김경종
435 lines
29 KiB
Markdown
435 lines
29 KiB
Markdown
<!-- source-page: 361 -->
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$R_{\theta} = \text{const.}$
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$R_{s} = x$
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Cylinder
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$R_{s} = x$
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Cone
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$R_{\theta} = R$
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$R_{s} = R$
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Sphere
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$R_{\theta} > 0$
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$R_{5} < 0$
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Hyperboloid
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Figure 12.1-1. Shells of revolution, showing principal radii of curvature. $R_{s}$ is the radius of curvature of the meridian.
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normal to the midsurface. Stresses in the shell are composed of a membrane component $\sigma_{m}$ and a bending component $\sigma_{b}$ ; for example, normal stress in the x direction is
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$$
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\sigma_ {x} = \sigma_ {m x} + \sigma_ {b x} = \frac {N _ {x}}{t} + \frac {M _ {x} z}{t ^ {3} / 1 2} \tag {12.1-2}
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$$
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Thus it is assumed that stresses vary linearly through the thickness. Stress on the midsurface z = 0 is zero if $N_{x} = N_{y} = N_{xy} = 0$ .
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A shell can carry a large load if membrane action dominates over bending, just as a thin wire can carry a large load in tension but only a small load in bending. Practically, it is not possible to have only membrane action in a shell. Bending action is also present if concentrated loads are applied, if supports apply moments or transverse forces, or if a radius of curvature changes abruptly. As an example of the latter, consider closing a cylindrical pressure vessel with a cap that is hemispherical or ellipsoidal: where the cylinder meets the cap, radius $R_{s}$ of the meridian changes abruptly from infinite to finite. Typically, bending action is quite localized—that is, bending stresses are large only quite near the load or discontinuity that produces them.
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Classical shell theory produces equations that are very difficult to solve. The governing equations in terms of displacements are complicated; they have relatively simple forms only if many approximations are made. Authorities do not agree on what approximations are acceptable, so various shell theories have been proposed (e.g., those of Donnell, Flügge, Sanders, Vlasov, etc.). Like Kirchhoff plate theory, shell theories are limited to small deflections unless higher-order terms are added to account for membrane strains associated with rotation of the shell midsurface.
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Classical shell theory is concerned with thin shells, in which transverse shear deformation is considered negligible. In practice one may also encounter thick shells. Then one must account for transverse shear deformation, and perhaps also for the effects of thickness-direction normal stress.
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Shell Elements. Finite elements for shells have been among the most difficult elements to devise. Three approaches to the problem have been pursued:
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1. Flat elements, formed by combining a plane membrane element with a plate bending element.
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2. Curved elements, formulated by use of a classical shell theory.
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3. Mindlin-type elements, similar to Mindlin plate elements described in Section 11.3. Such elements can be regarded as special forms of solid elements, made thin in one direction.
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Flat triangular elements model a shell as a faceted surface. Flat elements are easy to formulate. They pass patch tests and do not exhibit strain under rigid-body motion. However, although membrane-bending coupling is present throughout an actual curved shell, it is absent in individual flat elements. In the past, flat elements have not been particularly accurate, but have been useful because of the difficulties of other approaches.
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A curved element is necessarily more complicated than a flat element, first because its geometry is more complicated. Then, regardless of what classical shell theory is used, its approximations and complexities are incorporated in the element. Some curved elements cannot display rigid-body motion without strain, either because of defects in the shell theory or because of shortcomings in the element displacement field. Typically, the user of a curved element must supply data in addition to nodal coordinates in order to describe element geometry. Some curved shell elements include derivatives of membrane strain and curvature among their nodal d.o.f.
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Mindlin-type or “degenerated solid” elements can be curved and appear to occupy a middle ground between flat elements and curved elements formulated by use of shell theory, both in accuracy and in ease of use. Elements for thin
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<details>
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<summary>natural_image</summary>
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3D wireframe model of a mechanical component with intersecting surfaces (no text or symbols)
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</details>
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Figure 12.1-2. Intersecting pipes, modeled by quadrilateral and triangular shell elements. (Courtesy of Algor Interactive Systems, Inc., Pittsburgh, Pennsylvania.)
