Simply Supported Thick Square Plate Subjected to a Uniformly Distributed Transverse Load

Objective:

Determination of the strain state of a simply supported thick square plate subjected to a uniformly distributed transverse load.

Initial data files:

File name	Description
толстая_плита_a_h_2.spr	Design model for the plate side-to-thickness ratios a/h = 2.0
толстая_плита_a_h_4.spr	Design model for the plate side-to-thickness ratios a/h = 4.0
толстая_плита_a_h_8.spr	Design model for the plate side-to-thickness ratios a/h = 8.0

Problem formulation:

The simply supported thick square plate is subjected to a uniformly distributed transverse load p. Determine the deflection w in the center of the plate taking into account the transverse shear deformations.

References: L. G. Donnell, Beam, Plates, and Shells, Moscow, Nauka, 1982, p. 313-316.

Initial data:

 

Finite element model:

Three design models for the following plate side-to-thickness ratios a/h = 8.0; 4.0; 2.0 are considered. Four variants of each model with the following types of finite elements are considered: 44, 50 – quadrangular four-node and eight-node thin shell elements for the calculation according to the Kirchhoff-Love theory; 144, 150 – quadrangular four-node and eight-node thick shell elements for the calculation according to the Reissner–Mindlin theory.

Design models are created for the following meshes: 2x2; 4x4; 8x8; 16x16; 32x32; 64x64.

Boundary conditions are provided by imposing constraints in the directions of the degrees of freedom X, Y, Z, UY for the edges along the X axis of the global coordinate system, and X, Y, Z, UX for the edges along the Y axis of the global coordinate system.

 

Results in SCAD

Design models

Deflections w of plates with the ratio a/h = 8.0, m

Deflections w of plates with the ratio a/h = 4.0, m

Deflections w of plates with the ratio a/h = 2.0, m

 

Comparison of solutions:

 

Deflections w in the center of the plates with the ratio a/h = 8.0, m

Member type	SCAD, mesh						Theory	Deviation
	2x2	4x4	8x8	16x16	32x32	64x64
44	0.001065	0.001213	0.001262	0.001274	0.001277	0.001278	0.001278	0.00 %
50	0.001276	0.001278	0.001278	0.001278	0.001278	0.001278		0.00 %
144	0.001472	0.001472	0.001449	0.001443	0.001441	0.001441	0.001369	5.26 %
150	0.001314	0.001364	0.001368	0.001368	0.001368	0.001368		0.07 %

Deflections w in the center of the plates with the ratio a/h = 4.0, m

Member type	SCAD, mesh						Theory	Deviation
	2x2	4x4	8x8	16x16	32x32	64x64
44	0.000133	0.000152	0.000158	0.000159	0.000160	0.000160	0.000160	0.00 %
50	0.000159	0.000160	0.000160	0.000160	0.000160	0.000160		0.00 %
144	0.000258	0.000246	0.000242	0.000242	0.000241	0.000241	0.000205	17.56 %
150	0.000205	0.000205	0.000205	0.000205	0.000205	0.000205		0.00 %

Deflections w in the center of the plates with the ratio a/h = 2.0, m

Member type	SCAD, mesh						Theory	Deviation
	2x2	4x4	8x8	16x16	32x32	64x64
44	0.000017	0.000019	0.000020	0.000020	0.000020	0.000020	0.000020	0.00 %
50	0.000020	0.000020	0.000020	0.000020	0.000020	0.000020		0.00 %
144	0.000065	0.000062	0.000061	0.000061	0.000061	0.000061	0.000043	41.86 %
150	0.000043	0.000043	0.000043	0.000043	0.000043	0.000043		0.00 %

Notes: In the analytical solution the deflections w in the center of the plate are determined according to the following formulas:

without taking into account the transverse shear deformations

\[ w=\frac{12\cdot \left( {1-\nu^{2}} \right)\cdot a^{4}\cdot p}{E\cdot h^{3}}\cdot \left[ {\frac{5}{384}-\sum\limits_{m=1}^\infty {\frac{\sin \left( {\frac{m\cdot \pi }{2}} \right)\cdot \left( {4+m\cdot \pi \cdot th\left( {\frac{m\cdot \pi }{2}} \right)} \right)}{m^{5}\cdot \pi^{5}\cdot ch\left( {\frac{m\cdot \pi }{2}} \right)}} } \right] \quad or \] \[ w=\frac{192\cdot \left( {1-\nu^{2}} \right)\cdot a^{4}\cdot p}{\pi ^{6}\cdot E\cdot h^{3}}\cdot \sum\limits_{m=1}^\infty {\sum\limits_{n=1}^\infty {\left[ {\frac{1}{m\cdot n}\cdot \frac{1}{^{\left( {m^{2}+n^{2}} \right)^{2}}}\cdot \sin \left( {\frac{m\cdot \pi }{2}} \right)\cdot \sin \left( {\frac{n\cdot \pi }{2}} \right)} \right]} } ; \] \[ At \quad \nu \quad = 0.2 w\approx 0.004062\cdot \frac{a^{4}\cdot p}{D}, \quad where: \quad D=\frac{E\cdot h^{3}}{12\cdot \left( {1-\nu^{2}} \right)}. \]

taking into account the transverse shear deformations

\[w=\frac{12\cdot \left( {1-\nu^{2}} \right)\cdot a^{4}\cdot p}{E\cdot h^{3}}\cdot \left[ {\frac{5}{384}-\sum\limits_{m=1}^\infty {\frac{\sin \left( {\frac{m\cdot \pi }{2}} \right)\cdot \left( {4+m\cdot \pi \cdot th\left( {\frac{m\cdot \pi }{2}} \right)} \right)}{m^{5}\cdot \pi^{5}\cdot ch\left( {\frac{m\cdot \pi }{2}} \right)}+\frac{8-3\cdot \nu }{10\cdot \left( {1-\nu } \right)}\cdot \left( {\frac{h}{a}} \right)^{2}} \cdot \left( {\frac{1}{32}-\sum\limits_{m=1}^\infty {\frac{\sin \left( {\frac{m\cdot \pi }{2}} \right)}{m^{3}\cdot \pi^{3}\cdot ch\left( {\frac{m\cdot \pi }{2}} \right)}} } \right)} \right] \quad or \] \[ w=\frac{192\cdot \left( {1-\nu^{2}} \right)\cdot a^{4}\cdot p}{\pi ^{6}\cdot E\cdot h^{3}}\cdot \sum\limits_{m=1}^\infty {\sum\limits_{n=1}^\infty {\left[ {\frac{1}{m\cdot n}\cdot \frac{1}{^{\left( {m^{2}+n^{2}} \right)^{2}}}\cdot \left[ {1+\frac{\pi^{2}\cdot h^{2}}{5\cdot \left( {1-\nu } \right)\cdot a^{2}}\cdot \left( {m^{2}+n^{2}} \right)} \right]\cdot \sin \left( {\frac{m\cdot \pi }{2}} \right)\cdot \sin \left( {\frac{n\cdot \pi }{2}} \right)} \right]} } ; \] \[ At \quad \nu \quad = 0.2 \quad w\approx 0.004062\cdot \frac{a^{4}\cdot p}{D}\cdot \left[ {1+4.533786\cdot \left( {\frac{h}{a}} \right)^{2}} \right], \quad where: \] \[ D=\frac{E\cdot h^{3}}{12\cdot \left( {1-\nu^{2}} \right)}. \]