高級工程材料-作業(yè)答案-第二章晶體級別的應力

1、1. Show that “9 is for orthorhombic crystalsSolutio n:Con stitutive equati on:5i 1_CiiC12C13C14C15C22C23C24C25C32C33C34C35C42C43C44C45C52C53C54C55C62C63C64C65匡2 j-C61C26客22C3633C46C56丫13C66 _ ,.；12 _Ci6 11(1)Cij： 36 in depe ndent stiff ness con sta nts.Because of mechanical equilibriums, Cj =5 indep

2、endent stiffness constants change to 21.For orthotropic material, no interaction exists between normal stresses a 11, a 22, q 33 and shear strains y 23, y 13, y 12; i.e. normal stresses acting along principal material directions produce only normal strains. No interaction exists between shear stress

3、es t 23, t 13, t 12 and normal strains e 11, & 22, & 33； i.e. shear stresses acting on principal material planes produce only shear stra ins. No in teract ion exists betwee n shear stresses and shear stra ins on differe nt planes; i.e. a shear stress acting on a principal plane produces a shear stra

4、in only on that pla ne.Then, Eq. (1) cha nge to:r-niCHC12C130001-知122C21C22C23000芒2233C31C32C33000名3323000C4400丫23130000C550丫13卩12 一00000C66 _|The nu mber of in depe ndent con sta nts is reduced to 9.2.3 Plane stress and plane strain are important concepts,particularly in the realm of fracture. Plan

5、e 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 stressconditionslead to three principal strain componentsand that plane strain conditions result in three pri

6、ncipal stress components.Soluti on:From Eq. (2.3)-1-、- 2EE-2、c1EE-3 c1EE- 3- 一 2 、- 3EEC 2；3 八；_；2*_EThe n1 = E、i .E 2 E 3；-2 = E 12 - E13 _ E11；-3 = E 3 - E 二1 .E 2Application of a uniaxial stress leads to Iongitudinal extension along the tensile axis (；1 =、L1/L1) and to tran sverse con tract ions

7、along the two perpe ndicular axes ( 2 = ；3 0). For linear elastic deformation, the Iongitudinal and lateral strains are relative through Poisson s ratio2 = ；3 = -；1On the other hand, when apply a uni axial stress on two perpe ndicular axisl lead to transverse contractions along the axisl, thus total

8、 stain on axisl is .；i 八.；2；3 ; =- 1/ ；2= - 1/ 32 2 22.9 For iron, Cii=237 GN/m , Ci2=141 GN/m and C44=116 GN/m .a Determine E1oo, 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

9、 orientations are present in equal amounts, what is E for the wire?2c The polycrystalline elastic modulus of Fe is 209 GN/m . How does this compare with the modulus estimated in part (b)?d By changing the processingconditions,a different kind of texture can be produced in Fe wire; 70% of the grains

10、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 pertain

11、ing to parts (b) and (d)? Explain your reasoning.Soluti on:a:Si + S2(SISI2)(S112S12)02(S11 - S2)(Sii2SI2)Si宅2 血7.587*10 GN/m2(C11 2C12)(C11 -C12)一2.823*10 ”GN/m2(Gi 2Ci2)(Ci1 - C12)S44 = 1 = 1 =8.62*10 GN/m2C44116The modulus along an arbitrary hkl direct ion is obta ined as1ES11 2(S11 S12EhkiR2-：22；

12、2)For 100,I22-2 2=0, E100 =131.81GN/m2For 110,I222 2 =0.25,2E110 =220.57 GN/m2For 111, :-2 2-2 2=0.33,E111 =284.43 GN/m2b:(t2(r1*A= d 2*A=E*& =(E100+E110+E111)*E1=212.27 GN/m2Case 2(TE100E110E111(Td =E = E100 & 100=E110& 110=E111& 111& = & 100+ & 110+ & 1111/E=(1/3E100+1/3E110+1/3E111)2E2=191.86 GN/

13、m22c: The polycrystalline elastic modulus of Fe is 209 GN/m . It is in the range of2 2191.86212.27 GN/m2. close to 212.27 GN/md: 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

14、 from the modulus along the axis for the situations pertaining to parts (b) and (d). It2will be changed in the range of 168.02 is in the range of 131.81 284.43 GN/m .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). Th

15、e spring and dashpot in series alone are referred to as a Maxwell model. If a stress is applied at time t=0 and held constant,sketch the strain-time responseexpected for a Maxwell model.b Release the load in part (a) some time after its application. Sketch strain vs. time following this unloading.c

16、Show that the addition of the strains drawn in (a) and (b) to thoseschematized in Fig. 2.15 leads to the response illustrated in Fig. 2.17.Soluti on:2.20 Refer to Fig. 2.20b. Schematically plot二 max / max as a function of the appliedfrequency.(二 max is the maximum stress observed during the hysteres

17、is cycle and max is the maximum strain; note c max typically is not found when；二 max, cf. Fig. 2.20b).Soluti on:Fig. 2.20b shows the stress-stra in curve. We see that it is path-depe nden t, in that thestress-stra in curves on loadi ng and uni oadi ng differ. The averate modulus as defi ned, for exa

18、mple, by the ratio c max / ；max in Fig. 2.20b is between Er and Eu since some, butn ot all, of the pote ntial viscoelastic strain is mani fested at this in termediate freque ncy. The area between the unioading 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