Study Companion: S22 — Flexible Cables
A beam (S20–S21) carries load by bending. A cable cannot bend at all: it has no resistance to bending, so its internal force always points along it, and it can only pull. It carries load the other way round, by taking whatever shape makes pure tension enough. Suspension bridges, power lines, ski lifts and the washing line in a garden all work this way. This lesson finds that shape, and the tension, span, sag and length that go with it, from the equilibrium of one small piece of cable.
Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 5/8 ("Flexible Cables"), with Sample Problems 5/16–5/18. Harper, Solving Statics Problems in MATLAB, "Flexible Cables" (its Sample Problem 5/17, which replaces the parabolic cable of its Sample Problem 5/16 by a catenary; these are Meriam's 9th-edition Sample Problems 5/18 and 5/17). Meriam's Sample Problem 5/16 is in US units, converted here: 1 ft = 0.3048\,\mathrm m, 1 lbf = 4.448\,\mathrm N. Axes: x horizontal, y up, origin at the cable's lowest point unless stated.
How to use this companion
Each nugget is one idea. It opens with a claim, then a picture shows why it holds, a number checks it, and an equation states it, labelled with the kind of statement it is. A worked example with units, a real use, the most common wrong answer and a check question follow. A bridge says why the next idea is needed, and the Narration spine summarises the video beat. Traps and a self-check with full solutions close the lesson.
The puzzle
A surveyor measures with a 30.48\,\mathrm m steel tape (Meriam's 100 ft tape) weighing 2.669\,\mathrm N. Stretched between two pegs at the same level, it is pulled with 44.48\,\mathrm N (10 lb) at each end, seventeen times its own weight.
How far does it sag in the middle? The plausible guess is "pulled that hard, it is practically straight; a millimetre or two". The answer is in Nugget 04.
Nugget 01: A Cable Can Only Pull, Along Its Own Length
Claim. A flexible cable offers no resistance to bending, so the force inside it is always tension along the cable. Between concentrated loads it is straight; under a distributed load it curves. The horizontal component of its tension is the same everywhere when all the loads are vertical.
The picture. Meriam's Fig. 5/26a and b: a cable with concentrated loads, and one with a distributed load w per unit horizontal length.
With no load between them, two points of a cable are joined by a straight segment: a two-force member (S11). At each load the segment changes direction.
Check with numbers. Level supports 9\,\mathrm m apart, 2\,\mathrm{kN} hung at 3\,\mathrm m and at 6\,\mathrm m, and the cable 1\,\mathrm m below the supports at the loads. By symmetry each support carries 2\,\mathrm{kN} vertically. The end segment drops 1\,\mathrm m in 3\,\mathrm m, so its tension's components are in the ratio 1 : 3: horizontal T_0 = 2(3/1) = 6\,\mathrm{kN}, total \sqrt{6^2 + 2^2} = 6.32\,\mathrm{kN}. The middle segment is horizontal and carries the same 6\,\mathrm{kN}: the horizontal component did not change.
The idea. Meriam (5/8): cables are analysed as bodies in equilibrium, assuming any resistance to bending is negligible, so the force in the cable is always in its direction. A cable may carry distinct concentrated loads (Fig. 5/26a) or loads distributed continuously (5/26b); its own weight may be negligible against the loads, appreciable, or the only load; the equilibrium is formulated the same way in every case. For a load of intensity w per unit horizontal length, the resultant of the vertical load over an interval is R = \int w\,dx, placed by the moment principle, R\bar x = \int x\,dR: it passes through the centroid of the load diagram's area (S20).
The equation (material model: the perfectly flexible cable; with physical law, equilibrium of each segment):
Worked number. If the two loads above hang 0.5\,\mathrm m lower (sag 1.5\,\mathrm m), T_0 = 2(3/1.5) = 4\,\mathrm{kN} and the end tension \sqrt{4^2 + 2^2} = 4.47\,\mathrm{kN}: more sag, less tension.
Real example (one you can see). A washing line with a few heavy towels on it is a chain of straight segments with a kink at each peg.
Plausible wrong answer. "A cable can be pushed into place like a strut." A cable under compression simply goes slack: every support it acts on must be pulled, never pushed.
