where f is the stress on skin at S. Consider next the cantilever OK', taking into account the uniform load alone; a cantilever of length SK' would deflect, and taking account of its length, we have for the deflection of OK' for uniform load alone A load at end of OK', equal to the uniform load on it, would produce an additional deflection of v,; but the load at end of OK' is that on K'S, and it will produce a proportionate deflection; so that we have for the deflection of OK' for load at end = (√3-1), 3 Ο Κ SK – (OK) = (28 - 16 √/3), and (OK) = (6 √/3 – 10); 18 E Yo since M at the end and at the centre are in the ratio of 2 and 1. We thus have m = 1'; and n" = 1. From the above, we see that fixing the ends of an uniform beam which is loaded uniformly increases the strength in the ratio, or 3:2; As already shown at fig. 37, the strength is economised to the greatest extent if by means of hinges, or in some other way, we shift the point of contrary flexure to a distance c√2 from the centre; we then increase the Maximum Stiffness. For some position of the hinges, the stiffness will be a maximum; in order to find this point, let B be the distance of the hinges from the centre, and ƒ, the stress on the skin at centre; then as above we have— where f is the proof strength so long as Mmax. is at the centre, that is so long as B Bc, we have n". B= n" √2' When B< C √2' √2 (Hinges at 707c from centre.) then Mmax, is at the end, and it is neces sary to substitute for f, in terms of f, the proof stress on skin at end; thus-if The stiffness is a maximum when v is a minimum; this occurs when ẞc, sensibly giving Fixing the ends and placing the two hinges at the quarter points of span increases the stiffness in the ratio, or 4: 1 nearly. Some economy of a material whose strength to resist tension is, say, less than that to resist thrust, can be secured by making the constant cross section of such a form that the upper skin at end section, and under skin at central section, shall simultaneously come to their working stress. Thus for a material half as strong to resist tension as thrust, с and with the hinges fixed so that B = a section whose neutral axis is half as far from the upper as it is from the under skin, would give considerable economy; as the upper and under skin at end sections, and under skin at central section would all come to their working strength at one time. Beam of uniform strength, uniform depth, fixed at the ends. Suppose that OK'SKO', fig. 143, is the beam. From eqn. 1, page 300, the curvature is constant, so that OK' is part of a circle whose centre is on the vertical through O, and SK' is part of a circle of the same radius whose centre is on the vertical through S; by symmetry SK' = OK', and K and K' the points of contrary flexure are midway between the centre and the ends of the span. Since the beam is of uniform strength and depth, the breadth varies as the bending moment; hence the plan should correspond with the bending moment diagram. For the case of an uniform load, the plan will correspond with the bending moment diagram in fig. 143, the curve being drawn on both sides of a centre line, the two parabolas passing through the quarter points of the span; the breadths are then to be reduced till that at any point (say the end) is just sufficient to give the necessary resistance to bending, and we have the plan of the beam; see Rankine's "Applied Mechanics," fig. 144. Since O"B, fig. 143, is half of D'L, and since the load is uniform, O"F is one quarter, and O′′D' the maximum bending moment is three quarters, of FD' the amount of the maximum bending moment for the same beam not fixed at the ends. The plan must allow sufficient breadth at the points of contrary X flexure in order to be able to resist the shearing force at these points; on K', for instance, the shearing force will be the load on SK', and the breadth at K' must be sufficient to resist this amount. THIN HOLLOW CROSS SECTIONS. Let t be the uniform thickness of a thin hollow cross section of any form, and let n' and n be the numerical coefficients respectively of the moment of inertia and the moment of resistance to bending of a solid cross section of the same form; let B, H be the breadth and depth of the rectangle circumscribing the section, and b, h the breadth and depth of the rectangle circumscribing the hollow; thenI ̧= n'(BH3 – bh3). (See page 237.) = n'B(H3 – h3), since (B = b nearly). =n'B(H-h)(H2+Hh+h2). =n'B. 2t.3H2. = n'. 6BH2t. (H = h nearly). A closer approximation for any form is obtained thus: .. M = n'{(B-b)H3+b(H3 — h3) } n' {(B − b)H3 +b(H − h)(H2 + Hh +h3)} (2) where t is the thickness of each side, and t the thickness of the top or bottom. To design a thin hollow cross section ;-Choose the depth H the proper fraction of the span to give the required degree of stiffness; assume B a suitable fraction of H to give sufficient lateral stiffness, and the above is a simple equation from which to find t; or t may be assumed a multiple of the thickness that the metal plates are usually manufactured, and B found from the formula. Now find the moment of resistance to bending, and the resistance to shearing, by the accurate formulæ; and if this differs from M and F, alter t or B for further approximation. Examples. 175. Find the thickness of metal required for an aqueduct bridge 40 feet span, the waterway being 2 feet square, and the material wrought iron for which f = 4 tons per sq. in. The weight of water supported is 41 tons, and the weight of metal (assumed) 11⁄2 tons; making a further allowance of 2 tons, we have W 8 tons distributed, which gives 480 inch-tons. 24", H = 24", then = Let B = Allowing for rivets, the plates might be taken " thick; but as the plates are liable to rust, the thickness would require to be increased still further, and t might be taken as 3" in an actual case. 176. Find the moment of resistance to bending of a cast iron pipe 18" external diameter, metal 1" thick, and ƒ = 2 tons per square inch, by means of the exact formula; and compare the result with that obtained by the approximate formulæ 1 and 2. 177. Find the thickness of metal required for a cast iron pipe 24′′ external diameter, so that its moment of resistance to bending may be 50 foot-tons, and its resistance to shearing 10 tons; taking f = 2 tons per square inch. 600 = 6 × 2 × 242 × t; .. t=0′′.87 nearly. On checking the calculation by the accurate formula, and taking t=0·87, M = 705 inch-tons; so that t=0"·87 is more than sufficient; and for shearing this thickness will be |