김경종
435 lines
24 KiB
Markdown
435 lines
24 KiB
Markdown
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We next evaluate (5.75) to obtain (see Section 2.4)
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$$
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\mathbf {V} _ {t} ^ {k} = \det \left[ \begin{array}{l l l} \mathbf {e} _ {x} & \mathbf {e} _ {y} & \mathbf {e} _ {z} \\ \theta_ {x} ^ {k} & \theta_ {y} ^ {k} & \theta_ {z} ^ {k} \\ 0 & 0 & 1 \end{array} \right]
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$$
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or $\mathbf{V}_t^k = \theta_y^k\mathbf{e}_x - \theta_x^k\mathbf{e}_y$ (a)
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and $\mathbf{V}_s^k = \det \begin{bmatrix} \mathbf{e}_x & \mathbf{e}_y & \mathbf{e}_z \\ \theta_x^k & \theta_y^k & \theta_z^k \\ 0 & 1 & 0 \end{bmatrix}$
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or $\mathbf{V}_s^k = -\theta_z^k\mathbf{e}_x + \theta_x^k\mathbf{e}_z$ (b)
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The relations in (a) and (b) correspond to the three-dimensional action of the beam. We allow rotations only about the z-axis, in which case
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$$
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\mathbf {V} _ {t} ^ {k} = \mathbf {0}; \quad \mathbf {V} _ {s} ^ {k} = - \theta_ {z} ^ {k} \mathbf {e} _ {x}
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$$
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Furthermore, we assume that the nodal points can displace only in the $y$ direction. Hence, (5.73) yields the displacement assumptions
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$$
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u (r, s) = - \frac {s h}{2} \sum_ {k = 1} ^ {3} h _ {k} \theta_ {z} ^ {k} \tag {c}
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$$
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$$
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v (r) = \sum_ {k = 1} ^ {3} h _ {k} v _ {k} \tag {d}
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$$
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where we note that u is only a function of r, s and v is only a function of r. These relations are identical to the displacement assumptions used before, but with the more conventional beam displacement notation we identified the transverse displacement and section rotation at a nodal point with $w_{k}$ and $\theta_{k}$ instead of $v_{k}$ and $\theta_{z}^{k}$ .
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Now using (5.80), we obtain
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$$
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\left[ \begin{array}{l} \frac {\partial u}{\partial r} \\ \frac {\partial u}{\partial s} \end{array} \right] = \sum_ {k = 1} ^ {3} \left[ \begin{array}{l} - \frac {s h}{2} \frac {\partial h _ {k}}{\partial r} \\ - \frac {h}{2} h _ {k} \end{array} \right] \theta_ {z} ^ {k}
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$$
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$$
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\left[ \begin{array}{l} \frac {\partial v}{\partial r} \\ \frac {\partial v}{\partial s} \end{array} \right] = \sum_ {k = 1} ^ {3} \left[ \begin{array}{c} \frac {\partial h _ {k}}{\partial r} \\ 0 \end{array} \right] v _ {k}
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$$
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These relations could also be directly obtained by differentiating the displacements in (c) and (d). Since
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$$
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\mathbf {J} = \left[ \begin{array}{l l l} \frac {L}{2} & & 0 \\ 0 & & \frac {h}{2} \end{array} \right]; \quad \mathbf {J} ^ {- 1} = \left[ \begin{array}{l l l} \frac {2}{L} & & 0 \\ 0 & & \frac {2}{h} \end{array} \right]
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$$
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we obtain
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$$
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\left[ \begin{array}{c} \frac {\partial u}{\partial x} \\ \frac {\partial u}{\partial y} \end{array} \right] = \sum_ {k = 1} ^ {3} \left[ \begin{array}{c} - \frac {h}{2} \frac {2}{L} s \frac {\partial h _ {k}}{\partial r} \\ - h _ {k} \end{array} \right] \theta_ {z} ^ {k} \tag {e}
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$$
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and
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$$
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\left[ \begin{array}{l} \frac {\partial v}{\partial x} \\ \frac {\partial v}{\partial y} \end{array} \right] = \sum_ {k = 1} ^ {3} \left[ \begin{array}{c c} \frac {2}{L} & \frac {\partial h _ {k}}{\partial r} \\ & 0 \end{array} \right] v _ {k} \tag {f}
