2D MODELING OF THE DEFLECTION OF
A SIMPLY SUPPORTED BEAM UNDER
POINT OR DISTRIBUTED LOADS
YIN-YU CHEN
MANE4240 – INTRODUCTION TO FINITE ELEMENT ANALYSIS
PROFESSOR ERNESTO GUTIERREZ-MIRAVETE
APRIL 28, 2014
Page |1
TABLE OF CONTENTS
REFERENCES ........................................................................................................................ 2
ABSTRACT ........................................................................................................................... 3
INTRODUCTION .................................................................................................................... 3
FORMULATION AND SOLUTION ............................................................................................ 5
RESULTS AND DISCUSSION................................................................................................... 8
CONCLUSIONS .................................................................................................................... 13
Page |2
REFERENCES
(a) M1 Abrams; Wikipedia: http://en.wikipedia.org/wiki/M1_Abrams; 2014
(b) Mechanics of Materials – 2nd Edition; Craig, Roy; 2000
(c) Theory of Elasticity: Two Dimensional Problems; Timoshenko, Stephen; June
1970
(d) ANSYS Tips: ANSYS Quick Start – Belcan; Dufour, Paul;
http://www.ux.uis.no/~hirpa/KdB/FEM/ANSYS_quick_start.pdf; December 2003
Page |3
ABSTRACT
This report details the analysis of a simply supported elastic beam subject to a point or
uniformly distributed load and the beam’s maximum deflection under these loads. The
analytical method through the use of hand calculations and MAPLE determined the
maximum deflection of an 8m x 0.6m x 0.1m steel beam subjected to the force, due to
half the weight of an M1 Abrams tank, as a point load at the center of the beam and as a
uniformly distributed load across the beam. A modeling/numerical method through the
use of COMSOL also obtained the maximum deflection of the beam for both cases.
Results of this study show that in the numerical COMSOL method, using triangle cells
for the mesh, presents a 9% error in comparison to the analytical method. For additional
comparison, an ANSYS beam element model, using quadrilateral cells for the mesh, was
created and analyzed, which presented only a 0.03% error in comparison to the analytical
method. From this study, it may be concluded that the maximum deflection of a simply
supported elastic beam subject to point or distributed loads may be achieved using either
the modeling/numerical or analytical methods. However, the shape of the cells for the
mesh is a major factor in the accuracy of the maximum beam deflection results. A brief
slide presentation summarizing the study and results is attached in Appendix A.
INTRODUCTION
The system under this study is the maximum deflection of a simply supported elastic
beam subject to point or distributed loads. The specific scenario considered in this study
is that of an M1 Abrams tank equally supported by two simply supported steel beams that
are the length and width of the tank’s tracks. In this scenario, there is a land mine flush
with the ground under the center of each beam. This makes the maximum deflection of
the beam under the load of the M1 Abrams tank of specific interest to determine the
required height of the beam from the ground in order to avoid setting off the land mine,
as presented in Figure 1.
Figure 1 – Sketch of Side View of M1 Abrams Tank on Beam over Land Mine
Page |4
As provided in Reference (a), the properties of the M1 Abrams Tank include a weight of
61326kg, hull/track length of 7.93m and a track width of 0.635m. In this study, it is
assumed that the weight of the tank is evenly distributed, and that each beam supports
half the weight of the tank. Therefore, the study requires analysis of only one beam with
half the weight of the tank. Simplifying the study by rounding some of the specifications
of the tank, the properties of the tank and beam used in this study are presented in Tables
1 and 2 respectively below:
Table 1 - M1 Abrams Tank Properties
M1 Abrams Tank
Description
Value
Unit
Force 300000
N
Hull/Track Length
8
m
Track Width
0.6
m
Table 2 - Beam Properties
Beam
Description
Value
Unit
Length
8
m
Width
0.6
m
Thickness
0.1
m
Young's Modulus (E) 2.00E+11
Pa
Poisson's Ratio (n)
0.3
Density (r )
7800
kg/m3
Of additional interest is the difference in maximum deflection of the beam when
assuming all of the force due to the weight the tank as a point load at the center of the
beam versus a uniformly distributed load across the beam. These two cases are presented
in Figures 2 and 3 respectively.
