김경종
452 lines
22 KiB
Markdown
452 lines
22 KiB
Markdown
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<details>
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<summary>text_image</summary>
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A ← 25 ft → 50 ft → 25 ft → D
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B
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C
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10 ft
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E
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F
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15 ft
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</details>
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<details>
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<summary>text_image</summary>
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0.2 W
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0.8 W
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14 ft
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W = total weight of truck and load
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H20 - 44 8 k
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32 k
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H truck loading
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</details>
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Figure P5–71
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5.72 For the tripod space frame shown in Figure P5–72, determine standard steel pipe sections such that the maximum bending stress must not exceed 20 ksi, the compressive stress to prevent buckling must not exceed that given by the Euler buckling formula with a factor of safety of 2 and the maximum deflection will not exceed $\mathrm { L } / 3 6 0$ in any span, L. Assume the three bottom supports to be fixed. All coordinates shown in units of inches.
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<details>
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<summary>text_image</summary>
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1000 lb
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(-20, 30, 60)
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1000 lb
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(0, 10, 60)
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1000 lb
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(20, 30, 60)
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(-30, 40, 0)
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(0, 0, 0)
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(30, 40, 0)
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z
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y
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x
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</details>
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Figure P5–72
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# Development of the Plane Stress and Plane Strain Stiffness Equations
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# Introduction
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In Chapters 2–5, we considered only line elements. Two or more line elements are connected only at common nodes, forming framed or articulated structures such as trusses, frames, and grids. Line elements have geometric properties such as crosssectional area and moment of inertia associated with their cross sections. However, only one local coordinate x^ along the length of the element is required to describe a position along the element (hence, they are called line elements or one-dimensional elements). Nodal compatibility is then enforced during the formulation of the nodal equilibrium equations for a line element.
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This chapter considers the two-dimensional finite element. Two-dimensional (planar) elements are defined by three or more nodes in a two-dimensional plane (that is, x-y). The elements are connected at common nodes and/or along common edges to form continuous structures such as those shown in Figures 1–3, 1–4, 1–6, and 6–6(b). Nodal displacement compatibility is then enforced during the formulation of the nodal equilibrium equations for two-dimensional elements. If proper displacement functions are chosen, compatibility along common edges is also obtained. The two-dimensional element is extremely important for (1) plane stress analysis, which includes problems such as plates with holes, fillets, or other changes in geometry that are loaded in their plane resulting in local stress concentrations, such as illustrated in Figure 6–1; and (2) plane strain analysis, which includes problems such as a long underground box culvert subjected to a uniform load acting constantly over its length, as illustrated in Figure 1–3, a long, cylindrical control rod subjected to a load that remains constant over the rod length (or depth), as illustrated in Figure 1–4, and dams and pipes subjected to loads that remain constant over their lengths as shown in Figure 6–2.
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We begin this chapter with the development of the stiffness matrix for a basic two-dimensional or plane finite element, called the constant-strain triangular element. We consider the constant-strain triangle (CST) stiffness matrix because its derivation
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is the simplest among the available two-dimensional elements. The element is called a CST because it has a constant strain throughout it.
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We will derive the CST stiffness matrix by using the principle of minimum potential energy because the energy formulation is the most feasible for the development of the equations for both two- and three-dimensional finite elements.
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We will then present a simple, thin-plate plane stress example problem to illustrate the assemblage of the plane element stiffness matrices using the direct stiffness method as presented in Chapter 2. We will present the total solution, including the stresses within the plate.
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# 6.1 Basic Concepts of Plane Stress and Plane Strain
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In this section, we will describe the concepts of plane stress and plane strain. These concepts are important because the developments in this chapter are directly applicable only to systems assumed to behave in a plane stress or plane strain manner. Therefore, we will now describe these concepts in detail.
