彈塑性力學(xué)第四章概要

1、3CHAPTER 4 CONSTITUTIVE RELATIONS4. CONSTITUTIVE RELATIONS4.1. Generalized Hookes LawFor an elastic-plastic solid, the constitutive relation is assumed to connect the stress acting on an element of material with its deformation from an unstressed natural configuration. That is the constitutive relat

2、ion, in fact, is the relation between stress (and stress rate) and strain (or strain rate). Now we first give the stress-strain relation for the elastic material. Robert Hooke (1676) proposed on the basis of a simple tension experiment that the force and extension of an elastic body were proportiona

3、l to one another. That is the stress and strain has a linear relationship:= E (4.1)Where E is elastic modulus or Youngs modulus. This linear portion of the stress-strain relation ends at the proportional limit, and its general form is given byij = Cijkl kl (4.2)where Cijkl is the material elastic co

4、nstant tensor, equation (4.2) is so called the generalized Hookes law. Its expanding form is x = C11x+C12y+C13z+C14xy+C15yz+C16zx y= C21x+C22y+C23z+C24xy+C25yz+C26zx z= C31x+C32y+C33z+C34xy+C35yz+C36zx xy= C41x+C42y+C43z+C44xy+C45yz+C46zx (4.3) yz= C51x+C52y+C53z+C54xy+C55yz+C56zx zx= C61x+C62y+C63z

5、+C64xy+C65yz+C66zxAs we known, since both ij and kl are second-order tensors, it follows that Cijkl is a fourth-order tensor. In general, there are (3)4 = 81 constants for such a fourth-order tensor. But, since ij and kl are both symmetric, one has the following symmetry conditions: Cijkl = Cjikl =

6、Cijlk = Cjilk Hence, the independent constant is reduced to 36.In fact, these 36 constants are not always independently. We will verify that the number of the independently elastic constants is only 2 for the isotropic perfect elastic material in the following.For an isotropic material, the elastic

7、constants must be the same for all directions. Thus, if x, y, z are the principal direction, then the strain components xy, yz, xz all equal to zero. Hence, from (4.3) we have xy= C41x +C42y +C43z (a)Now we introduce a new coordinate system Oxyz, and let the initial coordinate system Oxyz coincide w

8、ith it after rotated 180o. The direction cosine angles between these two axes are n11= n33 = -1, n22= 1, n21= n31= n12= n32= n13= n23 = 0For isotropic material the elastic constants are independent of directions. Thus, for the new coordinate system we havex y = C41x + C42y + C43z (b)Referring coordi

9、nate transformation formula (2.34) of stress components, we can obtainx y = -x y , x =x , y =y , z =z (c)Considering Equations (b) and (c), we find -xy= C41x +C42y +C43z (d)Comparing Eq. (a) to Eq. (d), we obtain xy = - xy . Furthermore, it must, of course, leads toxy = 0Similarly, we getyz = zx = 0

10、For isotropic elastic material the strain principal axes must be coinciding with the stress principal axes.Let the coordinate system Oxyz coincides with the stress principal axes for an isotropic material, then we get the following relation between principal strains and principal stresses x = C11x+C

11、12y+C13z y= C21x+C22y+C23z z= C31x+C32y+C33z (e)As we know, the strain x have an influence upon x, which is the same like the y, z influence y and z. Therefore, we have C11 = C22 = C33.Similarly, C12 = C13, C21=C23, C31=C32, , etc. Thus C11 =C22 = C33 = a C12 =C21 =C13 =C31 =C23 =C32 = b (f )For ela

12、stic constants discussed above, it can be conveniently getting a result, for an isotropic linear elastic material, there are only two independent material constants. Let a-b = 2, b = , =1 +2 +3 , the strain principal axes are 1, 2, 3, then the elastic constitutive relations may be take the form (4.4

13、)Where , called Lames constants. After coordinate transformation, we can get the following constitutive relations for any coordinate system Oxyz. (4.5)orij = ij + 2ij (4.6)For some engineering materials, for example, the two-way reinforced concrete member, wood member, etc. are anisotropy material.

14、For these kind materials, the elastic properties always possess orthotropic symmetry. Considering the elastic property is symmetry about pre-selected coordinate planes x = 0, y = 0 and z = 0, the elastic constants will be reduced. If there are two planes of elastic symmetry and one of them is also s

15、ymmetry about the third orthogonal plane (orthotropic symmetry), then the number of elastic constants is reduced to 9, and for the transversely isotropic material (three orthogonal symmetry planes), the number is reduced to 5 (See problem 2).Conversely, strain ij can be expressed in term of stresses

16、 in the constitutive relation of Eq. (3.3). For which, one hasii = (3+2kk (4.7)orkk = ii / (3) (4.8 )Substituting this value of kk in to Eq. (4.6) and solving for ij, we get32CHAPTER 5 FORMULATION AND SOLUTION OF THE ELASTOPLASTIC PROBLEMSij = ij - ij (4.9)where ii. Eq. (4.3) and (4.9) are the gener

17、al forms of constitutive relation for an isotropic linear material, and so called the generalized Hookes law. Consider a simple tension case. The only nonzero stress component x causes axial strain x according to x = (4.10 ) and transverse strain according to y = z =- (4.11 )where is Poissons ratio.

