Calculation of Catenary7-Jul-05

JMTMetric Customary

Total linear load 0.28812 N/m 0.019348 lb/ftTotal Length 10 m 32.8 ft

sag 0.1 m 0.328 ft

Fh 36.000594 N 7.929646 lb

Span 9.99733312 m 32.79125 ftT 36.029406 N 7.935993 lb Calculation of size, given AWG

AWG diam inchcable diameter 2.05 mm 0.080709 inch 12 0.079433area 3.300636E-06 sq m 0.005116 sq inchStress 10.915898749 MPa 1.551212 kips

linear weight 0.28812 N/m 0.019348 lb/ft

Wind drag computation

Wind speed 0 m/sec 0 mi/hrrelative density to Std 1 1diameter 0.00205 m 0.006724 ftDrag/length 0 N/m 0 lb/ft

Temperature 10 C 50 FAltitude 1000 m 3280 ftstd tmp 8.5 C 47.30 F Rel density 0.8862888993 0.886035abs density 1.0857039017 kg/cu m

, Huygens, and Johann Bernoulliin 1691 in response to a challenge by Jakob Bernoulli

Calculation of size, given AWG .diam mm2.017594 Huygens

was the first to use the term catenary in a letter to Leibnizin 1690, and David Gregory

The parametric equations for the catenary are given by

The curve a hanging flexible wire or chain assumes when supported at its ends and acted upon by a uniform gravitational force. The word catenary is derived from the Latin word for "chain." In 1669, Jungius disproved Galileo's

claim that the curve of a chain hanging under gravity would be a parabola (MacTutor Archive). The curve is also called the alysoid and chainette. The equation was obtained by Leibniz

wrote a treatise on the catenary in 1690 (MacTutor Archive). If you roll a parabola along a straight line, its focus traces out a catenary. As proved by Euler

in 1744, the catenary is also the curve which, when rotated, gives the surface of minimum surface area (the catenoid) for the given bounding circle.

The CatenaryThe statics of a hanging chain

--------------------------------------------------------------------------------

The curve described by a uniform, flexible chain hanging under the influence of gravity is called the catenary. Let us choose the coordinates to describe the curve as shown in the Figure, with the origin at the lowest point of the chain, x to the right, and y upwards. Let one end of the chain be supported at x = L/2 and y = h, and let w be the weight per unit length of the chain. The tension T in the chain is in the direction of the derivative dy/dx. Its horizontal component is H, and its vertical component is V. If we cut the chain at some point as shown, and consider the equilibrium of the portion of the chain between the right-hand end and this point, we see that H is constant at all points of the chain, since the weight of the chain acts vertically downward. Therefore, the vertical component of the tension in the chain is Hy', where y' stands for dy/dx.

Consider the equilibrium of the short length of chain subtending a distance dx on the x-axis. The horizontal forces are equal, as we have already seen, and the sum of the vertical forces is V + dV - V - w ds = 0. Since V = Hy', dV = Hy" dx, and we obtain a differential equation that is first order in y'. This equation is integrated as shown in the Figure to find y'(x), and again to find y(x), using the conditions that y = y' = 0 at x = 0. The catenary is a hyperbolic cosine curve, and its slope varies as the hyperbolic sine. The arc length, measured from the origin, is s = (H/w) sinh (wx/H), which shows that the slope is proportional to arc length. The total length of the chain is S = (2H/w) sinh (wL/2H), and h = (H/w)[cosh (wL/2H) - 1]. We can use these two equations to find that H = (w/8h)(S2 - 4h2), in terms of the total length of the chain and the sag. Then, the span L = (2H/w) sinh-1 (Sw/2H). The vertical force at a support is V = wS/2, so the tension there is given by T2 = H2 + (wS/2)2. We have assumed that the chain is supported at equal heights on the two sides, but this is not necessary, and the equations can be modified to suit any case.

