김경종
483 lines
27 KiB
Markdown
483 lines
27 KiB
Markdown
<!-- source-page: 351 -->
|
||
|
||
TABLE 11.4-1. CENTER DEFLECTIONS OF SQUARE PLATES, COMPUTED BY THE DKT ELEMENT [11.10]. A TYPICAL MESH IS SHOWN IN FIG. 11.4-2. RESULTS ARE REPORTED AS THE RATIO OF COMPUTED DEFLECTION TO EXACT DEFLECTION ACCORDING TO THIN-PLATE THEORY USING $\nu = 0.3$ [11.1]. SIMPLY SUPPORTED CASES USE CLASSICAL BOUNDARY CONDITIONS (Eqs. 11.5-1).
|
||
|
||
<table><tr><td rowspan="2">Mesh Size</td><td colspan="2">Uniformly Loaded</td><td colspan="2">Concentrated Center Load</td></tr><tr><td>Simply Supported</td><td>Clamped</td><td>Simply Supported</td><td>Clamped</td></tr><tr><td> $N_{es} = 1$ </td><td>1.025</td><td>1.500</td><td>1.076</td><td>1.012</td></tr><tr><td> $N_{es} = 2$ </td><td>0.999</td><td>1.228</td><td>1.008</td><td>1.046</td></tr><tr><td> $N_{es} = 4$ </td><td>1.001</td><td>1.069</td><td>1.003</td><td>1.019</td></tr><tr><td> $N_{es} = 8$ </td><td>1.001</td><td>1.021</td><td>1.001</td><td>1.007</td></tr></table>
|
||
|
||
One can visualize the foregoing DKT plate element as a stack of plane linear-strain triangles, pinned together by a rigid thickness-direction rod at each vertex, and with additional constraints that impose Eqs. 11.4-4 and 11.4-5 at midsides.
|
||
|
||
If the DKT plate element is homogeneous and of constant thickness, [k] is integrated exactly by a three-point quadrature rule. Explicit formulas for [k] are also available [11.12,11.13]. After element d.o.f. {d} are known, element strains { $\epsilon$ } are computed by successive application of Eqs. 11.4-6 and 11.4-8. Hence, stresses are $\{\sigma\} = [\mathrm{E}](\{\epsilon\} - \{\epsilon_{0}\})$ .
|
||
|
||
The behavior of the DKT element is reported in Table 11.4-1 [11.10]. Distributed loads were lumped by assigning one-third the total element load to translational d.o.f. at each vetex. Additional results may be found in Refs. 11.11 and 11.12, which show that the element yields accurate bending moments, performs well even at large aspect ratios, and satisfactorily solves the “twisted ribbon” test case.
|
||
|
||
Fortran coding for the DKT element appears in Fig. 11.4-3. Required input consists of nodal x and y coordinates (X1,X2,X3,Y1,Y2,Y3) and rigidity matrix $[D_{K}]$ (the 3 by 3 array D). The stiffness matrix is delivered in array SE. Displacement w is positive in the +z direction. The arrangement of nodal d.o.f. used in this subroutine is
|
||
|
||
$$
|
||
\{\mathbf {d} \} = \left[ w _ {1} \quad w _ {, y 1} \quad - w _ {, x 1} \quad w _ {2} \quad w _ {, y 2} \quad - w _ {, x 2} \quad w _ {3} \quad w _ {, y 3} \quad - w _ {, x 3} \right] ^ {T} \tag {11.4-11}
|
||
$$
|
||
|
||
Thus $w_{,y}$ and $-w_{,x}$ are represented by rotation vectors in the $+x$ and $+y$ directions, respectively.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
Mesh Nes = 2
|
||
</details>
|
||
|
||
Figure 11.4-2. Typical mesh on one quadrant of a square plate (used for the results reported in Table 11.4-1). There is symmetry about the centerlines.
|
||
|
||
<!-- source-page: 352 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
SUBROUTINE DKT (D,X1,Y1,X2,Y2,X3,Y3,SE)
|
||
IMPLICIT DOUBLE PRECISION (A-H,O-Z)
|
||
DIMENSION D(3,3),DD(9,9),QQ(9,9),PP(3,3),PT(2,3),RS(2,3),Q(3)
|
||
DIMENSION GG(10,9),KOD(2,9),B(3),C(3),ALS(3),PX(3,3),SE(9,9)
|
||
DATA KOD /1,1,2,3,3,2,4,4,5,6,6,5,7,7,8,9,9,8/
|
||
DATA PP /12.D0,4.D0,4.D0,4.D0,2.D0,1.D0,4.D0,1.D0,2.D0/
|
||
B(1)=Y2-Y3
|
||
B(2)=Y3-Y1
|
||
B(3)=Y1-Y2
|
||
C(1)=X3-X2
|
||
C(2)=X1-X3
|
||
C(3)=X2-X1
|
||
DET=(B(1)*C(2)-B(2)*C(1))*24.