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shells and for thick shells are available. As with Mindlin plate elements, possible difficulties with locking must be addressed.
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Any of the foregoing three approaches might be used to provide elements for a particular shell, such as the shell shown in Fig. 12.1-2.
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The membrane stiffness of a thin shell is much larger than its bending stiffness. This is reflected in a large discrepancy between the associated stiffness coefficients in [K], regardless of how the shell elements are formulated. Numerical errors of the type discussed in Section 18.2 are possible.
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At present it is not clear whether the most cost-effective thin-shell elements will be flat or curved. In what follows, emphasis is placed on flat elements, elements for shells of revolution, and Mindlin elements. Curved elements for thin shells of general shape that are based on a classical shell theory are beyond the scope of this text.
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Test Cases for General Shell Elements. If assigned a flat geometry, shell elements can be used to solve problems of plane stress and plate bending. Accordingly, one can begin with patch tests and other commonly used test problems for plane problems and plates (e.g., Fig. 4.6-1, Tables 6.14-1, 6.14-2, and 11.4-1, and proposed test problems in Ref. 12.15). Singly curved shell elements can be tested on arch problems (e.g., Fig. 12.2-1a) and on cylindrical shell problems (e.g., Fig. 12.4-4). A good element will have good accuracy on these initial tests and will not have mechanisms. Various additional problems have been used as test cases for general shell elements, including a pinched cylinder, a pinched hemisphere, and a slit cylinder under twisting load. Details may be found in [12.8,12.14–12.16].
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# 12.2 CIRCULAR ARCHES AND ARCH ELEMENTS
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The study of arch elements provides insight into various aspects of shell element behavior. In what follows we consider an arch of constant mean radius R, loaded in its own plane.
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Equations for Thin Circular Arches. We assume that the arch is thin—that is, that $R \gg t$ in Fig. 12.2-1. A point on the arch midline has s-direction (tangential) displacement u and z-direction (radial) displacement w. Let $\epsilon_{s}$ represent tangential
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<details>
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<summary>text_image</summary>
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P
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t
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R
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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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z,w
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u
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t
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s
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1
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s = - L/2
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R
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s = L/2
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2
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</details>
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(b)
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Figure 12.2-1. (a) Semicircular arch with clamped ends and concentrated center load. (b) Arch element of arc length L. Coordinates s and z are respectively tangent and normal to the arch midline.
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strain at an arbitrary point, a distance z from the midline. With the aid of Fig. 12.2-2, and with R constant, we write
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$$
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\epsilon_ {s} = \frac {d}{d s} \left(\delta_ {a} + \delta_ {c}\right) + \frac {w}{R} = u _ {, s} + \frac {w}{R} + z \left(\frac {u _ {, s}}{R} - w _ {, s s}\right) \tag {12.2-1}
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$$
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If the approximation $\delta_{a} \approx u$ is introduced, the term $zu_{,s} / R$ disappears. In alternative notation, Eq. 12.2-1 is
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$$
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\epsilon_ {s} = \epsilon_ {m} + z \kappa , \quad \text { where } \quad \left\{ \begin{array}{l l} \epsilon_ {m} = u _ {, s} + \frac {w}{R} & \text {(12.2 - 2a)} \\ \kappa = \frac {u _ {, s}}{R} - w _ {, s s} & \text {(12.2 - 2b)} \end{array} \right.
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$$
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Membrane strain $\epsilon_{m}$ appears along the arch midline and is associated with s-direction (membrane) force. Curvature change $\kappa$ is associated with bending moment and is considered positive when the radius of curvature decreases. Transverse shear deformation $\gamma_{zs}$ is considered zero for a thin arch.