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$$
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To analyze the response of the beam in Fig E5.24 we now use the principle of virtual work [see (4.7)] with the appropriate strain measures:
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$$
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\int_ {- 1} ^ {+ 1} \int_ {- 1} ^ {+ 1} \left[ \overline {{{\epsilon}}} _ {x x} \quad \overline {{{\gamma}}} _ {x y} \right] \left[ \begin{array}{l l} E & 0 \\ 0 & G k \end{array} \right] \left[ \begin{array}{l} \epsilon_ {x x} \\ \gamma_ {x y} \end{array} \right] \det \mathbf {J} d s d r = - P \overline {{{v}}} | _ {r = 1 / 3} \tag {g}
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$$
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where
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$$
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\epsilon_ {x x} = \frac {\partial u}{\partial x}; \quad \overline {{{\epsilon}}} _ {x x} = \frac {\partial \overline {{{u}}}}{\partial x}
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$$
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$$
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\gamma_ {x y} = \frac {\partial u}{\partial y} + \frac {\partial v}{\partial x}; \quad \overline {{{\gamma}}} _ {x y} = \frac {\partial \overline {{{u}}}}{\partial y} + \frac {\partial \overline {{{v}}}}{\partial x}
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$$
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Considering the relations in (e), (f), (g), and (5.58), we recognize that (g) corresponds to (5.58) if we use $\beta \equiv \theta_{z}$ , and $w \equiv v$ .
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# Transition Elements
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In the preceding discussions, we considered continuum elements and beam elements separately. However, the very close relationship between these elements should be recognized; the only differences are the kinematic assumption that plane sections initially normal to the neutral axis remain plane and the stress assumption that stresses normal to the neutral axis are zero. In the beam formulation presented, the kinematic assumption was incorporated directly in the basic geometry and displacement interpolations and the stress assumption was used in the stress-strain law. Since these two assumptions are the only two basic differences between the beam and continuum elements, it is apparent that the structural element matrices can also be derived from the continuum element matrices by degeneration. Furthermore, elements can be devised that act as transition elements between continuum and structural elements. Consider the following example.
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EXAMPLE 5.27: Assume that the strain-displacement matrix of a four-node plane stress element has been derived. Show how the strain-displacement matrix of a two-node beam element can be constructed.
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Figure E5.27 shows the plane stress element with its degrees of freedom and the beam element for which we want to establish the strain-displacement matrix. Consider node 2 of the beam element and nodes 2 and 3 of the plane stress element. The entries in the strain-displacement matrix of the plane stress element are
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(a) Plane stress element
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<details>
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<summary>text_image</summary>
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v₂
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s
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v₁
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θ₂
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2
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u₂
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r
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θ₁
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1
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u₁
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t
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L
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</details>
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(b) Beam element
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Figure E5.27 Derivation of beam element from plane stress element
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$$
