EM102 Engineering Statics

Ch 8Chapter 8fCenter of Gravity and

Centroid(Part I)

Prepared by Lim CL

CENTER OF GRAVITY, CENTER OF MASS AND CENTROID OF A BODYCENTROID OF A BODY

Today’s Objective :

Students will:Students will:

a) Understand the concepts of center of gravity center of mass and centroidgravity, center of mass, and centroid.

b) Be able to determine the location of these points for a bodythese points for a body.

APPLICATIONS

To design the structure forTo design the structure for supporting a water tank, we will need to know the weights of the gtank and water as well as the locations where the resultant f i hforces representing these distributed loads act.

How can we determine these weights and their locations?

APPLICATIONS (continued)APPLICATIONS (continued)

One concern about a sport utility vehicle (SUV) is that it might tip over while taking a sharp turn.

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

Should it be higher or lower to make a SUV more stable?

d d i h l i f h fHow do you determine the location of the SUV’s center of mass?

APPLICATIONS (continued)APPLICATIONS (continued)

To design the ground support f h l i istructure for the goal post, it is

critical to find total weight of the structure and the center of gravity

I t ti t b d t

structure and the center of gravity location.

Integration must be used to determine total weight of the goal post due to the curvature of thepost due to the curvature of the supporting member.

H d d t i thHow do you determine the location of its center of gravity?gravity?

CONCEPT OF CENTER OF GRAVITY (CG)A body is composed of an infinite number ofparticles, and so if the body is located within agravitational field then each of these particlesgravitational field, then each of these particles will have a weight dW.

The center of gravity (CG) is a point, often shown as G, which locates the resultant weight of a system of particles or a solid body

From the definition of a resultant force, the sum of moments due to individual particle weighst

of a system of particles or a solid body.

of moments due to individual particle weighst about any point is the same as the moment due to the resultant weight located at G.

Also, note that the sum of moments due to the individual particle’s weights about point G is equal to zero.

CONCEPT OF CG (continued)

