김경종
489 lines
20 KiB
Markdown
489 lines
20 KiB
Markdown
<!-- source-page: 81 -->
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<details>
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<summary>text_image</summary>
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2000 N/m
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2000 N/m
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1
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2
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3
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F₃
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δ = 20 mm
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</details>
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Figure P2–11
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<details>
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<summary>text_image</summary>
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10,000 N/m
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450 N
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20,000 N/m
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10,000 N/m
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1
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2
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3
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4
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</details>
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Figure P2–12
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<details>
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<summary>text_image</summary>
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20 kN/m
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20 kN/m
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5 kN
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20 kN/m
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20 kN/m
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1
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2
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3
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4
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5
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</details>
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Figure P2–13
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<details>
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<summary>text_image</summary>
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400 N/m
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100 N
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400 N/m
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200 N
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1
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2
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3
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</details>
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Figure P2–14
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<details>
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<summary>text_image</summary>
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1 500 kN/m 3 1 kN
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2 500 kN/m 3 1000 kN/m 4
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1 kN
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1 kN
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</details>
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Figure P2–15
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<details>
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<summary>text_image</summary>
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k = 100 lb/in.
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1
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2
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k = 100 lb/in.
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100 lb
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3
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k = 100 lb/in.
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100 lb
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4
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</details>
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Figure P2–16
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2.17 Use the principle of minimum potential energy developed in Section 2.6 to solve the spring problems shown in Figure P2–17. That is, plot the total potential energy for variations in the displacement of the free end of the spring to determine the minimum potential energy. Observe that the displacement that yields the minimum potential energy also yields the stable equilibrium position.
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<!-- source-page: 82 -->
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<details>
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<summary>text_image</summary>
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1000 lb
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k = 2000 lb/in.
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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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k = 500 lb/in.
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1000 lb
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</details>
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(b)
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<details>
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<summary>text_image</summary>
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k = 2000 N/mm
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400 kg
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</details>
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<details>
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<summary>text_image</summary>
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k = 400 N/mm
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100 kg
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</details>
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(d)
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Figure P2–17
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2.18 Reverse the direction of the load in Example 2.4 and recalculate the total potential energy. Then use this value to obtain the equilibrium value of displacement.
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2.19 The nonlinear spring in Figure P2–19 has the force/deformation relationship $f = k \delta ^ { 2 }$ . Express the total potential energy of the spring, and use this potential energy to obtain the equilibrium value of displacement.
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<details>
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<summary>text_image</summary>
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k = 1000 lb/in.
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500 lb
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</details>
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Figure P2–19
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2.20–2.21 Solve Problems 2.10 and 2.15 by the potential energy approach (see Example 2.5).
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<!-- source-page: 83 -->
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# Development of Truss Equations
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# Introduction
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Having set forth the foundation on which the direct stiffness method is based, we will now derive the stiffness matrix for a linear-elastic bar (or truss) element using the general steps outlined in Chapter 1. We will include the introduction of both a local coordinate system, chosen with the element in mind, and a global or reference coordinate system, chosen to be convenient (for numerical purposes) with respect to the overall structure. We will also discuss the transformation of a vector from the local coordinate system to the global coordinate system, using the concept of transformation matrices to express the stiffness matrix of an arbitrarily oriented bar element in terms of the global system. We will solve three example plane truss problems (see Figure 3–1 for a typical railroad trestle plane truss) to illustrate the procedure of establishing the total stiffness matrix and equations for solution of a structure.
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Next we extend the stiffness method to include space trusses. We will develop the transformation matrix in three-dimensional space and analyze two space trusses. Then we describe the concept of symmetry and its use to reduce the size of a problem and facilitate its solution. We will use an example truss problem to illustrate the concept and then describe how to handle inclined, or skewed, supports.
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We will then use the principle of minimum potential energy and apply it to rederive the bar element equations. We then compare a finite element solution to an exact solution for a bar subjected to a linear varying distributed load. We will introduce Galerkin’s residual method and then apply it to derive the bar element equations. Finally, we will introduce other common residual methods—collocation, subdomain, and least squares to merely expose you to these other methods. We illustrate these methods by solving a problem of a bar subjected to a linear varying load.
