Double Clamped and Cantilever Beam Theoretical Solution and Numerical Solution by Finite Element Method

Double Clamped and Cantilever Beam Theoretical Solution and
Numerical Solution by Finite Element Method
Anastasios S. Lazaridis
M. Eng. in Electrical Engineering & Computer Science
tasoslazaridis13@gmail.com
Abstract—This paper presents a static analysis of a beam
made of Aluminum. Two cases are studied, the Double Clamped
Beam and the Cantilever Beam problem. The displacement of the
beam under a load is calculated theoretically using Calculus of
Variations and numerically using Finite Element Method (FEM).
Numerical results are obtained by ANSYS and SOLIDWORKS
software. All results are compared and relative differences are
calculated.
Keywords—static analysis; calculus of variations; FEM;
ANSYS Workbench; beam displacement; SolidWorks;
I. THEORETICAL PROBLEM AND SULUTION
A. Definition of the Problem
Suppose a beam of length l having uniform cross section
under a load. Let  3
: 0,y l   , describe the shape of the
axis of the beam and  3
: 0,l   be the load per unit
length on the beam. Assuming small deflections, the potential
energy from elastic forces [2] is the functional  3
: 0,eV l  
2
2
2
0 0
2 2
l l
e
EI d y EI
V dx y dx
dx
   (1)
where E is the modulus of elasticity and I is the moment of
inertia, both are constants. The potential energy from
gravitational forces [2] is the functional  3
: 0,gV l  
0
l
gV ydx  (2)
The total potential energy is the functional  3
: 0,V l  
  2
0
, ,
2
l
EI
V V x y y y y dx
 
   
   (3)
The shape of the beam is such that V is minimum. The
Lagrangian function  3
0,f l is
  2
, ,
2
EI
f x y y y y   (4)
must satisfy the Euler-Lagrange equation [1]. The Euler-
Lagrange Equation for a functional involving second-order
derivatives is written
2
2
0
d f d f f
dx y dx y y
  
  
    (5)
along with the condition
3
0
0 ,
x l
x
f d f f
y dx y y
  


   
     
   
 
  
(6)
The above equation spawns the four boundary conditions
0
0
x
f
y 



(7)
0
x l
f
y 



(8)
0
0
x
d f f
dx y y 
  
  
   
(9)
0
x l
d f f
dx y y 
  
  
   
(10)
This produces the differential equation
 
 
 (4)
, 0,
x
y x x l
EI

  (11)
The general solution of the differential equation (11) is the
function
     3 2
+ + , 0,y x h x Ax Bx Cx D x l    (12)
where , , ,A B C D are constants and  h x is a partial solution of
the differential equation (11), where
 
 (4)
=
x
h x
EI

(13)
B. Definition of the load
In this paper constant load case is studied
   , 0,x P x l     (14)
So the total impressed force on the beam is
 
0
l
F x dx Pl  (15)

C. Moment of Inertia
The moment of inertia of an area with respect to an axis is
the sum of the products obtained by multiplying each element
of the area dA by the square of its distance from the axis [3].
For a section in the yx plane, the moment of inertia, about y
and x axis respectively, is defined to be the double integral
2
y
A
I x dxdy  (16)
2
x
A
I y dxdy  (17)
In the case of rectangular area centered at the origin of yx
plane (Fig. 1), A is the region , ,
2 2 2 2
a a b b
A
   
      
   
. So yI
and xI are calculated
3
2
12
y
A
a b
I x dxdy  (18)
3
2
12
x
A
ab
I y dxdy  (19)
Fig. 1. Rectangular area centered at the origin of yx plane.
II. BEAM CAD MODEL
The CAD model of the beam is shown in Fig. 2. The
material of the beam is Aluminum Alloy AA380.0-F die. The
dimensions of the beam are shown in figure and the material
properties are shown in Table I.
Fig. 2. The CAD model of the beam made of Aluminum and the dimensions
of the beam.
TABLE I. MATERIAL PROPERTIES OF THE ALUMINUM BEAM
Material Properties
Property Value Unit
Elastic Modulus 71000 N/mm2
Poisson’s Ratio 0.33 N/A
Shear Modulus 26500 N/mm2
Mass Density 2760 kg/m3
Tensile Strength 317 N/mm2
Yield Strength 159 N/mm2
Thermal Conductivity 109 W/(m·K)
Specific Heat 963 J/(kg·K)
III. DOUBLE CLAMPED BEAM
A. Theoretical Solution
Suppose that the beam is clamped at each end (Fig. 3). The
beam is “fixed in the wall” at each end so that at 0x  and
x l we have respectively    0 0 0y y  and     0y l y l  .
Fig. 3. The beam is clamped at each end.
Here, there are four imposed boundary conditions. All the
allowable variations in this problem require that
   0 0 0   and     0l l   so that no natural boundary
conditions (6) arise.
So the problem is subscribed by the set of equations below
   
   
   
(4)
, 0,
0 0 0
0
Pl
y x x l
EI
y y
y l y l
 
  
 
  
  
 
 


(20)
The general solution of the system of equations (20) is the
polynomial function
   
