centroids and moment of inertia ... - university of - memphis and moment of... · centroids and...

14 January 2011

1

Mechanics of Materials CIVL 3322 / MECH 3322

Centroids and

Moment of Inertia Calculations

Centroid and Moment of Inertia Calculations 2

Centroids

x =xi Ai

i=1

n

∑

Aii=1

n

∑1

1

n

i iin

ii

y Ay

A

=

=

=∑

∑

z =zi Ai

i=1

n

∑

Aii=1

n

∑

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2


Parallel Axis Theorem

¢  If you know the moment of inertia about a centroidal axis of a figure, you can calculate the moment of inertia about any parallel axis to the centroidal axis using a simple formula

Iz = Iz + Ay2

P07_045


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P07_045


An Example ¢  Lets start with an example problem and

see how this develops

1

1

n

i iin

ii

x Ax

A

=

=

=∑

∑

1in

1 in

1 in

3 in

1 in

6 Centroid and Moment of Inertia Calculations

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4

An Example ¢ We want to locate both the x and y

centroids

1in

1 in

1 in

3 in

1 in

1

1

n

i iin

ii

x Ax

A

=

=

=∑

∑


An Example ¢ There isn’t much of a chance of

developing a function that is easy to integrate in this case

1in

1 in

1 in

3 in

1 in

1

1

n

i iin

ii

x Ax

A

=

=

=∑

∑


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An Example ¢ We can break this shape up into a series

of shapes that we can find the centroid of

1in

1 in

1 in

3 in

1 in

1

1

n

i iin

ii

x Ax

A

=

=

=∑

∑


An Example ¢ There are multiple ways to do this as long

as you are consistent

1in

1 in

1 in

3 in

1 in

1

1

n

i iin

ii

x Ax

A

=

=

=∑

∑


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6

An Example ¢ First, we can develop a rectangle on the

left side of the diagram, we will label that as area 1, A1

1in

1 in

1 in

3 in

1 in

A1

1

1

n

i iin

ii

x Ax

A

=

=

=∑

∑


An Example ¢ A second rectangle will be placed in the

bottom of the figure, we will label it A2

1in

1 in

1 in

3 in

1 in

A1

A2

1

1

n

i iin

ii

x Ax

A

=

=

=∑

∑


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7

An Example ¢ A right triangle will complete the upper

right side of the figure, label it A3

1in

1 in

1 in

3 in

1 in

A1

A2

A3

1

1

n

i iin

ii

x Ax

A

=

=

=∑

∑


An Example ¢ Finally, we will develop a negative area to

remove the quarter circle in the lower left hand corner, label it A4

1in

1 in

1 in

3 in

1 in

A1

A2

A3

A4

1

1

n

i iin

ii

x Ax

A

=

=

=∑

∑


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An Example ¢ We will begin to build a table so that

keeping up with things will be easier ¢ The first column will be the areas

1in

1 in

1 in

3 in

1 in

A2

A3A1

A4

1

1

n

i iin

ii

x Ax

A

=

=

=∑

∑ID Area

(in2)A1 2A2 3A3 1.5A4 -0.7854


An Example ¢ Now we will calculate the distance to the

local centroids from the y-axis (we are calculating an x-centroid)

1

1

n

i iin

ii

x Ax

A

=

=

=∑

∑

ID Area xi(in2) (in)

A1 2 0.5A2 3 2.5A3 1.5 2A4 -0.7854 0.42441

1in

1 in

1 in

3 in

1 in

A2

A3A1

A4


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An Example ¢ To calculate the top term in the expression

we need to multiply the entries in the last two columns by one another

1

1

n

i iin

ii

x Ax

A

=

=

=∑

∑ID Area xi xi*Area

(in2) (in) (in3)A1 2 0.5 1A2 3 2.5 7.5A3 1.5 2 3A4 -0.7854 0.42441 -0.33333

1in

1 in

1 in

3 in

1 in

A2

A3A1

A4


An Example ¢  If we sum the second column, we have the

bottom term in the division, the total area

1

1

n

i iin

ii

x Ax

A

=

=

=∑

∑

ID Area xi xi*Area(in2) (in) (in3)