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<!-- source-page: 84 -->
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<details>
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<summary>natural_image</summary>
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Black-and-white photo of a steel truss bridge spanning over stone pillars, with rural landscape in background (no visible text or symbols)
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</details>
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Figure 3–1 A typical railroad trestle plane truss. (By Daryl L. Logan)
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# d 3.1 Derivation of the Stiffness Matrix for a Bar Element in Local Coordinates
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We will now consider the derivation of the stiffness matrix for the linear-elastic, constant cross-sectional area (prismatic) bar element shown in Figure 3–2. The derivation here will be directly applicable to the solution of pin-connected trusses. The bar is subjected to tensile forces T directed along the local axis of the bar and applied at nodes 1 and 2.
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<details>
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<summary>text_image</summary>
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T̂ₓ = Cx̂ (force/length)
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T
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x
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y
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ŷ
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1
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θ
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L
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2
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T
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x̂, û
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d̂₂ₓ, f̂₂ₓ
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d̂₁ₓ, f̂₁ₓ
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</details>
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Figure 3–2 Bar subjected to tensile forces T; positive nodal displacements and forces are all in the local x^ direction
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Here we have introduced two coordinate systems: a local one ðx^; y^Þ with x^ directed along the length of the bar and a global one $( x , y )$ assumed here to be best suited with respect to the total structure. Proper selection of global coordinate systems is best demonstrated through solution of two- and three-dimensional truss problems as illustrated in Sections 3.6 and 3.7. Both systems will be used extensively throughout this text.
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The bar element is assumed to have constant cross-sectional area A, modulus of elasticity E, and initial length L. The nodal degrees of freedom are local axial displacements (longitudinal displacements directed along the length of the bar) represented by $\hat { d } _ { 1 x }$ and $\hat { d } _ { 2 x }$ at the ends of the element as shown in Figure 3–2.
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From Hooke’s law [Eq. (a)] and the strain/displacement relationship [Eq. (b) or Eq. (1.4.1)], we write
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$$
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\sigma_ {x} = E \varepsilon_ {x} \tag {a}
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$$
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$$
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\varepsilon_ {x} = \frac {d \hat {u}}{d \hat {x}} \tag {b}
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$$
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From force equilibrium, we have
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$$
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A \sigma_ {x} = T = \text { constant } \tag {c}
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$$
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for a bar with loads applied only at the ends. (We will consider distributed loading in Section 3.10.) Using Eq. (b) in Eq. (a) and then Eq. (a) in Eq. (c) and differentiating with respect to ${ \hat { x } } ,$ we obtain the differential equation governing the linear-elastic bar behavior as
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$$
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\frac {d}{d \hat {x}} \left(A E \frac {d \hat {u}}{d \hat {x}}\right) = 0 \tag {d}
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$$
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where u^ is the axial displacement function along the element in the x^ direction and A and E are written as though they were functions of x^ in the general form of the differential equation, even though A and E will be assumed constant over the whole length of the bar in our derivations to follow.
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The following assumptions are used in deriving the bar element stiffness matrix:
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1. The bar cannot sustain shear force or bending moment, that is,
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$$
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\hat {f} _ {1 y} = 0, \hat {f} _ {2 y} = 0, \hat {m} _ {1} = 0 \text { and } \hat {m} _ {2} = 0.
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$$
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2. Any effect of transverse displacement is ignored.
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3. Hooke’s law applies; that is, axial stress $\sigma _ { x }$ is related to axial strain $\varepsilon _ { x }$
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$$
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\text { by } \sigma_ {x} = E \varepsilon_ {x}.
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$$
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4. No intermediate applied loads.
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The steps previously outlined in Chapter 1 are now used to derive the stiffness matrix for the bar element and then to illustrate a complete solution for a bar assemblage.
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# Step 1 Select the Element Type
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Represent the bar by labeling nodes at each end and in general by labeling the element number (Figure 3–2).