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Strain energy U in an arch element is composed of a membrane contribution $U_{m}$ and a bending contribution $U_{b}$ . For an element of arc length L,
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$$
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U = U _ {m} + U _ {b} = \int_ {- L / 2} ^ {L / 2} \frac {E A}{2} \epsilon_ {m} ^ {2} d s + \int_ {- L / 2} ^ {L / 2} \frac {E I}{2} \kappa^ {2} d s \tag {12.2-3}
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$$
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where $E =$ elastic modulus, $A =$ cross-sectional area, and $I =$ moment of inertia of $A$ about the neutral axis of bending ( $I = bt^3/12$ for a rectangular cross section of width $b$ ). The expression for $U$ can be derived by integration of strain energy density $E\epsilon_s^2/2$ through the arch thickness $t$ . A term linear in $z$ disappears, and the terms shown in Eq. 12.2-3 remain. Note that membrane stiffness $EA$ becomes much larger than bending stiffness $EI$ as arch thickness $t$ becomes small.
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For rigid-body motion, $\epsilon_{m} = \kappa = 0$ . The displacement field for rigid-body motion is therefore
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$$
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u = b _ {1} \cos \phi + b _ {2} \sin \phi + b _ {3} \tag {12.2-4a}
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$$
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$$
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w = b _ {1} \sin \phi - b _ {2} \cos \phi \tag {12.2-4b}
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$$
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<details>
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<summary>text_image</summary>
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t
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z
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R
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δa
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u
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</details>
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$$
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\frac {\delta_ {n}}{R + z} = \frac {u}{R}
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$$
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(a)
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<details>
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<summary>text_image</summary>
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t
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R
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z
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w
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w
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</details>
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$$
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\epsilon = \frac {w}{R + z} \approx \frac {w}{R}
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$$
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(b)
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<details>
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<summary>text_image</summary>
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δc
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z
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R
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w1s
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w1s
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l
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</details>
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$$
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\delta_ {c} = - z w _ {1 s}
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$$
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{c}
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Figure 12.2-2. Axial, radial, and rotational motions of a thin arch (R >> t), used to formulate an expression for axial strain. Angle $w_{ss}$ is presumed small.
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where $\phi = s / R$ and the $b_{i}$ are constants. Here $b_{1}$ and $b_{2}$ represent translations in mutually perpendicular directions and $b_{3}$ represents a rotation about the center of curvature.
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A thin arch will bend but will have very little membrane strain. Accordingly, from Eq. 12.2-2, we obtain the inextensibility condition:
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$$
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\epsilon_ {m} = 0 \quad \text { implies } \quad u _ {, s} + \frac {w}{R} = 0 \tag {12.2-5}
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$$
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This condition is satisfied in the limit of thinness as L/t becomes infinite.
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Straight Arch Elements. The use of straight elements to model an arch is analogous to the use of flat elements to model a shell. A straight arch element is identical to a plane frame element. To obtain it, one merely combines a standard two-d.o.f. bar element (Eq. 2.4-5) with a standard four-d.o.f. beam element (Eq. 4.2-5). Thus, using d.o.f. in Fig. 12.2-3a, the element stiffness equation $[k]\{d\} = \{r\}$ is
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$$
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\left[ \begin{array}{c c} {[ \mathrm{k} _ {\text {bar}} ]} & {[ 0 ]} \\ {2 \times 2} & {2 \times 4} \\ {[ 0 ]} & {[ \mathrm{k} _ {\text {beam}} ]} \\ {4 \times 2} & {4 \times 4} \end{array} \right] \left\{ \begin{array}{l} u _ {1} \\ u _ {2} \\ w _ {1} \\ \beta_ {1} \\ w _ {2} \\ \beta_ {2} \end{array} \right\} = ^ {\prime} \{\mathbf {r} \} \tag {12.2-6}
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$$
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in which $[k_{bar}]$ and $[k_{beam}]$ come respectively from $U_{m}$ and $U_{b}$ in Eq. 12.2-3. Nodal rotations $\beta_{1}$ and $\beta_{2}$ are nodal values of $w_{ss}$ .
