김경종
415 lines
26 KiB
Markdown
415 lines
26 KiB
Markdown
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which is obtained by neglecting the load contributions in (5.65) and dividing by $\frac{1}{2}EI$ . The relation in (5.66) shows the relative importance of the bending and shearing contributions to the stiffness matrix of an element, where we note that the factor GAk/EI in the shearing term can be very large when a thin element is considered. This factor can be interpreted as a penalty number (see Section 3.4.1); i.e., we can write
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$$
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\tilde {\Pi} = \int_ {0} ^ {L} \left(\frac {d \beta}{d x}\right) ^ {2} d x + \alpha \int_ {0} ^ {L} \left(\frac {d w}{d x} - \beta\right) ^ {2} d x; \quad \alpha = \frac {G A k}{E I} \tag {5.67}
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$$
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where $\alpha \to \infty$ as $h \to 0$ . However, this means that as the beam becomes thin, the constraint of zero shear deformations (i.e., $dw / dx = \beta$ with $\gamma = 0$ ) will be approached.
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This argument holds for the actual continuous model which is governed by the stationarity condition of $\tilde{\Pi}$ .
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Considering now the finite element representation, it is important that the finite element displacement assumptions on $\beta$ and $w$ admit that for large values of $\alpha$ the shearing
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L = 10 m
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Square cross section; height = 0.1 m
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2-node beam elements (full integration)
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Young's modulus E
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Figure 5.20 Solution of cantilever beam problem; tip deflection as a number of elements used, showing locking of elements
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deformations can be small throughout the domain of the element. If by virtue of the assumptions used on w and $\beta$ the shearing deformations cannot be small—and indeed zero—everywhere, then an erroneous shear strain energy (which can be large compared with the bending energy) is included in the analysis. This error results into much smaller displacements than the exact values when the beam structure analyzed is thin. Hence, in such cases, the finite element models are much too stiff.
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This phenomenon is observed when the two-node beam element in Fig. 5.19 is used, which therefore should not be employed in the analysis of thin beam structures, and the conclusion is also applicable to the pure displacement-based low-order plate and shell elements discussed in Section 5.4.2. The very stiff behavior exhibited by the thin elements has been referred to as element shear locking. Figure 5.20 shows an example of locking using the two-node displacement-based element. We study the phenomenon of shear locking in the following example (see also Section 4.5.7).
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EXAMPLE 5.25: Consider a two-node isoparametric beam element modeling a cantilever beam that is subjected to only a moment end load (see Fig. E5.25). Determine what values of $\theta_{2}$ , $w_{2}$ would be obtained assuming that the shear strain is zero.
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<details>
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<summary>text_image</summary>
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L
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M
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h
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b = 1.0
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w1
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θ1
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r = -1
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r
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w2
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θ2
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r = +1
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</details>
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Figure E5.25 Two-node element representing a cantilever beam
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The interpolations for w and $\beta$ , for the given data, are
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$$
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\beta = \frac {1 + r}{2} \theta_ {2}
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$$
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$$
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w = \frac {1 + r}{2} w _ {2}
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$$
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Hence, the shearing strain is
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$$
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\gamma = \frac {w _ {2}}{L} - \frac {1 + r}{2} \theta_ {2}
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$$
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For an applied moment only, the shearing strain is to be zero. Imposing this condition gives
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$$
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0 = \frac {w _ {2}}{L} - \frac {1 + r}{2} \theta_ {2} \tag {a}
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$$
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However, for this expression to be zero all along the beam (i.e., for any value $-1 \leq r \leq +1$ ), we clearly must have $w_{2} = \theta_{2} = 0$ . Hence, a zero shear strain in the beam can be reached only when there are no deformations!
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Similarly, if we enforce (a) to hold at the two Gauss points $r = \pm 1 / \sqrt{3}$ , (i.e., if we use two-point Gauss integration), we obtain the two equations,
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$$
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\left[ \begin{array}{l l} \frac {1}{L} & - \frac {1 + 1 / \sqrt {3}}{2} \\ \frac {1}{L} & - \frac {1 - 1 / \sqrt {3}}{2} \end{array} \right] \left[ \begin{array}{l} w _ {2} \\ \theta_ {2} \end{array} \right] = \left[ \begin{array}{l} 0 \\ 0 \end{array} \right]
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$$
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Since the coefficient matrix is nonsingular, the only solution is $w_{2} = \theta_{2} = 0$ . This is of course the result obtained before, because setting the linearly varying shear strain equal to zero at two points means imposing a zero shear strain all along the element.
