---
title: "Suspension 50 m: max tension, BM at 10 m, BMD/SFD"
subject: "Theory of Structures I"
chapter: 7
key: "semester_4/theory_of_structures_i:suspension-50m-5m-dip-80kn"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/suspension-50m-5m-dip-80kn
site: BCT Engineering
---

# Suspension 50 m: max tension, BM at 10 m, BMD/SFD

Theory of Structures I · Chapter 7: Suspension Cable Systems

- Asked in: 2081 Chaitra, 12 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-50m-5m-dip-80kn
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7

## Question

A suspension cable of 50 m span and 5 m dip is stiffened by a three-hinged girder. The dead load is 8 kN/m. Determine maximum tension in the cable and maximum bending moment at section 10 m from left support in the girder due to concentrated load of 80 kN crossing the girder, assuming that whole dead load is carried by the cable without stressing the girder. Draw BMD and SFD for the girder.

## Short answer

Dead-load H_d = 500 kN, live-load H_L max = 200 kN at mid-span, so H_max = 700 kN; maximum cable tension T_max = 753.9 kN. At 10 m from left support, maximum bending moment is +384.0 kN m (load at section) and minimum is -240.0 kN m (load at mid-span.

## Answer

**Data:** span $L = 50$ m, dip $d = 5$ m, three-hinged stiffening girder (hinges at both ends and mid-span). Dead load $w = 8$ kN/m carried entirely by the cable. A moving load $W = 80$ kN crosses the girder. The cable is a parabola, $y = \dfrac{4d\,x(L-x)}{L^2} = 0.008\,x(50-x)$; $\tan\alpha_{support} = \dfrac{4d}{L} = 0.4$.

#### Horizontal pull in the cable
**Dead load:** $H_d = \dfrac{wL^2}{8d} = \dfrac{8\times 2500}{40} = 500$ kN.

**Live load** $W$ at distance $a$ from the left end (hinge at mid-span: girder moment at $C$ is zero):

$$
H_L = \frac{W a}{2d} \ (a\le 25),\qquad \frac{W(L-a)}{2d}\ (a\ge 25)
$$

The maximum is with $W$ at mid-span: $H_L = \dfrac{80\times 25}{10} = 200$ kN.

$$
H_{max} = 500 + 200 = 700\ \text{kN}
$$

#### Maximum tension in the cable
It occurs at the supports, where the cable is steepest:

$$
T_{max} = H_{max}\sqrt{1 + \left(\frac{4d}{L}\right)^2} = 700\sqrt{1+0.16} = 753.9\ \text{kN}
$$

(vertical component at the support $= 700\times 0.4 = 280$ kN).

#### Maximum bending moment at 10 m from the left end
Section $x_s = 10$ m, cable ordinate $y_s = 0.008\times 10\times 40 = 3.2$ m. For a unit load at $a$, $M = M_b - H\,y_s$:

- $a\le 10$: $M = 0.8a - \dfrac{a}{10}(3.2) = 0.48a$ (peak $+4.8$ m at $a=10$)
- $10\le a\le 25$: $M = 0.2(50-a) - 0.32a = 10 - 0.52a$ ($-3.0$ m at $a = 25$)
- $a\ge 25$: $M = -0.12\,(50-a)$

The ILD is $0 \to +4.8$ (at 10 m) $\to -3.0$ (at 25 m) $\to 0$.

$$
M_{max}^{+} = 80\times 4.8 = 384.0\ \text{kN m (load at the section)}, \qquad M_{max}^{-} = 80\times(-3.0) = -240.0\ \text{kN m (load at mid-span)}
$$

#### BMD and SFD (load at 10 m, giving the maximum positive moment)
With $W$ at 10 m: $R_A = 64$ kN, $R_B = 16$ kN, $H_L = \dfrac{80\times 10}{10} = 80$ kN.

$$
M(x) = 64x - 80\times 0.008x(50-x) = 32x + 0.64x^2 \ (x\le 10), \qquad M(x) = 800 - 48x + 0.64x^2 \ (x\ge 10)
$$

$$
V(x) = V_b - H_L\tan\alpha_x,\quad \tan\alpha_x = 0.4 - 0.016x
$$

| x (m) | M (kN m) | V (kN) |
|---|---|---|
| 0 (A) | 0 | +32.0 |
| 10 (left of load) | 384.0 | +44.8 |
| 10 (right of load) | 384.0 | -35.2 |
| 25 (hinge C) | 0.0 | -16.0 |
| 37.5 | -100.0 | 0.0 |
| 50 (B) | 0.0 | +16.0 |

```text
 BMD: A(0) -> +384 at x=10 -> 0 at hinge C (x=25) -> -100 at x=37.5 -> 0 at B
      sagging from A to C, hogging from C to B (parabolic segments)
 SFD: +32 at A -> +44.8 at 10-  | jump of -80 |  -35.2 at 10+ -> -16 at C -> +16 at B
```

**Answer: $T_{max} = 753.9$ kN; maximum BM at 10 m = $+384.0$ kN m (load at the section); at the same section the minimum is $-240.0$ kN m with the load at mid-span.**

## Check yourself

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

1. What horizontal pull does the 8 kN/m dead load produce (L = 50 m, d = 5 m)?
   A. H = wL^2/(8d) = 500 kN
   B. H = wL/2 = 200 kN
   C. H = wL^2/(2d) = 2000 kN
   D. H = wL/(8d) = 10 kN
   Answer: A. Dead load is carried entirely by the cable: H_d = 8 x 2500/40 = 500 kN. (Key point: Dead load thrust)

2. Where should the 80 kN load be for the maximum live-load thrust, and what is it?
   A. At the support, 0 kN
   B. At the quarter point, 100 kN
   C. At mid-span hinge, H_L = W x 25/(2d) = 200 kN
   D. At 10 m, 80 kN
   Answer: C. H_L = Wa/(2d) for a <= L/2 is greatest at a = 25 m, giving 200 kN, so H_max = 700 kN. (Key point: Live load thrust is maximum)

3. Where does the maximum tension occur and what is it?
   A. At mid-span, 700 kN
   B. At the supports, T = H_max sqrt(1 + (4d/L)^2) = 753.9 kN
   C. At the quarter point, 500 kN
   D. At the supports, 280 kN
   Answer: B. The cable is steepest at the supports (tan = 0.4), so T = 700 x 1.077 = 753.9 kN. (Key point: Maximum cable tension)

4. Where is the 80 kN load for maximum positive moment at the section 10 m from the left, and what is the value?
   A. At mid-span; 240 kN m
   B. At the right support; 0
   C. At the hinge; -240 kN m
   D. At the section itself; 80 x 4.8 = 384 kN m
   Answer: D. The M ILD peaks at +4.8 at the section; the minimum (-3.0 at mid-span) gives -240 kN m. (Key point: Maximum positive moment at 10 m)

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