Check. If a computed cable force comes out negative, what does it mean? (The cable would have to push, which it cannot: it goes slack, and the assumed shape is wrong.)
Narration spine. A rope hangs from two pegs; two weights clip on and the rope snaps into three straight segments. Tension arrows appear along each segment; their horizontal shadows line up, all the same length.
Bridge. Under a distributed load the segments shrink to nothing and the cable becomes a curve. One small element gives its equation.
Nugget 02: Every Element in Equilibrium Gives the Cable Equation
Claim. For a cable carrying a vertical load w per unit horizontal length, the horizontal component of tension T_0 is constant, and the shape y(x) satisfies d^2y/dx^2 = w/T_0. The shape follows by integrating twice with the end conditions.
The picture. Meriam's Fig. 5/26c: an element between x and x + dx.
Check with numbers. In Nugget 01's cable, the load between supports is all at two points, so w = 0 elsewhere and d^2y/dx^2 = 0: straight segments. At each load the slope jumps by 2\,\mathrm{kN}/6\,\mathrm{kN} = 1/3: from -1/3 to 0 at the first load, from 0 to +1/3 at the second. That is the cable equation summed across a point load, \Delta(dy/dx) = P/T_0.
The idea. Meriam (5/8): at x the tension is T at angle \theta to the horizontal; at x + dx it is T + dT at \theta + d\theta; the load is w\,dx. Vertical and horizontal equilibrium:
Expanding, with \sin d\theta = d\theta and \cos d\theta = 1 in the limit, and dropping second-order terms: T\cos\theta\,d\theta + dT\sin\theta = w\,dx and -T\sin\theta\,d\theta + dT\cos\theta = 0, that is
The second says the horizontal component is constant, T_0 = T\cos\theta. Substituting T = T_0/\cos\theta into the first, d(T_0\tan\theta) = w\,dx, and with \tan\theta = dy/dx:
the differential equation of the flexible cable. Its solution that also meets the conditions at the fixed ends (the boundary conditions) is the cable's shape.
The equation (physical law: equilibrium of the element, with the flexible cable's tension along its tangent):
Worked number. For a cable with T_0 = 6\,\mathrm{kN} under w = 0.3\,\mathrm{kN/m}, the curvature-like quantity d^2y/dx^2 = 0.05\,\mathrm{m^{-1}}: over 10\,\mathrm m the slope changes by 0.5.
Real example. Engineers stringing an overhead line set its horizontal tension T_0 with a dynamometer at one tower; the whole span's shape then follows from Eq. 5/13 and the line's weight.
Plausible wrong answer. "The tension is the same all along a cable." Only its horizontal component is; the total T = T_0/\cos\theta grows where the cable is steeper.
Check. Where along a hanging cable is the tension least? (At the lowest point, where \theta = 0 and T = T_0.)
Narration spine. A curved cable; a short piece is cut out and enlarged, with tension arrows at its ends and the load w\,dx hanging from its middle. The two arrows' horizontal parts light up equal; their vertical parts differ by exactly w\,dx, and the equation writes itself.
Bridge. Two limiting loadings bracket every real cable. The first: a load uniform along the horizontal.
Nugget 03: A Load Uniform Along the Horizontal Hangs in a Parabola
Claim. With w constant per unit horizontal length (a suspension bridge's deck, cable weight neglected), the cable is a parabola, y = wx^2/(2T_0). The tension is least at the lowest point and largest at the higher support; the length follows from a series in the sag-to-span ratio.
The picture. Meriam's Sample Problem 5/17: 12\,\mathrm{kg} per horizontal metre over a 300\,\mathrm m span with 60\,\mathrm m sag, and the free body of its right half.
Check with numbers. w = 12(9.81) = 117.7\,\mathrm{N/m} = 0.1177\,\mathrm{kN/m}, L = 300\,\mathrm m, h = 60\,\mathrm m. With l_A = L/2 in T_0 = wl_A^2/(2h_A):
The half-cable's free body checks it: T_0 = 22.1\,\mathrm{kN}, load R = 0.1177(150) = 17.66\,\mathrm{kN}, and \sqrt{22.1^2 + 17.66^2} = 28.3\,\mathrm{kN}. Length, h/L = 1/5 < 1/4 so the series converges:
(the exact length of this parabola is 329.5\,\mathrm m; the next term of the series is positive).
