Structural mechanics

written from the equations X 0 and Y 0. In the orthogonal di-

rection, vertical, the matrix appears to be divided into blocks columns, the number of which is equal to the number of bars. The width of the blocks column (the number of simple columns in it) is determined by the

length of the vector S for each bar.

Let's compose the equilibrium matrix using the example of a twospan frame (Figure 14.20).

Figure 14.20

The equilibrium matrix A is written in table. 14.3. For easier orientation in the matrix structure, explanatory notes are given in the upper part of the table and to the left of it. The blocks of columns of the table correspond to bars of the frame. The number of columns in the block corresponds to the conditions of adjacency of the separate bar with the nodes of the frame. In each block of columns, the matrix a of an individual bar is located. The upper part of this matrix is connected with the beginning of the bar, and the lower – with its end.

Table 14.3

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For example, the both ends of bar 2 are rigidly fixed at the nodes 2 and 4. Its matrix a has dimensions of 6x3.The first three lines refer to the beginning of the bar, that is, to node 2 (see the horizontal direction in the table), and the remaining lines to node 4.

If support links are imposed on one end of the bar (the bar is adjacent to the support node) and the forces in these links do not need to be calculated (no support reactions need to be determined), then the part of the bar equilibrium matrix a associated with the support node does not fit into the general equilibrium matrix of the A . So, the numerical values of the matrix a for the 3rd bar (5th, 6th and 7th columns) refer only to the 2nd node. A similar distribution of records takes place on the 1st and 4th bars.

The exclusion from the matrix of equations of equilibrium of the support nodes makes it possible to reduce the size of the matrix, which is expedient from a computational point of view..

For a deeper understanding of the physical meaning of the task, and also for the purpose of recording control during manual preparation of the initial data, one should sometimes check the record of individual equilibrium equations for the system nodes. So, for the same frame, the

equation Y 0 for the 2-nd node (Figure 14.21) is written in the form:

0.2425 N1 0.2 M B2 0.2 M E2 N3 20 103 ,

where the substitution

Q2 M E 2 l M B2

2

has already been taken into account.

A similar structure of matrix A takes place for other systems (beams, arches, trusses, etc.).

So, the number of rows in matrix A is equal to the number of equilibrium equations of the system nodes, the number of columns is the

number of unknown efforts (components of the vector S are indicated in the upper part of the table containing the equilibrium matrix).

Returning to the question of the degree of freedom of the system, we note that in our example m 6, n 9. The degree of static indetermina-

cy k n m 3.

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Figure 14.21

For a geometrically unchanged bars system, the rank of the equilibrium matrix is equal to the number of independent equilibrium equations

for the nodes of this system, that is r A m. Moreover, if m n and the determinant of the equilibrium matrix det A 0, then the investigated system is statically determinate; if m n, then the system is statically indeterminate.

each bar. The number of equilibrium equations, the coefficients of which are written in the matrix A , corresponds to the number of determined components of the displacement vector z both for an individual node and for the system as a whole.

The row number in the matrix A also indicates the number of the corresponding component of the vector z. For example, the second row

Y 0 of the matrix A corresponds to the vertical displacement of

node 2 and the second component in the displacement vector z. The total number of unknown displacements for the problem under consideration in the adopted formulation is six.

The matrix of internal stiffness of a single rod is square. Its size is determined by the number of components of the bar vector S. For the en-

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tire system, the internal stiffness matrix K has a quasi-diagonal structure; for the frame under consideration, it is presented in Table 14.4.

The indicated consistency of the matrices in the general equations of structural mechanics allows us to compose an algorithm for solving a mathematical model of the problem of verification calculation of bars systems.

Table 14.4

Note. All elements of the internal stiffness matrix have a factor of 30.24 106

Automated calculation of the bars system involves the formation of matrices of general equations and the solution of the latter based on the initial data on the system, which include:

–the number of nodes, including support, and their signs (properties);

–coordinates of nodes;

–the location of the bars connecting the nodes;

–stiffness of the bars;

–information about the load acting on the nodes.

Since the elements of the matrix are the sines and cosines of the angles of inclination of the bars to the coordinate axes, their calculation is reduced to determining the quotient of dividing the lengths of the projections of the bars on the coordinate axis (the difference in the coordinates of the end and beginning of the bar) by the lengths of the bars.

Depending on the tasks, calculation results can be values of the node displacements, the forces in the bars and their deformations, the matrix of the external rigidity of the system, the matrix of the influence for forces and displacements. This information allows you to identify the features of

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the system under load and can be used to plot diagrams of efforts and displacements, lines of influence of efforts and displacements.

For the frame under consideration, we determine the displacements of the nodes, the internal forces in the bars, and construct the force diagrams.

If we accept

F 0; 20; 4; 5; 10; 4 T 103,

where the dimension of concentrated forces and moments is N and Nm, then from the expression

0.1910; 0.4348; 1.2104; 0.3251; 0.2332; 2.1982 T 10 4.

The sixth component of the vector z corresponds to the angle of rotation of the section at the end of the 2nd bar.

The effort vector is determined by the ratio

S K A z.

S 3.58; 4.95; 0.12; 4.00;

Vertical lines separate the effort components related to specific bars. Diagrams of efforts in the frame are shown in Figure 14.22.

Figure 14.22

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To determine the transverse forces in the bars, the dependence had been used:

Q M E M B . l

Namely

Q4 0.418 0.045 .

E x a m p l e. Determine the forces in the rods of a statically indeterminate truss (Figure 14.23).

The rigidity of the rods are taken equal to:

EA1 4 EA1 3 EA2 4 EA, EA1 2 EA2 3 EA3 4 2EA.

Figure 14.23

The number of unknown efforts n = 6. The degree of kinematic indeterminacy m = 5. Degree of static indeterminacy k n m 1.

The calculation is performed by the displacement method using the general equations of structural mechanics. The primary system and the positive directions of the primary unknowns are shown in Figure 14.24.

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The system of equations of the displacement method is written as:

R z F .

If det R 0 , then the external stiffness matrix R is a nonsingular matrix and can be inverted.

The calculated values of the elements of the inverse matrix R 1 are given in the table below.

Load F = [5, –10, 0, –10, 0]T in the given truss causes node displacements the values of which can be determined with the matrix formula

z R 1F.

The calculated displacement values are also written in the table below.

Note. The first component of displacements for nodes 2 and 3 corresponds to the displacement along the X axis, and the second along the Y axis.

The vector of the truss rod forces N is calculated by expression:

N KAT z :

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The matrix R 1 can also be considered as a compliance matrix of the

given truss, i.e., of the original truss without additional nodal links. The compliance matrix allows expressing displacements through loads:

z R 1F DF.

The compliance matrix of the truss under consideration (Figure 14.23) can be written as

If the given truss (Figure 14.23) is loaded by a unit force F1 1 applied in the direction of the system displacement Z1 (Figure 14.25), then the values of the all node displacements shown on the last figure will make up the first column of the compliance matrix D R 1 .

Figure 14.25

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