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Chapter, , 12, , , , Cables and, , Suspension, , Bridges, , Introduction., , Equilibrium of cable under a given system, , of loading., , Equation of the cable., , Horizontal thrust on the cable., , Tension in the cable., , Tension in the cable supported at the same, , level., , Tension in the cable supported at different, , levels., , 8. Anchor cables., 9. Guide pulley support for suspension cable., , 10. Roller support for suspension cable., , 11. Length of the cable., , 12. Length of the cable, when supported at the, same level., , 13. Length of the cable, when supported at, different levels., , 14. Effect on the cable due to change in, temperature., , 15. Stiffening girders in the, bridges., , 16. Suspension bridges, stiffning girder., , 17. Influence lines for moving loads over the, suspension bridges with three-hinged, stiffening girders., , 18. Influence lines for a single concentrated, load rolling over the suspension bridge, with three-hinged stiffening girders., , 19. Influence lines for a uniformly distributed, load rolling over the suspension bridge, with three-hinged stiffening girders., , 20. Suspension bridges with two-hinged, stiffening girders., , 21. Influence lines for a single concentrated, , load rolling over the suspension bridge, , with two-hinged stiffening girders., , Na, , ove, , NN, , suspension, , with — three-hinged, , 12.1 Introduction, , A suspension bridge consists of two, cables, which are stretched over the span to be, bridged. Each cable, passing over two towers,, anchored by backstays to a firm foundation as, shown in Fig. 9.1. As the cable is flexible, throughout, therefore it cannot resist any, moment and can adopt any shape under the, loads ; that is why the bending moment at every, point of the cable is taken as zero. The roadway, is suspended from the cables by means of, hangers or suspenders. Since the hangers are, large in number, therefore the load transmitted, by the hangers, is taken as a uniformly, distributed load. The central sag or dip of the, cable generally varies from 1/10 to 1/15 of the, span. It may be noted, that since the cable is in, tension throughout, this type of construction is, most economical., , A suspension bridge is generally preferred, when its span is more than 200 metres for a, roadway or 300 metres for light traffic way i-e.,, pedestrians, cyclists, motorists etc. Due to, oscillations and sag, under loads, these types of, , bridges are not suitable for railways or very, heavy traffic., , 302

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Cables and Suspension Bridges ™ 303, , , , Fig. 12.1 Suspension Bridge, , 12.2. Equilibrium of cable under a given system of loading, , , , (d), , Fig. 12.2, , Consider a cable suspended at two points, A and B at the same level. Let the cable carry point, i“ W,, W, and W, at C, D and E. If the weight of the cable is negligible as compared to the loads, V,, W,, W, etc., the cable will take a shape as shown in Fig. 9.2 (a)., , Let T, = Tension in AC,, , T, = Tension in cD,, T, = Tension in DE, and, T, = Tension in EA., , Since all the joints of the cable are in equilibrium, therefore the funicular polygon drawn with, the help of loads and tensions in the cable must close. Now draw pq, qr and rs representing the, vertical loads W,, W, and W, respectively to some suitable scale. Through p draw a line parallel to, AC, and through q draw another line parallel to CD, meeting the first line at o. Join or and os. Now, Parsop is the force polygon, corresponding to the system of forces, which keeps the cable in, equilibrium. Through 0, draw om perpendicular to the load line p g rs. Now mp and sm represent, the vertical reactions at A and B respectively to the scale., , - Now, considering the equilibrium of the point A, we find that the reaction R, is equal to the, tension T,. Similarly, if we consider the equilibrium of C, we find that the forces acting at this point, are T\, T, and W,. From the law of triangle of forces, we find that W, is represented by pq, T, is, represented by op. Thus 7, is represented by qo. Similarly by considering the equilibrium of points

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304 m Theory of Structures, , D and E, we can find that T, and T, are represented by ro and so respectively. A little consideration, will show that the horizontal component of the tensions (i.e., T,, T>, T, and T,) will be given by, , om to the scale. The vertical component of the reactions V, and V, will be given by pm and ms to, the scale., , 12.3. Equation of the Cable, , , , Fig. 12.3, , Consider a cable ACB, supported at A and B. Let C be the lowest point of the cable as shown, in Fig. 12.3., , Let I, , span of the cable, and, Yc = central dip of the cable. ., Now consider a point P, having coordinates as x and y, with A as the origin as shown in Fig., , 12.3), We know that the equation of a parabola, , y = k(l-x) (i), where k is a constant., , 1, When x= >, y=Yo, , q Substituting these values of x and y in equation (i),, , if, 1) ke, = k—|1-—|=—, Ye ( | 4, , 4c, P, Now substituting the value of k in equation (i),, , k=, , 4, ye FEal-2), , This is the required equation for the dip y of the cable from its support at a distance x from, AtoB., , 12.4 Horizontal Thrust on the Cable, , , , , , , , , , , , , , , , , , Fig. 12.4, , Consider a cable ACB supported at A and B and carrying a uniformly distributed load as, shown in Fig. 12.4. .