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However, we can now also use (a) to investigate what happens when we enforce the shear strain to be zero only at the midpoint of the beam (i. e., at r = 0). In this case, (a) gives the relation
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$$
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w _ {2} = \frac {\theta_ {2}}{2} L \tag {b}
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$$
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Hence, if we were to assume a constant shear strain of value
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$$
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\gamma = \frac {w _ {2}}{L} - \frac {\theta_ {2}}{2}
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$$
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a more attractive element might be obtained. We actually used this assumption in Example 4.30.
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Various procedures may be proposed to modify this pure displacement-based beam element formulation—and the formulation of pure displacement-based isoparametric plate bending elements—in order to arrive at efficient nonlocking elements.
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The key point of any such formulation is that the resulting element should be reliable and efficient; this means in particular that the element stiffness matrix must not contain any spurious zero energy mode and that the element should have a high predictive capability under general geometric and loading conditions. These requirements are considerably more easy to satisfy with beam elements than with general plate and shell elements.
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An effective beam element is obtained by using a mixed interpolation of displacements and transverse shear strains. This mixed interpolation is an application of the more general procedure employed in the formulation of plate bending and shell elements (see Section 5.4.2).
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The discussion in Example 5.25 suggests that to satisfy the possibility of a zero transverse shear strain in the element, we may assume for an element with q nodes the interpolations (see also Example 4.30)
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$$
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\begin{array}{l} \left. \begin{array}{l} w = \sum_ {i = 1} ^ {q} h _ {i} w _ {i} \\ \beta = \sum_ {i = 1} ^ {q} h _ {i} \theta_ {i} \end{array} \right\} (5.68) \\ \gamma = \sum_ {i = 1} ^ {q - 1} h _ {i} ^ {*} \gamma \left| _ {G _ {i}} ^ {\mathrm{DI}} \right. (5.69) \\ \end{array}
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$$
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Here the $h_{i}$ are the displacement and section rotation interpolation functions for q nodes and the $h_{i}^{*}$ are the interpolation functions for the transverse shear strains. These functions are
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associated with the $(q - 1)$ discrete values $\gamma|_{G_{i}}^{DI}$ , where $\gamma|_{G_{i}}^{DI}$ is the shear strain at the Gauss point i directly obtained from the displacement/section rotation interpolations (i.e., by displacement interpolation); hence,
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$$
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\gamma \bigg | _ {G _ {i}} ^ {\mathrm{DI}} = \left(\frac {d w}{d x} - \beta\right) \bigg | _ {G _ {i}} \tag {5.70}
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$$
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Figure 5.21 shows the shear strain interpolations used for the two-, three-, and four-node beam elements. These mixed interpolated beam elements are very reliable in that they do not lock, show excellent convergence behavior, and of course do not contain any spurious zero energy mode. For the solution of the problem in Fig. 5.20 only a single element needs to be employed to obtain the exact tip displacement and rotation. We can easily prove this result for the two-node element by continuing the analysis presented in Example 5.25, and the three- and four-node elements contain the interpolations of the two-node element and must therefore also give the exact solution. Hence, there is a drastic improvement in element behavior resulting from the use of mixed interpolation.
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<details>
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<summary>text_image</summary>
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γ
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r
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</details>
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2-node element, constant $\gamma$ ; $G_{1}$ corresponds to $r = 0$
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<details>
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<summary>text_image</summary>
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-1/√3
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r
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+1/√3
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γ
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</details>
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3-node element, linearly varying $\gamma$ ; $G_{1}$ and $G_{2}$ correspond to $r = \pm \sqrt{\frac{1}{3}}$
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<details>
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<summary>text_image</summary>
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-√3/5
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r
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+√3/5
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</details>
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4-node element, parabolically varying $\gamma$ ;
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$G_{1}, G_{2}$ , and $G_{3}$ correspond to $r = \pm \sqrt{\frac{3}{5}}$ and $r = 0$
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Figure 5.21 Shear strain interpolations for mixed interpolated beam elements
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In addition, there is an attractive computational feature: the stiffness matrices of these elements can be evaluated efficiently by simply integrating the displacement-based model with one-point Gauss integration for the two-node element, two-point Gauss integration for the three-node element, and three-point Gauss integration for the four-node element. Namely, using one-point integration in the evaluation of the two-node element stiffness matrix, the transverse shear strain is assumed to be constant, and the contribution from the bending deformation is still evaluated exactly. A similar argument holds for the three- and four-node elements. This computational approach to evaluating the stiffness matrices of the elements may be called “reduced integration” of the displacement-based element but in fact is actually full integration of the mixed interpolated element. A mathematical analysis of the elements is presented in Section 4.5.7.
