---
title: "Truss: stiffness method, displacement at node A"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:truss-stiffness-200kn-displacement-a"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-stiffness-200kn-displacement-a
site: BCT Engineering
---

# Truss: stiffness method, displacement at node A

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2074 Chaitra, 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-stiffness-200kn-displacement-a
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyze the truss shown in figure below by the stiffness matrix method and find the vertical and horizontal displacement at node A. Given $A = 800\ \text{mm}^2$, $E = 200\ \text{kN/mm}^2$. [Figure: three bars AB, AC and AD meeting at joint A, 1 m above the supports B, C and D on a horizontal line; horizontal spacing 1.0 m, 0.5 m and 0.5 m as marked; 200 kN horizontal load at A.]

## Short answer

Stiffness matrix assembly for three bars gives the displacements of joint A: horizontal 1.570 mm to the right and vertical 0.395 mm upward.

## Answer

**Assumed geometry (the figure is only described):** joint A is 1.0 m above the support line. Measured from the vertical through A, support B is 1.0 m to the left, C is 0.5 m to the right and D is 1.0 m to the right (the spacings 1.0, 0.5 and 0.5 m run from B to the foot of A, to C, to D). So A is at (1.0, 1.0), with B(0, 0), C(1.5, 0) and D(2.0, 0). The 200 kN load acts horizontally to the right at A. $E=200$ kN/mm² $=2\times10^8$ kN/m², $A=800$ mm² $=8\times10^{-4}$ m², so $EA=1.6\times10^5$ kN.

#### Degrees of freedom
Joint A is free to move: $u$ (horizontal) and $v$ (vertical). The supports are fixed. Coordinates: 1 = $u$, 2 = $v$.

#### Member data (direction cosines taken from A towards the support)
| Bar | L (m) | c | s | k = EA/L (kN/m) | $kc^2$ | $kcs$ | $ks^2$ |
|---|---|---|---|---|---|---|---|
| AB | 1.4142 | -0.7071 | -0.7071 | 113137.1 | 56568.5 | 56568.5 | 56568.5 |
| AC | 1.1180 | 0.4472 | -0.8944 | 143108.4 | 28621.7 | -57243.3 | 114486.7 |
| AD | 1.4142 | 0.7071 | -0.7071 | 113137.1 | 56568.5 | -56568.5 | 56568.5 |

#### Structure stiffness matrix (sum over the bars of $k\begin{bmatrix}c^2&cs\\cs&s^2\end{bmatrix}$)
$$
[K]=\begin{bmatrix}141758.8&-57243.3\\-57243.3&227623.8\end{bmatrix}\ \text{kN/m}
$$

#### Displacements of A
$$
\begin{bmatrix}141758.8&-57243.3\\-57243.3&227623.8\end{bmatrix}\begin{Bmatrix}u\\v\end{Bmatrix}=\begin{Bmatrix}200\\0\end{Bmatrix}
\;\Rightarrow\; u=1.5703\ \text{mm},\quad v=0.3949\ \text{mm}
$$

#### Member forces
Elongation $\delta=-(uc+vs)$; force $F=k\delta$ (tension +):

| Bar | δ (mm) | Force (kN) |
|---|---|---|
| AB | 1.3896 | 157.22 |
| AC | -0.3491 | -49.95 |
| AD | -0.8311 | -94.03 |

Equilibrium check at A: $\sum F_x=200+\sum F c=0.000$ and $\sum F_y=\sum F s=0.000$ ✓.

**Answer: horizontal displacement of A $u=1.570$ mm (to the right), vertical displacement $v=0.395$ mm (up).**

## Check yourself

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

1. How many unknown displacement coordinates does the truss have, and what are they?
   A. Two: horizontal u and vertical v of joint A
   B. Six: u and v at B, C and D
   C. Three: u, v of A and rotation of A
   D. One: horizontal u of A only
   Answer: A. Supports B, C, D are fixed, so only joint A can move, in u and v. Truss joints are pins, so no rotation coordinate. (Key point: Degrees of freedom)

2. What contribution does each bar make to the structure stiffness matrix?
   A. 4EI/L on the diagonal and 2EI/L off it
   B. k[c^2 cs; cs s^2] with k = EA/L and c, s the direction cosines
   C. k[1 -1; -1 1] with no direction cosines
   D. EI/L times the bar length
   Answer: B. Each axial bar contributes (EA/L) times the direction-cosine matrix in global axes; the sum over AB, AC, AD gives [K]. (Key point: Member stiffness)

3. How are the displacements of A obtained once [K] is built?
   A. Multiply [K] by the load
   B. Take the average of the three bar elongations
   C. Solve [K]{u,v} = {200, 0}
   D. Divide 200 kN by EA only
   Answer: C. The equilibrium equation [K]{d} = {P} with P = {200, 0} (horizontal load only) is solved for u and v. (Key point: Solving for displacements)

4. What are the displacements of A?
   A. u = 0.395 mm right, v = 1.570 mm up
   B. u = 1.570 mm left, v = 0.395 mm down
   C. u = 1.570 mm right, v = 0
   D. u = 1.570 mm right, v = 0.395 mm up
   Answer: D. Solving the 2x2 system gives u = 1.5703 mm and v = 0.3949 mm; the load is to the right so u is positive. (Key point: Result)

## Related in Theory of Structures II

- [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
- [Stiffness matrix: frame with 100 kN, 15 kNm](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-100kn-20knm-15knm) — Theory of Structures II, chapter 3
- [Jib crane: matrix displacement method](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Ajib-crane-matrix-displacement-10kn) — Theory of Structures II, chapter 3
- [Stiffness matrix: frame, 25 kN, 100 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-25kn-30knm-100kn) — Theory of Structures II, chapter 3
- [Truss analysis by displacement method](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-displacement-method-cable-8cm2) — Theory of Structures II, chapter 3
- [Stiffness matrix: frame with 100 kN, end reactions](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-matrix-frame-100kn-end-reactions) — 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
- [Rigid bar AB: vertical displacement of F](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Arigid-bar-ab-steel-aluminium-rods) — Strength of Materials, chapter 3
- [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×
- [Two-bar plane truss displacement and member forces](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Atwo-bar-truss-member-forces) — Finite Element Method, chapter 3