MANE-4240: Introduction to Finite Elements Spring 2014
Class Project
Study of the Deformation of a Simply Supported
Square Plate Under Uniform Pressure Using the
Finite Element Method and Comparison with
Exact Solution
David Grant
April 27, 2014
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Table of Contents
Introduction .................................................................................................................................................. 3
Background ................................................................................................................................................... 3
Formulation................................................................................................................................................... 4
Results ........................................................................................................................................................... 6
Approach ....................................................................................................................................................... 9
Conclusions ................................................................................................................................................... 9
References .................................................................................................................................................. 10
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Introduction
This study explores the problem of a simply supported deformed plate under a uniform load by using
the finite element method and verifying the method with the exact solution. The finite element method
was performed using the computer software Abaqus from Dassault Systèmes reference (a). The
solution of the model created in Abaqus will be compared to the approximation of the exact solution
using the Navier solution as detailed in Timoshenko reference (b).
Background
The deformation of an elastic plate was historically a very difficult problem to solve and was first
considered by Sophie Germaine and through the years corrections and improvements were
made. However it wasn’t until 1850 that a breakthrough was achieved by Kirchhoff. Kirchhoff formed
the double partial differential equation of the form below.
The differential equation above is generally not what is going on behind the scenes in finite element
software. The equation above is usually modified to a variational or weak form and makes use of the
Galerkin finite element method. The variational/weak form is obtained by modifying the above
differential equation with a test function generally of the form of Lagrange Polynomials. To solve the
unique problem a set of boundary conditions will be necessary.
Finite element methods subdivide the domain (the plate in this study) into continuous elements. The
location where the elements meet are nodes. Each element will contain local finite element basis
functions that will be used for the global basis functions of the nodes.
After the part is meshed into elements the finite element software will introduce the Galerkin
Method. While u(x) is desired, an approximate solution uh(x) will be obtained. The test function
mentioned above, will be represented as the summation of the weighted nodal functions and uh(x) will
be the summation of the product of ui and the test function.
These equations form a collection of terms in a system of equations that can be solved with linear
algebra. The system of equations is of the form KU=F, where K is the stiffness, U the desired solution,
and F the forcing function.
After the analysis is complete the results are analyzed and verified for accuracy. Verification is either
through an exact solution, another approximation to the exact solution, data from the field, or through
mesh refinement. Exact solutions are generally not available to problems that use the finite element
method as finite element method problems can be quite complex and if the exact solution could be
obtained with little effort there would be no need for the finite element method to be used. The plate
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problem being studied will make use of the Navier solution to the problem. The Navier solution to the
problem assumes the deflection can be represented by Fourier Series.
Formulation
The method the finite element models use is described above in the background. The elements used in
the Abaqus finite element method analysis were S4 general purpose shell elements. General purpose
elements in Abaqus provide robust and accurate solutions in all loading conditions for thin and thick
shell problems. General shell elements do not suffer from transverse shear locking nor do they have
unconstrained hourglass modes. No hourglass controls are needed for fully integrated S4 elements for
membrane and bending response in Abaqus.
To verify the solution obtained in Abaqus a comparison will be made to the Navier solution to the
deflected plate problem as detailed in reference (b). The Navier solution and its use of the Fourier
Series is show below.
Starting with the differential equation:
And in the general case the load distributed over the plate is given by the expression:
𝑞 = 𝑞0 𝑠𝑖𝑛
𝜋𝑥
𝜋𝑦
𝑠𝑖𝑛
𝑎
𝑏
Where a, and b are dimensions of the plate and 𝑞0 is the intensity of the load at the center of the plate.
𝑞
𝐷
And 𝑓(𝑥, 𝑦) = , where
𝐷=
𝐸𝑡 3
12(1−𝑣 2 )
and t is thickness of the plate, E is young’s modulus and v is
poisson’s ratio.
Substituting back into the differential gives
a) ∆∆𝑤 =
𝑞0
𝐷
𝑠𝑖𝑛
𝜋𝑥
𝜋𝑦
𝑠𝑖𝑛 𝑏
𝑎
Boundary Conditions for the edges of the simply supported plate are as follows:
w=0, Mx=0 for x=0 and x=a
w=0, My=0 for y=0 and y=b
which leads to
1. w=0 for x=0 and x=a
2.