The idea. Meriam (5/8). Origin at the lowest point, where the tension is horizontal and equals T_0. Integrating Eq. 5/13 once, dy/dx = wx/T_0 + C, and C = 0 because the slope is zero at x = 0; again,
a vertical parabola. At the support A (x = l_A, y = h_A): T_0 = wl_A^2/(2h_A), so y = h_A(x/l_A)^2. The free body of a portion from the origin to x carries T_0, the load wx at x/2 and T along the cable: T = \sqrt{T_0^2 + w^2x^2} (5/15), = w\sqrt{x^2 + (l_A^2/2h_A)^2} (5/15a), largest at x = l_A: T_{\max} = wl_A\sqrt{1 + (l_A/2h_A)^2}. The length s_A = \int_0^{l_A}\sqrt{1 + (wx/T_0)^2}\,dx can be integrated in closed form, but a binomial series is handier:
convergent for h_A/l_A < 1/2. For level supports (L = 2l_A, h = h_A, S = 2s_A):
convergent for h/L < 1/4; three terms usually suffice since h is much smaller than L/4. Meriam's hint: check T_{\max} from the force polygon of the right half.
The equation (approximation: load uniform per horizontal metre, cable weight neglected):
Worked number. A suspension-bridge cable, L = 1000\,\mathrm m, h = 100\,\mathrm m, w = 100\,\mathrm{kN/m}: T_0 = 100(10^6)/800 = 125\,\mathrm{MN}, T_{\max} = 50\sqrt{1 + 6.25} = 134.6\,\mathrm{MN} at the towers.
Real example (one you can see). Look at a suspension bridge from the side: its main cables, loaded almost uniformly per metre of deck by the hangers, trace a parabola, not a catenary.
Plausible wrong answer. "Halve the sag and the tension doubles at the towers." T_0 doubles exactly (T_0 \propto 1/h); T_{\max} rises a little less, since the vertical part wL/2 stays the same.
Check. Where along the parabolic cable is the tension greatest? (At the higher support, the end farthest horizontally from the lowest point.)
Narration spine. Hangers drop from a level deck to a cable; the cable settles into a parabola. Its right half is cut free: a horizontal arrow at the bottom, a load arrow at the quarter point, a tangent arrow at the tower, and the three snap into a closed triangle.
Bridge. A power line or a hanging chain carries only its own weight, spread uniformly along its length, not the horizontal.
Nugget 04: A Cable Under Its Own Weight Hangs in a Catenary
Claim. A uniform cable of weight \mu per unit length, carrying only itself, hangs in a catenary, y = (T_0/\mu)[\cosh(\mu x/T_0) - 1]. Its tension grows linearly with height above the lowest point: T = T_0 + \mu y.
The picture. Meriam's Fig. 5/29b: a length s of cable from its lowest point.
The forces are T_0, the weight \mu s of that length, and T along the cable at its end; their triangle gives T^2 = \mu^2s^2 + T_0^2 and \tan\theta = \mu s/T_0.
Check with numbers (the puzzle, Meriam's Sample Problem 5/16). \mu = 2.669/30.48 = 0.08757\,\mathrm{N/m}; half the tape, s = 15.24\,\mathrm m, weighs \mu s = 1.334\,\mathrm N. From the force triangle, T_0 = \sqrt{44.48^2 - 1.334^2} = 44.46\,\mathrm N, and from T = T_0 + \mu y at the pegs:
The tape sags 229\,\mathrm{mm}, not a millimetre or two. (Meriam keeps an extra figure, T_0 = 9.995\,\mathrm{lb}, because h comes from the small difference T - T_0.) The chord between the pegs is 4.6\,\mathrm{mm} shorter than the tape: the sag correction surveyors subtract.