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# General Curved Beam Elements
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In the preceding discussion we assumed that the elements considered were straight; hence the formulation was based on equation (5.58). To arrive at a general three-dimensional curved beam element formulation, we proceed in a similar way but now need to interpolate the curved geometry and corresponding beam displacements. With these interpolations a pure displacement-based element is derived that, as for the straight elements, is very stiff and not useful. In the case of straight beam elements only spurious shear strains are generated (always for the two-node element, and for the three- and four-node elements when the interior nodes are not at their natural positions; see Exercise 5.34), but for curved elements also spurious membrane strains are obtained. Hence, a curved element also displays membrane locking (see, for example, H. Stolarski and T. Belytschko [A]).
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Efficient general beam elements are obtained by the mixed interpolation already introduced. However, now, in the case of general three-dimensional action, the transverse shear strains and the bending and membrane strains are interpolated using the functions in Fig. 5.21. These strain interpolations are tied to the nodal point displacements and rotations by evaluating the displacement-based strains and equating them to the assumed strains at the Gauss integration points.
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It follows that the mixed interpolated element stiffness matrices can be numerically obtained by evaluating the displacement-based element matrices with Gauss point integration at the points given in Fig. 5.21.
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Consider the three-dimensional beam of rectangular cross section in Fig. 5.22, and let us assume first that an accurate representation of the torsional rigidity is not required.
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The basic kinematic assumption in the formulation of the element is the same as that employed in the formulation of the two-dimensional element in Fig. 5.19: namely that plane sections originally normal to the centerline axis remain plane and undistorted under deformation but not necessarily normal to this axis. This kinematic assumption does not allow for warping effects in torsion (which we can introduce by additional displacement functions; see Exercise 5.37).
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Using the natural coordinates $r, s, t$ , the Cartesian coordinates of a point in the element with $q$ nodal points are then, before and after deformations,
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$$
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{ } ^ { \ell } x ( r , s , t ) = \sum _ { k = 1 } ^ { q } h _ { k } { } ^ { \ell } x _ { k } + \frac { t } { 2 } \sum _ { k = 1 } ^ { q } a _ { k } h _ { k } { } ^ { \ell } V _ { t x } ^ { k } + \frac { s } { 2 } \sum _ { k = 1 } ^ { q } b _ { k } h _ { k } { } ^ { \ell } V _ { s x } ^ { k }
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$$
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<details>
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<summary>text_image</summary>
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b₂
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a₂
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r, η
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s, ξ
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t, ξ
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b₁
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Node
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1
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a₁
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0V₁ₛ
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0V₁ₜ
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θz
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z, w
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θz
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θy
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θx
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x, u
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y, v
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θy
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θx
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Vectors ⁰V₁ₛ, ⁰V₁ₜ are normal to
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the neutral axis of the beam
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(and normal to each other)
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</details>
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Figure 5.22 Three-dimensional beam element
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$$
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{ } ^ { \ell } y ( r , s , t ) = \sum _ { k = 1 } ^ { q } h _ { k } { } ^ { \ell } y _ { k } + \frac { t } { 2 } \sum _ { k = 1 } ^ { q } a _ { k } h _ { k } { } ^ { \ell } V _ { t y } ^ { k } + \frac { s } { 2 } \sum _ { k = 1 } ^ { q } b _ { k } h _ { k } { } ^ { \ell } V _ { s y } ^ { k } \tag {5.71}
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$$
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$$
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{ } ^ { \ell } z ( r , s , t ) = \sum _ { k = 1 } ^ { q } h _ { k } { } ^ { \ell } z _ { k } + \frac { t } { 2 } \sum _ { k = 1 } ^ { q } a _ { k } h _ { k } { } ^ { \ell } V _ { t z } ^ { k } + \frac { s } { 2 } \sum _ { k = 1 } ^ { q } b _ { k } h _ { k } { } ^ { \ell } V _ { s z } ^ { k }
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$$
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where the $h_{k}(r)$ are the interpolation functions summarized in Fig. 5.3 and
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$^{e}x,\ ^{e}y,\ ^{e}z=$ Cartesian coordinates of any point in the element
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$^{e}x_{k}, ^{e}y_{k}, ^{e}z_{k} = \text{Cartesian coordinates of nodal point } k$
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$a_{k}, b_{k} =$ cross-sectional dimensions of the beam at nodal point $k$
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$^{\ell}V_{ix}^{k}, ^{\ell}V_{iy}^{k}, ^{\ell}V_{iz}^{k} = \text{components of unit vector } ^{\ell}\mathbf{V}_{i}^{k} \text{ in direction } t \text{ at nodal point } k$
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$^{\ell}V_{sx}^{k}, ^{\ell}V_{sy}^{k}, ^{\ell}V_{sz}^{k} = \text{components of unit vector } ^{\ell}\mathbf{V}_{s}^{k} \text{ in direction } s \text{ at nodal point } k; \text{ we call } ^{\ell}\mathbf{V}_{s}^{k} \text{ and } ^{\ell}\mathbf{V}_{s}^{k} \text{ the normal vectors or director vectors at nodal point } k,$
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and the left superscript $\ell$ denotes the configuration of the element; i.e., $\ell = 0$ denotes the original configuration, whereas $\ell = 1$ corresponds to the configuration in the deformed position.
