Thompson - Computing for Scientists and Engineers (523188), страница 51
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nFIGURE 8.8 Equilibrium shapes of chains hanging under gravity: (1) for a parabola, (2) for auniform-density catenary. (3) for uniform strength. and (4) for the arc of a circle. The formulas aregiven in Table 8.2.8.2 CATENARIES, CATHEDRALS, AND NUPTIAL ARCHES275FIGURE 8.9 Scaled linear densities for the catenaries shown in Figure 8.8: (1) for a parabola,(2) for a uniform-density catenary, (3) for uniform strength, and (4) for the arc of a circle.The uniform-density catenary, the usual example, can be solved by a similaranalysis to that for the parabola to show directly that the scaled shape is given by(8.49)which is just (8.40) with L = 1, as you might have guessed.
This shape, and thedensity, tension, weight, and strength distributions that follow from application of(8.43) through (8.46), are given in Table 8.2 and are shown in Figures 8.8 -8.11as curves (2). Notice that by comparison with the parabola, the usual catenary becomes steeper as x increases. Since for this catenary the chain density is constant,correspondingly the flatter parabola must have a decreasing density, which agreeswith the result in Table 8.2 and with curve (1) in Figure 8.9.FIGURE 8.10 Tension as a function of x for the catenaries shown in Figure 8.8: (1) for aparabola, (2) for a uniform-density catenary. (3) for uniform strength, and (4) for the arc of a circle.276SECOND-ORDER DIFFERENTIAL EQUATIONSFIGURE 8.11 Strength as a function of x for the catenaries shown in Figure 8.8: (1) for aparabola, (2) for a uniform-density catenary, (3) for a uniform-strength catenary, and (4) for the arcof a circle.
The formulas are given in Table 8.2.Exercise 8.15(a) For the uniform-density catenary calculate the first and second derivatives,then substitute into (8.43) through (8.46) for the scaled density, tension, weight,and strength, thus verifying the results for line (2) in Table 8.2.(b) Write a small program then calculate and graph the x dependence of the regular catenary shape, tension, and strength, thus verifying the curves (2) in Figures 8.8-8.11. nFor a circle-arc shape, that is, a semicircle concave upward, what is the corresponding density needed to produce this shape? It’s very simple, because the appropriate scaled circle, having zero value and unity second derivative at the x origin, isjust(8.50)with |x| < 1 to ensure that the chain is under tension.Exercise 8.16(a) For the circle-arc catenary calculate the first and second derivatives, thensubstitute into (8.43) through (8.46) for the scaled density, tension, weight, andstrength, thus verifying the results for line (4) of Table 8.2.(b) Write a program, then calculate and graph the x dependence of this catenaryshape, density, tension, and strength, thus verifying the curves (4) in Figures 8.8-8.11.
n8.2 CATENARIES, CATHEDRALS, AND NUPTIAL ARCHES277The circular arc is even steeper than a catenary, so its density must increase as xincreases, as Figure 8.9 shows. The tension also increases with x, but not as rapidly, so the strength of the chain steadily decreases with x. As |x| tends to unity, thedensity and tension must become indefinitely large in order to produce the nearlyvertical shape. We therefore show the density and tension only up to |x| = 0.8.
Forreal materials their elasticity would have to be considered for large densities and tensions, but this we have not included explicitly.In two of the above examples the strength of the chain (ratio of cable weight perunit length at a given x to the tension in the cable at x) was found to decrease with x,so that for a large enough span the chain will exceed its elastic limit.
For the thirdexample, the circle arc, the strength increases with x, as shown in Table 8.2 andFigure 8.11, curve (4).Consider a constant-strength catenary. If the strength is to be a constant with respect to x, what is its shape ? Presumably, it will be intermediate between a uniformdensity catenary and a circle, curves (2) and (4) in Figure 8.11. For the scaled unitsthat we are using, (8.42) shows that the constant strength, must be unity everywhere, since that is its value at x = 0. From (8.46) we deduce immediately the differential equation for the shape of the constant-strength catenary, namely(8.5 1)It is straightforward to verify that integrating this once produces the slope conditionin which the constant of integration is zero because the slope is to be zero at x = 0.Integrating (8.52) once more produces(8.53)as can be verified by substitution in (8.52).
