Ce 102 statics
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CE 102 Statics. Chapter 7 Distributed Forces: Centroids and Centers of Gravity. 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

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CE 102 Statics

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Ce 102 statics

CE 102 Statics

Chapter 7

Distributed Forces:

Centroids and Centers of Gravity


Contents

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


Introduction

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.


Center of gravity of a 2d body

  • Center of gravity of a plate

  • Center of gravity of a wire

Center of Gravity of a 2D Body


Centroids and first moments of areas and lines

  • Centroid of an area

  • Centroid of a line

Centroids and First Moments of Areas and Lines


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’.

  • 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).

First Moments of Areas and Lines

  • 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

  • The centroid of the area coincides with the center of symmetry.


Centroids of common shapes of areas

Centroids of Common Shapes of Areas


Centroids of common shapes of lines

Centroids of Common Shapes of Lines


Composite plates and areas

  • Composite plates

  • Composite area

Composite Plates and Areas


Sample problem 7 1

Sample Problem 7.1

  • 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.

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

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

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


Sample problem 7 11

Sample Problem 7.1

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


Sample problem 7 12

Sample Problem 7.1

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


Determination of centroids by integration

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

Determination of Centroids by Integration


Sample problem 7 2

Sample Problem 7.2

  • 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.


Sample problem 7 21

  • SOLUTION:

  • Determine the constant k.

  • Evaluate the total area.

Sample Problem 7.2


Sample problem 7 22

Sample Problem 7.2

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


Sample problem 7 23

Sample Problem 7.2

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


Sample problem 7 24

  • Evaluate the centroid coordinates.

Sample Problem 7.2


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.

Theorems of Pappus-Guldinus


Theorems of pappus guldinus1

  • 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.

Theorems of Pappus-Guldinus


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.

Sample Problem 7.3

  • 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.


Sample problem 7 31

  • 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.

Sample Problem 7.3


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).

  • 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.

Distributed Loads on Beams


Sample problem 7 4

Sample Problem 7.4

  • 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.

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


Sample problem 7 41

  • 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.

Sample Problem 7.4


Sample problem 7 42

Sample Problem 7.4

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


Center of gravity of a 3d body centroid of a volume

  • Center of gravity G

  • Results are independent of body orientation,

  • For homogeneous bodies,

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


Centroids of common 3d shapes

Centroids of Common 3D Shapes


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.

  • For homogeneous bodies,

Composite 3D Bodies


Sample problem 7 5

  • SOLUTION:

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

Sample Problem 7.5

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


Sample problem 7 51

Sample Problem 7.5


Sample problem 7 52

Sample Problem 7.5


Ce 102 statics

Problem 7.6

y

20 mm

30 mm

Locate the centroid of the plane

area shown.

36 mm

24 mm

x


Ce 102 statics

y

Problem 7.6

20 mm

30 mm

Solving Problems on Your Own

Locate the centroid of the plane area shown.

36 mm

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

24 mm

x

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.


Ce 102 statics

Problem 7.6 Solution

y

20 + 10

Decide how to construct the given area from common shapes.

C1

C2

24 + 12

30

x

10

Dimensions in mm


Ce 102 statics

Problem 7.6 Solution

y

20 + 10

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

C1

C2

24 + 12

30

x

10

Dimensions in mm

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

1 20 x 60 =1200 10 30 12,000 36,000

2 (1/2) x 30 x 36 =540 30 36 16,200 19,440

S 1740 28,200 55,440


Ce 102 statics

Problem 7.6 Solution

y

20 + 10

XSA = S xA

Then

X (1740) = 28,200

X = 16.21 mm

or

C1

C2

YSA = S yA

and

24 + 12

30

Y (1740) = 55,440

x

10

Y = 31.9 mm

or

Dimensions in mm

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

1 20 x 60 =1200 10 30 12,000 36,000

2 (1/2) x 30 x 36 =540 30 36 16,200 19,440

S 1740 28,200 55,440


Ce 102 statics

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.

24 kN

30 kN

a

0.3 m

A

B

wA

wB

1.8 m


Ce 102 statics

Problem 7.7

Solving Problems on Your Own

24 kN

30 kN

a

0.3 m

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.

A

B

wA

wB

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.


Ce 102 statics

1

2

1

2

Problem 7.7 Solution

24 kN

30 kN

a

0.3 m

Replace the distributed

load by a pair of

equivalent forces.

C

A

B

20 kN/m

wB

0.6 m

0.6 m

RI

RII

RI = (1.8 m)(20 kN/m) = 18 kN

We have

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


Ce 102 statics

+

+

Problem 7.7 Solution

24 kN

30 kN

a

0.3 m

C

A

B

wB

0.6 m

0.6 m

RII = 0.9 wB kN

RI = 18 kN

(a)

SMC = 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)

SFy = 0: -24 kN + 18 kN + (0.9 wB) kN - 30 kN= 0

or wB = 40 kN/m


Ce 102 statics

y

2 in

1 in

2 in

3 in

r = 1.25 in

x

z

0.75 in

2 in

r = 1.25 in

2 in

Problem 7.8

For the machine element

shown, locate the z coordinate

of the center of gravity.