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\mathbf {B} ^ {*} = \left[ \begin{array}{c c c c c c} & u _ {2} ^ {*} & v _ {2} ^ {*} & u _ {3} ^ {*} & v _ {3} ^ {*} \\ & \downarrow & \downarrow & \downarrow & \downarrow \\ & - \frac {1}{2 L} (1 + s) & 0 & - \frac {1}{2 L} (1 - s) & 0 \\ \dots & 0 & \frac {1}{2 t} (1 - r) & 0 & - \frac {1}{2 t} (1 - r) \\ & \frac {1}{2 t} (1 - r) & - \frac {1}{2 L} (1 + s) & - \frac {1}{2 t} (1 - r) & - \frac {1}{2 L} (1 - s) \end{array} \right] \quad \dots \tag {a}
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$$
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Using now the beam deformation assumptions, we have the following kinematic constraints:
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$$
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u _ {2} ^ {*} = u _ {2} - \frac {t}{2} \theta_ {2}
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$$
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$$
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u _ {3} ^ {*} = u _ {2} + \frac {t}{2} \theta_ {2} \tag {b}
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$$
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$$
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v _ {2} ^ {*} = v _ {2}; \quad v _ {3} ^ {*} = v _ {2}
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$$
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These constraints are now substituted to obtain from the elements of $B^{*}$ in (a) the elements of the strain-displacement matrix of the beam. Using the rows of $B^{*}$ , we have with (b),
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$$
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\begin{array}{l} - \frac {1}{2 L} (1 + s) u _ {2} ^ {*} - \frac {1}{2 L} (1 - s) u _ {3} ^ {*} = - \frac {1}{2 L} (1 + s) \left(u _ {2} - \frac {t}{2} \theta_ {2}\right) \\ - \frac {1}{2 L} (1 - s) \left(u _ {2} + \frac {t}{2} \theta_ {2}\right) \tag {c} \\ \end{array}
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$$
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<!-- source-page: 434 -->
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$$
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\frac {1}{2 t} (1 - r) v _ {2} ^ {*} - \frac {1}{2 t} (1 - r) v _ {3} ^ {*} = \frac {1}{2 t} (1 - r) v _ {2} - \frac {1}{2 t} (1 - r) v _ {2} \tag {d}
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$$
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$$
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\begin{array}{l} \frac {1}{2 t} (1 - r) u _ {2} ^ {*} - \frac {1}{2 L} (1 + s) v _ {2} ^ {*} - \frac {1}{2 t} (1 - r) u _ {3} ^ {*} - \frac {1}{2 L} (1 - s) v _ {3} ^ {*} \\ = \frac {1}{2 t} (1 - r) \left(u _ {2} - \frac {t}{2} \theta_ {2}\right) - \frac {1}{2 L} (1 + s) v _ {2} - \frac {1}{2 t} (1 - r) \left(u _ {2} + \frac {t}{2} \theta_ {2}\right) - \frac {1}{2 L} (1 - s) v _ {2} \tag {e} \\ \end{array}
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$$
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The relations on the right-hand side of (c) to (e) comprise the entries of the beam strain-displacement matrix
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$$
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\mathbf {B} = \left[ \begin{array}{c c c c} & u _ {2} & v _ {2} & \theta_ {2} \\ & \downarrow & \downarrow & \downarrow \\ & - \frac {1}{L} & 0 & \frac {t}{2 L} s \\ \dots & 0 & 0 & 0 \\ & 0 & - \frac {1}{L} & - \frac {1}{2} (1 - r) \end{array} \right]
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$$
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However, the first- and third-row entries are those that are also obtained using the beam formulation of (5.71) to (5.86). We should note that the zeros in the second row of B only express the fact that the strain $\epsilon_{yy}$ is not included in the formulation. This strain is actually equal to $-\nu\epsilon_{xx}$ because the stress $\tau_{yy}$ is zero. As pointed out earlier, we would use the entries in B at r=0.
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The formulation of a structural element using the approach discussed in Example 5.27 is computationally inefficient and is certainly not recommended for general analysis. However, it is instructive to study this approach and recognize that the structural element matrices can in principle be obtained from continuum element matrices by imposing the appropriate static and kinematic assumptions. Moreover, this formulation directly suggests the construction of transition elements that can be used in an effective manner to couple structural and continuum elements without the use of constraint equations [see Fig. E5.28(a)]. To demonstrate the formulation of transition elements, we consider in the following example a simple transition beam element.
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EXAMPLE 5.28: Construct the displacement and strain-displacement interpolation matrices of the transition element shown in Fig. E5.28.
