Material Properties and Testing: A Comprehensive Guide
.2.5Indicates the parameters that define the behavior of a plastic material: proportional elongation at fracture, defined by which is reached in the time of rupture of the specimen: A (r) = ((L (r)-L (0)) / L (0)) * 100. RESTRICTIONS: defined as the proportional decrease in cross section which has been localized fracture: E (of the sum) = ((S (0)-S (r)) / S (0)) * 1002.6. The following table illustrates three resistant materials with their characteristics. Justify: a) What is the highest ductility? B) What is the most stubborn? C) What would present more harshly? Material C.Rotura L.Elást Elong A 450 390 301 B 200 150 40 C 400 390 5 a) The solution is B because it has more alargamiento.B) The solution is because it has a high load. C) The solution is A. The higher hardness of a material is given by the product (elongation break load *) 2.7. Why in the trial of Rockwell C scale was Kp 150 load application sequence is 10 + 140? Because in all trials HR applies a preload of 10 kg as a reference L (0) 2.8. What are the causes that can not apply the theory of elasticity materials working at high temperature? A material when working at high temperature, or enters CREEP flows. You can not apply the theory of elasticity of a material when subjected to creep flow state because there is no elastic period because the curve that measures tension versus deformation has a slope of nearly 90 degrees so E = tan (90 degrees) = infinity with time everything is plastic.. 2 9Indicates that precautions be taken in designing a material with low toughness. The tenacity suffers a sharp diminution in value where the temperature drops below -10 degrees centigrade. Ensure that the material about certain values of Rm and A (percentage).2.10. Indicates that parameter depends on the level of tensions chosen to achieve a given service.Of the tensions engendered, the kind of work, frequency, provided the number of cycles without failure. 2.11. Can we recognize through the analysis of a fracture of the workpiece, the type of service that you had? Yes, because on the outside of the fracture is fibrous or kill zone, which will be greater the higher the toughness of the material, as this area indicates that has been absorbed in the fracture energy, ie the ductility of the same . Inside there is a bright crystalline area with geometric drawings in which no energy is absorbed and the material has become brittle. Given the dimensions of each zone, one can deduce the type of tension generated. 2.12. Justify the parameters that define the type of impact test. The primary parameters that define the field of resilience are: a) Impact velocity V in the test tube, cargo hub and height. B) kinetic energy at the point of reaching the specimen Eq These parameters are a function of test variables pendulum (M, H alpha) through the expressions: V = SQRT (2 * g * h) E = m * g * h 2.13. Explain if you could sort through the observation of the fracture if a material responds with a high or low toughness. If, in all sections of fracture are clearly differentiated two types of fractures: bright crystalline fracture, with geometric planes, and fibrous mate. Crystalline type fractures are achieved with low energy absorption. Type fractures gray, woody texture, have the highest energy absorption or resiliencia.La resilience is directly proportional to the energy absorbed before fracture. With the U-shaped notch, the energy absorbed is greater than the V-notch, so that their resilience will be greater. Resilience also increases the higher the radio background, high tenacity, high ductile area. 2.14. Justifies the possibility of calculating toughness values by extrapolation to the field of the tested temperatures below: On the correlation of resilience to the test temperature for Charpy test pieces, there is a sharp drop in the level of resilience between 0 and -20 degrees Celsius to the point of presenting a completely brittle behavior. Resilience less steep climbs between these grades and then the slope increases. 2.15. Justify the reasons for the correlation between hardness Rockwell, Brinell and Vickers, with the parameters of the resilience indicators traction. In all three trials, with increasing hardness, yield strength, R, and M tb increase and “epsilon” decreases (and vice versa). 2.16. Discuss the advantages and disadvantages between Rockwell hardness test, Brinell and Vickers. Brinell: For large deformations and soft materials. The drawback is that in this type of test, the elastic deformation is less appreciated. Rockwell and Vickers: For small deformations and stronger materials. Drawback, we must prepare the surface, so he makes a preload. 