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Problem 10.13
# Section 10.5
10.14 Assume that a certain axially symmetric pressure vessel can be adequately modeled by four-node elements (as in Fig. 10.1-1b), arranged in a 20 by 1 mesh, so that one element spans the wall thickness of the vessel and 20 elements span the axial dimension. Also assume that the loading is described by Eq. 10.5-2, with harmonics $n = 1, 2, 3, 4, 5$ , and 6. Alternatively, one could contemplate a fully three-dimensional analysis with a 20 by 1 by $m$ mesh of eight-node solid elements, where $m$ is the number of elements around the circumference.
(a) Estimate $m$ so that the three-dimensional model would be of adequate accuracy. Assume that three elements per half-wave of displacement are acceptable.
(b) Make an estimate of the cost ratio of the three-dimensional solution to the series solution. Base your estimate on the expense of generating stiffness matrices with an order 2 Gauss rule.
(c) Repeat part (b), but base your estimate on the expense of solving equations with a banded equation solver (see Appendix B).
10.15 If $\{\epsilon\} = \{\mathbf{0}\}$ , Eqs. 10.5-4 have the solution
$$
\left\{ \begin{array}{l} u \\ v \\ w \end{array} \right\} = \left[ \begin{array}{c c c c c c} 0 & \cos \theta & z \cos \theta & 0 & \sin \theta & z \sin \theta \\ 0 & - \sin \theta & - z \sin \theta & r & \cos \theta & z \cos \theta \\ 1 & 0 & - r \cos \theta & 0 & 0 & - r \sin \theta \end{array} \right] \left\{ \begin{array}{l} a _ {1} \\ a _ {2} \\ \cdot \\ \cdot \\ \cdot \\ a _ {6} \end{array} \right\}
$$
where the $a_{i}$ are constants. A displacement field must contain these terms if there is to be rigid-body motion without strain.
(a) Show that this field does in fact yield $\{\epsilon\} = \{0\}$ .
(b) Compare this field with Eqs. 10.5-9: identify columns of the rectangular matrix as to the value of $n$ and as to belonging to the single-barred or the double-barred series.
(c) For each of the six columns of the rectangular matrix, describe the physical meaning of the displacement mode it represents.
10.16 Specialize Eqs. 10.5-9 to represent the following rigid-body motions.
(a) Axial translation.
(b) Translation perpendicular to the z axis in the plane $\theta = 0$ .
(c) Translation perpendicular to the $z$ axis in the plane $\theta = \pi /2$ .
(d) Rotation about the z axis.
(e) Rotation about the line $\theta = z = 0$ .
(f) Rotation about the line $\theta = \pi / 2, z = 0$ .
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10.17 Seven forces are applied to one end of a cylindrical bar of outer radius c, as shown. Forces $P_{4}$ form the couple $2P_{4}c$ . Stresses at midheight z = h/2 could be calculated by elementary formulas $\sigma = Mc/I$ , and so on. However, imagine that, as an exercise, finite elements are to be used instead, so loads must be expressed in the form of Eqs. 10.5-8. For each of the six different loadings, use Eqs. 10.5-8 to write expressions for surface tractions on end z = h that are statically equivalent to the original loading. (This can be done using only terms for which n = 0 and/or n = 1). Express your answers in terms of c, the $P_{i}$ , $\sin \theta$ , and $\cos \theta$ .
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Problem 10.17
10.18 A flat plate containing a circular hole of radius R is loaded by axial force P and bending moment M, as shown. The region enclosed by the dashed line of radius c is to be isolated and analyzed as a solid of revolution. If t = plate thickness and $c >> R$ , what load terms from Eqs. 10.5-8 should be used in analysis? Express your answers in terms of P, M, h, t, n, and $\theta$ .
(a) Consider $P$ only (let $M = 0$ ).
(b) Consider $M$ only (let $P = 0$ ).
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Problem 10.18
# Section 10.6
10.19 Assume that [E] has the form shown in Eq. 10.5-1. Use the first partition of $[B]_{n}$ , as stated in Eq. 10.6-5, to demonstrate the following about the matrix product $[P] = [B]^{T}[E][B]$ . (Note: It is not necessary to write out details of terms extraneous to the question posed.)
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(a) Show that [P] contains $\sin^2 n\theta$ or $\cos^2 n\theta$ in each term of an on-diagonal submatrix.
