N. E. Strebeiko

A New Approach

to Frame Analysis

by the Method of Foci

(A New Method for Constructing

Influence Lines)

Tomsk

1929

N. E. Strebeiko.

A New Approach to Frame Analysis by the Method of Foci.

(A new method for constructing influence lines).

The requirements of the construction industry in recent years have brought to life a very large number of methods for analyzing frame structures. The general problem of the analysis of statically indeterminate systems, as is known, has long since received an exhaustive solution in principle (the theorems of Betti, Maxwell, Mohr, Castigliano, and Menabrea). Thus, for researchers of recent times there remained only the practical side of solving the frame problem, i.e., the search for such practically convenient methods of frame analysis which, with a minimum of time and labor, would provide an accurate and simple path to a solution accessible to the same broad circles of design engineers.

The enormous work done in this direction up to the present can in no way be considered completed; and any new procedure, provided only that it systematizes, facilitates, and accelerates the designer’s work, must in our day be of practical interest.

The present article, devoted to the method of foci, constitutes the substance of a report delivered by the author on November 18, 1928, for the members of the Bridge Section of the Scientific-Technical Circle of Builders of the STI.

The report had as its main purpose to propose a new method for constructing influence lines in frame systems; moreover, some conclusions and formulas, which resulted from a new interpretation of influence lines (influence lines as elastic lines), proved applicable also to particular cases of frame analysis under a fixed load—namely, to the determination of frame displacements without determining the displacing forces, which introduced simplifications into the computational part II of the method of foci. In addition, incidental clarification was given to the question of angular foci and their connection with foci for moments.

Fundamental Propositions.

The method of constructing influence lines which has been used up to now when applying the method of foci is very fully treated in the fundamental work of E. Suter, Die Methode der Festpunkte. In Russian—in the work of Prof. G. V. Ulyaninsky, Analysis of Frame Structures (3rd and 4th eds., Tomsk)—we have an exposition of this question with exhaustive completeness, with the presence of a detailed worked example of the analysis.

In the well-known article by engineer L. A. Samoilov, “Analysis of Frames by the Method of Foci” (Transactions of MIIT, issue III), we also encounter an example of constructing influence lines.

Referring those interested to the sources listed, I shall recall only that the essence of the method used consisted in constructing the so-called principal moment diagram under the assumption

the presence of a fictitious support, in the subsequent calculation, for each position of the moving load, of the displacing forces and in the determination, from the displacing forces, of the additional moments due to displacement (when the fictitious support is removed).

In the new method of constructing influence lines set forth below, the general course of the calculation remains unchanged. As usual in the method of foci, it is first assumed that by the introduction of redundant constraints (fictitious supports) the frame has been brought into an immovable state)—i.e., there can be no linear displacements of the joints. The computation of moments under this assumption constitutes the subject of the first part of the calculation*.

In the second calculation part, the linear displacements of the joints of the frame structure are determined for various positions of the moving load (after the frame has been freed from the fictitious supports), and from the displacements the additional moments are computed, which must be added to the corresponding moments from the first calculation part. In the proposed method the determination of the displacing forces is eliminated.

I. Construction of Influence Lines for Frame Structures with Immovable Joints.

§ 1. Notation.

In the further exposition, the following notation is used in all derivations**):

\(\alpha_1\) and \(\beta\) (Fig. 1)—the angles of rotation at the supports of a simple beam \(AB\) due to loading by the moment \(M^A = 1\); \(\beta\) is the angle of rotation of the unloaded end.

\(\alpha_2\) and \(\beta\) (Fig. 2)—the corresponding angles of rotation of the same simple beam \(AB\) when loaded by the support moment \(M^B = 1\). By the principle of reciprocity of deformations, the angle \(\beta\) (Fig. 2) is, obviously, equal to the angle \(\beta\) (Fig. 1).

\(\alpha^A\) and \(\alpha^B\) (Fig. 3)—the angles of rotation at the supports of the same simple beam \(AB\) under simultaneous loading by the support moments \(M^A = 1\) and \(M^B = 1\). In this case, obviously, \(\alpha^A = \alpha_1 + \beta\) and \(\alpha^B = \alpha_2 + \beta\).

\(\alpha_0^A\) and \(\alpha_0^B\) (Fig. 4)—the angles of rotation of the axis of a simple beam \(AB\) at the supports \((A)\) and \((B)\) under loading by any external load.

\(\gamma_n^A\) (Fig. 5)—the angle of rotation at the support \((A)\) of the axis of the \(n\)-th member of the frame structure when loaded by the moment \(M^A = 1\), it being assumed that the \(n\)-th member at the support \((A)\) is disconnected from the left-hand part of the structure and rotates freely (without translation), while at the support \((B)\) it retains its former connection with the rest of the structure.

\(\gamma_n^B\)—the angle of rotation of the end \((B)\) of the \(n\)-th member of the frame structure corresponding to the preceding case under the action of \(M^B = 1\).

\(\varphi_n^A\) (Fig. 6)—the angle of rotation of the joint \((A)\) of the frame, serving as the left support for member \(n\), when loaded by the moment \(M_n^A = 1\).

\(\varphi_n^B\) (Fig. 6)—the angle of rotation of the joint \((B)\), serving as the right support for member \(n\), under the action of the moment \(M_n^B = 1\).

\(\tau^N\)—the angle of rotation of joint \(N\) of the frame structure under the action of any load.

*) The change in length of the members due to axial compression and tension is, as usual, neglected.

) The letter notation corresponds to that adopted in Prof. G. V. Ulyaninsky’s Analysis of Frame Structures (1929, Tomsk).

\(a_n\) and \(b_n\)—the left and right focal distances of the \(n\)-th member.

\[ K_a=-\frac{M^A}{M^B}=-\frac{a}{l-a} \]
—the left moment focal ratio.

\[ K_b=-\frac{M^B}{M^A}=-\frac{b}{l-b} \]
—the right moment focal ratio.

\(\psi_{n-m}\)—the “distribution” coefficient = the coefficient of transmission of the support moment from the \(n\)-th member to the \(m\)-th.

\(\delta\) or \(\Delta\)—the linear relative displacement of the joints.

\[ \vartheta=\frac{\delta}{l}\quad \text{or}\quad \frac{\Delta}{l} \]
—the angle of rotation of the member due to the displacement of the joints.

\(y_a\) and \(y_b\)—the so-called “crossing segments” at the supports of the loaded member, denoted by Suter as \(k_a\) and \(k_b\).

Other notation not included in this list will be explained in the appropriate places in the subsequent exposition.

The sign convention for support bending moments is the one customary in the method of foci: a moment is considered positive if, under its action, the convexity of the member bent by it is directed to the right, when proceeding along the frame (Fig. 7).

§ 2. Relation between the influence lines of the rotation angles \(\tau\) of the joints and the influence lines of the support moments.

The essence of the new method of constructing influence lines will consist in the use of those angular quantities \(\gamma\) and \(\varphi\) (see § 1) which, up to the present time, have played only an auxiliary role in the method of foci in determining focal distances and distribution coefficients and, after these fundamental quantities had been computed, lost all significance.

Let us have a certain frame structure (Fig. 8) without closed contours. By introducing fictitious supports at joints I and VI, immobility of the frame joints is ensured.

Consider joint IV. Let a moving load \(P_m=1\) be located somewhere at point \(m\) of the frame girder—to the right of joint IV.

If, in this case, joint IV, participating in the general deformation of the structure, has rotated through some angle \(\tau^{IV}\), then from the definition of the angular quantities \(\gamma\) and \(\varphi\) (see § 1) we have the following relations:

\[ \tau^{IV}=\overline{M}_5^{IV}\varphi_5^{IV} \qquad \text{or}\qquad \overline{M}_5^{IV}=\tau^{IV}\cdot \frac{1}{\varphi_5^{IV}} \]

\[ \tau^{IV}=\overline{M}_4^{IV}\gamma_4^{IV} \qquad \text{or}\qquad \overline{M}_4^{IV}=\tau^{IV}\cdot \frac{1}{\gamma_4^{IV}} \]

\[ \tau^{IV}=\overline{M}_8^{IV}\gamma_8^{IV} \qquad \text{or}\qquad \overline{M}_8^{IV}=\tau^{IV}\cdot \frac{1}{\gamma_8^{IV}} \]

\[ \tau^{IV}=\overline{M}_{12}^{IV}\gamma_{12}^{IV} \qquad \text{or}\qquad \overline{M}_{12}^{IV}=\tau^{IV}\cdot \frac{1}{\gamma_{12}^{IV}} \]

The bar over the symbol for the moments indicates that, for the time being, we are dealing with absolute quantities.