A1 2 0.5 1A2 3 2.5 7.5A3 1.5 2 3A4 -0.7854 0.42441 -0.33333

5.714602

1in

1 in

1 in

3 in

1 in

A2

A3A1

A4


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An Example ¢ And if we sum the fourth column, we have

the top term, the area moment

1

1

n

i iin

ii

x Ax

A

=

=

=∑

∑


A1 2 0.5 1A2 3 2.5 7.5A3 1.5 2 3A4 -0.7854 0.42441 -0.33333

5.714602 11.16667

1in

1 in

1 in

3 in

1 in

A2

A3A1

A4


An Example ¢ Dividing the sum of the area moments by

the total area we calculate the x-centroid

1

1

n

i iin

ii

x Ax

A

=

=

=∑

∑


A1 2 0.5 1A2 3 2.5 7.5A3 1.5 2 3A4 -0.7854 0.42441 -0.33333

5.714602 11.16667x bar 1.9541

1in

1 in

1 in

3 in

1 in

A2

A3A1

A4


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An Example ¢ You can always remember which to divide

by if you look at the final units, remember that a centroid is a distance

1

1

n

i iin

ii

x Ax

A

=

=

=∑

∑


A1 2 0.5 1A2 3 2.5 7.5A3 1.5 2 3A4 -0.7854 0.42441 -0.33333

5.714602 11.16667x bar 1.9541

1in

1 in

1 in

3 in

1 in

A2

A3A1

A4


An Example ¢ We can do the same process with the y

centroid

1

1

n

i iin

ii

y Ay

A

=

=

=∑

∑


A1 2 0.5 1A2 3 2.5 7.5A3 1.5 2 3A4 -0.7854 0.42441 -0.33333

5.714602 11.16667x bar 1.9541

1in

1 in

1 in

3 in

1 in

A2

A3A1

A4


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An Example ¢ Notice that the bottom term doesn’t

change, the area of the figure hasn’t changed

1

1

n

i iin

ii

y Ay

A

=

=

=∑

∑


A1 2 0.5 1A2 3 2.5 7.5A3 1.5 2 3A4 -0.7854 0.42441 -0.33333

5.714602 11.16667x bar 1.9541

1in

1 in

1 in

3 in

1 in

A2

A3A1

A4


An Example ¢ We only need to add a column of yi’s

1

1

n

i iin

ii

y Ay

A

=

=

=∑

∑

ID Area xi xi*Area yi

(in2) (in) (in3) (in)A1 2 0.5 1 1A2 3 2.5 7.5 0.5A3 1.5 2 3 1.333333A4 -0.7854 0.42441 -0.33333 0.42441

5.714602 11.16667x bar 1.9541

1in

1 in

1 in

3 in

1 in

A2

A3A1

A4


14 January 2011

13

An Example ¢ Calculate the area moments about the x-

axis

1

1

n

i iin

ii

y Ay

A

=

=

=∑

∑

ID Area xi xi*Area yi yi*Area(in2) (in) (in3) (in) (in3)

A1 2 0.5 1 1 2A2 3 2.5 7.5 0.5 1.5A3 1.5 2 3 1.333333 2A4 -0.7854 0.42441 -0.33333 0.42441 -0.33333

5.714602 11.16667x bar 1.9541

1in

1 in

1 in

3 in

1 in

A2

A3A1

A4


An Example ¢ Sum the area moments

1

1

n

i iin

ii

y Ay

A

=

=

=∑

∑


A1 2 0.5 1 1 2A2 3 2.5 7.5 0.5 1.5A3 1.5 2 3 1.333333 2A4 -0.7854 0.42441 -0.33333 0.42441 -0.33333

5.714602 11.16667 5.166667x bar 1.9541

1in

1 in

1 in

3 in

1 in

A2

A3A1

A4


14 January 2011

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An Example ¢ And make the division of the area

moments by the total area

1

1

n

i iin

ii

y Ay

A

=

=

=∑

∑


A1 2 0.5 1 1 2A2 3 2.5 7.5 0.5 1.5A3 1.5 2 3 1.333333 2A4 -0.7854 0.42441 -0.33333 0.42441 -0.33333

5.714602 11.16667 5.166667x bar 1.9541 y bar 0.904117

1in

1 in

1 in

3 in

1 in

A2

A3A1

A4




¢  If you know the moment of inertia about a centroidal axis of a figure, you can calculate the moment of inertia about any parallel axis to the centroidal axis using a simple formula

2

2

= +

= +y y

x x

I I Ax

I I Ay

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¢ Since we usually use the bar over the centroidal axis, the moment of inertia about a centroidal axis also uses the bar over the axis designation

2

2

= +

= +y y

x x

I I Ax

I I Ay



¢  If you look carefully at the expression, you should notice that the moment of inertia about a centroidal axis will always be the minimum moment of inertia about any axis that is parallel to the centroidal axis.