(b) Show that [P] contains either $\sin m\theta$ $\sin n\theta$ or $\cos m\theta$ $\cos n\theta$ , where $m \neq n$ , in each term of an off-diagonal submatrix.
(c) Show that each term of an on-diagonal submatrix has the form $A + Bn^2$ or the form $Cn$ .
10.20 (a) Write the form of $[\mathbf{B}]_n$ in Eq. 10.6-5 appropriate to antisymmetric terms (the double-barred series in Eq. 10.5-9).
(b) Show that $[k]_{0}$ , $[k]_{1}$ , and so on, are identical to the corresponding matrices obtained for symmetric terms. Suggestion: See the note in Problem 10.19.
(c) Show that the conclusion reached in part (b) would not be true if the negative sign in Eq. 10.5-9b were changed to positive.
10.21 Consider the flat element analyzed in Problem 10.6. However, now allow circumferential displacement v as well as radial displacement u, so that loads without axial symmetry can be treated. Element nodal d.o.f. are now $u_{1}$ , $v_{1}$ , $u_{2}$ , and $v_{2}$ . Let $\theta = 0$ be a plane of symmetry. Formulate $[B]_{n}$ for this element (analogous to $[B]_{n}$ in Eq. 10.6-5, but including all partitions).
10.22 The element described in Problem 10.21 can be used to solve problems of disks and rings under concentrated loads (see sketch). Why do Fourier harmonics for displacement and stress form convergent series, although the series for concentrated loads are not convergent?
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Problem 10.22
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# 11 CHAPTER
# BENDING OF FLAT PLATES
Concepts and equations related to the bending of flat plates are reviewed. Elements for thin plates and plates having transverse shear deformation are discussed. Test problems for plate elements are presented.
# 11.1 PLATE-BENDING THEORY
Loads, Stresses, and Moments. A flat plate, like a straight beam, supports transverse loads by bending action. Figure 11.1-1a shows stresses that act on cross sections of a plate whose material is homogeneous and linearly elastic. Normal stresses $\sigma_{x}$ and $\sigma_{y}$ vary linearly with z and are associated with bending moments $M_{x}$ and $M_{y}$ . Shear stress $\tau_{xy}$ also varies linearly with z and is associated with twisting moment $M_{xy}$ . Normal stress $\sigma_{z}$ is considered negligible in comparison with $\sigma_{x}$ , $\sigma_{y}$ , and $\tau_{xy}$ . Transverse shear stresses $\tau_{yz}$ and $\tau_{zx}$ vary quadratically with z. Lateral load q includes surface load and body force, both in the z direction. Unless stated otherwise, “plate bending” means that external loads have no components parallel to the xy plane and that $\sigma_{x} = \sigma_{y} = \tau_{xy} = 0$ on the midsurface z = 0. Excepting stress $\tau_{xy}$ , the foregoing stress patterns are a direct extension of beam theory from one dimension to two.
Stresses in Fig. 11.1-1 produce the following bending moments M and transverse shear forces Q:
$$
M _ {x} = \int_ {- t / 2} ^ {t / 2} \sigma_ {x} z d z \quad M _ {y} = \int_ {- t / 2} ^ {t / 2} \sigma_ {y} z d z \quad M _ {x y} = \int_ {- t / 2} ^ {t / 2} \tau_ {x y} z d z \tag {11.1-1a}
$$
$$
Q _ {x} = \int_ {- t / 2} ^ {t / 2} \tau_ {z x} d z \quad Q _ {y} = \int_ {- t / 2} ^ {t / 2} \tau_ {y z} d z \tag {11.1-1b}
$$
The M's are moments per unit length and the Q's are forces per unit length. Differential total moments and forces are $M_{x}$ dy, $Q_{x}$ dy, and so on, as shown in Fig. 11.1-1b. Stresses $\sigma_{x}$ , $\sigma_{y}$ , and $\tau_{xy}$ are largest at the surfaces $z = \pm t/2$ , where they have the respective magnitudes $6M_{x}/t^{2}$ , $6M_{y}/t^{2}$ , and $6M_{xy}/t^{2}$ . At arbitrary values of z,
$$
\sigma_ {x} = \frac {M _ {x} z}{t ^ {3} / 1 2} \quad \sigma_ {y} = \frac {M _ {y} z}{t ^ {3} / 1 2} \quad \tau_ {x y} = \frac {M _ {x y} z}{t ^ {3} / 1 2} \tag {11.1-2}
$$
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Figure 11.1-1. (a) Stresses that act on a differential element of a homogeneous, linearly elastic plate. The distributed lateral load is q (force per unit area). (b) The same differential element, viewed normal to the plate. Forces $\odot$ and $\otimes$ act in the positive and negative z directions, respectively.
