Structural mechanics

THEME 14. GENERAL EQUATIONS OF STRUCTURAL

MECHANICS FOR BARS SYSTEMS

14.1. Concept of the Discrete Physical Model

The design schemes of bars systems, used in the classical methods of their analysis, have a pronounced continuity property. They are represented in the form of interconnected one-dimensional elements (bars), while the nodes are interpreted as the points at which the bars are joined or on which constreints (links) are superimposed. As a result of analysing the system for given exposures, dependencies are established. They describe the nature of the change in the internal forces and displacements along the axis of each bar. Information in this form about the stressdeformation state of the system is redundant for practical tasks. In the course of calculations, it is enough to find the internal forces or displacements only in a number of characteristic cross sections and then the forces or displacements can be found in any intermediate section of the bar, if necessary.

Design sections are usually assigned at the junction of the bars to the nodes; they separate the bars from the nodes. As a result, the design scheme of the investigated structure seems to be composed of bars and nodes. This scheme is called the discrete physical model of the structure. For example, the discrete model of the design schemes of a frame (Figure 14.1, a) is shown in Figure 14.1, b.

Each rigid node of a discrete model of a plane structure has three degrees of freedom (linear displacements along the coordinate axes and rotation angle), each hinged one has two degrees of freedom (linear displacements along the coordinate axes). The position of the nodes of the system determines the position of its bars. Therefore, the degree of freedom of a certain system is defined by the number of degrees of freedom of all its nodes.

In this chapter the degree of freedom of the system will be denoted through m . It is precisely the number of independent equations of equilibrium that can be compiled for all nodes of the system.

For plane trusses m is equal to twice the number of nodes minus the number of support rods, and the number of unknown internal forces n is equal to the number of truss rods. Using the notation introduced, the

372

degree of static indeterminacy of the system k can be calculated by the expression

Figure 14.1

In chapter 8, the degree of static indeterminacy of the system was denoted by (the number of redundant links), in chapter 9, the degree of kinematic indeterminacy (it is also the degree of freedom) of the system was denoted by n.

In this chapter, the degree of static indeterminacy is denoted by k, and the degree of kinematic indeterminacy is denoted by m. The number of unknown efforts is indicated through n.

The degree of freedom of a structural node also determines the dimension of the displacement vector of this node. The total number of components of the displacement vector of all nodes corresponds to the degree of kinematic indeterminacy of the system. Therefore, the relation (14.1) can be considered as the relationship between the degree of static indeterminacy k and the degree of kinematic indeterminacy m of the system.

373

Since any point of any bar can be declared a new node where two bars are joined. Therefore, for the same system, for example, a frame, several variants of its discrete model can be adopted. This means that the degree of freedom of a discrete model, in the general case, is a variable characteristic. Even in this case, relation (14.1) allows one to correctly find the degree of static indeterminacy of the system, since the number of unknown forces in each additional section coincides with the number of independent equilibrium equations that can be compiled for each additional node.

14.2. Loads and Displacements

To simplify the computational procedures, later in this section, the calculation of systems under only nodal forces effects will be considered.

Techniques for replacing distributed load with concentrated forces are well known. The essence of the conversion is as follows.

Initially, each element located between two adjacent nodes is considered as a bar with end (support) links corresponding to the type of a node (rigid or articulated). Calculating it as a single beam under local load, we can determine its support reactions and construct the diagrams of its efforts. For this, table 9.1 can be used.

Subsequently, having loaded the nodes of the design scheme of the system by forces equal in value and opposite in direction to the reactions of the single beams, we will calculate the system on action of these nodal forces.

The final diagrams of the internal forces are obtained by summing the corresponding diagrams from the calculation of the system as a whole and of the individual elements.

Figure 14.2 shows (symbolically) the transition from a system with distributed loads to a system with concentrated forces.

External forces acting on a rigid node i of a plane system are defined by a load vector in the form:

Where Fix, Fiy are components of external load along the axes x and y; mi is a concentrated moment in i -th node.

374

Figure 14.2

The rule of signs for the load: external forces are considered positive if their directions coincide with the directions of the corresponding coordinate axes; positive moments are directed counterclockwise.

The full load vector acting on the system is formed by sequential docking of the corresponding vectors for each node of the system:

F F1T , F2T , F3T ,..., FpT 1, FpT T .

The number of system nodes is denoted by p.

Under the load, the system takes a new (deformed) position. Frame nodes move.

The displacements of the rigid node “i” are characterized by a vector

zi zix , ziy , i T.

The displacements of the articulated node “j” are characterized by a vector

z j z xj , z yj T .

375

The full vector of displacements of the system nodes is represented as:

z z1T , z2T , z3T ,..., zTp T .