If we expand the hyperbolic cosine in powers of x, the lowest term yields y = wx2/2H. Therefore, to this approximation, we have h = wL2/8H, a familiar result for the parabolic curve of a weightless chain supporting a uniform load, easily obtained by taking moments about the right-hand support of the chain. If we assume the weight is uniformly distributed in x, the differential equation is w dx = Hy", which, when integrated twice, gives just the result above.

If a vertical downward force F acts at a point of the chain, the slope is discontinuous at that point, jumping by an amount F/H. This will add a term Fδ(x - x0) for a force F at x = x0 to the differential equation. The shape of the whole chain changes, illustrating a well-known problem with suspension bridges. If the deck is not sufficiently rigid, the bridge can oscillate, one portion sinking as another rises. This can happen in opposite phase on the two sides of the bridge, resulting in a torsional oscillation.

If the chain is imagined to become rigid, and is turned over, tensions will become compressions, and we will have an arch. The curve shows how the stress is propagated to the supports, provided the structure remains rigid. In this case, a thin structure tends to buckle, a problem not encountered with the chain. The results must be used with care for arches, since the loading condtions are usually quite different. The usual circular arch has heavy haunches, which helps to guide the force along the voussoirs. A segmental arch is helped by pierced spandrels, since this reduces the force on a flatter portion of the arch and distributes the load more uniformly.

-1-2-3

where

are given by

-4-5-6

The slope is proportional to the arc length as measured from the center of symmetry.

The Cesàro equation is

-7

The catenary also gives the shape of the road (roulette) over which a regular polygonal "wheel" can travel smoothly. For a regular n-gon, the Cartesian equation of the corresponding catenary is

-8

corresponds to the vertex and a is a parameter that determines how quickly the catenary "opens up." Catenaries for values of a ranging from 0.05 to 1.00 by steps of 0.05 are illustrated above.

The arc length, curvature, and tangential angle for

The St. Louis Arch closely approximates an inverted catenary, but it has a finite thickness and varying cross sectional area (thicker at the base; thinner at the apex). The centroid has half-length of

feet at the base, height of 625.0925 feet, top cross sectional area 125.1406 square feet, and bottom cross sectional area 1262.6651 square feet.

where

-9

REFERENCES:

Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, p. 214, 1987.

Geometry Center. "The Catenary." http://www.geom.umn.edu/zoo/diffgeom/surfspace/catenoid/catenary.html.

Gray, A. "The Evolute of a Tractrix is a Catenary." §5.3 in Modern Differential Geometry of Curves and Surfaces with Mathematica, 2nd ed. Boca Raton, FL: CRC Press, pp. 102-103, 1997.

Lawrence, J. D. A Catalog of Special Plane Curves. New York: Dover, pp. 195 and 199-200, 1972.

Lockwood, E. H. "The Tractrix and Catenary." Ch. 13 in A Book of Curves. Cambridge, England: Cambridge University Press, pp. 118-124, 1967.

MacTutor History of Mathematics Archive. "Catenary." http://www-groups.dcs.st-and.ac.uk/~history/Curves/Catenary.html.

National Park Service. "Arch History and Architecture: Catenary Curve Equation." http://www.nps.gov/jeff/equation.htm.

Pappas, T. "The Catenary & the Parabolic Curves." The Joy of Mathematics. San Carlos, CA: Wide World Publ./Tetra, p. 34, 1989.

Steinhaus, H. Mathematical Snapshots, 3rd ed. New York: Dover, pp. 247-249, 1999.

Wells, D. The Penguin Dictionary of Curious and Interesting Geometry. London: Penguin, pp. 26-27, 1991.

Yates, R. C. "Catenary." A Handbook on Curves and Their Properties. Ann Arbor, MI: J. W. Edwards, pp. 12-14, 1952.

SEE ALSO: Calculus of Variations, Catenoid, Lindelof's Theorem, Roulette, Surface of Revolution.

