forces parallel to all possible tangents of the ellipse, the rays will form a complete pencil of 360°. On the contrary, if the funicular be a hyperbola, the line of forces is reduced to a length subtending an angle at the pole equal to the supplement of the angle between the asymptotes. 56. The Elliptic Arch and Hyperbolic Chain.-Let x (Fig. 73) be the point of contact on the tangent parallel to the funicular bar 1'2'; then, since the ray O,,' (Fig. 75) is parallel to the tangent at x, it will also be parallel to My, the diameter conjugate to Mx. 23 12 Similarly, the ray O' is conjugate to the diameter drawn through M to the point of contact on the tangent parallel to bar 2'3. Hence the conjugate pencils M and O', formed in this way, are mutually projective; and, therefore, segments made through them in conjugate directions will also be mutually projective. Wherefore, since the force-curve becomes a straight line when the pole of the funicular curve passes to infinity, it follows that this punctuated line of forces or segment of the pencil issuing from the pole 8 8910 9 10 taken to represent an elliptic arch, the other a hyperbolic chain. We assume the vertical forces to consist of loads distributed over the arch or chain, which are divided into laminæ of equal breadth. The depth of each lamina must therefore be proportional to the corresponding incumbent load, inasmuch as the area of the lamina represents the total load. But, as already shewn, these loads or forces are also proportional to the segments cut off by the rays of a pencil projecting the points of contact of the funicular arch or chain upon a line whose direction is conjugate to the direction of the forces. Now the direction of the forces is vertical or parallel to AB. Wherefore, if from the centre B of the curve the segments of the arch or chain be projected upon the line CD, whose direction is conjugate to that of AB, the segments cut off upon CD will be proportional (here equal) to the forces and to the depths of the series of laminæ. If the range CD be then brought into a vertical position, it will constitute the line of forces, whose pole may be found by the rule that any two polar rays, such as O23 and 067, must be parallel respectively to the tangents between 2 and 3 or 6 and 7 on the funicular. Since, in the case of the arch, the centre lies within the funicular curve, the loads may increase to infinity. On the contrary, in the case of the chain, the loads decrease to zero from the centre towards the asymptotes, whilst the number of laminæ may be infinite. 57. Stress-Diagrams.—The theory of reciprocal figures has been usefully applied to finding the stresses set up in the bars of any wooden or metallic frame, such as the Warren girder shewn in Fig. 79. Assuming the vertical loads 1-8 on this girder to be applied at the joints of the lower boom, draw the vertical line of forces ai (Fig. 80) and set off along ai the given eight forces in their due order and magnitude. Next, pitch the pole O anywhere in the plane of the forces, and from O draw polar rays to the divisions between the forces on the line ai. Then draw the corresponding funicular 1'2'3' . . . 9', the first side 10'1' being made parallel to the polar ray O1-10 or Oa, and the last side 8'9' parallel to the polar ray Og, or Oi. 89 If now the closing sides 10'-1' and 9'8' of this funicular be produced, they will meet in a point R on the line of action of the resultant load (§ 50); whilst, if the polar ray 09-10 or Oz be drawn parallel to the completing side 9'10' of the funicular polygon, it will divide the line of forces into two parts, one of which (iz) represents the reaction 9 at the right abutment, and the other (za) the similar reaction at the left abutment. In conformity with Newton's third law, the sum of the forces is equal in magnitude but opposite in sign to the sum of the reactions. The rest of the reciprocal diagram of stress can be built up by constructing the reciprocal polygon of each joint, beginning at the left abutment. Thus, resolve the reaction za into two components, one ak parallel to the diagonal ak, the other kz parallel to the first segment kz of the top boom, each line in the frame being described by the lettered spaces between which it lies. Owing to the direction of za being upward, it will obviously produce a tension in ka and a compression in kz. Taking next in order the first joint of the lower boom, we find that of the four forces or stresses acting there two only, namely kl and lb, are unknown. Moreover, since no other force or stress can possibly influence the pressure at this joint, the stresses in kl and lb may be regarded as components of the resultant of the known stresses ka and ab. Now in the reciprocal diagram we already have the magnitude and direction of the force ab, together with the magnitude of the stress ka, the direction of which is determined by remembering that, since each bar is kept in equilibrium by the stresses acting at its ends, the resultants of those stresses, or the stresses along the bar in question, must be equal in magnitude but opposite in sign. Wherefore, since in the reciprocal figure of the first joint zakz the stress ak is directed downwards, it must be directed in the contrary direction ka at the lower joint kablk; so that the required resultant of the known stresses ka and ab will be represented by kb both in magnitude and direction. Resolving kb into two components parallel respectively to the bars kl and lb, we obtain the triangle klb in the reciprocal figure. Owing to its direction being upward, the resultant kb will produce a tension in lb and a thrust in kl. Proceeding now to the joint klmz, we find that the magnitudes and directions of all the stresses meeting there are known, except Im and mz. Composing, therefore, the two known stresses zk and kl into the single resultant zl, |