CONSTITUTIVE RELATIONS FOR LINEAR ELASTIC SOLIDS


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1 Chapter 9 CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS Figure 9.1: Hooke memorial window, St. Helen s, Bishopsgate, City of London 211
2 212 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS 9.1 Mechanical Constitutive quations Recall that in Chapters 2, 3 and 8 we briefly introduced the concept of a constitutive equation, which generally relates kinetic variables to kinematic variables in the application of interest. With respect to the application of the analysis of mechanical deformations in solids, the kinetic variable is the stress tensor, σ, whereas the kinematic variables are the displacements u x, u y, u z, and the strain tensor, ε which includes derivatives (sometimes called gradients) of the displacements. Since it is generally observed that rigid body displacements do not induce stresses, the displacement field u x, u y, u z, will not enter into a mechanical constitutive equation. Thus, the constitutive equations will in general relate stress, σ, to strains, ε, and temperature T. In 1660, Robert Hooke observed that for a broad class of solid materials called linear elastic (or Hookean), this relationship may be described by a linear relationship. Hooke originally considered the test of a uniaxial body with a force (stress) applied only in one direction and measured the corresponding elongation (strain) to obtain: y x A=crosssectional area yy = ε yy Figure 9.2: StressStrain Curve for Linear lastic Material Forageneral threedimensional state of stress, there are 6 independent stresses and 6 independent strains; therefore, the linear relationship between stress and strain can be written in matrix form as: σ =[C] ε (9.1) where [C] isa6 6 matrix of elastic constants that must be determined from experiments. expanded form, these 6 equations become: In σ xx = C 11 ε xx + C 12 + C 13 ε zz + C 14 ε yz + C 15 ε zx + C 16 ε xy = C 21 ε xx + C 22 + C 23 ε zz + C 24 ε yz + C 25 ε zx + C 26 ε xy σ zz = C 31 ε xx + C 32 + C 33 ε zz + C 34 ε yz + C 35 ε zx + C 36 ε xy σ yz = C 41 ε xx + C 42 + C 43 ε zz + C 44 ε yz + C 45 ε zx + C 46 ε xy (9.2) σ zx = C 51 ε xx + C 52 + C 53 ε zz + C 54 ε yz + C 55 ε zx + C 56 ε xy σ xy = C 61 ε xx + C 62 + C 63 ε zz + C 64 ε yz + C 65 ε zx + C 66 ε xy It is interesting to note that Robert Hooke first proposed the above law publicly in an anagram at Hampton Court (1676) given by the group of letters: ceiiinosssttuv.
3 9.1. MCHANICAL CONSTITUTIV QUATIONS 213 In 1678 he explained the anagram to be Ut tensio sic vis, which is Latin meaning as the tension so the displacement or in nglish the force is proportional to the displacement. Students may recall that during this time period, science and scientific writing was criticized and hence Hooke thought it necessary to discretely disclose his scientific finding with an anagram. Note that in the previous chapters stress and strain were represented as (3 3) matrices. It is convenient, however, here to represent them as (6 1) column vectors since they have only 6 independent components (stress due to conservation of angular momentum and strain by its definition). We then write {σ} = σ xx σ zz σ yz σ zx σ xy, {ε} = By adopting this representation for σ and ε, their linear relationship (9.2) can be easily written in matrix form: ε xx ε zz ε yz ε zx ε xy σ xx σ zz σ yz σ zx σ xy = C 11 C 12 C 13 C 14 C 15 C 16 C 21 C 22 C 23 C 24 C 25 C 26 C 31 C 32 C 33 C 34 C 35 C 36 C 41 C 42 C 43 C 44 C 45 C 46 C 51 C 52 C 53 C 54 C 55 C 56 C 61 C 62 C 63 C 64 C 65 C 66 ε xx ε zz ε yz ε zx ε xy (9.3) If a material is homogeneous then the constants (C ij, i =1,..., 6 and j =1,..., 6) are all independent of x, y, z for any time, t. If a material is isotropic, then for a given material point, C ij are independent of the orientation of the coordinate system (i.e., the material properties are the same in all directions). If a material is orthotropic, then for a given material point, C ij can be defined in