as may be verified by substituting Eqs. 11.1-2 into Eqs. 11.1-1a. Transverse shear stresses are usually small in comparison with $\sigma_{x}$ , $\sigma_{y}$ , and $\tau_{xy}$ . They have greatest magnitude at z = 0, where $\tau_{yz} = 1.5Q_{y}/t$ and $\tau_{zx} = 1.5Q_{x}/t$ .
Deformations (Kirchhoff Theory). Points on the midsurface z = 0 move in only the z direction as the plate deforms in bending. A line that is straight and normal to the midsurface before loading is assumed to remain straight and normal to the midsurface after loading (see line OP in Fig. 11.1-2). Thus transverse shear deformation is assumed to be zero. A point not on the midsurface has displacement components u and v in the x and y directions, respectively. From Fig. 11.1-2, with $w_{,x}$ and $w_{,y}$ small angles of rotation,
$$
\begin{array}{l} \epsilon_ {x} = u _ {, x} \quad = - z w _ {, x x} \\ u = - z w, x \quad \text { hence } \quad \epsilon_ {y} = v, y \quad = - z w, y y \tag {11.1-3} \\ \gamma_ {x y} = u _ {, y} + v _ {, x} = - 2 z w _ {, x y} \\ \end{array}
$$
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Figure 11.1-2. (a) Differential element of a thin plate before loading. (b) After loading: deformations associated with Kirchhoff plate theory. Point P displaces w units up and $zw_{,x}$ units leftward because of midsurface displacement w and small rotation $w_{,x}$ .
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These are the strain-displacement relations of Kirchhoff plate theory, which is applicable to a thin plate.
Deformations (Mindlin Theory). A line that is straight and normal to the midsurface before loading is assumed to remain straight but not necessarily normal to the midsurface after loading. Thus, transverse shear deformation is allowed. The motion of a point not on the midsurface is not governed by slopes $w_{,x}$ and $w_{,y}$ as in Kirchhoff theory. Rather, its motion depends on rotations $\theta_{x}$ and $\theta_{y}$ of lines that were normal to the midsurface of the undeformed plate (Fig. 11.1-3). Thus, with $\theta_{x}$ and $\theta_{y}$ small angles of rotation,
$$
\begin{array}{l} u = - z \theta_ {x} \quad \epsilon_ {x} = - z \theta_ {x, x} \quad \begin{array}{l} \gamma_ {x y} = - z \left(\theta_ {x, y} + \theta_ {y, x}\right) \\ \gamma_ {y z} = w _ {, y} - \theta_ {y} \end{array} \tag {11.1-4} \\ v = - z \theta_ {y} \quad \epsilon_ {y} = - z \theta_ {y, y} \quad \gamma_ {z x} = w _ {, x} - \theta_ {x} \\ \end{array}
$$
The foregoing expressions for strain are obtained by straightforward application of Eqs. 1.5-4 and 1.5-5. Equations 11.1-4 are the strain-displacement relations of Mindlin plate theory. This theory accounts for transverse shear deformation and is therefore especially suited to the analysis of thick plates and sandwich plates.
MomentCurvature Relations (Kirchhoff Theory). We begin with stressstrain relations. Let x and y be principal directions of an orthotropic material. Stress $\sigma_{z}$ is considered negligible in comparison with $\sigma_{x}$ , $\sigma_{y}$ , and $\tau_{xy}$ . Transverse shear strains are also considered ineligible, so stressstrain relations that involve them need not be written. What remains is the plane stressstrain relation $\{\sigma\} = [E](\{\epsilon\} - \{\epsilon_{0}\})$ ; that is [11.1],
$$
\left\{ \begin{array}{l} \sigma_ {x} \\ \sigma_ {y} \\ \tau_ {x y} \end{array} \right\} = \left[ \begin{array}{c c c} E _ {x} ^ {\prime} & E ^ {\prime \prime} & 0 \\ E ^ {\prime \prime} & E _ {y} ^ {\prime} & 0 \\ 0 & 0 & G \end{array} \right] \left(\left\{ \begin{array}{l} \epsilon_ {x} \\ \epsilon_ {y} \\ \gamma_ {x y} \end{array} \right\} - \left\{ \begin{array}{l} \alpha_ {x} T \\ \alpha_ {y} T \\ 0 \end{array} \right\}\right) \tag {11.1-5}
$$
where initial strains $\{\epsilon_{0}\}$ are presumed caused by thermal expansion with principal expansion coefficients $\alpha_{x}$ and $\alpha_{y}$ . For an isotropic material, with E = elastic modulus and $\nu =$ Poisson's ratio,
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Figure 11.1-3. Differential plate element after loading, analogous to Fig. 11.1-2b, but with transverse shear deformation allowed $w_{,x} \neq \theta_{x}$ , so that $\gamma_{zx} = w_{,x} - \theta_{x} \neq 0$ .
