---
title: "Forces in members of three-bar system"
subject: "Strength of Materials"
chapter: 3
key: "semester_3/strength_of_materials:bar-system-three-members-forces"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/bar-system-three-members-forces
site: BCT Engineering
---

# Forces in members of three-bar system

Strength of Materials · Chapter 3: Simple Stress and Strain

- Asked in: 2073 Shrawan, 8 marks
- Video: [[101] SIMPLE STRESS / NORMAL STRESS : Composite bar of different areas](https://www.youtube.com/watch?v=0nEfl2TOamk) — Engr Pogs
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/bar-system-three-members-forces
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3

## Question

Find the forces in each members of the bar system shown in figure below. Take cross sectional area of each bar as $6\ \text{cm}^2$ and modulus of elasticity E as $2\times10^5\ \text{N/mm}^2$. [Figure: three bars AD, BD and CD hinged to the ceiling at A, B and C and meeting at joint D; the bar BD is vertical with length 3 m, AD makes $50^\circ$ and CD makes $40^\circ$ with the vertical (angles as marked); a 30 kN load hangs at D.]

## Short answer

Using equilibrium and compatibility of small displacements, the forces are: F_AD = 8.01 kN, F_BD = 17.55 kN, F_CD = 9.54 kN, all tensile.

## Answer

The three bars are hinged to the ceiling at A, B, C (on one level) and meet at D, so there are 3 unknown forces but only 2 equilibrium equations. One compatibility equation (from the displacement of D) is needed.

**Data:** $A = 600\ \text{mm}^2$, $E = 2\times10^5\ \text{N/mm}^2$, $AE = 1.2\times10^8\ \text{N}$, $P = 30\ \text{kN}$. Bar BD is vertical: $L_{BD} = 3000$ mm. AD is at $50^\circ$ and CD at $40^\circ$ to the vertical. Assume A and C lie on opposite sides of B.

$$
L_{AD} = \frac{3000}{\cos 50^\circ} = 4667.2\ \text{mm}, \qquad L_{CD} = \frac{3000}{\cos 40^\circ} = 3916.2\ \text{mm}
$$

#### Equilibrium of joint D
$$
\begin{aligned}
\Sigma H = 0:&\quad F_{AD}\sin 50^\circ = F_{CD}\sin 40^\circ \\
\Sigma V = 0:&\quad F_{BD} + F_{AD}\cos 50^\circ + F_{CD}\cos 40^\circ = 30\,000
\end{aligned}
$$

#### Compatibility
Let D move down by $v$ and sideways (away from A) by $u$. Small-displacement elongations are:

$$
\delta_{BD} = v, \qquad \delta_{AD} = v\cos 50^\circ + u\sin 50^\circ, \qquad \delta_{CD} = v\cos 40^\circ - u\sin 40^\circ
$$

and each force is $F = \dfrac{AE}{L}\,\delta$. Substituting into the two equilibrium equations gives two equations in $u$ and $v$:

$$
u = 0.0384\ \text{mm}, \qquad v = 0.4386\ \text{mm}
$$

#### Forces (all tensile)
| Bar | Elongation (mm) | Force (kN) | Stress (N/mm²) |
|---|---|---|---|
| AD | 0.3114 | 8.01 | 13.34 |
| BD | 0.4386 | 17.55 | 29.24 |
| CD | 0.3114 | 9.54 | 15.90 |

**Check:** $\Sigma H = 6.13 - 6.13 = 0$ and $\Sigma V = 17.55 + 5.15 + 7.31 = 30.00\ \text{kN}$.

**Answer:** $F_{AD} = 8.01\ \text{kN}$, $F_{BD} = 17.55\ \text{kN}$, $F_{CD} = 9.54\ \text{kN}$ (all tension).

## Check yourself

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

1. Why is the system statically indeterminate?
   A. Three unknown forces but only two equilibrium equations at joint D
   B. It has too many loads
   C. Bars are inclined
   D. Bars are hinged
   Answer: A. A compatibility equation from the displacement of D is needed. (Key point: Indeterminacy)

2. What does the horizontal equilibrium give?
   A. F_AD cos50 = F_CD cos40
   B. F_AD sin50 = F_CD sin40
   C. F_AD = F_CD
   D. F_BD = 30 kN
   Answer: B. The horizontal components of AD and CD must balance. (Key point: Equilibrium)

3. What is used to relate the bar elongations?
   A. Elongations from the displacement (u, v) of joint D, with F = AE delta / L
   B. Equal elongation in all bars
   C. Equal forces in all bars
   D. Bar lengths are equal
   Answer: A. Each elongation is a geometric function of D's movement. (Key point: Compatibility)

4. Which force is the largest?
   A. F_BD = 17.55 kN
   B. F_AD = 8.01 kN
   C. F_CD = 9.54 kN
   D. All are equal
   Answer: A. The vertical bar BD carries most with 17.55 kN, all in tension. (Key point: Result)

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