|
||
DO 10 I=1,3
|
||
DO 10 J=1,3
|
||
10 PX(I,J)=PP(I,J)/DET
|
||
DO 25 I=1,3
|
||
DO 25 J=1,3
|
||
DO 25 K1=1,3
|
||
II=(I-1)*3+K1
|
||
DO 25 K2=1,3
|
||
JJ=(J-1)*3+K2
|
||
25 DD(II,JJ)=D(I,J)*PX(K1,K2)
|
||
DO 30 I=1,3
|
||
ALS(I)=B(I)*B(I)+C(I)*C(I)
|
||
PT(1,I)=6.*C(I)/ALS(I)
|
||
PT(2,I)=6.*B(I)/ALS(I)
|
||
RS(1,I)=3.*C(I)*C(I)/ALS(I)
|
||
RS(2,I)=3.*B(I)*B(I)/ALS(I)
|
||
30 Q(I)=3.*B(I)*C(I)/ALS(I)
|
||
DO 720 I=1,10
|
||
DO 720 J=1,9
|
||
720 GG(I,J)=0.
|
||
DO 730 I=1,2
|
||
II=(I-1)*5
|
||
P1=PT(I,1)
|
||
P2=PT(I,2)
|
||
P3=PT(I,3)
|
||
R1=RS(I,1)
|
||
R2=RS(I,2)
|
||
R3=RS(I,3)
|
||
GG(II+1,KOD(I,1))=P3
|
||
GG(II+2,KOD(I,1))=-P2
|
||
GG(II+3,KOD(I,1))=-P3
|
||
GG(II+4,KOD(I,1))=P2-P3
|
||
GG(II+5,KOD(I,1))=P2
|
||
GG(II+1,KOD(I,2))=-Q(3)
|
||
GG(II+2,KOD(I,2))=-Q(2)
|
||
GG(II+3,KOD(I,2))=Q(3)
|
||
GG(II+4,KOD(I,3))=-1.-R3
|
||
GG(II+5,KOD(I,3))=R3
|
||
GG(II+6,KOD(I,4))=-P3
|
||
GG(II+7,KOD(I,4))=P1+P3
|
||
GG(II+8,KOD(I,5))=-Q(3)
|
||
GG(II+9,KOD(I,6))=Q(3)
|
||
GG(II+10,KOD(I,7))=-Q(3)-Q(1)
|
||
GG(II+11,KOD(I,8))=-1.-R3
|
||
GG(II+12,KOD(I,9))=R3-R1
|
||
GG(II+13,KOD(I,10))=R3
|
||
GG(II+14,KOD(I,11))=R3-R1
|
||
GG(II+15,KOD(I,12))=-P2
|
||
GG(II+16,KOD(I,13))=-P2
|
||
GG(II+17,KOD(I,14))=-Q(2)
|
||
GG(II+18,KOD(I,15))=-Q(2)-Q(1)
|
||
GG(II+19,KOD(I,16))=-Q(2)
|
||
GG(II+20,KOD(I,17))=-P1-P2
|
||
GG(II+21,KOD(I,18))=-Q(2)
|
||
GG(II+22,KOD(I,19))=-1.-R2
|
||
GG(II+23,KOD(I,20))=-R2-R1
|
||
GG(II+24,KOD(I,21))=-R2
|
||
GG(II+25,KOD(I,22))=-R2
|
||
730 CONTINUE
|
||
DO 850 I=1,9
|
||
QQ(1,I)=B(2)*GG(1,I)+B(3)*GG(2,I)
|
||
QQ(2,I)=2.*B(2)*GG(3,I)+B(3)*GG(4,I)
|
||
QQ(3,I)=B(2)*GG(4,I)+2.*B(3)*GG(5,I)
|
||
QQ(4,I)=-C(2)*GG(6,I)-C(3)*GG(7,I)
|
||
QQ(5,I)=-2.*C(2)*GG(8,I)-C(3)*GG(9,I)
|
||
QQ(6,I)=-C(2)*GG(9,I)-2.*C(3)*GG(10,I)
|
||
QQ(7,I)=C(2)*GG(1,I)+C(3)*GG(2,I)
|
||
1 -B(2)*GG(6,I)-B(3)*GG(7,I)
|
||
QQ(8,I)=2.*C(2)*GG(3,I)+C(3)*GG(4,I)
|
||
1 -2.*B(2)*GG(8,I)-B(3)*GG(9,I)
|
||
QQ(9,I)=C(2)*GG(4,I)+2.*C(3)*GG(5,I)
|
||
1 -B(2)*GG(9,I)-2.*B(3)*GG(10,I)
|
||
850 CONTINUE
|
||
DO 855 I=1,9
|
||
DO 855 J=1,9
|
||
GG(I,J)=0.
