김경종
18 KiB
The element equations are given by
\left\{ \begin{array}{c} f _ {w i} \\ f _ {\theta x i} \\ f _ {\theta y i} \\ \vdots \\ f _ {\theta y n} \end{array} \right\} = \left[ \begin{array}{c c c c} k _ {1 1} & k _ {1 2} & \dots & k _ {1, 1 2} \\ k _ {2 1} & k _ {2 2} & \dots & k _ {2, 1 2} \\ k _ {3 1} & k _ {3 2} & \dots & k _ {3, 1 2} \\ \vdots & & \dots & \dots \\ k _ {1 2, 1} & & \dots & k _ {1 2, 1 2} \end{array} \right] \left\{ \begin{array}{c} w _ {i} \\ \theta_ {x i} \\ \theta_ {y i} \\ \vdots \\ \theta_ {y n} \end{array} \right\} \tag {12.2.22}
The rest of the steps, including assembling the global equations, applying boundary conditions (now boundary conditions on w; \theta _ { x } , \theta _ { y } ) , and solving the equations for the nodal displacements and slopes (note three degrees of freedom per node), follow the standard procedures introduced in previous chapters.
12.3 Some Plate Element Numerical Comparisons
We now present some numerical comparisons of quadrilateral plate element formulations. Remember there are numerous plate element formulations in the literature. Figure 12–5 shows a number of plate element formulation results for a square plate simply supported all around and subjected to a concentrated vertical load applied at the center of the plate. The results are shown to illustrate the upper and lower bound solution behavior and demonstrate the convergence of solution for various plate element formulations. Included in these results is the 12-term polynomial described in Section 12.2. We note that the 12-term polynomial converges to the exact solution from above. It yields an upper bound solution. Because the interelement continuity of slopes is not ensured by the 12-term polynomial, the lower bound classical characteristic of a minimum potential energy formulation is not obtained. However, as more elements are used, the solution converges to the exact solution [1].
Figure 12–6 shows comparisons of triangular plate formulations for the same centrally loaded simply supported plate used to compare quadrilateral element formulations in Figure 12–5. We can observe from Figures 12–5 and 12–6 a number of different formulations with results that converge from above and below. Some of these elements produce better results than others.
The Algor program [19] uses the Veubeke (after Baudoin Fraeijs de Veubeke) 16-degrees-of-freedom ‘‘subdomain’’ formulation [7] which converges from below, as it is based on a compatible displacement formulation. For more information on some of these formulations, consult the references at the end of the chapter.
Finally, Figure 12–7 shows results for some selected Mindlin plate theory elements. Mindlin plate elements account for bending deformation and for transverse shear deformation. For more on Mindlin plate theory, see Reference [6]. The ‘‘heterosis’’ element [10] is the best performing element in Figure 12–7.
Figure 12–5 Numerical comparisons: quadrilateral plate element formulations. (Gallagher, R. H., Finite Element Analysis Fundamentals, 1975, p. 345. Reprinted by permission of Prentice-Hall, Inc., Upper Saddle River, NJ.)
d 12.4 Computer Solution for a Plate Bending Problem
A computer program solution for plate bending problems [19] is now illustrated. The problem is that of a square steel plate fixed along all four edges and subjected to a concentrated load at its center as shown in Figure 12–8.
The plate element is a three- or four-noded element formulated in threedimensional space. The element degrees of freedom allowed are all three translations (u; v, and w) and in-plane rotations ( \theta _ { x } and \theta _ { y } ) . The rotational degrees of freedom normal to the plate are undefined and must be constrained. The element formulated in the computer program is the 16-term polynomial described in References [5] and [7]. This element is known as the Veubeke plate in the program. The 16-node formulation converges from below for the displacement analysis, as it is based on a compatible displacement formulation. This is also shown in Figure 12–5 for the clamped plate subjected to a concentrated center load.
line
| Method | X Value | % Error in central displacement |
|---|---|---|
| 6-term (quadratic) polynomial [15] | 4 | 16 |
| 21-term (quintic) polynomial [17] | 2 | 0 |
| 10-term (cubic) plus constraints [16] | 2 | -5 |
| 10-term (cubic) plus correction matrix [12] | 4 | -2 |
| Razzaque A-9 [14] | 1 | -15 |
| Bazeley et al., conforming [13] | 1 | -20 |
Mesh size (Fig. 12-5)
Figure 12–6 Numerical comparisons for a simply supported square plate subjected to center load triangular element formulations. (Gallagher, R. H., Finite Element Analysis Fundamentals, 1975, p. 350. Reprinted by permission of Prentice-Hall, Inc., Upper Saddle River, NJ.)
