---
title: "Truss: force method, 100 kN and 50 kN"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:truss-force-method-100kn-50kn-4x3"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-force-method-100kn-50kn-4x3
site: BCT Engineering
---

# Truss: force method, 100 kN and 50 kN

Theory of Structures II · Chapter 2: Force method

- Asked in: 2072 Kartik, 7 marks
- Video: [SA26: Force Method (Truss Analysis)](https://www.youtube.com/watch?v=AuBcV1oehD8) — Dr. Structure
- 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-100kn-50kn-4x3
- 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 in figure below by using the force method. AE is constant for all members. [Figure: rectangular truss ABCD with both diagonals, 4 m wide and 3 m high; 100 kN vertical at C and 50 kN horizontal at D; A hinged, B roller.]

## Short answer

Choose diagonal AC force as redundant. Compatibility gives X = 20.83 kN. Final member forces (kN, + tension): AB +33.33, BC –112.50, CD –16.67, DA +25.00, AC +20.83, BD –41.67.

## Answer

Reading of the figure: A (0, 0) hinged, B (4, 0) roller, C (4, 3) and D (0, 3) at the top; members AB, CD (4 m), BC, DA (3 m) and diagonals AC, BD (5 m). Loads: 100 kN downward at C and 50 kN horizontal at D (towards the right). $AE$ is constant. $D_s = 6 + 3 - 8 = 1$. Take the force in AC as the redundant $X$.

#### Reactions
- $\Sigma M_A = 0$: $V_B(4) = 100(4) + 50(3) \Rightarrow V_B = 137.50$ kN ($\uparrow$)
- $V_A = 100 - 137.50 = -37.50$ kN, i.e. $37.50$ kN ($\downarrow$)
- $H_A = 50$ kN ($\leftarrow$)

#### Force tables (tension +)
| Member | $L$ (m) | $S_0$ | $n_1$ | Final $S$ |
|---|---|---|---|---|
| AB | 4.000 | +50.000 | -0.800 | +33.333 |
| BC | 3.000 | -100.000 | -0.600 | -112.500 |
| CD | 4.000 | 0.000 | -0.800 | -16.667 |
| DA | 3.000 | +37.500 | -0.600 | +25.000 |
| AC | 5.000 | 0.000 | +1.000 | +20.833 |
| BD | 5.000 | -62.500 | +1.000 | -41.667 |

$$
f_{11} = \frac{\sum n^2L}{AE} = \frac{17.280}{AE},\qquad \Delta_{10} = \frac{\sum S_0nL}{AE} = \frac{-360.000}{AE}
$$

#### Compatibility
$$
f_{11}X + \Delta_{10} = 0 \ \Rightarrow\ X = -\frac{-360.000}{17.280} = 20.83\ \text{kN}
$$

#### Final forces $S = S_0 + nX$
| Member | Force (kN) | Nature |
|---|---|---|
| AB | +33.33 | tension |
| BC | -112.50 | compression |
| CD | -16.67 | compression |
| DA | +25.00 | tension |
| AC | +20.83 | tension |
| BD | -41.67 | compression |

**Answer: AB $= +33.33$, BC $= -112.50$, CD $= -16.67$, DA $= +25.00$, AC $= +20.83$, BD $= -41.67$ kN (positive = tension).**

## Check yourself

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

1. With 6 members, 4 joints, 3 reactions, which is true?
   A. Statically determinate
   B. D_s = 1; the force in AC is taken as redundant
   C. D_s = 3
   D. D_s = 2
   Answer: B. D_s = 6 + 3 - 8 = 1, so one member (AC) is the redundant. (Key point: Redundant)

2. With D10 = -360/AE and f11 = 17.28/AE, what is X and its nature?
   A. +20.83 kN tension
   B. -20.83 kN compression
   C. +360 kN tension
   D. -17.28 kN compression
   Answer: A. X = -D10/f11 = 360/17.28 = 20.83 kN, positive so tension. (Key point: Compatibility)

3. BC has S0 = -100 and n = -0.6. What is its final force?
   A. -87.5 kN
   B. -125 kN
   C. -100 kN
   D. -112.5 kN
   Answer: D. S = -100 + (-0.6)(20.83) = -112.5 kN, compression. (Key point: Final force)

## Close variants set in other exams

- [Truss: force method, 4 m x 3 m panel](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-force-method-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
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