Figure 2 – Simply Supported Beam with Point Load at the Center
Figure 3 – Simply Supported Beam with Uniformly Distributed Load
Page |5
FORMULATION AND SOLUTION
Two methods were utilized for solution of the maximum deflection of the beam in this
study. The analytical method through the use of hand calculations and MAPLE
determined the maximum deflection of the beam subjected to the force, due to half of the
weight of the M1 Abrams tank, as a point load at the center of the beam and as a
uniformly distributed load across the beam. A modeling/numerical method through the
use of COMSOL also obtained the maximum deflection of the beam for both cases. The
other program used to calculate and tabulate the results for this study was Microsoft
Excel 2010.
ANALYTICAL METHOD:
For the analytical method, the system was setup starting with the Moment-Curvature
Equation, provided in Reference (b) and expanded upon in Reference (c), is represented
in Equation 1 below:
𝑑2 𝑦
𝐸𝐼 𝑑𝑥 2 = 𝑀
Eq. 1
Where “E” is Young’s Modulus and “I” is the Moment of Inertia. To determine the
Moment of Inertia of the rectangular cross section of this beam, Figure 4 and Equation 2
below are utilized, where in this study b=0.6m and h=0.1m.
Figure 4 – Moment of Inertia of a Rectangular Cross Section of a Beam
𝐼=
𝑏ℎ3
Eq. 2
12
For the case of a simply supported beam with a point load at the center (Figure 2), the
bending moment at any position “x” up to the center of the beam is given by Equation 3
below:
𝑀=
𝑃𝑥
Eq. 3
2
Equation 4 below is the result of substituting Equation 3 into Equation 1. Integrating
Equation 4 with respect to x once and twice result in Equations 5 and 6 respectively:
𝑑2 𝑦
𝐸𝐼 𝑑𝑥 2 =
𝑃𝑥
2
Eq. 4
Page |6
𝑃𝑥 2
𝑑𝑦
𝐸𝐼 𝑑𝑥 =
𝐸𝐼𝑦 =
𝑃𝑥 3
12
4
+ 𝑐1
+ 𝑐1 𝑥 + 𝑐2
Eq. 5
Eq. 6
The constants of integration, c1 and c2, may be determined from the following boundary
conditions:
a) Due to no deflection at the end of the beam: At x=0, y=0
b) Tangent to the beam is horizontal at the center of the beam: At x=L/2, dy/dx=0
Therefore, substituting boundary condition “b” into Equation 5 results in Equation 7
below:
𝐿
𝑃( )2
2
𝐸𝐼(0) =
𝑐1 = −
4
+ 𝑐1
𝑃𝐿2
Eq. 7
Eq. 8
16
Thus, the constant c1 is solved in Equation 8. Substituting Equation 8 and boundary
condition “a” into Equation 6 results in Equations 9 and 10 below to determine the
constant c2:
𝐸𝐼(0) = 𝑐2
Eq. 9
𝑐2 = 0
Eq. 10
Substituting the solved constants of integration, c1 and c2, into Equation 6 results in
Equation 11 below:
𝐸𝐼𝑦 =
𝑃𝑥 3
12
−
𝑃𝐿2
16
𝑥
Eq. 11
Substituting x=L/2 into Equation 11 provides the standard equation of the maximum
deflection at the center of the beam due to a Point load, as presented in Equation 12
below and verified in Appendix E2 of Reference (b).
𝑃𝐿3
𝛿𝑚𝑎𝑥 = 𝑦 = − 48𝐸𝐼
Eq. 12
For the case of a simply supported beam with a uniformly distributed load (Figure 3), the
bending moment at any position “x” is given by Equation 13 below:
𝑀=
𝑤𝐿𝑥
2
−
𝑤𝑥 2
2
Eq. 13
Equation 14 below is the result of substituting Equation 13 into Equation 1. Integrating
Equation 14 with respect to x once and twice result in Equations 15 and 16 respectively.
