Structural mechanics

tions of possible independent linear displacements of nodes. The system thus obtained is called the primary system of the displacement method.

As an example, the Figure 9.6, a shows the primary system of the displacement method for the frame shown in Figure 9.4, a. The Figure 9.6, b contains the primary system for the frame shown in Figure 9.5, a.

The nodes displacements of the primary system are known: they are equal to zero. Therefore, the primary system can be called kinematically determinate.

An analysis of the structural interaction of the primary system elements (Figure 9.6) shows that the primary system of the displacement method consists of single-span independent beams with hinged and/or absolutely rigid supports at the ends.

Figure 9.6

When transferring any of the beams that make up the primary system into a state corresponding to its deformed position in a given frame under load (Figure 9.7), internal forces arise in it. In accordance with the principle of independence of the forces action (principle of superposition), these efforts can be represented as the sum of the efforts caused by:

1.The action of the load located on the beam (bar).

2.By turning the left end of the bar (and the right end if the bar is

pinched at both ends) by angle ZA equal to the true value of the angle of

rotation of node A .

3. Mutual linear displacement AB of the ends of the bar in a direc-

tion perpendicular to its axis.

Supporting reactions arising in a statically indeterminable beam of constant stiffness under various influences on it serve as auxiliary quantities in the calculation of frames by the displacement method. Their values can be found by the method of forces.

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Therefore,

X1 1F 3 ql .11 8

The final diagram of bending moments, constructed by expression M MF M1X1 , is shown in Figure 9.8, d, and in Figure 9.8, f the values of the support reactions are shown.

E x a m p l e 2. Consider a beam loaded with a concentrated force F (Figure 9.9, a). We construct the bending moments diagram M F using

the primary system from Example 1 (Figure 9.9, b).

Figure 9.9

We calculate the free term of the canonical equation

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From the equilibrium condition Y 0 it follows that

RA F X1 F2 v 3 v2 .

The final diagram of bending moments is shown in Figure 9.9, c. The values of the support reactions are given in Fugure. 9.9, d.

E x a m p l e 3. Let us plot the bending moments diagram from the rotation of the clamped end of the beam at angle (Figure 9.10, a).

Figure 9.10

We write the canonical equation of the force method in the form

11X1 1c 0.

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The free term can be calculated by the expression

1c Rk1ck ,

where Rk1 is the reactions in the link “k” of the primary system caused by force X1 1 (Figure 9.10, b).

1c l l .

The same value 1c can be obtained from the kinematic analysis of

the disign scheme (Figure 9.10, c): displacement of point B is opposite to the direction of force X1.

Then

X1 1c 3EJ .

11 l2

The bending moments diagram and the distribution of support reactions are shown in Figure 9.10, d and e.

E x a m p l e 4. Let us determine the efforts in the beam from the displacement of the fixed support by amount in the direction perpendicular to beam axis (Figure 9.11, a).

As in example 3, the displacement 1c can be found, using the formula:

1c Rk1ck (1 ) .

The support reaction at point B , equal to X1 , can be found as

X1 1c 3EJ .

11 l3

Diagram M and the values of the support reactions are shown in Figures 9.11, b, c.

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The bending moments diagram is shown in Figure 9.12, c and supports reactions are shown in Figure 9.12, d.

Figure 9.12

E x a m p l e 6. We show the calculation of a twice statically indeterminate beam at rotation of fixed support A by angle (Figure 9.13, a).

The primary system of the force method can be chosen as symmetric one (Figure. 9.13, b). The corresponding unit diagrams of bending moments are presented in Figures 9.13, c and d. The state of the primary system caused by the rotation of fixed support A at angle is shown in Figu-

re 9.13, e. Since 12 21 0 , the canonical equations for determining the primary unknowns are represented in the form:

11X1 1c 0 ,

22 X2 2c 0 .

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The coefficients at the unknown are:

2c Rk 2ck 1 .

Solving the equations gives

The bending moments diagram in the beam is represented in Figure 9.13, f. The support reactions are shown in Figure 9.13, g.

Figure 9.13

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