The idea. Meriam (5/8). Now the vertical load over dx is the weight of the cable's length ds, \mu\,ds, so Eq. 5/13 becomes
using (ds)^2 = (dx)^2 + (dy)^2. With p = dy/dx: dp/\sqrt{1 + p^2} = (\mu/T_0)\,dx, so \ln(p + \sqrt{1 + p^2}) = \mu x/T_0 (the constant is zero since p = 0 at x = 0), that is dy/dx = \sinh(\mu x/T_0). Integrating with y = 0 at x = 0:
the second from dy/dx = \tan\theta = \mu s/T_0. The force triangle then gives T^2 = T_0^2(1 + \sinh^2) = T_0^2\cosh^2:
The change of tension from the lowest point depends only on the height y.
The equation (approximation: uniform weight along the cable, no other load):
Worked number. Pull the tape with 70\,\mathrm N instead: T_0 = \sqrt{70^2 - 1.334^2} = 69.987\,\mathrm N, h = 0.0127/0.08757 = 0.145\,\mathrm m. Even at 26 times its weight it sags 145\,\mathrm{mm}.
Real example (one you can see). A power line between two pylons hangs in a catenary, sagging more on a hot day as the conductor lengthens; a necklace held by its two ends shows the same curve.
Plausible wrong answer. "Pull hard enough and the cable is straight." h = (T - T_0)/\mu shrinks with tension but never reaches zero: a straight cable would need infinite tension, since its weight would have no vertical support.
Check. At what height above the lowest point does a catenary's tension equal twice T_0? (y = T_0/\mu, from T = T_0 + \mu y.)
Narration spine. The tape between its pegs; a dial for the pull turns up and up, the tape flattens but never straightens, a sag readout creeping down from 229\,\mathrm{mm}. Then a half-tape free body with its three forces closing into a triangle.
Bridge. The catenary's equation for T_0, given a span and a sag, cannot be solved by algebra.
Nugget 05: The Catenary's Tension Is Found Numerically
Claim. Given span and sag, Eq. 5/19 at the support is a transcendental equation for T_0; it is solved graphically or by a root-finder. For small sag-to-span ratios the parabola gives nearly the same answer.
The picture. Harper's catenaries: 12\,\mathrm{kg/m} of cable over 300\,\mathrm m, sag 10, 30 and 60\,\mathrm m.
Check with numbers (Meriam's Sample Problem 5/18). \mu = 0.1177\,\mathrm{kN/m}, x = 150\,\mathrm m, y = 60\,\mathrm m:
Plotting both sides against T_0, or a root-finder, gives T_0 = 23.2\,\mathrm{kN}. Then T_{\max} = 23.2 + 0.1177(60) = 30.2\,\mathrm{kN} (Eq. 5/22) and 2s = 2(23.2/0.1177)\sinh(17.66/23.2) = 330\,\mathrm m (Eq. 5/20).
The idea. Meriam (5/8): most catenary problems involve Eqs. 5/19–5/22, solved
graphically (SP 5/18 plots each side of the equation and reads the intersection) or
by computer, writing it as f(T_0) = \cosh(17.66/T_0) - 7.06/T_0 - 1 = 0 (Appendix C/11 gives a
numerical method). A small sag-to-span ratio means a tight cable, whose weight per
unit length is hardly different from the same weight per unit horizontal length, so
the parabolic relations approximate it; Meriam's hint to SP 5/18 notes they are close
even at 60\,\mathrm m in 300\,\mathrm m. Harper (5.4) solves f = 0 with MATLAB's solve, which falls back on a
numerical root, for three sags:
| Sag h (m) | Catenary T_0 (kN) | T_{\max} (kN) | Length (m) | Parabola T_0 (kN) | T_{\max} (kN) | Length (m) |
|---|---|---|---|---|---|---|
| 10 | 132.6 | 133.8 | 300.9 | 132.4 | 133.6 | 300.9 |
| 30 | 44.7 | 48.2 | 307.9 | 44.1 | 47.5 | 307.8 |
| 60 | 23.2 | 30.2 | 329.9 | 22.1 | 28.3 | 329.5 |
(The parabola columns are ours, from Nugget 03's formulas with the same 0.1177\,\mathrm{kN/m} spread over the horizontal.) The catenary always needs a little more tension: its weight, spread along a cable longer than the span, is more than wL.
The equation (mathematical identity: the condition fixing T_0, solved numerically):
Worked number. For the 10\,\mathrm m sag, the catenary and the parabola differ by 0.2\,\mathrm{kN} in 132.6, 0.15\,\%: at a sag-to-span ratio of 1/30 the parabola is all that is needed.
