Example of Partially Restrained Bar:

Example:

Two parallel walls are stayed together by a steel rod of 5 cm diameter passing through metal plates & nuts at both ends. The nuts are tightened, while the rod is at 150ºC, to keep the walls 10 m apart. Calculate the stresses in 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 - (10/10 ⁴) × 2 × 10⁵

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