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For assembly with other elements having different orientation, a common set of structural d.o.f. is needed. The choice is not unique. One possibility is to use tangential and radial translational d.o.f. $D_{s}$ and $D_{r}$ at each node, as shown in Fig. 12.2-3b. Coordinate transformation (Section 7.4) is used to replace nodal values of u and w by nodal values of $D_{s}$ and $D_{r}$ . D.o.f. $\beta_{1}$ and $\beta_{2}$ are unchanged by this transformation. (A similar transformation is described in connection with Eq. 12.4-3.)
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The displacement field that produces Eq. 12.2-6 can be written
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$$
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u = a _ {1} + a _ {2} s \tag {12.2-7a}
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$$
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$$
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w = a _ {3} + a _ {4} s + a _ {5} s ^ {2} + a _ {6} s ^ {3} \tag {12.2-7b}
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$$
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<details>
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<summary>text_image</summary>
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w₁
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β₁
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t
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w₂
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β₂
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u₁
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u₂
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L
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2
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L
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2
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s
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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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D_{r1} D_{s1} D_{r2}
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β1 β2
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1 2 D_{s2}
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3 β3
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D_{r3}
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(b) D_{s3}
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</details>
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Figure 12.2-3. (a) Straight element, showing d.o.f. in the local coordinate system. (b) Possible choice of global d.o.f, where $D_{r}$ and $D_{s}$ are, respectively, radial and tangential.
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where $u$ and $w$ are respectively parallel and normal to the straight element, $s$ is a (straight) axial coordinate, and the $a_i$ are constants. Rigid-body motion is accounted for by $a_1, a_3$ , and $a_4$ . Since $R$ is infinite for a straight element, Eqs. 12.2-2 yield $\epsilon_m = a_2$ and $\kappa = -2a_5 - 6a_6s$ . Thus constant-strain states are possible, and these states are not coupled to rigid-body motion. The inextensibility condition, Eq. 12.2-5, is satisfied when $a_2 = 0$ . When an additional element is attached to elements already in place, three new d.o.f. are added to the mesh (e.g., the d.o.f. at node 3 in Fig. 12.2-3b), but only one inextensibility constraint is added ( $a_2 = 0$ ). Therefore the mesh will not lock, regardless of the values of $R, L,$ and $t$ . (Locking concepts are discussed in Sections 9.4 and 9.5.)
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Accordingly, the straight element of Eqs. 12.2-7 is free of major defects. It can be rigorously shown that the straight element is a valid model for a curved arch, and provides convergence to exact answers as the mesh is refined [12.1]. A drawback is that membrane and bending actions are not coupled within a single straight element, as evidenced by the off-diagonal null matrices in Eq. 12.2-6. Another drawback is that a distributed load must be modeled by nodal forces only. If nodal moment loads are also applied, as the consistent formulation of Eq. 4.1-6 would dictate, and if elements are of unequal lengths, then spurious bending moments appear. (These moments would be correct if the actual structure were a polygon of straight bars.) And, during stress computation, stresses caused by nodal displacements should not be adjusted to account for distributed load on the element (as discussed following Eq. 4.7-3).
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In a coarse mesh, accuracy is improved by taking element length L as the arc length between nodes, rather than the chord length as is suggested by Fig. 12.2-3b. For example, let R/t = 40 for the arch of Fig. 12.2-1a, and let the entire arch be modeled by four straight elements, each of length $L = \pi R/4$ . The error in the computed displacement of load P is approximately 2%. Use of the chord length $L = 2R \sin 22.5^{\circ}$ gives an error of approximately 10%.
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Curved Arch Elements. A great many types of curved arch element have been proposed, partly because a good curved element proved to be more elusive than first anticipated. Often, curved elements were far less accurate than straight elements: they were much too stiff, especially when applied to an arch that is thin and has a large rise-to-span ratio. Originally the difficulties were blamed on a lack of rigid-body motion capability. Subsequently the difficulties were attributed primarily to membrane locking [12.17], which is described as follows.