Page |7
𝑑2 𝑦
𝐸𝐼 𝑑𝑥 2 =
𝑑𝑦
𝐸𝐼 𝑑𝑥 =
𝐸𝐼𝑦 =
𝑤𝐿𝑥 2
4
𝑤𝐿𝑥 3
12
𝑤𝐿𝑥
2
−
−
𝑤𝑥 2
−
𝑤𝑥 3
6
𝑤𝑥 4
Eq. 14
2
+ 𝑠1
Eq. 15
+ 𝑠1 𝑥 + 𝑠2
24
Eq. 16
The constants of integration, s1 and s2, may again be determined from the following
boundary conditions:
a) Due to no deflection at the end of the beam: At x=0, y=0
b) Tangent to the beam is horizontal at the center of the beam: At x=L/2, dy/dx=0
Therefore, substituting boundary condition “b” into Equation 15 results in Equation 17
below:
𝐸𝐼(0) =
𝑠1 = −
𝐿
2
𝑤𝐿( )2
4
−
𝐿
2
𝑤( )3
6
+ 𝑠1 Eq. 17
𝑤𝐿3
Eq. 18
24
Thus, the constant s1 is solved in Equation 18. Substituting Equation 18 and boundary
condition “a” into Equation 16 results in Equations 19 and 20 below to determine the
constant s2:
𝐸𝐼(0) = 𝑠2
Eq. 19
𝑠2 = 0
Eq. 20
Substituting the solved constants of integration, s1 and s2, into Equation 16 results in
Equation 21 below:
𝐸𝐼𝑦 =
𝑤𝐿𝑥 3
12
−
𝑤𝑥 4
24
−
𝑤𝐿3 𝑥
24
Eq. 21
Substituting x=L/2 into Equation 21 provides the standard equation of the maximum
deflection at the center of the beam due to a uniformly distributed load, as presented in
Equation 22 below and verified in Appendix E2 of Reference (b).
5𝑤𝐿4
𝛿𝑚𝑎𝑥 = 𝑦 = − 384𝐸𝐼
Eq. 22
These equations and required parameters were input into a MAPLE worksheet solution
and also calculated by hand. Both the MAPLE worksheet solution and hand calculations
have been provided in Appendix B.
Page |8
MODELING/NUMERICAL METHOD USING COMSOL:
For the modeling/numerical method, the system was setup in COMSOL using the 2D
Structural Mechanics, Solid Mechanics and Stationary presets. Rectangular geometry
was used to build the length (8m) and thickness (0.1m) of the beam, with the width
(0.6m) specified as the “Thickness” in the “Solid Mechanics” tab. A material (steel) was
then created and assigned to the rectangle, with the critical properties of Young’s
Modulus, Poisson’s Ratio and Density set to the values in Table 2. Next, prescribed
displacements of 0m were set for both x and y directions on the bottom left corner of the
rectangle, and the y direction on the bottom right corner of the rectangle. The purpose of
this is to set the boundary conditions of a simply supported beam, as previously presented
in Figures 2 and 3.
For the case of the point load at the center of the beam, a point was created at the center
of the top boundary of the rectangle under the “Geometry” tab. The purpose of this is to
allow for an application point for the point load of 300000N. This was prescribed using
the point load feature under the “Solid Mechanics” tab, and was designated as negative to
reflect the downward direction.
For the case of the uniformly distributed load across the beam, the force per unit length
must be determined. The 300000N is first divided by the square area of length x width of
the beam, which is 4.8m2. This results in a force per unit area of 62500N/m2, which is a
value that would be useful in a 3D model. To determine the force per unit length, the
force per unit area is multiplied by the width of the beam (0.6m), which results in
37500N/m. This force per unit length was prescribed using the boundary load feature
under the “Solid Mechanics” tab on the top boundary of the rectangle, and was
designated as negative to reflect the downward direction.