Example 12.1
\mathbf { A } 2 \times 2 mesh was created to model the plate. The resulting displacement plot is shown in Figure 12–9.
The exact solution for the maximum displacement (which occurs under the concentrated load) is given in Reference [1] as w = 0 . 0 0 5 6 P L ^ { 2 } / D = 0 . 0 0 5 6 ( - 1 0 0 1 \mathrm { { b } } ) ( 2 0 \ \mathrm { i n . } ) ^ { 2 } / ( 2 . 7 4 7 \ \times \ 1 0 ^ { 3 } \ \mathrm { l b } { \cdot } \mathrm { i n . } ) \ = \ - 0 . 0 8 \mathrm { i } 5 in:, where D = ( 3 0 \times 1 0 ^ { 6 } \mathrm { p s i } ) ( 0 . 1 \mathrm { i n } ) ^ { 3 } / [ 1 2 ( 1 - 0 . 3 ^ { 2 } ) ] = 2 . 7 4 7 \times 1 0 ^ { 3 } lb-in.
Figure 12–10 (a) and (c) show models of plate and beam elements combined. Beams can be combined with plates by having the beams match the centerline of the plates as shown in Figure 12–10(a). This ensures compatibility between the plate and beam elements. The plate is the same as the one used in Figure 12–9. The beam elements reinforce the plate so the maximum deflection is reduced as shown in Figure 12–10(b).
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| L_T/t | Computed w_c ÷ (theoretical w_c) |
|---|---|
| 10 | 1.15 |
| 20 | 1.05 |
| 30 | 1.02 |
| 50 | 1.00 |
| 100 | 0.98 |
| 200 | 0.96 |
| 300 | 0.94 |
| 500 | 0.92 |
| 1000 | 0.90 |
| 10^6 | 0.88 |
Figure 12–7 Center deflection of a uniformly loaded clamped square plate of side length L _ { T } and thickness t. An 8
8 mesh is used in all cases. Thin plates correspond to large L _ { T } / t . Transverse shear deformation becomes significant for small L _ { T } / t . Integration rules are reduced (R), selective (S), and full (F) [18], based on Mindlin plate element formulations. (Cook, R., Malkus D., and Plesha, M. Concepts and Applications of Finite Element Analysis, 3rd ed., 1989, p. 326. Reprinted by permission of John Wiley & Sons, Inc., New York.)
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100 lb 20 in. 0.1 in. 20 in.
Figure 12–8 Displacement plot of the clamped plate of Example 12.1
The beam elements used in this model were 2 in. by 12 in. rectangular cross sections used to sti¤en the plate through the center, as indicated by the lines dividing the plate into four parts. Figure 1–5 also illustrates how a chimney stack was modeled using both beam and plate elements.
Another way to connect beam and plate elements is shown in Figure 12–10(c) where the beam elements are o¤set from the plate elements and short beam elements
radar
| Displacement |
|---|
| 0.07583 |
| 0.065 |
| 0.05417 |
| 0.04333 |
| 0.0325 |
| 0.02167 |
| 0.01083 |
Figure 12–9 Displacement plot of the clamped plate of Example 12.1
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Geometric diagram of a diamond-shaped figure divided into four quadrants with triangular markers, labeled (a) (no text or symbols within the diagram itself)
Figure 12–10 (a) Model of beam and plate elements combined at centerline of elements, (b) vertical deflection plot for model in part (a), and (c) model showing offset beam elements from the plate elements
Figure P12–10 ðContinuedÞ
are used to connect the beam and plate elements at the nodes. In this model, 2 in. by 2 in. by \textstyle { \frac { 1 } { 4 } } in. thick square tubing properties were selected for the beam elements.