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$$
E _ {x} ^ {\prime} = E _ {y} ^ {\prime} = \frac {E ^ {\prime \prime}}{\nu} = \frac {E}{1 - \nu^ {2}} \quad \text { and } \quad G = \frac {E}{2 (1 + \nu)} \tag {11.1-6}
$$
The moment-curvature relation is obtained by substitution of Eqs. 11.1-3 into Eq. 11.1-5 and the result into Eqs. 11.1-1a. This process yields
$$
\{\mathbf {M} \} = - [ \mathbf {D} _ {K} ] (\{\kappa \} - \{\kappa_ {0} \}) \tag {11.1-7}
$$
where moments and curvatures are
$$
\{\mathbf {M} \} = \left[ \begin{array}{l l l} M _ {x} & M _ {y} & M _ {x y} \end{array} \right] ^ {T} \quad \text { and } \quad \{\boldsymbol {\kappa} \} = \left[ \begin{array}{l l l} w _ {, x x} & w _ {, y y} & 2 w _ {, x y} \end{array} \right] ^ {T} \tag {11.1-8}
$$
In $[\mathbf{D}_K]$ we have $D_{K13} = D_{K31} = D_{K23} = D_{K32} = 0$ and the nonzero terms
$$
D _ {K 1 1} = \frac {E _ {x} ^ {\prime} t ^ {3}}{1 2} \quad D _ {K 1 2} = D _ {K 2 1} = \frac {E ^ {\prime \prime} t ^ {3}}{1 2} \quad D _ {K 2 2} = \frac {E _ {y} ^ {\prime} t ^ {3}}{1 2} \quad D _ {K 3 3} = \frac {G t ^ {3}}{1 2} \tag {11.1-9}
$$
If the material is isotropic, then
$$
\left[ \mathbf {D} _ {K} \right] = \left[ \begin{array}{c c c} D & \nu D & 0 \\ \nu D & D & 0 \\ 0 & 0 & (1 - \nu) D / 2 \end{array} \right], \quad \text { where } \quad D = \frac {E t ^ {3}}{1 2 (1 - \nu^ {2})} \tag {11.1-10}
$$
D is called “flexural rigidity” and is analogous to bending stiffness EI of a beam. Indeed, if the plate has unit width and $\nu = 0$ , then $D = EI = Et^{3}/12$ .
As a particular example of initial curvatures $\{\kappa_{0}\}$ , consider a temperature gradient $T = -2zT_{0}/t$ . This is a linear temperature variation from $T_{0}$ at z = -t/2 to $-T_{0}$ at z = t/2. Thus the process that yields Eq. 11.1-7 gives the initial curvatures
$$
\{\kappa_ {0} \} = \left[ \begin{array}{l l l} 2 \alpha_ {x} T _ {0} / t & 2 \alpha_ {y} T _ {0} / t & 0 \end{array} \right] ^ {T} \tag {11.1-11}
$$
Equation 11.1-7 shows that actions in the x and y directions are coupled, even for an isotropic plate. In Fig. 11.1-4a, $w_{,yy}$ is constant and $w_{,xx} = w_{,xy} = 0$ in the central portion, but $M_{x}$ is nonzero because of the Poisson effect. But $M_{x} = 0$ at the free edges $x = \pm a$ , so these edges curl a bit (Fig. 11.1-4b). Only if $a \approx t$ , so that $M_{x} \approx 0$ throughout, does the plate act like a beam, displaying the familiar anticlastic surface. The pure twist of Fig. 11.1-4c is associated with moments $-M_{xy}$ alone ( $M_{x} = M_{y} = 0$ ) if the plate is isotropic.