terms of properties in three orthogonal coordinate directions. If a material is anisotropic, then for a given material point, C ij are different for all orientations of the coordinate system. In order to determine the material constants in equation (9.3), consider a uniaxial tensile test using a test specimen of linear elastic isotropic material with crosssectional area A and subjected to a uniaxilly applied load F in the axial (y) direction as shown below. The crosssection may be any shape but generally a rectangular or cylindrical shape is chosen. For a rectangular specimen, assume a width W and thickness t so that the crosssectional area is A = Wt. Assume a small gauge length of L for which the axial deformation will be measured during the load application. During the uniaxial tensile test, weobserve that the gauge length changes from L to L and the gauge width decreases from W to W. We also observe a decrease (contraction) in the z dimension. We further observe no change in angular orientation of the vertical or horizontal elements and conclude that for uniaxial loading, no shear strains are produced. This leads us to postulate the following strain state: ε xx,,ε zz 0,ε xy,ε yz,ε xz =0. The axial stress and strain in the axial (y) direction are defined to be
4 214 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS F after deformation y x L L* z A=crosssectional area t test specimen F A=Wt W* W Figure 9.3: xperimental Measurement of Axial and Transverse Deformation = F A = L L = (L L) L The strain the in the transverse (x) direction due to the axial load is ε xx = W W = (W W ) W If we plot axial stress vs. axial strain and transverse strain vs. axial strain, weobtain the following two plots: = F A ε xx = W W 1 = ε xx = ν = L L 1 ν Figure 9.4: xperimental Results for Axial Stress vs. Axial Strain & Transverse Strain vs. Axial Strain From these two plots, we can write = and ε xx = ν for the uniaxial tension test. Consequently, we may define the following two material constants from this single uniaxial test: = slope of the uniaxial vs. curve = a material constant called Young s modulus
5 9.1. MCHANICAL CONSTITUTIV QUATIONS 215 ν = εxx = negative ratio of the strain normal to the direction of loading over the strain in the loading direction = a material constant called Poisson s ratio If the transverse strain were measured in the z direction, we would find the same ratio for transverse to axial strain: ν = εzz. Combining these equations, we can write the two transverse strains entirely in terms of the axial stress : ε xx = ν = ν ( ) and εzz = ν = ν ( ). In order to obtain a complete description of threedimensional constitutive behavior, consider a test where we apply normal tractions (stresses) in the x, y and z directions simultaneously and measure the strain only in the x direction. For a linear material response, we may use the principle of linear superposition and consider three separate cases as shown below: σ zz σ zz σ σ xx σ xx σ xx xx = + + σ zz σ zz Figure 9.5: xperimental Test with all Components of Normal Stresses Applied ε xx = normal strain in x direction due to σ xx + normal strain in x direction due to + normal strain in x direction due to σ zz = 1 σ xx ν ν σ zz or ε xx = 1 [σ xx ν( + σ zz )] (9.4) The stress in the x direction increases the strain in the x direction while the transverse stresses causes a contraction (decrease in ε xx ). Doing similar experiments in the y and z directions gives: = 1 [ ν(σ xx + σ zz )] (9.5) ε zz = 1 [σ zz ν(σ xx + )] xperiments with shear tractions will show that a shear stress σ xy in the xy plane produces only shear strain ε xy in the xy plane for a state of pure shear loading (i.e., no normal strain is observed so that the shear strain is uncoupled from the normal strain). 1 Thus, we obtain the following 1 Keep in mind that even for the case of pure shear, if one calculates shear stresses (or strains) at some angle θ from the xaxis (Mohr s circle), one may obtain nonzero normal stresses (or strains) for the offaxis planes.