2.17. He mentions the parameter with which could correlacionars reversal of the micrometer needle when the machine is canceled Rockwell load actuacion Home: By eliminating the load on the material that is recovered is elasticament and permanent deformation. The parameter is the recovery elastik 2.18. And justifying interpreted as the degree of toughness of a material from the observation of its fracture in a Charpy test: Crystalline type fractures are achieved with low energy absorption. Type fractures gray, woody texture, have the highest energy absorption and resilience.4.1. Justified because the alpha Fe (cc) s + q Fc gamma resistant (ccc). The alpha Fe (cc) has 4 sist d slip and gamma is 12 and therefore greater ability to yield 4.2. Justifies the paper d vacancies and dislocations in hardening d metals and alloys. Q favor plastic deformation gives pungency to metals and alloys, the mechanism is qThe atoms are sliding one to one occupying the vacancy left by the front of displacement 4.3. Explain the causes of landslides observed in the lab during the traction test d strain. The lines start at deslizamient d single crystals with dense sist d slip located at 45 deg dl dl D applied stress axis 4.5 can justice in the fact dq only a few grains of polycrystalline structure d … In a polycrystalline structure orienatcion d the grain is random, this means q can be oriented to any direction when the grains esfuerzosson pequenyos q q thrills are your guidance this close to 45 deg and q require little effort to stay in that orientation 4.6. A slip rear polished appearance d1metal dvuelv EBT NATURL eliminating roughness as warranted. The piece d polishing is to remove the piece d roughness after sliding the probe will have a rough surface by sanding pass the test piece in a direction and varying the same in 90 deg cda time to avoid breaks then will put it in the buff and this we will return an original external APPEARANCE d ia piece and the specimen sd q smooth surface to the top 4.7. Explain the cause justifies q dislocations n d l increase correlation with the plastic def. The main cause is the propagation of dislocations d theory this theory is based on the q disocacion an effort to expand the center leaving the other displacement d q is generated to repeat the ciclo.Cuantas more dsloccions have more plastic deformation will suffer the Material 4.8 . Explain how these characteristics influence resist q n have bean tuners d q processes are applied d d casting parts. Increase speed to facilitate the creation nucleation d d d minor tamany equiaxed grains 4.9. Describes destructivoy other non-destructive method to determine the state of tension mater. The destructive through bites d corrosion is a chemical attack surfaces metalicasq appropriate 1and is magnified in the intersection d pts d dislocations on the surface of the metal and not destructive the microhardness d 4.10. Reasonably induce the unranked q argments the proces grain refining d d d hardening as 1 technician. At the edges d grain dislocations are anchored at fixed sliding rising edges d grain, fine grain tamany d, d signific plasticity limit and hence single crystals comportamient d alloy hardened 4.11. consec d the dif plastic metals and alloys in terms evaluators N indices d d d endurecimiento.Como consec have the dif plastic hardens q absolute indicator strain increases achieved by Le = Lec/Leo-1, the indicator of Intrinsically d d hardening on the alloy deformed (elongation decreases) a = 1-Ai/Ao, if the indicator degree d d d hardening on the alloy deformed intrinsic increases (increasing Le and sigma sub r) 4.12. Concept d crystalline texture. D is the preferential orientation of certain crystallographic directions oriented application of esfuerzola Select d d quality orientation is a function of type d and d crystal structure conformation plastic made while the degree d q s orientation function of degree d deformation reached 4.13. Mechanisms twinned maclado.El d hardening hardening is a mechanism that enhances the inhibition of sliding by partition plasticopor flow of grain and 4.14 block metal edges obtained by processes d def .. The identification of this stage of annealing can be bitterly against it from hardening remaining d measures either from the heat flux by the decrease of internal energy or by medid to the grain shape and dimensions of 4.15. Stages of annealing vs riddled. Microstructure. You 3etapas.1-recovery in q is a slight