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<!-- source-page: 86 -->
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# Step 2 Select a Displacement Function
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Assume a linear displacement variation along the x^ axis of the bar because a linear function with specified endpoints has a unique path. These specified endpoints are the nodal values $\hat { d } _ { 1 x }$ and $\hat { d } _ { 2 x }$ . (Further discussion regarding the choice of displacement functions is provided in Section 3.2 and References [1–3].) Then
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$$
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\hat {u} = a _ {1} + a _ {2} \hat {x} \tag {3.1.1}
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$$
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with the total number of coefficients $a _ { i }$ always equal to the total number of degrees of freedom associated with the element. Here the total number of degrees of freedom is two—axial displacements at each of the two nodes of the element. Using the same procedure as in Section 2.2 for the spring element, we express Eq. (3.1.1) as
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$$
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\hat {u} = \left(\frac {\hat {d} _ {2 x} - \hat {d} _ {1 x}}{L}\right) \hat {x} + \hat {d} _ {1 x} \tag {3.1.2}
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$$
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The reason we convert the displacement function from the form of Eq. (3.1.1) to Eq. (3.1.2) is that it allows us to express the strain in terms of the nodal displacements using the strain/displacement relationship given by Eq. (3.1.5) and to then relate the nodal forces to the nodal displacements in step 4.
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In matrix form, Eq. (3.1.2) becomes
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$$
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\hat {u} = \left[ \begin{array}{l l} N _ {1} & N _ {2} \end{array} \right] \left\{ \begin{array}{l} \hat {d} _ {1 x} \\ \hat {d} _ {2 x} \end{array} \right\} \tag {3.1.3}
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$$
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with shape functions given by
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$$
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N _ {1} = 1 - \frac {\hat {x}}{L} \quad N _ {2} = \frac {\hat {x}}{L} \tag {3.1.4}
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$$
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These shape functions are identical to those obtained for the spring element in Section 2.2. The behavior of and some properties of these shape functions were described in Section 2.2. The linear displacement function u^ (Eq. (3.1.2)), plotted over the length of the bar element, is shown in Figure 3–3. The bar is shown with the same orientation as in Figure 3–2.
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<details>
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<summary>text_image</summary>
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y
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û
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d̂₁ₓ
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L
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θ
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1
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2
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x̂
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d̂₂ₓ
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</details>
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Figure 3–3 Displacement u^ plotted over the length of the element
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<!-- source-page: 87 -->
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# Step 3 Define the Strain= Displacement and Stress= Strain Relationships
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The strain/displacement relationship is
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$$
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\varepsilon_ {x} = \frac {d \hat {u}}{d \hat {x}} = \frac {\hat {d} _ {2 x} - \hat {d} _ {1 x}}{L} \tag {3.1.5}
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$$
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where Eqs. (3.1.3) and (3.1.4) have been used to obtain Eq. (3.1.5), and the stress/ strain relationship is
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$$
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\sigma_ {x} = E \varepsilon_ {x} \tag {3.1.6}
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$$
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# Step 4 Derive the Element Stiffness Matrix and Equations
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The element stiffness matrix is derived as follows. From elementary mechanics, we have
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$$
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T = A \sigma_ {x} \tag {3.1.7}
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$$
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Now, using Eqs. (3.1.5) and (3.1.6) in Eq. (3.1.7), we obtain
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$$
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T = A E \left(\frac {\hat {d} _ {2 x} - \hat {d} _ {1 x}}{L}\right) \tag {3.1.8}
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$$
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Also, by the nodal force sign convention of Figure 3–2,
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$$
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\hat {f} _ {1 x} = - T \tag {3.1.9}
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$$
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When we substitute Eq. (3.1.8), Eq. (3.1.9) becomes
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$$
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\hat {f} _ {1 x} = \frac {A E}{L} (\hat {d} _ {1 x} - \hat {d} _ {2 x}) \tag {3.1.10}
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$$
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Similarly, ${ \hat { f } } _ { 2 x } = T$ ð3:1:11Þ
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or, by Eq. (3.1.8), Eq. (3.1.11) becomes
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$$
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\hat {f} _ {2 x} = \frac {A E}{L} (\hat {d} _ {2 x} - \hat {d} _ {1 x}) \tag {3.1.12}
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$$
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Expressing Eqs. (3.1.10) and (3.1.12) together in matrix form, we have
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$$
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\left\{ \begin{array}{l} \hat {f} _ {1 x} \\ \hat {f} _ {2 x} \end{array} \right\} = \frac {A E}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} \hat {d} _ {1 x} \\ \hat {d} _ {2 x} \end{array} \right\} \tag {3.1.13}
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$$
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Now, because $\underline { { \hat { f } } } = \underline { { \hat { k } } } \underline { { \hat { d } } }$ , we have, from Eq. (3.1.13),
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$$
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\underline {{\hat {k}}} = \frac {A E}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \tag {3.1.14}
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$$
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<!-- source-page: 88 -->
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Equation (3.1.14) represents the stiffness matrix for a bar element in local coordinates. In Eq. (3.1.14), $A E / L$ for a bar element is analogous to the spring constant k for a spring element.
