---
title: "Truss: force method with EA = 10^5 kN"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:truss-force-method-ea-1e5-50kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-force-method-ea-1e5-50kn
site: BCT Engineering
---

# Truss: force method with EA = 10^5 kN

Theory of Structures II · Chapter 2: Force method

- Asked in: 2075 Asoj, 8 marks
- Video: [Force method – Indeterminate truss analysis Dr Jawed Qureshi](https://www.youtube.com/watch?v=eZ1LTYBCN1I) — Dr Jawed Qureshi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-force-method-ea-1e5-50kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Determine the forces in all members of the truss shown below, using the force method. Take $EA = 10^{5}$ kN. [Figure: truss ABDC, 4 m wide and 3 m high, A hinged, B roller, diagonal CB; 50 kN horizontal at D.]

## Short answer

Take force in AD as redundant X. From compatibility, X = 40.51 kN (tension). Final forces (kN): AB +17.59 (T), BD –24.31 (C), DC +17.59 (T), CA +13.19 (T), CB –21.99 (C), AD +40.51 (T).

## Answer

Reading of the figure: A (0, 0) hinged, B (4, 0) roller, D (4, 3) and C (0, 3) at the top; members AB, BD, DC, CA and the two diagonals CB and AD (a single diagonal would leave the truss determinate, so both are taken). The 50 kN load acts horizontally at D towards the right. $D_s = 6 + 3 - 8 = 1$. Take the force in AD as the redundant $X$. $EA = 10^5$ kN is the same for all members, so it cancels from the forces.

#### Reactions
$\Sigma M_A = 0$: $V_B(4) = 50(3) \Rightarrow V_B = 37.5$ kN ($\uparrow$); $V_A = 37.5$ kN ($\downarrow$); $H_A = 50$ kN ($\leftarrow$).

#### Tables (tension +)
| Member | $L$ (m) | $S_0$ | $n_1$ | Final $S$ |
|---|---|---|---|---|
| AB | 4.000 | +50.000 | -0.800 | +17.593 |
| BD | 3.000 | 0.000 | -0.600 | -24.306 |
| DC | 4.000 | +50.000 | -0.800 | +17.593 |
| CA | 3.000 | +37.500 | -0.600 | +13.194 |
| CB | 5.000 | -62.500 | +1.000 | -21.991 |
| AD | 5.000 | 0.000 | +1.000 | +40.509 |

$$
f_{11} = \sum\frac{n^2L}{AE} = \frac{17.280}{AE} = 1.728\times10^{-4}\ \text{m/kN}, \qquad \Delta_{10} = \sum\frac{S_0nL}{AE} = \frac{-700.000}{AE}
$$

#### Compatibility
$$
\Delta_{10} + f_{11}X = 0 \ \Rightarrow\ X = -\frac{-700.000}{17.280} = 40.51\ \text{kN}
$$

#### Final forces $S = S_0 + nX$
| Member | Force (kN) | Nature |
|---|---|---|
| AB | +17.59 | tension |
| BD | -24.31 | compression |
| DC | +17.59 | tension |
| CA | +13.19 | tension |
| CB | -21.99 | compression |
| AD | +40.51 | tension |

**Answer: AD $= 40.51$ kN (T); AB $= +17.59$ kN, BD $= -24.31$ kN, DC $= +17.59$ kN, CA $= +13.19$ kN, CB $= -21.99$ kN (positive = tension).**

## Check yourself

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

1. Why is D_s = 1 for this truss with six members, four joints and three reactions?
   A. D_s = m + r - 2j = 6 + 3 - 8
   B. D_s = m - j
   C. D_s = j + r - m
   D. D_s = 2j - r
   Answer: A. D_s = m + r - 2j = 6 + 3 - 8 = 1. (Key point: Indeterminacy)

2. What equation gives the redundant X (force in AD)?
   A. f11 X = Δ10
   B. Δ10 + f11 X = 0
   C. X = ΣS0 / Σn
   D. ΣS = 0
   Answer: B. The relative displacement at the cut in AD must vanish: Δ10 + f11 X = 0. (Key point: Compatibility equation)

3. How are f11 and Δ10 computed?
   A. f11 = Σ S0 n L/AE, Δ10 = Σ n^2 L/AE
   B. f11 = Σ S0 L/AE, Δ10 = Σ nL
   C. f11 = Σ n L, Δ10 = Σ S0
   D. f11 = Σ n^2 L/AE, Δ10 = Σ S0 n L/AE
   Answer: D. f11 = Σ n^2 L/AE and Δ10 = Σ S0 n L/AE, with S = S0 + nX. (Key point: Coefficients)

4. What is the force in AD, and does AE affect it?
   A. 40.51 kN tension; AE cancels because it is constant
   B. 40.51 kN compression; depends on AE
   C. 21.99 kN tension
   D. 17.59 kN tension; depends on AE
   Answer: A. X = 700/17.28 = 40.51 kN (tension); EA is the same for all members so it cancels. (Key point: Result)

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