Ce 102 statics

y

Problem 7.8

2 in

1 in

2 in

3 in

r = 1.25 in

x

z

0.75 in

2 in

r = 1.25 in

2 in

Solving Problems on Your Own

For the machine element

shown, locate the z coordinate

of the center of gravity.

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

the center of gravity coincides

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

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

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


Ce 102 statics

y

Problem 7.8 Solution

2 in

1 in

2 in

3 in

r = 1.25 in

x

z

0.75 in

2 in

r = 1.25 in

2 in

y

V

IV

I

III

x

II

z

Determine the center of gravity

of composite body.

First assume that the machine

element is homogeneous so

that its center of gravity will

coincide with the centroid of

the corresponding volume.

Divide the body into five common shapes.


Ce 102 statics

y

y

2 in

1 in

V

2 in

IV

3 in

r = 1.25 in

I

III

x

II

x

z

0.75 in

z

2 in

r = 1.25 in

2 in

V, in3z, in. z V, in4

I (4)(0.75)(7) = 213.5 73.5

II (p/2)(2)2 (0.75) = 4.71247+ [(4)(2)/(3p)] = 7.8488 36.987

III -p(11.25)2 (0.75)= -3.6816 7 -25.771

IV (1)(2)(4) = 8 2 16

V -(p/2)(1.25)2 (1) = -2.4533 2 -4.9088

S27.576 95.807

Z S V = S z V : Z (27.576 in3 ) = 95.807 in4Z = 3.47 in


Ce 102 statics

Problem 7.9

y

y = kx1/3

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

a

x

h


Ce 102 statics

y

Problem 7.9

y = kx1/3

Solving Problems on Your Own

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

a

The procedure for locating the

centroids of volumes by direct

integration can be simplified:

x

h

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

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.


Ce 102 statics

Problem 7.9 Solution

y

Use symmetry to help locate the centroid. Symmetry implies

x

dx

y = 0 z = 0

x

z

r

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

y = kx1/3

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

dV = p r2 dx xel = x


Ce 102 statics

Problem 7.9 Solution

y

x

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

dx

dV = p r2 dx xel = x

x

z

r

r = kx1/3

so that

Now

dV = p k2 x2/3dx

y = kx1/3

a = kh1/3

k = a/h1/3

or

At x = h, y = a :

a2

h2/3

dV = p x2/3dx

Then


Ce 102 statics

a2

h2/3

dV = p x2/3dx

3

5

h

ò

0

3

8

= p a2h2

Problem 7.9 Solution

y

Integrate and determine the centroid.

x

dx

a2

h2/3

h

ò

V = p x2/3dx

x

z

r

0

a2

h2/3

h

[ ]

3

5

= p x5/3

y = kx1/3

0

= p a2h

a2

h2/3

a2

h2/3

3

8

ò

xel dV = x (p x2/3 dx) = p [ x8/3 ]

Also


Ce 102 statics

3

5

3

5

5

8

x = h

Problem 7.9 Solution

y

Integrate and determine the centroid.

x

dx

V = p a2h

3

8

ò

xel dV = p a2h2

x

z

r

y = kx1/3

3

8

ò

xV = xdV:

x ( p a2h) = p a2h2

Now

y = 0 z = 0


Ce 102 statics

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

d

1.8 ft

B

30o


Ce 102 statics

Problem 7.10

Solving Problems on Your Own

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

d

1.8 ft

B

30o

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

1. First, determine the pressure distributionacting perpendicular

the surface of the submerged body. The pressure distribution

will be either triangular or trapezoidal.


Ce 102 statics

Problem 7.10

Solving Problems on Your Own

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

d

1.8 ft

B

30o

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.


Ce 102 statics

Problem 7.10 Solution

Determine the pressure distribution

acting perpendicular the surface of the submerged body.

1.7 ft

A

PA

PA = 1.7 rg

PB = (1.7 + 1.8 cos 30o)rg

(1.8 ft) cos 30o

B

PB


Ce 102 statics

1

2

1

2

1

2

1

2

Problem 7.10 Solution

Ay

A

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

Ax

1.7 rg

(1.8 ft) cos 30o

LAB/3

P1

FB

The force of the water on the gate is

LAB/3

P2

B

LAB/3

(1.7 + 1.8 cos 30o)rg

P = Ap = A(rgh)

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

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


Ce 102 statics

30o

Problem 7.10 Solution

Ay

A

Ax

Write the equations of

static equilibrium for the

problem, and solve them.

1.7 rg

(1.8 ft) cos 30o

LAB/3

P1

FB

S MA = 0:

+

LAB/3

P2

B

2

3

1

3

( LAB)P1 + ( LAB)P2

LAB/3

(1.7 + 1.8 cos 30o)rg

- LABFB = 0

P1 = 171.85 lb

P2 = 329.43 lb

2

3

1

3

(171.85 lb) + (329.43 lb) - FB = 0

FB = 276.90 lb

FB = 277 lb


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