10.5 MOMENT OF INERTIA FOR A COMPOSITE AREA

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10.5 MOMENT OF INERTIA FOR A COMPOSITE AREAA composite area is made by adding orsubtracting a series of “simple” shapedareas like rectangles, triangles, andcircles.For example, the area on the left can bemade from a rectangle minus a triangleand circle.The MoI of these “simpler” shaped areas about their centroidalaxes are found in most engineering handbooks as well as theinside back cover of the textbook.Using these data and the parallel-axis theorem, the MoI for acomposite area can easily be calculated.

STEPS FOR ANALYSIS1. Divide the given area into itssimpler shaped parts.2. Locate the centroid of each partand indicate the perpendiculardistance from each centroid tothe desired reference axis.3. Determine the MoI of each “simpler” shaped part about thedesired reference axis using the parallel-axis theorem( IX IX’ A ( dy )2 ) .4. The MoI of the entire area about the reference axis isdetermined by performing an algebraic summation of theindividual MoIs obtained in Step 3. (Please note that MoIof a hole is subtracted).

EXAMPLEGiven: The beam’s cross-sectional area.Find:The moment of inertia of the areaabout the y-axis and the radius ofgyration ky.Plan:Follow the steps for analysis.[1] [2] [3]Solution1. The cross-sectional area can be divided into three rectangles ([1], [2], [3] ) as shown.2. The centroids of these three rectangles are in their center.The distances from these centers to the y-axis are 0 mm,87.5 mm, and 87.5 mm, respectively.

EXAMPLE (continued)3. From the inside back cover of thebook, the MoI of a rectangle aboutits centroidal axis is (1/12) b h3.Iy[1] (1/12) (25mm) (300mm)3 56.25 (106) mm4[1] [2] [3]Using the parallel-axis theorem,IY[2] IY[3] IY’ A (dX)2 (1/12) (100) (25)3 (25) (100) ( 87.5 )2 19.27 (106) mm 4

EXAMPLE (continued)4.Iy Iy1 Iy2 Iy3 94.8 ( 106) mm 4ky ( Iy / A)A 300 (25) 25 (100) 25 (100) 12,500 mm 2ky ( 94.79) (106) / (12500) 87.1 mm

CONCEPT QUIZ1. For the area A, we know thecentroid’s (C) location, area, distancesbetween the four parallel axes, and theMoI about axis 1. We can determinethe MoI about axis 2 by applying theparallel axis theorem .A) directly between the axes 1 and 2.B) between axes 1 and 3 and thenbetween the axes 3 and 2.C) between axes 1 and 4 and thenaxes 4 and 2.D) None of the above.AxisAd3d2d1C 4321

CONCEPT QUIZ (continued)AxisAd3d2d1C 43212. For the same case, consider the MoI about each of the fouraxes. About which axis will the MoI be the smallest number?A)Axis 1B)Axis 2C)Axis 3D)Axis 4E)Can not tell.

GROUP PROBLEM SOLVINGGiven: The shaded area as shown in thefigure.Find: The moment of inertia for the areaabout the x-axis and the radius ofgyration kX.Plan: Follow the steps for analysis.(a) (b) (c)Solution1. The given area can be obtained by subtracting both the circle (b) andtriangle (c) from the rectangle (a).2. Information about the centroids of the simple shapes can beobtained from the inside back cover of the book. The perpendiculardistances of the centroids from the x-axis are: da 5 in , db 4 in, and dc 8 in.

GROUP PROBLEM SOLVING (continued)3. IXa (1/12) 6 (10)3 6 (10)(5)2 2000 in 4IXb (1/4) π (2)4 π (2)2 (4)2 213.6 in 4IXc(a) (b) (c)IX IXa (1 /36) (3) (6)3 (½) (3) (6) (8)2 594 in 4–IXb– IXc 1190 in 4kX ( IX / A )A 10 ( 6 ) –π (2)2 – (½) (3) (6) kX (1192 / 38.43) 5.57 in.38.43 in 2

A 10 cm2ATTENTION QUIZ1. For the given area, the moment of inertiaabout axis 1 is 200 cm4 . What is the MoIabout axis 3 (the centroidal axis)?A) 90 cm 4B) 110 cm 4C) 60 cm 4D) 40 cm 4d2A) 8 cmC) 24 cm 4 .4B) 56 cm .D) 26 cm 4 .32 d11d1 d2 2 cm3cm2. The moment of inertia of the rectangle aboutthe x-axis equals4.CC 2cm2cmx

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

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