In the degraded figure let h,, = 32 parts; from the top of this ordinate the slope down to the left will be half of the corresponding slope in the bending moment diagram; that is, 5 vert. to 1 hor., and it extends over 7 feet; hence h = h12 = 5 x 7 = 28.5 parts. The next slope from this down to left is at the rate 2.5 to 1, and extends over 5 feet; so that h2 = h ho - 2.5 x 5 16 parts. Beginning again at top of h the slope down to right is at the rate 1 and extends over 4 feet, so that he h12- 1 x 4 = 28 parts. The next slope down to right is at the rate 3 and extends over 4 feet; so that h2 = 3 × 4 = 16 parts; thus h 16 = 12 h1h h12 h1 h2:: 16:28·5:32:28:16. Putting h12 5 : 12 16 20 20 ins. a twelfth of the span for stiffness, and reducing the others in the proportion of 20 to 32, we have on the elevation To find the uniform breadth put 1 = × 1 × 6 × 202 384 inch-tons; and .. b = 5.76 inches. AREA, GEOMETRICAL MOMENT, AND MOMENT OF INERTIA. The functions of a plane surface which we require for our investigations regarding the moment of resistance to bending of a cross section in general, are the Area, Geometrical Moment, and Moment of Inertia. The Area of a rectangle is the product of its two adjacent sides; the area of any other surface is the sum of all the elementary rectangles into which it may be divided. We take it for granted that the area of the triangle, the circle, the ellipse, and parabolic quadrant are respectively half the product of the height into the base, π into radius. squared, into the product of the semi-major and semiminor diameters, and two-thirds of the product of the circumscribed rectangle. The area of any figure is quite independent of the position of the axis to which the figure is referred; and the word area is conveniently used as a name for a finite plane surface. DEFINITION.-The Geometrical Moment of a surface about any line in its plane as axis, is the sum of the products of each elemental area into its leverage or perpendicular distance from that axis; the leverages which lie to one side of the axis being reckoned positive, and those to the other side negative. Each elemental area is to be so small that the distance from the axis to every point in it is sensibly the same. If the surface were a plate of unit thickness and unit density, it is evident that the statical moment of the plate would be exactly the geometrical moment of the surface. Suppose the axis, for instance, to be a knife edge upon which the plate rests, the weights of the portions on different sides tend to cause the plate to rotate in opposite directions, and their statical moments are of different sign; the definition shows the geometrical moment of these portions of the surface also to be of opposite sign. It will be convenient for us always to choose a horizontal axis; and if we consider leverages up to be positive, then, when the axis is below the area, the geometrical moment is positive; when above it, negative; when the axis cuts the area, it will be positive or negative according as the axis is near one or other edge; and for one position of the axis, cutting the area, the positive and negative products will destroy each other, and the geometrical moment will be zero. An axis about which the geometrical moment of an area is zero passes through a point called the Geometrical Centre of the area. From this it appears that the geometrical centre of an area corresponds with the centre of gravity of a thin plate of uniform thickness and of that area; and for this reason the geometrical centre of an area is often called its centre of gravity. Theorem. The Geometrical Moment of a surface about any axis in its plane is equal to the area multiplied by the distance of the geometrical centre from the axis. Fig. 100. G s' S Suppose Gthe geometrical centre of the surface to be known; through G draw 00' parallel to the axis AA; let s and s' be a pair of elemental areas, one on each side of 00', such that the sum of their geometrical moments is zero; that is s'.a-s.b. It is evident that the whole area can be divided into such pairs from the definition of the geometrical centre. The geometrical moment of s' about AA is s' (d+a), that of s is s(d-b), and their joint moment is (s+8)d + (s'a-sb)=(s' + s)d, since the second term is zero. In the same way the moment of each pair is their sum multiplied by d; hence the geometrical moment of the area about AA is the sum of all the pairs into d, that is, the area multiplied by the distance of the geometrical centre from the axis. Fig.100. Corollary. The geometrical moment of an area which can be divided into simpler figures whose geometrical centres are