|
||
DO 855 K=1,9
|
||
855 GG(I,J)=GG(I,J)+DD(I,K)*QQ(K,J)
|
||
DO 960 L=1,9
|
||
DO 960 J=L,9
|
||
DUM=0.
|
||
DO 900 K=1,9
|
||
900 DUM=DUM+QQ(K,L)*GG(K,J)
|
||
SE(L,J)=DUM
|
||
960 SE(J,L)=DUM
|
||
RETURN
|
||
END
|
||
</details>
|
||
|
||
Figure 11.4-3. Fortran statements that generate the 9 by 9 stiffness matrix of a DKT plate element [adapted from Ref. 11.13]. See text for notation and order of d.o.f.
|
||
|
||
# 11.5 BOUNDARY CONDITIONS AND TEST CASES
|
||
|
||
Boundary Conditions. Conditions at the edge of a plate are classed as clamped, free, or simply supported. Typically, no single condition prevails along the entire plate boundary. In the notation of Fig. 11.5-1, plate boundary conditions are as follows.
|
||
|
||
<!-- source-page: 353 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
s
|
||
θₙ or wₙ
|
||
θₛ or wₛ
|
||
y
|
||
β
|
||
x
|
||
w
|
||
n
|
||
</details>
|
||
|
||
(n)
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
s
|
||
Mn
|
||
n
|
||
Mhs
|
||
Qn
|
||
y
|
||
β
|
||
x
|
||
</details>
|
||
|
||
(b)
|
||
|
||
$$
|
||
w _ {i, x} = w _ {i, n} \cos \beta - w _ {i, s} \sin \beta
|
||
$$
|
||
|
||
$$
|
||
w _ {, y} = w _ {, n} \sin \beta + w _ {, s} \cos \beta
|
||
$$
|
||
|
||
(c)
|
||
Figure 11.5-1. Coordinates n and s are edge-normal and edge-tangent, respectively. (a) Rotations and lateral displacement. (b) Moments and transverse shear force. (c) Transformation relations for rotations.
|
||
<table><tr><td>clamped</td><td>free</td><td>simply supported (finite element)</td><td>simply supported (classical theory)</td></tr><tr><td> $w = 0$ </td><td> $Q_n = 0^1$ </td><td> $w = 0$ </td><td> $w = 0$ </td></tr><tr><td> $\theta_n = 0$ </td><td> $M_n = 0$ </td><td> $M_n = 0$ </td><td> $M_n = 0$ </td></tr><tr><td> $\theta_s = 0$ </td><td> $M_{ns} = 0$ </td><td> $M_{ns} = 0$ </td><td> $\theta_s = 0$ </td></tr></table>
|
||
|
||
Since transverse shear strain is taken as zero in classical thin-plate theory, Eqs. 11.5-1 are modified for Kirchhoff and discrete Kirchhoff elements by replacing $\theta_{n}$ by $w_{,n}$ and $\theta_{s}$ by $w_{,s}$ .
|
||
|
||
A clamped edge prevents all motion along the edge. Along a free edge, nodal d.o.f. are unspecified, and remain part of the vector $\{D\}$ of unknown d.o.f. Conditions along a simply supported or “hinged” edge have been found troublesome and require more explanation, as follows.
|
||
|
||
In classical thin-plate theory, since $\gamma_{zs}=0$ , the boundary condition w=0 necessarily implies the boundary condition $w_{,s}=0$ as well. Thus, for a thin simply supported plate, we would expect good numerical results using the “classical theory” conditions in Eqs. 11.5-1, whether the element type is Kirchhoff, discrete Kirchhoff, or Mindlin. Such is indeed the case if boundaries intersect at right angles, as at the four corners of a rectangular plate. However, if the plate is skew, most elements give poor results. For example, if the plate of Fig. 11.5-2a is modeled by a uniform 14 by 14 mesh, the center displacement may be underestimated by more than 20%. If only the boundary conditions are changed, to the “finite element” simply supported conditions in Eqs. 11.5-1, the error may decline to less than 3%. Apparently, classical simply supported conditions overconstrain the mesh when interior corner angles exceed $\pi/2$ . The finite element simply supported conditions produce a plate model that is point-supported at its boundary nodes. Although this may appear to allow too little constraint, especially in a coarse mesh, good results are obtained in practice [11.14].