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# Plane Stress
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Plane stress is defined to be a state of stress in which the normal stress and the shear stresses directed perpendicular to the plane are assumed to be zero. For instance, in Figures 6–1(a) and 6–1(b), the plates in the x-y plane shown subjected to surface tractions T (pressure acting on the surface edge or face of a member in units of force/area) in the plane are under a state of plane stress; that is, the normal stress $\sigma _ { z }$ and the shear stresses $\tau _ { x z }$ and $\tau _ { y z }$ are assumed to be zero. Generally, members that are thin (those with a small z dimension compared to the in-plane x and y dimensions) and whose loads act only in the x-y plane can be considered to be under plane stress.
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# Plane Strain
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Plane strain is defined to be a state of strain in which the strain normal to the x-y plane ez and the shear strains $\gamma _ { x z }$ and $\gamma _ { y z }$ are assumed to be zero. The assumptions of plane strain are realistic for long bodies (say, in the z direction) with constant cross-sectional area subjected to loads that act only in the x and/or y directions and do not vary in the
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<details>
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<summary>text_image</summary>
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y
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T
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x
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</details>
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(a)
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<details>
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<summary>text_image</summary>
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T
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y
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x
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T
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</details>
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(b)
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Figure 6–1 Plane stress problems: (a) plate with hole; (b) plate with fillet
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Figure 6–2 Plane strain problems: (a) dam subjected to horizontal loading; (b) pipe subjected to a vertical load
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z direction. Some plane strain examples are shown in Figure 6–2 [and in Figures 1–3 (a long underground box culvert) and 1–4 (a hydraulic cylinder rod end)]. In these examples, only a unit thickness (1 in. or 1 ft) of the structure is considered because each unit thickness behaves identically (except near the ends). The finite element models of the structures in Figure 6–2 consist of appropriately discretized cross sections in the x-y plane with the loads acting over unit thicknesses in the x and/or y directions only.
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# Two-Dimensional State of Stress and Strain
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The concept of a two-dimensional state of stress and strain and the stress/strain relationships for plane stress and plane strain are necessary to understand fully the development and applicability of the stiffness matrix for the plane stress/plane strain triangular element. Therefore, we briefly outline the essential concepts of two-dimensional stress and strain (see References [1] and [2] and Appendix C for more details on this subject).
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First, we illustrate the two-dimensional state of stress using Figure 6–3. The infinitesimal element with sides dx and $d y$ has normal stresses $\sigma _ { x }$ and $\sigma _ { y }$ acting in the x and y directions (here on the vertical and horizontal faces), respectively. The shear stress $\tau _ { x y }$ acts on the x edge (vertical face) in the y direction. The shear stress $\tau _ { y x }$ acts on the y edge (horizontal face) in the x direction. Moment equilibrium of the element results in $\tau _ { x y }$ being equal in magnitude to $\tau _ { y x } .$ . See Appendix C.1 for proof of this equality. Hence, three independent stresses exist and are represented by the vector column matrix
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$$
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\{\sigma \} = \left\{ \begin{array}{l} \sigma_ {x} \\ \sigma_ {y} \\ \tau_ {x y} \end{array} \right\} \tag {6.1.1}
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$$
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The element equilibrium equations are derived in Appendix C.1.
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<details>
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<summary>text_image</summary>
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σy
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τyx
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σx ← τxy
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dy
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τxy
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σx
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dx
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τyx ← σy
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σy
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y
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x
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</details>
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Figure 6–3 Two-dimensional state of stress
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The stresses given by Eq. (6.1.1) will be expressed in terms of the nodal displacement degrees of freedom. Hence, once the nodal displacements are determined, these stresses can be evaluated directly.
