The statics of a hanging chain

The curve described by a uniform chain hanging from two supports in a uniform gravitational field is called a *catenary*, a name apparently coined by Thomas Jefferson. If the sag is mall, so that the weight is about uniformly distributed, the curve is close to a parabola, a quadratic curve, but the catenary is a hyperbolic cosine curve, y = a cosh(x/a), where x is measured from the lowest point. The tension in the chain increases from that at the lowest point toward the points of support. Because all the load is vertical, the horizontal component of the tension is constant. A uniform cable also hangs in a catenary; the essential thing is that there is negligible transverse stiffness.

The curve can be inverted to form an arch, in which the tension becomes a compression, least at the highest point and increasing downward. If the arch is uniform, the compressive stress will follow the catenary curve and there will be no tendency to buckle. If the load is horizontally uniform, the curve will be a parabola. The St. Louis Gateway Arch is an example. Since the arch is smaller in cross section at the top, the curve is a flattened catenary.

Let us refer the curve to rectangular axes x,y with the gravitational force in the -y direction. Consider the equilibrium of an infinitesimal length of chain in a distance dx of the x-axis. If the second derivative of the curve y = f(x) is y", then the change of slope from one end of dx to the other is y"dx, and the load supported by the length dx of the chain is wdx, so we have wdx = Hy"dx, where H is the constant horizontal component of the tension. Thus, y"=w/H relates the curvature of the chain to the load density w. If the load density is constant, then so is y". If this constant value of y" is 1/a, then integrating twice gives the form of the curve, y = x^{2}/2a, a parabola, where the constants of integration are chosen so that the origin is the lowest point on the curve. The quantity a is the radius of curvature of the curve at its lowest point, where the tangent is horizontal. An arc of this circle may be a useful approximation to the curve, when the sag is small.

In the case we are considering here, the chain is of uniform linear density w, so wds = Hy" instead. Since ds^{2}=dx^{2}+dy^{2}, we get the differential equation for the curve w√(1 + y'^{2}) = Hy". Taking p = y' and a = H/w, we find √(1 + p^{2}) = a(dp/dx), a separable first order equation that is easily integrated to p = sinh(x/a). A further integration gives y = a cosh(x/a). Note that the constants of integration are chosen so that y(0) = a. This is the catenary curve, whose properties we will discuss below.

Note that the greater the load density, the greater the curvature. If the load is applied at discrete points, the curve is a succession of straight lines with abrupt changes of slope at the points of load application. The change of slope will be proportional to the load applied at that point. The catenary is very commonly observed in pole lines of wires, usually with a small sag and well approximated by a circular arc or parabola.

In the diagram on the left, a flexible chain is shown suspended from points A and B at the same height, so the curve is symmetrical about the lowest point C. This is not essential, and the parts on either side of the lowest point may be treated separately if A and B are not at the same level. In the present case, 2s is the total length of chain suspended between A and B, which are at equal distances b on either side of C. The y-axis is taken vertically upward through C, while the x-axis is horizontal at a certain distance H/w below C, as will be determined later to make the equation of the curve as simple as possible. The tension T in the chain must be in the direction of the tangent to the curve. Its value at C is represented by H, which at C is horizontal and takes its minimum value. The gravitational force on a length ds of the chain is in the -y direction and is of magnitude wds, where w is the weight of the chain per unit length. Since this force is vertical, H will be constant everywhere along the chain, with the vertical component of T increasing towards A and B to support the weight of the chain. If the slope of the chain is dy/dx = tan θ, then T cos θ = H and T sin θ = wx. At B, T sin θ = ws. Therefore, dy/dx = tan θ = wx/H at any point, and ws/H at points A and B. It is convenient to define the quantity H/w = a, which gives the horizontal force as H = wa. We show below that if the origin is properly chosen, T = wy at any point. The quantity a is the radius of curvature of the catenary at its lowest point, as was shown above, and the parameter in the parabolic approximation y = a + x^{2}/2a.

Since ds^{2} = dx^{2} + dy^{2}, ds/dx = √(1 + y'^{2}). However, dy/dx = wx/H = x/a, so d^{2}y/dx^{2} = (1/a)√(1 + y'^{2}). Writing p = dy/dx, this is dp/dx = (1/a)√(1 + p^{2}), in which the variables are separable. The solution for which p = 0 at x = 0 is p = sinh (x/a). From this, we have s = a sinh(x/a), where s is the length of chain from C to the point x. At point B, s = a sinh(b/a), where s is the length of the chain from C to B. If s and b are known, which is usually the case, this equation can be solved for a. A practical method is by trying values of a until the equation is solved.

The equation for p can be integrated again to give y = a cosh(x/a) + C. If C is taken to be 0, then at x = 0, y = a = H/w, as in the figure. As an example, suppose a length of 50" of chain is suspended from two points 30" apart. Then, a is determined from 25 = a sinh(15/a). This is satisfied by a = 8.16, so the curve is y = 8.16 cosh(x/8.16). The sag in the chain is 8.16 (cosh 15/8.16 - 1) = 18.1". The slope of the chain at the point of support is tan θ = s/a, or tan θ = 15/8.16 = 3.063, θ = 71.9°. These figures agree with measurements on an actual chain.

It is only a little more difficult to find the parameters when the supports are not at the same level. The method is essentially the same, in which we assume a value of a, find the corresponding value of the chain length, and adjust a until this is the actual chain length. The first step is to find the lowest point on the curve, at x = x_{o}. We have y_{2} = a cosh (x_{2}-x_{o})/a + y_{o} and y_{1} = a cosh (x_{1}-x_{o})/a +y_{o}. The value of y_{o} can be made to vanish if the y=0 point is chosen as shown in the figure. Subtracting the second equation from the first, we find (y_{2}-y_{1})/a = 2 sinh(x_{m}-x_{o})/a sinh Δx/a, using the identity for the difference in hyperbolic cosines (Dwight 651.09) which we can solve for x_{m}-x_{o}, which locates the lowest point C at which the chain is horizontal.