known, may be found by multiplying the area of each such figure by the distance of the axis from its geometrical centre and summing algebraically. K Fig.101. Theorem. The Geometrical Moment of an area about an axis in its plane is expressed by the number which denotes the volume of that portion of a right-angled isosceles wedge whose sloping side passes through the axis, and which stands on the area as base. Let AEKFA', fig. 101, be the wedge, and AF its sloping side passing through the axis AA'; the angle at A is 45° and that at E is 90°; let BCDE be the area, then the geometrical moment of BCDE relatively to the axis AA' is represented by the volume of DBCKL. If s be an elemental portion of the area BCDE, its geometrical moment about AA' is 8 multiplied by its dis tance from AA'; but the column of the wedge standing on 8 is sensibly a parallelipiped whose height is the same as the distance of s from AA', and the volume of that column expresses the geometrical moment of 8; hence the volume of DBCKL, a portion of the isosceles wedge, is the geometrical moment of BCDE about AA'. When the axis cuts the area, the plane sloping at 45° will form a wedge above one portion and below the other; by considering these of different signs, their algebraic sum is still the geometrical moment. For example, if we wish to find the geometrical moment of the triangle, fig. 90, about the axis NA; let f=3h, then the plane will slope at 45°; in the figure, the volume on one side of the axis is equal to the volume on the other, that is, the geometrical moment is zero; the axis NA must pass therefore through the geometrical centre. To find an axis passing through the geometrical centre of a plane area then, it is only necessary to draw a plane sloping at 45° which will cut off an equal volume of the wedge above and below; the intersection of this plane with the area will be the axis required. By similar triangles a plane through that axis at any slope will cut off equal volumes above and below; the wedges which represent the normal stress, fig. 8, require to be equal, hence the neutral axis of a cross section passes through the geometrical centre (or centre of gravity) of the cross section. The distance of the neutral axis from the furthest away skin is, in each cross section, a definite fraction of h the depth of the circumscribing rectangle; for instance, for a triangular cross section, figs. 89, 90, OA = 3h. Rankine expresses this generally thus OA = m'h where m' is the fraction which the distance from the neutral axis to the farthest away skin is of the depth; for all cross sections, symmetrical above and below, as a rectangle, ellipse, hollow rectangle, &c., m'. DEFINITION. The Moment of Inertia of a surface, about a line in its plane as axis, is the sum of the products of each elemental area into the square of its distance from the axis. Whether the horizontal axis intersects, or is below or above the area, the moment of inertia will be positive; for, though the leverage be negative, the square of that quantity is always positive; it is also impossible that the sum can ever be zero. When the horizontal axis is at a great distance below, the moment of inertia is very great, since the leverage of each element is great; as the axis approaches, the moment of inertia decreases; when the axis has passed above the area and recedes, the moment of inertia again increases; hence for one position of the horizontal axis, the moment of inertia was less than when that axis was in any other position. Theorem.-The moment of inertia of a surface about any axis in its plane equals that about a parallel axis through its geometrical centre, together with the product of the area into the square of the deviation of the axis from the centre. Let s and s, fig. 100, be a pair of elemental areas, the sum of whose geometrical moments about 00′ is zero; and let x1 and x2 be their leverages about AA respectively. The sum of their moments of inertia about AA is sx2 + sx ̧2 = s′(d + a) + s(d — b)2 1 = (8′ + 8)d2 + 2(s'a — sb)d + (s'a2 + sb2) = (s′ + 8)d2 + (s'a2 + sb2), since (s'a + sh) = 0. Summing the left side for all pairs we have the moment of inertia of the area about AA. The first term on the right side is the area of each pair into the square of the deviation; and the sum of these for all pairs is the area into the square of the deviation; the second term on the right side is the moment of inertia of s and s' about 00', the sum will be the moment of inertia of the whole area about 00'. If I and I, repre sent the moments of inertia round the axes A and O tively, then ΙΑ IS.d2 + Io. where S is the area of the figure. respec |