decrease d the hardness without conformational change in crystals-2 d-q denotes the recrystallization characteristics d loss achieved in the acrimony and polycrystalline structure reconstrucció d – 3-d grain thickening, expands alternative treatment time d q lowers slightly resistant characteristics ngrosmient d result of grain has taken place q 4.16. Influence Timpo d nl d temperature recrystallization. Time and temperature have a correlation with. Exponential to the inverse being more sensitive to time variation d d recrystallization temperatures and show a correlation with bitterness. Inverse hyperbolic type acrimony d d influences inversely with the time required to produce the recrystallization 4.17. Effect of the recrystallization time required n variables. The increase in deform. Acrimony, and the recrystallization temperature d act in the sense d reduce the time required to recrystallize the whole mass 4.18 recristalizado.Para the same conditions of annealing and deformation tamany d one favors the recrystallized grain pequenyo more pequenyo and vice versa 4.19. Effect of time and the temperature at large d thickens. The grain tamany d d s growing exponentially with time d d grain thickness and the annealing temp d-d with time after recrystallization decreases sharply and the temperature (graphic) 4.20. Defin l l proces xa reduce grain tamany pro med acritu one alloy. The fine grain is obtained after successive deformation process sharply d recris maxima and without step d d grain thickening. D achieved after successive deformation processes with maximum sharpness and recrystallization without thickening step d 5.1. indicates the differences between homogeneous nucleation and heterogeneous nucleation heterogenea.La occurs when there is a lower critical radius of the embryo, this nucleation is formed attached to the surface extrañosolido nucleus in the liquid mass. Tuners are used grain. The critical radius in the homogeneous nucleation the nuclei are formed homogeneously and used grain tuners. 5.2. Indicates the parameters or conditions that facilitate the formation of dendritic structures: are usually characteristic of parts obtained by solidification of pure metals, also justified by the fastest growth of embryos in the preferred direction dela crystal structure, the size of dendrites depends on the cooling rate, the higher the smaller cooling rate of dendrites, the size of the dendrites will be diminished if used tuners grain, the number depends on the rate of nucleation of stable embryos. 5.3. Indicates areas where it is most likely find equiaxed grains and their causes: This type of grain is distributed in the central area of the surface of the casting and which are the first to solidify due to its higher cooling rate. 5.4. How can we encourage the isotropy of the pure metal castings? Tuners Adding extraneous grain or kernels tend to shape parts with structures more equiaxed structure favors isotrópicas.Tb isotropy. 5.5. Explain the conditions for an odd nucleus is a tuner of grain on loa must act on dendritic and equiaxed grains during solidification to reduce both types of grain, it must allow heterogeneous nucleation in the solidification of metals. They are usually transition elements or ionic form intermetallic compounds, carbides, nitrides, borides and finally present an isomorphic structure and atomic radius not too different. 5.6. Identification phase of the methods used briefly unaaleacion.Comentar “Metallographic Techniques: the existing phases displayed the temperatures are stabilized, are made at room temperature, in principle only seen those phases that are stable at that temperature, you could make a quenching from the temperature observed previously provided this does not result in other phase changes, can also be used microscopy with heating stage for observations at temperatures elevadas.-Chitosan: It allows us to quantify the temperature-resistant characteristics ambiente. X-ray diffraction: Identifies the stable phase by comparison of the characteristics obtained Expectra. 5.8. justifies the reasons why they can not exist aleen itersticialmente qse metals, with total soubilidad solido.Es been mainly due the similar atomic size (the atoms do not fit in the interstitial) 5.9. Anticipate problems can find us by heating an alloy below, but next, the solidus line, if this has been obtained by casting at high cooling rates: “That appears liquida.