---
title: "Rigid bars PQ and RS: maximum force W"
subject: "Strength of Materials"
chapter: 3
key: "semester_3/strength_of_materials:rigid-bars-pq-rs-two-rods"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/rigid-bars-pq-rs-two-rods
site: BCT Engineering
---

# Rigid bars PQ and RS: maximum force W

Strength of Materials · Chapter 3: Simple Stress and Strain

- Asked in: 2080 Bhadra, 8 marks
- Video: [Strength of Materials l Axial Deformation l Hooke's Law l Problem 214 l](https://www.youtube.com/watch?v=6agqRH-NGI8) — Mathematics and Engineering Videos
- 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-bars-pq-rs-two-rods
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3

## Question

The rigid bars PQ and RS shown in figure are supported by pins at P and R and the two rods. Determine the maximum force W that can be applied, if its vertical movement is limited to 5 mm under load. [Figure: two horizontal rigid bars; upper bar PQ pinned at P (3 m + 3 m), lower bar RS pinned at R (3 m + 3 m); an aluminium rod (L = 2 m, $A = 500\ \text{mm}^2$, $E = 70\ \text{GPa}$) hangs from the ceiling and connects to the upper bar at Q; a steel rod (L = 2 m, $A = 300\ \text{mm}^2$, $E = 200\ \text{GPa}$) connects Q to S; load W acts at the middle of the lower bar RS, 3 m from R and 3 m from S.]

## Short answer

From equilibrium each rod carries W/2. Deformation compatibility and geometry make load movement half the sum of rod elongations. Limiting to 5 mm yields Wmax = 221.05 kN.

## Answer

Assumed arrangement: both bars are 6 m long (3 m + 3 m) and pinned at P and R. The aluminium rod hangs from the ceiling to Q (upper bar), the steel rod joins Q to S (lower bar), and $W$ acts at the middle of RS.

#### Equilibrium
- Lower bar RS (moments about R): the steel rod pulls S up with $F_{st}$, so $F_{st}(6) = W(3)$ and $F_{st} = W/2$.
- Upper bar PQ (moments about P): the steel rod pulls Q down with $F_{st}$ and the aluminium rod pulls Q up with $F_{al}$, so $F_{al}(6) = F_{st}(6)$ and $F_{al} = F_{st} = W/2$.

Let $F = W/2$. Both rods are in tension.

#### Deformations
$$
\begin{aligned}
\delta_{al} &= \frac{F L}{AE} = \frac{F(2000)}{500\times70\times10^3} = 5.714\times10^{-5}\,F\ \text{mm}\\
\delta_{st} &= \frac{F(2000)}{300\times200\times10^3} = 3.333\times10^{-5}\,F\ \text{mm}
\end{aligned}
$$

#### Geometry of movement
- Q moves down by $\delta_{al}$ (bar PQ rotates about P).
- S moves down by $\delta_Q + \delta_{st} = \delta_{al} + \delta_{st}$ (bar RS rotates about R).
- The load point is at the middle of RS, so it moves $\tfrac12(\delta_{al}+\delta_{st})$.

Limiting the movement of W to 5 mm:

$$
\tfrac12(\delta_{al}+\delta_{st}) = 5 \Rightarrow (5.714\times10^{-5}+3.333\times10^{-5})F = 10
$$
$$
F = 110526.3\ \text{N},\qquad W = 2F = 221052.6\ \text{N} = 221.05\ \text{kN}
$$

**Answer: $W_{max} = 221.05$ kN** (force of 110.53 kN in each rod).

## Check yourself

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

1. What is the rod force in terms of W?
   A. Fst = W
   B. Fst = W/3
   C. Fst = W/2 from moments about R (Fst*6 = W*3)
   D. Fst = 2W
   Answer: C. The load acts at mid-length, so the steel rod takes half. (Key point: Lower bar)

2. Why is Fal = Fst?
   A. Both rods have equal area
   B. Moments about P: both forces act at Q, 6 m from P
   C. Both rods are the same length only
   D. The bars are equal in weight
   Answer: B. Fal(6) = Fst(6) gives equality. (Key point: Upper bar)

3. How far does S move?
   A. δal + δst (Q moves down δal, then the steel rod stretches)
   B. δst only
   C. δal only
   D. δal - δst
   Answer: A. S moves with Q and the steel rod elongation adds. (Key point: Geometry)

4. Why is the load-point movement (δal + δst)/2 = 5 mm?
   A. Load is at S
   B. Load is at R
   C. It moves the full amount
   D. Load is at mid-span of RS, so it moves half of S's movement
   Answer: D. The load point is at the middle of RS, so its movement is half. This gives F = 110.5 kN and W = 221.05 kN. (Key point: Limit)

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