2

2

= +

= +y y

x x

I I Ax

I I Ay

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Another Example

¢ We can use the parallel axis theorem to find the moment of inertia of a composite figure


Another Example

y

x

6" 3"

6"

6"

14 January 2011

17


Another Example

y

x

6" 3"

6"

6"

III

III

¢ We can divide up the area into smaller areas with shapes from the table


Another Example

Since the parallel axis theorem will require the area for each section, that is a reasonable place to start

y

x

6" 3"

6"

6"

III

III

ID Area (in2) I 36 II 9 III 27

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18


Another Example We can locate the centroid of each area with respect

the y axis.

y

x

6" 3"

6"

6"

III

III

ID Area xbari (in2) (in) I 36 3 II 9 7 III 27 6


Another Example

From the table in the back of the book we find that the moment of inertia of a rectangle about its y-centroid

axis is 31

12=yI b h y

x

6" 3"

6"

6"

III

III


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Another Example In this example, for Area I, b=6” and h=6”

( )( )3

4

1 6 612108

=

=

y

y

I in in

I in y

x

6" 3"

6"

6"

III

III



Another Example

For the first triangle, the moment of inertia calculation isn’t as obvious

y

x

6" 3"

6"

6"

III

III

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20


Another Example

The way it is presented in the text, we can only find the Ix about the centroid

x

b

h

y

x

6" 3"

6"

6"

III

III


Another Example The change may not seem obvious but it is

just in how we orient our axis. Remember an axis is our decision.

x

b

h

x

h

b

y

x

6" 3"

6"

6"

III

III

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21


Another Example So the moment of inertia of the II triangle can

be calculated using the formula with the correct orientation.

( )( )

3

3

4

1361 6 3364.5

=

=

=

y

y

y

I bh

I in in

I in

y

x

6" 3"

6"

6"

III

III


Another Example The same is true for the III triangle

( )( )

3

3

4

1361 6 936121.5

=

=

=

y

y

y

I bh

I in in

I in

y

x

6" 3"

6"

6"

III

III

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Another Example Now we can enter the Iybar for each sub-area into the

table

y

x

6" 3"

6"

6"

III

III

Sub-Area Area xbari Iybar

(in2) (in) (in4) I 36 3 108 II 9 7 4.5 III 27 6 121.5


Another Example We can then sum the Iy and the A(dx)2 to get

the moment of inertia for each sub-area y

x

6" 3"

6"

6"

III

III

Sub-Area Area xbari Iybar A(dx)2

Iybar + A(dx)2

(in2) (in) (in4) (in4) (in4)

I 36 3 108 324 432

II 9 7 4.5 441 445.5

III 27 6 121.5 972 1093.5

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Another Example And if we sum that last column, we have the

Iy for the composite figure

Sub-Area Area xbari Iy bar A(dx)2 Iy bar + A(dx)2

(in2) (in) (in4) (in4) (in4)I 36 3 108 324 432II 9 7 4.5 441 445.5III 27 6 121.5 972 1093.5

1971

y

x

6" 3"

6"

6"

III

III

Sub-Area Area xbari Iybar A(dx)2

Iybar + A(dx)2

(in2) (in) (in4) (in4) (in4) I 36 3 108 324 432 II 9 7 4.5 441 445.5 III 27 6 121.5 972 1093.5

1971


Another Example

We perform the same type analysis for the Ix

ID Area (in2) I 36 II 9 III 27

y

x

6" 3"

6"

6"

III

III

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Another Example Locating the y-centroids from the x-axis

y

x

6" 3"

6"

6"

III

IIISub-Area Area ybari

(in2) (in) I 36 3 II 9 2 III 27 -2


Another Example Determining the Ix for each sub-area

y

x

6" 3"

6"

6"

III

IIISub-Area Area ybari Ixbar

(in2) (in) (in4)

I 36 3 108

II 9 2 18

III 27 -2 54

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Another Example Making the A(dy)2 multiplications

y

x

6" 3"

6"

6"

III

IIISub-Area Area ybari Ixbar A(dy)2

(in2) (in) (in4) (in4)

I 36 3 108 324

II 9 2 18 36

III 27 -2 54 108


Another Example Summing and calculating Ix

y

x

6" 3"

6"

6"

III

III

Sub-Area Area ybari Ixbar A(dy)2

Ixbar + A(dy)2

(in2) (in) (in4) (in4) (in4) I 36 3 108 324 432 II 9 2 18 36 54 III 27 -2 54 108 162

648

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Homework Problem 1


Calculate the centroid of the composite shape in reference to the bottom of the shape and calculate the moment of inertia about the centroid z-axis.

Homework Problem 2



14 January 2011

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Homework Problem 3



centroids and moment of inertia ... - university of - memphis and moment of... · centroids and...

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