|
||
|
||
In Eqs. 11.5-1, the various quantities may have prescribed values other than zero. For example, a line load $(Q_{n} \neq 0)$ could be prescribed along an edge of a
|
||
|
||
<!-- source-page: 354 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
y
|
||
a
|
||
x
|
||
a sin (β/2)
|
||
β
|
||
</details>
|
||
|
||
(a)
|
||
|
||
|
||
|
||
<details>
|
||
<summary>line</summary>
|
||
|
||
| x/a | M_x / qa² (×10⁻³) | M_y / qa² (×10⁻³) |
|
||
| ------ | ----------------- | ----------------- |
|
||
| 0.00 | 20.0 | -10.0 |
|
||
| 0.05 | 10.0 | 0.0 |
|
||
| 0.10 | 10.0 | 10.0 |
|
||
| 0.15 | 15.0 | 10.0 |
|
||
| 0.20 | 18.0 | 11.0 |
|
||
| 0.25 | 20.0 | 11.5 |
|
||
</details>
|
||
|
||
(b)
|
||
Figure 11.5-2. (a) Skew plate with equal side lengths (rhombic plate). (b) Bending moments along the x axis in a simply supported rhombic plate [11.15].
|
||
|
||
plate and represented by concentrated lateral forces at nodes along the edge. A nonzero w along the edge may be prescribed instead of $Q_{n}$ . Thus the edge becomes simply supported or clamped, depending on whether $M_{n} = 0$ or $\theta_{n} = 0$ .
|
||
|
||
Test Cases. In Section 4.5 we argue that an element must be able to display a constant-strain state. Plate elements are no exception. Each layer z = constant of a plate element must be able to display constant $\epsilon_{x}$ , $\epsilon_{y}$ , and $\gamma_{xy}$ . Hence, in patch-testing a Kirchhoff or discrete Kirchhoff plate element, we look for constant curvatures $w_{,xx}$ and $w_{,yy}$ and for constant twist $w_{,xy}$ . In patch-testing a Mindlin plate element, we look for constant curvatures $\theta_{x,x}$ and $\theta_{y,y}$ , constant twist $\theta_{x,y} + \theta_{y,x}$ , and constant transverse shear strains $w_{,y} - \theta_{y}$ and $w_{,x} - \theta_{x}$ . Figure 11.5-3 depicts a patch test for constant $w_{,xx}$ (or for constant $\theta_{x,x}$ ). One must enforce $w_{,y} = 0$ (or $\theta_{y} = 0$ ) at nodes 1 through 4 in order to prevent curling of the edges (unless $\nu = 0$ ).
|
||
|
||
Popular test cases include square plates, with various support conditions and loads (see Table 11.4-1). Rectangular plates may be used as well, to test the effect of element aspect ratio. Circular plates can be used to test nonrectangular elements. Exact results are available for many such problems [11.1].
|
||
|
||
The simply supported rhombic plate, Fig. 11.5-2a, is a difficult test case. The obtuse corners are singular points, where moments are theoretically infinite. Some element types fail to show that $M_{x}$ and $M_{y}$ are of opposite sign near these corners.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
y
|
||
4
|
||
x
|
||
5
|
||
3
|
||
Mₙ
|
||
H
|
||
1
|
||
2
|
||
Mₙ
|
||
</details>
|
||
|
||
Figure 11.5-3. Patch test for constant curvature, with $M_{x} = 2M_{a}/H$ , where $M_{a}$ is a moment load on a node.
