Compound bar under axial compression: Two bars of equal length, one steel (cross-section 3500 mm²) and one brass (cross-section 3000 mm²), are rigidly connected to act as a compound bar. The assembly is subjected to a total compressive load of 100,000 N. Take Es = 0.2 MN/mm² and Eb = 0.1 MN/mm². Assuming equal strain (compatibility) and elastic behavior, determine the stresses developed in brass (σb) and steel (σs).

Introduction / Context:
This problem tests compound (composite) bars in axial loading. When different materials are rigidly connected and loaded axially, compatibility of deformation requires equal strain in each material. Load sharing therefore depends on both area and modulus of elasticity.

Given Data / Assumptions:

• Steel area As = 3500 mm², brass area Ab = 3000 mm².
• Total load P = 100,000 N (compression).
• Elastic moduli: Es = 0.2 MN/mm², Eb = 0.1 MN/mm².
• Bars have equal length and are perfectly composite (same strain), linear elasticity holds.

Concept / Approach:

Equal strain implies σs/Es = σb/Eb. Hence σs = (Es/Eb)σb. Force equilibrium requires σs As + σb Ab = P. Solve these two relations for σs and σb.

Step-by-Step Solution:

Verification / Alternative check:

Check load share: Steel carries 203500 = 70,000 N; Brass carries 10*3000 = 30,000 N; Sum = 100,000 N ✔. Steel takes more load due to higher Es and larger area.

Why Other Options Are Wrong:

Other pairs do not satisfy both compatibility (ratio 2:1) and total load 100,000 N simultaneously.

Common Pitfalls:

Using area ratio only (ignoring modulus), or summing stresses instead of forces. Always apply both compatibility and equilibrium.

Final Answer:

σb = 10 N/mm², σs = 20 N/mm²