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Recall from strength of materials [2] that the principal stresses, which are the maximum and minimum normal stresses in the two-dimensional plane, can be obtained from the following expressions:
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$$
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\sigma_ {1} = \frac {\sigma_ {x} + \sigma_ {y}}{2} + \sqrt {\left(\frac {\sigma_ {x} - \sigma_ {y}}{2}\right) ^ {2} + \tau_ {x y} ^ {2}} = \sigma_ {\max} \tag {6.1.2}
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$$
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$$
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\sigma_ {2} = \frac {\sigma_ {x} + \sigma_ {y}}{2} - \sqrt {\left(\frac {\sigma_ {x} - \sigma_ {y}}{2}\right) ^ {2} + \tau_ {x y} ^ {2}} = \sigma_ {\min}
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$$
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Also, the principal angle $\theta _ { p } ,$ which defines the normal whose direction is perpendicular to the plane on which the maximum or minimum principal stress acts, is defined by
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$$
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\tan 2 \theta_ {p} = \frac {2 \tau_ {x y}}{\sigma_ {x} - \sigma_ {y}} \tag {6.1.3}
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$$
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Figure 6–4 shows the principal stresses $\sigma _ { 1 }$ and $\sigma _ { 2 }$ and the angle $\theta _ { p }$ . Recall (as Figure 6–4 indicates) that the shear stress is zero on the planes having principal (maximum and minimum) normal stresses.
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In Figure 6–5, we show an infinitesimal element used to represent the general two-dimensional state of strain at some point in a structure. The element is shown to be displaced by amounts u and v in the x and y directions at point A, and to displace or extend an additional (incremental) amount $( \hat { o } u / \hat { o } x )$ dx along line AB, and $( \hat { o } v / \hat { o } y )$ dy along line AC in the x and y directions, respectively. Furthermore, observing lines AB and AC, we see that point B moves upward an amount $( \hat { o } v / \hat { o } x )$ dx with respect to A, and point C moves to the right an amount $( \hat { o } u / \hat { o } y )$ dy with respect to A.
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<!-- source-page: 326 -->
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<details>
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<summary>text_image</summary>
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σ₂
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σ₁
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σ₁
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σ₂
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θₚ
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x
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</details>
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Figure 6–4 Principal stresses and their directions
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<details>
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<summary>text_image</summary>
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v + \frac{\partial v}{\partial y} dy
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\frac{\partial u}{\partial y} dy
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C
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\frac{\partial u}{\partial y} \frac{\partial v}{\partial x}
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A
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B
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\frac{\partial v}{\partial x} dx
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v
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u
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dx
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u + \frac{\partial u}{\partial x} dx
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y, v
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x, u
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</details>
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Figure 6–5 Displacements and rotations of lines of an element in the x-y plane
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From the general definitions of normal and shear strains and the use of Figure $6 { - } 5 ,$ we obtain
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$$
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\varepsilon_ {x} = \frac {\partial u}{\partial x} \quad \varepsilon_ {y} = \frac {\partial v}{\partial y} \quad \gamma_ {x y} = \frac {\partial u}{\partial y} + \frac {\partial v}{\partial x} \tag {6.1.4}
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$$
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Appendix C.2 shows a detailed derivation of Eqs. (6.1.4). Hence, recall that the strains $\varepsilon _ { x }$ and $\varepsilon _ { y }$ are the changes in length per unit length of material fibers originally parallel to the x and y axes, respectively, when the element undergoes deformation. These strains are then called normal (or extensional or longitudinal ) strains. The strain $\gamma _ { x y }$ is the change in the original right angle made between dx and $d y$ when the element undergoes deformation. The strain $\gamma _ { x y }$ is then called a shear strain.
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The strains given by Eqs. (6.1.4) are generally represented by the vector column matrix
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$$
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\{\varepsilon \} = \left\{ \begin{array}{l} \varepsilon_ {x} \\ \varepsilon_ {y} \\ \gamma_ {x y} \end{array} \right\} \tag {6.1.5}
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$$
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<!-- source-page: 327 -->
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The relationships between strains and displacements referred to the x and y directions given by Eqs. (6.1.4) are sufficient for your understanding of subsequent material in this chapter.