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We assume here that the vectors $^{0}V_{s}^{k}$ , $^{0}V_{t}^{k}$ are normal to the neutral axis of the beam and are normal to each other. However, this condition can be relaxed, as is done in the shell element formulation (see Section 5.4.2).
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The displacement components at any point of the element are
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$$
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u (r, s, t) = ^ {1} x - ^ {0} x
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$$
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$$
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v (r, s, t) = ^ {1} y - ^ {0} y \tag {5.72}
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$$
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$$
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w (r, s, t) = ^ {1} z - ^ {0} z
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$$
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and substituting from (5.71), we obtain
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$$
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u (r, s, t) = \sum_ {k = 1} ^ {q} h _ {k} u _ {k} + \frac {t}{2} \sum_ {k = 1} ^ {q} a _ {k} h _ {k} V _ {t x} ^ {k} + \frac {s}{2} \sum_ {k = 1} ^ {q} b _ {k} h _ {k} V _ {s x} ^ {k}
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$$
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$$
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v (r, s, t) = \sum_ {k = 1} ^ {q} h _ {k} v _ {k} + \frac {t}{2} \sum_ {k = 1} ^ {q} a _ {k} h _ {k} V _ {t y} ^ {k} + \frac {s}{2} \sum_ {k = 1} ^ {q} b _ {k} h _ {k} V _ {s y} ^ {k} \tag {5.73}
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$$
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$$
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w (r, s, t) = \sum_ {k = 1} ^ {q} h _ {k} w _ {k} + \frac {t}{2} \sum_ {k = 1} ^ {q} a _ {k} h _ {k} V _ {t z} ^ {k} + \frac {s}{2} \sum_ {k = 1} ^ {q} b _ {k} h _ {k} V _ {s z} ^ {k}
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$$
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where $\mathbf{V}_t^k = {}^1\mathbf{V}_t^k - {}^0\mathbf{V}_t^k;$ $\mathbf{V}_s^k = {}^1\mathbf{V}_s^k - {}^0\mathbf{V}_s^k$ (5.74)
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Finally, we express the vectors $V_{t}^{k}$ and $V_{s}^{k}$ in terms of rotations about the Cartesian axes x, y, z:
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$$
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\mathbf {V} _ {t} ^ {k} = \boldsymbol {\theta} _ {k} \times {} ^ {0} \mathbf {V} _ {t} ^ {k} \tag {5.75}
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$$
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$$
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\mathbf {V} _ {s} ^ {k} = \boldsymbol {\theta} _ {k} \times {} ^ {0} \mathbf {V} _ {s} ^ {k}
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$$
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where $\theta_{k}$ is a vector listing the nodal point rotations at nodal point k (see Fig. 5.22):
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$$
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\boldsymbol {\theta} _ {k} = \left[ \begin{array}{l} \theta_ {x} ^ {k} \\ \theta_ {y} ^ {k} \\ \theta_ {z} ^ {k} \end{array} \right] \tag {5.76}
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$$
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Using (5.71) to (5.76), we have all the basic equations necessary to establish the displacement and strain interpolation matrices employed in evaluating the beam element matrices.
