Dear Alok

May I ask you to fully explain your case. What is the cylinder and its task, what is its mechanical load and how ist the applied thermal load like.

A homogeneous temperatur distribution only effects the properties.

A non-homogeneous temperature distribution within the cylinder will cause additional thermal stress which must be taken into account. A typical example is a cylinder liner of a big Diesel engine. During operation these cylinders are quite hot at thier upper regions and releatively cold at the central and bottom region. In addition there is the mechanical load due to the pressure of the fresh and the combustion gases.

Regards Nikolaus

I can wait for your answer to Nikolaus' question. Meanwhile, I can add a comment. Thermals stress is residual stress i.e. self-equilibrated. It can be shown in limit analysis (and by experiment) that residual stress has no influence on plastic collapse of ductile material under monotonous loading. Thermal stress is only important for brittle materials, or can contribute to fatigue.

This is a theorem in limit analysis. In pressure vessels and piping, this theorem is confused with a definition. Stress is classified in primary and secondary stress. The difference is: primary stress contributes to plastic collapse, and secondary stress does not. Thermal stress is secondary (because it is residual).

Manfred Staat, sir could you please share some literature related to your above statement "It can be shown in limit analysis (and by experiment) that residual stress has no influence on plastic collapse of ductile material under monotonous loading".

Thanks in advance.

Naveen Bhatt

Hello Nikolaus Hannoschöck, Thanks for your quick response.

Here is my detailed case.

Its basically a compressor cylinder which is loaded with an internal pressure of 250 bar and it would be seeing a temperature of 150 deg C due to compression.

The question is, how to accommodate this thermal effect in "Static structural Analysis" in ANSYS.?

If I include this temperature as a "thermal condition", it seems that the results are not looking right.

https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.1054.3916&rep=rep1&type=pdf

ASME Section III Design-ByAnalysis Criteria Concepts and Stress Limits.

ASME deals with primary and secondary stresses differently, specially for boilers and pressure vessels. Thermal conditions are secondary in nature, without any external force / moment. Thus, typical results from FEM under thermal conditions should not be directly addes as linear superpositions, to the stresses / strains due to physical / static loads. Starain based design of ASME gives some relief under such combined actions. Refer, above link for a technical paper, and ASME code (Section III), for more details.

Dear Naveen Bhatt, I have no access to the book in which I remember the theorem that self-equilibrated stress (such as thermal stress) does not contribute to plastic collapse under monotonous loading. I must be precise: Limit analysis does not apply to instability problems (buckling). Thermal stress can lead to buckling. We do not prove the theorem in our paper, but we refer to a classical experiment by Maier-Leibnitz. In this experiment the residual stress occurs because the bridge has too many supports so that it is statically indeterminate.Article LISA - A European project for FEM-based limit and shakedown analysis

Manfred Staat sir, thank you very much for your reply and explanation.

I could use an example to explain, that residual stress, load history and Young's modulus are all irrelevant for limit analysis. Consider a frame constructed from some beams so that it is n-times statically indeterminate (or n-times redundant). There is residual stress from the start, and residual stress is changed during loading. When load is increased, a section yields and finally becomes fully plastic if load increases further. The fully plastic cross-section is a plastic hinge, which reduces the redundancy to n-1. The stiffness of the frame is reduced.

When load is further increased, more and more plastic hinges develop, each reducing redundancy and stiffness. When redundancy is zero, the frame is statically determinate. This means that all stresses can be calculated from the equilibrium conditions alone. Therefore, there is no residual stress in a statically determinate structure and the stresses are independent of Young's modulus and load history. Everything what happens now is also independent of these quantities.

When load is now increased, a next plastic hinge develops and the frame becomes a mechanism with 1 degree of freedoms. The stiffness is zero and the displacement becomes infinite. This is the plastic collapse in limit analysis.

Manfred Staat

Sir, thank you very much for the explaination and I could understand it up to good extent.

In para 2 (When load is further increased.........also independent of these quantities) there is some gap in my understanding, I request you to please explain:

Let say at load F (structure is indeterminate)

• some of the beams have become totally plastic (means they are not taking any further load and are at stress level equal to yield strength),
• in some beams, yielding has started at outer surface (but they are ready to take further load and with every load increment their yielded zone will expand)
• remaining beams are elastic.

