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(BQ) Part 2 book The finite element method in engineering has contents: Basic equations and solution procedure; basic equations and solution procedure, analysis of plates, analysis of three dimensional problems, dynamic analysis, formulation and solution procedure,...and other contents.
APPLICATION TO SOLID MECHANICS PROBLEMS This Page Intentionally Left Blank BASIC EQUATIONS AND SOLUTION PROCEDURE 8.1 INTRODUCTION As stated in Chapter 1, the finite element method has been most extensively used in the field of solid and structural mechanics The various types of problems solved by the finite element method in this field include the elastic, elastoplastic, and viscoelastic analysis of trusses, frames, plates, shells, and solid bodies Both static and dynamic analysis have been conducted using the finite element method We consider the finite element elastic analysis of one-, two-, and three-dimensional problems as well as axisymmetric problems in this book In this chapter, the general equations of solid and structural mechanics are presented The displacement method (or equivalently the principle of minimum potential energy) is used in deriving the finite element equations The application of these equations to several specific cases is considered in subsequent chapters 8.2 BASIC EQUATIONS OF SOLID MECHANICS 8.2.1 Introduction The primary aim of any stress analysis or solid mechanics problem is to find the distribution of displacements and stresses under the stated loading and boundary conditions If an analytical solution of the problem is to be found, one has to satisfy the following basic equations of solid mechanics: Number of equations Type of equations Equilibrium equations Stress–strain relations Strain–displacement relations Total number of equations In 3-dimensional problems In 2-dimensional problems In 1-dimensional problems 6 3 1 15 279 280 BASIC EQUATIONS AND SOLUTION PROCEDURE The unknown quantities, whose number is equal to the number of equations available, in various problems are given below: In 3-dimensional problems Unknowns Displacements Stresses u, v, w σxx , σyy , σzz , σxy , σyz , σzx εxx , εyy , εzz , εxy εyz , εzx Strains Total number of unknowns 15 In 2-dimensional problems In 1-dimensional problems u, v σxx , σyy , σxy u σxx εxx , εyy , εxy εxx Thus, we have as many equations as there are unknowns to find the solution of any stress analysis problem In practice, we will also have to satisfy some additional equations, such as external equilibrium equations (which pertain to the overall equilibrium of the body under external loads), compatibility equations (which pertain to the continuity of strains and displacements), and boundary conditions (which pertain to the prescribed conditions on displacements and/or forces at the boundary of the body) Although any analytical (exact) solution has to satisfy all the equations stated previously, the numerical (approximate) solutions, like the ones obtained by using the finite element method, generally not satisfy all the equations However, a sound understanding of all the basic equations of solid mechanics is essential in deriving the finite element relations and also in estimating the order of error involved in the finite element solution by knowing the extent to which the approximate solution violates the basic equations, including the compatibility and boundary conditions Hence, the basic equations of solid mechanics are summarized in the following section for ready reference in the formulation of finite element equations 8.2.2 Equations (i) External equilibrium equations If a body is in equilibrium under specified static loads, the reactive forces and moments developed at the support points must balance the externally applied forces and moments In other words, the force and moment equilibrium equations for the overall body (overall or external equilibrium equations) have to be satisfied If φx , φy , and φz are the body forces, Φx , Φy , and Φz are the surface (distributed) forces, Px , Py , and Pz are the external concentrated loads (including reactions at support points such as B, C, and D in Figure 8.1), and Qx , Qy , and Qz are the external concentrated