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The simplest curved element has the geometry shown in Fig. 12.2-1b and the displacement field of Eqs. 12.2-7, where u and w are now circumferential and radial displacements. Coordinate s follows the arch midline. Element d.o.f. are nodal values of u, w, and $w_{,s}$ . (In contrast to straight elements, no coordinate transformation is needed to match d.o.f. prior to assembly of elements.) From Eqs. 12.2-2 and 12.2-7, membrane strain $\epsilon_{m}$ is
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$$
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\epsilon_ {m} = \left(a _ {2} + \frac {a _ {3}}{R}\right) + \frac {a _ {4}}{R} s + \frac {a _ {5}}{R} s ^ {2} + \frac {a _ {6}}{R} s ^ {3} \tag {12.2-8}
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$$
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Now imagine that thickness $t$ of the arch approaches zero. The inextensibility condition, $\epsilon_{m} = 0$ for all $s$ , demands that
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$$
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a _ {2} + \frac {a _ {3}}{R} = a _ {4} = a _ {5} = a _ {6} = 0 \tag {12.2-9}
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$$
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The condition $a_{2} + a_{3}/R = 0$ implies $\epsilon_{m} = 0$ at s = 0, the element center. This constraint causes no difficulty. The remaining conditions, $a_{4} = a_{5} = a_{6} = 0$ , imply that $w_{,s} = w_{,ss} = w_{,sss} = 0$ . These are severe and spurious constraints. They produce what is called membrane locking; that is, they make the model much too stiff. Numerical evidence supports this conclusion: for example, for the arch of Fig. 12.2-1a with R/t = 40, four identical elements predict a center displacement that is less than 5% of its correct value [12.2]. Equation 12.2-8 shows that the contribution of w terms to $\epsilon_{m}$ decreases as R becomes large. That is, the tendency to lock decreases as elements become more shallow, and disappears if elements become straight.
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One can also conclude that the mesh will lock by doing “constraint counting,” which is discussed in Section 9.5. Energy $U_{m}$ in Eq. 12.2-3 acts as a penalty function that enforces the inextensibility constraint as a curved element becomes thin—that is, as the ratio A/I becomes large. If numerically integrated, $U_{m}$ (and the membrane stiffness matrix it produces) enforces one constraint for every integration point used to evaluate $U_{m}$ . Each constraint effectively removes one d.o.f. from its intended role of modeling deformations. Accordingly, if there are three or more integration points per element, all d.o.f. of the structure are used to satisfy support conditions and constraints. No d.o.f. are left to model the bending deformations, so a mesh of curved elements tends to lock.
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This insight suggests the simple remedy of selective integration. One can use a two-point rule to evaluate the stiffness matrix that comes from $U_{b}$ , but use a one-point rule to evaluate the stiffness matrix that comes from $U_{m}$ . The single Gauss point is at the element center, s = 0. Accordingly, from Eq. 12.2-8, the constraint enforced is
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$$
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a _ {2} + \frac {a _ {3}}{R} = 0 \tag {12.2-10}
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$$
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Now only one d.o.f. per element is used in satisfying the constraint, and the mesh does not lock. Such elements are quite accurate: they have about the same accuracy as the aforementioned straight elements [12.2].