6 216 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS experimental observations for the shear strains: ε xy = 1+ν σ xy ε xz = 1+ν σ xz (9.6) ε yz = 1+ν σ yz Combining equations (9.4), (9.5) and (9.6), Hooke s law for a linear elastic isotropic solid with a threedimensional stress state becomes: Hooke s Law for a Linear lastic Isotropic Solid ε xx = 1 [σ xx ν( + σ zz )] = 1 [ ν(σ xx + σ zz )] ε zz = 1 [σ zz ν(σ xx + )] ε xy = 1+ν σ xy (9.7) ε xz = 1+ν σ xz ε yz = 1+ν σ yz where = Young s modulus and ν = Poisson s ratio. It should be noted that in materials that undergo permanent deformation, the above model is not accurate (such as metals beyond their yield point, or polymers that flow). A typical uniaxial stressstrain curve for a ductile metal is shown below: σ xx yield stress ultimate stress failure stress ε xx Figure 9.6: Typical StressStrain Curve for Ductile Metal An algebraic inversion of the strainstress relationship (9.7) provides the following relationship of stress in terms of strain:
7 9.1. MCHANICAL CONSTITUTIV QUATIONS 217 {σ} = 1+ν 1 ν 1 2ν ν 1 2ν ν 1 2ν ν 1 2ν 1 ν 1 2ν ν 1 2ν 1 2ν ν 1 2ν 1 ν 1 2ν ν ε xx ε zz ε yz ε zx ε xy (9.8) or, Hooke s Law for a Linear lastic Isotropic Solid {σ} = σ xx σ zz σ xy σ xz σ yz = (1+ν)(1 2ν) [(1 ν)ε xx + ν + νε zz ] (1+ν)(1 2ν) [νε xx +(1 ν) νε zz ] (1+ν)(1 2ν) [νε xx + ν +(1 ν)ε zz ] 1+ν ε xy 1+ν ε xz 1+ν ε yz (9.9) where = Young s modulus and ν = Poisson s ratio. The term (1+ν) 2G defines a shear modulus, G, relating shear strain and shear stress (similar to Young s modulus,, for extensional strain). Thus, the shear modulus is given by: G = 2(1 + ν) (9.10) 2( 1+ν Note that the shear strain ε xy is related to engineering shear strain γ xy by γ xy = 2ε xy = )σ xy = σxy G so that σ xy = Gγ xy =2Gε xy. σ xy G = shear modulus σ xy = Gγ xy = G2ε xy G = 2(1+ν) γ xy Figure 9.7: xperimental Results for Shear Stress vs. ngineering Shear Strain Note that G is defined in terms of and ν and consequently G is not a new material property. Thus, for a homogeneous linear elastic isotropic solid, we conclude that only two material properties (Young s modulus,, and Poisson s ratio, ν) are required to completely define the threedimensional constitutive behavior.
8 218 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS The stressstrain equations may also be written in terms of shear modulus to obtain: σ xx = = σ zz = σ yz = σ zx = σ xy = (1 + ν)(1 2ν) [(1 ν)ε xx + ν + νε zz ]= 2G 1 2ν [(1 ν) ε xx + ν + νε zz ] (1 + ν)(1 2ν) [νε xx +(1 ν) νε zz ]= 2G 1 2ν [νε xx +(1 ν) + νε zz ] (1 + ν)(1 2ν) [νε xx + ν +(1 ν)ε zz ]= 2G 1 2ν [νε xx + ν +(1 ν) ε zz ] 1+ν ε yz =2Gε yz (9.11) 1+ν ε zx =2Gε zx 1+ν ε xy =2Gε xy Side Note: Ifthe definition of ε xx and ε zz from ν = εxx = εzz is substituted into equation (9.9), we obtain for the uniaxial bar extension experiment described previously: σ xx = (1 + ν)(1 2ν) [(1 ν)( ν)+ν + ν( ν) ]=0 [ = ν 2 +(1 ν) ν 2 ] = εyy (1 + ν)(1 2ν) σ zz =0 σ xy = σ xz = σ yz =0 This result is consistent with all observations made regarding the nature of stress for the uniaxial test with an applied stress of. 9.2 Constitutive quations with Thermal Strain xperimentally, we observe for a linear isotropic metal that a temperature increase, T, produces a uniform expansion but no shear and the expansion is proportional to a material constant α (coefficient of thermal expansion). The additional strain due to heating is thus: ε xx = = ε zz = α T. Thus, the constitutive equation for a linear elastic isotropic solid (9.7) may be modified by the addition of the thermal strain to the normal strain components: ε xx = 1 [σ xx ν( + σ zz )] + α T = 1 [ ν(σ xx + σ zz )] + α T ε zz = 1 [σ zz ν(σ xx + )] + α T ε xy = ( 1+ν )σ xy (9.12) ε xz = ( 1+ν )σ xz ε yz = ( 1+ν )σ yz
