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condition for the extremum of a function of several variables is the zeroing of partial derivatives of the first order, i.e.:

E

0

i 1, 2 ,..., r .

(15.15)

a

 

 

 

i

 

 

 

Solving the system of equations (15.15), we find the values of the parameters ai , and hence obtain an approximate solution of the stationarity

conditions E 0 .

15.7. Calculating Elastic Systems Based on the Displacement Variation Principle

The displacement of any point (cross-section) of an element (Figure 15.10), taking into account generally accepted assumptions, can be unambiguously expressed through nodal (generalized) displacements. Thus, the horizontal displacement of cross-section C, as follows from Figure 15.11 can be determined by the formula:

u Z

1

 

x

 

Z

4

x

Z

f (x) Z

4

f

4

(x) ,

(15.16)

 

 

1

 

 

 

 

 

l

1

1

 

 

 

 

 

 

l

 

 

 

 

 

 

 

 

 

where f1(x) , f4 (x) are the basis (coordinate) functions.

Figure 15.10

Figure 15.11

To determine the displacements caused only by nodal displacements Z2 , Z3 , Z5 and Z6 , we use the differential equation:

d 4v 0 . dx4

461

Its general solution has the form:

v C1x3 C2 x2 C3x C4 .

We find, for example, the equation of the bended axis caused by loading the bar in the form Z2 1 . The boundary conditions for this case:

x 0

v 1;

x 0

v 0 ;

x l

v 0;

x l

v 0 .

Having solved the fourth-order system of equations, we obtain the values of arbitrary constants C1 , C2 , C3 , C4 . The equation of the bend-

ed axis is written as:

v 1

3x2

 

2x

3

l2

l3

.

 

 

 

For other unit nodal displacements of the bar clamped at the ends, the deflection curves are recorded in Table. 15.1.

 

 

 

 

 

 

 

 

Table 15.1

The scheme of the bar.

 

 

Bended axis

 

 

Kind of displacement

 

 

equation

 

 

 

 

 

 

 

 

 

1

 

2

 

 

3

 

 

 

 

 

f2 x

 

 

 

 

3x2

 

2x3

1

 

 

f2

(x) 1

 

1

x

 

l2

 

l3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

f3 x

 

 

 

 

 

2x

2

 

x3

2

 

 

f3

(x) x

 

1

x

l

 

l2

 

 

 

 

 

 

 

462

Table 15.1 (ending)

1

 

2

 

 

 

 

3

 

 

 

 

 

 

 

 

 

f5 x

 

 

 

 

3x2

 

 

 

2x3

3

 

1

x

 

f5

(x)

 

 

 

 

l

2

 

l3

 

 

 

 

 

 

 

 

 

 

 

 

f6 x

 

 

 

 

 

 

 

 

 

 

 

 

4

 

 

x

 

f6 (x)

x

2

 

x3

 

1

 

 

l

l

2

 

 

 

 

 

 

 

 

 

 

 

 

f7 x

 

 

 

 

 

 

 

 

 

 

 

 

5

1

 

 

f7

(x) 1

3x2

 

 

 

x3

 

 

x

2l

2

 

2l3

 

 

 

 

 

 

 

 

 

1

f8 x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3x2

 

 

 

x3

6

 

 

 

 

f8(x) x

 

 

 

 

x

 

2l

 

 

2l2

 

 

 

 

 

 

 

 

 

 

 

f9 x

1

 

f9

(x)

3x2

 

x3

7

 

x

2l2

2l3

 

 

 

 

 

Using the forces action independence principle, the vertical displacement of the cross-section C (Figure 15.9) can be represented as:

v Z2

 

 

3x2

 

2x3

 

Z3

 

 

 

2x

2

 

1

 

2

 

 

3

 

x

 

 

 

 

 

 

 

l

 

 

l

 

 

 

 

 

 

l

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3x2

 

2x

3

 

 

 

 

x

2

 

 

x3

Z5

 

 

 

 

 

 

 

 

Z6

 

 

 

 

 

 

 

 

 

2

 

3

 

 

 

 

 

 

2

 

 

l

 

l

 

 

 

 

 

 

l

 

 

 

l

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Z2 f 2 (x) Z3 f 3(x) Z5

f 5(x) Z6

x3 l2

(15.17)

f 6 (x).

463

The expression for determining the rotation angle of the cross-section is obtained by differentiating v v(x) relative to x .

For a bar clamped at one end and hinge-supported at the other end, the deflection curves for unit nodal displacements and the corresponding displacements functions are also shown in Table. 15.1.

In the general case, for a discrete system, the expression for determining the displacement of a certain point can be represented as:

Z Z1 f 1(s) Z2 f 2 (s) ... Zn f n (s) , (15.18)

where fi s are the basic functions corresponding to the generalized displacements Zi .