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We now present the stress/strain relationships for isotropic materials for both plane stress and plane strain. For plane stress, we assume the following stresses to be zero:
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$$
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\sigma_ {z} = \tau_ {x z} = \tau_ {y z} = 0 \tag {6.1.6}
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$$
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Applying Eq. (6.1.6) to the three-dimensional stress/strain relationship [see Appendix C, Eq. (C.3.10)], the shear strains $\gamma _ { x z } = \gamma _ { y z } = 0$ , but $\varepsilon _ { z } \neq 0$ . For plane stress conditions, we then have
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$$
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\{\sigma \} = [ D ] \{\varepsilon \} \tag {6.1.7}
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$$
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where
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$$
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[ D ] = \frac {E}{1 - \nu^ {2}} \left[ \begin{array}{c c c} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac {1 - \nu}{2} \end{array} \right] \tag {6.1.8}
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$$
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is called the stress/strain matrix (or constitutive matrix), E is the modulus of elasticity, and n is Poisson’s ratio. In Eq. (6.1.7), fsg and feg are defined by Eqs. (6.1.1) and (6.1.5), respectively.
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For plane strain, we assume the following strains to be zero:
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$$
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\varepsilon_ {z} = \gamma_ {x z} = \gamma_ {y z} = 0 \tag {6.1.9}
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$$
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Applying Eq. (6.1.9) to the three-dimensional stress/strain relationship [Eq. (C.3.10)], the shear stresses $\tau _ { x z } = \tau _ { y z } = 0$ , but $\sigma _ { z } \neq 0$ . The stress/strain matrix then becomes
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$$
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[ D ] = \frac {E}{(1 + \nu) (1 - 2 \nu)} \left[ \begin{array}{c c c} 1 - \nu & \nu & 0 \\ \nu & 1 - \nu & 0 \\ 0 & 0 & \frac {1 - 2 \nu}{2} \end{array} \right] \tag {6.1.10}
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$$
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The $\{ \sigma \}$ and $\{ \varepsilon \}$ matrices remain the same as for the plane stress case. The basic partial differential equations for plane stress, as derived in Reference [1], are
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$$
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\frac {\partial^ {2} u}{\partial x ^ {2}} + \frac {\partial^ {2} u}{\partial y ^ {2}} = \frac {1 + v}{2} \left(\frac {\partial^ {2} u}{\partial y ^ {2}} - \frac {\partial^ {2} v}{\partial x \partial y}\right) \tag {6.1.11}
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$$
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$$
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\frac {\partial^ {2} v}{\partial x ^ {2}} + \frac {\partial^ {2} v}{\partial y ^ {2}} = \frac {1 + v}{2} \left(\frac {\partial^ {2} v}{\partial x ^ {2}} - \frac {\partial^ {2} u}{\partial x \partial y}\right)
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$$
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<!-- source-page: 328 -->
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# 6.2 Derivation of the Constant-Strain Triangular Element Stiffness Matrix and Equations
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To illustrate the steps and introduce the basic equations necessary for the plane triangular element, consider the thin plate subjected to tensile surface traction loads $T _ { S }$ in Figure 6–6(a).
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<details>
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<summary>text_image</summary>
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y, v
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T_S
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T_S
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x, u
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</details>
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Figure 6–6(a) Thin plate in tension
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<details>
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<summary>text_image</summary>
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y
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m
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i
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j
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n
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x
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</details>
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Figure 6–6(b) Discretized plate of Figure 6–6(a) using triangular elements
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# Step 1 Select Element Type
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To analyze the plate, we consider the basic triangular element in Figure 6–7 taken from the discretized plate, as shown in Figure 6–6(b). The discretized plate has been divided into triangular elements, each with nodes such as $i , j ,$ and m. We use triangular elements because boundaries of irregularly shaped bodies can be closely approximated in this way, and because the expressions related to the triangular element are comparatively simple. This discretization is called a coarse-mesh generation if a few large elements are used. Each node has two degrees of freedom—an x and a y displacement. We will let $u _ { i }$ and $v _ { i }$ represent the node i displacement components in the x and y directions, respectively.
