Moment Of Inertia - Composite Areas
CIVL 2131- StaticsMomentof InertiaComposite AreasA math professor in an unheated roomis cold and calculating.Radius of GyrationThis actually sounds like some sort of rulefor separation on a dance floor. It actually is just a property of a shape andis used in the analysis of how someshapes act in different conditions. 2Moment of Inertia - Composite AreaMonday, November 26, 20121
Radius of Gyration The radius of gyration, k, is the square rootof the ratio of the moment of inertia to theareakx ky kO 3IxAIyAJO AIx IyAMoment of Inertia - Composite AreaMonday, November 26, 2012Parallel Axis Theorem If you know the moment of inertia about acentroidal axis of a figure, you cancalculate the moment of inertia about anyparallel axis to the centroidal axis using asimple formulaI y I y Ax2I x I x Ay 24Moment of Inertia - Composite AreaMonday, November 26, 20122
Parallel Axis Theorem Since we usually use the bar over thecentroidal axis, the moment of inertiaabout a centroidal axis also uses the barover the axis designationI y I y Ax 2I x I x Ay 25Moment of Inertia - Composite AreaMonday, November 26, 2012Parallel Axis Theorem If you look carefully at the expression, youshould notice that the moment of inertiaabout a centroidal axis will always be theminimum moment of inertia about any axisthat is parallel to the centroidal axis.I y I y Ax2I x I x Ay 26Moment of Inertia - Composite AreaMonday, November 26, 20123
Parallel Axis Theorem In a manner similar to that which we usedto calculate the centroid of a figure bybreaking it up into component areas, wecan calculate the moment of inertia of acomposite areaI y I y Ax 2I x I x Ay 27Moment of Inertia - Composite AreaMonday, November 26, 2012Parallel Axis Theorem Inside the back cover of the book, in thesame figure that we used for the centroidcalculations we can find calculations formoments of inertiaI y I y Ax2I x I x Ay 28Moment of Inertia - Composite AreaMonday, November 26, 20124
Parallel Axis TheoremI y I y Ax 2I x I x Ay 2HERE IS A CRITICAL MOMENT OFCAUTION REMEMBER HOW THE PARALLEL AXISIS WRITTEN IF THE AXIS SHOWN IN THE TABLE ISNOT THROUGH THE CENTROID, THENTHE FORMULA DOES NOT GIVE YOUTHE MOMENT OF INERTIA THROUGHTHE CENTROIDAL AXIS 9Moment of Inertia - Composite AreaParallel Axis TheoremMonday, November 26, 2012I y I y Ax 2I x I x Ay 2By example The Iy given for the Semicircular area inthe table is about the centroidal axis The Ix given for the same Semicirculararea in the table is not about the centroidalaxis 10Moment of Inertia - Composite AreaMonday, November 26, 20125
Using The Table We want to locate the moment of inertia inthe position shown of a semicircular areaas shown about the x and y axis, Ix and Iyy10"x11Moment of Inertia - Composite AreaMonday, November 26, 2012Using the Table First, we can look at the table and find theIx and Iy about the axis as showny10"x12Moment of Inertia - Composite AreaMonday, November 26, 20126
Using the Table In this problem, the y axis is 8” from the ycentroidal axis and x axis is 6” below thebase of the semicircle, this would beusually evident from the problemdescription5"y8"10"6"x13Moment of Inertia - Composite AreaMonday, November 26, 2012Using the Table Calculating the Iy you should notice thatthe y axis in the table is the centroid axisso we won’t have to move it yet1Iy π r4814I y π ( 5in )8I y 245.44in 414yMoment of Inertia - Composite Area5"10"xMonday, November 26, 20127
Using the Table A Next we can calculate the areaπ ( 5in )2y2A 39.27in 2155"10"xMoment of Inertia - Composite AreaMonday, November 26, 2012Using the Table If we know that distance between the yaxis and the ybar axis, we can calculatethe moment of inertia using the parallelaxis theorem5"y28 inyxI y I Ad166inI x I x Ad y 2Moment of Inertia - Composite Area10"2.12"xMonday, November 26, 20128
Using the Table I changed the notation for the distancesmoved to avoid confusion with thedistance from the originI y I y Ad x 28 in10"6inI x I x Ad y 2175"yMoment of Inertia - Composite Area2.12"xMonday, November 26, 2012Using the Table The axis we are considering may notalways be a the origin.I y I y Ad x 2188 in6inI x I x Ad y 25"yMoment of Inertia - Composite Area10"2.12"xMonday, November 26, 20129