The CatenaryThe statics of a hanging chain

--------------------------------------------------------------------------------

The curve described by a uniform, flexible chain hanging under the influence of gravity is called the catenary. Let us choose the coordinates to describe the curve as shown in the Figure, with the origin at the lowest point of the chain, x to the right, and y upwards. Let one end of the chain be supported at x = L/2 and y = h, and let w be the weight per unit length of the chain. The tension T in the chain is in the direction of the derivative dy/dx. Its horizontal component is H, and its vertical component is V. If we cut the chain at some point as shown, and consider the equilibrium of the portion of the chain between the right-hand end and this point, we see that H is constant at all points of the chain, since the weight of the chain acts vertically downward. Therefore, the vertical component of the tension in the chain is Hy', where y' stands for dy/dx.

Consider the equilibrium of the short length of chain subtending a distance dx on the x-axis. The horizontal forces are equal, as we have already seen, and the sum of the vertical forces is V + dV - V - w ds = 0. Since V = Hy', dV = Hy" dx, and we obtain a differential equation that is first order in y'. This equation is integrated as shown in the Figure to find y'(x), and again to find y(x), using the conditions that y = y' = 0 at x = 0. The catenary is a hyperbolic cosine curve, and its slope varies as the hyperbolic sine. The arc length, measured from the origin, is s = (H/w) sinh (wx/H), which shows that the slope is proportional to arc length. The total length of the chain is S = (2H/w) sinh (wL/2H), and h = (H/w)[cosh (wL/2H) - 1]. We can use these two equations to find that H = (w/8h)(S2 - 4h2), in terms of the total length of the chain and the sag. Then, the span L = (2H/w) sinh-1 (Sw/2H). The vertical force at a support is V = wS/2, so the tension there is given by T2 = H2 + (wS/2)2. We have assumed that the chain is supported at equal heights on the two sides, but this is not necessary, and the equations can be modified to suit any case.

If we expand the hyperbolic cosine in powers of x, the lowest term yields y = wx2/2H. Therefore, to this approximation, we have h = wL2/8H, a familiar result for the parabolic curve of a weightless chain supporting a uniform load, easily obtained by taking moments about the right-hand support of the chain. If we assume the weight is uniformly distributed in x, the differential equation is w dx = Hy", which, when integrated twice, gives just the result above.

If a vertical downward force F acts at a point of the chain, the slope is discontinuous at that point, jumping by an amount F/H. This will add a term Fδ(x - x0) for a force F at x = x0 to the differential equation. The shape of the whole chain changes, illustrating a well-known problem with suspension bridges. If the deck is not sufficiently rigid, the bridge can oscillate, one portion sinking as another rises. This can happen in opposite phase on the two sides of the bridge, resulting in a torsional oscillation.

If the chain is imagined to become rigid, and is turned over, tensions will become compressions, and we will have an arch. The curve shows how the stress is propagated to the supports, provided the structure remains rigid. In this case, a thin structure tends to buckle, a problem not encountered with the chain. The results must be used with care for arches, since the loading condtions are usually quite different. The usual circular arch has heavy haunches, which helps to guide the force along the voussoirs. A segmental arch is helped by pierced spandrels, since this reduces the force on a flatter portion of the arch and distributes the load more uniformly.

The catenary also gives the shape of the road (roulette) over which a regular polygonal "wheel" can travel smoothly. For a regular n-gon, the Cartesian equation of the corresponding catenary is

Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, p. 214, 1987.

Geometry Center. "The Catenary." http://www.geom.umn.edu/zoo/diffgeom/surfspace/catenoid/catenary.html.

Gray, A. "The Evolute of a Tractrix is a Catenary." §5.3 in Modern Differential Geometry of Curves and Surfaces with Mathematica, 2nd ed. Boca Raton, FL: CRC Press, pp. 102-103, 1997.

Lawrence, J. D. A Catalog of Special Plane Curves. New York: Dover, pp. 195 and 199-200, 1972.

Lockwood, E. H. "The Tractrix and Catenary." Ch. 13 in A Book of Curves. Cambridge, England: Cambridge University Press, pp. 118-124, 1967.