18、 It should be noted that the normal stress x produced no shear strain. On the other hand, in a pure shear case the shear stress xy produces no normal strain but only the shear strainxy asxy = (4.12)where G is the shear modulus of elasticity.Comparing the constitutive relation (4.9)-(4-12), Lames con

19、stants can be expressed in term of E and n as (4-13)In practice, Youngs modulus E and Poissons ratio n used to represent the generalized Hookes law as (4.14)where From which we know that G is not an independent elastic constant.Using the index notation, we can rewriting Eq. (4.9) in a concise form (

20、4.15)proceeding in a similar manner as above, we can solve Eq. (4.15) for stresses as (4.16)Considering , and and (4.17)the generalized Hookes law can be written as (4.18)where and are the mean normal stress and mean normal strain, (4.19)From Eqs. (4.15), (4.16), we can get the Hookes law of expandi

21、ng form two-dimensional and three-dimensional problems. For example, let , the stress-strain relations of plane stress problems are (4.20)or (4.21)For the plane strain problems are (4.22)or (4.23)Comparing the generalized Hookes law for plane stress problems and plane strain problems , we get that i

22、f the constants E and of stress-strain relation of plane stress problems change to E1 and 1 as E1 = E / (1- 2 ) 1 = / (1- )Then it can be given the stress-strain relation of plane strain problems.4.2. Yield conditionAs we known, when the stress is over beyond the yield stress s in a simple tension c

23、ase, as shown in Fig. (2.1), the stress-strain relation is no longer obeyed the Hookes law. In this case, the yield stress is shown in a figure of a simple tension test evidently. = s is the yield condition in a uniaxial tensile (compressive) case. In the complex stress state, the yield condition is

24、 not so simply. For example, there is a thin-wall tube subjected to internal pressure q , axial tension T and torque M, in this case, the stress state of tube wall can be considered accurately as the plane stress state. (Fig.4.1 ). When q = 0, the stress state is = 0, = , = Similarly, when M = 0, th

25、e stress state is =, =, =0Fig.4.1 Stress state of thin wall tubeIt is clear that yielding would occur in a combined state of stress for the above thin wall tube. We can get a yield condition from once complex stress experiment. It means that yield condition of materials should be determined experime

26、ntally. But for theoretical analysis we need a analytical formula, and many times experiments is a very inconvenient thing to us. Therefore we need the theory of yield condition based on experiments.In general, the yield condition is a function of the state of stress ij . Hence, the yield condition

27、can generally be expressed as f(ij , ki) = 0 (4.24)where ki( i =1, n) is the material constants, which are to be determined experimentally. f (ij ) is called the yield function .Equation (4.24 ) is expressed a hyper-surface in a 6-dimensional stress space. Every point in this stress space is express

28、ed a state of stress, and any point on this hyper-surface is expressed a yield stress state, so it is called the yield surface.The yield surface can be expressed in the stress space as above,. Similar, it can be also expressed the yield surface in the strain space ij .For isotropic materials, the ro

29、tation of axes is not influence the initial yielding. Thus, we can take the principal axes as the coordinate axes, in this case, Eq. (4.24 ) can be written as:f ( 1 , 2 , 3 ) = 0 (4.25 )Moreover, the hydrostatic pressure has no effect on yielding, Eq. (4.24 ) or (4.25 ) must be dependent only on dev

30、iator stress. It is easy to understand that hydrostatic pressure I1 representing a cylindrical surface whose generator is parallel to the hydrostatic axis. The hydrostatic axis On is a straight line passing through the origin of coordinates and equal inclined to the three principal stress axes ( Fig

31、. 4.2 ). The plane perpendicular to the straight line On is called plane, which must have the form 1 + 2 + 3 = 0 (4.26 )Fig. 4.2 The yield locus in planeIf the material (a) is isotropic, (b) is hydrostatic pressure independent, and (c) has equal yield stresses in tension and compression, then in the

32、 - plane, any initial yield locus must possess the following important characteristics:(1) The initial yield locus is a closed curve, and the origin of coordinates is inside of the locus.(2) The yield locus and any straight line starting from the origin of coordinates must intersect once, and only f

33、or once.(3) The yield locus is symmetry to the positive and negative directions of three axes of coordinates(4) The yield locus and yield surface all must be convexity.The convexity of yield locus or yield surface is a important characteristic of them, which we will proved in the following.The pract

34、ical yield condition will be discussed in the next section.4.2.1. The Tresca Yield ConditionThe first yield condition for a combined state of stress for metals was that proposed by H. Tresca in 1864. He suggested that yielding would occur when the maximum shearing stress at a point reaches a critica

35、l value k. Therefore it so called the maximum shearing stress condition. Thus, we have max = kwhere k is the yield stress in simple shear ( ks ). Stating this in terms of principal stresses , one-half of the greatest absolute value of the differences between the principal stresses taken in pairs mus

