or double integral.

dV = r2 dx xel = x

Now r = kx 1/3 so that

dV = k2 x2/3dx

At x = h, y = a : a = kh1/3 or k = a/h1/3

Then dV = x2/3dxa2

h2/3

50


x

y

z x

dx

y = kx1/3

r

dV = x2/3dxa2

h2/3

Integrate and determine the centroid.

0

h

V = x2/3dxa2

h2/3

= x5/3a2

h2/3

35[ ]

0

h

= a2h35

0

hxel dV = x ( x2/3 dx) = [ x8/3 ] a2

h2/3

a2

h2/3

38

Also

= a2h238

51


x

y

z x

dx

y = kx1/3

r

Integrate and determine the centroid.

V = a2h35

xel dV = a2h238

Now xV = xdV: 38x ( a2h) = a2h23

5

x = h58

y = 0 z = 0

52

Problem 7.10

The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water d = 3.5 ft, determine the force exerted on the gate by the shear pin.

A

B

d

1.8 ft30o

53



A

B

d

1.8 ft30o

Assuming the submerged body has a width b, the load per unit length is w = bgh, where h is the distance below the surface of the fluid.

1. First, determine the pressure distribution acting perpendicular the surface of the submerged body. The pressure distribution will be either triangular or trapezoidal.

Problem 7.10

54



A

B

d

1.8 ft30o

2. Replace the pressure distribution with a resultant force, and construct the free-body diagram.

3. Write the equations of static equilibrium for the problem, and solve them.

Problem 7.10

55


A

B

Determine the pressure distributionacting perpendicular the surface of the submerged body.

1.7 ft

(1.8 ft) cos 30o

PA

PB

PA = 1.7 gPB = (1.7 + 1.8 cos 30o)g

56


A

B

(1.8 ft) cos 30o

Replace the pressure distribution with a resultant force, and construct the free-body diagram.

Ay

Ax

FB

1.7 g

(1.7 + 1.8 cos 30o)g

LAB/3

LAB/3

LAB/3

P1

P2

The force of the water on the gate is

P = Ap = A(gh)12

12

P1 = (1.8 ft)2(62.4 lb/ft3)(1.7 ft) = 171.85 lb 12

P2 = (1.8 ft)2(62.4 lb/ft3)(1.7 + 1.8 cos 30o)ft = 329.43 lb12

57


A

B

(1.8 ft) cos 30o

Ay

Ax

FB

1.7 g

(1.7 + 1.8 cos 30o)g

LAB/3

LAB/3

LAB/3

P1

P2

P1 = 171.85 lb P2 = 329.43 lb

Write the equations of static equilibrium for theproblem, and solve them.

MA = 0:

( LAB)P1 + ( LAB)P2 13

23

- LABFB = 0

+

(171.85 lb) + (329.43 lb) - FB = 013

23 FB = 276.90 lb

30o

FB = 277 lb

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CE 102 Statics