10.5 MOMENT OF INERTIA FOR A COMPOSITE AREA

Transcription

1 10.5 MOMENT OF NERTA FOR A COMPOSTE 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 Mo 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 Mo for a composite area can easily be calculated.

2 Steps for calculating Mo for composite sections 1. 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 Mo of each simpler shaped part about the desired reference axis using the parallel-axis theorem ( X = X + A ( d y ) ).. Mo of entire area about the reference axis is determined by performing an algebraic summation of the individual Mos obtained in Step.

3 EXAMPLE 10. 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. [] [1] [] Solution 1. The cross-section can be divided into rectangles ( [1], [], [] ) 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.

4 EXAMPLE 10. (continued). From the inside back cover of the book, the Mo of a rectangle about its centroidal axis is (1/1) b h. y[1] = (1/1) (5mm)(00mm) = 56.5 (10 6 ) mm Using the parallel-axis theorem, Y[] = Y[] = Y + A (d X ) = (1/1) (100) (5) + (5) (100) ( 87.5 ) = 19.7 (10 6 ) mm

5 EXAMPLE 10. (continued). y = y1 + y + y = 9.8 ( 10 6 ) mm 5. Radius of gyration k y = ( y / A) A = 00 (5) + 5 (100) + 5 (100) = 1,500 mm k y = ( 9.79) (10 6 ) / (1500) = 87.1 mm

6 CONCEPT QUZ 1. For the area A, we know the centroid s (C) location, area, distances between the four parallel axes, and the Mo about axis 1. We can determine the Mo about axis by applying the parallel axis theorem. d d d 1 A C Axis 1 A) directly between the axes 1 and. B) between axes 1 and and then between the axes and. C) between axes 1 and and then axes and. D) None of the above.

7 CONCEPT QUZ (continued). For the same case, consider the Mo about each of the four axes. About which axis will the Mo be the smallest number? A) Axis 1 B) Axis C) Axis D) Axis d d A C Axis E) Can not tell. d 1 1

8 ATTENTON QUZ 1. For the given area, the moment of inertia about axis 1 is 00 cm. What is the Mo about axis (the centroidal axis)? A) 90 cm B) 110 cm C) 60 cm D) 0 cm d d 1 C C A=10 cm d 1 = d = cm 1. 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

9 Moments of nertia of simple shapes 9-9

10 Moments of nertia of standard sections (see steel design handbook for details) 9-10

11 Sample Problem 10. The strength of a W1x8 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. SOLUTON: 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. 9-11

12 Sample Problem 10. SOLUTON: 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 Beam Section A, in A y, in ya, in ya 50.1 Y ya A 50.1 in in.79 in. 9-1

13 Sample Problem 10. Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to composite section centroidal axis. x,beam section x,plate 7. in x x AY Ad 15. in x x, beam section x,plate x 618 in Calculate the radius of gyration from the moment of inertia of the composite section. k x A x in in k x 5.87 in. 9-1

14 Sample Problem 10.5 SOLUTON: 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. 9-1

15 Sample Problem 10.5 SOLUTON: Compute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis. 1 r 90 a 8. mm b 10 - a 81.8 mm A r mm Rectangle: x bh mm Half-circle: moment of inertia with respect to AA, AA r mm moment of inertia with respect to x, x AA Aa mm moment of inertia with respect to x, x Ab x mm

16 Sample Problem 10.5 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. x mm mm x mm 9-16