In order to determine the signs of these equalities, it is necessary to prescribe a positive direction for the angle of rotation \(\tau^{IV}\). We shall agree always to take the angles \(\tau\) as positive if the joint has rotated clockwise.

Under this condition, while preserving the sign convention for moments (§ 1) and with the direction of traversing the members shown in Fig. 8, we shall have:

\[ \begin{aligned} M_{5}^{\mathrm{IV}}&=-\tau^{\mathrm{IV}}\cdot\frac{1}{\varphi_{5}^{\mathrm{IV}}} =\tau^{\mathrm{IV}}\left(-\frac{1}{\varphi_{5}^{\mathrm{IV}}}\right)\\ M_{4}^{\mathrm{IV}}&=-\tau^{\mathrm{IV}}\cdot\frac{1}{\gamma_{4}^{\mathrm{IV}}} =\tau^{\mathrm{IV}}\left(-\frac{1}{\gamma_{4}^{\mathrm{IV}}}\right)\ldots\ldots\ldots (1)\\ M_{8}^{\mathrm{IV}}&=-\tau^{\mathrm{IV}}\cdot\frac{1}{\gamma_{8}^{\mathrm{IV}}} =\tau^{\mathrm{IV}}\left(-\frac{1}{\gamma_{8}^{\mathrm{IV}}}\right)\\ M_{12}^{\mathrm{IV}}&=+\tau^{\mathrm{IV}}\cdot\frac{1}{\gamma_{12}^{\mathrm{IV}}} =\tau^{\mathrm{IV}}\left(+\frac{1}{\gamma_{12}^{\mathrm{IV}}}\right) \end{aligned} \]

This is easy to verify if one imagines the scheme of the action of the right (loaded) part of the structure on node IV and the left part, and the resulting reaction of the left part of the structure—as shown in Fig. 9.

Next, suppose that the moving load \(P=1\) has moved to some point \(n\) of the girder—to the left of node IV.

In complete analogy with the preceding case we obtain the relations:

\[ \tau^{\mathrm{IV}}=\overline{M}_{4}^{\mathrm{IV}}\cdot\varphi_{4}^{\mathrm{IV}} \quad\text{or}\quad \overline{M}_{4}^{\mathrm{IV}}=\tau^{\mathrm{IV}}\cdot\frac{1}{\varphi_{4}^{\mathrm{IV}}} \]

\[ \tau^{\mathrm{IV}}=\overline{M}_{5}^{\mathrm{IV}}\cdot\gamma_{5}^{\mathrm{IV}} \quad\quad \overline{M}_{5}^{\mathrm{IV}}=\tau^{\mathrm{IV}}\cdot\frac{1}{\gamma_{5}^{\mathrm{IV}}} \]

\[ \tau^{\mathrm{IV}}=\overline{M}_{8}^{\mathrm{IV}}\cdot\gamma_{8}^{\mathrm{IV}} \quad\quad \overline{M}_{8}^{\mathrm{IV}}=\tau^{\mathrm{IV}}\cdot\frac{1}{\gamma_{8}^{\mathrm{IV}}} \]

\[ \tau^{\mathrm{IV}}=\overline{M}_{12}^{\mathrm{IV}}\cdot\gamma_{12}^{\mathrm{IV}} \quad\quad \overline{M}_{12}^{\mathrm{IV}}=\tau^{\mathrm{IV}}\cdot\frac{1}{\gamma_{12}^{\mathrm{IV}}} \]

To determine the signs of these equalities, let us turn to the scheme of interaction of the left and right parts of the structure for this case (Fig. 10). When node IV rotates in the positive direction (clockwise), we have the following signs of moments:

\[ \begin{aligned} M_{4}^{\mathrm{IV}}&=+\tau^{\mathrm{IV}}\cdot\frac{1}{\varphi_{4}^{\mathrm{IV}}} =\tau^{\mathrm{IV}}\cdot\left(+\frac{1}{\varphi_{4}^{\mathrm{IV}}}\right)\\ M_{5}^{\mathrm{IV}}&=+\tau^{\mathrm{IV}}\cdot\frac{1}{\gamma_{5}^{\mathrm{IV}}} =\tau^{\mathrm{IV}}\cdot\left(+\frac{1}{\gamma_{5}^{\mathrm{IV}}}\right)\ldots\ldots\ldots (2)\\ M_{8}^{\mathrm{IV}}&=-\tau^{\mathrm{IV}}\cdot\frac{1}{\gamma_{8}^{\mathrm{IV}}} =\tau^{\mathrm{IV}}\cdot\left(-\frac{1}{\gamma_{8}^{\mathrm{IV}}}\right)\\ M_{12}^{\mathrm{IV}}&=+\tau^{\mathrm{IV}}\cdot\frac{1}{\gamma_{12}^{\mathrm{IV}}} =\tau^{\mathrm{IV}}\cdot\left(+\frac{1}{\gamma_{12}^{\mathrm{IV}}}\right) \end{aligned} \]

Relations of an entirely similar form will be obtained for the cases in which the moving load is located on columns (8) and (12), and the signs of the moments can, without any difficulty, be obtained from the corresponding schemes.

A general consequence of the relations considered is that, knowing the values of the angle of rotation \(\tau\) of some joint of the structure for various positions of the moving load \(P\), we can always pass directly to the values of the fixed-end moments of all members meeting at this joint, by simply multiplying the values of \(\tau\) by the coefficients \(\left(\pm\dfrac{1}{\varphi}\right)\) or \(\left(\pm\dfrac{1}{\gamma}\right)\) corresponding to the moments under consideration.

The quantities \(\varphi\) and \(\gamma\) are known, since they are computed in advance when determining the basic focal quantities. As for the signs, as was said above, they are determined with extreme simplicity by the usual sign rule for support moments, once a positive direction has been assigned for the angles \(\tau\).

From what has just been set forth there follows the following proposition: knowing the influence line for the angle of rotation \(\tau\) of some joint, by a proportional change of the ordinates of this influence line in the ratio \(\dfrac{1}{\gamma}\) and \(\dfrac{1}{\varphi}\) (with the corresponding sign), we obtain the influence lines for the fixed-end moments of all members meeting at this joint.

Thus, the influence line for the angle \(\tau\) will be the initial line for constructing the influence lines of support moments, the ordinates of the latter differing from the ordinates of the influence line for \(\tau\) only by the scale numbers \(\left(\pm\dfrac{1}{\varphi}\right)\) and \(\left(\pm\dfrac{1}{\gamma}\right)\).

Let us return to the example in Fig. 8. Consider the support moment \(M_5^{\mathrm{IV}}\). When the moving load \(P=1\) is on the crossbar to the right of joint IV, the scale number for passing from the influence line of \(\tau\) to the influence line of the moment \(M_5^{\mathrm{IV}}\) will be \(\left(-\dfrac{1}{\varphi_5^{\mathrm{IV}}}\right)\); when the load \(P=1\) moves to the left of joint IV, the scale number will be \(\left(+\dfrac{1}{\gamma_5^{\mathrm{IV}}}\right)\). For the support moment \(M_4^{\mathrm{IV}}\) we shall have, respectively, \(\left(-\dfrac{1}{\gamma_4^{\mathrm{IV}}}\right)\) and \(\left(+\dfrac{1}{\varphi_4^{\mathrm{IV}}}\right)\). It will not be difficult to arrive at the general conclusion: to each support moment there correspond, in the general case, two scale numbers, \(\dfrac{1}{\varphi}\) and \(\dfrac{1}{\gamma}\) (each with its own sign), and in their application the frame structure is, as it were, divided into two parts: to the first part (scale number \(\dfrac{1}{\varphi}\)) belong the member itself, whose support moment is being determined, and the entire group of members which it connects with the joint under consideration. To the second part (scale number \(\dfrac{1}{\gamma}\)) belong, without exception, all the remaining members of the structure.