|
||
|
||
<!-- source-page: 355 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>line</summary>
|
||
|
||
| Element | L | w3 (×10³) |
|
||
|---------|----|-----------|
|
||
| Element X | 0 | 0 |
|
||
| Element X | 2 | 5 |
|
||
| Element X | 4 | 10 |
|
||
| Element X | 6 | 15 |
|
||
| Element X | 8 | 20 |
|
||
| Element X | 10 | 25 |
|
||
| Element X | 12 | 30 |
|
||
| Element DKT (both loadings) | 0 | 0 |
|
||
| Element DKT (both loadings) | 2 | 5 |
|
||
| Element DKT (both loadings) | 4 | 10 |
|
||
| Element DKT (both loadings) | 6 | 15 |
|
||
| Element DKT (both loadings) | 8 | 20 |
|
||
| Element DKT (both loadings) | 10 | 25 |
|
||
| Element DKT (both loadings) | 12 | 30 |
|
||
</details>
|
||
|
||
Figure 11.5-4. “Twisted ribbon” test case [11.12,11.16]. Here $E = 10^{7}$ , $\nu = 0.25$ , t = 0.05, $w_{3} = \text{deflection of corner indicated}$ .
|
||
|
||
[11.14-11.16]. Difficulties attendant to use of classical simply supported boundary conditions for this problem have already been noted.
|
||
|
||
The “twisted ribbon,” Fig. 11.5-4, is a test that shows the effect of aspect ratio $[11.12,11.16]$ . The twisting moment may be applied by corner forces or by corner couples, as shown. Usually the entire plate is modeled by one rectangular element or by two triangular elements. “Benchmark” values were obtained from a mesh of 16 rectangular Kirchhoff elements having 16 d.o.f. each. Many types of element fail this test, such as the one identified as “element X,” which is too stiff at large aspect ratio, and may fail even to produce a displacement of the correct algebraic sign. For element DKT, the two loadings product slightly different results, but the difference is scarcely noticeable when plotted. If triangulation is made along diagonal 2–4 rather than diagonal 1–3 as shown, elements DKT and X both product somewhat different results than shown in Fig. 11.5-4.
|
||
|
||
# PROBLEMS
|
||
|
||
# Section 11.1
|
||
|
||
11.1 (a) Verify the stress formulas of Eqs. 11.1-2.
|
||
(b) Similarly, verify the formulas for $\tau_{yz}$ and $\tau_{zx}$ given below Eqs. 11.1-2.
|
||
11.2 In elementary mechanics of materials, one derives equations for normal and shear stress at an arbitrary angle $\theta$ in the $xy$ plane, such as $\sigma_{n} = \frac{1}{2} (\sigma_{x} + \sigma_{y}) + \frac{1}{2} (\sigma_{x} - \sigma_{y})\cos 2\theta +\tau_{xy}\sin 2\theta$ . What analogous expressions relate bending and twisting moments $M_{n}$ and $M_{ns}$ to $M_{x},M_{y}$ , and $M_{xy}$ ? Suggestion: Use Eqs. 11.1-2.
|
||
11.3 (a) In Fig. 11.1-1b presume that the $M$ 's and $Q$ 's are functions of $x$ and $y$ , so that (for example) $M_x dy$ acts along the edge $x = 0$ and $(M_x + M_{x,x})$
|
||
|
||
<!-- source-page: 356 -->
|
||
|
||
$dx$ ) dy acts along the parallel edge. Show that the equilibrium equations are $Q_{x,x} + Q_{y,y} = -q, M_{x,x} + M_{xy,y} = Q_x$ , and $M_{xy,x} + M_{y,y} = Q_y$ .
|
||
|
||
(b) Hence, show that $M_{x,xx} + 2M_{xy,xy} + M_{y,yy} + q = 0$ .
|
||
(c) Use the result of part (b), and Eq. 11.1-7 for isotropic conditions, to show that $\nabla^4 w = q / D$ , where $\nabla^4$ is the biharmonic operator.
|
||
|
||
11.4 A rectangular plate of thickness t has dimensions a and b, as shown. The plate is simply supported along edges AB and CD. Edges BC and DA remain free. If a uniform downward pressure p is applied to the upper surface, what are the principal stresses at the middle of the lower surface, and what is the deflection at the center of the plate?
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
A
|
||
D
|
||
y
|
||
a
|
||
B
|
||
C
|
||
b
|
||
x
|
||
</details>
|
||
|
||
Problem 11.4
|
||
|
||
11.5 (a) Write Eq. 11.1-5 for an isotropic material. Then derive terms in $[\mathbf{D}_K]$ , using the procedure given above Eq. 11.1-7.
|
||
(b) Verify the correctness of Eq. 11.1-11.
|
||
11.6 Consider an isotropic thin square plate, with edges parallel to $x$ and $y$ axes, loaded only along its edges. Describe the edge loads if the lateral deflection is (a) $w = c_{1}(x^{2} + y^{2})$ , and (b) $w = c_{2}(y^{2} - x^{2})$ , where $c_{1}$ and $c_{2}$ are constants.
