--- title: "Suspension 100 m: SF and BM at 30 m, tension" subject: "Theory of Structures I" chapter: 7 key: "semester_4/theory_of_structures_i:suspension-100m-200-300kn-30m" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/suspension-100m-200-300kn-30m site: BCT Engineering --- # Suspension 100 m: SF and BM at 30 m, tension Theory of Structures I · Chapter 7: Suspension Cable Systems - Asked in: 2079 Chaitra, 10 marks - Video: [Cables - Problem No 13 ( Three Hinged Stiffening Girder)](https://www.youtube.com/watch?v=cf-8qzK6pMo) — 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/suspension-100m-200-300kn-30m - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7 ## Question A three-hinged stiffening girder of a suspension bridge of span 100 m is subjected to two point loads of 200 kN and 300 kN at distances of 25 m and 50 m from the left end. Find the shear force and bending moment for the girder at a distance of 30 m from the left end. The supporting cable has a central dip of 10 m. Find also the maximum tension and its slope in the cable. ## Short answer At 30 m from left: shear force = -60 kN, bending moment = -400 kN m (hogging). Maximum cable tension is 1077.0 kN at supports, inclined at 21.80°. ## Answer **Data:** three-hinged stiffening girder (hinges at both ends and at mid-span), span $L = 100$ m, loads 200 kN at 25 m and 300 kN at 50 m from the left end; cable dip $d = 10$ m. The cable is parabolic: $y = \dfrac{4d\,x(L-x)}{L^2} = 0.004\,x\,(100-x)$, slope $\tan\alpha = \dfrac{4d\,(L-2x)}{L^2}$. #### Horizontal pull in the cable Reaction of the simple beam (taking moments about the right end): $$ R_A = \frac{200\times 75 + 300\times 50}{100} = 300.0\ \text{kN} $$ Moment at the mid-span hinge $C$ of the simple beam: $$ M_C = 300\times 50 - 200\times 25 = 10000\ \text{kN m} $$ The girder has a hinge at $C$, so the cable thrust makes the net moment there zero: $$ H\,d = M_C \Rightarrow H = \frac{10000}{10} = 1000\ \text{kN} $$ #### Shear force and bending moment at 30 m from the left At $x = 30$ m: $y = 0.004\times 30\times 70 = 8.40$ m; $\tan\alpha = 0.004\,(100-60) = 0.16$. $$ M_b = 300\times 30 - 200\times 5 = 8000\ \text{kN m}, \qquad V_b = 300 - 200 = 100\ \text{kN} $$ $$ \begin{aligned} M &= M_b - H\,y = 8000 - 1000\times 8.40 = -400\ \text{kN m}\\ V &= V_b - H\tan\alpha = 100 - 1000\times 0.16 = -60\ \text{kN} \end{aligned} $$ #### Maximum tension in the cable and its slope The tension is greatest at the supports, where the slope is $\tan\alpha_s = \dfrac{4d}{L} = 0.4$: $$ \alpha_s = \tan^{-1}0.4 = 21.80^\circ, \qquad V_s = H\tan\alpha_s = 400\ \text{kN} $$ $$ T_{max} = \sqrt{H^2 + V_s^2} = \sqrt{1000^2 + 400^2} = 1077.0\ \text{kN} $$ **Answer: at 30 m from the left, $V = -60$ kN and $M = -400$ kN m (hogging); maximum cable tension $= 1077.0$ kN at the supports, inclined at 21.80° to the horizontal.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the cable thrust H found? A. H = total load / 2 B. H = R_A d / L C. H = 500 x 100 / (8 x 10) D. H d = M_C of the simple beam = 10000, so H = 1000 kN Answer: D. The hinge at C makes the net girder moment zero there, so H d equals the simple-beam moment at C: 300 x 50 - 200 x 25 = 10000. (Key point: Thrust from the mid-span hinge) 2. What is the bending moment in the girder at 30 m? A. M = M_b - H y = 8000 - 1000 x 8.4 = -400 kN m B. M = 8000 kN m C. M = 8400 kN m D. M = H y - M_b = +400 kN m Answer: A. The cable thrust moment H y (8400) exceeds the simple-beam moment (8000), giving a hogging moment of -400 kN m. (Key point: Bending moment at 30 m) 3. How is shear at 30 m found and what is it? A. V = V_b + H tan(alpha) = 260 kN B. V = V_b - H tan(alpha) = 100 - 1000 x 0.16 = -60 kN C. V = V_b = 100 kN D. V = H tan(alpha) = 160 kN Answer: B. The cable's vertical component H tan(alpha) at the section reduces the simple-beam shear; tan(alpha) = 0.004(100 - 60) = 0.16. (Key point: Shear in the girder) 4. Where is the cable tension maximum and what are its value and slope? A. At mid-span, 1000 kN, horizontal B. At the supports, 1077 kN at 21.80 deg to the horizontal (tan = 4d/L = 0.4) C. At the supports, 1400 kN at 45 deg D. At 30 m, 1077 kN at 9.1 deg Answer: B. At the supports V_s = H x 0.4 = 400 kN, so T = sqrt(1000^2 + 400^2) = 1077 kN with slope tan^-1 0.4 = 21.80 deg. (Key point: Maximum tension and slope) ## Related in Theory of Structures I - [Suspension 150 m: SF, BM at 30 m, tension, length](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-150m-dip15-180-150-dead-7-5) — Theory of Structures I, chapter 7, asked 2× - [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× - [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 120 m: SF, BM at 25 m, cable area](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-120m-road6m-5knm2-200-150) — Theory of Structures I, chapter 7 - [Suspension 120 m: BMD and BM at 25, 40, 80 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-120m-240kn-300kn-bmd) — Theory of Structures I, chapter 7 ## Related in other subjects - [Two-hinged parabolic arch: BM at D](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-point-d-30-50) — Theory of Structures II, chapter 2 - [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 - [Truss member forces with 25, 50, 10 kN loads](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-forces-25kn-50kn-10kn-hinge-roller) — Applied Mechanics, chapter 7 - [Two-hinged arch: BM at 10 m intervals](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-80kn-d-i-ic) — Theory of Structures II, chapter 2