In thick plates, PSO occurs, during which the core of the element is "trapped" by the enormous mass of surrounding steel. The material inside wants to contract but physically cannot. This results in a triaxial stress state – the third axis in the thickness direction. In this state, shear stresses are blocked, which are responsible for "slippage" in the crystal lattice, i.e., plasticity. Since the steel cannot flow plastically, stresses increase until they exceed the tensile strength of the crystal lattice itself. This results in brittle fracture at far the lowest permissible temperature. The steel breaks suddenly and with a tremendous bang, and with minimal energy absorption, just like glass. Therefore, thicker plates force designers to choose steels with a higher guaranteed fracture energy. This rule only applies to structures operating continuously at low temperatures (bridges), so it does not apply to galvanizing tanks operating at the temperature where creep begins.

How do you assess what value of work of fracture is sufficient? The Charpy impact test is a test dynamic on a notched sample. From it, the fracture work is determined, which is included in the standards. In turn, the most important material feature from the point of view of fracture mechanics, i.e. the critical stress intensity factor at plane strain PSO (K_IC), is determined under the conditions statycznych using a fatigue crack. There is no purely analytical formula that connects these two values. However, since K studies_IC are very expensive and time-consuming, empirical relationships have been developed that allow estimating K_IC Based on the cheaper and faster Charpy test. Thanks to these formulas, having studied the work of fracture, the designer can estimate the K_IC and calculate the critical defect size (e.g., a microcrack) that will lead to the destruction of the structure or apparatus. In other words, fracture mechanics will not answer the question of whether a low fracture energy value at room temperature is the primary cause of failure. However, it could answer whether, for a given fracture energy, a given initial microcrack can develop into a full crack through the thickness of the tank sheet. However, we will not have such data.

Why doesn't the work of fracture alone determine the breaking stress? The yield strength tells us at what stress a material will crack. idealny (without defects) will begin to permanently deform. The work of fracture, on the other hand, tells us how much energy the material can absorb once it has has a defect (notch) and we hit it dynamically. The breaking stress is proportional to K_IC and to the size of the notch according to the generally known formula:

The relationship between fracture toughness and yield strength is usually inversely proportional. There used to be a relationship for structural steels that the harder and more durable the steel, the more brittle it is. Today, with sophisticated heat treatment methods and alloying additives, this relationship is no longer so obvious. There is no single, magic formula. You have to follow the path: KV – K_IC – determining the defect size – calculating the breaking stress. In the power industry, there is a strictly observed rule regarding the commissioning and shutdown of pipelines operating at critical parameters, which are relatively evenly heated along the circumference of the internal surfaces. In the case of a galvanizing tank (despite being a non-pressurized device), this rule is absolutely crucial for the safe operation of the device. This is due to the fact that very intense heat sources are distributed in only a few locations. Temperature unevenness leads to uneven deformation, which results in an uneven stress field. To minimize these unevenness, stops should be made to standardize the gradients.    

The figure below shows a graph of the simulation convergence over time. It shows that the greatest convergence problems occurred in approximately the first 20% of the time. What does this mean? Although the simulation is hypothetical, it does illustrate a possible failure scenario. During the initially uneven heating (i.e., the first hour or two), a deformation state (buckling of the tank's side surface) occurs, which bends the tank's arcs. The arcs could have been at temperatures of several dozen degrees Celsius, but their impact strength was still low enough that the stresses resulting from the deformation led to brittle fracture. Unfortunately, the author has no hard evidence for such a scenario, and never will.

Model of the galvanizing tank used for simulation.

Results for the 5th hour of heating, i.e. for 10% of the entire heating time of the bathtub.

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This section presents simulations for various load combinations of galvanizing bathtub 50 mm thickness

Load case

This section presents simulations for several different load combinations

• Operation temperature +450C

The FEM Model

The galvanizing tank is constructed of 50 mm thick steel sheet, grade S235JR+N / 1.0038, according to the PN – EN 100025 – 2 /1/ standard. The tank model and its support scheme were created based on the drawings below. The zinc brick models were created based on data from a manufacturer of grade Z1 zinc (HCM SHG 99.995). [1] Huty Cynku „Miasteczko Śląskie” S.A.[2]

[1] https://hcm.com.pl/oferta/#cynk-z1

[2] https://hcm.com.pl/

The base model

For this purpose, a basic model of all the essential elements of the bathtub and the zinc brick charge was created, shown below. Depending on the needs, this model will be cut into smaller portions to complete a given simulation batch within a reasonable timeframe, i.e., no longer than one day.