|
||
11.7 (a) In a sandwich plate, Fig. 11.1-5, the average transverse shear strain $\gamma$ is related to the core shear strain $\gamma_{c}$ by $(c + h)\gamma = c\gamma_{c}$ . Derive this expression. Assume that the facings are much stiffer than the core.
|
||
|
||
(b) Derive the expression for $D_{M44}$ in Eqs. 11.1-13. Suggestion: Consider strain energy.
|
||
(c) Work from the bending stiffness $EI$ of a sandwich beam and derive the expression for $D_{M11}$ in Eqs. 11.1-13 (except for the $1 - \nu^2$ factor).
|
||
|
||
11.8 One sometimes wonders how wide a beam can be before it should be regarded as a plate. How would you decide? Or what would you do if unable to decide?
|
||
|
||
# Section 11.2
|
||
|
||
11.9 (a) Consider the rectangular plate element of Fig. 11.2-1 and the displacement field of Eq. 11.2-5. Show that interelement compatibility of normal slopes is lacking. For example, show that $w_{,y}$ along $y = b$ does not depend only on $w_{,y}$ at nodes 3 and 4.
|
||
|
||
(b) Similarly, what can be said about interelement compatibility of $w$ and $w_{xx}$ along edge 3-4?
|
||
(c) In Eq. 11.2-5, the terms $a_{11}x^3y$ and $a_{12}xy^3$ might be replaced by $a_{11}x^4$ and $a_{12}y^4$ , but it is not wise to do so. Why?
|
||
|
||
11.10 Imagine that the lateral displacement $w$ of a triangular thin-plate element
|
||
|
||
<!-- source-page: 357 -->
|
||
|
||
is taken as a complete quintic (21 terms). For each element shown, and without calculation, allocate d.o.f. to the nodes in a way that seems acceptable. Consider higher-order d.o.f. as needed. Is interelement compatibility achieved? Consider compatibility conditions on the edge x = 0.
|
||
|
||
|
||
Problem 11.10
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
Axis
|
||
w1
|
||
w2
|
||
w,r1
|
||
1
|
||
l
|
||
2
|
||
w,r2
|
||
r
|
||
r1
|
||
L
|
||
r2
|
||
</details>
|
||
|
||
Problem 11.11
|
||
|
||
11.11 The sketch represents the cross section of an annular (axisymmetric) element for analysis of thin plates ( $t << L$ ). Geometry, loads, and deformations are all axially symmetric. The material is isotropic. Derive the element stiffness matrix [k], to the extent of fully defining [B], [D $_{K}$ ], and all other terms used in your formula for [k], but do not integrate.
|
||
11.12 Imagine that the Mindlin beam element of Fig. 9.4-1 is uniform, uniformly loaded, and simply supported at nodes 1 and 2. With one-point quadrature for transverse shear terms, the stiffness matrix that operates of d.o.f. $\theta_{1}$ and $\theta_{2}$ is
|
||
|
||
$$
|
||
[ \mathrm{k} ] = \frac {E b t ^ {3}}{1 2 L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] + \frac {G b t L}{4 . 8} \left[ \begin{array}{l l} 1 & 1 \\ 1 & 1 \end{array} \right]
|
||
$$
|
||
|
||
where b = element width and t = element depth. Determine the rotation at node 2, and compare it with the exact value, if nodal loads produced by the distributed load are calculated (a) consistently, and (b) from a cubic field for w.
|
||
|
||
11.13 Imagine that the plate of Fig. 11.2-3 is square, simply supported, and modeled by a single finite strip (i.e., $L = b$ ). The lateral load is distributed and is described by $q = q_0 \sin (\pi x / b) \sin (\pi y / b)$ . Determine the center deflection of the finite strip. Suggestions: See Problem 11.3c. Arbitrarily elect to evaluate the equation at the center of the plate. Note, however, that this procedure would not be used in a finite strip computer program.
|
||
|
||
11.14 Let the rectangular plate element of Fig. 11.2-1 carry a uniformly distributed load q in the +z direction. Determine the resulting nodal moment loads by assuming that the plate element acts like a beam clamped at both ends, spanning first the dimension 2a and then the dimension 2b. Show these loads, properly directed, on a sketch of the plate.
|
||
|
||
11.15 (a) A constant moment $\overline{M}_{xy}$ is applied along edge 2--3 of the element in Fig. 11.2-1. What nodal loads result? Assume that $w$ along this edge is cubic in $y$ , and governed by $w$ and $w_{yy}$ at nodes 2 and 3.
