CIVL 2131Moment - Statics of Inertia Composite Areas
A math professor in an unheated room is cold and calculating.
Radius of Gyration
¢ This actually sounds like some sort of rule for separation on a dance floor. ¢ It actually is just a property of a shape and is used in the analysis of how some
shapes act in different conditions.
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1 Radius of Gyration
¢ The radius of gyration, k, is the square root of the ratio of the moment of inertia to the area I k = x x A I k = y y A J II+ k ==O xy O AA
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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 2 IIAxyy=+ 2 IIAyxx=+ 4 Moment of Inertia - Composite Area Monday, November 26, 2012
2 Parallel Axis Theorem
¢ 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=+ 5 Moment of Inertia - Composite Area Monday, November 26, 2012
Parallel Axis Theorem
¢ 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=+ 6 Moment of Inertia - Composite Area Monday, November 26, 2012
3 Parallel Axis Theorem
¢ In a manner similar to that which we used to calculate the centroid of a figure by breaking it up into component areas, we can calculate the moment of inertia of a
composite area 2 IIAxyy=+ 2 IIAyxx=+ 7 Moment of Inertia - Composite Area Monday, November 26, 2012
Parallel Axis Theorem
¢ Inside the back cover of the book, in the same figure that we used for the centroid calculations we can find calculations for
moments of inertia 2 IIAxyy=+ 2 IIAyxx=+ 8 Moment of Inertia - Composite Area Monday, November 26, 2012
4 2 IIAxyy=+ Parallel Axis Theorem 2 IIAyxx=+
¢ HERE IS A CRITICAL MOMENT OF CAUTION ¢ REMEMBER HOW THE PARALLEL AXIS IS WRITTEN ¢ IF THE AXIS SHOWN IN THE TABLE IS NOT THROUGH THE CENTROID, THEN THE FORMULA DOES NOT GIVE YOU THE MOMENT OF INERTIA THROUGH
THE CENTROIDAL AXIS
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2 IIAxyy=+ Parallel Axis Theorem 2 IIAyxx=+
¢ By example
¢ The Iy given for the Semicircular area in the table is about the centroidal axis
¢ The Ix given for the same Semicircular area in the table is not about the centroidal
axis
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5 Using The Table
¢ We want to locate the moment of inertia in the position shown of a semicircular area
as shown about the x and y axis, Ix and Iy
y
10"
x
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Using the Table
¢ First, we can look at the table and find the
Ix and Iy about the axis as shown
y
10"
x
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6 Using the Table
¢ In this problem, the y axis is 8” from the y centroidal axis and x axis is 6” below the base of the semicircle, this would be usually evident from the problem
description 5" y 8"
10" 6"
x
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Using the Table
¢ Calculating the Iy you should notice that the y axis in the table is the centroid axis
so we won’t have to move it yet 1 Ir= π 4 y 5" 8 y 1 4 Iy = π (5 in) 10" 8 x 4 Iy = 245.44 in
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7 Using the Table
¢ Next we can calculate the area
2 π (5in) A = 5" 2 y 2 10" Ain= 39.27 x
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Using the Table
¢ If we know that distance between the y axis and the ybar axis, we can calculate the moment of inertia using the parallel
axis theorem 5" 2 y Iyy=+ I Ad x 8 in
2 6in 10" Ixx=+ I Ad y 2.12" x
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8 Using the Table
¢ I changed the notation for the distances moved to avoid confusion with the
distance from the origin
5" 2 y Iyy=+ I Ad x 8 in
2 6in 10" Ixx=+ I Ad y 2.12" x
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Using the Table
¢ The axis we are considering may not
always be a the origin.
