P.1.9. A bar of mild steel has a diameter of 75 mm and is placed inside a hollow aluminum cylinder of internal diameter 75 mm and external diameter 100 mm; both bar and cylinder are the same length. The resulting composite bar is subjected to an axial compressive load of 1000 kN. If the bar and cylinder contract by the same amount, calculate the stress in each. The temperature of the compressed composite bar is then reduced by 150C but no change in length is permitted. Calculate the final stress in the bar and in the cylinder if E (steel) = 200,000 N/mm,E (aluminum) = 80,000 N/mm, a (steel) = 0.000012/"C and a (aluminum) = 0.000005/C. Answer: Due to load: o (steel) = 172.6 N/mm? (compression) o (aluminum) = 69.1 N/mm? (compression). Final stress: o (steel) = 187.4 N/mm? (tension) o (aluminum) = 9.i N/mm? (compression).