|
||
(b) What would be your answer if the element is instead a Mindlin element, Eqs. 11.2-7, with shape functions $N_{1}$ through $N_{4}$ ?
|
||
|
||
<!-- source-page: 358 -->
|
||
|
||
# Section 11.3
|
||
|
||
11.16 Imagine that the four-node plate element of Fig. 11.3-1a is to be obtained by specialization of the eight-d.o.f. solid element of Fig. 6.7-1. Describe the steps and substitutions that convert Eqs. 6.7-1 to the equations $w = \sum N_i w_i$ , $u = -z \sum N_i \theta_{xi}$ , and $v = -z \sum N_i \theta_{yi}$ , where $i = 1, 2, 3, 4$ .
|
||
11.17 Show that Eq. 11.3-8 follows from Eq. 11.3-6 when $[\mathbf{D}_M]$ , $[\mathbf{B}_b]$ , and $[\mathbf{B}_s]$ are defined as stated in the text.
|
||
11.18 A uniform load $q$ acts upward on a rectangular plate element of side lengths $2a$ and $2b$ . What are the consistent nodal loads for each of the four elements listed in Table 11.3-1?
|
||
11.19 A square plate under concentrated center load is to be analyzed. The boundary is clamped, meaning that all boundary d.o.f. are set to zero. Let the model consist of a single element, which occupies one quadrant. After imposing boundary conditions, how many unknown d.o.f. are left for each of the elements in Table 11.3-1?
|
||
11.20 The sketch shows more detail of the latter portion of Fig. 11.3-3b. Lateral displacement w is zero at the ends and at the center. Imagine that nothing varies with y; that is, beam action is to be modeled. Under bending moment that varies linearly with x, thin-beam theory shows that end sections rotate an amount $\theta_{b}$ and the middle rotates an amount $\theta_{b}/2$ in the opposite direction. On this must be superposed rotations $\theta_{s}$ caused by the constant transverse shear force. Thus
|
||
|
||
$$
|
||
\gamma_ {z x} = w _ {, x} - \theta_ {x} = \left(\frac {\theta_ {b}}{2} - \theta_ {s}\right) - \frac {3 \theta_ {b}}{2} \left(\frac {x}{a}\right) ^ {2}
|
||
$$
|
||
|
||
(a) Derive this expression for $\gamma_{zx}$ , using quadratic shape functions and nodal values of $w$ and $\theta_x$ at $A$ , $M$ , and $C$ .
|
||
(b) Determine the values of $x / a$ for which $\gamma_{zx}$ is correctly represented, even when $a >> t$ .
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
θb + θs
|
||
θ
|
||
θb/2 - θs
|
||
θb + θs
|
||
A
|
||
M
|
||
t
|
||
C
|
||
a
|
||
a
|
||
x
|
||
</details>
|
||
|
||
Problem 11.20
|
||
|
||
11.21 Consider a rectangular quadratic element (Table 11.3-1). Under what circumstances or deformation states will $\gamma_{yz}$ and $\gamma_{zx}$ be correctly evaluated along the line $\xi = 0$ ?
|
||
11.22 (a) Sketch a rectangular element in its deformed state if displacements are described by $u = v = 0$ , $w = 3\xi^2\eta^2 - \xi^2 - \eta^2$ .
|
||
(b) Show that this deformation mode yields zero strains at the Gauss points of a 2 by 2 rule.
|
||
|
||
<!-- source-page: 359 -->
|
||
|
||
(c) What kind of loads and support conditions would activate this mode, either for a single element or for a mesh of elements?
|
||
|
||
11.23 Sketch a 2 by 3 mesh of rectangular bilinear elements. Superposed on this sketch, show the mesh deformed into the w-hourglass mode of Fig. 11.3-4.
|
||
|
||
11.24 Use Eq. 11.3-4 to evaluate $\{\kappa\}$ for each of the modes in Fig. 11.3-4. Show that only the first two $\{\kappa\}$ 's are null for selective integration, and that all four $\{\kappa\}$ 's are null for reduced integration.
|
||
|
||
# Section 11.4
|
||
|
||
11.25 (a) Derive Eq. 11.4-2.
|
||
|
||
(b) In Eq. 11.4-2, express $w_{,s5}$ in terms of $w$ , $w_{,x}$ , and $w_{,y}$ at nodes 2 and 3. Let the $n$ axis in Fig. 11.4-1a make an angle $\alpha$ with the $x$ axis. What is $\alpha$ in terms of the $x$ and $y$ coordinates of nodes 2 and 3?
