--- title: "Suspension 120 m: girder with different support levels" subject: "Theory of Structures I" chapter: 7 key: "semester_4/theory_of_structures_i:suspension-120m-hinged-ends-middle-6m-diff" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/suspension-120m-hinged-ends-middle-6m-diff site: BCT Engineering --- # Suspension 120 m: girder with different support levels Theory of Structures I · Chapter 7: Suspension Cable Systems - Asked in: 2074 Bhadra, 12 marks - Video: [Cables - Problem No 8 ( With 4 concentrated loads and supports in different level )](https://www.youtube.com/watch?v=N7oFTlEZDbA) — 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-120m-hinged-ends-middle-6m-diff - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7 ## Question The stiffening girder of suspension bridge of span 120 m has hinged at the end and in the middle span, the cable is suspended between two points separated horizontally by 120 m and vertically by 6 m. The maximum dip of the cable is 12 m from upper end point. Two point loads 200 kN and 100 kN are concentrated at 30 m and 55 m from higher end. Calculate and draw shear force diagram and bending moment diagram for girder. ## Short answer Cable horizontal pull H = 657.7 kN from equilibrium at the central hinge. Bending moment M = M0 – H·y gave maximum sagging moment 1812.5 kN·m at 30 m under the 200 kN load, maximum hogging –1437.5 kN·m at 90 m, and zero at the hinge. ## Answer **Assumptions.** The girder is horizontal and the hangers are vertical and closely spaced, so the cable is parabolic. The "maximum dip of 12 m from the upper end" is taken as the depth of the lowest point of the cable below the higher end $A$; the lower end $B$ is then $12 - 6 = 6$ m above the lowest point. The self-weight of the girder is not given, so only the two point loads act (a uniform dead load would be carried wholly by the cable and would not change the girder forces). #### Dip of the cable below the chord at mid-span With heights $h_1 = 12$ m and $h_2 = 6$ m above the lowest point, $H = \dfrac{wL^2}{2(\sqrt{h_1}+\sqrt{h_2})^2}$, and the dip below the chord $AB$ at mid-span is $$ d = \frac{wL^2}{8H} = \frac{\left(\sqrt{h_1}+\sqrt{h_2}\right)^2}{4} = \frac{(3.464+2.449)^2}{4} = 8.743\ \text{m} $$ #### Cable pull $H$ Simple-beam reaction and moment at the centre (span 120 m, loads measured from the higher end): $$ \begin{aligned} R_A &= \frac{200(90)+100(65)}{120} = 204.17\ \text{kN} \\ M_{0C} &= 204.17(60) - 200(30) - 100(5) = 5750.0\ \text{kN m} \end{aligned} $$ Since the girder has a hinge at C ($M_C = 0$): $M_{0C} - H\,d = 0$ $$ H = \frac{M_{0C}}{d} = \frac{5750.0}{8.743} = 657.7\ \text{kN} $$ (If the 12 m is instead read as the cable depth below $A$ at mid-span, then $d = 12 - 3 = 9$ m and $H = 638.9$ kN. The girder SF and BM below are the same in either case, because $H\,y = M_{0C}\,\dfrac{4x(L-x)}{L^2}$ does not depend on $d$.) #### Girder moments and shears The parabolic cable below the chord is $y = \dfrac{4d\,x(L-x)}{L^2}$. For the girder: $$ M(x) = M_0(x) - H\,y(x), \qquad S(x) = S_0(x) - H\,\frac{dy}{dx}, \qquad \frac{dy}{dx} = \frac{4d(L-2x)}{L^2} $$ so that $H\,y = 1.5972\,x(120-x)$ and $H\,\dfrac{dy}{dx} = 1.5972\,(120-2x)$. | x from A (m) | $M_0$ (kN m) | y (m) | $Hy$ (kN m) | $M$ (kN m) | |---|---|---|---|---| | 0 | 0 | 0 | 0 | 0 | | 30 | 6125.0 | 6.557 | 4312.5 | 1812.5 | | 55 | 6229.2 | 8.682 | 5710.1 | 519.1 | | 60 (hinge C) | 5750.0 | 8.743 | 5750.0 | 0 | | 90 | 2875.0 | 6.557 | 4312.5 | -1437.5 | | 120 | 0 | 0 | 0 | 0 | Shear force (kN): at A = 12.50; at 30 m: 108.33 (left), -91.67 (right); at 55 m: -11.81 (left), -111.81 (right); at C: -95.83; zero at 90 m; just left of B: 95.83. The shear varies linearly between these points because $dy/dx$ is linear in $x$. Girder end reactions: $R_A = 12.50$ kN (up) and $R_B = -95.83$ kN (a downward anchor force at B). Check: $12.5 - 95.83 + 383.33\ (\text{total hanger pull } 8Hd/L) = 300$ kN, equal to the applied load. #### Diagrams (left end A at the top; $x$ from 0 to 120 m) Shear force diagram (kN), linear between the listed points (+ above, - below the zero line): ```text x = 0 | + 12.5 x = 30 left | ++++++++++ 108.3 x = 30 right| --------- -91.7 x = 55 left | - -11.8 x = 55 right| ---------- -111.8 x = 60 (C) | --------- -95.8 x = 90 | 0 x = 120 | +++++++++ 95.8 ``` Bending moment diagram (kN m, + sagging, - hogging; parabolic curves between points): ```text x = 0 | 0 x = 30 | ++++++++++++++++++ 1812.5 x = 55 | +++++ 519.1 x = 60 (C) | 0 x = 90 | -------------- -1437.5 x = 120 | 0 ``` **Answer:** $H = 657.7$ kN (638.9 kN if the 12 m is taken at mid-span); $M_{max}^{+} = 1812.5$ kN m under the 200 kN load (30 m); $M$ at 55 m $= 519.1$ kN m; $M_C = 0$; $M^{-}_{max} = -1437.5$ kN m at 90 m. The maximum shear force is $-111.8$ kN just right of the 100 kN load (55 m); the largest positive shear is $108.3$ kN just left of the 200 kN load. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The cable passes 12 m below the higher end at mid-span, and the ends differ in level by 6 m. What dip d is used below the chord at mid-span? A. 12 m B. 6 m C. 9 m D. 15 m Answer: C. The chord at mid-span is 3 m below the higher end, so the dip below the chord is 12 - 3 = 9 m. (Key point: Effective dip at mid-span) 2. How is H found? A. H = M0C/d, from M0C - H d = 0 at the central hinge, giving 638.89 kN B. H = total load / 2 = 150 kN C. H = RA x 60 = 12250 kN D. H = M0C x d = 51750 kN Answer: A. The girder hinge at C carries no moment, so H = 5750/9 = 638.89 kN. (Key point: Cable pull from the hinge) 3. How would a uniform self-weight of the girder affect the girder SF and BM? A. It would add a parabolic BM to the girder B. It would be carried by the cable and not change girder forces C. It would double the shear D. It would remove the hinge moment Answer: B. A uniform load is carried wholly by the parabolic cable, so the girder moments and shears are unchanged. (Key point: Dead load effect) 4. Why does the shear vary linearly between loads, and what is the reaction at B? A. dy/dx is linear in x; RB = -95.83 kN, a downward anchor force B. dy/dx is constant; RB is upward 95.83 kN C. The hangers are inclined; RB is zero D. dy/dx is quadratic; RB = 300 kN Answer: A. S = S0 - H dy/dx and dy/dx is linear, so S is linear between loads. The end reaction at B is negative, i.e. a downward anchor force. 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