Announcements - Walla Walla Universitycurt.nelson/engr222/spring...Center of Gravity and Centroid - Chapter 9 Today’s Objective: Students will: a) Understand the concepts of center

Engr 222 Spring 2004 Chapter 9 1

Announcements

• Web page: homepages.wwc.edu/staff/nelscu• Email: [email protected]• Course time change

Center of Gravity and Centroid - Chapter 9Today’s 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. In-Class Activities:

• Reading quiz• Applications• Center of gravity, etc.• Concept quiz• Group problem solving• Attention quiz


Reading Quiz

1. The _________ is the point defining the geometric center of an object.

A) Center of gravity B) Center of mass

C) Centroid D) None of the above

2. To study problems concerned with the motion of matter under the influence of forces, i.e., dynamics, it is necessary to locate a point called the ________.

A) Center of gravity B) Center of mass

C) Centroid D) None of the above

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.

How can we determine these weights and their locations?


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 for making a SUV more stable?

How do you determine its location?

Concept of CG and CM

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

From the definition of a resultant force, the sum of moments dueto individual particle weight about any point is the same as themoment due to the resultant weight located at G. For the figureabove, 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.

•• •

3m

4N

1m

A B1 N

3 NG


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.

CG / CM for a System of Particles - Section 9.1

Consider a system of n particles as shown in the figure. The net or the resultant weight is given as WR = �W.

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.