On increasing the load, this structure becomes determinant (from indeterminate)

• the beams which have become plastic hinge will not take any further load.
• but the beams which are not plastic hinge yet are at some stress state(s) and I think it is unknown

Let say load is increased by P now (Total load F+P). This increment P will get distributed among the non plastic beams (as they only can take any further load and it will be distributed among them) and change in stress state after becoming determinant could be calculated. But I could not get how the exact stress state will be determined (because it includes those stresses also which occured when structure was indeterminate)?

I could also not get "there is no residual stress in a statically determinate structure and the stresses are independent of Young's modulus and load history", why so?

Thanks in advance.

Regards

Naveen Bhatt

Consider your structure as a rigid body. Then you have the same number of equilibrium conditions as the degrees freedom (dof) of the free body. In the plane the dof is 3 and the equilibrium conditions are 2 sum of forces 1 sum of moments. Now you can reduce 3 dof by supporting the structure. Then it is statically determinate. This means that all 3 reactions at the supports can be calculated from equilibrium. If you want to know internal forces, moments or stress you cut at the section in which you need such a result, and you get it from equilibrium alone. The Young's modulus or load history cannot change the result because you have the result already. And the result is unique so that no residual stress can change it.

If you add 1 further support condition, the structure is statically indeterminate. There are unknown residuals stresses. Residual stress is in equilibrium with zero mechanical load, i.e. it is self-equilibrating. The indeterminate problem can be solved by considering the elastic deformation and depends on Young's modulus therefore. However, the solution of the indeterminate problem is not unique. You could add a solution of a zero load with its residual stress without changing the equilibrium of the indeterminate structure. Different load histories may lead to different residual stress.

Now back to the increasing load. 1 plastic hinge (fully plastic section) reduces the number of unknowns by 1 because now the plastic moment in this section is known. Then the structure is statically determinate and residual stress must disappear because there is only the unique equilibrium solution, which cannot depend on Young's modulus or load history (otherwise it would not be unique).

In further load increase, the plastic hinge still carries the plastic moment, which is constant for perfect plasticity. Elastic deformations increase, but keep the total deformations still small. If a next cross-section becomes fully plastic, the structures is a mechanism with 1 dof and large rotations in all plastic hinges, leading to plastic collapse. 1 dof means that FEM fails because a rigid body motion occurs, and the stiffness matrix becomes singular.

Manfred Staat

sir, once again thank you for your detailed and beautifully written reply.

In second last para (Now back .... be unique).), there is a sentence "Then the structure is statically determinate and residual stress must disappear because there is only the unique equilibrium solution". Does it mean residual stresses will have no effect on equilibrium solution (not that residual stress will disappear in the actual structure)...am I right?

Regards

Naveen

There is no residual stress in a statically determinate structure, and therefore residuals stress has no effect on the plastic collapse (this is different in buckling problems). Residual stress can occur only because of statically indeterminacy.

Consider a statically determinate bar. Heating will cause no thermal stress because the bar can deform freely. Now put the bar between 2 rigid walls. It becomes statically indeterminate and thermal stresses occur because the walls restrict deformations.

Removing the walls removes the thermal stress.

All this discussion holds approximately in real structures. Think of a thick-walled pipe. It is indeterminate, can have serious thermal stresses and stress analysis is the solution of a system of PDEs (by FEM or as approximate analytical solution).

A thin-walled is statically determinate, has a simple analytical solution of the equilibrium conditions, and has no thermals stress - all this in the limit of the wall-thickness going to zero.

But all discussed theorems of limit analysis hold exactly in the theory of classical plasticity for all structures.

Manfred Staat

Sir, I agree with you explanation above and can understand the disappearing of resudual stresses on removing the regid walls. But will it be true for every case?

My understanding of residual stress on a bar I have explained in the attached figure (assumed bending pattern of residual stress with magnitude equal to applied stress). Average resultant stress on bar will remain unchanged and will be unique also but distribution should change.

Regards

Naveen Bhatt