--- title: "Cable 20 m: three loads, length and area" subject: "Theory of Structures I" chapter: 7 key: "semester_4/theory_of_structures_i:cable-20m-three-loads-0-96m-dip" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cable-20m-three-loads-0-96m-dip site: BCT Engineering --- # Cable 20 m: three loads, length and area Theory of Structures I · Chapter 7: Suspension Cable Systems - Asked in: 2069 Poush, 10 marks - Video: [Cables - Problem No 5 ( With 3 concentrated loads )](https://www.youtube.com/watch?v=A8MyCtW0fNo) — Stan Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cable-20m-three-loads-0-96m-dip - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7 ## Question A cable is supported at two points 20 m apart at the same level. It is used to support three equidistant loads, first load is 40 kN, second is 30 kN and third is 20 kN. The central dip of the cable is 0.96 m. Find the length of the cable required and its sectional area if the safe tensile stress is 250 kN/mm². Also give the geometry (shape and dip) of the cable when it is hanging only with its weight (without the given loads). ## Short answer Cable length = 20.12 m, required cross-sectional area = 1266 mm². Under its own weight only, the cable forms a catenary (nearly parabolic) with a dip of approximately 0.94 m. ## Answer Loads 40, 30 and 20 kN are taken at 5 m, 10 m and 15 m from the left support (equal spacing on the 20 m span). The central dip (at the 30 kN load) is 0.96 m. #### Reactions and horizontal pull $$ \begin{aligned} V_A &= \frac{40(15)+30(10)+20(5)}{20} = 50.0\ \text{kN},\quad V_B = 90 - 50.0 = 40.0\ \text{kN} \\ M_{10} &= 50.0(10) - 40(5) = 300.0\ \text{kN m} \\ H &= \frac{M_{10}}{0.96} = 312.50\ \text{kN} \end{aligned} $$ #### Cable profile and segment lengths Depth of a load point below the supports is $y = M/H$ where $M$ is the beam moment there: $y_1 = 0.800$ m (5 m), $y_2$ = 0.960 m (10 m), $y_3 = 0.640$ m (15 m). | Segment | Horizontal (m) | Rise (m) | Length (m) | V (kN) | T (kN) | |---|---|---|---|---|---| | A-1 | 5 | 0.800 | 5.0636 | 50.0 | 316.47 | | 1-2 | 5 | 0.160 | 5.0026 | 10.0 | 312.66 | | 2-3 | 5 | 0.320 | 5.0102 | -20.0 | 313.14 | | 3-B | 5 | 0.640 | 5.0408 | -40.0 | 315.05 | Length of cable required = sum of segments = 20.117 m. #### Cross-sectional area The maximum tension occurs in the end segment A-1: $T_{max} = 316.47$ kN. The permissible stress is read as 250 N/mm² (250 kN/mm² is not physically possible for a cable): $$ A = \frac{T_{max}}{\sigma} = \frac{316.47\times 10^3}{250} = 1265.9\ \text{mm}^2 $$ #### Shape when hanging under its own weight only With no point loads, the cable carries only its own weight, uniformly distributed along its length, and hangs in a **catenary**, $y = \dfrac{H}{w}\left(\cosh\dfrac{wx}{H}-1\right)$. For a flat cable (dip/span below about 1/10) this is very close to a parabola. Its length is unchanged at 20.117 m, so the dip follows from $l \approx L\left[1+\frac{8}{3}\left(\frac{d}{L}\right)^2\right]$: $$ d = \sqrt{\frac{3L(l-L)}{8}} = \sqrt{\frac{3(20)(20.117-20)}{8}} = 0.94\ \text{m} $$ The cable hangs symmetrically with its lowest point at mid-span and a dip of about 0.94 m. The point loads make it a polygon whose dip (0.96 m at the 30 kN load) is slightly larger. **Answer:** length = 20.12 m; area = 1266 mm²; unloaded cable: catenary (nearly parabolic) with dip about 0.94 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the horizontal pull H found for the cable with three point loads? A. Take moments of the cable about the support to get V_A, then H = V_A / dip B. Take the beam bending moment at the 30 kN load (300 kN m) and divide by the dip: H = 300/0.96 C. H = total load / 2 = 45 kN D. H = w L^2 / (8 d) using the total load as a UDL Answer: B. At any point the cable moment equals the beam moment minus H times the dip, which is zero for a cable. So H = M/y = 300/0.96 = 312.5 kN. (Key point: Finding H from the central dip) 2. Where does the maximum tension occur, and what is its value? A. In the central segment 1-2, 312.66 kN B. In the segment 2-3, 313.14 kN C. In the end segment A-1, which has the steepest slope and the largest vertical component (V_A = 50 kN), 316.47 kN D. At the 30 kN load, where the dip is greatest Answer: C. H is constant, so T is largest where the vertical component is largest. This is the end segment next to support A with V = 50 kN, giving T = sqrt(312.5^2 + 50^2) = 316.47 kN. (Key point: Segment of maximum tension) 3. How is the required area found, and what is the result? A. A = T_max / permissible stress = 316.47 x 10^3 / 250 = about 1266 mm^2 B. A = H / stress = 312.5 x 10^3 / 250 = 1250 mm^2 using the horizontal pull only C. A = total load / stress = 90 x 10^3 / 250 = 360 mm^2 D. A = T_max x stress = 79 000 mm^2 Answer: A. The cable must be safe at its most heavily stressed section, so the area uses T_max. The stress is read as 250 N/mm^2, giving A = 1265.9 mm^2. (Key point: Cross-sectional area) 4. What shape does the cable take when hanging only under its own weight? A. A parabola exactly, since the weight is uniform per horizontal metre B. A catenary, since the weight is uniform along the cable length; it is nearly parabolic for a flat cable C. A polygon, as with the point loads D. A straight line between the supports Answer: B. Self-weight is uniform along the length of the cable, which gives a catenary. For small dip/span the catenary is very close to a parabola, so the approximate length formula applies, giving a dip of about 0.94 m. 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