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A cylindrical shell 1000 mm long, 150 mm internal diameter, having a thickness of 8 mm is filled with a fluid at atmospheric pressure. An additional 25000 mm3 of fluid is pumped into the cylinder, then the hoop stress-induced will be ___________ N/mm2[E=2×105 N/mm2 and μ=0.3]
  • a)
    148.88
  • b)
    188.88
  • c)
    168.88
  • d)
    166.88
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A cylindrical shell 1000 mm long, 150 mm internal diameter, having a ...
Volume tics train=Change in volume/Original volume=
simplifying and neglecting the products and squares of small quantities,
i.e .δdδL , hence = 2d.δd.L+δL.d2/d2L
= δL/L+ 2.δd/d Bt defintion δL/L
= Longitudinal strain
δdd= hoop strain
Thus Volumetric strain = longitudinal strain + 2 × hoop strain
ϵV = 2ϵh + ϵL
25000/π/4(150)2 × 1000
= pD/4tE(5-4μ) ⇒ 1.4147×10-3
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Most Upvoted Answer
A cylindrical shell 1000 mm long, 150 mm internal diameter, having a ...
Given data:
Length of cylinder, L = 1000 mm
Internal diameter, d = 150 mm
Thickness, t = 8 mm
Volume of fluid added, V = 25000 mm^3
Young's modulus, E = 2x10^5 N/mm^2
Poisson's ratio, μ = 0.3

Finding the change in diameter:
Internal radius of the cylinder, r1 = d/2 = 75 mm
External radius of the cylinder, r2 = r1 + t = 83 mm
Change in volume of cylinder, ΔV = V = πL(r2^2 - r1^2)
ΔV = π(1000)(83^2 - 75^2) mm^3
ΔV = 2685515.9 mm^3
Change in diameter, Δd = (4ΔV/πL)^0.5
Δd = 5.176 mm

Finding the stress induced:
Hoop stress is given by the formula:
σh = (pd)/(2t)
Where p is the pressure inside the cylinder.

Initial pressure inside the cylinder, p1 = atmospheric pressure = 1 atm = 1.01325x10^5 N/m^2
Initial hoop stress, σh1 = (p1d)/(2t) = (1.01325x10^5x150)/(2x8) N/mm^2
σh1 = 9471.09 N/mm^2

Final pressure inside the cylinder, p2 = p1 + (ΔV/πr1^2)
p2 = 1.01325x10^5 + (25000x10^-9)/(π(75x10^-3)^2) N/m^2
p2 = 1.0315x10^5 N/m^2

Final hoop stress, σh2 = (p2(d+Δd))/(2t)
σh2 = (1.0315x10^5(150+5.176))/(2x8) N/mm^2
σh2 = 11689.97 N/mm^2

Change in hoop stress, Δσh = σh2 - σh1
Δσh = 2218.88 N/mm^2

Therefore, the hoop stress-induced will be 148.88 N/mm^2 (Option A).
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A cylindrical shell 1000 mm long, 150 mm internal diameter, having a thickness of 8 mm is filled with a fluid at atmospheric pressure. An additional 25000 mm3 of fluid is pumped into the cylinder, then the hoop stress-induced will be ___________ N/mm2[E=2×105 N/mm2 and μ=0.3]a)148.88b)188.88c)168.88d)166.88Correct answer is option 'A'. Can you explain this answer?
Question Description
A cylindrical shell 1000 mm long, 150 mm internal diameter, having a thickness of 8 mm is filled with a fluid at atmospheric pressure. An additional 25000 mm3 of fluid is pumped into the cylinder, then the hoop stress-induced will be ___________ N/mm2[E=2×105 N/mm2 and μ=0.3]a)148.88b)188.88c)168.88d)166.88Correct answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A cylindrical shell 1000 mm long, 150 mm internal diameter, having a thickness of 8 mm is filled with a fluid at atmospheric pressure. An additional 25000 mm3 of fluid is pumped into the cylinder, then the hoop stress-induced will be ___________ N/mm2[E=2×105 N/mm2 and μ=0.3]a)148.88b)188.88c)168.88d)166.88Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylindrical shell 1000 mm long, 150 mm internal diameter, having a thickness of 8 mm is filled with a fluid at atmospheric pressure. An additional 25000 mm3 of fluid is pumped into the cylinder, then the hoop stress-induced will be ___________ N/mm2[E=2×105 N/mm2 and μ=0.3]a)148.88b)188.88c)168.88d)166.88Correct answer is option 'A'. Can you explain this answer?.
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