A bar of brass 20 mm is enclosed in a steel tube of 40 mm external diameter and 20 mm internal diameter. The bar and the tubes are initially 1.2 m long and are rigidly fastened at both ends. If the temperature is raised by 60C, find the stresses induced in the bar and tube.
Given:
Es= 2 × 105 N/mm2
Eb= 1 x 105 N/mm2
Qs= 11.6× 10-6/C
Cb= 18.7 x 106/C
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Explanation:
Given A bar of brass 20 mm is enclosed in a steel tube of 40 mm external diameter and 20 mm internal diameter. The bar and the tubes are initially 1.2 m long and are rigidly fastened at both ends. If the temperature is raised by 60 C, find the stresses induced in the bar and tube.
- Given Es = 2 x 10^5 N/mm^2
- Eb = 1 x 10^5 N/mm^2
- αs = 11.6 x 10^-6 / deg C
- αb = 18.7 x 10^-6 / deg C
- Also given length of the entire machine as 1200 mm
- Now A steel = π / 4 (40^2 – 20^2)
- = 942 mm^2
- A brass = π / 4 x 20^2
- = 314 mm^2
- So there will be a compressive force and tensile force.
- Under the equilibrium condition Rb = Rs (by taking R)
- Changes in length of steel and brass bar will be Δs and Δb
- So Δs + Δb = t L (αb – αs)
- Therefore R L (1/As Es + 1 / Ab Eb) = t L (αb – αs)
- R x 1200 (1 / 942 x 2 x 10^5 + 1/314 x 1 x 10^5) = 60 x 1200 (18.7 x 10^-6 – 11.6 x 10^-6)
- = 60 x 1200 x 7.1 x 10^-6
- 1200 R x 1/10^5 (1/942 x 2 + 1/314) = 511200 x 10^-6
- R x 1200 / 10^5 (1256 / 295788) = 511200 x 10^-6
- 1507200 R = 15120682560
- R = 10032.3 N
- Stress in steel = R / As
- = 10032.3 / 942
- = 10.65 N/mm^2
- Stress in brass = R / Ab
- = 10032.3 / 314
- = 31.95 N/mm^2
Reference link will be
https://brainly.in/question/19956716
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