Structural mechanics

If both ends of the bar are hinge-supported, then:

15.10. Bar Stiffness Matrix in the General Coordinate System

The bar stiffness matrix can be obtained the easiest way using the general equations of structural mechanics by the expression:

R a k aT ,

where a is the equilibrium matrix of the bar in the general coordinate system. We show the formation of the stiffness matrix for the truss rod (1–2) of the truss shown in Figure 15.14.

Figure 15.14

The rod is adjacent to the hinge nodes. We introduce the links at the rod ends that impede their displacements along the directions of the coordinate axes (Figure 15.15, a). We show in the same figure the positive directions of the nodes displacements.

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The reaction values in additional links caused by the displacement

Z 1 are given in the first column of the matrix R(1 2) .

In other cases, when it is necessary to organize the transition from the stiffness matrix in the local coordinate system to the stiffness matrix in the general coordinate system, it is necessary to use the transformation rules of the linear operator matrix in the transition from the old basis to the new one.

Linear displacements in the local and general coordinate systems (Figure 15.16) are related by the relations:

Z1 Z1 cos Z2 sin ,

Z2 Z1 sin Z2 cos .

Figure 15.16

The rotation angle of the bar end section does not change when the coordinate system is changed. Therefore, the matrix of the rotation operator has the form:

Since the directions of displacements at both ends of the bar coincide,

Z6 are performed using the same matrix. Consequently,

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The transformation of the stiffness matrix (matrix of the linear operator) during the transition to the basis Z is performed by the expression:

R V T R V .

For further discussion, we also present the matrix R in block form:

15.11. Stiffness Matrix Formation for the Entire System

The finite element model of the bars system, as already noted, is represented as a set of bars connected in nodes. The system node displacements cause the same end displacements of the bars (finite elements) adjacent to this node. The resulting forces in the bars of the primary system of the displacement method are determined using the stiffness matrices of the bars. The reactions in the links superimposed on the node of the system can be found as the sum of the terminal reactions in the links of the bars adjacent to the node. For example, the reaction force in the link along the X axis direction will be equal to the sum of the reaction forces in the links of the bars in the same direction. Reactions in other directions are defined in a similar way.

In the general case, the vector Ri of total reactions for the i-th node

of the system can be determined through the vectors ri(e) of end reactions in elements adjacent to this node, by expression:

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where ri1(e) , ri(2e) , , rin(e) – reaction vectors at the end of an element adjacent to a node caused by displacements Z1 1, Z2 1, , Zn 1. Symbol e i means summation over all elements adjacent to the node i .

For a rigid node, vector Ri has three components: the first component

of the vector indicates the value of the reactive force in the direction of the axis X , the second – in the direction of the axis Y , the third gives the value of the reactive moment.

If all the bars are connected at nodes rigidly, the vectors ri1 , ri2 , , rin

also contain three components. If some bar adjoins the node articulated, then for the operation of adding vectors in the i-th node, the third component of the end reaction vector should be taken equal to zero.

Having written expression (15.42) for each node of the structure, we present a system of equations connecting nodal reactions and displacements in the form:

where R is the stiffness matrix of the entire system.

From the presented form of recording the matrix R it follows that its elements are calculated through the elements of the stiffness matrices of individual finite elements. If the nodes i and j are not interconnected by

elements, then rij 0 ; if they are connected by several elements, then the corresponding element of the stiffness matrix is calculated as

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rij rij(e). e i, j

In block form, the matrix R is represented as:

where Rij is the reaction block in the links of the i-th node caused by the

unit displacements of the links of the j-th node.

In the stiffness matrix of the entire system formed according to the indicated principle, support nodes had not been taken into account. The displacements of the non-deformable supports are equal to zero. There-

fore, if it is known in advance that Z j 0 , then the j-th row and the j-th

column of the obtained matrix R should be deleted. The size of the matrix must be reduced. In the case of automated computing, a new numbering of unknowns will also be required. If the dimensions of the matrix will not be changed, then it is necessary to take the indicated row and

column as zero, but the element rjj of the matrix R must be taken equal

to one or to a number other than zero (so that det R 0 ).

We will show a graphical scheme of stiffness matrix formation of the entire system from the stiffness matrices of its elements.

For the frame shown in figure 15.17, the stiffness matrix in block form is written as:

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The numbers of degrees of freedom for each node of the frame correspond to the numbers of the components of the displacement vector for this node (1 – displacement along the axis X , 2 – displacement along the axisY , 3 – rotation about the axis Z ).

In Figures 15.18 symbol e i is the number of the corresponding bar element.

15.12. Stiffness Matrix of a Rectangular Finite Element for Calculating Thin Plates

The displacements of the finit element must correspond to the deformed scheme of the system under study at its location. In the general case, it is usually impossible to accurately describe the state of a continuum system through a finite set of nodal displacements using equations of type (15.18). Therefore, the FEM is classified as approximate. Nevertheless, it allows obtaining the calculation results of very high accuracy. Currently, FEM is the main method for solving the most diverse problems of statics, dynamics and stability of bars and continuum systems.

The procedure for obtaining stiffness matrices of FE for calculating plates, shells, and other continuum systems is in many respects similar to the method for obtaining a bar stiffness matrix (see Section 15.11). Let us explain this note by the example of constructing the FE stiffness matrix for plate calculation.

Brief information from the theory of plate calculation

A plate is a body whose thickness h is small compared with the dimensions of the sides of the base a and b (Figure 15.19, a).

The plane dividing the thickness of the plate in half is called the median. The intersection lines of the median plane with the lateral surface form the contour of the plate. According to the shape in plan, plates are distinguished rectangular, triangular, round, etc.

When calculating the plates, the origin of the coordinate axes is located in one of the points of the median plane. From the action of the transverse load, the plate bends, the median plane turns into the median surface. The displacements of the plate points in the direction of the axes x , y , z are denoted by u , v , w respectively.

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In turn, thin plates are divided into rigid thin plates (those for which tensile and shear forces in the middle surface are not taken into account during bending) and flexible thin plates (the shear and tensile stresses in the middle surface are taken into account during bending).

The theory of calculating thin plates is constructed using the following hypotheses.

1. The direct normals hypothesis, according to which a rectilinear element normal to the median plane before deformation of the plate remains normal to the median plane after deformation, and its length does

not change. According with this hypothesis, the shear angles xz andyz , as well as the linear deformation z are taken equal to zero:

2. The hypothesis of deformability of the middle layer, according to

3. The hypothesis of the absence of normal stresses on sites parallel to the middle layer, that is, stress z 0 .

In accordance with the first two hypotheses, the displacements u and

vof an arbitrary point (Figure 15.19, b) along the directions of the axes

xand y are equal

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