--- title: "Steel rod with two brass rods: stresses" subject: "Strength of Materials" chapter: 3 key: "semester_3/strength_of_materials:steel-brass-rods-60kn" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/steel-brass-rods-60kn site: BCT Engineering --- # Steel rod with two brass rods: stresses Strength of Materials · Chapter 3: Simple Stress and Strain - Asked in: 2075 Chaitra, 10 marks - Video: [Stresses in steel and brass rod/Elongation in rods/ simple stress and strain/strength of materials](https://www.youtube.com/watch?v=fo0qU2gNEYw) — Prof.Dr.Pravin Patil - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/steel-brass-rods-60kn - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3 ## Question A steel rod of cross sectional area $1000\ \text{mm}^2$ and two brass rod each of cross sectional area $800\ \text{mm}^2$ together support the load of 50KN. Calculate the stresses in the rod. Take E for steel as 200GPa and E for brass as 100 GPa. [Figure: a rigid plate carrying the load (marked 60 kN in the figure, 50 kN in the text) is supported by a central steel rod and two brass rods; the steel rod is 200 mm long and the brass rods are 250 mm long, standing on a common base.] ## Short answer Stresses are proportional to load. For 50 kN: σ_steel = 30.49 MPa, σ_brass = 12.20 MPa; for 60 kN: σ_steel = 36.59 MPa, σ_brass = 14.63 MPa. All stresses are compressive. ## Answer The rigid plate rests on all three rods, so each rod shortens by the same amount: $\delta_s = \delta_b$. The problem states 50 kN in the text (the figure shows 60 kN); the solution is worked for 50 kN, and the 60 kN values follow in proportion. #### Compatibility $$ \frac{\sigma_s L_s}{E_s} = \frac{\sigma_b L_b}{E_b} \Rightarrow \frac{\sigma_s(200)}{200\times10^3} = \frac{\sigma_b(250)}{100\times10^3} \Rightarrow \sigma_b = 0.40\,\sigma_s $$ #### Equilibrium $$ P = \sigma_sA_s + 2\sigma_bA_b = \sigma_s\left[1000 + 2(800)(0.40)\right] = \sigma_s\,(1640) $$ $$ 50\times10^3 = 1640\,\sigma_s \Rightarrow \sigma_s = 30.488\ \text{N/mm}^2,\quad \sigma_b = 12.195\ \text{N/mm}^2 $$ Load carried: steel 30.49 kN, each brass rod 9.76 kN (total 50.00 kN). **Answer: for 50 kN, $\sigma_{steel} = 30.49$ N/mm² and $\sigma_{brass} = 12.20$ N/mm² (compressive).** If the load is 60 kN, the stresses become 36.59 N/mm² and 14.63 N/mm². ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What condition applies to the rods under the rigid plate? A. All rods shorten equally B. Each carries equal load C. Each has equal stress D. Steel carries all the load Answer: A. The plate stays flat, so delta_s = delta_b. (Key point: Compatibility) 2. What is the relation between brass and steel stress? A. sigma_b = 2.5 sigma_s B. sigma_b = 0.40 sigma_s C. sigma_b = sigma_s D. sigma_b = 0.5 sigma_s Answer: B. sigma_s L_s/E_s = sigma_b L_b/E_b gives sigma_b = 0.4 sigma_s. (Key point: Stress ratio) 3. Which equilibrium equation is used? A. P = sigma_s A_s + 2 sigma_b A_b B. P = sigma_s A_s + sigma_b A_b C. P = 2 sigma_s A_s + sigma_b A_b D. P = sigma_s A_s - 2 sigma_b A_b Answer: A. There is one steel rod and two brass rods. (Key point: Equilibrium) 4. For 50 kN, what is the steel stress? A. 12.20 N/mm^2 B. 36.59 N/mm^2 C. 30.49 N/mm^2 D. 50 N/mm^2 Answer: C. 50000/1640 = 30.49 N/mm^2; brass 12.20. 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