The number of such equations corresponds to the number of deformable elements of the system and the number of displacements types (linear, angular).

For bars systems, these equations will be accurate, for continuum systems they will be approximate.

In connection with the above, the total energy of the system can be represented as a function of n generalized displacements (coordinates) and load:

E E(Z 1 , Z 2 ,..., Z n , F) .

Then the stationarity condition:

E

E

Z

 

E

Z

 

...

E

Z

Z

Z

 

 

Z

 

 

1

 

2

 

2

 

n

 

 

1

 

 

 

 

 

 

 

 

with independent variations Zi and constant load get n equations to determine Zi :

E

 

 

U

 

 

 

P

 

0,

Z

Z

Z

1

 

 

 

 

 

 

 

1

 

 

1

 

 

 

 

 

 

 

. . . . . . . .

E

 

 

U

 

P

 

0.

 

 

Z n

 

Zn

 

 

 

Zn

 

 

 

 

n 0

F will allow us to

(15.19)

464

For a linearly elastic system, the total energy is calculated by the formula (15.13), therefore, equations (15.19) in the expanded form of writing take the form of canonical equations of the displacement method:

r11Z 1 r12 Z 2 ... r1n Z n R1F 0,

. . . . . . . . . . .

rn1Z 1 r n2 Z 2 ... r nn Z n RnF 0.

For nonlinearly deformable systems, equations (15.19) will be nonlinear relative to Zi .

Due to the characters of equations (15.18) for continuum systems, only approximate values Zi may be determined. In this case, equations

(15.19) are called the equations of the Ritz method, which, as it was noted earlier, refers to direct methods of variations calculus.

15.8. The Essence of the Finite Element Method

The finite element method (FEM) is an effective numerical method for solving applied problems and is widely used to analyze various structures. This method is well adapted to computer implementation. According to a single methodology, bars, plates, massives and combined structural systems are calculated. Its essence is as follows. The system under study is mentally divided into many finite elements (disjoint areas), that is, a transition is made from a given design scheme to a discrete one.

The shape of the finite element (FE) is predetermined by the features of the analysed object (system, structure). In bars systems, FE is taken as a bar (as a rule) with constant longitudinal and bending rigidity. For plates and thin-walled spatial continuum systems, triangular or rectangular (ge-nerally quadrangular) finite elements are most often used; for solving three-dimensional problems, volume finite elements in the form of a tetrahedron or parallelepiped are used. The choice of the shape and sizes of finite elements has a significant effect on the calculation results. But the results, of course, should allow you correctly evaluate the stressstrain state of the initial system. Representation of the system under study as a sufficiently large set of finite elements leads to an increase in the calculation accuracy, but significantly increases the dimension of the problem, what is associated with a significant amount of calculations.

465

Readers who are interested in questions of estimation of discretization error can find relevant recommendations in the scientific literature.

The points at which FEs are connected are called nodes. We need to distinguish rigid and hinge nodes. In a rigid node, it is assumed that there are constraints ensuring the continuity of linear and angular FE displacements adjacent to this node. The constraints in the hinge node allow saving the continuity of linear movements. Nodal displacements and the corresponding nodal forces are taken as generalized.

The idea of the method is to describe the stress-strain state of the FE through generalized displacements Z of nodes and to establish their connection with the load acting on the system. To implement this idea, it is necessary to obtain a FE stiffness matrix.

Since the displacements function of the original system is unknown, it must be set. If in the Ritz method it was assumed that the basis functions are determined by one expression on the entire region of the system, then an alternative approach is implemented in the FEM. It consists in the fact that at each FE unknown functions of displacements are replaced by approximating ones so that the displacements of all element points are expressed through the nodal ones. For one-dimensional elements, taking into account the remark about the rigidity constancy, the displacement function is accurate (see Section 15.7), for twodimensional and three-dimensional FEs, these functions are written approximately, most often in the form of polynomials. Their selection is a rather difficult task. The accuracy of the final results substantially depends on the successful solution of this problem. Using approximating functions and based on the variational principles of structural mechanics, one of the main tasks of the FEM is solved – the determination of stiffness matrices of finite elements.

Since each bar in the composition of the system under study has its own orientation, stiffness matrices are first constructed in the local coordinate system, and then, when moving from the local system to the general, they are transformed. The stiffness matrix of the entire system is obtained by the corresponding combination of stiffness matrices of individual elements.

The resolving equations of the FEM are written in the form:

R Z RF 0 ,

(15.20)

466

where RF is the vector of reactions caused by the given load, which is

equal to the vector of nodal loads taken with the opposite sign.