References
[1] Timoshenko, S. and Woinowsky-Krieger, S., Theory of Plates and Shells, 2nd ed., McGraw-Hill, New York, 1969.
[2] Gere, J. M., Mechanics of Material, 5th ed., Brooks/Cole Publishers, Pacific Grove, CA, 2001.
[3] Hrabok, M. M., and Hrudley, T. M., ‘‘A Review and Catalog of Plate Bending Finite Elements,’’ Computers and Structures, Vol. 19, No. 3, 1984, pp. 479–495.
[4] Zienkiewicz, O. C., and Taylor R. L., The Finite Element Method, 4th ed., Vol. 2, McGraw-Hill, New York, 1991.
[5] Gallagher, R. H., Finite Element Analysis Fundamentals, Prentice-Hall, Englewood Cli¤s, NJ, 1975.
[6] Cook, R. D., Malkus, D. S., Plesha, M. E., and Witt, R. J., Concepts and Applications of Finite Element Analysis, 4th ed., Wiley, New York, 2002.
[7] Fraeijs De Veubeke, B., ‘‘A Conforming Finite Element for Plate Bending,’’ International Journal of Solids and Structures, Vol. 4, No. 1, pp. 95–108, 1968.
[8] Walz, J. E., Fulton, R. E., and Cyrus N.J., ‘‘Accuracy and Convergence of Finite Element Approximations,’’ Proceedings of the Second Conference on Matrix Method in Structural Mechanics, AFFDL TR 68-150, pp. 995–1027, Oct., 1968.
[9] Melosh, R. J., ‘‘Basis of Derivation of Matrices for the Direct Sti¤ness Method,’’ Journal of AIAA, Vol. 1, pp. 1631–1637, 1963.
[10] Hughes, T. J. R., and Cohen, M., ‘‘The ‘Heterosis’ Finite Element for Plate Bending,’’ Computers and Structures, Vol. 9, No. 5, 1978, pp. 445–450.
[11] Bron, J., and Dhatt, G., ‘‘Mixed Quadrilateral Elements for Bending,’’ Journal of AIAA, Vol. 10, No. 10, pp. 1359–1361, Oct., 1972.
[12] Kikuchi, F., and Ando, Y., ‘‘Some Finite Element Solutions for Plate Bending Problems by Simplified Hybrid Displacement Method,’’ Nuclear Engineering Design, Vol. 23, pp. 155–178, 1972.
[13] Bazeley, G., Cheung, Y., Irons, B., and Zienkiewicz, O., ‘‘Triangular Elements in Plate Bending—Conforming and Non-Conforming Solutions,’’ Proceedings of the First Conference on Matrix Methods on Structural Mechanics, AFFDL TR 66-80, pp. 547–576, Oct., 1965.
[14] Razzaque, A. Q., ‘‘Program for Triangular Elements with Derivative Smoothing,’’ International Journal for Numerical Methods in Engineering, Vol. 6, No. 3, pp. 333–344, 1973.
[15] Morley, L. S. D., ‘‘The Constant-Moment Plate Bending Element,’’ Journal of Strain Analysis, Vol. 6, No. 1, pp. 20–24, 1971.
[16] Harvey, J. W., and Kelsey, S., ‘‘Triangular Plate Bending Elements with Enforced Compatibility’’, AIAA Journal, Vol. 9, pp. 1023–1026, 1971.
[17] Cowper, G. R., Kosko, E., Lindberg, G., and Olson M., ‘‘Static and Dynamic Applications of a High Precision Triangular Plate Bending Element’’, AIAA Journal, Vol. 7, No. 10, pp. 1957–1965, 1969.
[18] Hinton, E., and Huang, H. C., ‘‘A Family of Quadrilateral Mindlin Plate Elements with Substitute Shear Strain Fields,’’ Computers and Structures, Vol. 23, No. 3, pp. 409–431, 1986.
[19] Linear Stress and Dynamics Reference Division, Docutech On-line Documentation, Algor, Inc., Pittsburgh, PA, 1999.
Problems
Solve these problems using the plate element from a computer program.