In special cases, when the movement of the moving load is assumed only over a definite group of members (for example, along the crossbars), it may happen that the influence line of the support moments for members not belonging to this group (for example, for columns) will require, over its entire extent, the application of only one scale number.

\(\left(\dfrac{1}{\gamma}\right)\). In the example in Fig. 8, in such a position there are the support moments of the columns \(M_8^{IV}\) and \(M_{12}^{IV}\), if the moving load is displaced only along the girder of the frame. Indeed, for any position of the load on the girder,

\[ M_8^{IV}=\tau^{IV}\left(-\frac{1}{\gamma_8^{IV}}\right);\quad M_{12}^{IV}=\tau^{IV}\left(+\frac{1}{\gamma_{12}^{IV}}\right). \]

In practice this means that, in a frame without displacement of joints, the construction of influence lines for such moments becomes unnecessary if the influence line for the angle of rotation \(\tau\) has been constructed.

The latter will also be the common influence line for these moments. In using it one need only keep in mind the scale factors, with their signs, in order to pass to the final result.

Fig. 11 gives an example of constructing influence lines for fixed-end moments in the case of a single-story viaduct. The influence line of the angle \(\tau^{III}\) is assumed known. By a proportional change in the magnitude of its ordinates in accordance with the scale factors, influence lines have been constructed for the support moments \(M_3^{III}\), \(M_4^{III}\), and \(M_8^{III}\).

The moving load is displaced only along the girders of the structure.

Under this restriction, as indicated above, the construction of a special influence line for the support moment \(M_8^{III}\) is unnecessary, since throughout its entire length the influence line for \(M_8^{III}\) differs from the influence line for \(\tau^{III}\) by one and the same scale factor \(\left(-\dfrac{1}{\gamma_8^{III}}\right)\).

In conclusion of the review of the dependencies between influence lines, we shall point out the use of the focal relations \(K_a\) and \(K_b\) in certain cases of finding influence lines for support moments.

Let us consider (Fig. 11) the lower support moment \(M_8^{VIII}\) of column (8). For all positions of the moving load on the girders of the frame, the upper and lower support moments of column (8) are connected by the focal relation

\[ M_8^{VIII}=-K_{a8}\cdot M_8^{III}. \]

It is therefore quite obvious that, if the influence line \(M_8^{III}\) has been constructed, then in order to find the influence line \(M_8^{VIII}\) it is sufficient to increase the ordinates of the former in the ratio \((-K_{a8})\).

Further, since \(M_8^{III}=\tau^{III}\left(-\dfrac{1}{\gamma_8^{III}}\right)\), it is obvious that \(M_8^{VIII}=\tau^{III}\left(+\dfrac{K_{a8}}{\gamma_8^{III}}\right)\), that is, the influence line for \(M_8^{VIII}\) can be obtained from the influence line for \(\tau^{III}\), the scale factor being \(\left(+\dfrac{K_{a8}}{\gamma_8^{III}}\right)\). Thus, the influence line for \(\tau^{III}\) will serve as the initial line not only in constructing the influence lines for \(M_3^{III}\), \(M_4^{III}\), \(M_8^{III}\), but also in finding the influence line \(M_8^{VIII}\).

§ 3. Construction of influence lines for the angles \(\tau\) of rotation of joints.

To solve this problem we shall use the general method from the theory of statically indeterminate systems, based on the application of Betti’s theorem. Suppose we have a frame (Fig. 12) for which it is required to...

construct the influence line of the angle of rotation \(\tau^I\) of joint I. If the moment \(M^I=1\), applied externally to joint I, causes at point \(x\) a deflection \(y_x\), and a load \(P=1\), situated at point \(x\), causes rotation of joint I through an angle \(\tau^I\), then, by Betti’s theorem:

\[ M^I \cdot \tau^I = P_x y_x, \]

or, for \(M^I=1\) and \(P_x=1\),

\[ \tau^I = y_x \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (3) \]

That is, the influence line for the angle of rotation \(\tau^I\) of joint I is the elastic line of the members of the frame under the action on the frame of the moment \(M^I=1\), applied at joint I.

The individual portions of the influence line over the length of a single member, obviously, represent the elastic line of a simple beam on two supports, loaded only by two end moments (we are considering a frame in which displacements of the joints do not occur).

The support (fixing) moments for all members can easily be found by using the usual procedures of the method of foci—the use of focal ratios and distribution coefficients—as soon as the support moments of the members meeting at the joint under consideration have been determined. The computation of the latter moments, i.e. the distribution of the externally applied moment \(M=1\) to the joint among the members meeting at this joint, presents not the slightest difficulty and is carried out with the aid of distribution coefficients, which, of course, differ from the ordinary distribution coefficients in the method of foci, since there the moment is distributed from one of the members meeting at the joint to the others, whereas here the moment applied externally is to be distributed among all the members. But the method of computing these coefficients from the angular quantities \(\gamma\) of the members remains exactly the same as usual (to make the analogy quite evident, it is only necessary to assume that the external moment \(M=1\) is transmitted to the joint through some fictitious additional member rigidly connected with the joint).

For example, for joint IV in Fig. 8, the coefficient of distribution of the external moment to member 5 will be expressed by the general formula:

\[ \mu_5 = \frac{\dfrac{1}{\gamma_5^{IV}}} {\dfrac{1}{\gamma_5^{IV}}+\dfrac{1}{\gamma_4^{IV}}+\dfrac{1}{\gamma_8^{IV}}+\dfrac{1}{\gamma_{12}^{IV}}} \]

Subsequently a formula will be given which greatly simplifies the determination of such distribution coefficients (Chap. III).

For constructing the elastic line of the members of a frame we must, in the general case, have the support moments for all members, which can be determined by the usual procedures of the method of foci.

But in practice, as will be shown below (Chap. III), it will be sufficient to find the moments (right and left) only for the members meeting at the given joint.

Having thus established that the influence line for angles of rotation can be constructed as the elastic line of the members of the frame, let us proceed to the method of constructing such elastic lines.

§ 4. Construction of the Elastic Line of a Simple Beam on Two Supports, Loaded by Support Moments.

The general case (Fig. 13), when the simply supported beam \(AB\) is loaded by the support moments \(M^A\) and \(M^B\), by the principle of the independence of the action of forces, can easily be decomposed into two separate cases (Fig. 13, \(a\) and \(b\)) of the one-sided action of the support moment \(M^A\) and of the moment \(M^B\); moreover, the deflection at any point due to the combined action of the moments will be equal to the sum of the separate deflections, i.e.

\[ y = y_A + y_B \ . . . . . . . . . . . . . (4) \]

Let us consider the one-sided action of the left support moment \(M^A\). The ordinates of the deflections \(y_A\) for this case, obviously, will be readily obtained if we have the ordinates \(\lambda_A\) for the particular case \(M_A = 1\) (Fig. 14) already computed. Namely, from the proportional relation we have:

\[ y_A = M^A \cdot \lambda_A \ . . . . . . . . . . . . . (5) \]

Therefore, the problem can be reduced to determining the ordinates of the deflections \(\lambda_A\) from loading the simply supported beam \(AB\) by the left support moment \(M^A = 1\) (Fig. 14).

The computation of the ordinates may be carried out by the ordinary methods of strength of materials: analytically or graphico-analytically (by Mohr’s method).