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Exactly integrated curved elements can also behave well if element displacement fields are properly designed. For example, consider the fields [12.3]
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$$
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u = a _ {1} + a _ {2} \phi + a _ {3} \phi^ {2} + a _ {4} \phi^ {3} + a _ {5} \phi^ {4} + a _ {6} \phi^ {5} \tag {12.2-11a}
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$$
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$$
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w = - a _ {2} - 2 a _ {3} \phi - 3 a _ {4} \phi^ {2} - 4 a _ {5} \phi^ {3} - 5 a _ {6} \phi^ {4} \tag {12.2-11b}
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$$
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where $\phi = s/R$ . These displacements satisfy the inextensibility condition $\epsilon_{m} = 0$ for all s. Since $\epsilon_{m} = 0$ , the resulting stiffness matrix comes entirely from the $U_{b}$ term in Eq. 12.2-3. Computed results are almost exact. We see that if displacement fields are to have enough d.o.f. for a six-d.o.f. element and are also to display $\epsilon_{m} = 0$ , the membrane field must be at least quintic and the lateral displacement field must be one degree lower.
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To allow the preceding element to display membrane strain, one can add a
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constant $a_7$ to Eq. 12.2-11b. Thus [k] becomes 7 by 7 and includes a contribution from $U_m$ in Eq. 12.2-3. D.o.f. $a_7$ is internal to each element. Again, results are excellent [12.3].
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We note that the rigid-body motion field of Eqs. 12.2-4 is not explicitly included in any of the preceding curved elements. One can say that such motion is implicitly approximated in Eqs. 12.2-11 because $\sin \phi$ and $\cos \phi$ can be expanded as power series, whose initial terms are contained in Eqs. 12.2-11. Another element that implicitly approximates rigid-body motion is based on a displacement field of the form [12.4]
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$$
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u = a _ {1} + a _ {2} s + a _ {7} \left(1 - \frac {4 s ^ {2}}{L ^ {2}}\right) \tag {12.2-12a}
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$$
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$$
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w = a _ {3} + a _ {4} s + a _ {5} s ^ {2} + a _ {6} s ^ {3} \tag {12.2-12b}
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$$
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The mode associated with d.o.f. $a_{7}$ is internal or “nodeless”; it vanishes at the element ends $s = \pm L/2$ . D.o.f. $a_{7}$ can be removed by condensation prior to assembly of elements. As compared with Eqs. 12.2-7, Eqs. 12.2-12 reduce spurious strain energy associated with rigid-body motion by factors of 122,000 and 18,000 for curved elements that subtend arcs of $12^{\circ}$ and $20^{\circ}$ , respectively [12.4].
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A curved element that explicitly includes rigid-body motion capability has been suggested. Known as the Cantin–Clough element, it is based on fields of the form [12.2]
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$$
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u = a _ {1} \cos \phi + a _ {2} \sin \phi + a _ {3} + a _ {4} s \tag {12.2-13a}
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$$
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$$
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w = a _ {1} \sin \phi - a _ {2} \cos \phi + a _ {5} s ^ {2} + a _ {6} s ^ {3} \tag {12.2-13b}
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$$
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where $\phi = s / R$ . Rigid-body motion, Eqs. 12.2-4, is displayed by the element when $a_4 = a_5 = a_6 = 0$ . From Eqs. 12.2-2 and 12.2-13,
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$$
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\epsilon_ {m} = a _ {4} + \frac {a _ {5}}{R} s ^ {2} + \frac {a _ {6}}{R} s ^ {3} \quad \text { and } \quad \kappa = \frac {a _ {4}}{R} - 2 a _ {5} - 6 a _ {6} s \tag {12.2-14}
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$$
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Inextensibility requires that $a_4 = a_5 = a_6 = 0$ , which in turn yields $\kappa = 0$ . Accordingly, we expect to encounter locking difficulties. Indeed, for the thin-arch problem of Fig. 12.2-1a, the central deflection is almost $50\%$ low when 20 Can-tin-Clough elements are used for the entire arch [12.2].
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From the foregoing examples we conclude that membrane locking is much more detrimental to thin curved elements than is a lack of explicit rigid-body motion capability.
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Mindlin Arch Elements. Displacements of a Mindlin arch element are described by tangential and normal displacements of the midline and by rotation of a normal to the midline—that is, by $u, w,$ and $\beta$ . Thus rotation $\beta$ may differ from rotation $w_{,s}$ . As with Mindlin beam and plate elements discussed in Section 9.4 and Chapter 11, Mindlin arch elements can account for transverse shear deformation, which here is $\gamma_{zs}$ . Only if $\gamma_{zs} = 0$ does $\beta = w_{,s}$ .