9 9.2. CONSTITUTIV QUATIONS WITH THRMAL STRAIN 219 These equations can be inverted to obtain stress in terms of strain: σ xx = = σ zz = σ xy = σ xz = σ yz = (1 + ν)(1 2ν) [(1 ν)ε xx + ν + νε zz ] α T (1 2ν) (1 + ν)(1 2ν) [νε xx +(1 ν) + νε zz ] α T (1 2ν) (1 + ν)(1 2ν) [νε xx + ν +(1 ν)ε zz ] α T (1 2ν) 2(1 + ν) ε xy (9.13) 2(1 + ν) ε xz 2(1 + ν) ε yz In the above, T = T ( x, y, z ) and represents the increase in temperature from a reference temperature where the thermal strain is zero. It should be noted that the first term in the extensional strain terms above (the[] term) is due to elastic behavior of the material (i.e., it has Young s modulus in it). The second part is due to thermal strain. We can separate the total strain into elastic and thermal strains: ε total xx = ε elastic xx + ε thermal ε total yy = ε elastic yy + ε thermal (9.14) ε total zz = ε elastic zz + ε thermal The elastic (also called mechanical) and thermal terms are given by: ε elastic xx = 1 [σ xx ν( + σ zz )] ε elastic yy = 1 [ ν(σ xx + σ zz )] (9.15) ε elastic zz = 1 [σ zz ν(σ xx + )] ε thermal = α T The terms ε total xx, ε total yy, ε total zz represent the total strain as measured or observed, and are thus equal to their deformation gradient definitions, i.e., for small strain, ε total xx = ε xx = u x x ε total yy = = u y y ε total zz = ε zz = u z z (9.16) We state once again that shear strains have no thermal component for an isotropic material. xamples of problems involving thermal strain will be considered in Chapter 10. Some typical values of material properties for isotropic metals are provided in the table below. Note that the values of (Young s modulus) are typically in the million psi or GPa range for engineering materials, while the values of ν are between zero and 0.5 (0 < ν < 0.5). The yield strength represents the stress level at which the metal yields (becomes inelastic). For ductile metals,
10 220 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS the ultimate tensile strength is typically 10 to 50% higher than the yield strength. It should be noted that material properties for commonly used metals must satisfy specifications established by regulatory agencies. Values in Tables 9.1 and 9.2 that are not provided are unknown for purposes of presentation herein and they should not to be interpreted as zero. The reader may with to consult other sources for the omitted values. For nonmetals, properties may vary significantly depending upon many variables (for example, wood has a Young s modulus varying from psi to psi depending upon the tree species and direction of wood grain; the modulus of concrete will depend on the concrete/aggregate ratio and curing process). xamples of nonmetals commonly used are concrete (ultimate compressive strength of 5 ksi but zero tensile strength; with an elastic modulus in compression of psi) and Douglas fir (parallel to grain, ultimate compressive strength of 7 ksi; with an elastic modulus of psi). Material Density ( lb in 3 ) Young s Modulus (10 6 psi) Poisson s Ratio Yield Strength (ksi) Ultimate Tensile Strength (ksi) Coefficient of Thermal xpansion Tension Shear Aluminum 2024T (200 F) 6061T ( F) Steel Structural (A36) AISI ( F) 5CrMoV ( F) Copper G3Heat Treated ( F) Titanium Ti5Al2.5Sn ( F) Ti6M4V ( F) Table 9.1: Structural Material Properties for Selected Metals (US Customary Units) ( 10 6 F ) Given: xample 91 [σ] = y Mz I zz 0 0 where M z and I zz are constants Required : (a) Verify that the stress tensor satisfies the Conservation of Linear Momentum. (b) Determine the components of the infinitesimal strain tensor, ε. (c) Determine the components of the displacement, u x, u y, and u z. (d) Describe the displacement and physical problem described by these equations. Use as reference the figure below.