For both cases, the mesh is addressed under the “Mesh” tab, and the sequence type was
left as Physics-controlled. The default element size provided by COMSOL is the Normal
element size. However, a total of five different element sizes were analyzed for both
cases as part of the mesh extension study. Along with the Normal element size, two finer
meshes (Finer and Extremely Fine) and two coarser meshes (Coarser and Extremely
Coarse) were also analyzed. After each mesh was built, the study was computed,
solution determined and results output. In addition to the default von Mises stress plot, a
1D plot was created to capture the displacement of the beam across the length of the
beam.
RESULTS AND DISCUSSION
The results of the modeling/numerical method using COMSOL to solve for the maximum
deflection of the beam when assuming all of the force of the tank as a point load at the
center of the beam and for a uniformly distributed load across the beam are summarized
in Tables 3 and 4 respectively.
Page |9
Table 3: Simply Supported Beam with Point Load at the Center
Degrees
Max
Min
of
Triangular Edge
Vertex Element Element Freedom
Size
Size
Solved x max (m)  max (m)  max (cm)
Mesh
Nodes Elements Elements Elements
Extremely Fine 511
414
208
5
0.08
1.60E-04
2074
4.0
-0.29131 -29.131
Finer 349
120
118
5
0.296
0.001
718
4.0
-0.29125 -29.125
Normal 349
120
118
5
0.536
0.0024
718
4.0
-0.29125 -29.125
Coarser 349
120
118
5
1.04
0.048
718
4.0
-0.29124 -29.124
Extremely Coarse 313
103
105
5
2.64
0.4
624
4.0
-0.29123 -29.123
Table 4: Simply Supported Beam with Uniformly Distributed Load
Degrees
Max
Min
of
Triangular Edge
Vertex Element Element Freedom
Size
Size
Solved x max (m)
Mesh
Nodes Elements Elements Elements
Extremely Fine 511
414
208
4
0.08
1.60E-04
2074
4.0
Finer 349
120
118
4
0.296
0.001
718
4.0
Normal 349
120
118
4
0.536
0.0024
718
4.0
Coarser 349
120
118
4
1.04
0.048
718
4.0
Extremely Coarse 313
103
105
4
2.64
0.4
624
4.0
 max (m)  max (cm)
-0.18206 -18.206
-0.18202 -18.202
-0.18202 -18.202
-0.18203 -18.203
-0.18202 -18.202
The modeling/numerical method shows that for both cases, the Nodes and Degrees of
Freedom solved at any specific mesh size level remained the same. The only difference
is that the case with the point load at the center has an extra vertex element due to the
point created for the application of the load. As expected, the maximum and minimum
element size decrease while the number of nodes, elements and degrees of freedom
solved increase as one moves from the extremely coarse mesh to the extremely fine
mesh. It is noted that while the maximum and minimum element size decreased between
the coarser, normal and finer meshes, the number of nodes (and likewise elements and
degrees of freedom) remained the same. An example of the COMSOL Normal mesh of
the beam is presented below in Figure 5. In addition, plots of the maximum deflection
(max) for the COMSOL Normal mesh for both cases have been provided in Figures 6 and
7. Results show that the maximum deflection of the beam was always located in the
center of the beam at x=4m. The COMSOL mesh extension study shows that the
difference between the extremely coarse mesh and extremely fine mesh is 0.027% and
0.022% for the point load at the center and uniformly distributed load respectively. With
the COMSOL Normal mesh, the maximum deflection of the beam was -0.291m and 0.182m for the point load at the center and uniformly distributed load respectively.
P a g e | 10
Figure 5 – COMSOL Normal Mesh of Simply Supported Beam
Figure 6 – COMSOL Maximum Deflection (max) of Simply Supported Beam with Point
Load at the Center
P a g e | 11
Figure 7 – COMSOL Maximum Deflection (max) of Simply Supported Beam with
Uniformly Distributed Load
The analytical results of the maximum deflection of the beam in both cases between the
hand calculations and MAPLE worksheet were identical (-0.32m and -0.20m for the point
load at the center of the beam and uniformly distributed load across the beam
respectively), as provided in Appendix B. A comparison of the results of the analytical
method and modeling/numerical method to solve for this study are summarized in Table
5 below.
