Solving the problem using finite element method

xb

xa

dx ) x ( q ) x (

v (5.6)

So, the total potential:

    



^{  }

 E I (x) v"(x) dx v(x)q(x)dx Fv(x ) M v'(x ) 2

)1 v

( _{i i j j}

x L

2 z

xb

a

. (5.7)

The 0 criterion of first variation of (5.7) potential leads to the basic equation and to the natural boundary conditions. The approximate solution of the task is the direct minimiza-tion of the total potential energy.

So, find the minimum of the potential (v), and the corresponding v(x) function. This minimization problem is solved by using the Ritz method, when the unknown v(x) function is looking as the following form:







 ⁿ

0 k

k kx a ) x ( ) x (

v (5.8)

where (x) the shape function, which satisfies the kinematical boundary conditions, i.e. the displacements at the supports are (x)0 and the angular displacements at the restrain are

0 ) ( ' x 

 . With this substitution the potential (v) became a multivariable function for a₁, a₂...a_n . This function has a minimum when:

a_k 0





 . (5.9)

Since the Ritz-method is an approximation procedure, the solution accuracy depends on how many members of the shape functions. For simple task enough a single tag, so the above equation depends on the a₀ only, i.e., univariate.

Matrix formulation and solution of the equation system leads to the base equation of the finite element method:

F u

K  . (5.10)

5.2. Solving the problem using finite element method

The problem shown in Figure 5.2, is a two-dimensional rod structure. The structure is over-loaded by two concentrated force lies in plane with the cantilever beam. Compression and

bending generated in the beam 1 of the structure and only bending stress generated in the beam 2, so this problem can not be solved by using TRUSS elements presented in the chapter 3.

Figure 5.2 A two-dimensional rod structure

Both beam of the structure are 60x40x4 box section. The properties of cross-sections are steel standards:

A = 8.69 cm2 I_z = 44.8 cm4.

The two forces are 200 N each.

During the solution we use bent beam elements according to Euler- Bernoulli’s beam theory.

We have seen that the finite element solution means the solution of an equations system:

F u

K  (5.11)

First we have to develop the element stiffness matrix, then assembly the stiffness matrix of total structure.

5.2.1. The element stiffness matrix

The previous equations system written to single element:



The physical interpretation of columns of stiffness matrix is forces and moment necessary to ensure the one unit displacement and the boundary conditions. Using this, we can easily produce the stiffness matrix in case of using beam elements. Let one member of the u vector one unit and all other is zero. In this case the k11 element of stiffness matrix belongs to u1=1 and according to the general procedure:



Solution of the equation system:

F1x=k11 (5.14)

The physical content of this case is illustrated in Figure 5.3.

Figure 5.3 The physical interpretation of the first column of the stiffness matrix

Based on the figure, the individual beam stressed by pure compression, so using the Hook's law:

L E l A E

k₁₁ F^1x     (5.15)

The value of dl is one unit, so that after rearrangement:

L F AE

k₁₁ _1x  (5.16)

To satisfy the boundary conditions still necessary that:

x

The other members of the first column of the stiffness matrix are zero.

We may act similarly with the second column of the stiffness matrix. In this case the equa-tion system:



Solution of the equation system:

F1x=k12 (5.20)

The physical content of this case is illustrated in figure 5.4. This state is produced super-position of cantilever beams. In the first case (see Figure 5.4 b) the end of the beam is loaded by concentrated force and in the other case (see Figure 5.4 c) loaded by concentrated bending moment.

Figure 5.4 The physical interpretation of the second column of the stiffness matrix

These cases are well known in the strength of materials, so we can write it using equations which from solution of the differential equation of the elastic curve:

IE

and the angular displacements:

IE

The solution of the multivariable equation system:

3 22

Furthermore, ensuring the equilibrium conditions:

52

and moments to the 2nd point:

2 62

The first and fourth members in the second column of the stiffness matrix are zero.

Elements in the third column of the stiffness matrix is determined similarly, so that the 1

in v(x) vector is one unit, and all other member 0.



The solution of the equation system is:

F1x=k13 (5.28) F_1y=k₂₃

M₁=k₃₃ F2x=k43

F2y=k53

M₂=k₆₃

The physical content of this case is illustrated in figure 5.5.

Figure 5.5 The physical interpretation of the third column of the stiffness matrix

The displacements presented in figure 5.5 is produced superposition of two displacements in this case too, so:

IE L M IE

L

v F^y

2 0 3

2 1 3 1 2 1

1     (5.29)

and the angular displacements:

IE L M IE

L F_1y ² ₁

2

1 2

1   



   (5.30)

The solution of the multivariable equation system is:

2 23 1

6 k

L

F_y  IE  (5.31)

33 1

4 k

L

M  IE  . (5.32)

Furthermore, ensuring the equilibrium conditions:

53 1 2

2

0 F1 F F F k

F_y   _y _y  _y  _y 

 (5.33)

and moments to the 2^nd point:

63 2 2

2 1

1 2

2 6

0 4 K

L M IE L L

IE L

M IE L F M M

M     _y      

 . (5.34)

The first and fourth members in the third column of the stiffness matrix is zero.

The members in 4^th-6^th column of stiffness matrix are defined similarly. Eventually the en-tire element stiffness matrix:



It should be noted that in generally the element stiffness matrix are generated by:



deformation-strain. This solution found in third chapter. The above presented solution would be difficult in case of using more complex elements. It is only for understanding of the concept of stiffness matrix.

The stiffness properties of the element were determined only in the element local coordi-nate system. In the global coordicoordi-nate system these stiffness values change depending on the position of elements. Elements properties in the global coordinate system are produced using the transformation matrix which was presented in chapter 3 (3.35 equation). However, in this case the transformation matrix is of order 6x6 according to degree of freedom of beam

The members of transformation matrix can be easily calculated by known element nodal coordinates:

i

So the stiffness matrix of 1^st element in global coordinate system:

5.2.2. The entire structure stiffness matrix

The size of the stiffness matrix of entire structure is equal to the number of degrees of free-dom of the whole structure. So now the stiffness matrix of entire system is of order 9x9, be-cause the system composed of two elements with 3 nodes, each with 3 degrees of freedom. In the whole stiffness matrix the elementary stiffness of the common nodes are added together, so:

5.2.3. The complete equations system and the solution

During the solution the displacement 0 locations (at the supports) are skipped. So we can delete rows and columns of the stiffness matrix in these places. In our case, this is the first three rows and columns. Thus we get the condensed stiffness matrix and the equation system to solve:

Substituting the data, solving the equations we obtain:

rad

The reaction forces can be calculated by the known results. From the equations of entire system, in this case, these are the first three lines:

N

Nm 800 k

v k u k

M_R  ¹₃₄ ₂ ¹₃₅ ₂ ¹₃₆₂ 

5.3. Remarks

The program systems based on finite element method can handle not only Euler-Bernoulli's beams. In such a case the shear factor of the section must be determined. It should be noted, this shear factor can only be reliably used in the case of linear static analysis.

6. ANALYSIS OF TWO-DIMENSIONAL BENT BARS USING

In document Finite Element Method (Page 77-87)