[Material Science] Thin Walled Cylinders & Yield Criterion

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SUMMARY

The discussion focuses on calculating the yield pressure for a thin-walled cylindrical boiler shell using various yield criteria: Rankine, Maximum Principal Strain, Tresca, and Von Mises. The shell has a diameter of 1.75 m and a wall thickness of 12 mm, with a yield strength of 300 MPa. The user, MarleyDH, initially miscalculated the yield pressure using the Von Mises criterion but later corrected the approach by applying the formula σyield = sqrt(3)·(P·D/4·t), leading to the correct answer of 4.75 MPa.

PREREQUISITES
  • Understanding of thin-walled cylinder theory
  • Familiarity with yield criteria: Rankine, Tresca, Von Mises
  • Basic knowledge of stress-strain relationships in materials
  • Proficiency in algebraic manipulation of equations
NEXT STEPS
  • Study the derivation of the Von Mises yield criterion in detail
  • Learn about the application of the Tresca criterion in practical scenarios
  • Explore the implications of material properties on thin-walled structures
  • Investigate computational tools for stress analysis in cylindrical shells
USEFUL FOR

Mechanical engineers, materials scientists, and students studying structural integrity of thin-walled components will benefit from this discussion.

MarleyDH
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Homework Statement



A long cylindrical boiler shell is 1.75 m in diameter and has a wall thickness of 12 mm. Treating the boiler as a thing shell, predict what internal pressures will produce yield in the shell according to the: (a) Rankine, (b) Max. Principal Strain, (c) Tresca and (d) Von Mises Criteria.

σyield = 300 MPa, E = 200 GPa, v = 1/3)

Homework Equations



The only issue I am having here is with the Von Mises criteria, which states for plane stress:

σ12 + σ1·σ2 + σ22 = σyield2

σhoop = P·D/2·t
σaxial = P·D/4·t

The Attempt at a Solution



So all good and well, I attempt the solution but get a value of 3.11 MPa, which is incorrect according to the answer. (4.75 MPa)

I then rechecked my results and did a google search. This lead me to a site which boiled the math down to a similar problem to:

σyield = sqrt(3)·(P·D/4·t)

When I use this equation, I get the correct answer.

I then reattempted my algebra and get to this:

(P·D/2·t)2 + (P·D/2·t)·(P·D/4·t) + (P·D/4·t)2 = σyield2

What am I missing here?

Any help would be appreciated.

Thank you,

- MarleyDH
 
Physics news on Phys.org
MarleyDH said:
σ12 + σ1·σ2 + σ22 = σyield2
Wrong. Read more carefully.
 
Ah right. Saw my mistake, the cross terms should be a preceded with a minus sign. Thanks for the assistance once again nvn. Now its on to failure criterion!
 

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