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We define the nodal point displacement vector of the element as
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$$
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\hat {\mathbf {u}} ^ {T} = \left[ \begin{array}{l l l l l l l} u _ {1} & v _ {1} & u _ {2} & v _ {2} & u _ {3} & v _ {3} & \theta_ {3} \end{array} \right] \tag {a}
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$$
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Since at $r = +1$ we have plane stress element degrees of freedom, the interpolation functions corresponding to nodes 1 and 2 are (see Fig. 5.4)
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$$
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h _ {1} = \frac {1}{4} (1 + r) (1 + s); \quad h _ {2} = \frac {1}{4} (1 + r) (1 - s)
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$$
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Node 3 is a beam node, and the interpolation function is (see Fig. 5.3)
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$$
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h _ {3} = \frac {1}{2} (1 - r)
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$$
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<!-- source-page: 435 -->
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<details>
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<summary>text_image</summary>
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Beam element
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Plane stress element
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Transition element
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</details>
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(a) Beam transition element connecting beam and plane stress elements
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<details>
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<summary>text_image</summary>
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y
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x
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v3
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θ3
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3
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u3
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s
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r
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v1
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1
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u1
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v2
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2
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u2
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L
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t
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</details>
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(b) Transition element
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Figure E5.28 Two-dimensional displacement-based beam transition element
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The displacements of the element are thus
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$$
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u (r, s) = h _ {1} u _ {1} + h _ {2} u _ {2} + h _ {3} u _ {3} - \frac {t}{2} s h _ {3} \theta_ {3}
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$$
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Hence, corresponding to the displacement vector in (a) we have
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$$
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\mathbf {H} = \left[ \begin{array}{c c c c c c c} h _ {1} & 0 & h _ {2} & 0 & h _ {3} & 0 & - \frac {t}{2} s h _ {3} \\ 0 & h _ {1} & 0 & h _ {2} & 0 & h _ {3} & 0 \end{array} \right]
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$$
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The coordinate interpolation is the same as that of the four-node plane stress element:
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$$
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x (r, s) = \frac {1}{2} (1 + r) L
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$$
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$$
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y (r, s) = \frac {s}{2} t
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$$
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Hence, $\mathbf{J} = \begin{bmatrix} \frac{L}{2} & & 0 \\ 0 & & \frac{t}{2} \end{bmatrix}$ ; $\mathbf{J}^{-1} = \begin{bmatrix} \frac{2}{L} & & 0 \\ 0 & & \frac{2}{t} \end{bmatrix}$
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<!-- source-page: 436 -->
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Using (5.25), we thus obtain
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$$
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\mathbf {B} = \left[ \begin{array}{c c c c c c c} \frac {1}{2 L} (1 + s) & 0 & \frac {1}{2 L} (1 - s) & 0 & - \frac {1}{L} & 0 & \frac {t}{2 L} s \\ 0 & \frac {1}{2 t} (1 + r) & 0 & - \frac {1}{2 t} (1 + r) & 0 & 0 & 0 \\ \frac {1}{2 t} (1 + r) & \frac {1}{2 L} (1 + s) & - \frac {1}{2 t} (1 + r) & \frac {1}{2 L} (1 - s) & 0 & - \frac {1}{L} & - \frac {1}{2} (1 - r) \end{array} \right]
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$$
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We may finally note that the last three columns of the B-matrix could also have been derived as described in Example 5.27.
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The isoparametric beam elements presented in this section are an alternative to the classical Hermitian beam elements (see Example 4.16), and we may ask how these types of beam elements compare in efficiency. There is no doubt that in linear analysis of straight, thin beams, the Hermitian elements are usually more effective, since for a cubic displacement description the isoparametric beam element requires twice as many degrees of freedom. However, the isoparametric beam element includes the effects of shear deformations and has the advantages that all displacements are interpolated to the same degree (which for the cubic element results in a cubic axial displacement variation) and that curved geometries can be represented accurately. The element is therefore used efficiently in the analysis of stiffened shells (because the element represents in a natural way the stiffeners for the shell elements discussed in the next section) and as a basis of formulating more complex elements, such as pipe and transition elements. Also, the generality of the formulation with all displacements interpolated to the same degree of variation renders the element efficient in geometric nonlinear analyses (see Section 6.5.1).