The total chain length is the sum of the lengths of chain on either side of the lowest point. The sum of the two sinh functions can be expressed as s = 2a sinh(Δx/a)cosh[(x_{m}-x_{o})/a]. Dividing the expression for y_{2}-y_{1} obtained in the preceding paragraph by this expression for s, we find s = (y_{2}-y_{1})/tanh[(x_{m}-x_{o})/a].

Knowing the coordinates of the two ends of the chain, we may assume a value of a and find (x_{m}-x_{o})/a. Then the length s of the chain may be calculated by using this quantity in the expression for s and comparing with the known length of chain. A new value of a is chosen that improves the value of s, and this is repeated until the length of the chain is correct. Making a smaller increases s, while making a larger decreases it. The chain becomes more taut as a increases.

As an example, let x_{1}=0, y_{1}=10, x_{2}=30, y_{2}=20, and chain length s=50.0. Then Δx=15. For a=8.00, (x_{m}-x_{o})/a=0.1950 and s=51.91. For a=9.00, (x_{m}-x_{o})/a=0.2159 and s=47.00. By linear interpolation, a=8.39 will give a chain length of 50.0.

The tension T in the chain at any point is in the direction of the tangent. At the lowest point, the tangent is horizontal and the tension is H. Because the weight of the chain is vertical, the horizontal component of T is equal to H at any point, and the tension is T = H/cos θ, where θ is the slope. Thus, T increases steadily as the points of support are approached. If the equation of the curve is y = a cosh[(x - x_{o})/a], so that at the lowest point y=a, dy/dx = sinh[(x - x_{o})/a] = tan θ, and 1/cos θ = √(1 + tan^{2}θ) = cosh[(x - x_{o})/a]. Therefore, T = H cosh[(x - x_{o})/a] = Hy/a = wy, a remarkably simple result. This is a consequence of the fact that the slope and the arc length are both sinh functions.

The vertical component of T at a point of support is V = H tan θ and is equal to ws, where s is the length of chain from the lowest point. Thus H sinh [(x - x_{o})/a] = wa sinh [(x - x_{o})/a], or H = wa, confirming the consistency of our analysis..

In many practical applications, the span between supports and the weight w of the cable per unit length are known. Selection of a value of a=H/w will determine the curve y = a cosh(x/a). Note that a is a scale factor and determines the length of the cosh curve that is used--the larger a is, the smaller the amount of the curve. Assuming a cable tension H will give a value for a. If a is considerably larger than the span, the cable will be tightly stretched and the sag small.

Supose the span is 150 ft and the cable weighs 5 lb/ft. Then, choosing a = 1000 makes H = wa = 5000 lb. Then, the y-value at a support is 1000 cosh(75/1000) = 1002.81 ft, so at the support T = wy1 = 5014 lb. The sag is 1002.81 - 1000 = 2.81 ft. The slope at the support is y'= sinh(75/1000) = 0.075070 = tan θ, so θ = 4.293° The length of cable is ay' = 75.07 times 2, or 150.14 ft. The vertical component of T at the support is T sin θ, or 375.33 lb, supporting half the cable weight of 5 x 75.07 = 375.35 lb.

If a = 500 ft is chosen instead, we find sag = 5.63 ft, θ = 8.56°, cable length 150.56 ft, H = 2500 lb and T = 2528 lb. For a = 150 ft, we get sag = 19.14 ft, θ = 27.52°, length = 156.32 ft, H = 750 lb, T = 845.7 lb. In the latter case, the cable is hanging rather loosely. It is easy to do these calculations with a pocket calculator with hyperbolic functions.

When a is large compared to the span, the power expansion of the cosh may be used. Then y = a cosh(x/a) = a + x^{2}/2a + ... . If only the lowest order is retained, we have y = x^{2}/2a if y is now measured from the lowest point on the curve. This is a parabola, which makes a good approximation to the catenary in this case. The slope dy/dx = x/a, and the second derivative d^{2}x/dy^{2} = 1/a is a constant. Since the second derivative is proportional to the load w dx. This implies that the load is uniform (not w ds as in the accurate catenary), the well-known result that the curve is a parabola in this case. If the span and the sag b are known, then a = x^{2}/2b. This formula may be used to estimate a for a catenary whose span and sag are known, if the sag is not too large. Note that it can also be found by taking moments about a point of support, where Hb = wx^{2}/2.

The length of a parabolic curve is easy to find by integrating ds = √(1 + y'^{2})dx, but the result is not pretty: s =(x/2)√(1 + x^{2}/a^{2}) + (a/2)ln[x/a + √(1 + x^{2}/a^{2})]. for x<< a, s = x. Using the parabolic approximation actually saves very little work compared to working with the accurate catenary when a calculator is used.

Another derivation of the catenary curve follows.

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)(S^{2} - 4h^{2}), 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 T^{2} = H^{2} + (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 = wx^{2}/2H. Therefore, to this approximation, we have h = wL^{2}/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 - x_{0}) for a force F at x = x_{0} 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.

There is more about the catenary in Curves.

A. L. Nelson, K. W. Folley and M. Coral, *Differential Equations* (Boston: D. C. Heath & Co., 1952). pp 160-161.

The Wikipedia article "Catenary" presents a number of examples of the curve.

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Composed by J. B. Calvert

Created 10 July 2000

Last revised 25 July 2011