Centroids and Moment of Inertia Calculation.Pptx

14 January 2011

Mechanics of Materials CIVL 3322 / MECH 3322 Centroids and Moment of Inertia Calculations

Centroids

n n n x A yA z A ∑ i i ∑ ii ∑ i i x i=1 i=1 i=1 = n y = n z = n A A A ∑ i ∑ i ∑ i i=1 i=1 i=1

2 Centroid and Moment of Inertia Calculations

1 14 January 2011

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

I I Ay2 z = z +

P07_045

2 14 January 2011

P07_045

An Example

¢ Lets start with an example problem and see how this develops n xA ∑ ii i=1 x = n 1 in A ∑ i i=1

1 in 1in

3 14 January 2011

An Example

¢ We want to locate both the x and y centroids n xA ∑ ii i=1 x = n 1 in A ∑ i i=1

1 in 1in

An Example

¢ There isn’t much of a chance of developing a function that is easy to integrate in this case n xA ∑ ii i=1 x = n 1 in A ∑ i i=1

1 in 1in

8 Centroid and Moment of Inertia3 Calculations in

4 14 January 2011

An Example

¢ We can break this shape up into a series of shapes that we can find the centroid of n xA ∑ ii i=1 x = n 1 in A ∑ i i=1

1 in 1in

9 Centroid and Moment of Inertia3 Calculations in

An Example

¢ There are multiple ways to do this as long

as you are consistent n xA ∑ ii i=1 x = n 1 in A ∑ i i=1

1 in 1in

3 in 10 Centroid and Moment of Inertia Calculations

5 14 January 2011

An Example

¢ First, we can develop a rectangle on the left side of the diagram, we will label that as area 1, A n 1 xA ∑ ii i=1 x = n 1 in A ∑ i i=1 A 1 in 1

1 in 1in

11 Centroid and Moment of Inertia3 in Calculations

An Example

¢ A second rectangle will be placed in the

bottom of the figure, we will label it A2

n xA ∑ ii 1 in i=1 x = n A A ∑ i 1 in 1 i=1

A 2 1 in 1in

6 14 January 2011

An Example

¢ A right triangle will complete the upper

right side of the figure, label it A3

n xA ∑ ii 1 in i=1 x = n A A ∑ i 1 in 1 i=1 A3 A 2 1 in 1in

An Example

¢ Finally, we will develop a negative area to remove the quarter circle in the lower left

hand corner, label it A4 n xA ∑ ii 1 in i=1 x = n A A ∑ i 1 in 1 i=1 A3 A A 4 2 1 in 1in

14 Centroid and Moment of Inertia3 Calculationsin

7 14 January 2011

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 n xA 1 in ∑ ii i=1 ID Area x = n 2 (in ) A A 2 1 in ∑ i 1 A i 1 1 A3 = A2 3

A3 1.5 A A 1 in A4 -0.7854 4 2 1in

An Example

¢ Now we will calculate the distance to the local centroids from the y-axis (we are

calculating an x-centroid) n

xAii ID Area xi 1 in ∑ 2 i 1 (in ) (in) = x = n A1 2 0.5 Ai A2 3 2.5 1 in ∑ A1 A i=1 A3 1.5 2 3

A4 -0.7854 0.42441 A 1 in A4 2 1in

8 14 January 2011

An Example

¢ To calculate the top term in the expression we need to multiply the entries in the last

two columns by one another n xA ∑ ii i 1 ID Area xi xi*Area 1 in = x = n (in2) (in) (in3) A A1 2 0.5 1 ∑ i A2 3 2.5 7.5 1 in i=1 A1 A A3 1.5 2 3 3

A4 -0.7854 0.42441 -0.33333 A 1 in A4 2 1in

17 Centroid and Moment of Inertia Calculations 3 in

An Example

¢ If we sum the second column, we have the

bottom term in the division, the total area

ID Area xi xi*Area xA ∑ ii 2 3 (in ) (in) (in ) x = i=1 A1 2 0.5 1 1 in n

A2 3 2.5 7.5 A ∑ i A3 1.5 2 3 i=1 A -0.7854 0.42441 -0.33333 1 in A 4 1 A3 5.714602 A 1 in A4 2 1in