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Here all formulations are based on this counterclockwise system of labeling of nodes, although a formulation based on a clockwise system of labeling could be used. Remember that a consistent labeling procedure for the whole body is necessary
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<details>
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<summary>text_image</summary>
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y
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vi
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i(xi, yi)
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ui
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vj
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m(xm, ym)
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vm
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um
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j(xj, yj)
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uj
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x
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</details>
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Figure 6–7 Basic triangular element showing degrees of freedom
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<!-- source-page: 329 -->
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to avoid problems in the calculations such as negative element areas. Here $( x _ { i } , y _ { i } )$ , $( x _ { j } , y _ { j } )$ , and $\left( x _ { m } , y _ { m } \right)$ are the known nodal coordinates of nodes $i , j ,$ and $m ,$ respectively.
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The nodal displacement matrix is given by
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$$
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\{d \} = \left\{ \begin{array}{l} \underline {{d}} _ {i} \\ \underline {{d}} _ {j} \\ \underline {{d}} _ {m} \end{array} \right\} = \left\{ \begin{array}{l} u _ {i} \\ v _ {i} \\ u _ {j} \\ v _ {j} \\ u _ {m} \\ v _ {m} \end{array} \right\} \tag {6.2.1}
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$$
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# Step 2 Select Displacement Functions
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We select a linear displacement function for each element as
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$$
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u (x, y) = a _ {1} + a _ {2} x + a _ {3} y \tag {6.2.2}
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$$
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$$
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v (x, y) = a _ {4} + a _ {5} x + a _ {6} y
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$$
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where $u ( x , y )$ and $v ( x , y )$ describe displacements at any interior point $( x _ { i } , y _ { i } )$ of the element.
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The linear function ensures that compatibility will be satisfied. A linear function with specified endpoints has only one path through which to pass—that is, through the two points. Hence, the linear function ensures that the displacements along the edge and at the nodes shared by adjacent elements, such as edge $i \mathrm { - } j$ of the two elements shown in Figure $6 { \ - } - 6 ( \mathtt { b } )$ , are equal. Using Eqs. (6.2.2), the general displacement function $\{ \psi \}$ , which stores the functions u and $v ,$ can be expressed as
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$$
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\{\psi \} = \left\{ \begin{array}{l} a _ {1} + a _ {2} x + a _ {3} y \\ a _ {4} + a _ {5} x + a _ {6} y \end{array} \right\} = \left[ \begin{array}{c c c c c c} 1 & x & y & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & x & y \end{array} \right] \left\{ \begin{array}{l} a _ {1} \\ a _ {2} \\ a _ {3} \\ a _ {4} \\ a _ {5} \\ a _ {6} \end{array} \right\} \tag {6.2.3}
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$$
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To obtain the $\boldsymbol { a ^ { \prime } \mathrm { s } }$ in Eqs. (6.2.2), we begin by substituting the coordinates of the nodal points into Eqs. (6.2.2) to yield
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$$
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u _ {i} = u (x _ {i}, y _ {i}) = a _ {1} + a _ {2} x _ {i} + a _ {3} y _ {i}
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$$
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$$
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u _ {j} = u (x _ {j}, y _ {j}) = a _ {1} + a _ {2} x _ {j} + a _ {3} y _ {j}
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$$
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$$
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u _ {m} = u \left(x _ {m}, y _ {m}\right) = a _ {1} + a _ {2} x _ {m} + a _ {3} y _ {m} \tag {6.2.4}
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$$
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$$
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v _ {i} = v (x _ {i}, y _ {i}) = a _ {4} + a _ {5} x _ {i} + a _ {6} y _ {i}
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$$
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$$