𝜕2 𝑤
𝜕𝑥 2
= 0 for x=0 and x=a
3. w=0 for y=0 and y=b
4.
𝜕2 𝑤
𝜕𝑦 2
= 0 for y=0 and y=b
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All boundary conditions are satisfied if we take for deflections the expression
b) 𝑤 = 𝐶𝑠𝑖𝑛
𝜋𝑥
𝜋𝑦
𝑠𝑖𝑛 𝑏
𝑎
Where constant C must be chosen to satisfy equation a) above. Substituting b) into a) above yields
1
1
𝑞0
𝜋 4 ( 2 + 2 )2 𝐶 =
𝑎
𝑏
𝐷
Which leads to the deflection equation below
𝑐) 𝑤 =
𝑞0
1
1
𝜋4 𝐷( 2 + 2 )2
𝑎
𝑏
𝑠𝑖𝑛
𝜋𝑥
𝑎
𝑠𝑖𝑛
𝜋𝑦
𝑏
However the equation C) above is for a sinusoidal load and for the application of a uniform load the load
will be represented as a sum of partial sinusoidal loadings.
To represent any kind of load 𝑞 = 𝑓(𝑥, 𝑦)
∞
∞
𝑑) 𝑓(𝑥, 𝑦) = ∑ ∑ 𝑎𝑚𝑛
𝑠𝑖𝑛
𝑚=1 𝑛=1
𝑚𝜋𝑥
𝑛𝜋𝑦
𝑠𝑖𝑛
𝑎
𝑏
Using the methods in reference (b), the term 𝑎𝑚𝑛 can be obtained.
𝑎𝑚𝑛 =
4 𝑎 𝑏
𝑚𝜋𝑥
𝑛𝜋𝑦
∫ ∫ 𝑓(𝑥, 𝑦)𝑠𝑖𝑛
𝑠𝑖𝑛
𝑑𝑥𝑑𝑦
𝑎𝑏 0 0
𝑎
𝑏
Substituting 𝑎𝑚𝑛 back into f(x,y) and substituting that back into equation c) results in the deflection
below
𝑤=
1
𝜋4 𝐷
∞
∞
∑∑
𝑚=1 𝑛=1
𝑎𝑚𝑛
2
2
𝑛2
(𝑚2 + 2 )
𝑎
𝑏
𝑠𝑖𝑛
𝑚𝜋𝑥
𝑛𝜋𝑦
𝑠𝑖𝑛
𝑎
𝑏
To represent the uniform load, f(x,y) is set equal to 𝑞0 .
This results in the coefficient below
𝑎𝑚𝑛 =
4𝑞0 𝑎 𝑏
𝑚𝜋𝑥
𝑛𝜋𝑦
16𝑞0
∫ ∫ 𝑠𝑖𝑛
𝑠𝑖𝑛
𝑑𝑥𝑑𝑦 = 2
𝑎𝑏 0 0
𝑎
𝑏
𝜋 𝑚𝑛
It should be noted that m and n are odd integers, as m or n being even results in 𝑎𝑚𝑛 = 0
The deflection for the uniform load then becomes
𝑤=
16𝑞0
∞
∞
∑∑
𝜋6 𝐷 𝑚=1 𝑛=1
𝑠𝑖𝑛 𝑚𝜋𝑥
𝑎 𝑠𝑖𝑛
𝑛𝜋𝑦
𝑏
2
2
(𝑚2 + 𝑛2 )
𝑎
𝑏
2
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Where m=1,3,5.. and n=1,3,5…
And substituting the location of maximum deflection at x=a/2 and y=b/2 we obtain
𝑒) 𝑤𝑚𝑎𝑥 =
16𝑞0
∞
𝑚+𝑛
−1
2
∞
∑∑
(−1)
𝜋6 𝐷 𝑚=1 𝑛=1
2
2
𝑚𝑛(𝑚2 + 𝑛2 )
𝑎
𝑏
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Results
Results for the various mesh densities are shown in Table 1 below. The results show weak correlation to
the rectangular plate theory offered in reference (b). The coarsest mesh seems to correlate the best,
but is still more than 10 percent off the Navier Solution. There are two possible reasons for this and will
be discussed in the conclusions section of this report. For the square plate model, parameters used were
a and b=1 meter; t= 0.1 meter; Young’s modulus, E=2E11 Pa, Poisson’s ratio, v=0.3, and a uniform load,
q of 1E6 Pa. Deflection plots for each case are also shown in Figures 1 through 4.