Summing the moments about the y-axis, we get

x WR = x1W1 + x2W2 + ……….. + xnWn

where x1 represents x coordinate of W1, etc.

~~~

~


CG / CM / Centroid of a Body - Section 9.2A rigid body can be considered as composed 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.

Steps for Determining the Centroidal Location

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).

Note: Similar steps are used for determining CG, CM, etc. These steps will become clearer by doing a few examples.

4. Express all the variables and integral limits in the formulausing either x or y, depending on whether the differential element is in terms of dx or dy, respectively, and integrate.

3. Determine coordinates (x , y) of the centroid of the rectangular element in terms of the general point (x,y).

~ ~


Example

Solution

1. Since y is given in terms of x, choose dA as a vertical rectangular strip.

•

•x,y

x , y~~2. dA = y dx = (9 – x2) dx

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

Given: The area as shown.

Find: The centroid location (x , y)

Plan: Follow the steps.

Example - continued

4. x = ( �A x dA ) / ( �A dA )~

0

0

0 � x ( 9 – x2) d x [ 9 (x2)/2 – (x4) / 4] 3

0 � x ( 9 – x2) d x [ 9 x – (x3) / 3 ] 3

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

= 1.13 ft

3= =

3

33.60 ft�A y dA ½ 0 � ( 9 – x2) ( 9 – x2) dx

�A dA 0 � ( 9 – x2) d x

3

=y = = ~


Concept Quiz1. The steel plate with known weight and non-

uniform 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) (center of gravity, no)B) (center of gravity, yes)C) (centroid, yes)D) (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

Group Problem SolvingGiven: 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) (x2,y)

2. dA = ( x2 – x1) dy

= ((2 – y) – y2) dy

3. x = ( x1 + x2) / 2

= 0.5 (( 2 – y) + y2 )

~


4. x = ( �A x dA ) / ( �A dA )~

�A dA = 0� ( 2 – y – y2) dy

[ 2 y – y2 / 2 – y3 / 3] 1 = 1.167 m2

1

0

�A x dA = 0� 0.5 ( 2 – y + y2 ) ( 2 – y – y2 ) dy

= 0.5 0� ( 4 – 4 y + y2 – y4 ) dy

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

= 1.067 m30

1

1

~

x = 1.067 / 1.167 = 0.914 m

Group Problem Solving - continued

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)

~ ~


Textbook Problem 9-16

Locate the centroid of the shaded area shown below.


Locate the centroid of the solid.


Announcements

Composite Bodies - Section 9.3Today’s 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.

In-Class Activities:• Reading quiz• Applications• Method of composite bodies.• Concept quiz• Group problem solving• Attention quiz


Reading Quiz1. A composite body in this section refers to a body made of ____.

A) carbon fibers and an epoxy matrixB) steel and concreteC) a collection of “simple” shaped parts or holesD) a collection of “complex” shaped parts or holes

2. The composite method for determining the location of the center of gravity of a composite body requires _______.

A) integration B) differentiation

C) simple arithmetic D) All of the above.

Applications

The I - beam is commonly used in building structures. When doing a stress analysis on an I - beam, the location of the centroid is very important.

How can we easily determine the location of the centroid for a given beam shape?


Applications - continued

Cars, SUVs, bikes, etc., are assembled using many individual components.

When designing for stability on the road, it is important to know the location of the bikes’ center of gravity (CG).

If we know the weight and CG of individual components, how can we determine the location of the CG of the assembled unit?

Concept of a Composite Body

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.

a

be

d

ab

ed


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.

a

be

d

ab

ed

Concept of a Composite Body - continued

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 centerof 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 xi Ai ) / ( Σ Ai ) or x = ( Σ xi Wi ) / ( Σ Wi )

This approach will become clear by doing examples.

∼ ∼


ExampleGiven: The part shown.

Find: The centroid of the part.

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)

abc

d

Example - continued

39.8376.528.0ΣΣΣΣ

274.59

- 2/3

5431.5– 9 0

1.51

4(3) / (3 π)4(1) / (3 π)

37

– 4(3) / (3 π)0

184.5

9 π / 4– π / 2

RectangleTriangleQ. CircleSemi-Circle

A y( in3)

A x( in3)

y(in)

x(in)

Area A(in2)

Segment ∼ ∼ ∼ ∼

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

abc

d


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

∼

∼

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

C·

Example - continued

Concept Quiz

1. Based on the typical centroid information available in handbooks, what are the minimum number of segments you will have to consider for determing the centroid of the given area?

A) 1 B) 2 C) 3 D) 4

2. A storage box is tilted up to clean the rug underneath the box. It is tilted up by pulling the handle C, with edge A remaining on the ground. What is the maximum angle of tilt (measured between bottom AB and the ground) possible before the box tips over?

A) 30° B) 45 ° C) 60 ° D) 90 °

3cm 1 cm

1 cm

3cm

30º

G

C

AB


Group Problem SolvingGiven: Two blocks of different materials are assembled as shown. The weight densities of the materials are

ρA = 150 lb / ft3 andρB = 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 twosegments.

Weight = w = ρ (Volume in ft3)

wA = 150 (0.5) (6) (6) (2) / (12)3 = 3.125 lb

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

62.559.3831.2521.88ΣΣΣΣ

6.2556.25

3.12556.25

12.518.75

23

13

41

3.12518.75

AB

wz(lb·in)

w y (lb·in)

w x (lb·in)

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



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

~


Attention Quiz

1. A rectangular area has semicircular and triangular cuts as shown. For determining the centroid, what is the minimum number of pieces that you can use?

A) Two B) Three

C) Four D) Five

2. For determining the centroid of the area, two square segments are considered; square ABCD and square DEFG. What are the coordinates (x, y ) of the centroid of square DEFG?

A) (1, 1) m B) (1.25, 1.25) m

C) (0.5, 0.5 ) m D) (1.5, 1.5) m

~ ~

2cm 2cm

2cm

4cm

x

y

A1m

1m

y

E

F G

CB x

1m 1m

D



Determine the location, ybar, of the centroid of the beam’s cross-sectional area. Neglect the size of the corner welds at A and B for the calculation.

Documents

Announcements - Walla Walla Universitycurt.nelson/engr222/spring...Center of Gravity and Centroid - Chapter 9 Today’s Objective: Students will: a) Understand the concepts of center