Structural mechanics

If the displacement method is taken as the main method, then the kinematically indeterminate fragment of the primary system is calculated by the force method (auxiliary method).

The features of such calculation will be explained in the following examples.

For the frame (Figure 10.2, a), we will take the displacement method as the main calculation method, and select the primary system according to the variant shown in Figure 10.2, b, i.e. we will calculate the given frame as a twice kinematically indeterminate system.

Figure 10.2

To construct diagrams of bending moments caused by the given loads and by the unit values of unknowns, first it is necessary to first calculate a statically indeterminate fragment ABC D on the action of the given

load and the rotation of the support constraint at point D through angle Z1 1. The frame ABC D is statically indeterminate once (by the dis-

placement method the number of unknowns is equal to two). Therefore, its calculation is performed by the force method, which in this version of

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its use is considered as auxiliary. The corresponding final moments diagrams caused by the mentioned loads are shown in Figures 10.3, a, b.

Figure 10.3

Further, following the well-known algorithm for calculating frames by the displacement method, we construct load diagram M F (Figu-

re 10.4, a), unit diagrams M1 and M2 (Figures 10.4, b, c) in the primary

system. Ultimately, the final bending moments diagram (Figure 10.4, d) is constructed in the given system.

Figure 10.4

Let us consider another example. The frame (Figure 10.5, a) contains seven redundant constraints. However, we will calculate this system under the action of given load as a system containing three unknowns.

The primary system of the force method (this method is the main here) is shown in Figure 10.5, b. It includes statically indeterminate fragment

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ABC and symmetric to it fragment A B C . In order to build the mo-

ment diagrams in the primary system of the force method, first it is necessary to calculate these fragments due to the loads that they perceive.

Figure 10.5

Frame ABC contains one unknown of the displacement method. The bending moment diagrams due to the action of a unit distributed load and a unit moment are shown in Figures 10.6, a, b.

m

Figure 10.6

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With their help, using the properties of linearly deformable systems, we construct a load diagram (Figure 10.7, a) and, as an example, the second unit diagram (Figure 10.7, b) of bending moments in the primary system of the force method.

Two other unit diagrams are constructed taking into account the distribution of moments on fragment ABC due to the M 1 (Figure 10.6, b).

The further calculation course corresponds to the algorithm of the force method.

Figure 10.7

10.3. Mixed method

When calculating the frame by the mixed method, the main unknowns in one part are efforts in the redundant constraints. In the other, the remaining part, the unknowns are the displacements of the nodes. That is, during the calculation, both groups of unknowns (force method unknowns and displacement method unknowns) are determined simultaneously. The choice of unknowns, of course, is determined by the structure of the given frame. As a rule, in the part where a small number of redundant constraints are observed, the redundant links are removed and

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the primary unknowns of the force method are introduced, and in the other part, the additional constraints are introduced that prevent the angular and linear displacements of the nodes. These displacements are primary unknowns of the displacement method.

The system of equations from which these unknowns are determined is written on the basis of conditions similar to those used to write the canonical equations of the force method and the displacement method. We will give more complete explanations of the essentially mixed method by the example of calculation the frame shown in Figure 10.8, a.

Fragment AD of this frame (there is rigidly fixed support in node D) contains only two redundant constraints; it is convenient to calculate it by the force method; to calculate the rest part of the frame (its nodes are located at points B, C, D, E, G) it is more convenient to use the displacement method. Based on these considerations, we accept the primary system as it is shown in Figure 10.8, b.

Figure 10.8

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In this primary system the bending moment diagrams caused by the unit value of primary unknowns and by the given load are shown in Fig-

ure 10.9. From diagram M1 (Figure 10.9, a) it can be seen that force X1 1 causes in the third additional link (its number corresponds to the number of the primary unknown) reactive force r31 (note: the reaction

(force) is caused by the force).

In the displacement method, the notation r31 would indicate the reaction in the third constraint caused by the displacement Z1 1, that is, the causes of the reactions r31 and r31 are different, therefore they are denoted differently. Similarly, the physical meaning of the reaction r32 should be understood too.

Figure 10.9

There is diagram M3 in Figure 10.9, c. The displacement of application point of force X1 in its direction caused by displacement Z3 1, is

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denoted by 13 . As in the case of the notation of reactions, writing 13

with the dash emphasizes the difference between this displacement and the displacement 13 , caused by force X3 1 (see the force method).

In accordance with the theorem of reciprocity of reactions and displacements (9.8) r31 13 . Indeed, from the equilibrium equation of

node D (Figure 10.9, a) it follows that r31 3,0 , Indeed, from the equi-

librium equation of node D (Figure 10.9, a) it follows that displacement13 occurs in the direction opposite to force X1 1.

The value 13 can also be found by the rules for determining the dis-

placements caused by the support settlements.

We now write down the canonical equations of the mixed method:

The first equation from this system expresses the condition that the displacement of the application point of force X1 in its direction is equal

to zero, where the first and second terms are the displacements caused by the forces X1 and X 2 , the third and fourth are displacements caused by

the rotations of the nodes at angles Z3 and Z4 , and the fifth is the dis-

placement caused by the load. The meaning of the second equation is revealed in a similar way.

The third and the fourth equations have the meaning of the displacement method equations: the total reactions in the third and fourth addi-

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Coefficients rik and free terms RiF are determined by the ways used in the displacement method. For example, from the equilibrium of forces

in the node D (Figure 10.9, d) we find r34 E3J .

From the equation of equilibrium of moments in the same node on the bending moment diagram, caused by the given load, we obtain

R3F 123.27 .

The coefficients rik and ki , as it has already been noted, are related by the ratio:

rik ki .

Analyzing the distribution of moments in Figures 10.9, a, b note that

r41 0 and r42 0 .

Having determined the coefficients and free terms, we obtain:

Final moment diagram is formed according to the following formula:

M MF M1X1 M2 X2 M3Z3 M4Z4.

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It is shown in Figure 10.10.

Figure 10.10

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