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    "source_title": "Encyclopaedia Britannica (1911)",
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    "chunk_id": "1911:exaf:f40132a47628",
    "title": "EXAF",
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    "verified_text": "exaf, the reciprocal figure of the lines meeting at that point in the frame, and representing the forces at the point exaf; the direction of the forces on eh and xa being known determines the direction of the forces due to the elastic reaction of the members af and ef, showing af to push as a strut, while ef is a tie. we have been guided in the selection of the particular quadrilateral adopted by the rule of arranging the order of the sides so that the same letters indicate corresponding sides in the diagram of the frame and its reciprocal. continuing the construction of the diagram in the same way, we arrive at fig. 67 d as the complete reciprocal figure of the frame and forces upon it, and we see that each line in the reciprocal figure measures the stress on the corresponding member in the frame, and that the polygon of forces acting at any point, as ijky, in the frame is represented by a polygon of the same name in the reciprocal figure. the direction of the force in each member is easily ascertained by proceeding in the manner above described. a single known force in a polygon determines the direction of all the others, as these must all correspond with arrows pointing the same way round the polygon. let the arrows be placed on the frame round each joint, and so as to indicate the direction of each force on that joint; then when two arrows point to one another on the same piece, that piece is a tie; when they point from one another the piece is a strut. it is hardly necessary to say that the forces exerted by the two ends of any one member must be equal and opposite. this method is universally applicable where there are no redundant members. the reciprocal figure for any loaded frame is a complete formula for the stress on every member of a frame of that particular class with loads on given joints. [illustration: fig. 68] [illustration: fig. 69] consider a warren girder (fig. 68), loaded at the top and bottom joints. fig. 69 b is the polygon of external forces, and 69 c is half the reciprocal figure. the complete reciprocal figure is shown in fig. 69 a. the method of sections already described is often more convenient than the method of reciprocal figures, and the method of influence lines is also often the readiest way of dealing with braced girders. 35. _chain loaded uniformly along a horizontal line._--if the lengths of the links be assumed indefinitely short, the chain under given simple distributions of load will take the form of comparatively simple mathematical curves known as catenaries. the true catenary is that assumed by a chain of uniform weight per unit of length, but the form generally adopted for suspension bridges is that assumed by a chain under a weight uniformly distributed relatively to a horizontal line. this curve is a parabola. remembering that in this case the centre bending moment [sigma]wl will be equal to wl2/8, we see that the horizontal tension h at the vertex for a span l (the points of support being at equal heights) is given by the expression 1 . . . h = wl2/8y, or, calling x the distance from the vertex to the point of support, h = wx2/2y, the value of h is equal to the maximum tension on the bottom flange, or compression on the top flange, of a girder of equal span, equally and similarly loaded, and having a depth equal to the dip of the suspension bridge. [illustration: fig. 70.] consider any other point f of the curve, fig. 70, at a distance x [v.04 p.0556] from the vertex, the horizontal component of the resultant (tangent to the curve) will be unaltered; the vertical component v will be simply the sum of the loads between o and f, or wx. in the triangle fdc, let fd be tangent to the curve, fc vertical, and dc horizontal; these three sides will necessarily be proportional respectively to the resultant tension along the chain at f, the vertical force v passing through the point d, and the horizontal tension at o; hence h : v = dc : fc = wx2/2y : wx = x/2 : y, hence dc is the half of oc, proving the curve to be a parabola. the value of r, the tension at any point at a distance x from the vertex, is obtained from the equation r2 = h2+v2 = w2x^4/4y2+w2x2, or, 2 . . . r = wx[root](1+x2/4y2). let i be the angle between the tangent at any point having the co-ordinates x and y measured from the vertex, then 3 . . . tan i = 2y/x. let the length of half the parabolic chain be called s, then 4 . . . s = x+2y2/3x. the following is the approximate expression for the relation between a change [delta]s in the length of the half chain and the corresponding change [delta]y in the dip:-- s+[delta]s = x+(2/3x) {y2+2y[delta]y+([delta]y)2} = x+2y2/3x+4y[delta]y/3x+2[delta]y2/3x, or, neglecting the last term, 5 . . . [delta]s = 4y[delta]y/3x, and 6 . . . [delta]y = 3x[delta]s/4y. from these equations the deflection produced by any given stress on the chains or by a change of temperature can be calculated. [illustration: fig. 71.] 36. _deflection of girders._-- let fig. 71 represent a beam bent by external loads. let the origin o be taken at the lowest point of the bent beam. then the deviation y = de of the neutral axis of the bent beam at any point d from the axis ox is given by the relation d2y m --- = -- , dx2 ei where m is the bending moment and i the amount of inertia of the beam at d, and e is the coefficient of elasticity. it is usually accurate enough in deflection calculations to take for i the moment of inertia at the centre of the beam and to consider it constant for the length of the beam. then dy 1 -- = ---[integral]mdx dx ei 1 y = ---[integral][integral]mdx2. ei the integration can be performed when m is expressed in terms of x. thus for a beam supported at the ends and loaded with w per inch length m = w(a2-x2), where a is the half span. then the deflection at the centre is the value of y for x = a, and is 5 wa^4 [delta] = --- ----. 24 ei the radius of curvature of the beam at d is given by the relation r = ei/m. [illustration: fig. 72.] 37. _graphic method of finding deflection._--divide the span l into any convenient number n of equal parts of length l, so that nl = l; compute the radii of curvature r_1, r_2, r_3 for the several sections. let measurements along the beam be represented according to any convenient scale, so that calling l_1 and l_1 the lengths to be drawn on paper, we have l = al_1; now let r_1, r_2, r_3 be a series of radii such that r_1 = r_1/ab, r_2 = r_2/ab, &c., where b is any convenient constant chosen of such magnitude as will allow arcs with the radii, r_1, r_2, &c., to be drawn with the means at the draughtsman's disposal. draw a curve as shown in fig. 72 with arcs of the length l_1, l_2, l_3, &c., and with the radii r_1, r_2, &c. (note, for a length 1⁄2l_1 at each end the radius will be infinite, and the curve must end with a straight line tangent to the last arc), then let v be the measured deflection of this curve from the straight line, and v the actual deflection of the bridge; we have v = av/b, approximately. this method distorts the curve, so that vertical ordinates of the curve are drawn to a scale b times greater than that of the horizontal ordinates. thus if the horizontal scale be one-tenth of an inch to the foot, a = 120, and a beam 100 ft. in length would be drawn equal to 10 in.; then if the true radius at the centre were 10,000 ft., this radius, if the curve were undistorted, would be on paper 1000 in., but making b = 50 we can draw the curve with a radius of 20 in. the vertical distortion of the curve must not be so great that there is a very sensible difference between the length of the arc and its chord. this can be regulated by altering the value of b. in fig. 72 distortion is carried too far; this figure is merely used as an illustration. 38. _camber._--in order that a girder may become straight under its working load it should be constructed with a camber or upward convexity equal to the calculated deflection. owing to the yielding of joints when a beam is first loaded a smaller modulus of elasticity should be taken than for a solid bar. for riveted girders e is about 17,500,000 lb per sq. in. for first loading. w.j.m. rankine gives the approximate rule working deflection = [delta] = l2/10,000h, where l is the span and h the depth of the beam, the stresses being those usual in bridgework, due to the total dead and live load. (w. c. u.) [1] for the ancient bridges in rome see further rome: _archaeology_, and such works as r. lanciani, _ruins and excavations of ancient rome_ (eng. trans., 1897), pp. 16 foll.",
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