11 9.2. CONSTITUTIV QUATIONS WITH THRMAL STRAIN 221 Material Density ( Mg m 3 ) Young s Modulus (GPa) Poisson s Ratio Yield Strength (MPa) Ultimate Tensile Strength (MPa) Tension Shear Aluminum 2024T T Steel Structural A Stainless Copper Alloy Bronze C Titanium Ti6Al4V , Ti6M4V Table 9.2: Structural Material Properties for Selected Metals (SI Units) Coefficient of Thermal xpansion ( 10 6 C ) y y z h x t L Figure 9.8: (a) xcomponent of linear momentum: x σ xx x + σ xy y + σ xz z + ρg x =0 ( ) y Mz I zz x =0  Stress tensor satisfies the Conservation of Linear Momentum (b) The strains are given by: ε xx = σ xx, = ε zz = ν σ xx, ε xy = ε xz = ε yz =0 ε xx = y M z I zz, = νym z I zz, ε zz = νym z I zz
12 222 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS (c) Integrate the displacement equations and apply boundary conditions: ε xx = u x x = y M z I zz u x = y M z = u y y = νym ν z u y = I zz x + C 1 I ) zz M z ( y 2 2 I zz + C 2 ε zz = u z z = νym z I zz u z = νym z I zz z + C 3 u x x=0 =0, C 1 =0 u y y=0 =0, C 2 =0 u z z=0 =0, C 3 =0 u x = yx M z I zz u y = νv M 2 y2 z I zz u z = ν yz M z I zz (d) The displacement in the xdirection is negative (shortening) when y is positive due to the negative u x term. If y is negative the displacement in the xdirection is positive (expanding). In the ydirection and zdirection the displacement is expanding when y is greater than zero and vice versa. Displacement in the ydirection changes with respect to y 2, and in the zdirection it changes with respect to y.
13 9.2. CONSTITUTIV QUATIONS WITH THRMAL STRAIN 223 Deep Thought Ut tensio sic vis! Ut tensio sic vis!! Ut tensio sic vis!!!
14 224 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS 9.3 Questions 9.1 Which conservation laws are especially useful for describing stresses and strains? How are stress and strain related? 9.2 Write the equations that result from an inversion of the stressstrain relationship. 9.3 Describe in your own words the meanings of the state of plane stress and the state of plane strain? 9.4 Describe the two types of problems which when solved using the theory of plane elasticity provide exact solutions. 9.5 Consider small shear strain for a moment. It is often given in terms of an angle. xplain why this is done. 9.6 What is a constitutive relation? Write down the general constitutive relation in terms of Cauchy stress and strain. 9.7 For an elastic, isotropic solid material, how many constants are required to define the constitutive relations? Name these and define their meaning. 9.4 Problems 9.8 Structural steel is subjected to the deformation defined by u x ( x, y, z )=0.002x, u y ( x, y, z )= 0, u z ( x, y, z )=0(displacements in inches). Determine the following in US units: a) Infinitesimal strain tensor. b) Stress tensor. c) Draw Mohr s Circle for the given state of stress. d) Principal Stresses and Strains. 9.9 Repeat steps a and b in 9.8 for u x ( x, y, z ) = 0.002x x, u y ( x, y, z ) = 0.002xy, u z ( x, y, z )=0.001z GIVN :AHookean material with = psi and ν =0.5 experiences the following deformation: u x ( x, y, z )=0,u y ( x, y, z )=0.004x, u z ( x, y, z )=0 RQUIRD: a) Sketch u x versus x, u y versus y, and u z versus z, and calculate ux x, uy y, uz z. b) Calculate the infinitesimal strain tensor. c) Calculate the stress tensor GIVN : ν =0.25 and = Pa, and strain tensors as follows (1) , (2) RQUIRD: (1) Calculate the stress tensors; (2) How much is the relative volume change (the dilatation) for this deformation, and compare the results obtained by using both finite strain formula and the small strain formula.