The vector of generalized displacements must correspond to the vec-

tor of generalized load. Dimensions of vectors F and z coincide. The scalar product of these vectors determines the work of external forces. Such vectors are called dual.

14.3. Internal Forces (Efforts) and Deformations

Generally, a bending moment M , a transverse force Q and a longi-

tudinal force N arise in the cross section of the bar. Together they form a vector of efforts (internal forces) in the section

S M , Q, N T.

The components of this vector must be determined.

In special cases, this vector may contain two components, for example:

In the first case, the longitudinal force is not included in the number of unknowns, and in the second case, the transverse force is.

It is possible that only the bending moment can be an unknown factor in the cross section. Then:

S M .

Under the action of the nodal load on the system, the stress state of ith bar can be characterized by the vector:

Si Ni , M Bi , M Ei , Qi T ,

where Ni is longitudinal force in the bar;

M Bi – bending moment at the beginning of the bar;

376

MEi – bending moment at the end of the bar; Qi – transverse force in the bar.

Since the load does not act directly on the rod, the transverse force along its length does not change and, as is known (8.17), is calculated by the formula:

Q M E M B .

l

Therefore, having saved the first three components in the vector Si , we rewrite it in the form:

Si Ni , M Bi , M Ei T.

Each type of effort corresponds to a certain deformation. The longitudinal force causes elongation or shortening of the element, bending moments – rotations of the cross sections, transverse forces – mutual shear of the cross sections.

The deformation vector i corresponding to the effort vector Si will have the form:

i li , Bi , Ei T ,

where li is linear deformation of an element;

Bi , Ei are the angles of rotation of the cross sections at be-

ginning and at end of the bar relative to the straight line which connectes the nodes in the deformed state.

The vector of efforts S and deformation vectors for the entire sys-

tem, as well as the vectors F and z , are formed by sequential joining of the efforts and deformation vectors for individual bars.

The vectors S and are dual; their scalar product gives the work of internal forces.

For a spatial system, the force vector in the cross section is, as a rule, six-dimensional.

377

14.4. Equilibrium Equations

Let us consider an arbitrary, for example, statically indeterminate frame (Figure 14.1, a), which is in equilibrium under the action of a given load. The corresponding discrete model in the form of a set of nodes and bars is shown in Figure 14.1, b.

We compose the equilibrium equations for the 2d node of the frame:

The written down three equations contain six unknown efforts. Consider the equilibrium of the frame bars. Each of them is under the

action of the end forces shown in the same figure.

Composing the three equilibrium equations for the first bar, we obtain:

x 0, NB1 NE1 N1,

y 0, QB1 QE1 Q1,

Similar relations can be obtained for the second bar, that is:

Substituting the expressions of efforts from the equations of equilibrium of the bars into the equations of equilibrium of the nodes, after simple transformations we get:

378

The matrix form of this system of equations will be as follows:

or abbreviated:

In equations (14.2), the signs of the components of the vector F correspond to the accepted positive directions of the nodal loads. To numerically solve these equations, together with others, we rewrite them in the form:

where A A* is equilibrium matrix:

is the vector of effort;

is the vector of loads.

379

To easily navigate in the structure of matrix A , we remember that in its first line there are the coefficients at unknown efforts in the end sections of the bars from the equilibrium equation X 0. In the second -

from the equation Y 0, and in the third from the equation M2 0.

Moreover, in the first three columns of the matrix, the coefficients are recorded for the efforts N, M B , M E in the first bar, i.e. in the bar 1–2,

and in the next three columns for the corresponding efforts in the second bar (in the bar 2–3).

The shown method of forming the equilibrium matrix is called the matrix forming method “by nodes”. For frames with a large number of nodes, it is laborious and therefore is not often used in practice. Another, more effective method, which allows to organize the formation of a matrix “by bars”, will be described in section 14.14.

14.5. Geometric Equations

Let us imagine the process of “transition" of a frame (Figure 14.3) to a deformed state as a result of the successive influence of first longitudinal deformations of its elements, and then bending deformations. The first stage of deformation is equivalent to loading of the corresponding hinged system by nodal forces, which cause the same values of the internal longitudinal forces. Then, with the positions of the nodes at points 1, 2 and 3, bending moments are applied which transfare the bars into a curved state. According to the assumption of small displacements, the second stage of deformation does not change the position of the frame nodes. Therefore, the deformation of each bar fixed at the ends can be characterized by three components:

li is an absolute elongation (shortening) of the ith bar,Bi , Ei are angles of rotation of the end sections.

Consequently, the strain vectors of the 1st and 2nd bars of the frame (Figure 14.3) have the form:

1 l1, B1, E1 ,

2 l2 , B2 , E2 .

380