The analytical method gives a solution by double integration of the fundamental equation of bending

\[ \lambda_A = - \iint \frac{M_x}{EJ_x}\,dx^2, \quad \text{where } x \text{ is the distance from the left support.} \]

For our case the moment

\[ M_x = \frac{l - x}{l}, \]

i.e. the deflection

\[ \lambda_A = - \iint \frac{l - x}{EJ_x \cdot l}\,dx^2. \]

Having the law of variation of the moment of inertia \(J_x\) along the length of the beam, \(J_x = f(x)\), we can obtain the solution in general form.

For a beam of constant stiffness, that is, when \(J_x = \text{const}\), integration gives the formula for computing the deflection

\[ \lambda_A = \frac{1}{6EJl}\,x(2l - x)(l - x) = \frac{l^2}{6EJ}\cdot \frac{x}{l} \left(2 - \frac{x}{l}\right) \left(1 - \frac{x}{l}\right) = \frac{l^2}{6EJ}\cdot \eta_A . \tag{6} \]

The coefficient

\[ \eta_A = \frac{x}{l} \left(2 - \frac{x}{l}\right) \left(1 - \frac{x}{l}\right) \quad \text{or} \quad \eta_A = m(2 - m)(1 - m), \]

where \(m = \frac{x}{l}\), can easily be computed for the case of dividing the bar into a specified number of equal segments, and its values may be found in most handbooks.

We give the values of the coefficients \(\eta_A\) for division of the bar into 10 and 12 equal parts:

for division into 10 parts

\(m=\dfrac{x}{l}\)	0	0,1	0,2	0,3	0,4	0,5	0,6	0,7	0,8	0,9	1
\(\eta_A\)	0	0,171	0,288	0,357	0,384	0,375	0,336	0,273	0,192	0,099	0

for division into 12 parts

\(m=\dfrac{x}{l}\)	0	\(\dfrac{1}{12}\)	\(\dfrac{2}{12}\)	\(\dfrac{3}{12}\)	\(\dfrac{4}{12}\)	\(\dfrac{5}{12}\)	\(\dfrac{6}{12}\)	\(\dfrac{7}{12}\)	\(\dfrac{8}{12}\)	\(\dfrac{9}{12}\)	\(\dfrac{10}{12}\)	\(\dfrac{11}{12}\)	1
\(\eta_A\)	0	0,146	0,255	0,328	0,370	0,385	0,375	0,344	0,296	0,234	0,162	0,083	0

It will not be uninteresting to point out that the maximum value of the coefficient \(\eta_A\) corresponds to the abscissa

\[ x=\frac{\sqrt{3}-1}{\sqrt{3}}\,l=0{,}422\,l,\qquad \text{with } \max \eta_A=0{,}385. \]

It is not difficult to see that, if we wish to use the coefficients given in the table for constructing the elastic line under the action of the moment \(M=1\) at the right support, we must take the tabular values in the reverse order, i.e., read the table from the end.

In the case where the moment of inertia \(J_x\) is not constant along the entire length of the bar—there are haunches, or the outline of the member changes continuously along its entire length—the analytical method of solution, in view of the difficulties that arise, becomes disadvantageous and must give way to graphical-analytical procedures. The method of constructing the elastic line of a bar with variable stiffness according to Mohr is well known, and we shall not touch upon it here. It may be taken as the basis for computing the coefficients \(\eta_A\) for the cases, most commonly encountered in practice, of members with variable moment of inertia.

However, such computations become superfluous if one takes into account that the quantities \(\eta\) that interest us may be obtained directly from Suter’s tables) (in his Methode der Festpunkte*) for the so-called “crossed” segments \(y_a\) and \(y_b\) (see § 1).

The possibility of using the indicated tables is easily verified by simple reasoning.

Let (Fig. 15) we have a beam \(AB\), whose deflection under the action of the support moment \(M=1\) is to be determined.

By Betti’s theorem, assuming the beam first loaded by the support moment \(M^A=1\), under whose action at some point \(x\) the deflection \(\eta_A\) is obtained, and in the second case by the load \(P_x=1\) at point \(x\), whose influence causes a rotation of the end of the beam at support \(A\) through the angle \(\alpha_v^A\), we have the equality:

\[ M^A\cdot \alpha_v^A=P_x\cdot \lambda,\quad \text{that is, } 1\cdot \alpha_v^A=1\cdot \lambda \text{ or } \lambda_A=\alpha_v^A \ldots \tag{7} \]

) Similar tables are also given by Strassner in Neuere Methoden. Suter’s tables are reproduced in Calculation of Frame Structures* by Prof. G. V. Ulyaninsky, 3rd and 4th editions.

According to the theory of foci, the “cross” segment is

\[ y_{\mathrm{B}}=-\frac{\alpha_0^{\mathrm{A}}}{\beta}. \]

In Suter’s table, for “cross” segments \(y_{\mathrm{B}}=-l\nu_2\), where \(\nu_2\) is the coefficient in the lower line of one of the horizontal rows of the table.

From the last two equalities,

\[ \alpha_0^{\mathrm{A}}=-y_{\mathrm{B}}\cdot\beta=l\beta\nu_2. \]

Consequently,

\[ \lambda_{\mathrm{A}}=\alpha_0^{\mathrm{A}}=l\beta\nu_2 . . . . . . . . . . . . . . . . . . . . . . . . (8) \]

For a member with constant stiffness we previously obtained

\[ \lambda_{\mathrm{A}}=\frac{l^2}{6EJ}\cdot\eta_{\mathrm{A}}. \]

Taking into account that, for \(J_x=\mathrm{const.}\),

\[ \frac{l}{6EJ}=\beta, \]

this equality may be represented in the form

\[ \lambda_{\mathrm{A}}=l\beta\eta_{\mathrm{A}}, . . . . . . . . . . . . . . . . . . . . . . (9) \]

We note that, for the case of a variable moment of inertia, formula (8) is completely identical with the formula for a constant moment of inertia (9), with the coefficient \(\nu_2=\eta_{\mathrm{A}}\).

Thus, the formula for the deflection

\[ \lambda_{\mathrm{A}}=l\beta\eta_{\mathrm{A}} . . . . . . . . . . . . . . . . . . . . . . (10) \]

is universal for all cases, both constant and variable \(J\), with \(\eta_{\mathrm{A}}=\nu_2\) from Suter’s tables.

By entirely similar reasoning we shall derive that, for the deflection under the action of the right support moment \(M^{\mathrm{B}}=1\), there exists the universal formula

\[ \lambda_{\mathrm{B}}=l\beta\eta_{\mathrm{B}}, . . . . . . . . . . . . . . . . . . . . . . (10') \]

where \(\eta_{\mathrm{B}}=\nu_1\), the coefficient in the upper line of Suter’s tables.

In the general case, for arbitrary support moments \(M^{\mathrm{A}}\) and \(M^{\mathrm{B}}\), by formula (5) of this § we have

\[ y_{\mathrm{A}}=M^{\mathrm{A}}\lambda_{\mathrm{A}} =M^{\mathrm{A}}l\beta\eta_{\mathrm{A}} \]

\[ y_{\mathrm{B}}=M^{\mathrm{B}}\lambda_{\mathrm{B}} =M^{\mathrm{B}}l\beta\eta_{\mathrm{B}} . \]

With the joint action of the moments, the deflection is

\[ y=y_{\mathrm{A}}+y_{\mathrm{B}} =M^{\mathrm{A}}l\beta\eta_{\mathrm{A}} +M^{\mathrm{B}}l\beta\eta_{\mathrm{B}} =l\beta\left(M^{\mathrm{A}}\eta_{\mathrm{A}}+M^{\mathrm{B}}\eta_{\mathrm{B}}\right). . \quad (11) \]

This formula is general for determining the deflections of a simple beam on two supports when it is loaded by its support moments.

It is valid for both constant and variable stiffness of the member, and its coefficients \(\eta_{\mathrm{A}}=\nu_2\) and \(\eta_{\mathrm{B}}=\nu_1\) may be taken from Suter’s tables.

The angular quantity \(\beta\), for cases of variable stiffness, is likewise taken from the corresponding Suter tables.

The moments \(M^{\mathrm{A}}\) and \(M^{\mathrm{B}}\) must be taken with their signs (see § 1).

