and doubling, we have for semi-circle, 0 If we divide G by the area of the semi-circle, we obtain the distance from the centre of the circle to the centre of gravity of the semi-circle, Elliptic Section.-The Ellipse is immediately derived from the circle thus:-Let the ellipse, fig. 117, have the same minor diameter as the circle above it, so that b2r; all the vertical dimensions of the circle are, however, to be altered in the constant ratio a: r, since the vertical radius of the circle is altered to a. Let the circle be divided into rectangular elements, each with an edge lying on the neutral axis; when the circle is changed into the ellipse, each element remains of the same breadth bas before, but its vertical dimension is changed from y to Y, where Y: y::a:r. K r Y:y::a:r k---- Fig.117. For each element, and therefore for the whole figure, the moment of inertia has changed in the ratio For a semi-ellipse about a diameter as axis, we have Hollow Circular Section.-For a hollow circle or ellipse the reduction is the same as for the hollow rectangle, page 237, 136. Find the moment of resistance to bending of the symmetrical section, the upper half of which is shown in fig. 118; the material is wrought iron, for which the weaker working strength is fa= 4 tons per square inch. Consider the two triangles as one of breadth 8 inches; for a triangle about its own neutral axis, I = bh3 = 48; hence for semi-section 江。 = (48+24 × 82) for triangle+ × 2(103-03) forrectangle. 137. For the cross section shown in fig. 119, and which is the semi-section shown in fig. 118, find the working value of M, if the working strengths of the material be f = 4, and fo 5 tons per square inch. = Choose BB as an axis of reference; then S=24+20=44; G2 = (24 × 8) + 1⁄2 × 2(102 −02) = 292. II of example 136, = 2251. 138. Find M for the section, the upper half of which is shown in fig. 120; the material is cast iron, for which f 2 tons per square inch. For the cross section, 11 = (201+50·26 × 82) + } × 3(43 — 03). fo -I = 126964 1161 inch-tons. 0 139. Find M for the section shown in fig. 121, the working strengths fa and f, being 4 and 5 tons per square inch respectively. It is convenient to choose CC the diameter of the ellipse as the axis of reference, and we have Ellipse. Middle rect. + Lower rect. π = 75·7; 0 + 1 × 3(92 − 32) + 1⁄2 × 10(112 — 92) = 308; Ic= × 8 × 63 + 1 × 3(93 − 33) + 1⁄2 × 10(113— 93) = 2793; 64 = 4 inches sensibly; so that y. = 7, and = 7 inches. The neutral axis being sensibly in the middle, take ƒ = 4 the smaller strength, and M = 1, 904 inch-tons. 140. Find the resistance to bending for the section shown in fig. 122; the dimensions are in inches, and the strengths R = of the material are fa 4 tons (thrust), and f = 5 tons (tension) per square inch. Choose CC the diameter of the ellipse as an axis of reference; and for semi-ellipse S= 236; Go-30; I, 53. = For semi-circle about its diameter, I 245, and distance. from centre to g its centre of gravity is 2:12 inches; hence about its own neutral axis, that is an axis through g, we have I ̧ = I – S(2·12)2 = 69; and for semi-circle— S = 393; G = 398; I = 4093. For the two triangles, reckoned as one about its own neutral axis I= 427; hence for double triangle y= 8.3 ins., and y. = 77 ins.; and I ̧ = 5 = '602; Уз 8.3 taking the smaller value, we have M5191 1167 inch-tons. As shown in the figure, the section is lying in a position suitable for a cantilever; if intended for a beam, it should be turned upside down; should it be kept as it stands, however, then for a beam we take the smaller of the two viz., 482; multiplying I, by this ratios, 5 7.7 and 4 8.3' quantity gives a value of M, which is less than the above. |