1312 - Ch9 - 1208471587

Chapter 9 Center of Gravity and Centroid

CENTER OF GRAVITY AND CENTROID (Chapter 9) Objective : Students will: a)

Understand the concepts of center of gravity, center of mass, and centroid.

b) Be able to determine the location of these points for a system of particles or a body.

APPLICATIONS To design the structure for supporting a water tank, we will need to know the weights of the tank and water as well as the locations where the resultant forces representing these distributed loads are acting---center of gravity (CG). .

One of the important factors in determining SUV’s stability is its center of mass (CM).

4N 3m A • 1N

CONCEPT OF CG and CM

1m • G

•

B

The center of gravity (G) is a point which locates the resultant weight of a system of particles or body.

3N

From the definition of a resultant force, the sum of moments due to individual particle weight about any point is the same as the moment due to the resultant weight located at G. For the figure above, try taking moments about A and B. Also, note that the sum of moments due to the individual particle’s weights about point G is equal to zero. Similarly, the center of mass is a point which locates the resultant mass of a system of particles or body. Generally, its location is the same as that of G.

CG / CM FOR A SYSTEM OF PARTICLES Consider a system of n particles as shown in the figure. The net or the resultant weight is given as WR = ∑W. Summing the moments about the y-axis, we get ~ ~ ~ x WR = x1W1 + x2W2 + ……….. + xnWn ~ where x1 represents x coordinate of W1, etc.. Similarly, we can sum moments about the x and z-axes to find the coordinates of G.

By replacing the W with a M in these equations, the coordinates of the center of mass can be found.

CG / CM / CENTROID OF A BODY (Section 9.2) A rigid body can be considered as made up of an infinite number of particles. Hence, using the same principles as in the previous slide, we get the coordinates of G by simply replacing the discrete summation sign ( ∑ ) by the continuous summation sign ( ∫ ) and W by dW.

Similarly, the coordinates of the center of mass and the centroid of volume, area, or length can be obtained by replacing W by m, V, A, or L, respectively.

CONCEPT OF CENTROID The centroid C is a point which defines the geometric center of an object. The centroid coincides with the center of mass or the center of gravity only if the material of the body is homogenous (density or specific weight is constant throughout the body). If an object has an axis of symmetry, then the centroid of object lies on that axis. In some cases, the centroid is not located on the object.

Centroids of volumes

• If the density ρ (mass per unit volume) of the material that makes up a body is constant, then m = ρV,dm = ρdV V = ∫∫∫ dV V

VxC = ∫ dM yz = ∫∫∫ x dV ,

xC =

∫∫∫ xdV V

V

V

VyC = ∫ dM xz = ∫∫∫ ydV ,

yC =

∫∫∫ ydV V

V

V

VzC = ∫ dM xy = ∫∫∫ z dV , V

zC =

∫∫∫ zdV V

V

Centroids of Areas

• If the body consists of a homogeneous, thin plate of uniform thickness t and surface area A, then m = ρtA, dm = ρtdA A = ∫∫ dA A

AxC = ∫ dM yz = ∫∫ x dA,

xC =

∫∫ xdA A

A

A

AyC = ∫ dM xz = ∫∫ ydA,

yC =

∫∫ ydA A

A

A

AzC = ∫ dM xy = ∫∫ z dA, A

zC =

∫∫ zdA A

A

Centroid of Lines

• If the body consists of a homogeneous curved wire with a small uniform crosssectional area A and length L, then m = ρLA, dm = ρAdL L = ∫ dL L

LxC = ∫ dM yz = ∫ x dL,

xC =

∫ xdL L

L

L

LyC = ∫ dM xz = ∫ ydL,

yC =

∫ ydL L

L

L

LzC = ∫ dM xy = ∫ z dL, L

zC =

∫ zdL L

L

STEPS FOR DETERMING AREA CENTROID 1. Choose an appropriate differential element dA at a general point (x,y). Hint: Generally, if y is easily expressed in terms of x (e.g., y = x2 + 1), use a vertical rectangular element. If the converse is true, then use a horizontal rectangular element. 2. Express dA in terms of the differentiating element dx (or dy).

~ , ~y ) of the centroid of the rectangular 3. Determine coordinates (x element in terms of the general point (x,y). 4. Express all the variables and integral limits in the formula using either x or y depending on whether the differential element is in terms of dx or dy, respectively, and integrate. Note: Similar steps are used for determining CG, CM, etc.. These steps will become clearer by doing a few examples.

EXAMPLE Given: The area as shown. Find: The centroid location (x , y)

Plan: Follow the steps. Solution x,y • •

~ ~ x,y

1. Since y is given in terms of x, choose dA as a vertical rectangular strip. 2. dA = y dx = (9 – x2) dx 3. x~ = x and ~ y = y/2

EXAMPLE (continued) ~

4. x = ( ∫A x dA ) / ( ∫A dA ) 3

=

2) d x ∫ x ( 9 – x 0 3

2) d x ∫ ( 9 – x 0

=

[ 9 (x2)/2 – (x4) / 4] 03 [ 9 x – (x3) / 3 ] 3

0

= ( 9 ( 9 ) / 2 – 81 / 4 ) / ( 9 ( 3 ) – ( 27 / 3 ) ) = 1.13 ft 3

~

y =

∫A y dA ∫A dA

=

½ 0 ∫ ( 9 – x2) ( 9 – x2) dx 3

2) d x ∫ ( 9 – x 0

= 3.60 ft

CONCEPT QUIZ 1. The steel plate with known weight and nonuniform thickness and density is supported as shown. Of the three parameters (CG, CM, and centroid), which one is needed for determining the support reactions? Are all three parameters located at the same point? A) B) C) D)

(center of gravity, no) (center of gravity, yes) (centroid, yes) (centroid, no)

2. When determining the centroid of the area above, which type of differential area element requires the least computational work? A) Vertical

B) Horizontal

C) Polar

D) Any one of the above.