-Segregation dendritic phase 5.10. Explains the phenomenon of dendritic segregation: Rapid cooling out of equilibrium, it generates a higher temperature range in which the liquid and solid are present at the same solidification occurs last tiempo.La at a temperature lower than that predicted by the equilibrium diagram. The last liquid to solidify will have a higher concentration of metal that has a lower temperature fusion.Cuanto higher the cooling rate, the greater are the effects mentioned. 5.11. Describes the coring effect: Is the difference of concentration in the Componenet successive layers from the core to the exterior of a grain monofasico.El grain is richer in the core component of higher melting point and the lower crust melting point. 5.12. Characteristics of a segregated structure: They are not suitable for industrial, segregated structures, grain boundaries can act as a plane of weakness, and that acts as a matrix effect or assembler granos.Soncausa enlo lack of uniformity with respect to the physical and mechanical properties, and in some cases, an increased susceptibility to intergranular corrosion, due to preferential corrosivo.Tiene half would exert mechanical cararteristicas Le and R minor lasestructuras uniforms. 5.13. homogenization annealing stages: the piece is introduced into the furnace, heated to the temperature of homogenization and then cooled. 5.14 Indicates resistant characteristics that are obtained after a homogenization anneal in a segregated structure: Phase segragadamantiene invariant hardness and therefore its primary phase composicion.La suffers a gradual decrease of the microhardness, which means an enrichment of the metal lower melting point from the segregated phase.5.15.1. Owing to the diffusion of metals: it happens in any state and is subject to the diffusion of atoms or molecules between areas that show lasdiversas concentracion.La differential diffusion is caused by innumerable trips desordenados.Si a large number of atoms involved in these systematic movimientosproducen inthe flow direction of the gradient of concentration.5.15. Laws governing the phenomenon of diffusion: -> Pick Laws: (1) The flow of atoms is proportional to the concentration gradient and is directed in the opposite direction of the gradient. (2) The variation of concentration on time is a direct function of the derivative with respect to the space of linear concentration gradient. 5.16 microstructure of the eutectic constituent. How this affects the behavior mecanicode alloys? Identified by:-uniformity macroestructural.-level forms of alternating lamellar phases present, at microestructural.-The maximum hardening is observed precisely at the eutectic composition and increased up to this point when we left the composition of the pure metal 5.19. How can avoid the effect Coring in an alloy that has a wide range of solidification? How do I correct? We can help Shutter decreasing cooling using a mold with a more correct fino.Lo homogenization annealing. 5.20. justifies the possibility that appears in a segragacion dendritic alloy composition eutectica.Se dendritic segregation can occur by rapid cooling. 5.21. Justifies resistant characteristics of a structure that has dendritic segregation: The structures are segregated because of lack of uniformity in terms the physical and mechanical properties, and in some cases, an increased susceptibility to intergranular corrosion due to attack referential exercise corrosivo.La half segregated structure is mechanical characteristics, Ler below the standard structure. 5.22. Microstructure of hypoeutectic alloys: grain are monophasic, biphasic immersed in the eutectic structure. 5.23. Microstructure of hypereutectic alloys: The hypoeutectic microstructure and hpereutectica are similar with just replacing the Cu-rich phase of hypoeutectic base for rich in the hypereutectic Ag. Are similar 7.1. Processing martensitica.Justificacion: It is a process of hardening alloys have some metal alloy alotropico.Este as is the case of iron steel. Yan transformaciGraph To obtain this we must use a tempering process that involves heating the steel to a temperature above the 723 ANDC for steel components (ferrite and pearlite in the case of steels hipoteuctoides) perlite and cementite (in the If the hypereutectoid) is transformed into austenite. 