Real example. Line designers' software iterates exactly this equation for every span of a transmission line, for each temperature and ice load, to check ground clearance at the sag.
Plausible wrong answer. "Halving the sag halves the tension." It roughly doubles it: from 30\,\mathrm m to 10\,\mathrm m sag, T_0 triples, 44.7 to 132.6\,\mathrm{kN}.
Check. In SP 5/18, at what height above the lowest point is the tension 25\,\mathrm{kN}? (y = (25 - 23.16)/0.1177 = 15.6\,\mathrm m.)
Narration spine. Two curves, 7.06/T_0 and \cosh(17.66/T_0) - 1, cross at 23.2\,\mathrm{kN}. Then the 300\,\mathrm m span with three cables at different sags; a tension gauge beside each reads 23, 45 and 133\,\mathrm{kN}, the tight one straining.
Bridge. Real supports are rarely level.
Nugget 06: Supports at Different Heights: Work From the Lowest Point
Claim. When the supports are at different levels, apply the relations to each side of the lowest point separately: the same T_0, different l, h and s on each side. For a parabola, h is proportional to l^2, which locates the lowest point.
The picture. A parabolic cable under 2\,\mathrm{kN} per horizontal metre between supports 100\,\mathrm m apart, A 10\,\mathrm m and B 2.5\,\mathrm m above the lowest point.
Check with numbers. h_A = wl_A^2/2T_0 and h_B = wl_B^2/2T_0 give l_A/l_B = \sqrt{h_A/h_B} = 2, with l_A + l_B = 100\,\mathrm m: l_A = 66.7\,\mathrm m, l_B = 33.3\,\mathrm m. Then T_0 = 2(66.7)^2/(2 \times 10) = 444.4\,\mathrm{kN}, the same from the B side: 2(33.3)^2/(2 \times 2.5) = 444.4\,\mathrm{kN}. Support tensions: T_A = \sqrt{444.4^2 + 133.3^2} = 464.0\,\mathrm{kN}, T_B = \sqrt{444.4^2 + 66.7^2} = 449.4\,\mathrm{kN}.
The idea. Meriam (5/8): the relations for l_B, h_B, s_B follow from those for A by replacing the subscripts (Figs. 5/27, 5/29), and many cable problems, parabolic or catenary, have supports at different levels: the relations are applied to the part of the cable on each side of the lowest point. If the lowest point falls outside the span, the whole cable is on one side of it and the cable slopes the same way throughout.
The equation (mathematical identity: Eq. 5/14 on each side, one T_0):
Worked number. The longer side carries more load (133.3\,\mathrm{kN} against 66.7\,\mathrm{kN}) and its support more tension, 464.0\,\mathrm{kN}.
Real example (one you can see). A ski lift or a cable car climbing a slope: the cable's lowest point between two towers is shifted toward the lower tower, and the upper tower's anchorage carries the larger tension.
Plausible wrong answer. "The lowest point is at mid-span." Only for level supports; here it is two thirds of the way across, toward the lower support.
Check. If h_A = h_B, where is the lowest point? (At mid-span.)
Narration spine. One tower rises while the other stays put; the cable's low point slides toward the shorter tower. The two halves on either side of it highlight in different colours, each a piece of the same parabola with the same T_0 at the bottom.
Bridge. Cables are loaded per metre of length. A fluid loads a surface per square metre, and its intensity grows with depth.
Synthesis
The surveyor's tape sags 229\,\mathrm{mm} under a pull of seventeen times its weight, and it would still sag at any finite pull: a cable can carry its own weight only by hanging in a curve, because its tension acts along it and something must hold the weight up. That one fact, tension along the cable, made each element's horizontal tension the same and gave d^2y/dx^2 = w/T_0. A load uniform along the horizontal makes the curve a parabola; a load uniform along the cable makes it a catenary, whose T_0 is found numerically; for a tight cable the two agree. A flexible cable carries vertical load by its shape: constant horizontal tension, d^2y/dx^2 = w/T_0, a parabola or a catenary, and tension rising with height as T = T_0 + \mu y.