36、t be equal to k at yield ,namely,max ( | |, 3 |, 1| ) = k (4.27)where the material constant k may be determined from the simple tension test. Thusks=s / 2 (4.28)It means, that according to the Tresca condition, the yield stress of simple shearing stress equal to half yield stress of simple tension.E

37、quation (4.27) is represented regular hexagonal prism in stress space, and the center axis line of this prism is equally inclined to the three axes, so that its direction cosines are ( 1/31/2, 1/31/2, 1/31/2 ),see Fig. (4.1 ). For two-dimensional case, the Tresca hexagonal prism will be changed to a

38、 plane yield hexagon (Fig. 4.2 ). Suppose 3 = 0, we have, if , if , if , if (4.29), if , if Eq. (4.29) is represented a yield hexagon in plane ( see Fig. 4.3 )Fig. 4.3, Yield locus stress in p planeIn the above discussion we are using the concepts of stress space. As we just seen, the principal stre

39、ss space will be degenerated to a plane, so called principal stress plane. Every stress state is a determinate point (so called stress point ) in the stress space ( or stress plane, in plane stress state ). If the stress point is laid inside the yield surface (yield hexagon ), then it means the mate

40、rial is in the elastic state. If the stress point is on the yield surface (yield curve ), then the material begins to yield .For the perfect elastic-plastic material, the stress point can not be run beyond the yield surface. For the strain-hardening elastic-plastic material, we will discussing later

41、. 4.2.2. The Mises yield conditionAlthough the Tresca condition is simple, it does not reflect any influence of the intermediate principal stress. Von Mises (1913) proposed a yield condition based on thinking of the strain energy of .distortion. It states that yielding begins when the strain energy

42、of distortion reaches a critical value k2. The strain energy of distortion Uod is the strain energy stored in a unit volume due to the form change only (not include the volume change). Thus Uod = sijeij (4.30)Which can be written in terms of principal stresses, Uod must have the form (4.31)or 2GUod

43、=J2 Therefore the Mises yield condition can be written as J2 = k2 (4.32)where k is yield stress in pure shear and can be determined by simple tension test. In this case, from Eq. (4.31),we have, Then we getand k = (4.33)Fig.4.4 yield locus matched in principal stress plane s1-s2In the two-dimensiona

44、l stress case, the Mises condition must have the form (4.34)Eq. (4.34)is expressed an ellipse in the principal stress plane (Fig. 4.4).The Mises yield condition for three-dimensional stress state is a circular cylinder inscribing the Tresca hexagonal prism in stress space (see Fig. 4.5). It is clear

45、ly, for an isotropic material whose yielding is independent of hydrostatic pressure, the yield function must be a cylinder with generator parallel to the hydrostatic axis (i.e. equal inclined line to the three principal stress axes).There have been many experimental results showing that the yield po

46、ints fall between the Tresca hexagon and the Mises circle and closer to latter. Taylor G. I.and Quinney H. in their classical experiments observed this fact (Fig. 4.5).Fig. 4.5 Tresca and Mises ellipses on the () plane together with the experimental resultsof Taylor and Quinney.4.2.3. The Mohr-Coulo

47、mb ConditionMohr O. (1900) considers the limiting shear stress in a plane to be a function of the normal stress in the same plane at a point, this case is especially available for soil, i.e.| | = f ()where f () is an experimentally determined function.The simplest form of the Mohr function f () is a

48、 straight line known as Coulombs equation (proposed in 1733):| |= ctan (4.35)in which c is the cohesion and is the angle of internal friction; both are material constants must determined by experiment.From Eq.(4.35) and for , the Mohr-Coulomb condition can be written as (4.36)Fig.4.6 Mohr-Coulomb co

49、ndition in the s-t plane.In principal stress space the Mohr-Coulomb condition is represented a hexagonal pyramid see Fig.(4.6). Similarly to what we have done for the Tresca condition in the plane for the case , we have the yield loci are irregular hexagon as shown in Fig.(4.7), in which m=fc / ft a

50、nd fc, ft are the strength in simple compression and simple tension respectively.Fig.4.7 Kupfers experimental results for concreteKupfer H. et al. (1969 ) have given them experimental results in plane stress state for concrete, which can be considered as a smooth approximation to the Mohr-Coulomb co

51、ndition(Fig. 4.7)4.2.4.The Drucker-Prager Condition Drucker and Prager (1952) proposed a yield condition for soils that is a simple modification of the Mises condition just as Mohr-Coulomb condition is related to that of Tresca condition. Drucker-Prager condition is written as a function of the inva

52、riants of the stress tensor and of the deviator. That is f (I1 , J2) =k (4.37)wherek =(a) (b)Fig.4.8 Drucker-Prager and Mohr-Coulomb conditions in principal stress spaceEq.(4.37)is so called Drucker-Prager yield condition. Here and k are material constants. Corresponding to the constants the Drucker

53、-Prager condition is represented a circular con circumscribed the hexagonal pyramid and represents an outer bound on the Mohr-Coulomb yield condition (See Fig. 4.8). When is zero, Eq. (4.37) reduces to the von Mises condition. The Drucker-Prager yield condition for a biaxial stress state is represented by the intersection of circular con with coordinate plane of Substituting it into Eq.(4.37) leads