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Further applications of the general beam formulation given here lie in the use for plane strain situations (see Exercise 5.40) and the development of axisymmetric shell elements.
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# Axisymmetric Shell Elements
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The isoparametric beam element formulation presented above can be directly adapted to the analysis of axisymmetric shells. Figure 5.24 shows a typical three-node element.
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In the formulation, the kinematics of the beam element is used as if it were employed in two-dimensional action (i.e., for motion in the x, y plane), but the effects of the hoop strain and stress are also included. Hence, the strain-displacement matrix of the element is the matrix of the beam amended by a row corresponding to the hoop strain u/x. This evaluation is quite analogous to the construction of the B-matrix of the two-dimensional axisymmetric element when compared with the two-dimensional plane stress element. In that case, also only a row corresponding to the hoop strain was added to the B-matrix of the plane stress element in order to obtain the B-matrix for the axisymmetric element. In addition of course the correct stress-strain law needs to be used (allowing for the Poisson effect coupling between the hoop and the r-direction and for the stress to be zero in the s-direction), and the integration is performed corresponding to axisymmetric conditions over 1 radian of the structure (see Example 5.9 and Exercise 5.41). Of course, using the procedures in Example 5.28, transition elements for axisymmetric shell conditions can also be designed (see Exercise 5.42).
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<!-- source-page: 437 -->
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<details>
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<summary>text_image</summary>
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y, v
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0V₁s
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0V₃s
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0V₂s
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1
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s
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r
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3
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v₂
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θ₂
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2
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u₂
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Axis of
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revolution
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x, u
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Fig
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</details>
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Figure 5.24 Axisymmetric shell element
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# 5.4.2 Plate and General Shell Elements
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The procedures we have employed in the previous section to formulate beam elements can also be directly used to establish effective plate and shell elements. In the following presentation we first discuss the formulation of plate elements, and then we proceed to summarize the formulation of general shell elements.
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# Plate Elements
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The plate element formulation is a special case of the general shell element formulation presented later and is based on the theory of plates with transverse shear deformations included. This theory, due to E. Reissner [B] and R. D. Mindlin [A], uses the assumption that particles of the plate originally on a straight line that is normal to the undeformed middle surface remain on a straight line during deformation, but this line is not necessarily normal to the deformed middle surface. With this assumption, the displacement components of a point of coordinates x, y, and z are, in the small displacement bending theory,
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$$
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u = - z \beta_ {x} (x, y); \quad v = - z \beta_ {y} (x, y); \quad w = w (x, y) \tag {5.88}
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$$
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where w is the transverse displacement and $\beta_{x}$ and $\beta_{y}$ are the rotations of the normal to the undeformed middle surface in the x, z and y, z planes, respectively (see Fig. 5.25). It is instructive to note that in the Kirchhoff plate theory excluding shear deformations, $\beta_{x} = w_{,x}$ and $\beta_{y} = w_{,y}$ (and indeed we have selected the convention for $\beta_{x}$ and $\beta_{y}$ so as to have these Kirchhoff relations).
