Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering Structural Mechanics. Chapter 2 SECTION PROPERTIES

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1 Section Properties Centroid The centroid of an area is the point about which the area could be balanced if it was supported from that point. The word is derived from the word center, and it can be though of as the geometrical center of an area. For three-dimensional bodies, the term center of gravity, or center of mass, is used to define a similar point. Centroid of Simple Areas For simple areas, the location of the centroid is easy to visualize. B B/2 x C H/2 H H/3 x C H B y C D/2 D C S/2 S/2 Page 2-1

2 Centroid of Complex Areas Most complex shapes can be considered to be made up by combining several simple shapes together. If the area has an axis of symmetry, the controid will be on that axis. Some complex shapes have two axes of symmetry, and therefore the centroid is at the intersection of these two axes. C C C C C C Page 2-2

3 Where two axes of symmetry do not occur, the method of composite areas can be used to locate the centroid. For example, consider the following shape. It has a vertical axis of symmetry but not a horizontal axis of symmetry. Such areas can be considered to be a composite of two or more simple areas for which the centroid can be found by applying the following principle: The product of the total area times the distance to the centroid of the total area is equal to the sum of the products of the area of each component part times the distance to its centroid, with the distances measured from the same reference axis. This can be stated mathematical as A T Y = Σ(A i y i ) Where A T = total area of the composite shape Y = distance to the centroid of the composite shape measured from some reference axis A i = area of one component part of the shape Y i = distance to the centroid of the component part from the reference axis. Page 2-3

4 Example Find the location of the centroid of following composite area. Solution 2 1 iyi Y A = A T 40x50x25 Y = 40x x10x55 60x10 = 31.9 mm Page 2-4

5 Example Find the location of centroid of the following T-section. Solution Y = (100*10*60+100*10*5) / (100*10+100*10) = 32.5 mm Page 2-5

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9 Moment of Inertia In the study of strength of materials, the property of moment of inertia is an indication of the stiffness of a particular shape. That is, a shape having a higher moment of inertia would deflect less when subject to bending moments than one having a lower moment of inertia. Moment of inertia of simple shapes I = y 2 da Where y = distance from an element of area to the reference axis A = Area of an element Example Determine the moment of inertia, I, of the following area with respect to its centroidal axis. I = y 2 da da = b * dy I = h / 2 2 y h / 2 ( b * dy) I = 3 bh 12 Page 2-9

10 Moment of inertia of complex shapes If the component parts of a composite area all have the same centroidal axis, the total moment of inertia can be found by adding or subtracting the moments of inertia of the components parts with respect to the centroidal axis. Example The cross-section of a beam shown in the following figure has its centroid at the intersection of its axes of symmetry. Compute the moment of inertia of the section with respect to the horizontal axis x-x. Then, I x = I 1 + I 2 +I 3 I 2 = 30*80 3 /12 = 1.28 x 10 6 mm 4 I 1 = I 3 = 30*40 3 /12 = 0.16 x 10 6 mm 4 I x = I 1 + I 2 +I 3 = 1.6 x 10 6 mm 4 Page 2-10

11 Example Compute the moment of inertia for the following section with respect to the axis x-x. Then, I x = I 1 - I 2 I 1 = 50 4 /12 = x 10 3 mm 4 I 2 = π*35 4 /64 = 73.7 x 10 3 mm 4 I x = I 1 - I 2 = x 10 3 mm 4 Page 2-11

12 Polar Moment of Inertia The moment of inertia for an area relative to a line or axis perpendicular to the plane of the area is called the polar moment of inertia and is denoted by the symbol J. The moment of inertia of an area in the xy plane with respect to the z axis is J Z = r 2 da = (x 2 + y 2 ) da = x 2 da + y 2 da Therefore, J Z = I x + I y Expressed in words, this equation states that the polar moment of inertia for an area with respect to an axis perpendicular to its plane is equal to the sum of the moments of inertia about any two mutually perpendicular axes in its plane that interest on the polar axis. Page 2-12

13 Parallel axis theorem The theorem states that the second moment of an area with respect to any axis is equal to the second moment of the area with respect to a parallel axis through the centroid of the area added to the product of the area and the square of the distance between the two axes. The second moment of the area in the following figure about the b axis is I b = (y + d) 2 da = y 2 da + 2d y da + d 2 da =I c + 2d y da + d 2 A Where, 2d y da = 0 Therefore, I b = I c +Ad 2 Page 2-13

14 Example Compute the moment of inertia for the following inverted T-section with respect to its centroidal axis. Solution Y = (100*10*60+100*10*5)/(100*10+100*10) = 32.5 mm By using the Parallel Axis Theorem, I = 10*100 3 / *10*( ) *10 3 / *10*(32.5-5) 2 = 2.35 x 10 6 mm 4 Page 2-14

15 Example Compute the moment of inertia for the following section with respect to its centroidal axis. Solution Method 1, We divide the section into 3 small sections and apply the Parallel Axis Theorem, I = 15*300 3 /12 + 2*[215*15 3 / *15*( ) 2 ] = 1.94 x 10 8 mm 4 Page 2-15

16 Method 2, Moment Inertia of the I-section, I = I area 1 I area 2 I area 3 = 215*330 3 /12 100*300 3 /12 100*300 3 /12 = 1.94 x 10 8 mm 4 Page 2-16

17 Example Compute the moment of inertia for the following section with respect to its centroidal axis. Solution The x-x centroidal axis is shown. It is the axis of symmetry. Only one transfer distance is required (part 1). I xx for parts 2 and 4 = 50*200 3 /12 = 33.3 x 10 6 mm 4 I xx for part 3 = 150*50 3 /12 = 1.56 x 10 6 mm 4 I xx for parts 1 and 5 = 250*50 3 / *50*(150-25) 2 = x 10 6 mm 4 Total I xx = 2*33.3 x x *197.9 x 10 6 = 464 x 10 6 mm 4 Page 2-17

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