Tools for assessing stress state

Principal stresses were used to assess the stress level. These stresses, designated S1 to S3 in the simulation, have positive or negative values, indicating the nature of the material's work at a given point. Sign convention: positive: the material is in tension; negative: the material is in compression. Bending of the walls also occurs, during which one side of the wall is stretched (positive stress) and the other side is compressed (negative stress). Depending on which side of the tank is viewed (inside or outside) or at which point in the cross-section, different values ​​and signs will be observed.

Principal stresses are perpendicular stresses acting on the walls of an element rotated so that the shear stresses disappear.

S1 (Maximum Principal Stress) is the most "positive" stress. It shows the maximum tension at a given point. Even if S1 is negative, it means the element is compressed from all sides, i.e., a so-called triaxial compression state occurs. Maximum positive values ​​(red/yellow zones) indicate that these areas could fracture under tension. Blue zones are where the "maximum" tension is actually compression (or close to zero).

S3 (Minimum Principal Stress) is dominated by blue (negative values). It shows how strongly the material is being crushed, especially over supports.

S2 (Middle Principal Stress) is useful in shell structures because a flat sheet metal surface typically experiences a so-called biaxial stress state. This means that the material is being pulled or compressed in two directions simultaneously. A good analogy is a stretched drum membrane. One might think that since S2 is "medium," it can be ignored, but this is not the case. S2 is crucial in calculating the equivalent stress. Steel fails through shear (crystal slippage), and shear depends on the difference in stress. If S1 is large and positive, and S2 is large and negative, the equivalent stress will be large. If S1 and S2 are both large and positive, meaning stretching occurs in multiple directions, the equivalent stress will be smaller.

It is common to confuse principal stresses with Huber's reduced stresses.[1]The latter is always positive because it actually reflects the scalar energy of shear displacement. The reduced stress is excellent for assessing the strain in steel, as it indicates the moment when the steel transitions into the plastic state. In the language of linear algebra, a single principal stress is a stress tensor matrix. Looking at the principal stress field, we see a scalar number derived from the tensor field.

[1] Maksymilian Tytus Huber (1872–1950), an outstanding Polish scientist. In 1904, he published the hypothesis of the specific energy of shear deformation.

The results⁰C

The first principal stress, S1, represents the maximum tensile stress. The highest values ​​occur at the upper flange (flange) and at the corners. This is the result of stress concentration at the points of geometry change and the effect of hydrostatic pressure from the zinc, which "pushes" the tank. The value of principal stress S1 suggests that the material has likely exceeded its yield point and is flowing in these areas.

The second principal stress S2 acts perpendicular to S1 and S3. It usually describes the stresses along the plane of the wall [1]Negative values ​​(blue) on the long walls suggest that in certain directions the material is being “squeezed” by thermal expansion constraints from the rigid bottom.

The third principal stress, S3, represents the maximum compressive stress, which in this case primarily represents compressive forces. High compressive stresses can lead to local buckling of the walls if they are too thin, although in this case, this effect is unlikely to occur. However, it can occur in the final phase of the bathtub's service life, when the wall is much thinner than the initial 50 mm. Unfortunately, this combination is not a matter of dispute, as the failure occurred when the bathtub was new.    

[1] In pipeline design, this is the most important circumferential stress

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Let's consider the composite stiffness itself. In pipeline technology, it is used in the case of equipment on flexible support structures, such as those shown below. The stiffness determined for the equipment's nozzles in isolation from the entire device is a mistake. In an ideal situation, the stiffness of the nozzles should be issued by the manufacturer. Of course, this is a very rare case. If someone is interested in how to determine the stiffness of the nozzle themselves, I encourage you to read Part I - HERE

Due to lack of time I will not model such a complex support structure, which can be seen in the above photo. Let's assume that we have a much simplified support structure of the camera, as in the drawing below. For the description of the concept of complex stiffness it does not matter.

From the above figure, the directional stiffness of the equipment's frame can be determined, e.g. KFx = 1 kN/mm. We proceed in this way for all six degrees of freedom. For example, in the X direction, the stiffness KNx = 3 kN/mm.

Compound stiffness nozzle-equipment-foundation system is the equation of a simple relationship. For each degree of freedom, it is the reciprocal of the sum of the reciprocals of the stiffness of the stub pipe and the frame.

