---
title: "Truss: displacement method (stiffness matrix)"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:truss-stiffness-method-400kncm"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-stiffness-method-400kncm
site: BCT Engineering
---

# Truss: displacement method (stiffness matrix)

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2069 Asar, 5 marks
- Video: [SA45: Matrix Displacement Method: Introduction](https://www.youtube.com/watch?v=kFkU1M7xVbg) — 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-method-400kncm
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Use the displacement method (stiffness matrix) to find forces in members of the truss shown in figure below. Take axial stiffness for each member to be $400\ \text{kN cm}^{-1}$. [Figure: truss with wall supports A (top) and C (bottom, 3 m below A), joint B at 4 m from the wall; horizontal member AB, inclined member CB; 50 kN downward and 100 kN horizontal at B.]

## Short answer

Stiffness method yields u=0.4167 cm, v=-0.9028 cm at B. Member forces: F_AB=166.67 kN (tension), F_CB=83.33 kN (compression). Axial stiffness 400 kN/cm.

## Answer

**Setup:** A is on the wall at the top and C is on the wall 3 m below A; B is 4 m from the wall, so AB is horizontal (4 m) and CB is inclined (3-4-5 triangle, length 5 m). The loads at B are 100 kN horizontal (to the right) and 50 kN vertical (downward). The axial stiffness $EA/L$ of each member is 400 kN/cm $=40\,000$ kN/m. The only free joint is B, with two displacements: $u$ (horizontal) and $v$ (vertical).

#### Member stiffness in global coordinates
For a bar with direction cosines $(c,s)$ taken from B towards the support, its contribution to the joint stiffness is $k\begin{bmatrix}c^2&cs\\cs&s^2\end{bmatrix}$.

| Member | k (kN/m) | c | s | Contribution |
|---|---|---|---|---|
| BA | 40000 | -1 | 0 | $40000\begin{bmatrix}1&0\\0&0\end{bmatrix}$ |
| BC | 40000 | -0.8 | -0.6 | $40000\begin{bmatrix}0.64&0.48\\0.48&0.36\end{bmatrix}$ |

#### Structure stiffness matrix and displacements
$$
[K]=40000\begin{bmatrix}1.64&0.48\\0.48&0.36\end{bmatrix}=\begin{bmatrix}65600&19200\\19200&14400\end{bmatrix}
$$
$$
\begin{bmatrix}65600&19200\\19200&14400\end{bmatrix}\begin{Bmatrix}u\\v\end{Bmatrix}=\begin{Bmatrix}100\\-50\end{Bmatrix}
\;\Rightarrow\; u=0.4167\ \text{cm},\quad v=-0.9028\ \text{cm}
$$

#### Member forces
Elongation $\delta=-(uc+vs)$, force $F=k\delta$:
- AB: $\delta=u=0.4167$ cm $\Rightarrow F_{AB}=400\times0.4167=166.67$ kN (tension)
- CB: $\delta=-(u(-0.8)+v(-0.6))=-0.2083$ cm $\Rightarrow F_{CB}=-83.33$ kN (compression)

Equilibrium of joint B (AB in tension pulls B towards A; CB in compression pushes B away from C): horizontal $100-166.67+0.8\times83.33=0.000$ ✓; vertical $-50+0.6\times83.33=0.000$ ✓.

**Answer: $F_{AB}=166.67$ kN (tension), $F_{CB}=-83.33$ kN (compression); joint B displaces 0.4167 cm horizontally and -0.9028 cm vertically.**

## Check yourself

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

1. What are the unknown displacements of the truss?
   A. Horizontal u and vertical v of joint B, since A and C are supported
   B. Rotations at A and C
   C. Axial forces of both members
   D. Only the vertical displacement of B
   Answer: A. Only joint B is free, with two translations u and v. (Key point: Unknowns)

2. For member CB (c = -0.8, s = -0.6, k = 40000 kN/m), which contribution is added to [K]?
   A. 40000 [0.64 0.48; 0.48 0.36]
   B. 40000 [0.8 0.6; 0.6 0.8]
   C. 40000 [1 0; 0 0]
   D. 40000 [0.36 0.48; 0.48 0.64]
   Answer: A. The contribution is k[c^2 cs; cs s^2] = 40000 [0.64 0.48; 0.48 0.36]. (Key point: Member stiffness matrix)

3. What are the member forces found from the displacements?
   A. AB = 166.67 kN tension, CB = 83.33 kN compression
   B. AB = 100 kN tension, CB = 50 kN compression
   C. AB = 83.33 kN tension, CB = 166.67 kN compression
   D. Both members in tension
   Answer: A. Elongation of AB is u = 0.4167 cm giving 400 x 0.4167 = 166.67 kN; CB shortens by 0.2083 cm giving -83.33 kN. (Key point: Member forces)

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