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Strain energy in a Mindlin arch element is $U = U_{m} + U_{b} + U_{s}$ , in which the respective contributions to U are due to membrane strain $\epsilon_{m}$ , curvature change
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$\kappa$ , and transverse shear strain $\gamma_{zs}$ [12.5]. For an element of length $L$ and constant radius $R$ ,
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$$
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U _ {m} = \int_ {- L / 2} ^ {L / 2} \frac {E A}{2} \epsilon_ {m} ^ {2} d s, \quad \text {where} \quad \epsilon_ {m} = u _ {, s} + \frac {w}{R} \tag {12.2-15a}
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$$
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$$
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U _ {b} = \int_ {- L / 2} ^ {L / 2} \frac {E I}{2} \kappa^ {2} d s, \quad \text { where } \quad \kappa = \frac {u _ {, s}}{R} - \beta_ {, s} \tag {12.2-15b}
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$$
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$$
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U _ {s} = \int_ {- L / 2} ^ {L / 2} \frac {G A}{2} \gamma_ {z s} ^ {2} d s, \quad \text { where } \quad \gamma_ {z s} = w _ {, s} - \beta \tag {12.2-15c}
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$$
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To account for a parabolic variation of $\gamma_{zs}$ through the thickness, we may replace G with 5G/6. A two-node element can be based on linear interpolations. With $a_{i} = \text{constants and } N_{1} = 0.5 - s/L$ , $N_{2} = 0.5 + s/L$ , linear fields are
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$$
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u = a _ {1} + a _ {2} s \quad \text { or } \quad u = N _ {1} u _ {1} + N _ {2} u _ {2} \tag {12.2-16a}
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$$
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$$
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w = a _ {3} + a _ {4} s \quad \text { or } \quad w = N _ {1} w _ {1} + N _ {2} w _ {2} \tag {12.2-16b}
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$$
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$$
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\beta = a _ {5} + a _ {6} s \quad \text { or } \quad \beta = N _ {1} \beta_ {1} + N _ {2} \beta_ {2} \tag {12.2-16c}
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$$
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Arguments concerning locking in this element are very similar to arguments made in connection with Eqs. 12.2-8 through 12.2-10 and are summarized as follows. As thickness t approaches zero, all strain energy should be in bending; that is, $\epsilon_{m}$ and $\gamma_{zs}$ should each vanish for all s, which implies, for a curved element,
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$$
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a _ {2} + \frac {a _ {3}}{R} = a _ {4} = a _ {5} = a _ {6} = 0 \tag {12.2-17}
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$$
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or which implies, for a straight element $(R = \infty)$ ,
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$$
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a _ {2} = a _ {4} - a _ {5} = a _ {6} = 0 \tag {12.2-18}
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$$
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In either case, an element added to a thin arch brings three d.o.f. with it, but all d.o.f. are occupied in satisfying constraints, and the mesh locks as t approaches zero. However, reduced integration can produce a workable element. If $U_{m}$ and $U_{s}$ are integrated with a one-point rule, only two constraints are imposed per element, one each on $\epsilon_{m}$ and $\gamma_{zs}$ , and the mesh does not lock.
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A curved element based on Eqs. 12.2-16 does not explicitly contain the rigid-body motion field, Eqs. 12.2-4. A straight element $(R = \infty)$ can display rigid-body motion; that is, it displays $\epsilon_{m} = \kappa = \gamma_{zs} = 0$ if $a_{2} = a_{6} = 0$ and $a_{4} = a_{5}$ .
