CVL 131 - Statics Moment of nertia Composite Areas A math professor in an unheated room is cold and calculating. Radius of Gration This actuall sounds like some sort of rule for separation on a dance floor. t actuall is just a propert of a shape and is used in the analsis of how some shapes act in different conditions. Monda, November 6, 01 1
Radius of Gration The radius of gration, k, is the square root of the ratio of the moment of inertia to the area k = A k k = A J A O O = = + A 3 Monda, November 6, 01 Parallel Ais Theorem f ou know the moment of inertia about a centroidal ais of a figure, ou can calculate the moment of inertia about an parallel ais to the centroidal ais using a simple formula = + A = + A Monda, November 6, 01
Parallel Ais Theorem Since we usuall use the bar over the centroidal ais, the moment of inertia about a centroidal ais also uses the bar over the ais designation 5 = + A = + A Monda, November 6, 01 Parallel Ais Theorem 6 f ou look carefull at the epression, ou should notice that the moment of inertia about a centroidal ais will alwas be the minimum moment of inertia about an ais that is parallel to the centroidal ais. = + A = + A Monda, November 6, 01 3
Parallel Ais Theorem 7 n a manner similar to that which we used to calculate the centroid of a figure b breaking it up into component areas, we can calculate the moment of inertia of a composite area = + A = + A Monda, November 6, 01 Parallel Ais Theorem 8 nside 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 = + A = + A Monda, November 6, 01
Parallel Ais Theorem = + A = + A HERE S A CRTCAL MOMENT OF CAUTON REMEMBER HOW THE PARALLEL AXS S WRTTEN F THE AXS SHOWN N THE TABLE S NOT THROUGH THE CENTROD, THEN THE FORMULA DOES NOT GVE YOU THE MOMENT OF NERTA THROUGH THE CENTRODAL AXS 9 Monda, November 6, 01 Parallel Ais Theorem = + A = + A B eample The given for the Semicircular area in the table is about the centroidal ais The given for the same Semicircular area in the table is not about the centroidal ais 10 Monda, November 6, 01 5
Using The Table We want to locate the moment of inertia in the position shown of a semicircular area as shown about the and ais, and 11 Monda, November 6, 01 Using the Table First, we can look at the table and find the and about the ais as shown 1 Monda, November 6, 01 6
Using the Table n this problem, the ais is 8 from the centroidal ais and ais is 6 below the base of the semicircle, this would be usuall evident from the problem description 8" 13 Monda, November 6, 01 Using the Table Calculating the ou should notice that the ais in the table is the centroid ais so we won t have to move it et 1 = π r 8 1 = π ( 5in) 8 = 5.in 1 Monda, November 6, 01 7
Using the Table Net we can calculate the area A A ( 5in) π = = 39.7in 15 Monda, November 6, 01 Using the Table f we know that distance between the ais and the bar ais, we can calculate the moment of inertia using the parallel ais theorem = + Ad = + Ad.1" 16 Monda, November 6, 01 8
Using the Table changed the notation for the distances moved to avoid confusion with the distance from the origin = + Ad = + Ad.1" 17 Monda, November 6, 01 Using the Table The ais we are considering ma not alwas be a the origin. = + Ad = + Ad.1" 18 Monda, November 6, 01 9
Using the Table f the ais is ches to the left of the centroidal ais, then the moment of inertia about the ais would be = + Ad ( )( ) = 5.in + 39.7in 8in = 758.7in.1" 19 Monda, November 6, 01 Using the Table The moment of inertia about the ais is a slightl different case since the formula presented in the table is the moment of inertia about the base of the semicircle, not the centroid.1" 0 Monda, November 6, 01 10
Using the Table To move it to the moment of inertia about the -ais, we have to make two steps = A d = + A d base ( ) base to centroid ( ) centroid to -ais.1" 1 Monda, November 6, 01 Using the Table We can combine the two steps = A d = + A d ( ) base to centroid ( ) centroid to -ais ( ) ( ) = A d + A d base base base to centroid centroid to -ais.1" Monda, November 6, 01 11
