Determine the thermal stress in the bar:

A stepped bar made of aluminum is held between two supports. The bar is 900 mm in length at 38ºC, 600 mm of which is having a diameter of 50 mm, while the rest is of 25 mm diameter. Determine the thermal stress in the bar at a temperature of 21ºC, if

(a) The supports are unyielding, and

(b) When the supports move towards each other by 0.1 mm.

Given E for aluminum = 74 kN/mm²

α for aluminum = 23.4 × 10^-6 K-1

Solution

When the Supports are Unyielding

We have, L₁ = 600 mm; diameter d₁ = 50 mm

Thus,  A₁ = π(50) ²/4 mm2

Also,   L₂ = 300 mm; diameter d₂ = 25 mm

Therefore, A₂ = π(25) ²/4 mm2

and     ΔT = 38 - 21 = 17ºC

Therefore, free contraction, ΔL = L α ΔT = 900 × 23.4 × 10^{- 6} × 17 = 0.358 mm. Let σ₁ be the stress in 50 mm φ part and σ₂ be the stress in 25 mm φ part.

Then,

σ₁ =   ΔTEA₂ αL/A₁L₂ +A₂L₁ = 900 × 23.4 × 10 ^-6  × 17 × 74 × 103 × π × (25) ²/4(π × (50)2/4× 300 + π × (25) 2/4 × 600)

= 14.72 N/mm²

σ₂ = σ_{1 (}A₁ + A₂) =14.72 × (50) ²/ (25) ²= 58.88 N/mm²

These stresses are tensile in nature

When the Supports Move towards Each Other by 0.1 mm

Here,  ΔL´ = 0.1 mm.

∴          σ₁ =   EA₂ [ΔL - ΔL′]/ A₁L₂ + A₂ L₁ = 74 × 10³ × π × (25) ² × (0.358 - 0.1)/4(π × (50) ²/4 × 300 + π × (25) ² /4× 600)

= 10.61 N/mm²

∴          σ₂ = 10.61 × (50) ²/ (25) ² = 42.44 N/mm²

Both these stresses are tensile.