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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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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Kształt izolinii naprężenia dla spoiny czołowej charakteryzują się one łagodnym przejściem
na linii: ścianka belki – spoina – blacha. Jest to wynikiem tego, że przy założonym w symulacji typie kontaktu, siły i momenty są przekazywane przez węzły elementów skończonych bez żadnych przeszkód. W przypadku spoiny pachwinowej nie ma trwałego kontaktu pomiędzy ścianką belki a blachą, zatem nie ma takiego typu kontaktu. Jest jedynie styk pomiędzy tymi elementami. To zjawisko właśnie jest podstawową przyczyną kształtów izolinii naprężenia pokazanych na rysunku 11.

Jeżeli w warunkach statycznego rozciągania następuje lokalne uplastycznienie się materiału, to  należy przypuszczać, że w warunkach normalnej eksploatacji o charakterze zmęczeniowym spoina pachwinowa w złączu teowym będzie pękać w jednym z trzech newralgicznych punktów. Zjawisko to przedstawiono na rysunku poniżej.

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Takie ćwiczenie:

o ile trzeba zmniejszyć wydatek w gazociągu, aby można rozgrzać strefę spawania do 100C ? 

W gazociągu płynie 1 000 000 Nm3/h.

Trójnik jest na lewo od namiotu.

Trójnik, aby go dospawać na gazociąg trzeba stosować ogrzewanie matami, bo w przepływ gazu w tak dużej ilości skutecznie odbiera ciepło na skutek konwekcji wymuszonej w strefie ograniczanej. Od strony zewnętrznej jest za to konwekcja niewymuszona do przestrzeni nieograniczonej oraz wymiana ciepła przez promieniowanie.

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Badania mikroskopowe wykazały, że kształty zaobserwowanych przekrojów pęcherzy, znacznie odbiegały od idealnych – koła lub elipsy. Postać ich była nieregularna z fraktalnym rozwinięciem powierzchni obwodowej. Ponadto zaobserwowano liczne pęknięcia rozwijające się od powierzchni tych pęcherzy.

Kruchość wodorowa związana jest z wnikaniem wodoru do metalu w postaci atomowej, gdzie na ściankach mikroszczelin np. powierzchni międzyfazowych siarczek/osnowa, ulega on rekombinacji w wodór cząsteczkowy. Generuje to  znaczne ciśnienie dochodzące do 1 GPa. W materiale uszkodzonych rur oszacowana zawartość wodoru dochodzi niekiedy do 100 mg/dm³. Rozpuszczony w żelazie wodór dyfundując może reagować z zawartym w stali cementytem. Reakcja ta przebiegająca w następujący sposób: Fe₃C+ 4H = CH₄ + 3Fe powoduje otrzymanie metanu. Wytworzone cząsteczki metanu ze względu na swoją wielkość nie mogą dyfundować w stali. Metan gromadząc się w pęcherzach powstających na granicach ziaren powoduje powstanie wysokich naprężeń, które doprowadzają do powstania pęknięć na granicach ziaren.

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