- Central idea: a cable has no bending stiffness; its equilibrium shape carries the load in pure tension.
- Key relationship: d^2y/dx^2 = w/T_0; parabola T_0 = wL^2/8h; catenary T = T_0 + \mu y.
- Most important assumption: a perfectly flexible, inextensible cable; for the two limits, load uniform per horizontal metre or per metre of cable.
- Open question: a fluid presses on a dam or a tank wall with an intensity that grows with depth. Where does its resultant act? (S23.)
Computed views
The same calculations, written twice so you can work in whichever language you prefer. The MATLAB follows Harper's programs; the Python is a translation, and both print the same numbers.
Python (numpy, scipy). S22_cables.py does:
- the cable with two point loads;
- the parabolic cable, with its length both from the series and by integrating the arc;
- the catenary equation, solved for T_0 with
brentqat Harper's three sags and set beside the parabola; - the puzzle's tape;
- the cable between supports at different heights.
Part 1: two 2 kN loads, supports 9 m apart
sag 1.0 m: T0 = 6.00 kN, end tension 6.32 kN
sag 1.5 m: T0 = 4.00 kN, end tension 4.47 kN
Part 2: parabolic cable, w = 0.1177 kN/m, L = 300 m, h = 60 m
T0 = 22.1 kN, T_max = 28.3 kN, length 328.9 m (series), 329.5 m (arc integral)
bridge L = 1000 m, h = 100 m, 100 kN/m: T0 = 125 MN, T_max = 134.6 MN
Part 3: catenaries, Harper's 12 kg/m cable over 300 m
sag (m) T0 (kN) parabola T_max (kN) length (m)
10 132.63 132.44 133.81 300.9
30 44.72 44.15 48.25 307.9
60 23.16 22.07 30.23 329.9
tape pulled with 44.48 N: T0 = 44.460 N, sag 228.7 mm
tape pulled with 70.0 N: T0 = 69.987 N, sag 145.3 mm
Part 4: supports at different heights, w = 2 kN/m, 100 m, sags 10 and 2.5 m
l_A = 66.7 m, T0 = 444.4 kN, T_A = 464.0 kN, T_B = 449.4 kN
The three-term series stops at 328.9\,\mathrm m; the arc integral gives the parabola's full 329.5\,\mathrm m. At h/L = 1/5 the dropped terms still add 0.6\,\mathrm m.
MATLAB (Octave-compatible). S22_cables.m uses integral and
fzero where Harper uses solve.
parabola: T0 = 22.1 kN, T_max = 28.3 kN, length 329.5 m
sag (m) T0 (kN) parabola T_max (kN) length (m)
10 132.63 132.44 133.81 300.9
30 44.72 44.15 48.25 307.9
60 23.16 22.07 30.23 329.9
tape: T0 = 44.460 N, sag 228.7 mm
Traps
- Letting a cable push. It can only pull; a negative tension means slack. (From Nugget 01.)
- Taking the tension as constant along the cable. Only its horizontal component is. (From Nugget 02.)
- Using the parabola for a cable loaded only by its own weight with a large sag. That is a catenary; the parabola underestimates T_0. (From Nugget 03.)
- Using the length series beyond its range. Eq. 5/16a needs h/L < 1/4. (From Nugget 03.)
- Rounding T_0 before taking T - T_0. The sag comes from a small difference; keep an extra figure. (From Nugget 04.)
- Expecting a cable to pull straight. The sag shrinks with tension but never vanishes. (From Nugget 04.)
- Looking for an algebraic T_0 in a catenary. It needs a graph or a root-finder. (From Nugget 05.)
- Putting the lowest point at mid-span for unequal supports. Use h_A/h_B = (l_A/l_B)^2. (From Nugget 06.)
Self-check
- Nugget 01's cable (supports 9\,\mathrm m apart, 2\,\mathrm{kN} at 3\,\mathrm m and 6\,\mathrm m) is let down so the loads hang 1.5\,\mathrm m below the supports. Find T_0 and the end tension.
- Why is the horizontal component of a cable's tension constant along it when all loads are vertical?
- Meriam's Sample Problem 5/17 with a 30\,\mathrm m sag: find T_0, T_{\max} and the length.
