Structural mechanics

Expression (9.8) is a formal notation of the theorem on reciprocity of reactions and displacements (second theorem of J. Rayleigh): the displacement of the point of application of force Fi 1 in its direction,

caused by the unit displacement of link “ k ”, is numericly equal (with the opposite sign) to the reaction in link “ k ” due to unit force Fi 1.

The dimensions of reactions and displacements in this expression are the same. They are installed like this:

To determine the free terms, we consider the primary system of the displacement method in state “ k ” and in state “ F ” (load state, Figure 9.16, d).

On the basis of the reciprocity theorem, there is equality:

WkF WFk .

Revealing this equality, we obtain:

RkF 1 F ik 0 .

From the last equality it follows

RkF F ik .

281

To determine the displacement ik (Figure 9.16, a) in a statically

indeterminate system using the Mohr formula, as you know, one of two “multiplied” diagrams of bending moments can be constructed in a statically determinate system obtained from the given system. In this case, it is the primary system of the displacement method.

Then, denoting the diagram of bending moments due to F 1 in a

ments diagram constructed in the primary system of the force method due to a generalized unit force corresponding to the nature of a given exposure.

Substituting the value of ik into the expression for RkF , we obtain:

So, the calculation of the load reaction RkF is reduced to the calculation of the expression (9.10), in which:

Mk is a unit diagram of bending moments constructed in the primary system of the displacement method;

M F0 is a diagram of bending moments from the given load, built in a

statically determinate system, obtained either from the primary system of the displacement method with the mandatory rejection of link “k”, or

282

obtained from the given statically indeterminate system, i.e. constructed in the primary system of the force method .

E x a m p l e. Let us determine by kinematic method the reactions

obtained from the primary system of the method of displacements (Fi-

Figure 9.17

283

To determine R2F , we select a statically determinate system obtained from the given system (Figure 9.14, a), and construct a diagram of the M F0(b) (Figure 9.17, b):

9.6. Building and Checking Internal Forces Diagrams Due to External Loads

Having solved the system of canonical equations (9.4), we find the values of the primary unknowns of the displacements method. To build the final diagram of bending moments, it is necessary to first build the

adjusted unit diagrams MiZi (they are called “corrected" unit diagrams

of moments). The final diagram M of bending moments from an external load in the given statically indeterminate system is constructed by summing the load diagram M F with the “corrected" unit diagrams, i. e.,

the ordinate of the diagram M in each concrete section “k” of the frame bar is calculated by the formula:

Mk MkF Mk1 Z1 Mk 2 Z2 Mkn Zn .

The main verification of the correctness of the final diagram of bending moments M in the displacement method is a static check, which, as is known, reduces to checking the equilibrium of moments in the frame nodes.

In addition, as in the force method, a kinematic check can be applied to check the correctness of the final diagram M: the result of “multiplying”

each unit moments diagram (or total unit diagram) of the force method by the final moments diagram should be zero

The diagram of transverse forces Q is constructed, as in the method of forces, according to diagram M , and the diagram of the longitudinal

(in the displacement method, static check refers to the main one).

284

E x a m p l e 1. Let us construct a diagram of bending moments in the frame shown in Figure 9.18, a.

To reduce the number of unknowns, when calculating the degree of linear mobility of the frame nodes, the console, as a statically determinate fragment, is discarded. Then the degree of freedom W of the

hinge-rod system (Figure 9.18, b) will be equal to unity, that is nl 1 .

The total number of the displacement method unknowns is equal n na nl 2 1 3 . At Figure 9.18, c the primary system and the pos-

itive directions of the primary unknowns are shown, and Figure 9.18, d– g shows the bending moments diagram in the primary system, due to load, and unit diagrams of bending moments.

The system of canonical equations has the form:

We note some features of calculating the coefficients at the unknowns and free terms of the canonical equations. To determine coefficient r32 ,

we write the condition for the equilibrium of the fragment (Figure 9.18, h) of the design scheme taken from Figure 9.18, f:

Using the data (Figure 9.18, g), we can verify that the reciprocity of the reactions is observed: r23 r32. Indeed, from the condition of the

equilibrium of moments in the node (Figure 9.18, i) it follows that

Free term R3F can be determined from the equilibrium equation for the fragment (Figure 9.18, j) obtained from (Figure 9.18, d):

285

In numerical form, the system of canonical equations is written as follows:

The final diagram of bending moments is shown in Figure 9.18, k.

E x a m p l e 2. Let us construct a diagram of bending moments for the frame shown in Figure 9.19, a.

To determine the degree of linear mobility of the nodes, we use a hinged-rod system (Figure 9.19, b). The total number of unknowns is

equal n na nl 2 1 3 . The primary system of the displacement

method and the positive directions of the primary unknowns are shown in Figure 9.19, c. The load and unit diagrams of bending moments are shown in Figure 9.19, d,…,g.

The system of canonical equations in numerical form has the form:

Z3 371.778 E1J m.

287

The final diagram of bending moments is shown in Figure 9.19, h.

Figure 9.19

9.7. Calculating Frames with Inclined Elements

In frames with inclined elements, the displacement of the linear link of a node to a predetermined value, for example, equal to one causes linear displacements of other nodes, which depend not only on the given displacement, but also on the geometry of the frame (the location of its elements). Therefore, to build diagrams of bending moments in the primary system, it is necessary, first of all, to find the mutual displacements

288

of the ends of the bars forming the frame. The displacement values are found from the displacement analysis of the hinge-rod system corresponding to the given frame.

When building any unit diagram only one additional link is displaced by a distance equal to unity, the rest remain motionless. In this case the hinge-rod system is a kinematic mechanism with one degree of freedom. With a known displacement of one node, the displacements of the others can be determined from the movement diagram of the mechanism. Let us explain this with the following examples.

Let us consider a frame with one inclined column (Figure 9.20, a).

The degree of its kinematic indeterminacy is equal to n 2 . The link that impedes the linear movement of nodes we place perpendicular to the rod 2–3 (Figure 9.20, b).

To determine the mutual displacements of the ends of the frame rods at Z1 1, we give the hinged-rod system position 0–1 – 2 – 3, which is

possible under the conditions of its fastening (Figure 9.20, c). The arcs described by points 1 and 2 when the rods rotate around the reference points 0 and 3 are replaced by their tangents in points 1 and 2. From the

right triangle 2-k-2 it follows that the mutual displacement of the ends of rod 1–2 is cos , of rod 0–1 is sin , of rod 2–3 is 1.

The same displacement values are obtained from the displacement diagram (Figure 9.20, d). Let us explain its construction.

The support nodes 0 and 3 are fixed. The corresponding point on the displacement diagram is called the pole. From this pole (point 0,3 (Figure 9.20, d)) we draw rays perpendicular to the rods 0–1 and 2–3, i.e., along the directions of possible movements of nodes 1 and 2. On the

ray perpendicular to rod 2–3, at a distance equal to unity, point 2 will lie. To determine the position of point 1 , it is necessary to draw a line

from point 2 perpendicular to rod 1–2. Segments 1 – 2 and 1 – 0 in the displacement diagram are equal to the mutual displacements of the

ends of rods 1–2 and 0–1 for Z1 1.

E x a m p l e 1. Let us construct a diagram of bending moments for the frame shown in Figure 9.20, a. The rigidity in bending of all bars are assumed to be the same. Angle = / 3 rad. The length of bar 2–3 is

l2 3 8 / 3 m.

289