--- title: "Suspension 100 m: max/min tension, SFD, BMD" subject: "Theory of Structures I" chapter: 7 key: "semester_4/theory_of_structures_i:suspension-100m-dip10-3x200kn-dead-40" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/suspension-100m-dip10-3x200kn-dead-40 site: BCT Engineering --- # Suspension 100 m: max/min tension, SFD, BMD Theory of Structures I · Chapter 7: Suspension Cable Systems - Asked in: 2072 Asoj, 10 marks - 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/suspension-100m-dip10-3x200kn-dead-40 - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7 ## Question A suspension bridge, 100 m span, has two three-hinged stiffening girders supported by two cables with a central dip of 10 m. The dead load is a uniformly distributed load of 40 kN/m for the entire span and in addition, it supports three point loads of 200 kN each placed along the center line of the roadway, dividing the span in four equal parts. Calculate the maximum tension and minimum tension with their locations in the cable and the length of the cable, also draw shear force and bending moment diagrams for the girders. ## Short answer Per girder, H = 3000 kN. Maximum cable tension is 3231.10 kN at the supports; minimum is 3000.0 kN at mid-span. Cable length is 102.60 m. Girder maximum hogging moment is -312.5 kN m at 12.5, 37.5, 62.5 and 87.5 m; maximum shear is 50 kN. ## Answer **Assumptions.** The loads given are for the whole bridge, so each of the two girders and cables carries half: $w$ = 40/2 = 20 kN/m and $P$ = 200/2 = 100 kN (if the data are taken for a single cable, double all forces below; the method does not change). Span $L$ = 100 m, dip $d$ = 10 m. #### Cable pull $H$ A three-hinged girder gives the cable a uniform load, so the cable stays parabolic. - Dead load part: $H_d = \dfrac{wL^2}{8d} = \dfrac{20(100)^2}{8(10)} = 2500.0$ kN - Point loads: $R_A = 1.5P = 150.0$ kN; $M_{0C} = 150.0(50) - 100(25) = 5000.0$ kN m, so $H_p = \dfrac{M_{0C}}{d} = 500.0$ kN $$ H = H_d + H_p = 2500.0 + 500.0 = 3000.0\ \text{kN} $$ #### Maximum and minimum tension End slope: $\tan\theta = 4d/L = 0.40$, $\sec\theta = 1.0770$. - **Maximum tension** at the supports A and B: $T_{max} = H\sec\theta = 3000.0(1.0770) = 3231.10$ kN - **Minimum tension** at mid-span (crown, where the cable is horizontal): $T_{min} = H = 3000.0$ kN #### Length of the cable Using $l = L\left[1+\dfrac{8}{3}\left(\dfrac{d}{L}\right)^2-\dfrac{32}{5}\left(\dfrac{d}{L}\right)^4\right]$ with $d/L = 0.1$: $$ l = 100\left[1 + \frac{8}{3}(0.1)^2 - \frac{32}{5}(0.1)^4\right] = 102.60\ \text{m} $$ #### Girder shear force and bending moment (per girder) The dead load is carried wholly by the cable, so it produces no SF or BM in the girder. For the three loads, with $y = 0.004x(100-x)$ and $H_p$ = 500 kN: $$ M = M_0 - H_p\,y,\qquad S = S_0 - H_p\,\frac{dy}{dx} $$ - $0 \le x \le 25$: $M = 2x^2 - 50x$, $S = 4x - 50$ - $25 \le x \le 50$: $M = 2x^2 - 150x + 2500$, $S = 4x - 150$ - The right half is the mirror image (SF with opposite sign). So $M = 0$ at the load points (0, 25, 50, 75, 100 m), and $M = -312.5$ kN m at 12.5 m, 37.5 m, 62.5 m and 87.5 m (the greatest hogging value). Shear force is $-50$ to $+50$ kN in each quarter, changing sign at the mid-point of each panel (a saw-tooth). ```text SFD (kN), +-50 in each quarter span * * * * ** ** ** ** * * * * ** ** ** ** ------*-----------*-----------*-----------*------ ** ** ** ** * * * * ** ** ** ** * (top = 50.0, bottom = -50.0; '-' is zero line) ``` ```text BMD (kN m), all hogging, zero at the loads *-----------*-----------*-----------*-----------* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *** *** *** *** (top = 0.0, bottom = -312.5; '-' is zero line) ``` **Answer:** $T_{max} = 3231.10$ kN at the supports; $T_{min} = 3000.0$ kN at the crown (mid-span); cable length = 102.60 m; girder $M_{max} = -312.5$ kN m (hogging) at 12.5, 37.5, 62.5 and 87.5 m; $S_{max} = 50$ kN. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The dead load is 40 kN/m and three loads of 200 kN are for the whole bridge with two cables. What does each cable-girder carry? A. 40 kN/m and 200 kN B. 20 kN/m and 100 kN C. 80 kN/m and 400 kN D. 10 kN/m and 50 kN Answer: B. The loads are shared equally between the two systems. (Key point: Load per cable) 2. What is H? A. 2500 kN B. 500 kN C. 3000 kN D. 3500 kN Answer: C. Hd = 20 x 100^2/80 = 2500 kN and Hp = 5000/10 = 500 kN, so H = 3000 kN. (Key point: Cable pull) 3. Where do Tmax and Tmin occur? A. Tmax at the crown, Tmin at the supports B. Tmax at the supports (3231.10 kN), Tmin at the crown (3000 kN) C. Both at the supports D. Tmax at the quarter points, Tmin at the crown Answer: B. T = H at the crown where the cable is horizontal. At the supports T = H sec(theta) with tan = 4d/L = 0.4, giving 3231.10 kN. (Key point: Maximum and minimum tension) 4. Why is the girder BM zero at the load points and at most -312.5 kN m between them? A. The dead load gives a girder BM of 312.5 B. With H_p y equal to M0 under the loads, M = M0 - Hp y is zero there, and hogging between them C. The hinge moment is -312.5 D. The shear is zero at the loads Answer: B. The dead load gives no girder moment. The point-load pull Hp = 500 kN makes the net moment vanish at the load points, with hogging in between, e.g. M = 2x^2 - 50x gives -312.5 at 12.5 m. (Key point: Girder BM pattern) ## Close variants set in other exams - [Suspension 200 m: max/min tension, cable length](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-200m-dip20-3x800kn) — Theory of Structures I, chapter 7, asked 3× - [Suspension 400 m: max/min tension, cable length](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-400m-dip30-3x300kn) — Theory of Structures I, chapter 7, asked 3× ## Related in Theory of Structures I - [Suspension 50 m: max tension, BM at 10 m, BMD/SFD](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-50m-5m-dip-80kn) — Theory of Structures I, chapter 7 - [Suspension 100 m: UDL on right half, SF and BM at 20 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-100m-5knm2-10knm2-right-half) — Theory of Structures I, chapter 7, asked 2× - [Suspension 100 m: SF and BM at 30 m, tension](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-100m-200-300kn-30m) — Theory of Structures I, chapter 7 - [Suspension 150 m: four moving loads, BM at 40 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-150m-four-moving-loads-200-150-300-100) — Theory of Structures I, chapter 7 - [Suspension 120 m: UDL on left half, SF and BM at 30 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-120m-live-load-left-half) — Theory of Structures I, chapter 7, asked 2× - [Suspension 150 m: BM and SF at 30 m, tension, length](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-150m-dip15-180-120-dead-5) — Theory of Structures I, chapter 7, asked 2× ## Related in other subjects - [Frame SFD, BMD (100 kN, 20 kN/m, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-sfd-bmd-100kn-hinge-60deg) — Strength of Materials, chapter 1 - [Suspension bridge triangle ABC; span AB](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Asuspension-bridge-triangle-abc-span-ab) — Surveying I, chapter 7 - [Castigliano: central deflection of simply supported beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-central-deflection-ss-beam-udl) — Theory of Structures II, chapter 2 - [Frame AFD, SFD, BMD (100 kN, inclined member)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-100kn-20knm) — Strength of Materials, chapter 1