5" 2 y Iyy=+ I Ad x 8 in
2 6in 10" Ixx=+ I Ad y 2.12" x
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9 Using the Table
¢ If the y axis is 8 inches to the left of the centroidal axis, then the moment of inertia
about the y axis would be
IIAd2 5" yy=+ x y 422 8 in Iy =+245.44 in( 39.27 in)( 8 in) 4 6in 10" 2.12" Iy = 2758.72 in x
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Using the Table
¢ The moment of inertia about the x axis is a slightly different case since the formula presented in the table is the moment of inertia about the base of the semicircle,
not the centroid 5" y 8 in
6in 10" 2.12" x
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10 Using the Table
¢ To move it to the moment of inertia about
the x-axis, we have to make two steps
5" 2 y IIx= base − Ad( base to centroid ) 8 in
2 6in 10" 2.12" IIAdxx=+( centroid to x-axis ) x
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Using the Table
¢ We can combine the two steps
2 IIx= base − Ad( base to centroid ) 2 IIAdxx=+( centroid to x-axis ) 22 IIx= base − Ad( base to centroid) + Ad( centroid to x-axis )
5" y 8 in
6in 10" 2.12" x 22 Moment of Inertia - Composite Area Monday, November 26, 2012
11 Using the Table
¢ Don’t try and cut corners here
¢ You have to move to the centroid first 2 IIx= base − Ad( base to centroid ) 2 IIAdxx=+( centroid to x-axis ) 22 IIx= base − Ad( base to centroid) + Ad( centroid to x-axis )
5" y 8 in
6in 10" 2.12" x
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Using the Table
¢ In this problem, we have to locate the y centroid of the figure with respect to the base
¢ We can use the table to determine this This ybar is with respect the base of 4r 45( in) the object, not the x-axis. y == 33ππ 5" y yin= 2.12 8 in
6in 10" 2.12" x
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12 Using the Table
¢ Now the Ix in the table is given about the bottom of the semicircle, not the centroidal axis ¢ That is where the x axis is shown in the table
5" y 8 in
6in 10" 2.12" x
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Using the Table
¢ So you can use the formula to calculate
the Ix (Ibase) about the bottom of the semicircle 1 Ir= π 4 5" base 8 y 8 in 1 4 Ibase = π (5 in) 6in 10" 2.12" 8 x 4 Ibase = 245.44 in
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13 Using the Table
¢ Now we can calculate the moment of inertia about the x centroidal axis
2 IIAdbase=+ x base to centroid 2 IIxbase=− Adbase to centroid I245.44 in42 39.27 in 2.12 in 2 x =−( )( ) 5" y 4 Ix = 68.60 in 8 in
6in 10" 2.12" x
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Using the Table
¢ And we can move that moment of inertia the the x-axis
2 Ixx=+ I Adcentroid to x-axis 42 2 Iininininx =+68.60( 39.27)( 6 + 2.12 ) 4 5" y Iinx = 2657.84 8 in
6in 10" 2.12" x
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14 Using the Table
¢ The polar moment of inertia about the origin would be
JIIOxy=+ 44 JO =+2657.84 in 2758.72 in 4 5" JO = 5416.56 in y 8 in
6in 10" 2.12" x
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Another Example
¢ We can use the parallel axis theorem to find the moment of inertia of a composite figure
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15 Another Example
y 6" 3"
6"
x
6"
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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"
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16 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"
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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"
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17 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 35 Moment of Inertia - Composite Area Monday, November 26, 2012
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 36 Moment of Inertia - Composite Area Monday, November 26, 2012
18 Another Example
For the first triangle, the moment of inertia calculation isn’t as obvious
y 6" 3"
6" I II x III 6"
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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"
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19 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"
b
x 6" I II x b III 6"
h
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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"
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20 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"
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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"
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21 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
2 2 Sub-Area Area xbari Iybar A(dx) Iybar + A(dx) 43 Moment of Inertia - Composite Area Monday, November 26, 2012 (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
1971
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22 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
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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
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23 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
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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
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24 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
648
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Homework
¢ Problem 10-27 ¢ Problem 10-29 ¢ Problem 10-47
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