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Two copper, one steel rod: 5000 kg

Short answer

Compatibility gives σ_steel = 1.50 σ_copper. From equilibrium, stress in each copper rod is 202.10 kg/cm² and in the steel rod 303.15 kg/cm² (compressive).

Two copper rods and one steel rod is of 3 cm diameter, together support a load of 5000 kg as shown in figure below. Find the stresses in each rod. Take 'E' for steel and copper as 2×106 kg/cm22\times10^6\ \text{kg/cm}^2 and 106 kg/cm210^6\ \text{kg/cm}^2. [Figure: a rigid plate carrying a 5000 kg load rests on two outer copper rods (3 m long) and a central steel rod standing in a recess 1 m deeper.]

The rigid plate rests on three rods, so all rods shorten equally. The steel rod stands in a recess 1 m deeper: copper is 3 m long, steel 4 m.

Compatibility

σcLcEc=σsLsEs⇒σc(300)106=σs(400)2×106⇒σs=1.50 σc\frac{\sigma_c L_c}{E_c} = \frac{\sigma_s L_s}{E_s} \Rightarrow \frac{\sigma_c(300)}{10^6} = \frac{\sigma_s(400)}{2\times10^6} \Rightarrow \sigma_s = 1.50\,\sigma_c

Equilibrium

A=π4(3)2=7.069A = \dfrac{\pi}{4}(3)^2 = 7.069 cm² for each rod.

5000=σcA(2+1.50)⇒σc=50007.069×3.50=202.102 kg/cm25000 = \sigma_cA(2+1.50) \Rightarrow \sigma_c = \frac{5000}{7.069\times3.50} = 202.102\ \text{kg/cm}^2 σs=1.50×202.102=303.152 kg/cm2\sigma_s = 1.50\times202.102 = 303.152\ \text{kg/cm}^2

Answer: each copper rod 202.10 kg/cm²; steel rod 303.15 kg/cm² (compressive).

✓ Answer checked by Claude Opus 5.5 (high effort)

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