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The terms in the displacement interpolation matrix H are obtained by substituting (5.75) into (5.73). To evaluate the strain-displacement matrix, we recognize that for the beam the only strain components of interest are the longitudinal strain $\epsilon_{\eta\eta}$ and transverse shear strains $\gamma_{\eta\xi}$ and $\gamma_{\eta\xi}$ , where $\eta$ , $\xi$ , and $\zeta$ are convected (body-attached) coordinates axes (see Fig. 5.22). Thus, we seek a relation of the form
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$$
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\left[ \begin{array}{l} \epsilon_ {\eta \eta} \\ \gamma_ {\eta \xi} \\ \gamma_ {\eta \zeta} \end{array} \right] = \sum_ {k = 1} ^ {q} \mathbf {B} _ {k} \hat {\mathbf {u}} _ {k} \tag {5.77}
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$$
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where $\hat{\mathbf{u}}_k^T = [u_k v_k w_k \theta_x^k \theta_y^k \theta_z^k]$ (5.78)
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and the matrices $\mathbf{B}_k, k = 1, \ldots, q$ , together constitute the matrix $\mathbf{B}$ ,
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$$
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\mathbf {B} = \left[ \begin{array}{l l l} \mathbf {B} _ {1} & \dots & \mathbf {B} _ {q} \end{array} \right] \tag {5.79}
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$$
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Following the usual procedure of isoparametric finite element formulation, we have, using (5.73),
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$$
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\left[ \begin{array}{l} \frac {\partial u}{\partial r} \\ \frac {\partial u}{\partial s} \\ \frac {\partial u}{\partial t} \end{array} \right] = \sum_ {k = 1} ^ {q} \left[ \begin{array}{l l l l l} \frac {\partial h _ {k}}{\partial r} & [ 1 & (g) _ {1 i} ^ {k} & (g) _ {2 i} ^ {k} & (g) _ {3 i} ^ {k} ] \\ h _ {k} & [ 0 & (\hat {g}) _ {1 i} ^ {k} & (\hat {g}) _ {2 i} ^ {k} & (\hat {g}) _ {3 i} ^ {k} ] \\ h _ {k} & [ 0 & (\overline {{g}}) _ {1 i} ^ {k} & (\overline {{g}}) _ {2 i} ^ {k} & (\overline {{g}}) _ {3 i} ^ {k} ] \end{array} \right] \left[ \begin{array}{l} u _ {k} \\ \theta_ {x} ^ {k} \\ \theta_ {y} ^ {k} \\ \theta_ {z} ^ {k} \end{array} \right] \tag {5.80}
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$$
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and the derivatives of v and w are obtained by simply substituting v and w for u. In (5.80) we have i = 1 for u, i = 2 for v, and i = 3 for w, and we employ the notation
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$$
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(\hat {\mathbf {g}}) ^ {k} = \frac {b _ {k}}{2} \left[ \begin{array}{c c c} 0 & ^ {- 0} V _ {s z} ^ {k} & ^ {0} V _ {s y} ^ {k} \\ ^ {0} V _ {s z} ^ {k} & 0 & ^ {- 0} V _ {s x} ^ {k} \\ ^ {- 0} V _ {s y} ^ {k} & ^ {0} V _ {s x} ^ {k} & 0 \end{array} \right] \tag {5.81}
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$$
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$$
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(\overline {{{\mathbf {g}}}}) ^ {k} = \frac {a _ {k}}{2} \left[ \begin{array}{c c c} 0 & ^ {- 0} V _ {t z} ^ {k} & ^ {0} V _ {t y} ^ {k} \\ ^ {0} V _ {t z} ^ {k} & 0 & ^ {- 0} V _ {t x} ^ {k} \\ ^ {- 0} V _ {t y} ^ {k} & ^ {0} V _ {t x} ^ {k} & 0 \end{array} \right] \tag {5.82}
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$$
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$$
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(g) _ {i j} ^ {k} = s (\hat {g}) _ {i j} ^ {k} + t (\overline {{g}}) _ {i j} ^ {k} \tag {5.83}
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$$
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To obtain the displacement derivatives corresponding to the coordinate axes x, y, and z, we employ the Jacobian transformation
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$$
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\frac {\partial}{\partial \mathbf {x}} = \mathbf {J} ^ {- 1} \frac {\partial}{\partial \mathbf {r}} \tag {5.84}
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$$
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where the Jacobian matrix J contains the derivatives of the coordinates x, y, and z with respect to the natural coordinates r, s, and t. Substituting from (5.80) into (5.84), we obtain
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$$