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Considering the plate in Fig. 5.25 the bending strains $\epsilon_{xx}$ , $\epsilon_{yy}$ , $\gamma_{xy}$ vary linearly through the plate thickness and are given by the curvatures of the plate using (5.88),
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$$
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\left[ \begin{array}{l} \epsilon_ {x x} \\ \epsilon_ {y y} \\ \gamma_ {x y} \end{array} \right] = - z \left[ \begin{array}{c} \frac {\partial \beta_ {x}}{\partial x} \\ \frac {\partial \beta_ {y}}{\partial y} \\ \frac {\partial \beta_ {x}}{\partial y} + \frac {\partial \beta_ {y}}{\partial x} \end{array} \right] \tag {5.89}
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$$
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<!-- source-page: 438 -->
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<details>
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<summary>text_image</summary>
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z, w
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βx
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βy
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y, v
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θy
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x, u
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θx
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h
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</details>
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Figure 5.25 Deformation assumptions in analysis of plate including shear deformations
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whereas the transverse shear strains are assumed to be constant through the thickness of the plate
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$$
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\left[ \begin{array}{l} \gamma_ {x z} \\ \gamma_ {y z} \end{array} \right] = \left[ \begin{array}{l} \frac {\partial w}{\partial x} - \beta_ {x} \\ \frac {\partial w}{\partial y} - \beta_ {y} \end{array} \right] \tag {5.90}
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$$
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We may note that each transverse shear strain component is of the form (5.56) used in the description of the beam deformations. The state of stress in the plate corresponds to plane stress conditions (i.e., $\tau_{zz} = 0$ ). For an isotropic material, we can thus write
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$$
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\left[ \begin{array}{l} \tau_ {x x} \\ \tau_ {y y} \\ \tau_ {x y} \end{array} \right] = - z \frac {E}{1 - \nu^ {2}} \left[ \begin{array}{c c c} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac {1 - \nu}{2} \end{array} \right] \left[ \begin{array}{c} \frac {\partial \beta_ {x}}{\partial x} \\ \frac {\partial \beta_ {y}}{\partial y} \\ \frac {\partial \beta_ {x}}{\partial y} + \frac {\partial \beta_ {y}}{\partial x} \end{array} \right] \tag {5.91}
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$$
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$$
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\left[ \begin{array}{l} \tau_ {x z} \\ \tau_ {y z} \end{array} \right] = \frac {E}{2 (1 + \nu)} \left[ \begin{array}{l} \frac {\partial w}{\partial x} - \beta_ {x} \\ \frac {\partial w}{\partial y} - \beta_ {y} \end{array} \right] \tag {5.92}
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$$
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To establish the element equilibrium equations we now proceed as in the formulation of the two-dimensional beam element of rectangular cross section [see (5.58) to (5.64)].
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Considering the plate, the expression for the principle of virtual work is, with p equal to the transverse loading per unit of the midsurface area A,
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$$
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\int_ {A} \int_ {- h / 2} ^ {h / 2} \left[ \overline {{\epsilon}} _ {x x} \quad \overline {{\epsilon}} _ {y y} \quad \overline {{\gamma}} _ {x y} \right] \left[ \begin{array}{l} \tau_ {x x} \\ \tau_ {y y} \\ \tau_ {x y} \end{array} \right] d z d A + k \int_ {A} \int_ {- h / 2} ^ {h / 2} \left[ \overline {{\gamma}} _ {x z} \quad \overline {{\gamma}} _ {y z} \right] \left[ \begin{array}{l} \tau_ {x z} \\ \tau_ {y z} \end{array} \right] d z d A = \int_ {A} \overline {{w}} p d A \tag {5.93}
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$$
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where the overbar denotes virtual quantities and k is again a constant to account for the actual nonuniformity of the shearing stresses (the value usually used is $\frac{5}{6}$ ; see Example 5.23). Substituting from (5.89) to (5.92) into (5.93), we thus obtain
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$$
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\int_ {A} \overline {{{\mathbf {K}}}} ^ {T} \mathbf {C} _ {b} \mathbf {K} d A + \int_ {A} \overline {{{\boldsymbol {\gamma}}}} ^ {T} \mathbf {C} _ {s} \boldsymbol {\gamma} d A = \int_ {A} \overline {{{w}}} p d A \tag {5.94}
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$$
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where the internal bending moments and shear forces are $C_{b}\kappa$ and $C_{s}\gamma$ , respectively, and
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$$
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\boldsymbol {\kappa} = \left[ \begin{array}{c} \frac {\partial \beta_ {x}}{\partial x} \\ \frac {\partial \beta_ {y}}{\partial y} \\ \frac {\partial \beta_ {x}}{\partial y} + \frac {\partial \beta_ {y}}{\partial x} \end{array} \right]; \quad \boldsymbol {\gamma} = \left[ \begin{array}{c} \frac {\partial w}{\partial x} - \beta_ {x} \\ \frac {\partial w}{\partial y} - \beta_ {y} \end{array} \right] \tag {5.95}