7.3. depends on what parameters the hardness of a martensitic steel structure: “The cooling rate in the temperate “The C content in the steel and alloy-TamaGraph and 7.4. Why is it necessary to apply a tempering treatment after quenching with martensitic transformation: Because the temper and conduct a steel trans.Martensitica, the steel becomes very static properties produced by high internal stresses created in the crystals and therefore very pequeGraph dynamic properties and 7.5. Define the concept of critical velocity of nerve. That depends parLinemetros: The minimum speed of cooling to which we must submit a mass of austenite to be transformed in its entirety martensita.Depende:-The grain-tamany The alloy steel7.7. Under what conditions: material temperature austenitizing … We find that, after quenching, hardness is not achieved for the 100% martensite, observLinendose mixtures of M + A: You can not be achieved by:-The composiciGraph Yan Material C and alloying, as in alloys with a high content of alloying austenite is often left after the warm-Speed templado.-Tammany _.En Part 7.8 tempering an industrial process in steels with 0.4% C begins to detect a more harshly than in the pieces for that same right treatment. A study of defective parts and determined: a) The steel is composiciGraph Yan correcta.B) The microstructure of the temple presents a mixture of martensite with a 10-20% ferrita.Senyale and explain the causes of failure? That the cooling rate has been lower than the critical velocity of templey that a percentage of the austenite has evolved in the form of ferrite. 7.10 -. As can be 100% martensitic structure obtained by cooling the air a piece of steel? Justificar.Si we get the critical velocity of quenching is less than the rate of cooling air, which can be achieved with steel alloying components such as Mn, Mo, Cr, Si or Ni. 7.11 -. Establishes the differences that might exist between the products in the hyper-hipoeutectoide.La Yan transformaciGraph difference between hyper-and hypoeutectoid bainite lies in the carbon Yan composiciGraph these. 7.12 -. Justifies the adverse effects on the posiciGraph Yan curves of the elements forming aceros.Los Elemene carbides in the carbide formers reduced hardenability of steels, and high temperatures quenecesitan austenitising to achieve their full disoluciGraph Yan, undissolved quedandocarburos if not met, which implies Yan disminuciGraph percentage of C in the austeniticay mass therefore obtain a lower hardness martensite. However, these Braun S right so we need more time for the austenite is transformed and therefore the critical velocity of quenching is reduced. 7.13 -. Indicates items that are distorting effects of the S curves in steels Mn, Mo, Cr, Si or Ni. 7.14 -. Reasoning that allows the correlations support the TTT diagram of isothermal transformation as suitable to define the critical velocity of temple: If we draw on the TTT diagram the kinetics of cooling at different rates, we note that some of these areas cut to transformaciGraph Yan ferrite + perlite or bainite, interms of making the percentage of austenite not transformed to martensite at those speeds. But there are others that do not cut any of these areas and go directly to the martensitic Yan transformaciGraph area, so the whole mass of austenite is transformed to martensita.Y among these there will be one that will be tangent to the curve called speed up processing temper criticism. 7.18 Justify permetirian processes that eliminate the retained austenite in steel structures steel aleados.Deberemos submit to an isothermal process that allows the martensite surrounding the retained austenite can deform and allow the austenite to martensite change, with more estrucuta voluminous. Classifies 7.19 alloys and more general applications of shape memory transformations. The shape memory alloys are those that after a manufacturing process, return to the form that had a temperature alloys as determinada.Son: Cu.14, 2 al.4, 2 ‘, is characterized by: having martensitic transformation reversible. Martensitic transformation of the type of deformation by twinning. Aplicaiones valves or closures on materials needed to vary their mechanical properties when the temperature inside increases.7.23 Analyze the characteristics that can be expected in a high-content steel alloy that after quenching tempering is applied to transform retained austenite for long periods of time. If a hardened steel alloy high we will apply a tempered for long periods of time to eliminate the retained austenite can expect: Transformation of austenite to martensite, the tempering of martensite which improves the dynamic characteristics, and if tempering lengthen the time we get to superannuation and the consequent loss of C of martensite, for which mechanical properties will be lost. Conclusion, instead of a transformed proucir of retained austenite to martensite to get more strength, have diminished considerably the mechanical properties of steel. |