9 14 January 2011

An Example

¢ And if we sum the fourth column, we have

the top term, the area moment

ID Area xi xi*Area n (in2) (in) (in3) xAii A1 2 0.5 1 ∑ i=1 A2 3 2.5 7.5 x = 1 in n A3 1.5 2 3 A A -0.7854 0.42441 -0.33333 ∑ i 4 i=1 5.714602 11.16667 1 in A 1 A3

A 1 in A4 2 1in

An Example

¢ Dividing the sum of the area moments by

the total area we calculate the x-centroid

n ID Area x x*Area i i xAii (in2) (in) (in3) ∑ x = i=1 A1 2 0.5 1 1 in n A2 3 2.5 7.5 A ∑ i A3 1.5 2 3 i=1 1 in A4 -0.7854 0.42441 -0.33333 A1 A 5.714602 11.16667 3

x bar 1.9541 A 1 in A4 2 1in

10 14 January 2011

An Example

¢ You can always remember which to divide by if you look at the final units, remember

that a centroid is a distance n ID Area x x*Area i i xAii (in2) (in) (in3) ∑ x = i=1 A1 2 0.5 1 1 in n A2 3 2.5 7.5 A ∑ i A3 1.5 2 3 i=1 1 in A4 -0.7854 0.42441 -0.33333 A1 A 5.714602 11.16667 3

x bar 1.9541 A 1 in A4 2 1in

An Example

¢ We can do the same process with the y

centroid

n ID Area x x*Area i i yAii (in2) (in) (in3) ∑ y = i=1 A1 2 0.5 1 1 in n A2 3 2.5 7.5 A ∑ i A3 1.5 2 3 i=1 1 in A4 -0.7854 0.42441 -0.33333 A1 A 5.714602 11.16667 3

x bar 1.9541 A 1 in A4 2 1in

11 14 January 2011

An Example

¢ Notice that the bottom term doesn’t change, the area of the figure hasn’t

changed n ID Area x x*Area i i yAii (in2) (in) (in3) ∑ y = i=1 A1 2 0.5 1 1 in n A2 3 2.5 7.5 A ∑ i A3 1.5 2 3 i=1 1 in A4 -0.7854 0.42441 -0.33333 A1 A 5.714602 11.16667 3

x bar 1.9541 A 1 in A4 2 1in

An Example

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

ID Area xi xi*Area yi (in2) (in) (in3) (in) n A1 2 0.5 1 1 yA A2 3 2.5 7.5 0.5 ∑ ii i=1 A3 1.5 2 3 1.333333 y = n A4 -0.7854 0.42441 -0.33333 0.42441 A ∑ i 5.714602 11.16667 i=1

x bar 1.9541 1 in

1 in A 1 A3

A 1 in A4 2 1in

12 14 January 2011

An Example

¢ Calculate the area moments about the x-

ID Area xi xi*Area yi yi*Area n (in2) (in) (in3) (in) (in3) yAii A1 2 0.5 1 1 2 ∑ A 3 2.5 7.5 0.5 1.5 i=1 2 y = n A3 1.5 2 3 1.333333 2 Ai A4 -0.7854 0.42441 -0.33333 0.42441 -0.33333 ∑ 5.714602 11.16667 i=1

x bar 1.9541 1 in

1 in A 1 A3

A 1 in A4 2 1in

An Example

¢ Sum the area moments

ID Area xi xi*Area yi yi*Area (in2) (in) (in3) (in) (in3) n A1 2 0.5 1 1 2 yA A2 3 2.5 7.5 0.5 1.5 ∑ ii i 1 A3 1.5 2 3 1.333333 2 = y = n A4 -0.7854 0.42441 -0.33333 0.42441 -0.33333 A 5.714602 11.16667 5.166667 ∑ i i=1 x bar 1.9541 1 in

1 in A 1 A3

A 1 in A4 2 1in

13 14 January 2011

An Example

¢ And make the division of the area

moments by the total area

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

A1 2 0.5 1 1 2 yA ∑ ii A2 3 2.5 7.5 0.5 1.5 y = i=1 A 1.5 2 3 1.333333 2 n 3 A A4 -0.7854 0.42441 -0.33333 0.42441 -0.33333 ∑ i 5.714602 11.16667 5.166667 i=1

x bar 1.9541 y bar 0.904117 1 in

1 in A 1 A3

A 1 in A4 2 1in

¢ 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 IIAxyy=+ 2 IIAyxx=+ 28 Centroid and Moment of Inertia Calculations

