---
title: "Jib crane: matrix displacement method"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:jib-crane-matrix-displacement-10kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/jib-crane-matrix-displacement-10kn
site: BCT Engineering
---

# Jib crane: matrix displacement method

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2070 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/jib-crane-matrix-displacement-10kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

A jib-crane is carrying a vertical load of 10 kN at A as shown in figure below. Determine by the matrix displacement method, the displacement of joint A and hence calculate the forces in members AB and AC. Take cross-sectional area of members AB and AC as 10000 mm² and 20000 mm² respectively and $E = 200\ \text{kN/mm}^2$. [Figure: wall support with hinges B (top) and C (3 m below B); horizontal member BA of 4 m; inclined member CA; 10 kN vertical load at A.]

## Short answer

Using matrix displacement method, joint A displaces 0.027 mm right and 0.070 mm down. Member forces: AB = 13.33 kN (tension), AC = -16.67 kN (compression).

## Answer

**Setup:** B and C are hinges on the wall, B at the top and C 3 m below it. Member AB is horizontal (4 m) and member AC joins A to C, so AC $=\sqrt{4^2+3^2}=5$ m. Take A at (4, 0), B at (0, 0) and C at (0, -3). A carries a 10 kN vertical load (downward). $E=200$ kN/mm² $=2\times10^8$ kN/m², $A_{AB}=10000$ mm² $=0.01$ m², $A_{AC}=20000$ mm² $=0.02$ m².

#### Degrees of freedom
Joint A: $u$ (horizontal, right +) and $v$ (vertical, up +). B and C are fixed.

#### Member stiffness and direction cosines (from A towards the support)
| Member | L (m) | $EA$ (kN) | $k=EA/L$ (kN/m) | c | s |
|---|---|---|---|---|---|
| AB | 4 | 2000000 | 500000 | -1 | 0 |
| AC | 5 | 4000000 | 800000 | -0.8 | -0.6 |

For each bar the contribution to the joint stiffness is $k\begin{bmatrix}c^2&cs\\cs&s^2\end{bmatrix}$:

- AB: $500000\begin{bmatrix}1&0\\0&0\end{bmatrix}$
- AC: $800000\begin{bmatrix}0.64&0.48\\0.48&0.36\end{bmatrix}$

$$
[K]=\begin{bmatrix}1012000&384000\\384000&288000\end{bmatrix}\ \text{kN/m}
$$

#### Displacement of joint A
$$
\begin{bmatrix}1012000&384000\\384000&288000\end{bmatrix}\begin{Bmatrix}u\\v\end{Bmatrix}=\begin{Bmatrix}0\\-10\end{Bmatrix}
\;\Rightarrow\; u=0.0267\ \text{mm},\quad v=-0.0703\ \text{mm}
$$
A moves 0.027 mm to the right and 0.070 mm downward.

#### Member forces
Elongation $\delta=-(uc+vs)$ and force $F=k\delta$:
- AB: $\delta=-(u\times(-1))=0.0267$ mm, $F_{AB}=13.333$ kN (tension)
- AC: $\delta=-(u(-0.8)+v(-0.6))=-0.0208$ mm, $F_{AC}=-16.667$ kN (compression)

Check by equilibrium at A: the vertical component of AC is $16.667\times0.6=10.000$ kN, which balances the 10 kN load ✓; its horizontal component $16.667\times0.8=13.333$ kN balances $F_{AB}=13.333$ kN ✓.

**Answer: displacement of A: $u=0.027$ mm, $v=-0.070$ mm; $F_{AB}=13.33$ kN (tension), $F_{AC}=-16.67$ kN (compression).**

## Check yourself

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

1. How many unknown displacements does the jib-crane analysis have, and which?
   A. Four: both ends of each member
   B. Two: u and v of joint A, since B and C are fixed on the wall
   C. One: the vertical deflection of A
   D. Three including rotation of A
   Answer: B. Truss joints have only translations, and B and C are fixed, so only u and v of A are unknown. (Key point: Degrees of freedom)

2. What is the contribution of a bar to the joint stiffness matrix?
   A. k [c^2, cs; cs, s^2], with k = EA/L
   B. k [c, s; s, c]
   C. EI/L times identity
   D. k [s^2, cs; cs, c^2] with k = EI/L^3
   Answer: A. For a pin-jointed bar the 2x2 joint stiffness is (EA/L) times [c^2 cs; cs s^2] using the direction cosines. (Key point: Member stiffness contribution)

3. Using the displacements found, what are the forces in AB and AC?
   A. AB 13.33 kN tension, AC 16.67 kN compression
   B. AB 16.67 kN tension, AC 13.33 kN compression
   C. Both in tension
   D. AB compression, AC tension
   Answer: A. Elongation of AB is 0.0267 mm giving +13.33 kN; AC shortens by 0.0208 mm giving -16.67 kN (compression). (Key point: Member force from displacement)

4. How is the answer F_AC = 16.67 kN checked at joint A?
   A. Its vertical component 0.6 x 16.67 = 10 kN balances the load
   B. Its horizontal component equals 10 kN
   C. F_AC equals F_AB
   D. Its component 0.8 x 16.67 equals the 10 kN load
   Answer: A. AC has slope 3/5, so its vertical component 0.6 x 16.67 = 10 kN balances the downward 10 kN load; the horizontal component 13.33 kN balances AB. (Key point: Equilibrium check)

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