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Figure 11.1-4. (a) Bending to a cylindrical surface by moments $M_{y}$ on the edges y = constant. (b) Cross section cut by the xz plane. (c) The w = xy state of pure twist: $w_{xx} = w_{yy} = 0$ , $w_{xy} > 0$ .
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Moment-Curvature Relations (Mindlin Theory). Again let $x$ and $y$ be principal material directions. The moment-curvature relations of Mindlin plate theory are obtained by essentially the same procedure as used to obtain Eq. 11.1-7. However, we must use Eqs. 11.1-4 instead of Eqs. 11.1-3 and include the shear stress-strain relations $\tau_{yz} = G_{yz}\gamma_{yz}$ and $\tau_{zx} = G_{zx}\gamma_{zx}$ . The resulting moment-curvature relation is abbreviated as $\{\mathbf{M}\} = -[\mathbf{D}_M](\{\boldsymbol{\kappa}\} - \{\boldsymbol{\kappa}_0\})$ . Written out, this relation is
$$
\left\{ \begin{array}{l} M _ {x} \\ M _ {y} \\ M _ {x y} \\ Q _ {y} \\ Q _ {x} \end{array} \right\} = - \underbrace {\left[ \begin{array}{c c c c c} & & 0 & 0 \\ \left[ \mathbf {D} _ {K} \right] & & 0 & 0 \\ 3 \times 3 & & 0 & 0 \\ 0 & 0 & 0 & G _ {y z} t & 0 \\ 0 & 0 & 0 & 0 & G _ {z x} t \end{array} \right]} _ {[ \mathbf {D} _ {M} ]} \left(\underbrace {\left\{ \begin{array}{c} \theta_ {x , x} \\ \theta_ {y , y} \\ \theta_ {x , y} + \theta_ {y , x} \\ \theta_ {y} - w _ {, y} \\ \theta_ {x} - w _ {, x} \end{array} \right\}} _ {\{\boldsymbol {\kappa} \}} - \{\boldsymbol {\kappa} _ {0} \}\right) \tag {11.1-12}
$$
where $[\mathbf{D}_K]$ is the same as in Eq. 11.1-7. The shear stiffness terms $G_{yz}t$ and $G_{zx}t$ in Eq. 11.1-12 may be replaced by $G_{yz}t / 1.2$ and $G_{zx}t / 1.2$ to permit the parabolic distributions of $\tau_{yz}$ and $\tau_{zx}$ (shown in Fig. 11.1-1a) to be replaced by uniform distributions, as explained in Section 9.4. If represented as rotation vectors by the right-hand rule, $\theta_x$ and $\theta_y$ point in the $-y$ and $+x$ directions, respectively. Initial curvatures $\{\kappa_0\}$ are those of Kirchhoff theory, augmented by zeros in positions 4 and 5.
Initial curves are not possible. If the plate is isotropic, then $G_{yz} = G_{zx} = G$ and Eqs. 11.1-10 apply to submatrix $[\mathbf{D}_K]$ in Eq. 11.1-12. For an isotropic sandwich plate, Fig. 11.1-5, with thin facings, $G$ the shear modulus of the core, and $E$ and $\nu$ the elastic modulus and Poisson ratio of each facing,
$$
D _ {M 1 1} = D _ {M 2 2} = \frac {D _ {M 1 2}}{\nu} = \frac {D _ {M 2 1}}{\nu} = \frac {E h (c + h) ^ {2}}{2 \left(1 - \nu^ {2}\right)} \tag {11.1-13}
$$
$$
D _ {M 3 3} = \frac {E h (c + h) ^ {2}}{4 (1 + \nu)} \quad D _ {M 4 4} = D _ {M 5 5} = \frac {G (c + h) ^ {2}}{c}
$$
and all other entries in the 5 by 5 matrix $[\mathbf{D}_M]$ of Eq. 11.1-12 are zero [11.2]. If principal material directions are $x'$ and $y'$ rather than $x$ and $y$ , as in Fig. 7.3-1, then coordinate transformation is required. Arrays in Eq. 11.1-7 transform as $\{\mathbf{M}\} = [\mathbf{T}_\epsilon]^T\{\mathbf{M}'\}$ , $\{\boldsymbol{\kappa}'\} = [\mathbf{T}_\epsilon]\{\boldsymbol{\kappa}\}$ , and $[\mathbf{D}_K] = [\mathbf{T}_\epsilon]^T [\mathbf{D}_K'] [\mathbf{T}_\epsilon]$ , where $[\mathbf{T}_\epsilon]$ is given by Eq. 7.3-11. Coefficients in the southeast corner of $[\mathbf{D}_M]$ in Eq. 11.1-12 are $D_{K44} = m_1^2 G_{z'x'}t + m_2^2 G_{y'z'}t$ , $D_{K55} = \ell_1^2 G_{z'x'}t + \ell_2^2 G_{y'z'}t$ , and $D_{K45} = D_{K54} = \ell_1m_1G_{z'x'}t + \ell_2m_2G_{y'z'}t$ , where the $\ell$ 's and $m$ 's are given in Fig. 7.3-1. Remarks. A plate can be loaded by distributed lateral load of intensity $q$ and by initial curvatures $\{\kappa_0\}$ , as just discussed. Concentrated forces and line loads may
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Figure 11.1-5. Cross section of a sandwich plate. Typically the core resists little but transverse shear strains, so almost all bending stiffness is provided by membrane action in thin facings.