14 14 January 2011

¢ 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 IIAxyy=+ 2 IIAyxx=+ 29 Centroid and Moment of Inertia Calculations

¢ 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 IIAxyy=+ 2 IIAyxx=+ 30 Centroid and Moment of Inertia Calculations

15 14 January 2011

Another Example

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

Another Example

y 6" 3"

16 14 January 2011

Another Example

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

y 6" 3"

6" I II x III 6"

Another Example

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

y 6" 3" ID Area (in2) I 36 6" I II 9 II III 27 x III 6"

17 14 January 2011

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

y 6" 3" ID Area xbari (in2) (in) I 36 3 6" I II 9 7 II x III 27 6 III 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 1 3 Ibh= y 6" 3" y 12

ID Area xbar 6" I i II (in2) (in) x III I 36 3 6" II 9 7 III 27 6 36 Centroid and Moment of Inertia Calculations

18 14 January 2011

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

1 3 Iininy = (66)( ) 12 4 y 6" 3" Iiny =108

ID Area xbar 6" I i II (in2) (in) x III I 36 3 6" II 9 7 III 27 6 37 Centroid and Moment of Inertia Calculations

Another Example

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

y 6" 3"

6" I II x III 6"

19 14 January 2011

Another Example

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

the Ix about the centroid

y 6" 3" h x 6" I II x III b 6"

Another Example The change may not seem obvious but it is just in how we orient our axis. Remember an axis is our decision.

h x y 6" 3"

x 6" I II x b III 6"

20 14 January 2011

Another Example So the moment of inertia of the II triangle can be calculated using the formula with the correct orientation.

1 3 Ibh= y 6" 3" y 36

1 3 6" I Iy = (63 in)( in) 36 II x 4 III Iy = 4.5 in 6"

Another Example The same is true for the III triangle

1 3 Ibh= y 6" 3" y 36

1 3 6" I Iy = (69 in)( in) 36 II x 4 III Iy =121.5 in 6"

21 14 January 2011

Another Example

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

Sub- y 6" 3" Area Area xbari Iybar (in2) (in) (in4) 6" I I 36 3 108 II x II 9 7 4.5 III III 27 6 121.5 6"

Another Example 2 We can then sum the Iy and the A(dx) to get the moment of inertia for each sub-area y 6" 3"

6" I II x III 6" Sub- Iybar + 2 2 Area Area xbari Iybar A(dx) A(dx) (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

22 14 January 2011 2 2 Sub-Area Area xbari Iybar A(dx) Iybar + A(dx) (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 And if we sum that last column, we have the

Iy for the composite figure y 6" 3"

6" I II x III 6" Sub- Iybar + 2 2 Area Area xbari Iybar A(dx) A(dx) (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

Another Example

We perform the same type analysis for the Ix y 6" 3"

6" I II x ID Area III (in2) 6" I 36 II 9 III 27

23 14 January 2011

Another Example

Locating the y-centroids from the x-axis

y 6" 3"

6" I Sub-Area Area ybari II x 2 III (in ) (in) 6" I 36 3 II 9 2 III 27 -2

Another Example

Determining the Ix for each sub-area

y 6" 3"

6" I II x Sub-Area Area ybari Ixbar III 6" (in2) (in) (in4) I 36 3 108 II 9 2 18 III 27 -2 54

24 14 January 2011

Another Example 2 Making the A(dy) multiplications

y 6" 3"

6" I II Sub- x 2 III Area Area ybari Ixbar A(dy) 6" (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 6" 3" Sub- I + xbar 6" I 2 2 Area Area ybari Ixbar A(dy) A(dy) II x 2 4 4 4 III (in ) (in) (in ) (in ) (in ) 6" I 36 3 108 324 432 II 9 2 18 36 54 III 27 -2 54 108 162

25 14 January 2011

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

26 14 January 2011

Homework Problem 3