Using the Table If the y axis is 8 inches to the left of thecentroidal axis, then the moment of inertiaabout the y axis would beI y I y Ad x2I y 245.44in 4 ( 39.27in 2 ) (8in )28 in10"6inI y 2758.72in 4195"yMoment of Inertia - Composite Area2.12"xMonday, November 26, 2012Using the Table The moment of inertia about the x axis is aslightly different case since the formulapresented in the table is the moment ofinertia about the base of the semicircle,not the centroid5"y8 in6in20Moment of Inertia - Composite Area10"2.12"xMonday, November 26, 201210
Using the Table To move it to the moment of inertia aboutthe x-axis, we have to make two stepsI x I base A ( d base to centroid )2125"y8 in10"6inI x I x A ( d centroid to x-axis )2Moment of Inertia - Composite Area2.12"xMonday, November 26, 2012Using the Table We can combine the two stepsI x I base A ( d base to centroid )I x I x A ( d centroid to x-axis )22I x I base A ( d base to centroid ) A ( d centroid to x-axis )225"y8 in6in22Moment of Inertia - Composite Area10"2.12"xMonday, November 26, 201211
Using the TableDon’t try and cut corners here You have to move to the centroid first I x I base A ( d base to centroid )I x I x A ( d centroid to x-axis )22I x I base A ( d base to centroid ) A ( d centroid to x-axis )225"y8 in6in23Moment of Inertia - Composite Area10"2.12"xMonday, November 26, 2012Using the TableIn this problem, we have to locate the ycentroid of the figure with respect to thebase We can use the table to determine this 4r 4 ( 5in )y 3π3πy 2.12inThis ybar is with respect the base ofthe object, not the x-axis.5"y8 in6in24Moment of Inertia - Composite Area10"2.12"xMonday, November 26, 201212
Using the TableNow the Ix in the table is given about thebottom of the semicircle, not the centroidalaxis That is where the x axis is shown in thetable 5"y8 in6in10"25Moment of Inertia - Composite Area2.12"xMonday, November 26, 2012Using the Table So you can use the formula to calculatethe Ix (Ibase) about the bottom of thesemicircle265"yMoment of Inertia - Composite Area8 in6in1I base π r 4814I base π ( 5in )8I base 245.44in 410"2.12"xMonday, November 26, 201213
Using the Table Now we can calculate the moment ofinertia about the x centroidal axis2I base I x Ad baseto centroid2I x I base Ad baseto centroidI x 245.44in 4 ( 39.27in 2 ) ( 2.12in )2I x 68.60in 45"y8 in6in10"27Moment of Inertia - Composite Area2.12"xMonday, November 26, 2012Using the Table And we can move that moment of inertiathe the x-axis2I x I x Ad centroidto x-axisI x 68.60in 4 ( 39.27in 2 ) ( 6in 2.12in )I x 2657.84in 45"y8 in6in28Moment of Inertia - Composite Area210"2.12"xMonday, November 26, 201214
Using the Table The polar moment of inertia about theorigin would beJO I x I yJ O 2657.84in 4 2758.72in 4J O 5416.56in 45"y8 in6in29Moment of Inertia - Composite Area10"2.12"xMonday, November 26, 2012Another Example 30We can use the parallel axis theorem tofind the moment of inertia of a compositefigureMoment of Inertia - Composite AreaMonday, November 26, 201215
Another Exampley6"3"6"x6"31Moment of Inertia - Composite AreaMonday, November 26, 2012Another Example We can divide up the area into smallerareas with shapes from the tabley6"6"I3"IIxIII6"32Moment of Inertia - Composite AreaMonday, November 26, 201216
Another ExampleSince the parallel axis theorem will require the area foreach section, that is a reasonable place to t of Inertia - Composite AreaMonday, November 26, 2012Another ExampleWe can locate the centroid of each area with respectthe y III6"34Moment of Inertia - Composite AreaMonday, November 26, 201217
Another ExampleFrom the table in the back of the book we find that themoment of inertia of a rectangle about its y-centroidaxis is3y6"3"y1I 6Moment of Inertia - Composite AreaMonday, November 26, 2012Another ExampleIn this example, for Area I, b 6” and h 6”136in6in( )( )12I y 108in 4Iy "6Moment of Inertia - Composite AreaMonday, November 26, 201218
Another ExampleFor the first triangle, the moment of inertia calculationisn’t as obviousy6"6"I3"IIxIII6"37Moment of Inertia - Composite AreaMonday, November 26, 2012Another ExampleThe way it is presented in the text, we can only findthe Ix about the centroidhy6"x 6"I3"IIxIIIb386"Moment of Inertia - Composite AreaMonday, November 26, 201219
Another ExampleThe change may not seem obvious but it isjust in how we orient our axis. Remember anaxis is our decision.hxy6"6"I3"bxIIxIIIb6"h39Moment of Inertia - Composite AreaMonday, November 26, 2012Another ExampleSo the moment of inertia of the II triangle canbe calculated using the formula with thecorrect orientation.1 3bh3613I y ( 6in )( 3in )36I y 4.5in 4Iy 40y6"6"IMoment of Inertia - Composite Area3"IIxIII6"Monday, November 26, 201220