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also be present. Edge moments M and transverse shears Q may be applied as known loads or as support reactions. Except for line loads, these loads are analogous to loads present in beam theory. Nodal equivalents of loads q and $\{\kappa_{0}\}$ can be computed by means of Eq. 4.1-6.
In finite element analysis of plates, whether by Kirchhoff or Mindlin theory, d.o.f. at a node i are typically one lateral displacement $(w_{i})$ and two rotations $(w_{,xi}$ and $w_{,yi}$ or $\theta_{xi}$ and $\theta_{yi})$ . At a free edge none of the three d.o.f. is restrained. At a clamped edge all d.o.f. are restrained. Further discussion of boundary conditions appears in Section 11.5.
Full compatibility of interelement displacements requires that, in any $z =$ constant layer, displacements $u, v,$ and $w$ be the same in adjacent elements where the elements meet. Accordingly, from Eqs. 11.1-3, compatible Kirchhoff elements are $C^1$ elements, as they must display interelement continuity of $w, w_{xx}$ , and $w_{yy}$ . Note that along (say) a $y$ -parallel interelement boundary, continuity of $w$ ensures continuity of $w_{yy}$ but not continuity of the boundary-normal slope $w_{xx}$ . From Eqs. 11.1-4, compatible Mindlin elements are $C^0$ elements, as the fields, $w, \theta_x,$ and $\theta_y$ (but not their derivatives) must be interelement-continuous. Note that along (say) a $y$ -parallel interelement boundary, continuity of $\theta_x$ does not imply continuity of $w_{xx}$ unless the plate is so thin that $\gamma_{zx} = 0$ .
We have tacitly assumed that material properties are either independent of z or symmetric with respect to the midsurface z = 0. If not, bending may produce forces in the xy plane so that the midsurface is not a surface where $\sigma_{x} = \sigma_{y} = \tau_{xy} = 0$ . This effect is pronounced in two-layer laminated plates [11.3].
Appreciable in-plane forces may also arise if deflections w are more than a few tenths of the plate thickness. This happens even when supports apply no in-plane forces, because the deflected shape of the plate requires stretching or shortening in the midsurface (unless deflections are small or the deflected shape is cylindrical or conical). In-plane forces act to support part of the load. Thus the stiffness of a plate effectively increases as deflection increases, which makes the problem nonlinear. In some problems linear theory may overestimate displacements by 50% if deflection w equals thickness t [11.1].
# 11.2 FINITE ELEMENTS FOR PLATES
A great many finite elements for plates have been proposed: an incomplete survey lists 154 references and 88 different elements $[11.4]$ . In what follows we briefly consider some options in element formulation. Further details may be found in Ref. 11.5 and in papers cited by Ref. 11.4.
Kirchhoff Elements. Kirchhoff theory is applicable to thin plates, in which transverse shear deformation is neglected. Strain energy in the plate is determined entirely by in-plane strains $\epsilon_{x}$ , $\epsilon_{y}$ and $\gamma_{xy}$ . In turn, strains are determined entirely by the lateral displacement field $w = w(x, y)$ , as shown by Eqs. 11.1-3.