If between the moments \(M^{\mathrm{A}}\) and \(M^{\mathrm{B}}\) there exists a focal dependence

\[ M^{\mathrm{B}}=-K_b\cdot M^{\mathrm{A}}, \]

the general formula assumes the form

\[ y=M^{\mathrm{A}}\beta l\left(\eta_{\mathrm{A}}-K_b\eta_{\mathrm{B}}\right) . . . . . . . . . . . . . . . . (12) \]

In calculating frames by influence lines, we always also have to take into account a uniformly distributed load (self-weight or equivalent load), for which, of course, the areas of the portions of the influence line must be determined.

We shall show that, in the proposed method of constructing influences as elastic lines, the areas of individual portions of influence lines can be determined with extraordinary ease, and the computation will be carried out with complete mathematical precision.

Let us determine the area bounded by the elastic line of a beam loaded by the support moment \(M^{A}=1\).

Earlier in this paragraph, by applying Betti’s theorem, it was shown that this elastic line is the influence line for the angle of rotation of the left end of the beam at support \(A\), \(-a_{0}^{A}\).

Therefore, loading the beam by a uniform load of intensity \(p=\dfrac{\mathrm{t}}{\mathrm{m}}\), we must obtain an angle \(a_{0}^{A}\) equal to

\[ a_{0}^{A}=p\cdot \Omega_{A}, \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (13) \]

where \(\Omega_{A}\) is the area bounded by the elastic line.

According to Suter’s tables for “crossed” segments under loading of the member by a uniformly distributed load) we have

\[ y_{B}=-\frac{a_{0}^{A}}{\beta}=-p\,l^{2}\cdot \nu_{2}, \quad \text{whence} \]

\[ a_{0}^{A}=p\beta l^{2}\nu_{2}. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (14) \]

Here \(\nu_{2}\), as before, is the coefficient in the lower row of the table.
From comparison of equalities (13) and (14) we have

\[ \Omega_{A}=\beta l^{2}\nu_{2}\ **). \]

We shall adopt for the coefficient \(\nu_{2}\) in this formula another notation, so as not to confuse it with the corresponding coefficient in the formula for deflection; namely, we shall take \(\nu_{2}=\nabla_{A}\); then we obtain the formula for the area

\[ \Omega_{A}=\beta l^{2}\nabla_{A}\ . \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (15) \]

For any moment \(M^{A}\), obviously, we obtain the general formula

\[ \Omega_{A}=M^{A}\beta l^{2}\nabla_{A}. \]

By entirely analogous reasoning, for the elastic line due to the action of the right support moment we obtain the expression for the area

\[ \Omega_{B}=M^{B}\beta l^{2}\nabla_{B}, \]

where \(\nabla_{B}=\nu_{1}\) from Suter’s tables.

*) See this table in Calculation of Frame Structures by Prof. G. V. Ulyaninsky, 4th ed., Tomsk, pp. 182 and 184.

*) In the particular case of a member with constant stiffness*, the area \(\Omega_{A}\) can easily be determined by simple quadrature:

\[ \Omega_{A}=\int_{0}^{l}\lambda_{A}\,dx = \int_{0}^{l}\frac{l^{2}}{6EJ}\cdot \eta_{A}\cdot dx, \]

With

\[ \eta_{A}=\frac{x}{l}\left(2-\frac{x}{l}\right)\left(1-\frac{x}{l}\right), \]

carrying out the integration, we have

\[ \Omega^{A}=\frac{l^{2}}{6EJ}\cdot \frac{l}{4} = \frac{l}{6EJ}\cdot \frac{l^{2}}{4} = \frac{1}{4}\beta l^{2}. \]

The area of the elastic line under the joint action of the right and left support moments is expressed as

\[ \Omega=\Omega_A+\Omega_B=M^A\beta l^2\nabla_A+M^B\beta l^2\nabla_B =\beta l^2\left(M^A\nabla_A+M^B\nabla_B\right) \ldots \ldots \ldots \ldots \tag{16} \]

This is the general formula for determining the area of the elastic line of a simple beam bent by the support moments \(M^A\) and \(M^B\).

The coefficients \(\nabla_A\) and \(\nabla_B\) are taken from Suter’s tables for “cross” segments in the case of a uniformly distributed load, with \(\nabla_A=\nu_2\) and \(\nabla_B=\nu_1\).

The moments must be taken with their signs (see § 1).

For the members of symmetrical outline most frequently encountered in practice (for example, members with equal double-sided haunches, members of constant stiffness), the coefficients are

\[ \nabla_A=\nabla_B=\nu_2=\nu_1=0.25=\frac{1}{4}. \]

In this case the general formula is greatly simplified and has the form

\[ \Omega=\frac{1}{4}\beta l^2\left(M^A+M^B\right) \ldots \ldots \ldots \ldots \ldots \tag{17} \]

This formula is applicable in the majority of cases encountered in practice.

It is quite obvious that, if all ordinates of the elastic line are changed in a known ratio, the area bounded by the elastic line will change in the same ratio; i.e., for example, if the changed ordinate is

\[ y'=y\cdot\frac{1}{\gamma}, \]

then the new area is

\[ \Omega'=\Omega\cdot\frac{1}{\gamma}. \]

Let us summarize everything that has been set forth concerning the construction of influence lines in frames with fixed joints.

1. The proposed method may be applied to all types of frame structures without closed contours, in particular, obviously, to continuous beams.

2. The influence lines of the fixing moments of members meeting at a certain joint of the structure may be obtained from the influence lines for the angle of rotation of this joint by proportionally changing the ordinates of the latter in the ratios of the coefficients \(\left(\pm\frac{1}{\varphi}\right)\) and

\[ \left(\pm\frac{1}{\gamma}\right), \]

corresponding to the moments under consideration, the sign being established according to the general rule of signs for moments in the theory of foci. In known cases the influence line of the angle of rotation of a joint may serve as the influence line for certain fixing moments.

1. The influence line for the angle of rotation of a joint of a frame structure is constructed as the elastic line of the members under the action on the frame of a moment \(M=1\) applied at the joint under consideration.

The indicated moment \(M=1\) is distributed through the structure by means of distribution coefficients and focal ratios, using the usual procedures of the method of foci.

1. The elastic lines of the individual members, both in the case of constant and in the case of variable stiffness, are constructed from the fixed-end moments by means of coefficients from Suter’s tables for finding the so-called “cross” segments.

2. The areas bounded by the elastic line of the members (i.e., the areas of the portions of the influence line) can at the same time be determined with full mathematical accuracy by means of an extremely simple formula, both for the case of constant stiffness of the members and for the case of variable stiffness; moreover, the coefficients entering into the area formula are taken from the same above-mentioned Suter tables for the “cross” segments.

II. Construction of influence lines in frame structures with movable joints.

As was already indicated in the introduction, the construction of influence lines for the general case of a frame with movable joints, when using the method proposed in the present article, is divided into 2 independent computational parts. In the first computational part, by introducing fictitious supports, according to the number of stories of the system, linear immobility is ensured for the joints of the frame, after which the basic fixed-end moments are determined, i.e., the influence lines for the basic fixed-end moments are constructed under the assumption that the joints of the structure are immovable.

In the second computational part, entirely independent of the first, influence lines are constructed for the additional moments arising as a result of displacement of the frame joints under different positions of the moving load.

The influence lines of the total fixed-end moments are obtained by summing the corresponding ordinates of the influence lines for the basic and for the additional moments.

§ 1. Construction of influence lines for additional fixed-end moments arising as a result of displacement of the frame joints.

In order to understand the subsequent derivations and formulas, it is appropriate to recall briefly the essence of the second computational part of the general method of foci.

We shall adhere to the method of exposition adopted by Prof. G. V. Ulyaninsky in Analysis of Frame Structures (3rd ed., 1928, and 4th ed., 1929).

Consider a frame system with 3 independent displacements, i.e., a 3-story system in which, in order to make the joints immovable, the introduction of 3 fictitious supports is required.

