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Compatibility Condition:
Total strain in steel = Total strain in copper
i.e. 3 × 10-4 + 1.965 × 10-5 Ps = 4.5 × 10-4 + 1.8421 × 10-5 Pc
Recognising that Pc = - Ps, and dividing both sides by 10-5, we get
30 + 1.965 Ps = 45 - 1.8421 Ps
or Ps = 45 - 30/ (1.965 + 1.8421) = 3.94 kN
∴ Stress in steel = 3.94 × 1000/ (π/4) × 182 = 15.5 N/mm2
∴ Stress in copper = - 3.94 × 1000/ (π /4)(302 - 182) = - 8.71 N/mm2
is shear stress distribution on I and T sections are same????
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