Lösungen Übung Verformung


 Constance Hamilton
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1 Lösungen Übung Verformung 1. (a) What is the meaning of T G? (b) To which materials does it apply? (c) What effect does it have on the toughness and on the stress strain diagram? 2. Name the four main hardening mechanisms in metals discussed in the lecture. Briefly discuss each principle and explain the corresponding formula. 3. How is a dislocation (Versetzung) defined (vector, plane)? How do these parameters contribute to the energy of a dislocation? 4. Sketch the atomic arrangement and Burgers vector orientations in the slip plane of a bcc metal. (Note the shaded area of Table 6.9, Shackelford p. 216) 5. (a) Different planes in a crystal lattice are differently dense packed. In which plane does plastic deformation take place? Consider the energy of a dislocation. (b) A crystalline grain of aluminium in a metal plate is situated so that a tensile load (Zugspan nung) is oriented along the [111] crystal direction. If the applied stress (Spannung) is 0.5 MPa, what will be the resolved shear stress (Schubspannung) τ, along the [101] direction within the (111) plane? (Equation 6.14, Shackelford p. 218) (c) What does it mean if the applied shear stress is above the resolved shear stress? 6. Consider the slip systems for aluminium shown in Figure 624 (Shackelford page 216). For an applied tensile stress in the [111] direction, which slip system(s) would be most likely to operate? Additional for interest: 7. Calculate the length of a Burgers vector in Cu (Kupfer) fcc (kubisch flächenzentriert). The lattice parameter is a = nm. 8. You are provided an unknown alloy with a measured Brinell hardness value of 100. Having no other information than the data of Figure 628a (Shackelford, p. 220), estimate the tensile strength (Zugfestigkeit) of the alloy. Express your answer in the form x ± y where y is the max. and min. deviation (Abweichung) from x. 9. Identification of preferred slip planes The planar density of the (112) plane in bcc iron is atoms/cm 2. Calculate (a) the planar density of the (110) plane and (b) the interplanar spacings for both the (112) and (110) planes. On which plane would slip normally occur?
2 Lösungen 1. (a) T G describes the temperature below which the molecules/atoms have little relative mobility. This is shown in the following figure: (b) It applies to all amorphous and partially amorphous materials (glasses, polymers, bulk metallic glasses, etc.) (c) Above T G the toughness increases significantly (area under the stress strain curve). The stress strain diagram nicely illustrates that the E Modulus is lower, the elongation is higher and the up taking forces of the materials are much lower at temperatures above the glass transition temper ature. 2. Plastic deformation in crystals is being carried out by dislocations, which are generated upon ex ternal mechanical load. Generally, the yield stress R p0.2 can be increased (hardening) by hindering the movement of dislocations. The relevant formulas are given in the script of the lecture De formation. It is important to know the proportionality (linear, inverse, square root dependence...).  Solid solution hardening (Mischkristallhärtung): The creation of extrinsic atomic defects, i.e. in troduction of substitution or interstitial impurity atoms (Fremdatome), results in a lattice distor tion (lattice strain, Gitterverzerrung) and thus stress fields (Spannungsfelder) are created. These stress fields decrease the mobility of dislocations, e.g. C in Fe, steel.  Precipitation hardening (Teilchenhärtung): Precipitates that cannot be cut through and dis persed particles in the microstructure are obstacles for dislocations.
3  Grain boundary / Grain size hardening (Korngrenzen / Feinkornhärtung): Grain boundaries are obstacles for dislocations. The finer the grains are the more effective they are in limiting the dislo cation movements.  Dislocation hardening (Versetzungshärtung): Cold work increases the dislocation density in a sample. The more dislocation there are the more difficult it is to increase the number of disloca tions because of the strain fields created by the existing dislocations. 3. A dislocation is defined by the Burgers vector and the dislocation line (Linienvektor), which togeth er define the slip plane (Gleitebene). The energy of a dislocation is given by: E = G b 2. Therefore the shortest Burgers vector b represents the dislocation with the lowest energy and therefore the most favoured dislocation. 4. For example, two of the 12 systems would be: 5. (a) The energy of a dislocation is E = G b 2, therefore proportional to the length of the Burgers vec tor b. The length of b is given by the distance between two atoms in the slip plane. This is shown in the following figure. This means: since the length of the Burgers vector is at least the distance between two lattice positions, it is easier to form a dislocation in a slip plane where the separation between two at
4 oms is small. Therefore, dislocation movement in such a plane is also easier because of the com paratively low energy. The binding energy also plays a role: it gives information about the Peierls potential which de scribes the energy barrier that has to be overcome to pass by an atom. Further points that influ ence the dislocation movement are: Kinks, steps, climbing (Klettern), splitting in partial disloca tions... (b) From crystallography you should be familiar with crystal planes, directions and their indices. F along [111] with σ = 0.5 MPa λ: angle between [101] and [111]: cos λ = φ: angle between [111] and [111]: cos φ = τ = σ cos λ cos φ = 0.5 MPa = MPa (c) Applied shear stresses above the resolved shear stress would initiate plastic deformation. 6. From equation 6.14 in the Shackelford: τ~ cos λ cos φ. For each angle, the cosine is determined by the dot product of 111 with 111 (for φ) or with 110 (for λ). The most likely slip systems are those with maximum τ. For the twelve systems in Figure 624, cos λ cos φ = 2 or 0. The six for which cos λ cos φ = 2 are: The lattice parameter for Cu is a = nm. The Burgers vector is along the closest packed direction and therefore of the form 110. The distance between two atoms is ½ of the diagonal length (see exercise 1 in h4): d = a = nm 8. All data in Figure 628 (a) fall within the band:
5 Average estimated tensile strength: TS = "#"# MPa = 400 MPa Error bar : MPa = 140 MPa Or: estimated tensile strength: 400 ± 140 MPa 9. The lattice parameter of bcc iron is nm or cm. The (110) plane is shown in the figure, with the portion of the atoms lying within the unit cell being shaded. Note that one fouth of the four corner atoms plus the centre atom lie within an area of a 0 times 2 a 0. (a) The planar density is: Planar density (110): atoms area =."" " cm = " atoms/cm 2 Planar density (112): " atoms/cm 2 (from problem statement) (b) The interplanar spacings are: d " =."" " = cm d " =."" " = cm The planar density and interplanar spacing of the (110) are larger than those for the (112) plane; therefore, the (110) plane would be the preferred slip plane.
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