Structural mechanics

THEME 11. CALCULATING CONTINUOUS BEAMS

11.1. General information

A beam that overlaps two or more spans, not interrupted along its length by hinges is called a continuous beam.

The degree of static indeterminacy of continuous beams can be determined according to general rules (section 8.2). Since the beam is a single disk that overlaps several spans, the formula (8.2) is converted to:

Beams shown in Figures 11.1, a, b contain two and three redundant constraints, respectively.

Figure 11.1

The reader is already familiar with the methods of calculation of statically indeterminate frames for various types of external actions (see the Themes 8, 9, 10). Features of the application of continuous beams calculation are considered in the next section.

11.2. Examples of Continuous Beam Calculation

E x a m p l e 1. Construct bending moments and shear forces diagrams for the continuous beam (Figure 11.2, a) using the force method.

The degree of static indeterminacy of the beam is equal to three. The primary system of the force method can be obtained by eliminating of support rods (Figure 11.3, b). Then, support reactions will be accepted as unknowns. It is easy to notice that in this case, none of the secondary

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coefficients of the canonical equations is equal to zero (Figure 11.3, c). The same can be seen, "multiplying" the corresponding unit moment diagrams. This means that such a primary system is not rational.

Figure 11.2

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knowns, and all intermediate equations have three unknowns (the equation i contains unknowns Xi 1, Xi , Xi 1).

Having solved the system of equations, we find:

The final bending moment diagram (Figure 11.2, g) is based on the expression:

M M F M1X1 M2 X2 M3X3 .

The analytical expression for determining the bending moment in the cross-section, which is located between the support points of the beam, can be obtained from formula (8.16).

A kinematic check of the correctness of the diagram M consists in checking the displacements in the directions of the primary unknowns and is performed according to the formula (8.23).

The shear forces diagram is shown in Figure 11.2, h.

To determine the reaction in the support with the number n (Figure 11.4), we cut out an infinitely small section of the beam using two sections located on both sides of the support, and show the shear forces in these cross-sections. From the equation Y 0 it follows that:

Rn Qn 1 Qn .

In particular, in the fixed support (Figure 11.5) and the first intermediate support (Figure 11.6), the vertical reactions are 12.625 kN and 45.155 kN, respectively (Figure 11.6).

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E x a m p l e 2. Calculate the same beam (Figure 11.7, a) using the displacement method.

The degree of kinematic indeterminacy of a continuous beam is a variable characteristic. Indeed, any cross-section of the beam can be declared as a node in which two rods are joined. Such a node, in the general case, will have two degrees of freedom: vertical displacement and the angle of rotation (displacement along the beam axis according to the accepted assumptions for a linearly deformable system is not taken into account). In the primary system of the displacement method, such a node must be fixed with two additional constraints. As a result, the dimension of the beam calculating problem increases.

In order to reduce the dimension of the problem, it is advisable to consider only support nodes. Each section above the supports of the beam has only one degree of freedom – the angle of rotation.

For the given beam (Figure 11.7, a), we choose the primary system of the displacement method shown in Figure 11.7 a. The primary unknowns are the angles of rotation of the support sections of the beam.

In Figures 11.7, b–d, the bending moment diagrams caused by unit value of primary unknowns are shown, and in Figure 11.7, d the bending moment diagram caused by the acting load is shown.

Having calculated the coefficients and free terms of the canonical equations according to well-known rules, we obtain a system of equations in the following form:

Having solved the equations system (11.2), we find:

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Final bending moment diagram is constructed from the expression:

M M F M1Z1 M2Z2 M3Z3 ,

It has the form shown in Figure 11.7, g.

Figure 11.7

The same beam (Figure 11.7, a) also can be calculated using the displacement method as a beam with two unknowns. In this case, the primary system (Figure 11.8, a) include the “non-standard” element shown in Figure 11.9, a (it is not in the set of elements in table. 9.1). The bending moment diagrams caused by unit value of two primary unknowns are shown in Figures 11.8, b, c.

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The canonical equations, after determining the coefficients and free terms, are written in the form:

Solving them, we get:

Naturally, the final moment diagram will be the same as in Figure 11.7, g.

Note the following. Removing the additional link in the primary system (Figure 11.7, a) allowed us to move from the system of equations (11.2) to the system (11.3). This transition could be carried out without calculating the beam as a twice kinematically indeterminate system.

To solve the system of equations (11.2), we apply Jordan eliminations (Gauss method). The coefficients and free terms of system (11.2) are written in the form of table 11.1 (the factors EJ in front of the unknowns

Zi are not included in the table) and we take one step of ordinary Jordanian eliminations, taking the coefficient r33 as the resolving element.

Table 11.1 Table 11.2

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Transition from Table 11.1 to table 11.2 is carried out according to the following rules.

value.

2. The remaining elements of the resolving column s are divided by

the resolving element.

3. The remaining elements of the resolving row r are divided by

the resolving element and change signs.

4. Other elements are calculated by the formula

bij aijarsa aisarj , rs

at i r, j s (according to the rule of the rectangle).

In Table 11.2, the coefficients and free terms of the system of equations (11.3) are written. Zero columns could not be written to the table. From this table it follows that:

EJ Z3 12 EJ Z2 152 .

E x a m p l e 3. It shows the calculation of the beam (Figure 11.2, a) by the mixed method.

There are many variations of the primary systems of the mixed method. Some of them are shown in Figure 11.10. To demonstrate the features of the mixed method, we choose the primary system shown in Figure 11.11, a.

The bending moment diagrams caused by unit value of primary unknowns and acting load are shown. in Figures 11.11, b–e.

The system of canonical equations for the accepted primary unknowns has the following form:

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