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Suspension Bridges and Calculus

Suspension bridges are among the most commonly built bridges in the world. Such bridges feature long horizontal decks which are hung below suspension cables on vertical rods or hangers. Assuming a negligible weight as compared to that of the deck and the vehicles being supported, the main cables of a suspension bridge will take the shapes of parabolas (which closely resemble that of catenaries, i.e., the shapes of the free cables before the hangers are attached). The following analysis of a suspension bridge cable, supporting a uniform load, is excerpted from Morris Kline’s Caclulus: An Intuitive and Physical Approach Calculus enables us to determine the shape of the bridge cable when the total weight of the load, the cable and the roadway per horizontal foot is constant. The cable is assumed to be perfectly flexible (i.e., it offers no resistance to bending) and is inelastic (i.e., it does not stretch or contract). Both properties are possessed by a piece of ordinary rope. Consider the cable Q’AQ in the figure below. Because the cable supports the roadway, there is tension in the cable. Let us first satisfy ourselves that at any point on the cable the tension that any part of the cable exerts on the adjoining part is along the cable. Specifically we should see that the tension exerted by arc QP on the rest of the cable has the direction of the tangent at P. Intuitively this fact is rather obvious, for if the pull at P were not along the tangent then cable would bend (because it is perfectly flexible) inward or outward.

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Now let us consider any section of the flexible cable, say the section from A to P. Because the pull at A by the section to the left of A is tangential to the cable, at A the pull or tension is horizontal and to the left. The tension is some constant which we shall denote by To . The tension on AP at P exerted by the section PQ of the cable is along the tangent at P. Let us call the magnitude of this tension T. Because this tension also has a direction, namely, the direction of the tangent at P, it is represented geometrically by a directed line segment or a vector. We shall let be the angle which the direction of the tension (or the tangent) at P makes with the horizontal.

There is one more force acting on the section AP of the cable, namely, the pull of that portion of the cable, load and the roadway which extends from O to P’. Because the total pull or weight is assumed to be the same per horizontal foot, then if w is the weight per horizontal foot, the load supported by the arc AP is wx, where x is the abscissa of P. (We are here using the fact that the y-axis passes through A.) The pull of this load is actually distributed along AP, but all of the pulls are downward, and we need to know only that there is a total downward pull on AP amounting to wx. Thus, there are three forces acting on AP: the horizontal pull To to the left, the downward pull wx of the total load, and the tension or pull T tangential at P. Because the arc AP is at rest or in equilibrium, the three forces must somehow offset one another, for if there were some net force the cable would move or bend under the action of that force. Alternatively, we may say that the horizontal forces must offset one another and that the vertical forces must do likewise. The tension T is equivalent to a horizontal and a vertical force acting simultaneously, for a tension is a force and, any force may be replaced by the proper horizontal and vertical components. Specifically, the tension of the magnitude T is equivalent to the combined action of the horizontal component: T cos and the vertical component: T sin . The horizontal component, T cos , must be offset by the only other horizontal force that is acting, namely, To. Hence, T cos = To . (1)

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The vertical component, T sin , of the tension must offset the only other vertical force that is acting, namely, wx. Hence, T sin = wx. (2) If we divide (2) by (1) , we obtain (3) Now tan is the slope of the tangent line at P, and this slope is y’ where y’ is the derivative of whatever function represents the shape of the cable. Thus, (4) By antidifferentiation, we obtain (5) To fix the constant C we can choose the x-axis so as to pass through the point A of the main figure on page 2. Then y = 0 when x = 0, so that C = 0. Thus, (6) Let us note one or two implications of the result (6). Although the equation seems to involve two unknown quantities w and To , only their ratio is involved. We can in fact write the equation as y = k 𝑥2 where k = 𝑤 (2𝑇𝑜)⁄ . To determine the equation of the parabola, we need to know the coordinates of one point through which the parabola must go. Suppose that we do specify the coordinates of one point, say the point (2 , 3). Because the curve is symmetric with respect to the y-axis, the point (–2 , 3) must also be on the curve. Moreover, the lowest point on the cable must be the origin if (6) is presupposed. Hence, to specify the coordinates of one point means really that the width from (–2 , 3) to (2 , 3) and the depth, 3, are specified. We know, however, that the cable must have an equation of the form y = k 𝑥2 , and there is only one parabola of this form through a given point. Hence, it would seem that if we fix one point, that is, fix the width and depth, then there is no freedom to choose the load that the cable will support. However, this is not true. The quantity k = 𝑤 (2𝑇𝑜)⁄ . So fixing k, fixes only the ration of w to To . If we increase w, the load per horizontal foot, then To must also increase in the same ratio. To sum up, if the width of a cable and the depth at that width are specified, there is only one shape for the cable regardless of the load, provided that the load is constant over each horizontal foot.

tan 𝜃 = 𝑤

𝑇𝑜 𝑥.

𝑦′ = 𝑤

𝑇𝑜 𝑥.

𝑦 = 𝑤

2 𝑇𝑜 𝑥2 + 𝐶 .

𝑦 = 𝑤

2 𝑇𝑜 𝑥2 .

𝒚 = 𝒌 𝒙𝟐