Table 5: Comparison of COMSOL Modeling/Numerical and Analytical Method Results
Analytical
Mesh
Nodes Total Elements  max (m) Result  max (m) % Error
COMSOL
Normal Mesh Point Load 349
243
-0.29125
-0.32
-8.985%
COMSOL
Normal Mesh Uniformly
Distributed
Load 349
242
-0.18202
-0.20
-8.988%
While the percentage of error is near identical between the two cases, there is
approximately a 9% error between the modeling/numerical method and analytical
method. In both cases for the modeling/numerical results, the magnitude of maximum
deflection of the beam is lower than the analytical result. For comparison, a 1 meter long
P a g e | 12
beam with a 0.1m x 0.1m rectangular cross section was also modeled in COMSOL, and
both cases produced a 5% error from the analytical method using an extremely fine mesh.
This appears to stipulate that the COMSOL modeling/numerical method may be more
accurate for a proportionally smaller length to cross section ratio of the beam.
For additional comparison, an ANSYS beam element model was created and analyzed,
with the assistance of the tutorial provided in Reference (d), for the case of a simply
supported beam with a uniformly distributed load. The results of this ANSYS beam
element model analysis are provided in Table 6. A plot of the beam’s deflection and
mesh with a similar number of nodes compared to the COMSOL normal mesh is
provided in Figure 8. Through the mesh extension validation in the ANSYS model, the
results show percentage of error from the analytical result to be on an order of 50 to 200
times smaller than that of the COMSOL results. With the 5 element sizes meshed in the
ANSYS model, the largest magnitude of error was 0.154%, with a majority of the
magnitudes of error at 0.038%. It is observed that while both the ANSYS and COMSOL
meshes are in a structured grid, the cells of the ANSYS mesh is quadrilateral while the
cells of the COMSOL mesh are triangle. It was also observed that when an odd number
of elements resulted due to the input element size, the result would fluctuate a minuscule
amount from the converging maximum beam deflection value of 0.200077m.
Table 6: Comparison of ANSYS Modeling/Numerical and Analytical Method Results
Analytical
Result
Element Size Nodes Total Elements  max (m)  max (m) % Error
0.05
481
160
-0.20008
-0.20
0.038%
0.075
322
107
-0.20006
-0.20
0.028%
0.1
241
80
-0.20008
-0.20
0.038%
0.33
76
25
-0.19969
-0.20
-0.154%
1
25
8
-0.20008
-0.20
0.038%
P a g e | 13
Figure 8 – ANSYS Maximum Deflection (max) of Simply Supported Beam with Uniformly
Distributed Load
CONCLUSIONS
From this study, it may be concluded that the maximum deflection of a simply supported
elastic beam subject to point or distributed loads may be achieved using either the
modeling/numerical or analytical methods. An ANSYS model using a quadrilateral cell
mesh was within 0.028% of the analytical result. However, it appears that the shape of
the cells for the mesh is a major factor in the accuracy of the maximum beam deflection
results. This was apparent when the COMSOL models using a triangle cell mesh
reported results with a 9% error from the analytical solutions. Therefore, when using the
modeling/numerical approach with beams and deflection, a quadrilateral cell mesh may
offer the most accurate solution.
In this scenario of an M1 Abrams tank equally supported by two simply supported steel
beams with a land mine flush with the ground under the center of each beam, uniformly
distributing the load across the beam gives a more accurate representation of the
maximum deflection of the beam. The beam requires a minimum height of 0.2m from
the ground for the tank to avoid setting off the land mine. However, using a point load to
determine the maximum deflection of the beam would be a more conservative approach
P a g e | 14
to determine the minimum height of the beam from the ground to avoid setting off the
land mine. It is noted that if the results from the COMSOL analysis of the uniformly
distributed load across the beam were used without a factor of safety greater than 1.1 for
the height of the beam from the ground, the maximum deflection due to the tank would
set off the land mine. This highlights the necessity for verifying the reliability of the
approximate solution by comparing the results to a theoretical/exact solution, a different
modeling approach or a mesh extension validation.