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PRINCIPAL STRESSES AND STRESS HYPOTHESES

Let us imagine a material element subjected to tension as shown in the figure below. The a – a plane is directed at an angle alpha to the direction of the force F acting on the plane A, generating a tensile stress S. The force F can be resolved into two components: F_n – normal and F_s – tangent to the plane a – a. These components cause normal stresses sigma and tangential stresses theta with values ​​described by the formulas below:

It is obvious that for a constant value of the force F, the normal and tangential stresses depend only on the angle of inclination of the plane. The maximum tangential stress theta will be for sin 2 alfa = 1 (which corresponds to the angle a = 45⁰) half the value of the tensile stress of the sample. The maximum of the normal stress sigma will of course occur for the angle alpha = 0⁰, and its minimum for angle alpha = 90⁰. Stresses in such planes are characterized by the absence of a tangential component and are called principal stress.

Traditionally, principal stresses have been determined using a series of simple constructions from analytical geometry called Mohr's circle, an example of which is given below.

Let us assume that the normal stress sigma_x and shear theta_xz are greater than zero and can be shown in the sigma-tetea coordinate system below. We put the value of the normal stress sigma_x on the s-axis, and at its end the tangential stress value theta_xz.

We draw a circle of radius R, the beginning of which is half the value of the normal stress sigma_x. We determine the maximum and minimum principal stresses, where the circle intersects at the points of zero tangential stress theta. In the last step, we determine the directions of their action. 

Principal stresses (here sigma₁, sigma_2, sigma₃₎ are the starting point for the two hypotheses used in the pipeline calculations.

The first one is the hypothesis of the specific work of purely shear deformation formulated and published in 1904 by the outstanding Polish professor M.T Hubera (1872-1950) more widely known as the hypothesis R. Misesa (1883-1953), who published it only nine (!) years later, in 1913. Both were under the resumption of the same ruler, Franz Joseph I, except that the former was a Pole born in a tiny town in Podhale, and the latter an Austrian living in the capital of the Empire in Vienna. Whether this was the theft of the century that the Austrian aristocrat got away with is difficult to decide now. However, there was something to fight for, because the hypothesis of the specific work of purely shear deformation, which was the crowning achievement of many European scientists throughout the 19th century, is the same as Maxwell's four equations for our electromagnetic civilization. This is the fudamental basis of the science of strength of materials.

The general form of the equation for the triaxial stress state is as follows:

The second hypothesis, the Tresca-Coloumba greatest tangential stress hypothesis, which postulates that the material's stress is determined by the material's achievement of a critical tangential stress. For simple tension, this value is half of the tensile stress. For a triaxial stress state, the stress value is: 

PRACTICAL USE OF THE MOHR CIRCLE

Let us assume a local coordinate system with the names characteristic of a pipeline system: longitudinal stress (stress along the pipe), circumferential stress (or circumferential stress on the circumference) and radial stress passing through the pipe wall. They are shown in the figure below.

In the next step, a cube was isolated from the pipe wall and the principal stresses were applied. We thus have the stresses: longitudinal sigma_L, shear theta , peripheral sigma_H and radial sigma_R.

In the next step we define the above stresses, which are the result of displacement, mainly thermal, and not internal pressure:

• Longitudinal: force / cross-sectional area and/or bending moment / bending modulus of cross-sectional strength
• on the wall thickness: negligible for thin walls
• Radial: Zero on the outer surface
• Shear: Shear force / 2 x shear section modulus.

From the above table it can be seen that if the cause is displacement, only the longitudinal and shear stresses are different from zero. Additionally, they are balanced because the system remains at rest. 

After being cut by a plane, the system remains in equilibrium. To determine the stresses in the new plane, we can use Mohr's circle construction. 

As we know, the principal stress occurs when the shear stresses are equal to zero. The circle intersects the ordinate axis at two such points, which are marked with stress pairs, at which the theta stress is equal to zero. Any angle phi of intersection with the plane can be found from the scheme below, and then the normal and shear stresses can be determined graphically.

The principal stresses in the displacement loaded pipe occur at three points from S1 to S3:

• S1 with the highest value corresponds to stretching
• S2 is equal to zero, which corresponds to radials
• S3 corresponds to compression    

From Mohr's circle, the maximum shear stress can be determined, the value of which can be determined by the simple formula below. In design standards, e.g. ASME B31.3, this is the strength limit of the material.

The above formula can be compared with the known standard dependence, which turns out to be identical in principle. The topic of standard calculations will be developed in the next thread 

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There is nothing more practical than a good theory!

Suppose we have a pipe or cylindrical pressure vessel that has a slot that is inclined to make it more difficult at a certain angle. To determine if the component can work, the stress intensity factor (SIF) must be determined and compared to the critical value for the material. The pressure is constant and there are no vibrations.