- A suspension-bridge cable: span 1000\,\mathrm m, sag 100\,\mathrm m, 100\,\mathrm{kN/m} along the horizontal. Find T_0, T_{\max} and the cable length.
- The surveyor's tape (puzzle) pulled with 70\,\mathrm N: find the sag.
- Compare Harper's catenary at h = 30\,\mathrm m with the parabola for the same span and sag.
- In SP 5/18, at what height above the lowest point is the tension 25\,\mathrm{kN}?
- A 10\,\mathrm m washing line of 0.1\,\mathrm{kg/m}, between level hooks. Estimate the pull needed for a sag of 0.1\,\mathrm m, and of 0.01\,\mathrm m.
- Nugget 06's cable: find the support tensions, and say which support carries more.
- A cable has h/L = 0.3. May Eq. 5/16a be used for its length?
- Why is a cable straight between two concentrated loads?
- For the same span and sag, why does the catenary need more tension than the parabola in Harper's table?
Fully worked solutions
1. Each support carries 2\,\mathrm{kN} vertically. The end segment drops 1.5\,\mathrm m in 3\,\mathrm m: T_0 = 2(3/1.5) = 4\,\mathrm{kN}; end tension \sqrt{4^2 + 2^2} = 4.47\,\mathrm{kN}.
2. Horizontal equilibrium of any element: the only horizontal forces are the horizontal components of the tension at its two ends, so they are equal: d(T\cos\theta) = 0.
3. T_0 = wL^2/8h = 0.1177(300)^2/240 = 44.1\,\mathrm{kN}; T_{\max} = 17.66\sqrt{1 + (300/120)^2} = 17.66(2.693) = 47.5\,\mathrm{kN}; S = 300[1 + \frac83(0.1)^2 - \frac{32}{5}(0.1)^4] = 300(1.02603) = 307.8\,\mathrm m.
4. T_0 = 100(1000)^2/800 = 125\times10^{3}\,\mathrm{kN} = 125\,\mathrm{MN}; T_{\max} = 50\,\mathrm{MN}\sqrt{1 + 2.5^2} = 134.6\,\mathrm{MN}; S = 1000[1 + \frac83(0.01) - \frac{32}{5}(10^{-4})] = 1026\,\mathrm m.
5. T_0 = \sqrt{70^2 - 1.334^2} = 69.987\,\mathrm N; h = (70 - 69.987)/0.08757 = 0.145\,\mathrm m.
6. Catenary: T_0 = 44.7\,\mathrm{kN}, T_{\max} = 48.2\,\mathrm{kN}, length 307.9\,\mathrm m. Parabola: 44.1\,\mathrm{kN}, 47.5\,\mathrm{kN}, 307.8\,\mathrm m. The tensions differ by about 1.5\,\%, the lengths by 0.1\,\mathrm m.
7. T = T_0 + \mu y: y = (25 - 23.16)/0.1177 = 15.6\,\mathrm m.
8. w = 0.1(9.81) = 0.981\,\mathrm{N/m}; with the parabola, T_0 = wL^2/8h: 0.981(100)/0.8 = 122.6\,\mathrm N for 0.1\,\mathrm m sag, and 1226\,\mathrm N for 0.01\,\mathrm m. (The catenary also gives 122.6\,\mathrm N for the first: the sag is tiny against the span.) Taking out the last centimetre of sag costs ten times the pull.
9. T_A = \sqrt{T_0^2 + (wl_A)^2} = \sqrt{444.4^2 + 133.3^2} = 464.0\,\mathrm{kN}; T_B = \sqrt{444.4^2 + 66.7^2} = 449.4\,\mathrm{kN}. The higher support, A, carries more.
10. No: the series converges only for h/L < 1/4. Integrate \sqrt{1 + (wx/T_0)^2} directly instead.
11. Between the loads there is no load on the cable, so w = 0, d^2y/dx^2 = 0, and y is linear: each segment is a two-force member pulling along its own line.
12. For the same span and sag the catenary's load is its own weight along a length longer than the span, spread more toward the supports, while the parabola spreads wL uniformly; the catenary's larger and more outward load needs a larger horizontal tension to hold the same sag.