Another ExampleThe same is true for the III triangle1 3bh3613I y ( 6in )( 9in )36I y 121.5in 4Iy 41y6"6"I3"IIxIII6"Moment of Inertia - Composite AreaMonday, November 26, 2012Another ExampleNow we can enter the Iybar for each sub-area into 08II974.5III276121.5y6"6"IMoment of Inertia - Composite Area3"IIxIII6"Monday, November 26, 201221
Another ExampleWe can then sum the Iy and the A(dx)2 to getthe moment of inertia for eachy xbariIybarA(dx)2Iybar )376A(dx )2 Iy bar A(dx )2Iy barMoment of Inertia - Composite Area(in4)1084.5121.5Monday, November 26, 2012(in4)324441972(in4)432445.51093.51971Another ExampleAnd if we sum that last column, we have theIy for the composite ar 41445.5III276121.59721093.5I6"197144Moment of Inertia - Composite AreaMonday, November 26, 201222
Another ExampleWe perform the same type analysis for the IxIDy6"6"IIIArea(in2)36927IIIIII453"xIII6"Moment of Inertia - Composite AreaMonday, November 26, 2012Another ExampleLocating the y-centroids from the oment of Inertia - Composite Areay6"6"I3"IIxIII6"Monday, November 26, 201223
Another ExampleDetermining the Ix for each 8II9218III27-25447Moment of Inertia - Composite Areay6"6"I3"IIxIII6"Monday, November 26, 2012Another ExampleMaking the A(dy)2 108Moment of Inertia - Composite AreaxIII6"Monday, November 26, 201224
Another ExampleSumming and calculating IxSubAreaArea)2A(dyIxbar oment of Inertia - Composite AreaMonday, November 26, 2012HomeworkProblem 10-27 Problem 10-29 Problem 10-47 50Moment of Inertia - Composite AreaMonday, November 26, 201225
3 Moment of Inertia - Composite Area Monday, November 26, 2012 Radius of Gyration ! The radius of gyration, k, is the square root of the ratio of the moment of inertia to the area I x x y y O xy O k A I k A J II k AA 4 Moment of Inertia - Composite Area Monday, November 26, 2012 Para
31 Moment of Inertia by Integraion Monday, November 19, 2012 An Example ! The polar moment of inertia, J O, is the sum of the moments of inertia about the x and y axis y x yx2 4 1 2 4 yx 4m 4m 44 4 21.94 21.94 43.88 Ox y O O JII Jm m Jm 32 Moment of Inertia by Integraion Monday, N
I : moment of inertia about any parallel axis I CM: moment of inertia about an axis through its center of mass M : total mass h : distance from a parallel axis to the center of mass. BTW: The moment of inertia of any object about an axis through its center of mass is the minimum moment of
10.5 MOMENT OF INERTIA FOR A COMPOSITE AREA A composite area is made by adding or subtracting a series of “simple” shaped . CONCEPT QUIZ 1. For the area A, we know the centroid’s (C) location, area, distances . The shaded area as shown in the figure. Find: The moment of inertia for the area about the x-axis and the radius of
MOMENT OF INERTIA The moment of inertia, or the second moment of area, is defined as I y x2 dA I x y2 dA The polar moment of inertia J of an area about a point is equal to the sum of the moments of inertia of the area about any two point. I z J I y I x (x 2 y2) dA
Moment of Inertia is always positive. Units of Moment of Inertia are length raised to the fourth power, such as in4 or m4. Polar Moment of Inertia The second moment of Area A with respect to the pole O or the z-ax
dependent on the wheel inertia. So it is crucial that the inertia of the tire-wheel combination is simulated. The typical problem is that the dynamometer inertia is greater than the tire-wheel inertia. So the simulation must apply a force to the dyno inertia to compensate for the inertia difference
Moment of Inertia Calculation: We again split up the area into sub-areas 1 and 2 as shown previously. The moment of inertia of the composite area is determined by Where (Ix)i is the moment of inertia of the sub-area i about the x-axis passing through the centroid of t
ANSI A300 standards are the accepted industry standards for tree care practices. ANSI A300 Standards are divided into multiple parts, each focusing on a specific aspect of woody plant management. Tree Selection and Planting Recommendations Evaluation of the Site The specific planting site should be evaluated closely as it is essential to understand how the chemical, biological and physical .