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A three-node Mindlin element, having nodes at $s = -L / 2$ , $s = 0$ , and $s = +L / 2$ , would be called a quadratic element. With $\xi = s / (L / 2)$ , its displacement field has the form
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$$
|
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u = a _ {1} + a _ {2} \xi + a _ {3} \xi^ {2} \tag {12.2-19a}
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$$
|
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$$
|
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w = a _ {4} + a _ {5} \xi + a _ {6} \xi^ {2} \tag {12.2-19b}
|
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$$
|
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$$
|
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\beta = a _ {7} + a _ {8} \xi + a _ {9} \xi^ {2} \tag {12.2-19c}
|
||
$$
|
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As the arch becomes extremely thin, the inextensibility condition $\epsilon_{m} = 0$ for all $s$ implies
|
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|
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$$
|
||
\frac {2 a _ {2}}{L} + \frac {a _ {4}}{R} = \frac {4 a _ {3}}{L} + \frac {a _ {5}}{R} = a _ {6} = 0 \tag {12.2-20}
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||
$$
|
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|
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The constraint $a_{6}=0$ implies membrane locking, as it prevents a thin arch from displaying a constant value of $w_{,ss}$ . Reduced integration offers a remedy [12.5,12.6]. Membrane strain can be written in the form
|
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|
||
$$
|
||
\epsilon_ {m} = \left(\frac {2 a _ {2}}{L} + \frac {a _ {4}}{R} + \frac {a _ {6}}{3 R}\right) + \left(\frac {4 a _ {3}}{L} + \frac {a _ {5}}{R}\right) \xi + \frac {a _ {6}}{R} \left[ \xi^ {2} - \frac {1}{3} \right] \tag {12.2-21}
|
||
$$
|
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If integration of membrane energy $U_{m}$ is performed by two-point Gauss quadrature, for which Gauss points are at $\xi = \pm 1/\sqrt{3}$ , then the bracketed expression vanishes. Thus no constraint is placed on $a_{6}$ ; rather, $\epsilon_{m} = 0$ implies only the vanishing of the two parenthetic expressions in Eq. 12.2-21.
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A similar argument can be applied to the condition $\gamma_{zs} = 0$ , which should prevail in an extremely thin arch. The conclusion is that $\gamma_{zs} = 0$ enforces the constraint $\beta_{ss} = 0$ in a quadratic element. This is not a locking condition, but it degrades element performance. Again the remedy is to use reduced integration: a two-point Gauss rule should be used to evaluate shear energy $U_{s}$ . (This advice was previously given in connection with a quadratic plate element, Fig. 11.3-3b.)
|
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The quadratic arch element does not explicitly contain the rigid-body motion capability described by Eqs. 12.2-4. However, Eqs. 12.2-4 pertain to a circular arch. Imagine now that the element shape is defined by the coordinates of its three nodes. Thus the arch element has parabolic shape. Then, since displacement fields are also second degree, the parabolic element is of the isoparametric family, and arguments given in Section 6.10 demonstrate that the capability for rigid-body motion is present. Similarly, a quadratic shell element, adapted from a quadratic isoparametric solid element, is able to display rigid-body motion without strain. (Some cautions about thickness-direction integration should be noted; see Ref. 12.7.)
|
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Remarks. During stress computation in numerically integrated elements, one should evaluate strains at the Gauss points of the reduced quadrature rule appropriate to the element: for example, for Mindlin elements, at the center in linear elements and at $\xi = \pm 1/\sqrt{3}$ in quadratic elements. Large spurious strains may appear at other locations, as shown in Fig. 6.13-2. Figure 6.13-2 depicts transverse shear strain, but membrane strain can display similar behavior. Accurate stress computation in extremely thin elements may require a restriction on thickness t, as noted in the following paragraph.
|
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|
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Use of reduced integration avoids the imposition of spurious constraints, but not all constraints. The constraints that remain may cause numerical difficulty if the element is extremely thin. Difficulty is avoided by simply not allowing thickness t to fall below a certain limit in the computation of stiffness matrix coefficients associated with strain energies $U_{m}$ and $U_{s}$ . This matter is discussed at the end of Section 9.4.
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