---
title: "Truss: force method, 4 m x 3 m panel"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:truss-force-method-50kn-4x3"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-force-method-50kn-4x3
site: BCT Engineering
---

# Truss: force method, 4 m x 3 m panel

Theory of Structures II · Chapter 2: Force method

- Asked in: 2074 Chaitra, 10 marks
- 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-50kn-4x3
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Determine forces in all members of the truss shown in figure below using the force method. AE for all members is constant. [Figure: rectangular truss ABCD with both diagonals, 4 m wide and 3 m high; 50 kN horizontal at C; A and D supported.]

## Short answer

Treating force in AC as redundant, reactions are VA = 37.5 kN downward at A, VD = 37.5 kN upward, HA = 50 kN leftward. Final forces (kN, + tension): AB +13.19, BC +17.59, CD -24.31, AD +17.59, AC +40.51, BD -21.99.

## Answer

Reading of the figure: A (0, 0) hinged and D (4, 0) roller, B (0, 3) and C (4, 3) at the top; members AB, CD (3 m), BC, AD (4 m) and diagonals AC, BD (5 m). The 50 kN load acts horizontally at C (towards the right). $AE$ is constant. $D_s = 6 + 3 - 8 = 1$. Take the force in AC as the redundant $X$.

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

#### Force tables (tension +; $S_0$ with AC removed; $n$ for $X=1$)
| Member | $L$ (m) | $S_0$ | $n_1$ | Final $S$ |
|---|---|---|---|---|
| AB | 3.000 | +37.500 | -0.600 | +13.194 |
| BC | 4.000 | +50.000 | -0.800 | +17.593 |
| CD | 3.000 | 0.000 | -0.600 | -24.306 |
| AD | 4.000 | +50.000 | -0.800 | +17.593 |
| AC | 5.000 | 0.000 | +1.000 | +40.509 |
| BD | 5.000 | -62.500 | +1.000 | -21.991 |

$$
f_{11} = \frac{\sum n^2L}{AE} = \frac{17.280}{AE}, \qquad \Delta_{10} = \frac{\sum S_0nL}{AE} = \frac{-700.000}{AE}
$$

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

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

**Answer: AC $= +40.51$ kN, BD $= -21.99$ kN, AB $= +13.19$ kN, BC $= +17.59$ kN, CD $= -24.31$ kN, AD $= +17.59$ kN (positive = tension).**

## Check yourself

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

1. What is D_s for this truss with 6 members, 4 joints, 3 reactions?
   A. 0
   B. 2
   C. 1
   D. 3
   Answer: C. D_s = m + r - 2j = 6 + 3 - 8 = 1; the force in AC is taken as redundant. (Key point: Degree of indeterminacy)

2. What are the support reactions?
   A. VD = 50 kN, HA = 0
   B. VD = 37.5 kN up, VA = 37.5 kN down, HA = 50 kN left
   C. VA = VD = 25 kN
   D. VD = 62.5 kN, HA = 37.5 kN
   Answer: B. Moments about A: VD x 4 = 50 x 3, so VD = 37.5 kN; HA = 50 kN left and VA = 37.5 kN down. (Key point: Reactions)

3. How are final member forces obtained?
   A. S = S0 + n X
   B. S = S0 x n
   C. S = n / X
   D. S = S0 - X
   Answer: A. S0 is the force with the redundant removed and n the force for X = 1, so S = S0 + nX. (Key point: Superposition)

4. What are the force in AC and in BD?
   A. AC = +17.59 kN, BD = -24.31 kN
   B. AC = +40.51 kN (T), BD = -21.99 kN (C)
   C. AC = -40.51 kN, BD = +21.99 kN
   D. AC = +13.19 kN, BD = +17.59 kN
   Answer: B. X = 700/17.28 = 40.51 kN tension; BD = -62.5 + 40.51 = -21.99 kN compression. (Key point: Result)

## Close variants set in other exams

- [Truss: force method, 100 kN and 50 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-force-method-100kn-50kn-4x3) — Theory of Structures II, chapter 2, asked 2×

## Related in Theory of Structures II

- [Truss: force method, EA and 2EA members](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-force-method-ea-2ea-50kn) — Theory of Structures II, chapter 2
- [Truss forces: member AC 10 mm too long](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-lack-of-fit-ac-10mm) — Theory of Structures II, chapter 2
- [Truss: force method, double-diagonal truss](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-force-method-50kn-20kn-6m) — Theory of Structures II, chapter 2
- [Indeterminate truss: bar forces and reactions](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aindeterminate-truss-bar-forces-400kn) — Theory of Structures II, chapter 2
- [Truss: force method with EA = 10^5 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-force-method-ea-1e5-50kn) — Theory of Structures II, chapter 2
- [Castigliano: square truss member forces](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-square-truss-120kn-90kn) — Theory of Structures II, chapter 2

## Related in other subjects

- [Forces in all members of truss with 10.8 kN loads](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-forces-10-8kn-d-support) — Applied Mechanics, chapter 7
- [Member forces in truss with 20, 10, 12, 50 kN](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-forces-20kn-10kn-12kn-50kn) — Applied Mechanics, chapter 7
- [Truss member forces with 25, 50, 10 kN loads](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-forces-25kn-50kn-10kn-hinge-roller) — Applied Mechanics, chapter 7
- [Truss ILDs for FG and BC; max forces from 100 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-ild-fg-bc-100kn-roll) — Theory of Structures I, chapter 5