--- title: "Rigid bar pinned at B with two vertical rods" subject: "Strength of Materials" chapter: 3 key: "semester_3/strength_of_materials:rigid-bar-pinned-two-rods-500n" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/rigid-bar-pinned-two-rods-500n site: BCT Engineering --- # Rigid bar pinned at B with two vertical rods Strength of Materials · Chapter 3: Simple Stress and Strain - Asked in: 2081 Baisakh, 8 marks - Video: [The rigid bar AB, attached to two vertical rods as shown in](https://www.youtube.com/watch?v=E-Zw-Nsw12o) — MECHANICAL PROBLEMS - 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-pinned-two-rods-500n - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3 ## Question A rigid bar ABCD pinned at B and connected to two vertical rods as shown in the figure below. Assuming that the bar was initially horizontal and the rods stress free, determine the stress in each rod after the load $P = 500\ \text{N}$ is applied. $E_{steel} = 200\ \text{GPa}$ and $E_{al} = 100\ \text{GPa}$. [Figure: horizontal rigid bar A-B-C-D with pin support at B; A to B 1.5 m, B to C 0.75 m, C to D 0.75 m; aluminium rod at A (area $300\ \text{mm}^2$, length 1.5 m) fixed to the ground below; steel rod at D (area $200\ \text{mm}^2$, length 0.9 m) hanging from the ceiling; load $P = 500\ \text{N}$ downward at C.] ## Short answer Moment equilibrium about B gives Fal + Fst = 250 N. Compatibility requires equal deflection, leading to Fal = 0.45 Fst. Solving yields σsteel = 0.862 N/mm² (tension) and σaluminium = 0.259 N/mm² (tension). ## Answer The load $P$ at C rotates the rigid bar clockwise about the pin B: A moves up (stretching the aluminium rod fixed to the ground below) and D moves down (stretching the steel rod hung from the ceiling). Both rods are in tension; let the forces be $F_{al}$ and $F_{st}$. ```text F_al F_st | ^ v | A---1.5m---B---0.75---C---0.75---D (Al, 1.5 m) pin P=500 N (steel, 0.9 m) ``` #### Equilibrium (moments about B) Aluminium pulls A down and steel pulls D up; both give anticlockwise moments about B, balancing the clockwise moment of $P$: $$ F_{al}(1.5) + F_{st}(1.5) = P(0.75) = 500\times0.75 $$ $$ F_{al} + F_{st} = 250\ \text{N} $$ #### Compatibility The bar is rigid, and A and D are both 1.5 m from the pin, so the upward movement of A equals the downward movement of D: $\delta_{al} = \delta_{st}$. $$ \frac{F_{al}(1500)}{300\times100\times10^3} = \frac{F_{st}(900)}{200\times200\times10^3} \;\Rightarrow\; 5\times10^{-5}F_{al} = 2.25\times10^{-5}F_{st} \;\Rightarrow\; F_{al} = 0.450\,F_{st} $$ #### Solution $$ \begin{aligned} F_{st}(1+0.450) &= 250 \Rightarrow F_{st} = 172.41\ \text{N}\\ F_{al} &= 77.59\ \text{N}\\ \sigma_{st} &= \frac{172.41}{200} = 0.862\ \text{N/mm}^2\\ \sigma_{al} &= \frac{77.59}{300} = 0.259\ \text{N/mm}^2 \end{aligned} $$ **Answer: $\sigma_{steel} = 0.862$ N/mm² (tension) and $\sigma_{aluminium} = 0.259$ N/mm² (tension).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What moment equation about B is used? A. Fal + Fst = 500 B. Fal(1.5) = Fst(0.75) C. Fal(1.5) + Fst(1.5) = 500(0.75) D. Fal(1.5) - Fst(1.5) = 500(0.75) Answer: C. Both rod forces give anticlockwise moments on arms of 1.5 m; they balance P's clockwise moment, so Fal + Fst = 250 N. (Key point: Equilibrium) 2. Why is δal = δst? A. Both rods have the same length B. The bar is rigid and A and D are equidistant (1.5 m) from the pin C. Both have the same area D. Both have the same E Answer: B. A rotates about B, so equal arms give equal vertical displacements. (Key point: Compatibility) 3. Why does steel carry more force than aluminium? A. Steel is stiffer per length: AE/L is larger so it needs more force for the same deflection B. Steel is nearer to the load C. It has a larger area D. It is longer Answer: A. Fal = 0.45 Fst gives Fst = 172.41 N and Fal = 77.59 N. (Key point: Force split) 4. Which stresses result? A. σst = 0.259, σal = 0.862 B. Both 0.5 C. Compressive in both D. σst = 0.862 N/mm^2 and σal = 0.259 N/mm^2, both tensile Answer: D. 172.41/200 = 0.862 and 77.59/300 = 0.259; both rods are stretched. 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