For stresses, the typical formulas apply:

The formula for a crack in a shell undergoing biaxial tension can be obtained from fracture mechanics tables. The coordinate system is attached at the center of the crack.

After substituting the shell stress formulas into the stress intensity factor formulas and using simple trigonometric transformations, we can finally write that:

Below is the calculation for a DN 250 pipe (273 x 10 mm) under pressure of 2 MPa.

Regarding the temptation to obtain K-factors by simulation. In Ansys at the current stage of its development there is a certain limitation, namely the crack must be perpendicular to the surface. In addition, a local coordinate system for the crack must be created, in which the X-axis is directed towards the crack axis. This can be cleverly achieved by selecting "Hit Point Normal" and clicking on the surface of the pipe.

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In the 1980s, Carrucci and Muller proposed a method for assessing the susceptibility of pipelines to AIV fatigue, which has been widely published. One of the main elements is a graph (valid only for a specific type of steel) estimating the range of safe operation of pipelines. It should be noted that according to it, the occurrence of the AVI phenomenon can be said to occur above 150 dB of acoustic power and for diameters greater than DN 250. This does not mean that in the case of a specific installation, the costs of which can be enormous for an emergency shutdown, these two boundary conditions should not be significantly reduced. As always, this depends on the knowledge and experience of the designer.

The whole problem is in correctly determining the value of the internal pressure of the acoustic wave generated inside the pipeline during the AIV phenomenon.

Below is a simple simulation of acoustic fatigue. This is a section of a pipeline fixed at both ends and subjected to internal pressure. The reduced stress in the pipeline sections varies with frequency. It has a maximum value at 1240 Hz.

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I thought I would do a simulation showing how a column behaves in a wind of 25 m/s - 90 km/h. Of course, I did not know the exact construction of the column, but it is about the principle and showing that it is possible. The result of the total deformation during 10 seconds of the simulation is shown in the graph below. From this graph, it can be estimated that the column was swinging with an average frequency of about 0.6 Hz. There are two videos of the simulation under these links: https://drive.google.com/file/d/1s0cpuLbniuHyTJmWCWa3-US9Q0NSmDm7/view?usp=drive_link i https://drive.google.com/file/d/1Nn5imJvNfpjzF_n8ZQs2kZFnc1a50PeJ/view?usp=drive_link.

Here is the speed of the wind around the column.

The natural frequencies of the column for the first six deformation modes ranged from 0.5 Hz to 13 Hz.

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Poniżej przykład projektu izolacji aparatu pracującego powyżej temperatury pełzania. Grubość ścianki to 100 mm.

According to the mechanical design, the temperature of the external surface of the device was assumed to be +4290C. The maximum possible insulation thickness was assumed to be 150 mm. The use of thicker insulation is not possible due to the specific design of the device and the shape of the connection stubs.
Wyznaczenie kluczowej danej, to jest współczynnika przejmowania ciepła z powierzchni poziomego aparatu przy konwekcji swobodnej w przestrzeni nieograniczonej wyznaczono według ogólnie znanych reguł teorii podobieństwa.

Due to the significant dependence of the thermal conductivity coefficient of mineral wool on temperature, the finite element method was used. The quality level of the mesh used on the external surface is close to unity, so the mesh error is negligible.  

Poniżej przedstawiono wyniki.

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Poniżej wygląd całego modelu odcinak rury DN250 (10′) z kompensacją o długości 56m.

Wykonano symulację rurociągu położonego na zewnątrz na estakadzie. Temperatura pracy 350C. Izolacja to wełna mineralna o grubości 100 mm, której współczynnik przewodzenia ciepła jest silnie zależny od temperatury. Użyto danych systemu otulin produkcji Rockwool. Medium to ciecz.

Jak wiadomo wyznaczenie temperatury wody na końcu długiego rurociągu można było kiedyś obliczyć bardzo pracochłonną w tym wypadku metodą kolejnych przybliżeń. Po zastosowaniu symulacji MES i zastosowaniu właściwych warunków brzegowych jest to w miarę proste, choć symulacja trwa godzinami.

Poniżej wygląd całego modelu odcinak rury DN250 (10′) z kompensacją o długości 56m. Wykonano może trochę za dużą ilość iteracji bo widać na ostatnim wykresie, że po 160 sekundach temperatura się ustaliła.

Co z tej symulacji można się dowiedzieć ? Mianowicie to, że skuteczność izolacji jest zaskakująco wysoka.

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