---
title: "Cable area, BMD of girder, cable length"
subject: "Theory of Structures I"
chapter: 7
key: "semester_4/theory_of_structures_i:suspension-girder-hinge-8m-cable-area"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/suspension-girder-hinge-8m-cable-area
site: BCT Engineering
---

# Cable area, BMD of girder, cable length

Theory of Structures I · Chapter 7: Suspension Cable Systems

- Asked in: 2071 Bhadra, 13 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-girder-hinge-8m-cable-area
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7

## Question

Determine the cross sectional area required for the cable loaded as shown in figure below if the permissible tensile stress of the cable material is 1500 N/mm². The self weight of the girder is 10 kN/m. Draw bending moment diagram of three hinged stiffening girder and also calculate the length of cable. [Figure: suspension bridge, span 100 m, central dip 8 m, stiffening girder hinged at mid span C; 50 kN load 20 m from the left support and 80 kN load 30 m from the left support (as read from the scan).]

## Short answer

The required cross-sectional area of the cable is 1242.4 mm². The maximum sagging moment in the girder is 952.0 kN m under the 80 kN load, maximum hogging -425.0 kN m. The cable length is 101.68 m.

## Answer

Taking the figure as read: span 100 m, dip 8 m, central hinge C, loads 50 kN at 20 m and 80 kN at 30 m from A, girder weight 10 kN/m. A single cable and a single girder are assumed.

#### Cable pull $H$
$$
\begin{aligned} R_A &= \frac{50(80)+80(70)}{100} = 96.0\ \text{kN} \\ M_{0C} &= 96.0(50) - 50(30) - 80(20) = 1700.0\ \text{kN m} \\ H_p &= \frac{1700}{8} = 212.5\ \text{kN}, \qquad H_d = \frac{10(100)^2}{8(8)} = 1562.5\ \text{kN} \\ H &= 1562.5 + 212.5 = 1775.0\ \text{kN} \end{aligned}
$$

#### Cable area
At the supports $\tan\theta = 4d/L = 0.32$, $\sec\theta = 1.0500$:
$$
T_{max} = H\sec\theta = 1775.0(1.0500) = 1863.67\ \text{kN}
$$
$$
A = \frac{T_{max}}{\sigma} = \frac{1863.67\times 10^3}{1500} = 1242.4\ \text{mm}^2
$$

#### BMD of the girder
The self-weight is carried by the cable (no girder moment). For the point loads, $M = M_0 - H_p y$, $y = \dfrac{4(8)x(100-x)}{100^2}$:

| x (m) | $M_0$ (kN m) | y (m) | $H_p y$ (kN m) | $M$ (kN m) |
|---|---|---|---|---|
| 20 | 1920.00 | 5.120 | 1088.00 | 832.00 |
| 30 | 2380.00 | 6.720 | 1428.00 | 952.00 |
| 50 | 1700.00 | 8.000 | 1700.00 | 0.00 |
| 70 | 1020.00 | 6.720 | 1428.00 | -408.00 |
| 80 | 680.00 | 5.120 | 1088.00 | -408.00 |

$M$ = 0 at both ends and at the hinge C. The BMD is sagging from A to C (peak 952.0 kN m under the 80 kN load) and hogging from C to B (peak -425.0 kN m at 75 m).

```text
            ****
         ***    *
        *        *
       *          *
     **            **
   **                *
 **                   **
*-----------------------**---------------------**
                          **                 **
                            ****         ****
                                *********
(top = 952.0, bottom = -425.0; '-' is zero line)
```

#### 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 $d/L = 0.08$:
$$
l = 100\left[1+\frac{8}{3}(0.08)^2-\frac{32}{5}(0.08)^4\right] = 101.68\ \text{m}
$$

**Answer:** Cable area = 1242.4 mm²; $M_{20} = 832.0$ kN m, $M_{30} = 952.0$ kN m, $M_C = 0$, maximum hogging $-425.0$ kN m at 75 m; cable length = 101.68 m.

## Check yourself

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

1. What is H for this cable?
   A. 212.5 kN
   B. 1562.5 kN
   C. 1775 kN
   D. 1350 kN
   Answer: C. Hd = 10 x 100^2/64 = 1562.5 kN and Hp = 1700/8 = 212.5 kN, so H = 1775 kN. (Key point: Total cable pull)

2. How is the required cable area found?
   A. A = H / stress
   B. A = Tmax / stress, Tmax = H sec(theta) = 1863.67 kN at the supports, A = 1242.4 mm^2
   C. A = Hp / stress
   D. A = RA / stress
   Answer: B. The cable must resist the maximum tension, found at the supports where tan(theta) = 4d/L = 0.32. (Key point: Cable area)

3. What is the girder BM at the central hinge C?
   A. 952 kN m
   B. Zero
   C. -425 kN m
   D. 1700 kN m
   Answer: B. A hinge carries no moment: M0C - Hp d = 1700 - 1700 = 0. (Key point: BM at the hinge)

4. Which gives the cable length with d/L = 0.08?
   A. L(1 + 8/3 (d/L)^2 - 32/5 (d/L)^4) = 101.68 m
   B. L + 8d = 164 m
   C. L(1 + d/L) = 108 m
   D. sqrt(L^2 + 4d^2)= 100.5 m
   Answer: A. The parabolic length series gives 100 x (1 + 0.017067 - 0.000262) = 101.68 m. (Key point: Cable length)

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