moments (including reactions at support points such as B, C, and D in Figure 8.1), the external equilibrium equations can be stated as follows [8.1]: ⎫ Φx ds + φx dv + Px = ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ S V ⎪ ⎪ ⎪ ⎬ Φy ds + φy dv + Py = (8.1) ⎪ ⎪ S V ⎪ ⎪ ⎪ ⎪ ⎪ Φz ds + φz dv + Px = ⎪ ⎪ ⎭ S V BASIC EQUATIONS OF SOLID MECHANICS 281 Figure 8.1 Force System for Macroequilibrium for a Body For moment equilibrium: (Φz y − Φy z) ds + S (φz y − φy z) dv + V (Φx z − Φz x) ds + S (φx z − φz x) dv + V (Φy x − Φx y) ds + S (φy x − φx y) dv + ⎫ Qx = 0⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ Qy = ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Qz = 0⎪ ⎪ ⎭ (8.2) V where S is the surface and V is the volume of the solid body (ii) Equations of internal equilibrium: Due to the application of loads, stresses will be developed inside the body If we consider an element of material inside the body, it must be in equilibrium due to the internal stresses developed This leads to equations known as internal equilibrium equations Theoretically, the state of stress at any point in a loaded body is completely defined in terms of the nine components of stress σxx , σyy , σzz , σxy , σyx , σyz , σzy , σzx , and σxz , where the first three are the normal components and the latter six are the components of shear stress The equations of internal equilibrium relating the nine components of stress can be derived by considering the equilibrium of moments and forces acting on the elemental volume shown in Figure 8.2 The equilibrium of moments about the x, y, and z axes, assuming that there are no body moments, leads to the relations σyx = σxy , σzy = σyz , σxz = σzx (8.3) These equations show that the state of stress at any point can be completely defined by the six components σxx , σyy , σzz , σxy , σyz , and σzx The equilibrium of forces in x, y, 282 BASIC EQUATIONS AND SOLUTION PROCEDURE Figure 8.2 Elemental Volume Considered for Internal Equilibrium (Only the Components of Stress Acting on a Typical Pair of Faces Are Shown for Simplicity) and z directions gives the following differential equilibrium equations: ⎫ ∂σxx ∂σxy ∂σzx + + + φx = 0⎪ ⎪ ⎪ ⎪ ∂x ∂y ∂z ⎪ ⎪ ⎪ ⎪ ⎬ ∂σxy ∂σyy ∂σyz + + + φy = ⎪ ∂x ∂y ∂z ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂σzx ∂σyz ∂σzz ⎭ + + + φz = 0⎪ ∂x ∂y ∂z (8.4) where φx , φy , and φz are the body forces per unit volume acting along the directions x, y, and z, respectively For a two-dimensional problem, there will be only three independent stress components (σxx , σyy , σxy ) and the equilibrium equations, Eqs (8.4), reduce to ⎫ ∂σxx ∂σxy ⎪ + + φx = 0⎪ ⎪ ⎬ ∂x ∂y ∂σxy ∂σyy + ∂x ∂y ⎪ ⎪ ⎭ + φy = 0⎪ (8.5) In one-dimensional problems, only one component of stress, namely σxx , will be there and hence Eqs (8.4) reduce to ∂σxx + φx = ∂x (8.6) BASIC EQUATIONS OF SOLID MECHANICS 283 (iii) Stress–strain relations (constitutive relations) for isotropic materials Three-dimensional case In the case of linearity elastic isotropic three-dimensional solid, the stress–strain relations are given by Hooke’s law as ⎧ ⎫ ⎫ ⎧ ⎫ ⎧ εxx0 ⎪ σxx ⎪ εxx ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ε σ ε ⎪ ⎪ ⎪ ⎪ ⎪ yy yy yy 0⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎬ ⎨ ⎬ ⎨ εzz σzz εzz0 ε= = [C]σ + ε0 ≡ [C] + εxy ⎪ σxy ⎪ εxy0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ε σ ε ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ yz yz yz ⎪ ⎭ ⎪ ⎪ ⎪ ⎪ ⎩ ⎩ ⎭ ⎪ ⎭ ⎩ εzx σzx εzx0 (8.7) where [C] is a matrix of elastic coefficients given by ⎡ ⎢−v ⎢ ⎢ ⎢−v [C] = E⎢ ⎢ ⎣ 0 −v −v 0 −v −v 0 0 0 2(1 + v) 0 0 0 2(1 + v) ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ 2(1 + v) (8.8) ε0 is the vector of initial strains, E is Young’s modules, and v is Poisson’s ratio of the material In the case of heating of an isotropic material, the initial strain vector is given by ⎫ ⎧ ε ⎪ xx ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ε ⎪ ⎪ yy ⎪ ⎪ ⎬ ⎨ εzz0 ε0 = = αT εxy0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ εyz0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ εzx0 ⎧ ⎫ 1⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ ⎪ 0⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ ⎪ (8.9) where α is the coefficient