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Cables and Suspension Bridges ™@ 305, , Let C be the lowest point of the cable. A little consideration will show, that as a result of, loading, the two supports A and B will tend to come nearer to each other. Since these two supports, are in equilibrium, therefore an outward force must act, on both the supports to keep them in, equilibrium. As the cable is supporting vertical load only, therefore the horizontal thrust at A must, be equal to the horizontal thrust at B., , Let w = Uniformly distributed load per unit length on the cable,, 1 = Span of the cable, and, Ye = Central dip of the cable., , Now consider a point P on the cable, having co-ordinates as x and y as shown in Fig. 12.4., Draw a tangent to the cable at P meeting the horizontal line CQ though C at R. From the geometry, , of the figure, we know that CR = RQ = =, A little consideration will show that the part CP of the, , cable is in equilibrium under the action of the following forces :, , 1. Horizontal pull (equal to H)., , 2. Downward load (equal to wx), and, , 3. Tension in the cable at P (equal to 7)., , Now we see that the triangle PQR represents, to some scale, the above mentioned three, forces, which keep the part CP of the cable in equilibrium., , . P PO _ RQ_NP, “ | wx H L :, y ats fees pQ=y and RO- 5), oO —_= - ), " low % wx 2H \ 2, , © AOE aa wx, , or an. we *H = >, ; 2y, , H 1, We know that when x = 2 = Yo _ therefore, , (3) wt, H= = Bye, This is the required equation for the horizontal thrust on the cable., Note: The Bending moment at the lowest point C of the cable,, , , , Me = bending moment due to loading — bending moment due to H, = Ue- Aye, or Hye = We (‘bending moment at C is zero), He, He Ye, , 12.5 Tension in the Cable, , We have already discussed that the cable is always subjected to some tension, whose, magnitude can be conveniently found out by the laws of statics. Here we shall discuss the, Magnitude of tension in the cable in the following two cases :, , (a) When supported at the same level, and, , (b) When supported at different levels., , wx?, , * This equation may also be written as =o, , , , Since H is a constant quantity, therefore the above equation is that of a parabola. It is thus obvious, that, the cable hangs in the form of a parabola.

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306 m Theory of Structures, 12.6 Tension in the Cable Supported at the Same Level, , , , , , Fig. 12.5, , Consider a cable ACB, supported at A and B, at the same level and carrying a uniformly, distributed load as shown in Fig. 9.5, , Let w = Uniformly distributed load per unit length, over the cable,, | = Span of the cable,, Ye = Central dip of the cable, and, H = Horizontal thrust at A and B., Since the cable is symmetrical about the central point C, therefore the vertical reaction at A,, R, = R,= 2, , We have already discussed in Art. 9.1 that the cable, being flexible, can not resist any, moment. Therefore equating the clockwise moment and the anticlockwise moment at C we get, , | (2 “| wi?, x= |-| —x-— | =—, 2°4) 8, , = =a (where W = wi), , *Hy, =, , !, —>, nlz, tv, , A little consideration will show, that the maximum tension in the cable will be at the supports, A and B. Therefore tension in the cable at either of support,, 2, 1+ i 7, l6ye, , Note: If the cable is subjected to point loads, with or without uniformly distributed load, then the, magnitude of tension in the cable will be different at the two supports. In such a case, first of all, find out the two vertical reactions V, and V,, considering the cable as a simply supported beam, of length /. Now the tension in the cable at A,, , T, = (Ri +H?, VR} +H?, , aged, * The horizontal thrust may also be found out by considering C as the origin. Now the co-ordinates of the, , , , , , ", , Similarly Ts,, , I, support B are > and Ye., , 2, ; : 2 (5) 2, We have discussed in Art. 12.4 that the horizontal thrust, A =“ okey WEE, 2y 2xye Bye