---
title: "Rigid bar AB hinged at A: 100 kN"
subject: "Strength of Materials"
chapter: 3
key: "semester_3/strength_of_materials:rigid-bar-ab-hinged-a-copper"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/rigid-bar-ab-hinged-a-copper
site: BCT Engineering
---

# Rigid bar AB hinged at A: 100 kN

Strength of Materials · Chapter 3: Simple Stress and Strain

- Asked in: 2069 Chaitra, 8 marks
- Video: [Composite Material - Stress, deflection and load on wires attached to a rigid bar hinged at a point](https://www.youtube.com/watch?v=GUWZ_AGwCWk) — Prof Nawaz E Patshala
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/rigid-bar-ab-hinged-a-copper
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3

## Question

A rigid bar AB is hinged at A and supported by a 2m long copper rod and a 1m long steel rod. It carries a load of 100 KN at the free end B as shown in figure below. If the area of cross-section of the steel and copper rods be $10\ \text{cm}^2$ and $8\ \text{cm}^2$ respectively and their respective values of E be $200\ \text{GN/m}^2$ and $100\ \text{GN/m}^2$, find stresses in each rod and reaction at A (assume no bending in steel and copper rods). [Figure: horizontal rigid bar B-A with hinge at A (right end); a copper rod (2 m long) and a steel rod (1 m long) connect the bar to the ceiling, spaced 1 m, 2 m and 1 m along the bar from B; 100 kN load at B.]

## Short answer

By moment equilibrium about hinge A and compatibility δ_c = 3 δ_s, forces are 142.86 kN in steel and 85.71 kN in copper. Stresses: σ_s = 142.86 MPa, σ_c = 107.14 MPa. Reaction at A is 128.57 kN downward.

## Answer

**Assumptions (from the figure):** measured from B the supports are at 1 m (copper rod, 2 m long), then 3 m (steel rod, 1 m long), and the hinge A is at 4 m. So the lever arms from A are: steel 1 m, copper 3 m, load 4 m. Both rods are in tension and the bar is rigid.

**Data:** $A_s = 1000\ \text{mm}^2$, $A_c = 800\ \text{mm}^2$, $E_s = 200\ \text{GPa}$, $E_c = 100\ \text{GPa}$, $L_s = 1$ m, $L_c = 2$ m.

#### Equilibrium (moments about A)
$$
F_s(1) + F_c(3) = 100(4) = 400\ \text{kN·m}
$$

#### Compatibility
The rigid bar rotates about A, so $\delta_c = 3\delta_s$:

$$
\frac{F_c L_c}{A_cE_c} = 3\,\frac{F_s L_s}{A_sE_s} \Rightarrow \frac{F_c(2)}{800\times10^{-6}(100\times10^9)} = 3\,\frac{F_s(1)}{1000\times10^{-6}(200\times10^9)}
$$

$$
F_c = 0.6\,F_s
$$

#### Forces and stresses
$F_s + 3(0.6F_s) = 400 \Rightarrow 2.8F_s = 400$

$$
F_s = 142.86\ \text{kN}, \qquad F_c = 85.71\ \text{kN}
$$

$$
\sigma_s = \frac{142857}{1000} = 142.86\ \text{MPa}, \qquad \sigma_c = \frac{85714}{800} = 107.14\ \text{MPa}
$$

#### Reaction at A
$$
\Sigma V = 0: \quad R_A = F_s + F_c - 100 = 142.86 + 85.71 - 100 = 128.57\ \text{kN}
$$

The rods pull the bar up with more than the 100 kN load, so $R_A$ acts **downward**.

**Answer:** $\sigma_s = 142.86\ \text{MPa}$, $\sigma_c = 107.14\ \text{MPa}$ (both tensile), $R_A = 128.57\ \text{kN}$ downward.

## Check yourself

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

1. What are the lever arms from A?
   A. Steel 1 m, copper 3 m, load 4 m
   B. Steel 3 m, copper 1 m, load 4 m
   C. All 4 m
   D. Steel 1 m, copper 2 m, load 4 m
   Answer: A. Measured from B: copper 1 m, steel 3 m, hinge 4 m. (Key point: Lever arms)

2. What does the rotation about A give?
   A. delta_c = 3 delta_s
   B. delta_c = delta_s
   C. delta_s = 3 delta_c
   D. delta_c = delta_s / 3
   Answer: A. Copper is 3 times farther from A. (Key point: Compatibility)

3. What is F_c in terms of F_s?
   A. 1.67 F_s
   B. 0.6 F_s
   C. 3 F_s
   D. F_s
   Answer: B. From deformations: F_c = 0.6 F_s. (Key point: Force ratio)

4. What is R_A?
   A. 128.57 kN upward
   B. 228.57 kN downward
   C. 128.57 kN downward
   D. 100 kN
   Answer: C. Rods pull up 228.57 kN, load 100 kN down, so the hinge pulls down 128.57 kN. (Key point: Reaction)

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