We give the 1st fictitious support a displacement \(\Delta x = 0.001\) m, while preserving the immobility of the other two supports*). The reactions of the fictitious supports for this case shall be denoted by \(Z_{\mathrm{I}-\mathrm{I}}\), \(Z_{\mathrm{II}-\mathrm{I}}\), and \(Z_{\mathrm{III}-\mathrm{I}}\). The fixed-end moments computed for this case shall be called \(\mathfrak{M}_{\mathrm{I}}\).

In exactly the same way*), by giving the 2nd fictitious support a displacement \(\Delta x = 0.001\) m, we find the reactions of the fictitious supports \(Z_{\mathrm{I}-\mathrm{II}}\), \(Z_{\mathrm{II}-\mathrm{II}}\), and \(Z_{\mathrm{III}-\mathrm{II}}\) for this case. We denote the fixed-end moments by \(\mathfrak{M}_{\mathrm{II}}\).

Giving a displacement \(\Delta x = 0.001\) m to the 3rd fictitious support, in turn, we have the reactions \(Z_{\mathrm{I}-\mathrm{III}}\), \(Z_{\mathrm{II}-\mathrm{III}}\), and \(Z_{\mathrm{III}-\mathrm{III}}\) and the moments \(\mathfrak{M}_{\mathrm{III}}\).

*) The separate, independent phases of displacement may also have another interpretation, which in no way affects the subsequent derivations. See on this the theory of the method of foci.

If, under the action of some external load, the actual displacements of the 1st, 2nd, and 3rd tiers are expressed by the quantities

\[ \delta_l' = K_{\mathrm{I}}\cdot \Delta x,\qquad \delta_l^{\mathrm{II}} = K_{\mathrm{II}}\cdot \Delta x,\qquad \delta_l^{\mathrm{III}} = K_{\mathrm{III}}\cdot \Delta x, \]

then, by the principle of independence of the action of forces and owing to the proportionality between force actions and displacements, the additional fixing moments due to the displacement of the joints will have the magnitude

\[ M' = K_{\mathrm{I}}\cdot \mathfrak{M}_{\mathrm{I}} + K_{\mathrm{II}}\cdot \mathfrak{M}_{\mathrm{II}} + K_{\mathrm{III}}\cdot \mathfrak{M}_{\mathrm{III}} \ldots\ldots\ldots\ldots (18) \]

Thus, in order to determine the additional fixing moments, it is necessary to know the displacements of the 1st, 2nd, and 3rd tiers, or, what is the same, the coefficients \(K_{\mathrm{I}}, K_{\mathrm{II}}\), and \(K_{\mathrm{III}}\). These displacement coefficients are computed for each separate case of external loading by setting up and solving the system of linear equations

\[ \begin{aligned} K_{\mathrm{I}} Z_{\mathrm{I}-\mathrm{I}} + K_{\mathrm{II}} Z_{\mathrm{I}-\mathrm{II}} + K_{\mathrm{III}} Z_{\mathrm{I}-\mathrm{III}} &= V_{\mathrm{I}},\\ K_{\mathrm{I}} Z_{\mathrm{II}-\mathrm{I}} + K_{\mathrm{II}} Z_{\mathrm{II}-\mathrm{II}} + K_{\mathrm{III}} Z_{\mathrm{II}-\mathrm{III}} &= V_{\mathrm{II}},\\ K_{\mathrm{I}} Z_{\mathrm{III}-\mathrm{I}} + K_{\mathrm{II}} Z_{\mathrm{III}-\mathrm{II}} + K_{\mathrm{III}} Z_{\mathrm{III}-\mathrm{III}} &= V_{\mathrm{III}}, \end{aligned} \qquad\ldots\ldots (19) \]

where \(V_{\mathrm{I}}, V_{\mathrm{II}}\), and \(V_{\mathrm{III}}\) are the displacing forces for the 1st, 2nd, and 3rd fictitious supports, computed for the given case of external loading.

In the static sense these equations, after transferring the quantities \(V\) to the left-hand side, mean that the total reactions of the fictitious supports, after the simultaneous realization of the displacements \(K_{\mathrm{I}}\Delta x\), \(K_{\mathrm{II}}\Delta x\), and \(K_{\mathrm{III}}\Delta x\), are equal to zero, as of course they must be, since in reality there are no fictitious supports.

Thus, if the system contains \(n\) independent displacements, then for each case of loading we must solve \(n\) linear equations with \(n\) unknowns.

In the case of a single-tier system it is necessary to solve only one equation with one unknown; that is, in fact the solution of equations drops out, namely,

\[ KZ = V, \]

whence

\[ K = \frac{V}{Z}, \]

and the additional moments from displacement are

\[ M' = K\mathfrak{M} = \frac{V}{Z}\cdot \mathfrak{M} = V\cdot \frac{\mathfrak{M}}{Z} = V\cdot m,\ldots\ldots\ldots (20) \]

where

\[ m = \frac{\mathfrak{M}}{Z} \]

are the moments from the action of the displacing force \(V=1\).

Let us pass to the question of constructing influence lines for the additional moments due to displacement of the joints.

The general expression for the additional fixing moments is

\[ M' = K_{\mathrm{I}}\cdot \mathfrak{M}_{\mathrm{I}} + K_{\mathrm{II}}\mathfrak{M}_{\mathrm{II}} + K_{\mathrm{III}}\mathfrak{M}_{\mathrm{III}} \ldots\ldots\ldots (18) \]

The moments \(\mathfrak{M}_{\mathrm{I}}, \mathfrak{M}_{\mathrm{II}}\), and \(\mathfrak{M}_{\mathrm{III}}\) are known. The coefficients \(K_{\mathrm{I}}, K_{\mathrm{II}}\), and \(K_{\mathrm{III}}\), proportional to the displacements, are variable quantities, depending on the position of the moving load \(P=1\).

Consequently, in order to construct influence lines for additional moments, one must first construct the influence lines for the coefficients \(K_{\mathrm{I}}, K_{\mathrm{II}}\), and \(K_{\mathrm{III}}\), or for the displacements \(\delta_l^{\mathrm{I}} = K_{\mathrm{I}}\Delta x\), \(\delta_l^{\mathrm{II}} = K_{\mathrm{II}}\Delta x\), and \(\delta_l^{\mathrm{III}} = K_{\mathrm{III}}\Delta x\).

Having these latter influence lines, we multiply their ordinates by the known quantities \(\mathfrak{M}_{\mathrm{I}}, \mathfrak{M}_{\mathrm{II}}\), and \(\mathfrak{M}_{\mathrm{III}}\), and by the corresponding addition obtain the ordinates for the additional moments \(M'\).

Thus, one must be able to construct influence lines for displacements.

Let us have (Fig. 16) a certain frame system for which it is necessary to construct the influence line for the displacement of the 2nd tier. Let us use Betti’s theorem. If the displacing force \(V_{\mathrm{II}} = 1\), producing a general deformation of the frame, creates at point \(x\) a deflection \(y_x\), and the load \(P_x = 1\), applied at point \(x\), produces a displacement of the 2nd tier \(\delta_l^{\mathrm{II}} = K_{\mathrm{II}}\Delta x\), then the equality must hold

\[ P_x \cdot \delta_l^{\mathrm{II}} = V_{\mathrm{II}} \cdot y_x, \]

whence, for \(P_x = 1\) and \(V_{\mathrm{II}} = 1\),

\[ \delta_l^{\mathrm{II}} = y_x, \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (21) \]

in other words, to construct the influence line of the displacement \(\delta_l^{\mathrm{II}}\) one must apply to the frame (in the direction of this displacement) a displacing force \(V_{\mathrm{II}} = 1\) and construct, for this load, the elastic line of the frame members, which will be the required influence line.

The construction of this elastic line presents no difficulty. The deformation of each separate member under loading of the frame only by an external displacing force, in the general case (Fig. 17), consists of:

1) the displacement of the joints serving as supports of the member; the line connecting the joints \(AB\), after deformation, assumes the new position \(A'B'\).

2) the bending of the member relative to the line \(A'B'\) under the influence of the support moments \(m^A\) and \(m^B\).