The starting point for formulating an element stiffness matrix is the strain energy term of Eq. 4.1-1,
$$
U ^ {\prime} = \int_ {V} \frac {1}{2} \{\boldsymbol {\epsilon} \} ^ {T} [ \mathbf {E} ] \{\boldsymbol {\epsilon} \} d V, \quad \text { where } \quad \{\boldsymbol {\epsilon} \} ^ {T} = \left[ - z w _ {, x x} - z w _ {, y y} - 2 z w _ {, y y} \right] \tag {11.2-1}
$$
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z, w
1
w₃
4
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Figure 11.2-1. Twelve-d.o.f. rectangular Kirchhoff plate element, with typical d.o.f. shown at node 3.
and [E] is given by Eq. 11.1-5. With $dV = dz \, dA$ , where $dA = dx \, dy$ is an increment of midsurface area $A$ , integration through thickness $t$ yields
$$
U = \int_ {A} \frac {1}{2} \{\boldsymbol {\kappa} \} ^ {T} [ \mathbf {D} _ {K} ] \{\boldsymbol {\kappa} \} d A, \quad \text { where } \quad \{\boldsymbol {\kappa} \} ^ {T} = \left[ w _ {, x x} \quad w _ {, y y} \quad 2 w _ {, x y} \right] \tag {11.2-2}
$$
and $[D_{K}]$ is the momentcurvature relation of Eq. 11.1-7. An interpolation of w from element nodal d.o.f. $\{d\}$ is devised, then differentiated to yield curvatures $\{\kappa\}$ . For an element having N nodes,
$$
w = \left\lfloor \mathbf {N} \right\rfloor_ {1 \times 3 N} \{\mathbf {d} \} \quad \text { hence } \quad \{\kappa \} = \left[ \mathbf {B} \right] _ {3 \times 3 N} \{\mathbf {d} \} \tag {11.2-3}
$$
D.o.f. of a Kirchhoff element are $\{\mathbf{d}\} = \left[w_1 \quad w_{,x1} \quad w_{,y1} \ldots w_N \quad w_{,xN} \quad w_{,yN}\right]^T$ . Finally, by substitution of Eq. 11.2-3 into Eq. 11.2-2, the element stiffness matrix [k] appears:
$$
U = \frac {1}{2} \{\mathbf {d} \} ^ {T} [ \mathbf {k} ] \{\mathbf {d} \}, \quad \text { where } \quad \underset {3 N \times 3 N} {[ \mathbf {k} ]} = \int_ {A} [ \mathbf {B} ] ^ {T} [ \mathbf {D} _ {K} ] [ \mathbf {B} ] d A \tag {11.2-4}
$$
For example, consider the twelve-d.o.f. rectangular element of Fig. 11.2-1. Typical d.o.f. $w_{,x3}$ and $w_{,y3}$ are slopes (i.e., rotations) of the plate midsurface at node 3. Their vector representations, shown in Fig. 11.2-1, are determined according to the right-hand rule. Lateral displacement w of this element has the form [11.5]
$$
w = \left[ 1, x, y, x ^ {2}, x y, y ^ {2}, x ^ {3}, x ^ {2} y, x y ^ {2}, y ^ {3}, x ^ {3} y, x y ^ {3} \right] \{\mathrm{a} \} \tag {11.2-5}
$$
This element does not preserve interelement continuity of boundary-normal slopes. Vector $\{a\}$ contains twelve generalized coordinates, which must be exchanged for the twelve nodal d.o.f. $\{d\}$ by the usual process (e.g., Eqs. 3.13-3). Thus Eqs. 11.2-3 are established, and [k] follows from Eq. 11.2-4.
Early efforts to formulate triangular Kirchhoff elements in the same way met with unexpected difficulties. A nine-term field for w is appropriate to the element of Fig. 11.2-2a. Unfortunately, as seen in Eq. 11.2-5, a complete cubic contains 10 terms. Candidate nine-term fields include
$$
w = \left[ 1, x, y, x ^ {2}, y ^ {2}, x ^ {3}, x ^ {2} y, x y ^ {2}, y ^ {3} \right] \{\mathbf {a} \} \tag {11.2-6a}
$$
$$
w = \left\lfloor 1, x, y, x ^ {2}, x y, y ^ {2}, x ^ {3}, x ^ {2} y + x y ^ {2}, y ^ {3} \right\rfloor \{\mathrm{a} \} \tag {11.2-6b}
$$