EXAMPLE

Given: The area as shown. Find:

The x of the centroid.

Plan:

Follow the steps.

Solution (x1,,y)

1. Choose dA as a horizontal rectangular (x2,y) strip. 2. dA = ( x2 – x1) dy = ((2 – y) – y2) dy ~ 3. x = ( x1 + x2) / 2 = 0.5 (( 2 – y) + y2 )

EXAMPLE(continued)

4.

x

~

= ( ∫A x dA ) / ( ∫A dA )

∫A dA =

0∫

1

( 2 – y – y2) dy

[ 2 y – y2 / 2 – y3 / 3] 01 =

1.167 m2

1 ~ ∫A x dA = 0∫ 0.5 ( 2 – y + y2 ) ( 2 – y – y2 ) dy 1

= 0.5 0∫ ( 4 – 4 y + y2 – y4 ) dy = 0.5 [ 4 y – 4 y2 / 2 + y3 / 3 – y5 / 5 ] 10 = 1.067 m3 x =

1.067 / 1.167 = 0.914 m

ATTENTION QUIZ 1.

If a vertical rectangular strip is chosen as the differential element, then all the variables, including the integral limit, should be in terms of _____ . A) x

B) y

C) z

D) Any of the above.

2. If a vertical rectangular strip is chosen, then what are the values of ~ ~ x and y? A) (x , y)

B) (x / 2 , y / 2)

C) (x , 0)

D) (x , y / 2)

homework

• 9-7, 9-9, 9-13, 9-34

COMPOSITE BODIES (Section 9.3) Objective: Students will be able to determine the location of the center of gravity, center of mass, or centroid using the method of composite bodies.

CONCEPT OF A COMPOSITE BODY

b

a e

b d

a

e d

Many industrial objects can be considered as composite bodies made up of a series of connected “simpler” shaped parts or holes, like a rectangle, triangle, and semicircle. Knowing the location of the centroid, C, or center of gravity, G, of the simpler shaped parts, we can easily determine the location of the C or G for the more complex composite body.

CONCEPT OF A COMPOSITE BODY (continued)

b

a e

a

b e d

d

This can be done by considering each part as a “particle” and following the procedure as described in Section 9.1. This is a simple, effective, and practical method of determining the location of the centroid or center of gravity.

Centroids of common shapes of areas and volumes

STEPS FOR ANALYSIS 1. Divide the body into pieces that are known shapes. Holes are considered as pieces with negative weight or size. 2. Make a table with the first column for segment number, the second column for weight, mass, or size (depending on the problem), the next set of columns for the moment arms, and, finally, several columns for recording results of simple intermediate calculations. 3. Fix the coordinate axes, determine the coordinates of the center of gravity of centroid of each piece, and then fill-in the table. 4. Sum the columns to get x, y, and z. Use formulas like x = ( Σ x∼i Ai ) / ( Σ Ai ) or x = ( Σ x∼i Wi ) / ( Σ Wi ) This approach will become clear by doing examples.

EXAMPLE Given: The part shown. a

c

Find:

The centroid of the part.

b d

Plan: Follow the steps for analysis.

Solution: 1. This body can be divided into the following pieces rectangle (a) + triangle (b) + quarter circular (c) – semicircular area (d)

EXAMPLE (continued) a

Steps 2 & 3: Make up and fill the table using parts a, b, c, and d.

c

b d

Segment

Area A (in2)

x∼ (in)

∼y (in)

A∼ x ( in3)

A∼ y ( in3)

Rectangle Triangle Q. Circle Semi-Circle

18 4.5 9π/4 –π/2

3 7 – 4(3) / (3 π) 0

1.5 1 4(3) / (3 π) 4(1) / (3 π)

54 31.5 –9 0

27 4.5 9 - 2/3

76.5

39.83

Σ

28.0

EXAMPLE (continued)

·C

4. Now use the table data and these formulas to find the coordinates of the centroid.

x = ( Σ x∼ A) / ( Σ A ) = 76.5 in3/ 28.0 in2 = 2.73 in y = ( Σ y∼ A) / (Σ A ) =

39.83 in3 / 28.0 in2 = 1.42 in

EXAMPLE - 2 Given: Two blocks of different materials are assembled as shown. The specific weights of the materials are ρA = 150 lb / ft3 and ρB = 400 lb / ft3. Find: The center of gravity of this assembly. Plan: Follow the steps for analysis Solution 1. In this problem, the blocks A and B can be considered as two segments.

EXAMPLE – 2 (continued) Weight = w = ρ (Volume in ft3) wA = 150 (0.5) (6) (6) (2) / (12)3 = 3.125 lb wB = Segment w (lb) x∼ (in) A B

3.125 18.75

Σ

21.88

4 1

450 (6) (6) (2) / (12)3 = 18.75 lb

y∼ (in)

∼z (in)

w x∼ (lb·in)

∼ w ∼y wz (lb·in) (lb·in)

1 3

2 3

12.5 18.75

3.125 56.25

6.25 56.25

31.25

59.38

62.5

EXAMPLE – 2(continued)

x = (Σ x~ w) / ( Σw ) = 31.25/21.88 = 1.47 in y = (Σ y~ w) / ( Σw ) = 59.38/21.88 = 2.68 in z = (Σ z~ w) / ( Σw ) = 62.5 /21.88 = 2.82 in

homework

• 9-44, 9-55, 9-61, 9-83