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 × 10⁵ N/mm² and α = 12 × 10^-6 K^-1.

Solution

Given Length of the rod, L = 10 m =10⁴ mm

Diameter of the rod, d = 5 cm = 50 mm

Change in temperature, ΔT = 150 - 50 = 100ºC

E = 2 × 10⁵ N/mm²

α = 12 × 10^-6 K^-1

(a)        When the ends do not yield (let the stress be σ₁)

Thermal stress, σ1, in the rod = E α ΔT

= 2 ×105 × 12 × 10- 6 × 100 = 240 N/mm²

(b)       When the ends yield by 1 cm (let the stress be σ₂)

 Thermal Stress, σ₂ = ( α ΔT -ΔL′ /L)  × E

= (12 × 10 ^-6 × 100 -10/10⁴) × 2 × 10⁵

= (0.0012 - 0.001) × 2 × 10⁵ = 40 N/mm²