Using the Table Don t tr and cut corners here You have to move to the centroid first = A d base = + A d ( ) base to centroid ( ) centroid to -ais ( ) ( ) = A d + A d base base to centroid centroid to -ais.1" 3 Monda, November 6, 01 Using the Table n this problem, we have to locate the centroid of the figure with respect to the base We can use the table to determine this ( in) r 5 = = 3π 3π =.1in This bar is with respect the base of the object, not the -ais..1" Monda, November 6, 01 1
Using the Table Now the in the table is given about the bottom of the semicircle, not the centroidal ais That is where the ais is shown in the table.1" 5 Monda, November 6, 01 Using the Table So ou can use the formula to calculate the ( base ) about the bottom of the semicircle base base base 1 = π r 8 1 = π ( 5in) 8 = 5.in.1" 6 Monda, November 6, 01 13
Using the Table Now we can calculate the moment of inertia about the centroidal ais = + Ad base = Ad base base to centroid base to centroid ( )( ) = 5.in 39.7in.1in = 68.60in.1" 7 Monda, November 6, 01 Using the Table And we can move that moment of inertia the the -ais = + Ad centroid to -ais ( )( ) = 68.60in + 39.7in +.1in = 657.8in.1" 8 Monda, November 6, 01 1
Using the Table The polar moment of inertia about the origin would be J = + J = 657.8in + 758.7in J O O O = 516.5.1" 9 Monda, November 6, 01 Another Eample We can use the parallel ais theorem to find the moment of inertia of a composite figure 30 Monda, November 6, 01 15
Another Eample 3" 31 Monda, November 6, 01 Another Eample We can divide up the area into smaller areas with shapes from the table 3" 3 Monda, November 6, 01 16
Another Eample Since the parallel ais theorem will require the area for each section, that is a reasonable place to start D Area (in ) 36 9 7 3" 33 Monda, November 6, 01 Another Eample We can locate the centroid of each area with respect the ais. D Area bar i (in ) (in) 36 3 9 7 7 6 3" 3 Monda, November 6, 01 17
Another Eample From the table in the back of the book we find that the moment of inertia of a rectangle about its -centroid 1 ais is 3 = b h 3" 35 1 D Area bar i (in ) (in) 36 3 9 7 7 6 Monda, November 6, 01 Another Eample n this eample, for Area, b=6 and h=6 1 = 6 in 6 in 1 = 108in ( )( ) 3 3" D Area bar i (in ) (in) 36 3 9 7 7 6 36 Monda, November 6, 01 18
Another Eample For the first triangle, the moment of inertia calculation isn t as obvious 3" 37 Monda, November 6, 01 Another Eample The wa it is presented in the tet, we can onl find the about the centroid 3" h b 38 Monda, November 6, 01 19
Another Eample The change ma not seem obvious but it is just in how we orient our ais. Remember an ais is our decision. h 3" b b h 39 Monda, November 6, 01 Another Eample So the moment of inertia of the triangle can be calculated using the formula with the correct orientation. 1 3 = bh 36 1 = 6 in 3 in 36 =.5in ( )( ) 3 3" 0 Monda, November 6, 01 0
Another Eample The same is true for the triangle = 1 36 bh 3 3" 1 = 6 in 9 in 36 = 11.5in ( )( ) 3 1 Monda, November 6, 01 Another Eample Now we can enter the bar for each sub-area into the table Sub- Area Area bar i bar (in ) (in) (in ) 36 3 108 9 7.5 7 6 11.5 3" Monda, November 6, 01 1
Another Eample We can then sum the and the A(d ) to get the moment of inertia for each sub-area 3" Sub- Area Area bar i bar A(d ) bar + A(d ) (in ) (in) (in ) (in ) (in ) 36 3 108 3 3 9 7.5 1 5.5 7 6 11.5 97 1093.5 3 Monda, November 6, 01 Another Eample 1971 And if we sum that last column, we have the for the composite figure 3" Sub- Area Area bar i bar A(d ) bar + A(d ) (in ) (in) (in ) (in ) (in ) 36 3 108 3 3 9 7.5 1 5.5 7 6 11.5 97 1093.5 1971 Monda, November 6, 01
Another Eample We perform the same tpe analsis for the 3" D Area (in ) 36 9 7 5 Monda, November 6, 01 Another Eample Locating the -centroids from the -ais 3" Sub-Area Area bar i (in ) (in) 36 3 9 7-6 Monda, November 6, 01 3
Another Eample Determining the for each sub-area 3" Sub-Area Area bar i bar (in ) (in) (in ) 36 3 108 9 18 7-5 7 Monda, November 6, 01 Another Eample Making the A(d ) multiplications 3" Sub- Area Area bar i bar A(d ) (in ) (in) (in ) (in ) 36 3 108 3 9 18 36 7-5 108 8 Monda, November 6, 01
Another Eample Summing and calculating Sub- Area Area bar i bar A(d ) bar + A(d ) (in ) (in) (in ) (in ) (in ) 36 3 108 3 3 9 18 36 5 7-5 108 16 3" 68 9 Monda, November 6, 01 Homework Problem 10-7 Problem 10-9 Problem 10-7 50 Monda, November 6, 01 5