--- title: "Rigid bar ABC: aluminium wire, steel bar" subject: "Strength of Materials" chapter: 3 key: "semester_3/strength_of_materials:rigid-bar-abc-aluminium-steel-wire" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/rigid-bar-abc-aluminium-steel-wire site: BCT Engineering --- # Rigid bar ABC: aluminium wire, steel bar Strength of Materials · Chapter 3: Simple Stress and Strain - Asked in: 2072 Chaitra, 8 marks - Video: [The rigid bar ABC shown in Fig is hinged at A and supported by a steel](https://www.youtube.com/watch?v=l9t3PqsA-7A) — 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-abc-aluminium-steel-wire - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3 ## Question ABC is a rigid bar, wire BD is made of aluminum and EC is made of steel. Determine the stresses in rods and reactions at A. Take $A_{al} = 4\ \text{mm}^2$, $A_{st} = 2\ \text{mm}^2$, $E_{al} = 72\ \text{KN/m}^2$, $E_{st} = 210\ \text{KN/m}^2$. [Figure: rigid bar A-B-C hinged at A, horizontal; aluminium wire BD from B up to the ceiling at D; steel bar EC from the ceiling at E to C, length 1 m; 30 kN load hangs at 1 m beyond C; A to B 2 m, B to C 1 m, C to load 1 m.] ## Short answer Using moment equilibrium about A and compatibility δ_C = 1.5 δ_B, forces are 14.01 kN in the aluminum wire and 30.66 kN in the steel bar. Stresses: σ_al = 3503.6 N/mm², σ_st = 15328.5 N/mm²; reaction at A is 14.67 kN downward. ## Answer **Assumptions:** the figure gives only the length of the steel bar (1 m), so the aluminium wire BD is also taken as 1 m long. $E_{al} = 72\ \text{GPa}$ and $E_{st} = 210\ \text{GPa}$ (the units in the question are a misprint). The bar ABC is rigid and hinged at A; both rods are in tension, holding the bar against the 30 kN load. Let $F_B$ = force in the aluminium wire (2 m from A) and $F_C$ = force in the steel bar (3 m from A); the 30 kN load is 4 m from A. #### Equilibrium Taking moments about the hinge A: $$ 2F_B + 3F_C = 30\times 4 = 120\ \text{kN·m} $$ #### Compatibility The rigid bar rotates about A, so the deflection is proportional to the distance from A: $$ \frac{\delta_C}{\delta_B} = \frac{3}{2} $$ $$ \delta_B = \frac{F_B L}{A_{al}E_{al}} = \frac{F_B(1000)}{4(72\,000)} = 0.003472\,F_B, \qquad \delta_C = \frac{F_C L}{A_{st}E_{st}} = \frac{F_C(1000)}{2(210\,000)} = 0.002381\,F_C $$ ($F$ in N, $\delta$ in mm). Hence $$ 0.002381\,F_C = 1.5(0.003472)F_B \Rightarrow F_C = 2.1875\,F_B $$ Substituting in the moment equation: $2F_B + 3(2.1875)F_B = 120$ kN, so $$ F_B = 14.01\ \text{kN}, \qquad F_C = 30.66\ \text{kN} $$ #### Stresses $$ \sigma_{al} = \frac{14015}{4} = 3503.6\ \text{N/mm}^2, \qquad \sigma_{st} = \frac{30657}{2} = 15328.5\ \text{N/mm}^2 $$ These are very large because the given areas are very small (if the areas were in cm², the stresses would be 100 times smaller: 35.0 and 153.3 N/mm²). #### Reaction at A $\Sigma V = 0$, taking upward forces as positive: $R_A + F_B + F_C = 30$, so $$ R_A = 30 - 14.01 - 30.66 = -14.67\ \text{kN} $$ The negative sign means $R_A = 14.67\ \text{kN}$ acts **downward**. **Answer:** $\sigma_{al} = 3503.6\ \text{N/mm}^2$, $\sigma_{st} = 15328.5\ \text{N/mm}^2$, $R_A = 14.67\ \text{kN}$ downward. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What moment equation about A is used? A. 2F_B + 3F_C = 30 x 4 B. 3F_B + 2F_C = 30 x 4 C. F_B + F_C = 30 D. 2F_B + 3F_C = 30 Answer: A. B is at 2 m, C at 3 m and the load at 4 m from A. (Key point: Moment equation) 2. Which relation holds? A. delta_C / delta_B = 3/2 B. delta_C = delta_B C. delta_C / delta_B = 2/3 D. delta_C = 2 delta_B Answer: A. The rigid bar rotates about A, so deflection is proportional to distance. (Key point: Compatibility) 3. What is the reaction at A and direction? A. 14.67 kN upward B. 14.67 kN downward C. 30 kN upward D. Zero Answer: B. R_A = 30 - 14.01 - 30.66 is negative, so downward. (Key point: Reaction) 4. Why are the stresses so large? A. E is too large B. Load is too high in kg C. Given areas are very small (mm^2, not cm^2) D. Wires are too long Answer: C. With 4 and 2 mm^2 areas, stress is force/area, giving thousands of N/mm^2. 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