In certain special cases, when the line \(AB\), connecting the joints after deformation, retains its original position (for example, horizontal transoms of frames with vertical posts), the first part of the deformation is absent, and the elastic line of the frame for these members can be constructed directly according to the instructions of Chapter I of this article.

For all other members, however, it is first necessary to construct the picture of the displacement of the frame joints, i.e., the scheme of the frame with the changed position of the joints after deformation (shown by dashed lines in Fig. 16), and only after this, according to the instructions of Chapter I of this article, to construct the deflection curves of the individual members under the action of the support moments.

The procedure for constructing the influence line for a displacement will be as follows.

1) First it is necessary to determine the fixed-end moments and the reactions of the fictitious supports \(\mathfrak{M}, Z_{\mathrm{I}}, Z_{\mathrm{II}}\), and \(Z_{\mathrm{III}}\) from a displacement by \(\Delta x = 0.001\) m of the 1st, 2nd, and 3rd fictitious supports.

2) We apply, in the direction of the displacement whose influence line is to be constructed (for example, \(\delta_l^{\mathrm{II}}\)), a displacing force \(V = 1\).

We compose the system of linear equations (19):

\[ \left. \begin{aligned} K_{\mathrm{I}}'' Z_{\mathrm{I}-\mathrm{I}} + K_{\mathrm{II}}'' Z_{\mathrm{I}-\mathrm{II}} + K_{\mathrm{III}}'' Z_{\mathrm{I}-\mathrm{III}} &= 0,\\ K_{\mathrm{I}}'' Z_{\mathrm{II}-\mathrm{I}} + K_{\mathrm{II}}'' Z_{\mathrm{II}-\mathrm{II}} + K_{\mathrm{III}}'' Z_{\mathrm{II}-\mathrm{III}} &= 1,\\ K_{\mathrm{I}}'' Z_{\mathrm{III}-\mathrm{I}} + K_{\mathrm{II}}'' Z_{\mathrm{III}-\mathrm{II}} + K_{\mathrm{III}}'' Z_{\mathrm{III}-\mathrm{III}} &= 0 \end{aligned} \right\} \qquad \begin{gathered} \text{The prime mark }('')\text{ means that}\\ \text{the displacements are determined}\\ \text{for the case of a displacing}\\ \text{force } V_{\mathrm{II}}=1 \text{ in the 2nd tier.} \end{gathered} \]

and solve it with respect to the unknown displacement coefficients \(K''\):

\[ K_{\mathrm{I}}'' = \frac{ \begin{vmatrix} 0 & Z_{\mathrm{I}-\mathrm{II}} & Z_{\mathrm{I}-\mathrm{III}}\\ 1 & Z_{\mathrm{II}-\mathrm{II}} & Z_{\mathrm{II}-\mathrm{III}}\\ 0 & Z_{\mathrm{III}-\mathrm{II}} & Z_{\mathrm{III}-\mathrm{III}} \end{vmatrix} }{ \begin{vmatrix} Z_{\mathrm{I}-\mathrm{I}} & Z_{\mathrm{I}-\mathrm{II}} & Z_{\mathrm{I}-\mathrm{III}}\\ Z_{\mathrm{II}-\mathrm{I}} & Z_{\mathrm{II}-\mathrm{II}} & Z_{\mathrm{II}-\mathrm{III}}\\ Z_{\mathrm{III}-\mathrm{I}} & Z_{\mathrm{III}-\mathrm{II}} & Z_{\mathrm{III}-\mathrm{III}} \end{vmatrix} } = - \frac{ \begin{vmatrix} Z_{\mathrm{I}-\mathrm{II}} & Z_{\mathrm{I}-\mathrm{III}}\\ Z_{\mathrm{III}-\mathrm{II}} & Z_{\mathrm{III}-\mathrm{III}} \end{vmatrix} }{D} \]

\[ K_{\mathrm{II}}'' = \frac{ \begin{vmatrix} Z_{\mathrm{I}-\mathrm{I}} & 0 & Z_{\mathrm{I}-\mathrm{III}}\\ Z_{\mathrm{II}-\mathrm{I}} & 1 & Z_{\mathrm{II}-\mathrm{III}}\\ Z_{\mathrm{III}-\mathrm{I}} & 0 & Z_{\mathrm{III}-\mathrm{III}} \end{vmatrix} }{D} = + \frac{ \begin{vmatrix} Z_{\mathrm{I}-\mathrm{I}} & Z_{\mathrm{I}-\mathrm{III}}\\ Z_{\mathrm{III}-\mathrm{I}} & Z_{\mathrm{III}-\mathrm{III}} \end{vmatrix} }{D} \qquad \cdots \cdot (22) \]

\[ K_{\mathrm{III}}'' = \frac{ \begin{vmatrix} Z_{\mathrm{I}-\mathrm{I}} & Z_{\mathrm{I}-\mathrm{II}} & 0\\ Z_{\mathrm{II}-\mathrm{I}} & Z_{\mathrm{II}-\mathrm{II}} & 1\\ Z_{\mathrm{III}-\mathrm{I}} & Z_{\mathrm{III}-\mathrm{II}} & 0 \end{vmatrix} }{D} = - \frac{ \begin{vmatrix} Z_{\mathrm{I}-\mathrm{I}} & Z_{\mathrm{I}-\mathrm{II}}\\ Z_{\mathrm{III}-\mathrm{I}} & Z_{\mathrm{III}-\mathrm{II}} \end{vmatrix} }{D} \]

As we see, the solution of the equations for the case of a single displacing force (with the other two equal to zero) is facilitated by the fact that the numerators turn into minors of the determinant of the system \((D)\).

Having the coefficients \(K_{\mathrm{I}}''\), \(K_{\mathrm{II}}''\), and \(K_{\mathrm{III}}''\), we obtain the magnitudes of the fixed-end moments under the action of the displacing force \(V_{\mathrm{II}}=1\)

\[ m_{\mathrm{II}} = K_{\mathrm{I}}'' \cdot \mathfrak{M}_{\mathrm{I}} + K_{\mathrm{II}}'' \mathfrak{M}_{\mathrm{I}} + K_{\mathrm{III}}'' \mathfrak{M}_{\mathrm{III}} . \]

The displacements of the tiers of the system in this case will be

\[ \delta_{\mathrm{I}-\mathrm{II}}=K_{\mathrm{I}}'' \Delta x,\qquad \delta_{\mathrm{II}-\mathrm{II}}=K_{\mathrm{II}}'' \Delta x,\qquad \delta_{\mathrm{III}-\mathrm{II}}=K_{\mathrm{III}}'' \Delta x. \]

3) From the obtained displacements of the joints and the fixed-end moments, we construct the elastic line of the frame, which will also be the desired influence line for displacement.

In exactly the same way, the influence lines for the displacements of the remaining tiers can be constructed.

It is not difficult to see that, in solving the corresponding systems of linear equations for determining the coefficients \(K\), with one and the same determinant of the system \((D)\) in the denominator of the formulas,

we shall have in the numerator, as in the case analyzed, the minors of the determinant \((D)\).

Therefore the computations in solving the equations will not present much difficulty.

Having constructed the influence lines for the displacements of the separate stories, we can, by formula (18) of this section, construct the influence line for the additional moments \(M'\) due to displacement of the joints of the frame.

In single-story systems the construction of influence lines for additional moments due to displacement is greatly simplified.

The additional moments are

\[ M' = K \cdot \mathfrak{M} \ldots \ldots \ldots \ldots \ldots \ldots (20) \]

The quantity \(\mathfrak{M}\)—equal to the moment when the fictitious support is displaced by \(\Delta x = 0.001\ \text{m}\)—is determined by a preliminary computation and is a constant quantity for each restraining moment under consideration.

The displacement coefficient \(K\), common to all the moments, is a variable quantity for which it is necessary to construct an influence line.

The influence line for \(K\), or for the displacement \(\delta_i = K \Delta x\), may serve as the common influence line for all additional restraining moments.

