---
title: "Truss analysis by displacement method"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:truss-displacement-method-cable-8cm2"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-displacement-method-cable-8cm2
site: BCT Engineering
---

# Truss analysis by displacement method

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2075 Asoj, 8 marks
- Video: [SA48: Matrix Displacement Method: Truss Analysis](https://www.youtube.com/watch?v=iqdxu8gEZ4c) — 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-displacement-method-cable-8cm2
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyse the truss by the displacement method. Take $E = 2\times10^{5}$ MPa, $A = 8\ \text{cm}^2$. [Figure: joint hanging 5 m below a horizontal support line, connected by four bars at angles 60°, vertical, 60° and 45°; loads 50 kN horizontal and 100 kN vertical at the joint.]

## Short answer

The joint stiffness matrix is formed by summing bar contributions using direction cosines and axial stiffness. Solving the 2×2 system gives joint displacements u=2.676 mm, v=-1.535 mm. Bar forces: F1=73.92, F2=49.11, F3=-0.25, F4=-18.26 kN (tension positive).

## Answer

**Setup:** the joint O hangs 5 m below the horizontal line of supports and is held by four bars. Measured from the +x axis (towards the support), the bars make angles $120^\circ$ (bar 1), $90^\circ$ (bar 2), $60^\circ$ (bar 3) and $45^\circ$ (bar 4). This reading of the figure is assumed. Loads at O: $P_x=50$ kN (to the right) and $P_y=-100$ kN (downward). $E=2\times10^5$ MPa $=2\times10^8$ kN/m², $A=8$ cm² $=8\times10^{-4}$ m², so $EA=1.6\times10^5$ kN. The degrees of freedom are the joint displacements $u$ (right) and $v$ (up). All support joints are fixed.

#### Bar data ($L=5/\sin\alpha$, $k=EA/L$, direction cosines $c=\cos\alpha$, $s=\sin\alpha$ from O towards the support)
| Bar | α | c | s | L (m) | k = EA/L (kN/m) | $kc^2$ | $kcs$ | $ks^2$ |
|---|---|---|---|---|---|---|---|---|
| 1 | 120° | -0.5000 | 0.8660 | 5.7735 | 27712.8 | 6928.2 | -12000.0 | 20784.6 |
| 2 | 90° | 0.0000 | 1.0000 | 5.0000 | 32000.0 | 0.0 | 0.0 | 32000.0 |
| 3 | 60° | 0.5000 | 0.8660 | 5.7735 | 27712.8 | 6928.2 | 12000.0 | 20784.6 |
| 4 | 45° | 0.7071 | 0.7071 | 7.0711 | 22627.4 | 11313.7 | 11313.7 | 11313.7 |

#### Stiffness matrix of the joint
For one bar the contribution to the joint is $k\begin{bmatrix}c^2&cs\\cs&s^2\end{bmatrix}$. Summing the four bars:
$$
[K]=\begin{bmatrix}25170.1&11313.7\\11313.7&84882.9\end{bmatrix}\ \text{kN/m}
$$

#### Solution of $[K]\{d\}=\{P\}$
$$
\begin{bmatrix}25170.1&11313.7\\11313.7&84882.9\end{bmatrix}\begin{Bmatrix}u\\v\end{Bmatrix}=\begin{Bmatrix}50\\-100\end{Bmatrix}
\;\Rightarrow\; u=2.6764\ \text{mm},\quad v=-1.5348\ \text{mm}
$$

#### Bar forces
Elongation of a bar: $\delta=-(uc+vs)$ (the joint moving away from the support lengthens it), force $F=k\delta$ (tension +).

| Bar | δ (mm) | Force (kN) |
|---|---|---|
| 1 | 2.6674 | 73.92 |
| 2 | 1.5348 | 49.11 |
| 3 | -0.0090 | -0.25 |
| 4 | -0.8072 | -18.26 |

Equilibrium check at O: $\sum F_x=0.000$, $\sum F_y=0.000$ ✓.

**Answer: joint displacement $u=2.676$ mm, $v=-1.535$ mm; bar forces $F_1=73.92$, $F_2=49.11$, $F_3=-0.25$, $F_4=-18.26$ kN (+ tension).**

## Check yourself

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

1. One bar contributes to the joint stiffness matrix as:
   A. kL
   B. k[c^2 cs; cs s^2]
   C. k[1 0; 0 1]
   D. EA only
   Answer: B. Resolving the axial stiffness k = EA/L along x and y gives k times the direction-cosine matrix. (Key point: Bar contribution)

2. What are the unknowns?
   A. Bar forces
   B. Support reactions
   C. Joint displacements u and v
   D. Bar lengths
   Answer: C. Supports are fixed; only joint O moves. (Key point: Degrees of freedom)

3. Why is L = 5/sin α?
   A. Bars are equal
   B. Angle is 90°
   C. E varies
   D. The joint is 5 m below the support line, so vertical depth is L sin α
   Answer: D. Vertical depth 5 m equals L sin α. (Key point: Bar length)

4. Bar force is found from:
   A. F = k x elongation, with δ = -(uc + vs)
   B. F = P/A
   C. F = kL
   D. F = EA
   Answer: A. Axial force is stiffness times elongation from joint displacement. (Key point: Bar force)

## Related in Theory of Structures II

- [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×
- [Truss: stiffness method, displacement at node A](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-stiffness-200kn-displacement-a) — Theory of Structures II, chapter 3
- [Truss: force method, 60 kN at D](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-force-method-60kn-3m-4m) — Theory of Structures II, chapter 2
- [Truss: displacement method (stiffness matrix)](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-stiffness-method-400kncm) — Theory of Structures II, chapter 3
- [Truss: lack of fit, temperature, bars BC BF](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-lack-of-fit-temp-bc) — Theory of Structures II, chapter 2
- [Moment distribution: frame, 150 kN, 60 kN/m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-150kn-50kn-60knm) — Theory of Structures II, chapter 3

## Related in other subjects

- [Truss horizontal displacement at C by virtual work](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-virtual-work-horizontal-c) — Theory of Structures I, chapter 3
- [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 joint B: temperature and vertical members long](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-defl-b-temp-lackfit-2073) — Theory of Structures I, chapter 3, asked 2×
- [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