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\left[ \begin{array}{l} \frac {\partial u}{\partial x} \\ \frac {\partial u}{\partial y} \\ \frac {\partial u}{\partial z} \end{array} \right] = \sum_ {k = 1} ^ {q} \left[ \begin{array}{l l l l} J _ {1 1} ^ {- 1} \frac {\partial h _ {k}}{\partial r} & (G 1) _ {i 1} ^ {k} & (G 2) _ {i 1} ^ {k} & (G 3) _ {i 1} ^ {k} \\ J _ {2 1} ^ {- 1} \frac {\partial h _ {k}}{\partial r} & (G 1) _ {i 2} ^ {k} & (G 2) _ {i 2} ^ {k} & (G 3) _ {i 2} ^ {k} \\ J _ {3 1} ^ {- 1} \frac {\partial h _ {k}}{\partial r} & (G 1) _ {i 3} ^ {k} & (G 2) _ {i 3} ^ {k} & (G 3) _ {i 3} ^ {k} \end{array} \right] \left[ \begin{array}{l} u _ {k} \\ \theta_ {x} ^ {k} \\ \theta_ {y} ^ {k} \\ \theta_ {z} ^ {k} \end{array} \right] \tag {5.85}
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$$
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and again, the derivatives of $v$ and $w$ are obtained by simply substituting $v$ and $w$ for $u$ . In (5.85) we employ the notation
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$$
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(G m) _ {i n} ^ {k} = \left[ J _ {n 1} ^ {- 1} (g) _ {m i} ^ {k} \right] \frac {\partial h _ {k}}{\partial r} + \left[ J _ {n 2} ^ {- 1} (\hat {g}) _ {m i} ^ {k} + J _ {n 3} ^ {- 1} (\overline {{{g}}}) _ {m i} ^ {k} \right] h _ {k} \tag {5.86}
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$$
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Using the displacement derivatives in (5.85), we can now calculate the elements of the strain-displacement matrix at the element Gauss point by establishing the strain components corresponding to the x, y, z axes and transforming these components to the local strains $\epsilon_{\eta\eta}$ , $\gamma_{\eta\xi}$ , and $\gamma_{\eta\xi}$ .
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The corresponding stress-strain law to be employed in the formulation is (using k as the shear correction factor, possibly different for different directions)
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<!-- source-page: 429 -->
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TABLE 5.2 Performance of isoparametric beam elements in the analysis of the problem in Fig. 5.23
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(a) $\theta_{finite element}/\theta_{analytical}$ at tip of beam, one three-node element solution
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<table><tr><td rowspan="2">h/R</td><td colspan="2">Midnode at $\alpha = 22.5^{\circ}$ </td><td colspan="2">Midnode at $\alpha = 20^{\circ}$ </td></tr><tr><td>Displacement-based</td><td>Mixed interpolated</td><td>Displacement-based</td><td>Mixed interpolated</td></tr><tr><td>0.50</td><td>0.92</td><td>1.00</td><td>0.91</td><td>1.00</td></tr><tr><td>0.10</td><td>0.31</td><td>1.00</td><td>0.31</td><td>1.00</td></tr><tr><td>0.01</td><td>0.004</td><td>1.00</td><td>0.005</td><td>1.00</td></tr><tr><td>0.001</td><td>0.00004</td><td>1.00</td><td>0.00005</td><td>1.00</td></tr></table>
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(b) $\theta_{finite element}/\theta_{analytical}$ at tip of beam, one four-node element solution
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<table><tr><td rowspan="2">h/R</td><td colspan="2">Internal nodes at $\alpha_{1} = 15^{\circ}$ , $\alpha_{2} = 30^{\circ}$ </td><td colspan="2">Internal nodes at $\alpha_{1} = 10^{\circ}$ , $\alpha_{2} = 25^{\circ}$ </td></tr><tr><td>Displacement-based</td><td>Mixed interpolated</td><td>Displacement-based</td><td>Mixed interpolated</td></tr><tr><td>0.50</td><td>1.00</td><td>1.00</td><td>0.97</td><td>0.997</td></tr><tr><td>0.10</td><td>0.999</td><td>1.00</td><td>0.70</td><td>0.997</td></tr><tr><td>0.01</td><td>0.998</td><td>1.00</td><td>0.37</td><td>0.997</td></tr><tr><td>0.001</td><td>0.998</td><td>1.00</td><td>0.37</td><td>0.997</td></tr></table>$$
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\left[ \begin{array}{l} \tau_ {\eta \eta} \\ \tau_ {\eta \xi} \\ \tau_ {\eta \zeta} \end{array} \right] = \left[ \begin{array}{c c c} E & 0 & 0 \\ 0 & G k & 0 \\ 0 & 0 & G k \end{array} \right] \left[ \begin{array}{l} \epsilon_ {\eta \eta} \\ \gamma_ {\eta \xi} \\ \gamma_ {\eta \zeta} \end{array} \right] \tag {5.87}
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$$
|
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The stiffness matrix of the element is then obtained by numerical integration, using for the r-integration the Gauss points shown in Fig. 5.21 and for the s- and t-integrations either the Newton-Cotes or Gauss formulas (see Section 5.5).
