0.5 MOMENT OF INERTIA FOR A COMPOSITE AREA A composite area is made by adding or subtracting a series of simple shaped areas like rectangles, triangles, and circles. For example, the area on the left can be made from a rectangle minus a triangle and circle. The MoI of these simpler shaped areas about their centroidal axes are found in most engineering handbooks as well as the inside back cover of the textbook. Using these data and the parallel-axis theorem, the MoI for a composite area can easily be calculated. Steps for calculating MoI for composite sections. Divide the given area into its simpler shaped parts.. Locate the centroid of each part and indicate the perpendicular distance from each centroid to the desired reference axis.. Determine the MoI of each simpler shaped part about the desired reference axis using the parallel-axis theorem ( I X = I X + A ( d y ) ).. MoI of entire area about the reference axis is determined by performing an algebraic summation of the individual MoIs obtained in Step. EXAMPLE 0. Given: Find: Plan: The beam s cross-sectional area. The moment of inertia of the area about the y-axis and the radius of gyration k y. Follow the steps for analysis. Solution []. The cross-section can be divided into rectangles ( [], [], [] ) as shown.. 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. [] [] Statics: Lecture Notes for Sections 0.5
EXAMPLE 0. (continued). From the inside back cover of the book, the MoI of a rectangle about its centroidal axis is (/) b h. I y[] = (/) (5mm)(00mm) = 56.5 (0 6 ) mm Using the parallel-axis theorem, I Y[] = I Y[] = I Y + A (d X ) = (/) (00) (5) + (5) (00) ( 87.5 ) = 9.7 (0 6 ) mm EXAMPLE 0. (continued). I y = I y + I y + I y = 9.8 ( 0 6 ) mm 5. Radius of gyration k y = ( I y / A) A = 00 (5) + 5 (00) + 5 (00) =,500 mm k y = ( 9.79) (0 6 ) / (500) = 87. mm CONCEPT QUIZ. For the area A, we know the centroid s (C) location, area, distances between the four parallel axes, and the MoI about axis. We can determine the MoI about axis by applying the parallel axis theorem. d d d A C Axis A) directly between the axes and. B) between axes and and then between the axes and. C) between axes and and then axes and. D) None of the above. Statics: Lecture Notes for Sections 0.5
CONCEPT QUIZ (continued). For the same case, consider the MoI about each of the four axes. About which axis will the MoI be the smallest number? A) Axis B) Axis C) Axis D) Axis E) Can not tell. d d d A C Axis ATTENTION QUIZ. For the given area, the moment of inertia about axis is 00 cm. What is the MoI about axis (the centroidal axis)? A) 90 cm B) 0 cm C) 60 cm D) 0 cm d d C C A=0 cm d = d = cm. The moment of inertia of the rectangle about the x-axis equals A) 8 cm. B) 56 cm. cm cm C) cm. D) 6 cm. cm x Moments of Inertia of simple shapes 9-9 Statics: Lecture Notes for Sections 0.5
Moments of Inertia of standard sections (see steel design handbook for details) 9-0 Sample Problem 0. The strength of a Wx8 rolled steel beam is increased by attaching a plate to its upper flange. Determine the moment of inertia and radius of gyration with respect to an axis which is parallel to the plate and passes through the centroid of the section. Determine location of the centroid of composite section with respect to a coordinate system with origin at the centroid of the beam section. Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to composite section centroidal axis. Calculate the radius of gyration from the moment of inertia of the composite section. 9 - Sample Problem 0. Determine location of the centroid of composite section with respect to a coordinate system with origin at the centroid of the beam section. Section Plate BeamSection A, in y, in. ya, in 6.75 7.5 50..0 0 0 A = 7.95 ya = 50. Y = ya 50. in = =.79 in. A 7.95 in 9 - Statics: Lecture Notes for Sections 0.5
Sample Problem 0. Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to composite section centroidal axis. I x,beam section = Ix + AY = 85 + (.0)(.79) = 7. in I x,plate = Ix + Ad = = 5. in I x = I x, beam section + I x,plate = 7. + 5. ( 9)( ) + ( 6.75)( 7.5.79) = 68 in I x Calculate the radius of gyration from the moment of inertia of the composite section. I 67.5 in k = x x = A 7.95 in = 5.87 in. k x 9 - Sample Problem 0.5 Determine the moment of inertia of the shaded area with respect to the x axis. Compute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis. The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle. 9 - Sample Problem 0.5 r ( )( 90) a = = = 8. mm π π b = 0 - a = 8.8 mm A = πr = π ( 90) =.7 0 mm Compute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis. Rectangle: 6 I x = bh = ( 0 )( 0) = 8. 0 mm Half-circle: moment of inertia with respect to AA, 6 = πr = ( 90 ) = 5.76 0 mm I AA π 8 8 moment of inertia with respect to x, I x = I AA Aa = 6 = 7.0 0 mm 6 ( 5.76 0 )(.7 0 ) moment of inertia with respect to x, 6 I x = Ix + Ab = 7.0 0 + (.7 0 )( 8.8) 6 = 9. 0 mm 9-5 Statics: Lecture Notes for Sections 0.5 5
Sample Problem 0.5 The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle. I x = 8. 0 6 mm 9. 0 6 mm I = 5.9 0 6 mm x 9-6 Statics: Lecture Notes for Sections 0.5 6