2
4 3 2
+ + , 0,
24 12 24
P Pl Pl
y x x x x x l
EI EI EI
  (21)
The values of Table II are used to plot the polynomial (21),
which describes the displacement of a double clamped beam.
TABLE II. NUMERICAL VALUES OF THE VARIABLES
Variable Values
Variable Symbol Value Unit
Elastic Modulus E 71000 N/mm2
Area Moment of Inertia I 6666.67 mm4
Length of Beam L 500 mm
Load per unit Length P 10 N/mm
x
y
0x  x l
 p x P
x
y
a
b

0 100 200 300 400 500
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
x - axis [mm]
y-axis-Displacement[mm]
D ouble Clamped Beam - D isplacement
Fig. 4. The displacement of the double clamped Aluminum beam.
B. FEM Solution
ANSYS and SOLIDWORKS software is used to make the
simulations and calculate the solution of the FEM. The same
CAD model is used for both simulations and the same mesh
too (Standard Mesh with Global Size: 1.5mm and Tolerance:
0.075mm).
1) ANSYS Solution
ANSYS solution is shown below in Fig.5. ANSYS
Structural Toolbox was used to simulate the model.
Fig. 5. The displacement of the double clamped Aluminum beam using
ANSYS.
2) SOLIDWORKS Solution
SOLIDWORKS solution is shown below in Fig.6.
SOLIDWORKS Static Simulation Toolbox was used for the
simulation.
Fig. 6. The displacement of the double clamped Aluminum beam using
SolidWorks.
IV. CANTILEVER BEAM
A. Theoretical Solution
Suppose that the beam is clamped at 0x  (Fig. 7). The
boundary conditions    0 0 0y y  are imposed. No
boundary conditions are imposed at the other end of the beam
and consequently, the natural boundary conditions (8) and (10)
must be satisfied. The assumption here is made that the
unclamped endpoint of the beam still lies on the line x l
(small deflections).
Fig. 7. Cantilever beam.
Equation (8) yields the relation
 0 0
x l
f
EI y l
y 

  



(22)
Equation (10) gives the condition
 0 0
x l
d f f
EI y l
dx y y 
  
    
  

 
(23)
In this case the problem is described by the set of equations
   
   
   
(4)
, 0,
0 0 0
0
Pl
y x x l
EI
y y
y l y l
 
  
 
  
  
 
 

 
(24)
The general solution of the system of equations (24) is the
polynomial function
   
2
4 3 2
+ , 0,
24 6 4
P Pl Pl
y x x x x x l
EI EI EI
   (25)
The values of Table II are used to plot the polynomial (25),
which describes the displacement of a cantilever beam.
x l0x 
y
 p x P
x

0 100 200 300 400 500
-180
-160
-140
-120
-100
-80
-60
-40
-20
0
x - axis [mm]
y-axis-Displacement[mm]
Cant ilever Beam - D isplacement
Fig. 8. The displacement of the cantilever Aluminum beam.
B. FEM Solution
ANSYS and SOLIDWORKS software is used to make the
simulations and calculate the solution of the FEM. The same
CAD model is used for both simulations and the same mesh
too (Standard Mesh with Global Size: 1.5mm and Tolerance:
0.075mm).
1) ANSYS Solution
ANSYS solution is shown below in Fig.5. ANSYS
Structural Toolbox was used to simulate the model.
Fig. 9. The displacement of the cantilever Aluminum beam using ANSYS.
2) SOLIDWORKS Solution
SOLIDWORKS solution is shown below in Fig.6.
SOLIDWORKS Static Simulation Toolbox was used for the
simulation.
Fig. 10. The displacement of the cantilever Aluminum beam using
SolidWorks.
V. RESULTS
A. Double Clamped Beam
TABLE III. DOUBLE CLAMPED BEAM – DISPLACEMENT
Displacement of Double Clamped Beam [mm]
Theoretical ANSYS SolidWorks
3.438 3.489 3.489
TABLE IV. DOUBLE CLAMPED BEAM – RELATIVE PERCENT
DIFFERENCES
Relative Percent Differences
ANSYS Theoretical
SolidWorks 0% 1.462%
ANSYS - 1.462%
B. Cantilever Beam
TABLE V. CANTILEVER BEAM - DISPLACEMENT
Displacement of Cantilever Beam [mm]
Theoretical ANSYS SolidWorks
165.05281 164.88011 157.86049
TABLE VI. CANTILEVER BEAM – RELATIVE PERCENT DIFFERENCES
Relative Percent Differences
ANSYS Theoretical
SolidWorks 4.257% 4.357%
ANSYS - 0.104%
REFERENCES
[1] I. M. Gelfand and S.V. Fomin, “The Canonical Form of the Euler
equation and related topics,” in Calculus of Variations, (Translated and
Edited by Richard A. Silverman), New Jersey: Prentc Hall, 1963.
[2] Bruce van Brant, “Problems with Variable Endpoints,” in The Calculus
of Variations, New York: Springer-Verlag, 2004, pp. 140-145.
[3] Walter D. Pilkey, “Geometric Properties of Plane Areas,” in Formulas
for Stress, Strain and Structural Matrices, 2nd ed., Ney Jersey:John
Wiley & Sons, 2005, pp. 19-21.

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