of thermal expansion, and T is the temperature charge Sometimes, the expressions for stresses in terms of strains will be needed By including thermal strains, Eqs (8.7) can be inverted to obtain ⎧ ⎧ ⎫ ⎫ ⎧ ⎫ σ 1⎪ ε ⎪ ⎪ ⎪ ⎪ xx xx ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ σ ε ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ yy yy ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎨ ⎪ ⎬ ⎨ ⎬ EαT σzz εzz σ= = [D](ε − ε0 ) ≡ [D] − 0⎪ σxy ⎪ εxy ⎪ ⎪ ⎪ − 2v ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ σ ε ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ yz ⎪ yz ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎭ ⎩ ⎭ ⎩ ⎪ σzx εzx (8.10) 284 BASIC EQUATIONS AND SOLUTION PROCEDURE where the matrix [D] is given by ⎡ 1−v ⎢ v ⎢ ⎢ v ⎢ ⎢ ⎢ E ⎢ [D] = (1 + v)(1 − 2v) ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ v 1−v v v v 1−v 0 0 0 0 − 2v 0 0 0 0 − 2v 0 − 2v ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ (8.11) In the case of two-dimensional problems, two types of stress distributions, namely plane stress and plane strain, are possible Two-dimensional case (plane stress) The assumption of plane stress is applicable for bodies whose dimension is very small in one of the coordinate directions Thus, the analysis of thin plates loaded in the plane of the plate can be made using the assumption of plane stress In plane stress distribution, it is assumed that σzz = σzx = σyz = (8.12) where z represents the direction perpendicular to the plane of the plate as shown in Figure 8.3, and the stress components not vary through the thickness of the plate (i.e., in z direction) Although these assumptions violate some of the compatibility conditions, they are sufficiently accurate for all practical purposes provided the plate is thin In this case, the stress–strain relations, Eqs (8.7) and (8.10), reduce to ε = [C]σ + ε0 (8.13) y y x z Figure 8.3 Example of a Plane Stress Problem; A Thin Plate under Inplane Loading BASIC EQUATIONS OF SOLID MECHANICS 285 where ⎧ ⎫ ⎧ ⎫ ⎨σxx ⎬ ⎨εxx ⎬ σ = σyy ε = εyy , ⎩ ⎭ ⎩ ⎭ εxy σxy ⎤ ⎡ −v ⎣ ⎦ −v [C] = E 0 2(1 + v) ⎧ ⎫ ⎧ ⎫ ⎨1⎬ ⎨εxx0 ⎬ in the case of thermal strains ε0 = εyy0 = αT ⎩ ⎭ ⎩ ⎭ εxy0 (8.14) (8.15) and ⎧ ⎫ ⎨1⎬ EαT σ = [D](ε − ε0 ) = [D]ε − 1−v ⎩ ⎭ (8.16) with ⎡ ⎢ E ⎢v [D] = − v2 ⎣ v 0 1−v ⎤ ⎥ ⎥ ⎦ (8.17) In the case of plane stress, the component of strain in the z direction will be nonzero and is given by (from Eq 8.7) εzz = − v −v 1+v (σxx + σyy ) + αT = (εxx + εyy ) + αT E 1−v 1−v (8.18) while εyz = εzx = (8.19) Two-dimensional case (plane strain) The assumption of plane strain is applicable for bodies that are long and whose geometry and loading not vary significantly in the longitudinal direction Thus, the analysis of dams, cylinders, and retaining walls shown in Figure 8.4 can be made using the assumption of plane strain In plane strain distribution, it is assumed that w = and (∂w/∂z) = at every cross section Here, the dependent variables are assumed to be functions of only the x and y coordinates provided we consider a cross section of the body away from the ends In this case, the three-dimensional stress–strain relations given by Eqs (8.7) and (8.10) reduce to ε = [C]σ + ε0 (8.20) 286 BASIC EQUATIONS AND SOLUTION PROCEDURE x z y (a) Dam z x x z y y (c) Retaining wall (b) Long cylinder Figure 8.4 Examples of Plane Strain Problems where ⎧ ⎫ ⎧ ⎫ ⎨σxx ⎬ ⎨εxx ⎬ σ = σyy , ε = εyy , ⎩ ⎩ ⎭ ⎭ εxy σxy ⎤ ⎡ 1−v −v 1+y ⎣ −v − v 0⎦ , [C] = E 0 ⎧ ⎫ ⎧ ⎫ ⎨1⎬ ⎨εxx0 ⎬ ε0 = εyy0 = (1 + v)αT ⎩ ⎭ ⎩ ⎭ εxy0 (8.21) in the case of thermal strains (8.22) and ⎧ ⎫ ⎨1⎬ EαT σ = [D](ε − ε0 ) = [D]ε − − 2v ⎩ ⎭ (8.23) ... 2 = Q 12 ⎩ ⎭ σ 12 Q 12 Q 22 ⎤⎧ ⎫ ⎨ ε1 ⎬ ⎦ 2 ≡ [Q]ε ⎩ ⎭ Q66 ε 12 (8. 42) where the elements of the matrix [Q] are given by ⎫ E 22 ⎪ ⎪ − v 12 v21 ⎪ ⎪ ⎪ ⎪ ⎬ Q11 = E11 , − v 12 v21 Q 12 = v21 E11 v 12 E 22. .. constants The elements of the compliance matrix, in this case, can be expressed as E11 = E 22 v 12 v21 =− =− E11 E 22 = G 12 C11 = C 22 C 12 C66 (8.41) 29 0 BASIC EQUATIONS AND SOLUTION PROCEDURE The stress–strain... cylinder is given by σrr = b2 a2 p − , b2 − a2 r2 σθθ = b2 a2 p + , b2 − a2 r2 σrθ = where a, b, and p denote the inner radius, outer radius, and internal pressure, respectively, determine whether