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v _ {j} = v (x _ {j}, y _ {j}) = a _ {4} + a _ {5} x _ {j} + a _ {6} y _ {j}
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$$
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$$
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v _ {m} = v \left(x _ {m}, y _ {m}\right) = a _ {4} + a _ {5} x _ {m} + a _ {6} y _ {m}
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$$
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<!-- source-page: 330 -->
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We can solve for the $\boldsymbol { a ^ { \prime } \mathrm { s } }$ beginning with the first three of Eqs. (6.2.4) expressed in matrix form as
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$$
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\left\{ \begin{array}{l} u _ {i} \\ u _ {j} \\ u _ {m} \end{array} \right\} = \left[ \begin{array}{c c c} 1 & x _ {i} & y _ {i} \\ 1 & x _ {j} & y _ {j} \\ 1 & x _ {m} & y _ {m} \end{array} \right] \left\{ \begin{array}{l} a _ {1} \\ a _ {2} \\ a _ {3} \end{array} \right\} \tag {6.2.5}
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$$
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$_ { \mathrm { o r , } }$ solving for the $\curvearrowleft$ we have
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$$
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\{a \} = [ x ] ^ {- 1} \{u \} \tag {6.2.6}
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$$
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where $[ x ]$ is the $3 \times 3$ matrix on the right side of Eq. (6.2.5). The method of cofactors (Appendix $\mathbf { A } )$ is one possible method for finding the inverse of ½x . Thus,
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$$
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[ x ] ^ {- 1} = \frac {1}{2 A} \left[ \begin{array}{c c c} \alpha_ {i} & \alpha_ {j} & \alpha_ {m} \\ \beta_ {i} & \beta_ {j} & \beta_ {m} \\ \gamma_ {i} & \gamma_ {j} & \gamma_ {m} \end{array} \right] \tag {6.2.7}
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$$
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where
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$$
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2 A = \left| \begin{array}{c c c} 1 & x _ {i} & y _ {i} \\ 1 & x _ {j} & y _ {j} \\ 1 & x _ {m} & y _ {m} \end{array} \right| \tag {6.2.8}
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$$
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is the determinant of ½x , which on evaluation is
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$$
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2 A = x _ {i} (y _ {j} - y _ {m}) + x _ {j} (y _ {m} - y _ {i}) + x _ {m} (y _ {i} - y _ {j}) \tag {6.2.9}
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$$
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Here A is the area of the triangle, and
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$$
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\alpha_ {i} = x _ {j} y _ {m} - y _ {j} x _ {m} \quad \alpha_ {j} = y _ {i} x _ {m} - x _ {i} y _ {m} \quad \alpha_ {m} = x _ {i} y _ {j} - y _ {i} x _ {j}
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$$
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$$
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\beta_ {i} = y _ {j} - y _ {m} \quad \beta_ {j} = y _ {m} - y _ {i} \quad \beta_ {m} = y _ {i} - y _ {j} \tag {6.2.10}
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$$
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$$
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\gamma_ {i} = x _ {m} - x _ {j} \quad \gamma_ {j} = x _ {i} - x _ {m} \quad \gamma_ {m} = x _ {j} - x _ {i}
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$$
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Having determined $[ x ] ^ { - 1 }$ , we can now express Eq. (6.2.6) in expanded matrix form as
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$$
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\left\{ \begin{array}{l} a _ {1} \\ a _ {2} \\ a _ {3} \end{array} \right\} = \frac {1}{2 A} \left[ \begin{array}{c c c} \alpha_ {i} & \alpha_ {j} & \alpha_ {m} \\ \beta_ {i} & \beta_ {j} & \beta_ {m} \\ \gamma_ {i} & \gamma_ {j} & \gamma_ {m} \end{array} \right] \left\{ \begin{array}{l} u _ {i} \\ u _ {j} \\ u _ {m} \end{array} \right\} \tag {6.2.11}
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$$
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Similarly, using the last three of Eqs. (6.2.4), we can obtain
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$$
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\left\{ \begin{array}{l} a _ {4} \\ a _ {5} \\ a _ {6} \end{array} \right\} = \frac {1}{2 A} \left[ \begin{array}{c c c} \alpha_ {i} & \alpha_ {j} & \alpha_ {m} \\ \beta_ {i} & \beta_ {j} & \beta_ {m} \\ \gamma_ {i} & \gamma_ {j} & \gamma_ {m} \end{array} \right] \left\{ \begin{array}{l} v _ {i} \\ v _ {j} \\ v _ {m} \end{array} \right\} \tag {6.2.12}
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$$
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We will derive the general x displacement function $u ( x , y )$ of $\{ \psi \}$ (v will follow analogously) in terms of the coordinate variables x and $y ,$ known coordinate variables
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