Table 1 Summary of Results
Deformation of Simply Supported Square Plate under Uniform Load
Mesh Density :
Width x Height,
Type
6x6 1st order quad
S4
10x10 1st order
quad S4
20x20 1st order
quad S4
40x40 1st order
quad S4
Side Studies
20x20 S4 change
thickness to 0.01
Solid Model
6x30x30
Model with
p= 1E6 Pa,
Deflection
(m) absolute
value
Navier
Solution
deflection
(m)
Error ((exactmodel)/exact)*100%
0.000245 0.0002218
-10.46
0.000249 0.0002218
-12.44
0.000251 0.0002218
-13.30
0.000252 0.0002218
-13.57
0.2201 0.2218042
0.768
0.0002637 0.0002218
-18.89
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Figure 1: Deflection of 6 by 6 mesh, Type S4
Figure 2Deflection of 10 by 10 Mesh, Type S4
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Figure 3: Deflection of 20 by 20 Mesh, Type S4
Figure 4: Deflection with 40 by 40 element mesh, Type S4
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Approach
Each finite element model was created with the following given properties:
Parameter
a=b
t
E
v
q,
pressure
Value
1m
0.1 m
2E11 Pa
0.3
1E6 Pa
Each model was meshed using general purpose S4 elements. According to the Abaqus manual, general
purpose elements in Abaqus provide robust and accurate solutions in all loading conditions for thin and
thick shell problems.
The problem analyzed required the plate to be simply supported. Boundary conditions can be seen in
Figure 5.
Figure 5: Loads and Boundary Conditions Applied
Four different mesh sizes were used to indicate a convergence. Mesh sizes involved edge mesh seeds of
6, 10, 20 and 40 elements. Pictures of the meshes can be seen in the result Figures 1-4 above. As
shown in Table 1, it can be seen that after an edge mesh seed of 10 elements, the model deflections
change very little. And no further meshing was required.
Conclusions
The results of this study indicate that the model is converging to a solution of about 2.52E-4 meters.
While the Navier Solution gives a result of 2.22E-4 meters. This results in an error greater than 13
percent, which is considered high in most engineering problems. In the study the Navier solution
converged quickly after m and n equal to 2. The result tabulated is m and n equal to 10. The error is not
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likely in the number of terms of the Navier solution. There are two possible reasons for the differences.
The first and least likely is that Abaqus elements are not representing the stiffness accurately. The
second is that the thickness of 0.1 meter is two large in comparison to the width of the plate of 1 meter
for the Navier Solution to work. Thin shell theory may be violated here and the given value of 0.1 meter
is too thick. A side study was performed with a plate thickness of 0.01meter in both the finite element
model and the Navier Solution and the error was found to be less than one percent. Different sources
indicate that the thickness to width ratio should either be less than 0.25 or 0.1. Reference () indicates
0.25 as thickness to width ratio while some internet sources indicate 0.1. The thickness of 0.1 meter
with a width of 1 meter is right at the boundary and this may be the reason for the greater than 13
percent error. If this study was performed again the finite element models should be compared to
another given solution that allows thick shells.
It should be noted another side study was performed with solid elements in abaqus. The same load and
boundary conditions were applied. And the model was of the same dimensions as the study outlined
above. This resulted in about 18 percent error and indicated that something may be inaccurate with the
Navier solution for the problem described.
References
a) Abaqus/ CAE 6.13-1, Dassault Systèmes, 2013
b) Timoshenko, Stephen P., Krieger, S. Woinowsky, Theory of Plates and Shells, 2nd Edition,
McGraw-Hill International Editions, 1959
c) Young, Warren C., Budynas, Richard G., Roark’s Formulas for Stress and Strain, 7th Edition,
McGraw- Hill, 2002.
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