高级工程材料-作业答案-第二章晶体级别的应力.docx

1. Show that “9 is for orthorhombic crystals”.Solution:Constitutive equation: (1)Cij: 36 independent stiffness constants.Because of mechanical equilibriums, Cij=Cji, independent stiffness constants change to 21.For orthotropic material, no interaction exists between normal stresses 11, 22, 33 and shear strains 23, 13, 12; i.e. normal stresses acting along principal material directions produce only normal strains. No interaction exists between shear stresses 23,13,12 and normal strains 11,22,33; i.e. shear stresses acting on principal material planes produce only shear strains. No interaction exists between shear stresses and shear strains on different planes; i.e. a shear stress acting on a principal plane produces a shear strain only on that plane.Then, Eq. (1) change to:The number of independent constants is reduced to 9.2.3 Plane stress and plane strain are important concepts, particularly in the realm of fracture. Plane stress is defined by two finite principal stress components, one principal stress component being zero. Plane strain is defined analogously. Use Eq. (2.3) and its analogs to show that plane stress conditions lead to three principal strain components and that plane strain conditions result in three principal stress components.Solution:From Eq. (2.3) ThenApplication of a uniaxial stress leads to longitudinal extension along the tensile axis (e1=dL1/L1) and to transverse contractions along the two perpendicular axes (e2 = e3 0). For linear elastic deformation, the longitudinal and lateral strains are relative through Poissons ration, e2 = e3 = -ne1On the other hand, when apply a uniaxial stress on two perpendicular axis1 lead to transverse contractions along the axis1, thus total stain on axis1 is e1 - ne2 -ne3 ; n=-e1/e2= -e1/e3132-e2,-e31E1es1e12.9 For iron, C11=237 GN/m2, C12=141 GN/m2 and C44=116 GN/m2.a Determine E100, E110 and E111 of Fe.b Suppose a polycrystalline Fe wire is composed of grains with either 100, 110, or 111 crystal directions lying along the wire axis. If all the three orientations are present in equal amounts, what is E for the wire?c The polycrystalline elastic modulus of Fe is 209 GN/m2. How does this compare with the modulus estimated in part (b)?d By changing the processing conditions, a different kind of texture can be produced in Fe wire; 70% of the grains are oriented so that a direction is aligned with the wire axis and the remaining 30% are evenly split between and directions lying along this axis. What is the modulus along the axis for this case.e Is E transverse to the wire axis different from the modulus along the axis for the situations pertaining to parts (b) and (d)? Explain your reasoning.Solution:010100110001111a: The modulus along an arbitrary hkl direction is obtained asFor 100, =0, E100 =131.81GN/m2For 110, =0.25, E110 =220.57 GN/m2For 111, =0.33, E111 =284.43 GN/m2b:E100Case 1E1101 2E111 1*A=2*A=E*=(E100+E110+E111)*/3E1=212.27 GN/m2Case 2E111E110E100 =E= E100100=E110110=E111111=100+110+ 1111/E=(1/3E100+1/3E110+1/3E111)E2=191.86 GN/m2c: The polycrystalline elastic modulus of Fe is 209 GN/m2. It is in the range of 191.86212.27 GN/m2. close to 212.27 GN/m2.d: E=70%E100+30%(E110+E111)/2=70%*131.81+15%(220.57+284.43)=168.02 GN/m2. e: under different condition, the value of E is different. E is transverse to the wire axis different from the modulus along the axis for the situations pertaining to parts (b) and (d). It will be changed in the range of 168.02 is in the range of 131.81284.43 GN/m2.2.18 a The additional spring and dashpot in series with the voigt model of Fig. 2.14 constitute a standard linear solid (Fig. 2.16). The spring and dashpot in series alone are referred to as a Maxwell model. If a stress s is applied at time t=0 and held constant, sketch the strain-time response expected for a Maxwell model.b Release the load in part (a) some time after its application. Sketch strain vs. time following this unloading.c Show that the addition of the strains drawn in (a) and (b) to those schematized in Fig. 2.15 leads to the response illustrated in Fig. 2.17.Solution:2.20 Refer to Fig. 2.20b. Schematically plot as a function of the applied frequency. (is the maximum stress observed during the hysteresis cycle and is the maximum strain; note typically is not found when , cf. Fig. 2.20b).Solution: Fig. 2.20b shows the stress-strain curve. We see that it is path-dependent, in that the stress-strain curves on loading and unloading differ. The averate modulus as defined, for example, by the ratio in Fig. 2.20b is between Er and Eu since some, but not all, of the potential viscoelastic strain is manifested at this intermediate frequency. The area between the unloading the loading curves (the cross-hatched region in Fig. 2.20b) represents an irreversible or hysteretic loss of work or energy per stress cycle. Analogous phenomena are observed in magnetic materials when the internal magnetic field does not respond instantaneously to the applied external field, and in electrical capacitors subjected to time-varying electrical fields. In all cases, the irreversible energy loss serves to heat the material unless some external means to cool it are provided. That irreversible permanent deformation is accompanied by internal heating is easily visualized. Hammering thin a metal either manually or by cold rolling is accompanied by