--- title: "Rigid bar AB hinged at C: 40 kN" subject: "Strength of Materials" chapter: 3 key: "semester_3/strength_of_materials:rigid-bar-ab-hinged-at-c" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/rigid-bar-ab-hinged-at-c site: BCT Engineering --- # Rigid bar AB hinged at C: 40 kN Strength of Materials · Chapter 3: Simple Stress and Strain - Asked in: 2068 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-at-c - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3 ## Question A rigid bar 'AB' is hinged at 'C' and connected with a steel rod and a copper rod at 'A' and 'B' respectively as shown in figure. Both the rods are rigidly fixed with the ceiling at the upper ends. A load of 40KN is applied at 'B'. Find the magnitude of stresses in the steel rod and the copper rod. Cross sectional area of steel is $400\ \text{mm}^2$ and copper is $600\ \text{mm}^2$. Take $E_s = 200\ \text{KN/mm}^2$ and $E_c = 110\ \text{KN/mm}^2$. [Figure: horizontal rigid bar A-C-B with the hinge at C; A to C 4 m, C to B 2 m; steel rod at A and copper rod at B, both 1 m long, fixed to the ceiling; 40 kN load W at B.] ## Short answer Using moment equilibrium and strain compatibility (δ_A = 2δ_B), forces in steel and copper are 16.58 kN (compressive) and 6.84 kN (tensile). Dividing by areas yields stresses: steel 41.45 N/mm² (compressive), copper 11.40 N/mm² (tensile). ## Answer **Assumptions:** the bar is rigid and hinged at C; A is 4 m and B is 2 m from C, on opposite sides. The load at B pulls B down, so the copper rod at B is in **tension**. The bar lifts at A, so the steel rod at A is in **compression** (it pushes the bar down at A). Both rods are 1 m long. **Data:** $A_s = 400\ \text{mm}^2$, $E_s = 200\ \text{kN/mm}^2$, $A_c = 600\ \text{mm}^2$, $E_c = 110\ \text{kN/mm}^2$. #### Equilibrium (moments about hinge C) $$ F_c(2) + F_s(4) = 40(2) \Rightarrow F_c + 2F_s = 40\ \text{kN} $$ #### Compatibility If the bar rotates through a small angle $\phi$, then $\delta_B = 2\phi$ (elongation of the copper rod) and $\delta_A = 4\phi$ (shortening of the steel rod), so $\delta_A = 2\delta_B$: $$ \frac{F_sL}{A_sE_s} = 2\,\frac{F_cL}{A_cE_c} \Rightarrow \frac{F_s}{400(200)} = \frac{2F_c}{600(110)} $$ $$ F_s = \frac{2\times 400\times200}{600\times110}F_c = 2.4242\,F_c $$ Substituting: $F_c + 2(2.4242F_c) = 40$, which gives $$ F_c = 6.839\ \text{kN}, \qquad F_s = 16.580\ \text{kN} $$ #### Stresses $$ \sigma_s = \frac{16580}{400} = 41.45\ \text{N/mm}^2\ \text{(compressive)}, \qquad \sigma_c = \frac{6839}{600} = 11.40\ \text{N/mm}^2\ \text{(tensile)} $$ **Answer:** steel $\approx 41.45\ \text{N/mm}^2$, copper $\approx 11.40\ \text{N/mm}^2$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the rod states? A. Copper in tension, steel in compression B. Both in tension C. Steel in tension, copper in compression D. Both in compression Answer: A. B is pulled down so copper stretches; A lifts and steel is compressed. (Key point: Rod states) 2. Which moment equation about C is used? A. F_c(4) + F_s(2) = 40(2) B. F_c(2) + F_s(4) = 40(2) C. F_c + F_s = 40 D. F_c(2) = F_s(4) Answer: B. B is 2 m and A is 4 m from the hinge; the load is at B. (Key point: Equilibrium) 3. What is the deformation relation? A. delta_A = 2 delta_B (steel shortening twice copper elongation) B. delta_A = delta_B C. delta_B = 2 delta_A D. delta_A = 4 delta_B Answer: A. Rotation phi gives 4 phi and 2 phi. (Key point: Compatibility) 4. What is the steel stress? A. 11.40 N/mm^2 tensile B. 41.45 N/mm^2 tensile C. 41.45 N/mm^2 compressive D. 16.58 N/mm^2 Answer: C. F_s = 16.58 kN over 400 mm^2 = 41.45 N/mm^2. 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