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# Step 5 Assemble Element Equations to Obtain Global or Total Equations
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Assemble the global stiffness and force matrices and global equations using the direct stiffness method described in Chapter 2 (see Section 3.6 for an example truss). This step applies for structures composed of more than one element such that (again)
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$$
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\underline {{K}} = [ K ] = \sum_ {e = 1} ^ {N} \underline {{k}} ^ {(e)} \quad \text { and } \quad \underline {{F}} = \{F \} = \sum_ {e = 1} ^ {N} \underline {{f}} ^ {(e)} \tag {3.1.15}
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$$
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where now all local element stiffness matrices $\underline { { \hat { k } } }$ must be transformed to global element stiffness matrices k (unless the local axes coincide with the global axes) before the direct stiffness method is applied as indicated by Eq. (3.1.15). (This concept of coordinate and stiffness matrix transformations is described in Sections 3.3 and 3.4.)
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# Step 6 Solve for the Nodal Displacements
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Determine the displacements by imposing boundary conditions and simultaneously solving a system of equations, $\underline { { F } } = \underline { { K } } \underline { { d } }$ .
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# Step 7 Solve for the Element Forces
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Finally, determine the strains and stresses in each element by back-substitution of the displacements into equations similar to Eqs. (3.1.5) and (3.1.6).
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We will now illustrate a solution for a one-dimensional bar problem.
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# Example 3.1
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For the three-bar assemblage shown in Figure 3–4, determine (a) the global stiffness matrix, (b) the displacements of nodes 2 and 3, and (c) the reactions at nodes 1 and 4. A force of 3000 lb is applied in the x direction at node 2. The length of each element is 30 in. Let $E = 3 0 \times 1 0 ^ { 6 }$ psi and $A = 1 \mathrm { i n } ^ { 2 }$ for elements 1 and 2, and let $E = 1 5 \times 1 0 ^ { 6 }$ psi and $A = 2 \mathrm { i n } ^ { 2 }$ for element 3. Nodes 1 and 4 are fixed.
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<details>
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<summary>text_image</summary>
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3000 lb
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1 ① 2 ② 3 ③ 4
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30 in. 30 in. 30 in.
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90 in. x
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</details>
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Figure 3–4 Three-bar assemblage
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<!-- source-page: 89 -->
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(a) Using Eq. (3.1.14), we find that the element stiffness matrices are
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$$
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\begin{array}{r l} \underline {{k}} ^ {(1)} = \underline {{k}} ^ {(2)} & = \frac {(1) (3 0 \times 1 0 ^ {6})}{3 0} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] = 1 0 ^ {6} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \frac {\mathrm{lb}}{\mathrm{in.}} \\ \underline {{k}} ^ {(3)} & = \frac {(2) (1 5 \times 1 0 ^ {6})}{3 0} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] = 1 0 ^ {6} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \frac {\mathrm{lb}}{\mathrm{in}.} \end{array} \tag {3.1.16}
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$$
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where, again, the numbers above the matrices in Eqs. (3.1.16) indicate the displacements associated with each matrix. Assembling the element stiffness matrices by the direct stiffness method, we obtain the global stiffness matrix as
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$$
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\underline {{{K}}} = 1 0 ^ {6} \left[ \begin{array}{c c c c} 1 & - 1 & 0 & 0 \\ - 1 & 1 + 1 & - 1 & 0 \\ 0 & - 1 & 1 + 1 & - 1 \\ 0 & 0 & - 1 & 1 \end{array} \right] \frac {\mathrm{lb}}{\text { in. }} \tag {3.1.17}
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$$
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(b) Equation (3.1.17) relates global nodal forces to global nodal displacements as follows:
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$$
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\left\{ \begin{array}{l} F _ {1 x} \\ F _ {2 x} \\ F _ {3 x} \\ F _ {4 x} \end{array} \right\} = 1 0 ^ {6} \left[ \begin{array}{r r r r} 1 & - 1 & 0 & 0 \\ - 1 & 2 & - 1 & 0 \\ 0 & - 1 & 2 & - 1 \\ 0 & 0 & - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {2 x} \\ d _ {3 x} \\ d _ {4 x} \end{array} \right\} \tag {3.1.18}
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$$