|
||
(c) Similarly, in Eq. 11.4-5, what is $\theta_{n5}$ in terms of $w$ , $w_{,x}$ , and $w_{,y}$ at nodes 2 and 3 and angle $\alpha$ ?
|
||
|
||
11.26 What can be said about interelement compatibility of the DKT element?
|
||
11.27 Why are three Gauss points adequate for exact integration of a DKT element?
|
||
11.28 In Table 11.4-1, consider the clamped, uniformly loaded test case with mesh N = 1. Do you think the computed result would be more accurate if nodal moments were included in the element load vectors? Why?
|
||
11.29 Imagine that the serendipity element of Table 11.3-1 is to be given a “discrete Kirchhoff” treatment by explicitly enforcing zero transverse shear strain at the Gauss points of a 2 by 2 rule. How may d.o.f. do these constraints eliminate? What d.o.f. do you think it appropriate to retain in $\{d\}$ , and why?
|
||
|
||
# Section 11.5
|
||
|
||
11.30 Demonstrate the free-edge condition $Q_{n} + M_{ns,s} = 0$ , which is stated in the footnote for Eqs. 11.5-1. Suggestion: Consider couple forces $M_{ns} \Delta s$ and $(M_{ns} + M_{ns,s} \Delta s) \Delta s$ in adjacent “cells” of length $\Delta s$ along the edge.
|
||
11.31 Consider the mesh shown in Fig. 11.4-2. The mesh models one quadrant of a symmetrically loaded and symmetrically supported square plate. Under each of the following support conditions, how many unknown d.o.f. remain in $\{D\}$ after boundary conditions have been imposed? Of these, which do you expect will have the same magnitude?
|
||
|
||
(a) Clamped.
|
||
(b) Simply supported (finite element conditions).
|
||
(c) Simply supported (classical theory conditions).
|
||
|
||
<!-- source-page: 360 -->
|
||
|
||
# SHELL'S
|
||
|
||
The physics of shell behavior is summarized. Advantages and disadvantages of various displacement fields are illustrated by means of singly curved (arch) elements. Element formulations are presented for general shells and for shells of revolution.
|
||
|
||
# 12.1 SHELL GEOMETRY AND BEHAVIOR. SHELL ELEMENTS
|
||
|
||
A shell forms a curved surface in space. Usually a shell is thin in comparison with its span. Geometrically, a shell is described by its thickness t and the shape of the shell midsurface. At every point on the midsurface, one can draw two small arcs that lie in the midsurface, and orient the arcs so as to fit the largest and smallest curvatures of the midsurface at that point. These arcs will be mutually perpendicular. They define the principal radii of curvature at that point. In general, principal radii vary from point to point. The centers of curvature lie on a normal to the midsurface.
|
||
|
||
Examples of shell geometry appear in Fig. 12.1-1. The cylinder and cone are singly curved. In addition, they are developable, which means that if slit lengthwise they can be unrolled to form flat sheets without having to stretch their midsurfaces. The sphere and hyperboloid are doubly curved and are not developable. Radii of curvature are constant in the cylinder and sphere but are variable in the cone and hyperboloid.
|
||
|
||
Each shell in Fig. 12.1-1 happens to be a shell of revolution, meaning that the midsurface is generated by rotating a straight or curved generating line about an axis of revolution in the plane of the generator. A meridian is the intersection of the midsurface with a plane that contains the axis of revolution. A parallel is the intersection of the midsurface with a plane perpendicular to the axis.
|
||
|
||
In general, a shell simultaneously displays bending stresses and membrane stresses. Bending stresses in a shell correspond to bending stresses in a plate (Fig. 11.1-1a) and produce bending and twisting moments (Eqs. 11.1-1a). Membrane stresses correspond to stresses in a plane stress problem: they act tangent to the midsurface, and produce midsurface-tangent forces per unit length. These are the membrane forces $N_{x}$ , $N_{y}$ , and $N_{xy}$ given by
|
||
|
||
$$
|
||
N _ {x} = \int_ {- t / 2} ^ {t / 2} \sigma_ {x} d z \quad N _ {y} = \int_ {- t / 2} ^ {t / 2} \sigma_ {y} d z \quad N _ {x y} = \int_ {- t / 2} ^ {t / 2} \tau_ {x y} d z \tag {12.1-1}
|
||
$$
|
||
|
||
where $x$ and $y$ are orthogonal coordinates in the midsurface and $z$ is a direction
|