A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = $$2.0 × 10^{–5} K^{–1}$$, steel = $$1.2 × 10^{–5} K^{–1}$$ ).

Asked by Pragya Singh | 1 year ago |  120

##### Solution :-

Length of the brass rod = length of the steel rod = L0 = 50 cm

Diameter of the brass rod = diameter of the steel rod = 3 mm

Initial temperature ( T1) = 40°C
Final temperature ( T2) = 250°C
Therefore, the increase in temperature (∆T)

= 250 – 40 =210°C

Coefficient of linear expansion of brass, ∝ = 2 × 10-5 K-1
Coefficient of linear expansion of steel, β = 1.2 × 10-5 K-1

Final length of brass, L1

L1 = Lo(1 + ∝∆T)
= 50( 1 + (2 × 10-5 × 210))
=50( 1 + 420 × 10-5)
= 50× 1.00420
= 50.21 cm

Increase in length of brass ( ∆L)

= L2 – L1
= 50.21 – 50
= 0.21 cm

Final length of steel, L2= Lo(1 + β∆T)
= 50( 1 + (1.2 × 10-5 × 210))
= 50 × 1.00252
= 50.126 cm

Increase in length ( ∆L’)

= 50.126 cm – 50 cm

= 0.126 cm
Total increase in the length = ∆L + ∆L’
= 0.21 + 0.126
= 0.336 cm

Answered by Pragya Singh | 1 year ago

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