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Invoking the boundary conditions, we have
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$$
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d _ {1 x} = 0 \quad d _ {4 x} = 0 \tag {3.1.19}
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$$
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Using the boundary conditions, substituting known applied global forces into Eq. (3.1.18), and partitioning equations 1 and 4 of Eq. (3.1.18), we solve equations 2 and 3 of Eq. (3.1.18) to obtain
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$$
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\left\{ \begin{array}{c} 3 0 0 0 \\ 0 \end{array} \right\} = 1 0 ^ {6} \left[ \begin{array}{c c} 2 & - 1 \\ - 1 & 2 \end{array} \right] \left\{ \begin{array}{c} d _ {2 x} \\ d _ {3 x} \end{array} \right\} \tag {3.1.20}
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$$
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Solving Eq. (3.1.20) simultaneously for the displacements yields
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$$
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d _ {2 x} = 0. 0 0 2 \text { in. } \quad d _ {3 x} = 0. 0 0 1 \text { in. } \tag {3.1.21}
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$$
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<!-- source-page: 90 -->
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(c) Back-substituting Eqs. (3.1.19) and (3.1.21) into Eq. (3.1.18), we obtain the global nodal forces, which include the reactions at nodes 1 and 4, as follows:
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$$
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F _ {1 x} = 1 0 ^ {6} (d _ {1 x} - d _ {2 x}) = 1 0 ^ {6} (0 - 0. 0 0 2) = - 2 0 0 0 \mathrm{lb}
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$$
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$$
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F _ {2 x} = 1 0 ^ {6} \left(- d _ {1 x} + 2 d _ {2 x} - d _ {3 x}\right) = 1 0 ^ {6} [ 0 + 2 (0. 0 0 2) - 0. 0 0 1 ] = 3 0 0 0 \mathrm{lb} \tag {3.1.22}
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$$
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$$
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F _ {3 x} = 1 0 ^ {6} \left(- d _ {2 x} + 2 d _ {3 x} - d _ {4 x}\right) = 1 0 ^ {6} \left[ - 0. 0 0 2 + 2 (0. 0 0 1) - 0 \right] = 0
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$$
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$$
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F _ {4 x} = 1 0 ^ {6} \left(- d _ {3 x} + d _ {4 x}\right) = 1 0 ^ {6} (- 0. 0 0 1 + 0) = - 1 0 0 0 \mathrm{lb}
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$$
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The results of Eqs. (3.1.22) show that the sum of the reactions $F _ { 1 x }$ and $F _ { 4 x }$ is equal in magnitude but opposite in direction to the applied nodal force of 3000 lb at node 2. Equilibrium of the bar assemblage is thus verified. Furthermore, Eqs. (3.1.22) show that $F _ { 2 x } = 3 0 0 0$ lb and $F _ { 3 x } = 0$ are merely the applied nodal forces at nodes 2 and 3, respectively, which further enhances the validity of our solution.
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# 3.2 Selecting Approximation Functions for Displacements
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Consider the following guidelines, as they relate to the one-dimensional bar element, when selecting a displacement function. (Further discussion regarding selection of displacement functions and other kinds of approximation functions (such as temperature functions) will be provided in Chapter 4 for the beam element, in Chapter 6 for the constant-strain triangular element, in Chapter 8 for the linear-strain triangular element, in Chapter 9 for the axisymmetric element, in Chapter 10 for the three-noded bar element and the rectangular plane element, in Chapter 11 for the three-dimensional stress element, in Chapter 12 for the plate bending element, and in Chapter 13 for the heat transfer problem. More information is also provided in References [1–3].
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1. Common approximation functions are usually polynomials such as the simplest one that gives the linear variation of displacement given by Eq. (3.1.1) or equivalently by Eq. (3.1.3), where the function is expressed in terms of the shape functions.
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2. The approximation function should be continuous within the bar element. The simple linear function for u^ of Eq. (3.1.1) certainly is continuous within the element. Therefore, the linear function yields continuous values of u^ within the element and prevents openings, overlaps, and jumps because of the continuous and smooth variation in u^ (Figure 3–5).
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3. The approximating function should provide interelement continuity for all degrees of freedom at each node for discrete line elements and along common boundary lines and surfaces for two- and threedimensional elements. For the bar element, we must ensure that nodes
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