MacTutor History of Mathematics Archive. "Catenary." http://www-groups.dcs.st-and.ac.uk/~history/Curves/Catenary.html.

National Park Service. "Arch History and Architecture: Catenary Curve Equation." http://www.nps.gov/jeff/equation.htm.

Pappas, T. "The Catenary & the Parabolic Curves." The Joy of Mathematics. San Carlos, CA: Wide World Publ./Tetra, p. 34, 1989.

Wells, D. The Penguin Dictionary of Curious and Interesting Geometry. London: Penguin, pp. 26-27, 1991.

Yates, R. C. "Catenary." A Handbook on Curves and Their Properties. Ann Arbor, MI: J. W. Edwards, pp. 12-14, 1952.

The CatenaryThe statics of a hanging chain

--------------------------------------------------------------------------------

The curve described by a uniform, flexible chain hanging under the influence of gravity is called the catenary. Let us choose the coordinates to describe the curve as shown in the Figure, with the origin at the lowest point of the chain, x to the right, and y upwards. Let one end of the chain be supported at x = L/2 and y = h, and let w be the weight per unit length of the chain. The tension T in the chain is in the direction of the derivative dy/dx. Its horizontal component is H, and its vertical component is V. If we cut the chain at some point as shown, and consider the equilibrium of the portion of the chain between the right-hand end and this point, we see that H is constant at all points of the chain, since the weight of the chain acts vertically downward. Therefore, the vertical component of the tension in the chain is Hy', where y' stands for dy/dx.

Consider the equilibrium of the short length of chain subtending a distance dx on the x-axis. The horizontal forces are equal, as we have already seen, and the sum of the vertical forces is V + dV - V - w ds = 0. Since V = Hy', dV = Hy" dx, and we obtain a differential equation that is first order in y'. This equation is integrated as shown in the Figure to find y'(x), and again to find y(x), using the conditions that y = y' = 0 at x = 0. The catenary is a hyperbolic cosine curve, and its slope varies as the hyperbolic sine. The arc length, measured from the origin, is s = (H/w) sinh (wx/H), which shows that the slope is proportional to arc length. The total length of the chain is S = (2H/w) sinh (wL/2H), and h = (H/w)[cosh (wL/2H) - 1]. We can use these two equations to find that H = (w/8h)(S2 - 4h2), in terms of the total length of the chain and the sag. Then, the span L = (2H/w) sinh-1 (Sw/2H). The vertical force at a support is V = wS/2, so the tension there is given by T2 = H2 + (wS/2)2. We have assumed that the chain is supported at equal heights on the two sides, but this is not necessary, and the equations can be modified to suit any case.

If we expand the hyperbolic cosine in powers of x, the lowest term yields y = wx2/2H. Therefore, to this approximation, we have h = wL2/8H, a familiar result for the parabolic curve of a weightless chain supporting a uniform load, easily obtained by taking moments about the right-hand support of the chain. If we assume the weight is uniformly distributed in x, the differential equation is w dx = Hy", which, when integrated twice, gives just the result above.

If a vertical downward force F acts at a point of the chain, the slope is discontinuous at that point, jumping by an amount F/H. This will add a term Fδ(x - x0) for a force F at x = x0 to the differential equation. The shape of the whole chain changes, illustrating a well-known problem with suspension bridges. If the deck is not sufficiently rigid, the bridge can oscillate, one portion sinking as another rises. This can happen in opposite phase on the two sides of the bridge, resulting in a torsional oscillation.

If the chain is imagined to become rigid, and is turned over, tensions will become compressions, and we will have an arch. The curve shows how the stress is propagated to the supports, provided the structure remains rigid. In this case, a thin structure tends to buckle, a problem not encountered with the chain. The results must be used with care for arches, since the loading condtions are usually quite different. The usual circular arch has heavy haunches, which helps to guide the force along the voussoirs. A segmental arch is helped by pierced spandrels, since this reduces the force on a flatter portion of the arch and distributes the load more uniformly.