15 9.4. PROBLMS GIVN : ν =0.33 and = MPa, and stress tensors as follows 10 MPa 4 MPa 0 20 MPa 50 MPa 0 (1) 4MPa 30MPa 0 (2) 50 MPa MPa RQUIRD: Calculate the strain tensors GIVN : u x = z10 4, u z = x10 4, and u y =0,and material constants = , and ν =0.3. (1) Compute infinitesimal strain tensor. (2) Compute the corresponding stress tensor. (3) What are the principal stresses and principal strains? (4) Are the principal stresses and strains acting in the same directions? 9.14 A steel plate lies flat in the xy plane and has dimensions 20 cm 40 cm. If the plate is uniformly heated throughout at 1000 C and the thermal expansion coefficient is given by α = , calculate the new dimensions of the plate due to thermal expansion. m (m  C) 9.15 A thin rectangular sheet of linearly elastic material has an xy coordinate system located at its lower left corner. The body extends 15 in. in the x direction and 8 in. in the y direction. The material is isotropic with an =35, 000, 000 psi and ν =0.33. A plane stress condition has been created by forces acting along the edges of the body with a displacement field of: u x = x 2 y u y = xy 2 Write expressions for the surface force normal to and for the tangential surface force along the upper 15 in. boundary as functions of x. Write expressions for the surface force normal to and for the tangential surface force along the right 8 in. boundary as functions of y. Draw the distribution of normal surface force along these two boundaries on a sketch of the body Use web resources to determine the following material properties. Provide the URL (http address) that you used. (a) Yield strength in shear of 2024T4 and 2014T6 aluminum. (b) Poisson s ratio and yield strength in shear for Ti5Al2.5Sn. (c) All of the table values as presented in Table 9.1 for 4130 heat treated alloy steel. (d) All of the table values as presented in Table 9.1 for balsa wood GIVN : The isothermal (no temperature gradient) uniaxial bar specimen of 2024T4 Aluminum (isotropic) shown below: The axial displacements are measured to be: RQUIRD: u x = 0.02x in u y = x in x in inches!! 1. Sketch the deformed configuration of the test section boundary (using the displacements given above). 2. Calculate the infinitesimal strain tensor for the test section.
16 226 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS Problem Calculate the stress tensor for the test section GIVN :Alinear isotropic Thermolastic plate of Stainless 304 is subjected to a uniform temperature change of T and is assumed to be in a state of stress as shown below. At equilibrium, the T is known. σ = MPa RQUIRD: a) Calculate the infinitesimal strain tensor when T =0 C. (review equations 9.12 and 9.13 in the notes) b) Calculate the infinitesimal strain tensors for the two cases: when T = 100 C, and when T =25 C. c) Find the temperature change T necessary to produce zero strain GIVN : Consider the state of stress called plane stress in which nonzero stresses exist in only one plane. RQUIRD: a) For a state of plane stress in the xy plane, show that the constitutive equations for an elastic isotropic material (isothermal case) reduce to the following. Hint: start with the constitutive equations for the general 3D elastic, isotropic case and reduce to plane stress; see equation 10.6): SHOW ALL STPS. σ xx = = σ xy = (1 ν 2 ) [ε xx + ν ] (1 ν 2 ) [νε xx + ] (1 + ν) ε xy
17 9.4. PROBLMS 227 b) Starting with the above relations, show that the strains for plane stress in the xy plane become those shown below (see equation 10.7). You must show all steps necessary to obtain the relations below. ε xx = 1 [σ xx ν ] = 1 [ νσ xx ] ε xy = ( 1+ν )σ xy ε zz = ν (σ xx + ) You may use Scientific Workplace to do the matrix algebra GIVN : Given that for a general orthotropic elastic material there are 12 unique coeffecients such that: [D] = 1 11 ν12 11 ν13 11 ν23 22 ν µ µ µ 12 ν ν31 33 The constitutive equation for this form would then be: where the stress have the following values RQUIRD: {σ} = σ xx =5ksi =10ksi σ zz =20ksi σ yz =0ksi σ zx =0ksi σ xy =7.5 ksi {ε} =[D] {σ} ; {ε} = a) Write the stress tensor in its more common form (i.e., as a tensor or matrix). Does this constitute generalized plane stress? Why or why not? ε xx ε zz ε yz ε zx ε xy Recall that generalized plane stress is a requirement for Mohr s Circle b) Suppose that the 12 material coefficients have the following values: 11 = 10 6 psi 22 = psi 33 = psi
18 228 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS ν 12 = 0.2 ν 13 = 0.25 ν 21 = 0.33 ν 23 = 0.43 ν 31 = 0.05 ν 32 = 0.06 µ 23 = 10 4 psi µ 31 = psi µ 12 = psi Calculate the infinitesimal strain tensor. c) Write the strain tensor in its more common form. Does this constitute generalized plane strain? Why or why not?
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