Determine the thermal stress in the bar, Civil Engineering

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 = σ1 (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.

Posted Date: 1/30/2013 5:57:51 AM | Location : United States







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