The location of the center of gravity, measured from the y axis, is determined by equating the moment of W about the y axis to the sum of the moments of the weights of the particles about this same axissame axis.

~~~

∫ d~If dW is located at point (x, y, z), then

x W = ∫ x dW~_

Similarly, y W = ∫ y dW~_z W = ∫ z dW~_

y, y W ∫ y dW z W ∫ z dW

Therefore, the location of the center of gravity G with respect to the x, y z axes becomesy,z axes becomes

CM & CENTROID OF A BODY

By replacing the W with a m in these equations, the coordinates y p g q ,of the center of mass can be found.

Similarly, the coordinates of the centroid of volume, area, or l h b b i d b l i blength can be obtained by replacing W by 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

g j

mass or the center of gravity only if the material of the body is homogenous (density or specific weight is constant throughout the p g gbody).

If an object has an axis of symmetry thenIf an object has an axis of symmetry, then the centroid of object lies on that axis.

I th t id i t l t dIn some cases, the centroid is not located on the object.

STEPS TO DETERME THE CENTROID OF AN AREA

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).

3 Determine coordinates ( x y) of the centroid of the rectangular3. Determine coordinates ( x , y) of the centroid of the rectangular 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 the CG or CM. These t ill b l b d i f l

terms of dx or dy, respectively, and integrate.

steps will become clearer by doing a few examples.

EXAMPLE

Given: The area as shown.

Fi d Th t id l ti ( )Find: The centroid location (x , y)

Plan: Follow the steps.

SolutionSolution

1. Since y is given in terms of x, choose dA ti l t l t i

2 dA = y dx = x3 dx

dA as a vertical rectangular strip.

2. dA = y dx = x dx

3. x = x and y = y / 2 = x3 / 2~~

EXAMPLE(continued)( )

4. x = ( ∫A x dA ) / ( ∫A dA )~

0∫ x (x3 ) d x 1/5 [ x5 ]11

0

0∫ (x3 ) d x 1/4 [ x4 ]1

( 1/5) / ( 1/4) 0 8

1= = 0

0= ( 1/5) / ( 1/4) = 0.8 m

1∫A y dA 0 ∫ (x3 / 2) ( x3 ) dx 1/14[x7]1

=y =~

= 0

∫A dA 0 ∫ x3 dx 1/41=y = =

(1/14) / (1/4) 0 2857= (1/14) / (1/4) = 0.2857 m

GROUP PROBLEM SOLVING

Given: The area as shown.

Find: The x of the centroid.

Plan: Follow the steps.

Solution

1. Choose dA as a horizontal rectangular strip.(x1 ,y) (x y)( 1,,y) (x2,y)

2. dA = ( x2 – x1) dy

= ((2 y) y2) dy= ((2 – y) – y ) dy

3. x = ( x1 + x2) / 2

= 0.5 (( 2 – y) + y2 )

GROUP PROBLEM SOLVING (continued)GROUP PROBLEM SOLVING (continued)

4. x = ( ∫A x dA ) / ( ∫A dA )~

∫A dA = 0∫ ( 2 – y – y2) dy

[ 2 2 / 2 3 / 3] 1 1 167 2

1

[ 2 y – y2 / 2 – y3 / 3] 1 = 1.167 m20

∫A x dA = 0∫ 0.5 ( 2 – y + y2 ) ( 2 – y – y2 ) dy1~∫A x dA 0∫ 0.5 ( 2 y y ) ( 2 y y ) dy

= 0.5 0∫ ( 4 – 4 y + y2 – y4 ) dy

0 5 [ 4 4 2 / 2 + 3 / 3 5 / 5 ] 1

1

= 0.5 [ 4 y – 4 y2 / 2 + y3 / 3 – y5 / 5 ] 1

= 1.067 m30

x = 1.067 / 1.167 = 0.914 m

EM102 Engineering Statics

Ch 8Chapter 8fCenter of Gravity and

Centroid(Part II)

Prepared by Lim CL

COMPOSITE BODIESToday’s Objective: COMPOSITE BODIESy j

Students will be able to determine:

) Th l ti f th t fa) The location of the center of gravity,

b) The location of the center of massb) The location of the center of mass,

c) And, the location of the centroid using the method of compositeusing the method of composite bodies.

APPLICATIONSAPPLICATIONS

The I-beam or tee-beam shown are commonly used in building various types of t tstructures.

When doing a stress or gdeflection analysis for a beam, the location of the centroid is

How can we easily determine

very important.

ow c we e s y de e ethe location of the centroid for a given beam shape?

APPLICATIONS (continued)

The compressor is assembled with many individual components.

In order to design the ground h i

y p

support structures, the reactions at blocks A and B have to be found To do this easily it isfound. To do this easily, it is important to determine the location of the compressor’s pcenter of gravity (CG).

If we know the weight and CG of individual components, we need a simple way to determine the location of the CG of the assembled unitassembled unit.

CG / CM OF A COMPOSITE BODY

Consider a composite body which consists of a series of particles(or bodies) as shown in the fi Th h l i h i i

Summing the moments about the y axis we get

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 ythe coordinates of G.

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

CONCEPT OF A COMPOSITE BODY

Many industrial objects can be considered as composite bodies made up of a series of connected “simple” shaped parts or holes, like a rectangle, triangle, and semicircle.

Knowing the location of the centroid, C, or center of gravity, G, of the simple shaped parts, we can easily determine the location f th C G f th l it b dof the C or G for the more complex composite body.

CONCEPT OF A COMPOSITE BODY(continued)

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 of a complex

t t t hipart, structure or machine.

STEPS FOR ANALYSISSTEPS FOR ANALYSIS

1. Divide the body into pieces that are known shapes. H l id d i ith ti i ht iHoles 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.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 tablegravity 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 xi Ai ) / ( ΣAi ) or x = ( Σ xi Wi ) / ( Σ Wi )

This approach will become clear by doing examples!

∼ ∼

pp y g p

EXAMPLE

Given: The part shown.

Find: The centroid of the part.

Plan: Follow the steps for analysis.

Solution:

1 Thi b d b di id d i h f ll i i

y

1. This body can be divided into the following pieces: rectangle (a) + triangle (b) + quarter circular (c) –semicircular area (d) Note the negative sign on the hole!semicircular area (d). Note the negative sign on the hole!

EXAMPLE (continued)

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

y Ax AyxArea ASegment ∼ ∼ ∼ ∼

274 5

5431 5

1.51

37

184 5

RectangleT i l

( in3)( in3)(in)(in)(in2)

4.59

- 2/3

31.5– 9 0

14(3) / (3 π)4(1) / (3 π)

7– 4(3) / (3 π)

0

4.59 π / 4– π / 2

TriangleQ. CircleSemi-Circle

39.8376.528.0Σ

2/3( ) ( )Semi Circle

EXAMPLE (continued)

C

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

Area A x A y A28 0 76 5 39 83

∼∼

3 2

28.0 76.5 39.83

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

y = ( Σ y A) / (ΣA ) = 39.83 in3 / 28.0 in2 = 1.42 in

∼

∼

GROUP PROBLEM SOLVING

Given: Two blocks of different materials are assembled as shown.

The densities of the materials are:The densities of the materials are:

ρA = 150 lb / ft3 andρ = 400 lb / ft3ρB = 400 lb / ft .

Find: The center of gravity of this assemblyassembly.

Plan: Follow the steps for analysis.Solution

1. In this problem, the blocks A and B can be considered as two pieces (or segments).

GROUP PROBLEM SOLVING (continued)( )

Weight = w = ρ (Volume in ft3)3wA = 150 (0.5) (6) (6) (2) / (12)3 = 3.125 lb

wB = 400 (6) (6) (2) / (12)3 = 16.67 lb

zA (lb·in)

yA (lb·in)

xA (lb·in)

z (in)y (in)x (in)w (lb)Segment ∼ ∼ ∼ ∼∼ ∼

6.2550.00

3.12550.00

12.516.67

23

13

41

3.12516.67

AB

56.2553.1229.1719.79Σ

GROUP PROBLEM SOLVING (continued)( )

Table Summary

ΣW (lb) x w y w z w(lb·in) (lb·in) (lb·in)

∼∼ ∼

19.79 29.17 53.12 56.25

~x = (Σ x w) / ( Σw ) = 29.17/19.79 = 1.47 in

Substituting into the equations:

( ) ( )

y = (Σ y w) / ( Σw ) = 53.12/ 19.79 = 2.68 in

(Σ ) / ( Σ ) 56 25 / 19 79 2 84 i~

~

z = (Σ z w) / ( Σw ) = 56.25 / 19.79 = 2.84 in