Thank you for sharing your nice work!

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I have a query in linearization of streses of shell and solid element. In Shell element, mid surface stresses are membrane and top/bottom surface stress are membrane + bending and peak stresses were not accounted.

But in solid element, stresses were linearized through the thickness and every point along the thickness, we will get M, M+B, peak stresses. Hence for solid element, providing M, M+B, M+B+Q at top and bottom surface is not required.

Confirm my understanding and provide me further clarity on this subject.

Thanks. ]]>

You stated that, values of deformations have no physical meaning for buckling analysis, then what it really signifies?

For your question, Whether code calculated Max Allowed External Pressure has some inbuilt factor of safety?

Refer STP-PT-029, Where knock down factor of 5.0 is used by ASME for effect of geometrical imperfections on axial compression. In addition, a design factor DF, of 2.0 is also added to account for reduced modulus in the inelastic range and variation of material properties that cause test to deviate from theory.

]]>Upon question 1, I would like to know how who fix the initial true stress, (Proportional limit. Adjust to get plastic strain) in your case 17.4 & 19.4 for SA-105 & SA516-70 respectively. This initial value would be used in determining true strain in accordance with Annex 3-D. Your guidance is appreciated.

]]>2. To determine the plastic collapse load, I’ll typically apply a high load and solve the model. The non linear solution will proceed by incrementing the load gradually from zero to 10%, 20% etc of full value. Finally beyond a point, say 70% load, the solution won’t converge. The collapse load in this case is 70% of applied load. In this methodology I am not using the tangent intersection or the twice yield method to arrive at the collapse load, rather I am getting the collapse load from the FEA directly.

3. The local acceptance criteria does not check for strength of the material, rather its a limit on the local deformations. Hence factor of 2.4 which is on strength is not applicable here. ]]>