--- title: "Suspension 150 m: BM and SF at 30 m, tension, length" subject: "Theory of Structures I" chapter: 7 key: "semester_4/theory_of_structures_i:suspension-150m-dip15-180-120-dead-5" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/suspension-150m-dip15-180-120-dead-5 site: BCT Engineering --- # Suspension 150 m: BM and SF at 30 m, tension, length Theory of Structures I · Chapter 7: Suspension Cable Systems - Asked in: 2072 Magh, 12 marks - Video: [Cables - Problem No 15 ( Three Hinged Stiffening Girder)](https://www.youtube.com/watch?v=sXjJR4-9lLo) — 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-150m-dip15-180-120-dead-5 - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7 ## Question A suspension cable having central dip 15 m supports a three hinged stiffening girder 150 m long which supports point loads 180 kN at 50 m from left support and 120 kN at 30 m from the right support. The dead load of the girder is 5 kN/m. Determine bending moment and shear force at a section 30 m from the left support. Also determine the maximum tension in the cable and length of cable. ## Short answer Using H = Hd + Hp = 1357.5 kN, at 30 m from left support M = 288.0 kN m sagging and S = 43.2 kN. Maximum cable tension is 1462.07 kN at the supports. Cable length is 153.90 m. ## Answer Single cable and three-hinged stiffening girder: span $L$ = 150 m, dip $d$ = 15 m, dead load $w$ = 5 kN/m, loads 180 kN at 50 m from A and 120 kN at 30 m from B (that is, 120 m from A). #### Cable pull $H$ $$ \begin{aligned} R_A &= \frac{180(100)+120(30)}{150} = 144.0\ \text{kN} \\ M_{0C} &= 144.0(75) - 180(25) = 6300.0\ \text{kN m} \\ H_p &= \frac{M_{0C}}{d} = 420.0\ \text{kN}, \qquad H_d = \frac{wL^2}{8d} = \frac{5(150)^2}{8(15)} = 937.5\ \text{kN} \\ H &= H_d + H_p = 1357.5\ \text{kN} \end{aligned} $$ #### BM and SF at 30 m from the left support The dead load is carried entirely by the cable (parabolic), so it gives no moment or shear in the girder; only the point loads act on it. $$ y_{30} = \frac{4(15)(30)(120)}{150^2} = 9.60\ \text{m},\qquad \frac{dy}{dx} = \frac{4d(L-2x)}{L^2} = \frac{4(15)(90)}{150^2} = 0.24 $$ $$ \begin{aligned} M_{30} &= R_A(30) - H_p\,y_{30} = 4320.0 - 420(9.60) = 288.0\ \text{kN m} \\ S_{30} &= R_A - H_p\,\frac{dy}{dx} = 144.0 - 420(0.24) = 43.2\ \text{kN} \end{aligned} $$ #### Maximum tension in the cable At the supports $\tan\theta = 4d/L = 0.40$, so $\sec\theta = 1.0770$: $$ T_{max} = H\sec\theta = 1357.5(1.0770) = 1462.07\ \text{kN} $$ #### 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 $d/L = 0.1$: $$ l = 150\left[1+\frac{8}{3}(0.1)^2-\frac{32}{5}(0.1)^4\right] = 153.90\ \text{m} $$ **Answer:** $M_{30} = 288.0$ kN m (sagging), $S_{30} = 43.2$ kN; $T_{max} = 1462.07$ kN; cable length = 153.90 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Loads are 180 kN at 50 m and 120 kN at 120 m from A. What is RA? A. 144 kN B. 156 kN C. 150 kN D. 300 kN Answer: A. RA = (180 x 100 + 120 x 30)/150 = 144 kN. (Key point: Reaction and hinge moment) 2. How is the total H made up? A. Hp = 420 kN from the point loads only B. Hd = 937.5 kN from dead load plus Hp = 420 kN, total 1357.5 kN C. Hd = 937.5 kN only D. H = 5 x 150 = 750 kN Answer: B. Hd = wL^2/8d = 937.5 kN, Hp = M0C/d = 6300/15 = 420 kN; they add. (Key point: Horizontal pull) 3. What is the girder BM at 30 m and why does dead load not appear? A. 288 kN m; the dead load is carried entirely by the cable B. 4320 kN m; the dead load is neglected for safety C. 288 kN m; the dead load is balanced by RA D. -288 kN m; the dead load acts upwards Answer: A. M = 144 x 30 - 420 x 9.6 = 288 kN m. A UDL on the girder is taken by the parabolic cable so only point loads act on the girder. (Key point: BM at 30 m) 4. How is Tmax found? A. Tmax = H = 1357.5 kN B. Tmax = H sec(theta) at the supports, with tan(theta) = 4d/L = 0.4, giving 1462.07 kN C. Tmax = H tan(theta) D. Tmax = H/ sec(theta) Answer: B. The tension is greatest where the slope is greatest, at the supports: 1357.5 x 1.0770 = 1462.07 kN. (Key point: Maximum cable tension) ## Close variants set in other exams - [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× ## Related in Theory of Structures I - [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 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: 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 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× - [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 150 m: UDL right half, SF and BM at 50 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-150m-two-cables-udl-right-half) — Theory of Structures I, chapter 7