Chapter 9 Moment of Inertia

Transcription

1 Chapter 9 Moment of Inertia Dr. Khairul Salleh Basaruddin Applied Mechanics Division School of Mechatronic Engineering Universiti Malaysia Perlis (UniMAP) khsalleh@unimap.edu.my

2 PARALLEL-AXIS THEOREM, RADIUS OF GYRATION & MOMENT OF INERTIA FOR COMPOSITE AREAS Today s Objectives: Students will be able to: 1. Apply the parallel-axis theorem. 2. Determine the moment of inertia (MoI) for a composite area. In-Class Activities: Applications Parallel-Axis Theorem Radius of Gyration Method for Composite Areas Group Problem Solving

3 APPLICATIONS Cross-sectional areas of structural members are usually made of simple shapes or combination of simple shapes. To design these types of members, we need to find the moment of inertia (MoI). It is helpful and efficient if you can do a simpler method for determining the MoI of such cross-sectional areas as compared to the integration method. Do you have any ideas about how this problem might be approached?

4 APPLICATIONS (continued) This is another example of a structural member with a composite cross-area. Such assemblies are often referred to as a built-up beam or member. Design calculations typically require use of the MoI for these crosssectional areas.

5 PARALLEL-AXIS THEOREM FOR AN AREA (Section 10.2) This theorem relates the moment of inertia (MoI) of an area about an axis passing through the area s centroid to the MoI of the area about a corresponding parallel axis. This theorem has many practical applications, especially when working with composite areas. Consider an area with centroid C. The x' and y' axes pass through C. The MoI about the x-axis, which is parallel to, and distance d y from the x' axis, is found by using the parallel-axis theorem.

6 PARALLEL-AXIS THEOREM (continued) I X = A y 2 da = A (y' + d y ) 2 da = A y' 2 da + 2 d y A y'da + d y2 A da Using the definition of the centroid: y' = ( A y' da) / ( A da). Now since C is at the origin of the x' y' axes, y' = 0, and hence A y' da = 0. Thus I X = I X ' + A d y 2 Similarly, I Y = I Y ' + A d X 2 and J O = J C + A d 2

7 RADIUS OF GYRATION OF AN AREA (Section 10.3) y A For a given area A and its MoI, I x, imagine that the entire area is located at distance k x from the x axis. k x y A x 2 Then, I x = k x A or k x = ( I x / A). This k x is called the radius of gyration of the area about the x axis. k Y Similarly; 2 k Y = ( I y / A ) and k O = ( J O / A ) x The radius of gyration has units of length and gives an indication of the spread of the area from the axes. This characteristic is important when designing columns.

8 MOMENT OF INERTIA FOR A COMPOSITE AREA (Section 10.4) 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 plus a triangle, minus the interior rectangle. The MoI about their centroidal axes of these simpler shaped areas are found in most engineering handbooks. Using these data and the parallel-axis theorem, the MoI for a composite area can easily be calculated.

9 STEPS FOR ANALYSIS 1. Divide the given area into its simpler shaped parts. 2. Locate the centroid of each part and indicate the perpendicular distance from each centroid to the desired reference axis. 3. 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 ) 2 ). 4. The MoI of the entire area about the reference axis is determined by performing an algebraic summation of the individual MoIs obtained in Step 3. (Please note that MoI of the hole is subtracted).

10 EXAMPLE Given: The beam s cross-sectional area. Find: Plan: The moment of inertia of the area about the y-axis and the radius of gyration, k y. Follow the steps for analysis. Solution: [2] [3] [1] 1. 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, 37.5 mm, and 37.5 mm, respectively.

11 EXAMPLE (continued) 3. From the inside back cover of the book, the MoI of a rectangle about its centroidal axis is (1/12) b h 3. I y[1] = (1/12) (50 mm) (150 mm) 3 [1] = 14,062,500 mm 4 [2] [3] Using the parallel-axis theorem, I Y[2] = I Y[3] = I Y + A (d X ) 2 = (1/12) (100) (25) 3 + (100) (25) ( 37.5 ) 2 = 3,645,833 mm 4

12 EXAMPLE (continued) Summing these three MoIs: 4. I y = I y1 + I y2 + I y3 I y = 21,354,166 mm 4 Now, finding the radius of gyration: k y = ( I y / A) A = 150 (50) (25) (25) = mm 2 k y = (21,354,166) / (12500) = 41.3 mm

13 GROUP PROBLEM SOLVING (continued) (c) 3. I Xa = (1/12) (6) (10) 3 + (6) (10) (5) 2 = 2000 m 4 I Xb = (1/4) (2) 4 + (2) 2 (4) 2 = m 4 (b) (a) The radius of gyration: k X = ( I X / A ) I Xc = (1/36) (3) (6) 3 + (1/2) (3) (6) (8) 2 = 594 m 4 Summing: I X = I Xa I Xb I xc = 1192 m 4 A = 6 (10) (2) 2 (1/2) 3 (6) = m 2 k X = 1192 / = 5.57 m

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