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$$
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$$
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\mathbf {C} _ {b} = \frac {E h ^ {3}}{1 2 (1 - \nu^ {2})} \left[ \begin{array}{c c c} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac {1 - \nu}{2} \end{array} \right]; \quad \mathbf {C} _ {s} = \frac {E h k}{2 (1 + \nu)} \left[ \begin{array}{c c} 1 & 0 \\ 0 & 1 \end{array} \right] \tag {5.97}
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$$
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Let us note that the variational indicator corresponding to (5.93) is given by (see Example 4.4)
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$$
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\begin{array}{l} \Pi = \frac {1}{2} \int_ {A} \int_ {- h / 2} ^ {h / 2} \left[ \epsilon_ {x x} \quad \epsilon_ {y y} \quad \gamma_ {x y} \right] \frac {E}{1 - \nu^ {2}} \left[ \begin{array}{c c c} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac {1 - \nu}{2} \end{array} \right] \left[ \begin{array}{l} \epsilon_ {x x} \\ \epsilon_ {y y} \\ \gamma_ {x y} \end{array} \right] d z d A \tag {5.98} \\ + \frac {k}{2} \int_ {A} \int_ {- h / 2} ^ {h / 2} \left[ \gamma_ {x z} \quad \gamma_ {y z} \right] \frac {E}{2 (1 + \nu)} \left[ \begin{array}{c} \gamma_ {x z} \\ \gamma_ {y z} \end{array} \right] d z d A - \int_ {A} w p d A \\ \end{array}
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$$
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with the strains given by (5.89) and (5.90). The principle of virtual work corresponds to invoking $\delta\Pi = 0$ with respect to the transverse displacement w and section rotations $\beta_{x}$ and $\beta_{y}$ .
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We emphasize that in this theory w, $\beta_{x}$ , and $\beta_{y}$ are independent variables. Hence, in the finite element discretization using the displacement method, we need to enforce inter-element continuity only on w, $\beta_{x}$ , and $\beta_{y}$ and not on any derivatives thereof, which can readily be achieved in the same way as in the isoparametric finite element analysis of solids.
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Let us consider the pure displacement discretization first. As in the analysis of beams, the pure displacement discretization will not yield efficient lower-order elements but does provide the basis for the mixed interpolation that we shall discuss afterward.
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In the pure displacement discretization we use
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$$
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w = \sum_ {i = 1} ^ {q} h _ {i} w _ {i}; \quad \beta_ {x} = - \sum_ {i = 1} ^ {q} h _ {i} \theta_ {y} ^ {i}
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$$
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$$
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\beta_ {y} = \sum_ {i = 1} ^ {q} h _ {i} \theta_ {x} ^ {i} \tag {5.99}
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$$
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where the $h_{i}$ are the interpolation functions and q is the number of nodes of the element. With these interpolations we can now proceed in the usual way, and all concepts pertaining to the isoparametric finite elements discussed earlier are directly applicable. For example, some interpolation functions applicable to the formulation of plate elements are listed in Fig. 5.4, and triangular elements can be established as discussed in Section 5.3.2. Since the interpolation functions are given in terms of the isoparametric coordinates r, s, we can also directly calculate the matrices of plate elements that are curved in their plane (to model, for example, a circular plate).
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We demonstrate the formulation of a simple four-node element in the following example.
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EXAMPLE 5.29: Derive the expressions used in the evaluation of the stiffness matrix of the four-node plate element shown in Fig. E5.29.
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<details>
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<summary>text_image</summary>
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z
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2 cm
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3
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2
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y
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3 cm
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r
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s
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w1
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1
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4
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x
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h
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θx1
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θy1
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</details>
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Figure E5.29 A four-node plate element
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The calculations are very similar to those performed in the formulation of the two-dimensional plane stress element in Example 5.5.
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For the element in Fig. E5.29 we have (see Example 5.3)
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$$
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\mathbf {J} = \left[ \begin{array}{c c} \frac {3}{2} & 0 \\ 0 & 1 \end{array} \right]
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$$
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