Lecture 6: Distributed Forces Part 2 Second Moment of Area

Transcription

1 Lecture 6: Distributed Forces Part Second Moment of rea The second moment of area is also sometimes called the. This quantit takes the form of The phsical representation of the above integral can be described as follows. When are distributed continuousl over, the of these forces about some axis is proportional to the distance of the line of action of the force from the moment axis. Considering an elemental area on the surface, the elemental moment is proportional to the distance squared times the elemental area. Example Figure 1 reservoir wall denoted b a surface BCD is holding water on the inside. The water applies a which increases linearl with the depth on the wall. The moment about the axis B due to the pressure on the element of area d is given b Page 1 of 6

2 dm = ( p) d= ( k )d Hence, the total moment over the surface BCD can be found b integrating the above elemental moment over the area. This is given b M = k d Definitions of second moment of area Figure Figure shows an area in the x- plane with a small element of area d. B definition, the second moments of area of the element about the x- and -axes are given b = d, and d = x d, respectivel. d x Therefore, the second moments of area of the entire area about the x- and -axes are defined as Consider a rectangular lamina as shown in figure 3. We will determine the following 1. (Example 1). (Example ) Page of 6

3 Figure 3 Example 1 We will calculate the moment of inertia about the x-axis, x, which lies along the bottom edge of the lamina. Let us define the elemental area d using horiontal strip elements as shown in figure 3, so that The integration begins of the lamina and finishes at, so the integration limits are, respectivel. Finall, the moment of inertia about the x-axis is given b (1) Similarl, the moment of inertia about the -axis is found using vertical strip elements so the elemental area is given b The integration is now bounded between 0 and b, therefore the moment of inertia about the -axis is given b () Example Let us now shift the x- and -axes from the corner of the lamina to. From inspection, the centre of the rectangular lamina also coincides with its centre of gravit. Usuall, moments of inertia taken about the x- and -axes through the are written as, respectivel. Page 3 of 6

4 The moment of inertia about the horiontal axis labeled x 0 in figure 3 is given b (3) Similarl, the moment of inertia about the vertical axis 0 is given b (4) Parallel xis Theorem Note that equations (1) and (3) are ver similar. Onl their integration limits are different. The difference in their phsical meaning is the from x 0-0 (axes through the centroid). Therefore, moments of inertia of the same bod taken about axes can be related using where is the area of the lamina d x is the between x 0 - and x-axes d is the between 0 - and -axes (See figure 4) Figure 4 Page 4 of 6

5 Radius of Gration The radius of gration for the x-axis, x, is defined b the relationship where is the area of the lamina or x is the second moment of area about the x-axis (horiontal axis) Consider an irregularl shaped lamina of area in figure 5. ts moment of inertia about the x-axis is given b x. Now, consider a long thin (negligible width) strip in figure 5B whose area is also. The moment of inertia of the thin strip in figure 5B is also given b x. The distance k x is known as the radius of gration. Figure 5 Similar expressions for moments of inertia about the - and -axes are given b = k = k or k k = = Perpendicular xis Theorem Suppose we need to determine the moment of inertia about the -axis,, of the rectangular lamina shown in figure 3. Let us define the -axis to be perpendicular to the plane which contains x- and -axes. Page 5 of 6

6 We can make use of known quantities found in equations (1) and () to determine the moment of inertia about the -axis. The perpendicular axis theorem states that This relationship is useful when determining the moments of inertia of a lamina about an axis which is perpendicular to its surface. Composite Bodies Figure 6 Figure 6 shows the cross section of an -beam, which can be easil disassembled into three parts, block, block B and block C, respectivel. The moment of inertia of the entire shape about the central axis (x b axis) can be found b using the relationship where d represents the perpendicular distance between the axes. n general, the moments of inertia of a composite area about the x- and -axes are given b Page 6 of 6