12.1 A square steel plate of dimensions 20 in. by 20 in. with thickness of 0.1 is clamped all around. The plate is subjected to a uniformly distributed loading of 1 lb/in2. Using a 2 by 2 mesh and then a 4 by 4 mesh, determine the maximum deflection and maximum stress in the plate. Compare the finite element solution to the classical one in [1].
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3D coordinate system diagram with x, y, z axes and a rectangular grid structure (no text or symbols)
Figure P12–1
12.2 An L-shaped plate with thickness 0.1 in. is made of ASTM A-36 steel. Determine the deflection under the load and the maximum principal stress and its location using the plate element. Then model the plate as a grid with two beam elements with each beam having the sti¤ness of each L-portion of the plate and compare your answer.
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10" 30 lb 25" 15"
Figure P12–2
12.3 A square simply supported 20 in. by 20 in. steel plate with thickness 0.15 in. has a round hole of 4 in. diameter drilled through its center. The plate is uniformly loaded with a load of 2 lb/in2. Determine the maximum principal stress in the plate.
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z y x
Figure P12–3
12.4 A C-channel section structural steel beam of 2 in. wide flanges, 3 in. depth and thickness of both flanges and web of 0.25 in. is loaded as shown with 100 lb acting in the y direction on the free end. Determine the free end deflection and angle of twist. Now move the load in the z direction until the rotation (angle of twist) becomes zero. This distance is called the shear center (the location where the force can be placed so that the cross section will bend but not twist).
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z y 100 lb 12" 3" x
Figure P12–4
12.5 For the simply supported structural steel W 14 61 wide flange beam shown, compare the plate element model results with the classical beam bending results for deflection and bending stress. The beam is subjected to a central vertical load of 22 kip. The cross-sectional area is 17.9 in.2, depth is 13.89 in., flange width is 9.995 in., flange thickness of 0.645 in., web thickness of 0.375 in., and moment of inertia about the strong axis of 640 in.4
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22 kip 20 ft x y z
Figure P12–5
12.6 For the structural steel plate structure shown, determine the maximum principal stress and its location. If the stresses are unacceptably high, recommend any design changes. The initial thickness of each plate is 0.25 in. The left and right edges are simply supported. The load is a uniformly applied pressure of 10 lb/in.2 over the top plate.
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z y x 8" 10" 6" 8" 10"
Figure P12–6
12.7 Design a steel box structure 4 ft wide by 8 ft long made of plates to be used to protect construction workers while working in a trench. That is, determine a recommended thickness of each plate. The depth of the structure must be 8 ft. Assume the loading is from a side load acting along the long sides due to a wet soil (density of 62.4 lb/ft3) and varies linearly with the depth. The allowable deflection of the plate type structure is 1 in. and the allowable stress is 20 ksi.
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3D wireframe model of a rectangular prism with x, y, z axes labeled (no text or symbols on the model itself)
Figure P12–7
12.8 Determine the maximum deflection and maximum principal stress of the circular plate shown in Figure P12–8. The plate is subjected to a uniform pressure p = 7 0 0 kPa and fixed along its outer edge. Let E ¼ 200 GPa, n ¼ 0:3, radius r = 5 0 0 mm, and thickness t = 5 mm.
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r o x y
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p z x
Figure P12–8
12.9 Determine the maximum deflection and maximum principal stress for the plate shown in Figure P12–9. The plate is fixed along all three sides. A uniform pressure of 100 psi is applied to the surface. The plate is made of steel with E = 2 9 \times 1 0 ^ { 6 } psi, \nu = 0 . 3 , and thickness t = 0 . 5 0 in. a = 3 0 in. and b = 4 0 in.
12.10 An aircraft cabin window of circular cross section and simple supports all around as shown in Figure P12–10 is made of polycarbonate with E = 0 . 3 4 5 \times 1 0 ^ { 6 } psi, \nu = 0 . 3 6 , radius ¼ 20 in., and thickness t = 0 . 7 5 in. The safety of the material is tested at a uniform pressure of 10 psi. Determine the maximum deflection and maximum principal stress in the material. The yield strength of the material is 9 ksi. Comment on the potential use of this material in regard to strength and deflection.