---
title: "Suspension 150 m: SF, BM at 30 m, tension, length"
subject: "Theory of Structures I"
chapter: 7
key: "semester_4/theory_of_structures_i:suspension-150m-dip15-180-150-dead-7-5"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/suspension-150m-dip15-180-150-dead-7-5
site: BCT Engineering
---

# Suspension 150 m: SF, BM at 30 m, tension, length

Theory of Structures I · Chapter 7: Suspension Cable Systems

- Asked in: 2077 Chaitra, 10 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-dip15-180-150-dead-7-5
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7

## Question

A suspension cable having dip 15 m supports a three hinged stiffening girder 150 m long which supports point loads 180 kN and 150 kN at distance 50 m and 100 m respectively from the left support. The dead load of the girder is 7.5 kN/m. Determine the values of SF and BM at a section 30 m from the left support. Also determine the maximum tension and length of the cable.

## Short answer

At 30 m from left: V = 38.0 kN, M = -180.0 kN m. Max cable tension = 2106.9 kN, cable length = 153.90 m.

## Answer

**Data:** three-hinged stiffening girder, span $L = 150$ m, cable dip $d = 15$ m. Point loads 180 kN at 50 m and 150 kN at 100 m from the left support. Dead load of the girder $w = 7.5$ kN/m, taken wholly by the cable (so the girder carries no dead-load moment). Cable: $y = \dfrac{4d\,x(L-x)}{L^2} = \dfrac{x(150-x)}{375}$.

#### Horizontal pull
**Dead load:** $H_d = \dfrac{wL^2}{8d} = \dfrac{7.5\times 150^2}{8\times 15} = 1406.25\ \text{kN}$

**Point loads** (hinge at mid-span):
$$
R_A = \frac{180\times 100 + 150\times 50}{150} = 170.0\ \text{kN}, \qquad M_C = 170\times 75 - 180\times 25 = 8250\ \text{kN m}
$$

$$
H_p = \frac{M_C}{d} = \frac{8250}{15} = 550\ \text{kN}
$$

**Total:** $H = 1406.25 + 550 = 1956.25$ kN.

#### SF and BM in the girder at 30 m from the left
Cable ordinate: $y = \dfrac{30\times 120}{375} = 9.60$ m; slope $\tan\alpha = \dfrac{4\times 15\,(150-60)}{22500} = 0.24$. Only the point loads act on the girder (the dead load is carried by the cable):

$$
\begin{aligned} M &= R_A\times 30 - H_p\,y = 170\times 30 - 550\times 9.60 = -180.0\ \text{kN m}\\ V &= R_A - H_p\tan\alpha = 170 - 550\times 0.24 = 38.0\ \text{kN} \end{aligned}
$$

#### Maximum tension in the cable
At the supports, $\tan\alpha_s = \dfrac{4d}{L} = 0.4$:

$$
V_s = H\tan\alpha_s = 782.50\ \text{kN}, \qquad T_{max} = \sqrt{H^2 + V_s^2} = 2106.9\ \text{kN}
$$

#### Length of the cable
$$
S = L\left[1 + \frac83\left(\frac{d}{L}\right)^2 - \frac{32}{5}\left(\frac{d}{L}\right)^4\right] = 150\left[1 + \frac83(0.1)^2 - \frac{32}{5}(0.1)^4\right] = 153.90\ \text{m}
$$

**Answer: at 30 m from the left support, $V = 38.0$ kN and $M = -180.0$ kN m; maximum cable tension $= 2106.9$ kN; length of cable $= 153.90$ m.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How does the 7.5 kN/m dead load of the girder affect the girder SF and BM?
   A. It adds a full simply supported UDL moment to the girder
   B. It causes no girder moment or shear, as the cable takes it, but it adds to H
   C. It reduces the horizontal pull H
   D. It only affects the shear, not the moment
   Answer: B. A uniformly distributed load on a three-hinged girder is carried wholly by the parabolic cable. It adds to H but gives no girder BM or SF. (Key point: Dead load carried by the cable)

2. What is the total H for the cable?
   A. 550 kN (point loads only)
   B. 1406.25 kN (dead load only)
   C. 1956.25 kN
   D. 856.25 kN
   Answer: C. H = Hd + Hp = wL^2/8d + M0C/d = 1406.25 + 550 = 1956.25 kN. (Key point: Total horizontal pull)

3. Which expression gives the girder BM at 30 m?
   A. RA x 30 - H_total x y = 5100 - 18780
   B. RA x 30 - Hp x y = 5100 - 5280 = -180 kN m
   C. RA x 30 - Hd x y
   D. RA x 30 only, 5100 kN m
   Answer: B. Only the point-load pull Hp acts on the girder, so M = 170 x 30 - 550 x 9.6 = -180 kN m. (Key point: Girder BM at 30 m)

4. Where does the maximum cable tension occur, and what is the cable length formula?
   A. At mid-span, with S = L(1 + 8/3 (d/L)^2 - 32/5 (d/L)^4)
   B. At the supports, with S = L(1 + 8/3 (d/L)^2 - 32/5 (d/L)^4)
   C. At the supports, with S = L + 8d
   D. At mid-span, with S = sqrt(L^2 + d^2)
   Answer: B. The slope is greatest at the supports, so T = sqrt(H^2 + Vs^2) = 2106.9 kN there. The parabolic length series gives 153.90 m. (Key point: Max tension and cable length)

## Close variants set in other exams

- [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 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 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 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
- [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×