Again, the constant of integration iszero, because in our coordinate systemy (0) = 0.Exercise 8.17Carry out the steps indicated to produce the uniform-strength catenary formula(8.53). nThe result (8.53) for the chain of uniform strength is attributed to British engineerDavies Gilbert, who reported in 1826 his use of it in designing the first suspensionbridge across the Menai Strait in north Wales.
Because he did not use scaled variables, much of his article consists of very long tables of bridge-design parameters.From (8.53) we see that, just as for the circular arc, the uniform-strength catenary is possible only for a limited range of the scaled x variable; here |x| < /2. We278SECOND-ORDER DIFFERENTIAL, EQUATIONStherefore show in Figure 8.8 the uniform-strength catenary, curve (3), only up to|x| = 1.4. Notice that for uniform strength the shape is intermediate between thatfor the uniform-density catenary, curve (2), which has decreasing strength with increasing x, and the circle-arc catenary, curve (4).For the uniform-strength catenary given by (8.53) it is straightforward to use(8.43) to (8.45) and to calculate the scaled density (which equals the scaled tension)and the scaled weight of chain below x , to produce the formulas in Table 8.2 andcurves (3) in Figure 8.9 or 8.10.
Generally, the uniform-strength catenary has mechanical properties intermediate between those for the uniform-density catenary andthe circle-arc catenary.Demonstrating archesHow can you demonstrate the relation between forces on a catenary and the catenaryshape? A simple method is to follow up an idea described in an article by Honig andto use light helium balloons attached at equal horizontal intervals to a relatively lightstring. Thereby the weight distribution of the chain is replaced by the buoyancy ofthe balloons, and the analysis is identical to that above, except that the sign of y isreversed, so that inverted curves are obtained. For each shape, the size of the balloons will need to be adjusted for the appropriate buoyancy, the analog of p (x).We illustrate the setup for the circle-arc catenary because this has the most rapidvariation of density with x, as curve (4) in Figure 8.9 shows. If the balloons arefilled so that their volumes increase as 1/(1 - x2 ) and if the ends are fixed at horizontal points, then a curve which is almost a semicircle will be formed.
The shapewill be slightly different near the endpoints, where very large balloons would beneeded to pull the endpoints straight up.FIGURE 8.12 Arc of a circle made of a light string supported by helium balloons with diameters increasing to satisfy the weight relations, w(x), for line (4) in Table 8.2.8.3 NUMERICAL METHODS FOR SECOND-ORDER DIFFERENTIAL, EQUATIONS279A sketch of such an arrangement is shown in Figure 8.12, where the outermostballoons at |x| = 0.9 have diameters more than 1.7 times greater than that of the center balloon. When the system is in static equilibrium, the midpoint of the catenaryshould be above the base by half the separation of the ends. If helium is carefullyreleased from the outer balloons until all the balloons have the same shape, then thegraceful uniform-density “nuptial arch” referred to by Honig should form.
If slightly more helium is released from the outer balloons, then a parabola will form.Practical arches and catenariesThe history of the catenary and its relation to the stability of arches and domes inlarge buildings, such as cathedrals, is very interesting. Historically, the uniformdensity catenary problem was pondered by Galileo, who explained (incorrectly) that“This [uniform] chain curves in a parabolic shape.” The first correct solution waspublished (in Latin) by Johannis Bernoulli in the late seventeenth century, and healso indicated the general relation (8.43) between the shape and the density distribution. Robert Hooke, better known for his research on elasticity (Hooke’s law) andin microscopy, solved the uniform-density catenary problem in 1676.
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