The total load on the node is defined as the sum of the loads from the elements adjacent to the node. Since the non-nodal load is replaced by

the equivalent nodal load in the directions Z , then the vector of rections

RF R1F , R2F ,..., RnF T F ,

where

F F1 , F2 ,..., Fn T is the vector of nodal loads.

After solving the system of equations (15.20), the displacements Z of nodes in the general coordinate system become known. To calculate the forces in a finite element, it is convenient to first find the nodes displace-

ment vector Z in the local coordinate system, and then determine the reactions at the ends of the FE using the stiffness matrix R . For-mulas for the corresponding transformations are given in sections 15.9 and 15.10.

This form of calculation corresponds to the FEM variant “in displacements”. It is the most common form.

Another approach to solving the problem with the help of FEM is also possible. The stress-strain state of the FE must be described by a finite set of generalized nodal forces, and then establish their relationship with the load. This form of calculation corresponds to the FEM “in efforts”.

15.9. Bar Stiffness Matrix in the Local Coordinate System

There are several ways to obtain stiffness matrices of separate bars. One of the simplest is a method based on the known conditions of the displacement method.

Each end of the bar adjacent to the rigid node has three degrees of freedom: linear displacements in the horizontal and vertical directions and the angle of rotation. The force factors corresponding to these dis-

placements are the reactive forces R1 , R2 , R4 , R5 and the moments R3, R6 , located at the FE edges. (Displacements, reactions, stiffness

matrix and its elements in the local coordinate system are indicated by letters with strokes). The stiffness matrix (the matrix of unit reactions) converts the displacement vector

467

Z Z1, Z2 , Z3, Z4, Z5, Z6 T

into the vector of reactions at the ends of FE, i.e., there is the relation:

 

 

 

 

R1

r11

R

 

r

 

2

 

21

 

 

 

.

R

 

r

 

6

 

61

r12 r22

.

r62

 

 

 

 

 

 

 

 

r16

Z1

 

 

 

r

 

Z

 

 

.

 

26

 

 

2

 

R Z

. .

 

 

 

 

 

 

r

 

Z

 

 

 

 

66

 

 

6

 

 

 

The positive directions of the reactions Ri correspond to the positive directions of Zi .

Elements of the matrix R are reactions in constraints caused by unit displacements Zi 1 (Figure 15.12).

In the first column, the reaction values caused by Z1 1 are recorded, in the second column – caused by Z2 1 etc. Therefore, to calculate the

elements of R matrix, we can use the data from the table used in the displacement method to determine the reactions in the supports of a bar with constant cross-section.

The matrix R has the form:

 

EA

 

 

 

 

 

EA

 

 

 

 

 

 

 

 

 

l

 

 

 

 

l

 

 

 

 

 

 

 

 

 

12EJ

6EJ

 

 

12EJ

6EJ

 

 

 

 

 

 

 

 

 

 

 

l3

 

l2

 

 

l3

 

l2

 

 

 

 

 

6EJ

4EJ

 

 

 

6EJ

2EJ

 

 

 

 

 

 

 

 

 

 

 

 

l2

 

l

 

l2

 

l

 

 

R

 

EA

 

 

 

 

EA

 

 

 

 

 

 

. (15.21)

 

l

 

 

 

 

 

l

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

12EJ

 

6EJ

 

12EJ

 

6EJ

 

 

 

 

l3

l2

 

 

 

l3

l2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6EJ

2EJ

 

 

6EJ

4EJ

 

 

 

 

 

l

2

 

l

 

l

2

 

l

 

 

 

 

 

 

 

 

 

 

 

 

 

 

468

Figure 15.12

For bars with other fixing conditions, matrix R elements are calculated similarly.

469

Consider methods based on the use of approximating displacement functions. The necessary calculation procedure for one of them, for example, for a bar fixed at the ends is as follows.

1. In a linearly deformable bar the longitudinal and transverse displacements of the cross-sections are not interconnected. Therefore, the functions that describe the character of the change in displacements along the length of the bar will be different for them. In accordance with the differential equation

N E A u ,

we will approximate the displacements of the bar cross-sections along its axis by a linear function:

u(x) a1 a4 x .

(15.22)

The bent axis of the bar in the absence of distributed load along its length is described by a third-order curve, which is a consequence of the differential equation

IV 0 .

Therefore, the approximating polynomial of the third degree allows us to precisely set the function of the displacements.

Let us assume that

v(x) a

2

a

3

x a

x2 a x3 .

(15.23)

 

 

5

6

 

There are unknown parameters ai in expressions (15.22) and (15.23);

their number is equal to the number of degrees of freedom. Functions u(x) and (x) are called form functions

The rotation angle of the bar cross-section is determined by the value of the first derivative:

d a

3

2a x 3a

6

x2 .

(15.24)

dx

5

 

 

 

 

 

 

 

470

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