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Table 5.2 illustrates the performance of the mixed interpolated elements in the analysis of the curved cantilever in Fig. 5.23 and shows the efficiency of the elements.
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As pointed out already, this element does not include warping effects, which can be significant for the rectangular cross-sectional beam elements and of course for beam ele-
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<details>
|
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<summary>text_image</summary>
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|
||
h
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α
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||
R
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45°
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M, θ
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||
</details>
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Figure 5.23 Curved cantilever problem to test isoparametric beam elements
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<!-- source-page: 430 -->
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ments of general cross sections. Warping displacements can be introduced by adding appropriate displacement interpolations to those given in (5.73). If additional degrees of freedom are also introduced, corresponding to the warping deformations, continuity of warping between elements can be imposed. However, it may be sufficient to allow for “free” warping in each element without enforcing continuity of warping between elements. This is achieved by adding warping deformations to the displacement interpolations and then statically condensing out the intensity of these deformations (see K. J. Bathe and A. B. Chaudhary [A] and Exercise 5.37).
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In another application of the general curved beam formulation, the cross section is circular and hollow, that is, a pipe cross section is considered. In this case, ovalization and warping deformations can be important when the element is curved, and once again the displacement interpolations given in (5.73) must be amended. In this case it is important to impose continuity in the ovalization and warping deformations, and for this reason additional nodal degrees of freedom must be introduced (see K. J. Bathe and C. A. Almeida [A]).
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Considering the basic formulation in (5.71) to (5.87), we recognize that the element can be arbitrarily curved and that the cross-sectional dimensions can change along its axis. The beam width and height and the location of the element axis are interpolated along the element. In the given formulation, the axis of the element coincides with the element geometric midline, but this is not necessary, and more general elements can be formulated directly (see Exercise 5.38).
|
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|
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In addition to representing a general formulation for linear analysis, this approach is particularly useful for the nonlinear large displacement analysis of beam structures. As discussed in Section 6.5.1, in such analyses initially straight beam elements become curved and distorted, and these deformations can be modeled accurately.
|
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|
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Of course, if linear analysis is pursued and the element is straight and has a constant cross-sectional area, the formulation reduces to the formulation given in (5.56) to (5.70). We illustrate this point in the following example.
|
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EXAMPLE 5.26: Show that the application of the general formulation in (5.71) to (5.87) to the beam element in Fig. E5.24 reduces to the use of (5.58).
|
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For the application of the general relations in (5.71) to (5.87), we refer to Figs. E5.24 and 5.22 and thus have
|
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|
||
$$
|
||
{ } ^ { 0 } \mathbf { V } _ { s } = \left[ \begin{array} { l } 0 \\ 1 \\ 0 \end{array} \right] ; \qquad { } ^ { 0 } \mathbf { V } _ { t } = \left[ \begin{array} { l } 0 \\ 0 \\ 1 \end{array} \right] ; \qquad a _ { k } = 1 ; \quad b _ { k } = h ; \quad k = 1 , 2 , 3
|
||
$$
|
||
|
||
Hence, the relations in (5.71) reduce to
|
||
|
||
$$
|
||
{ } ^ { 0 } x = \sum _ { k = 1 } ^ { 3 } h _ { k } { } ^ { 0 } x _ { k }
|
||
$$
|
||
|
||
$$
|
||
{ } ^ { 0 } y = \frac { s } { 2 } h
|
||
$$
|
||
|
||
$$
|
||
{ } ^ { 0 } z = \frac { t } { 2 }
|
||
$$
|