Determine the stresses within the rod:
Two parallel walls are stayed together through a steel rod of 5 cm diameter passing through metal plates and nuts at both ends. The nuts are tightened, while the rod is at 150ºC, to remain the walls 10 m apart. Determine the stresses within the rod while the temperature falls down to 50ºC, if
(a) The ends do not yield, and
(b) The ends yield by 1 cm.
Take E = 2 × 10^{5} N/mm^{2} and α = 12 × 10^{-6} K^{-1}.
Solution
Given Length of the rod, L = 10 m =10^{4} mm
Diameter of the rod, d = 5 cm = 50 mm
Change in temperature, ΔT = 150 - 50 = 100ºC
E = 2 × 10^{5} N/mm^{2}
α = 12 × 10^{-6 }K^{-1}
(a) When the ends do not yield (let the stress be σ_{1})
Thermal stress, σ1, in the rod = E α ΔT
= 2 ×105 × 12 × 10- 6 × 100 = 240 N/mm^{2}
(b) When the ends yield by 1 cm (let the stress be σ_{2})
Thermal Stress, σ_{2} = ( α ΔT -ΔL′ /L) × E
= (12 × 10 ^{-6} × 100 -10/10^{4}) × 2 × 10^{5}
= (0.0012 - 0.001) × 2 × 10^{5} = 40 N/mm^{2}