To construct the influence line for \(\delta_i = K \cdot \Delta x\), we apply, in the direction of the possible displacement, a displacing force \(V = 1\), and from the restraining moments for this case

\[ m = \frac{\mathfrak{M}}{Z} \]

(see above), and from the displacement

\[ \delta_{v=1} = \frac{\Delta x}{Z} = \frac{0.001}{Z} \]

we construct the elastic line of the frame.

It will be the required influence line for \(\delta_i = K \cdot \Delta x\).

Having the influence line for the displacement \(\delta_i\), we, by multiplying its ordinates by the quantities

\[ \frac{\mathfrak{M}}{\Delta x}, \]

obtain the magnitudes of the additional restraining moments, since

\[ \delta_i \cdot \frac{\mathfrak{M}}{\Delta x} = K \Delta x \cdot \frac{\mathfrak{M}}{\Delta x} = K \cdot \mathfrak{M} = M'. \]

Practical examples of constructing influence lines of additional moments are given in the appendix to the present article.

§ 2. Application of the stated method of constructing influence lines of additional moments to the analysis of frames under a fixed load.

The described method of constructing influence lines for additional moments is valid for all possible cases of frame structures, since it is based on the application of Betti’s general theorem.

It is well known that, in the analysis of frames (especially those with closed contours) by the method of foci, the greatest difficulty and complexity is presented by the determination of additional moments due to displacement, since for each case of external loading we must compute the displacing forces and, after substituting the found quantities into the linear equations for the displacements \(K\), must each time solve a system of \(n\) equations with \(n\) unknowns, if \(n\) is the number of stories of the system.

The computation of the displacing forces is an especially cumbersome task.

In the proposed method of constructing influence lines, the determination of the displacing forces is eliminated; as for the solution of the linear equations, it is greatly simplified if, as indicated above, one uses the method of determinants: all the elements of the determinants are the quantities \(Z\), and the denominator of the formulas for \(K\) is the same for all

cases, the numerator consists of the minors of the determinant in the denominator.

It is true that one must solve \(n\) times a system of \(n\) equations, but, when determinants are used, this solution reduces to the application of formulas of the same type, in which the same known quantities \(Z\) appear.

Therefore, if in certain cases of calculating frames under a fixed load it is possible to use ready-made results from the theory of constructing influence lines (without carrying out the construction itself), then we shall obtain substantial simplifications in the calculation.

We have this case in frames with horizontal girders and vertical columns, in the absence of load within the columns. Indeed, the influence lines of displacements within the girders of such frames are constructed only from the fixed-end moments (see above), and therefore all the conclusions and formulas of § 4 of Chapter I of this article are applicable to them. The calculation may be carried out in the following order. After the obligatory determination in all cases of \(\mathfrak{M}\) and \(Z\), we find the displacements of the 1st, 2nd, etc. tiers for the cases of application of the displacing forces \(V=1\), successively in the 1st, 2nd, etc. tiers, i.e., we solve the above-mentioned equations of the same type.

From the obtained displacements:

\[ K'_{\mathrm{I}},\ K'_{\mathrm{II}},\ K'_{\mathrm{III}}\ .\ .\ . \quad \text{for the case } V_{\mathrm{I}}=1 \]

\[ K''_{\mathrm{I}},\ K''_{\mathrm{II}},\ K''_{\mathrm{III}}\ .\ .\ . \quad ” \quad ” \quad V_{\mathrm{II}}=1 \]

\[ K'''_{\mathrm{I}},\ K'''_{\mathrm{II}},\ K'''_{\mathrm{III}}\ .\ .\ . \quad ” \quad ” \quad V_{\mathrm{III}}=1 \]

\[ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ . \]

we compute the fixed-end moments only for the girders:

\[ m_{\mathrm{I}} = K'_{\mathrm{I}}\mathfrak{M}_{\mathrm{I}} + K'_{\mathrm{II}}\mathfrak{M}_{\mathrm{II}} + K'_{\mathrm{III}}\mathfrak{M}_{\mathrm{III}} + .\ .\ .\ .\ .\ .\ . \quad \text{for } V_{\mathrm{I}}=1 \]

\[ m_{\mathrm{II}} = K''_{\mathrm{I}}\mathfrak{M}_{\mathrm{I}} + K''_{\mathrm{II}}\mathfrak{M}_{\mathrm{II}} + K''_{\mathrm{III}}\mathfrak{M}_{\mathrm{III}} + .\ .\ .\ .\ .\ .\ . \quad ” \quad V_{\mathrm{II}}=1 \]

\[ m_{\mathrm{III}} = K'''_{\mathrm{I}}\mathfrak{M}_{\mathrm{I}} + K'''_{\mathrm{II}}\mathfrak{M}_{\mathrm{II}} + K'''_{\mathrm{III}}\mathfrak{M}_{\mathrm{III}} + .\ .\ .\ .\ .\ .\ . \quad ” \quad V_{\mathrm{III}}=1 \]

\[ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ . \]

\[ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ . \]

From these moments one can construct the elastic lines of the girders, which will be the influence lines for the displacements \(\delta_l^{\mathrm{I}}, \delta_l^{\mathrm{II}}\), and \(\delta_l^{\mathrm{III}}\) under the action of a moving load on the girders. Without constructing the elastic lines, let us use the general formulas from § 4, Chapter I, to compute ordinates and areas. It is not difficult to see that, under the action of a vertical concentrated load \(P\) within a certain girder, we have, by formula (11):

\[ \delta_l^{\mathrm{I}} = P\beta l \left( m_{\mathrm{I}}^{A}\eta_A + m_{\mathrm{I}}^{B}\eta_B \right), \qquad K_{\mathrm{I}} = P\cdot \frac{\beta l}{\Delta x} \left( m_{\mathrm{I}}^{A}\eta_A + m_{\mathrm{I}}^{B}\eta_B \right) \]

\[ \delta_l^{\mathrm{II}} = P\beta l \left( m_{\mathrm{II}}^{A}\eta_A + m_{\mathrm{II}}^{B}\eta_B \right), \quad \text{or} \quad K_{\mathrm{II}} = P\cdot \frac{\beta l}{\Delta x} \left( m_{\mathrm{II}}^{A}\eta_A + m_{\mathrm{II}}^{B}\eta_B \right) \ .\ .\ (23) \]

\[ \delta_l^{\mathrm{III}} = P\beta l \left( m_{\mathrm{III}}^{A}\eta_A + m_{\mathrm{III}}^{B}\eta_B \right), \qquad K_{\mathrm{III}} = P\cdot \frac{\beta l}{\Delta x} \left( m_{\mathrm{III}}^{A}\eta_A + m_{\mathrm{III}}^{B}\eta_B \right) \]

\[ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ . \]

\[ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ . \]

where \(\beta\) and \(l\) are quantities pertaining to the given girder, \(m^A\) and \(m^B\) are the left and right fixed-end moments, and \(\eta_A\) and \(\eta_B\) are coefficients corresponding to the position of the load on the girder, i.e., to its distance from the supports \(A\) and \(B\).

Under the action of a uniformly distributed load \(p\ \dfrac{\text{tons}}{\text{lin. meter}}\):

\[ K_{\mathrm{I}}=p\cdot\frac{\beta l^{2}}{\Delta x}\left(m_{\mathrm{I}}^{A}\nabla_{A}+m_{\mathrm{I}}^{B}\nabla_{B}\right) \]

\[ K_{\mathrm{II}}=p\cdot\frac{\beta l^{2}}{\Delta x}\left(m_{\mathrm{II}}^{A}\nabla_{A}+m_{\mathrm{II}}^{B}\nabla_{B}\right)\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (24) \]

\[ K_{\mathrm{III}}=p\cdot\frac{\beta l^{2}}{\Delta x}\left(m_{\mathrm{III}}^{A}\nabla_{A}+m_{\mathrm{III}}^{B}\nabla_{B}\right) \]

\[ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ . \]

\[ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ . \]

Any additional moment due to displacements, when the \(K\)’s are known, may be obtained from formula (18)

\[ M'=K_{\mathrm{I}}\mathfrak{M}_{\mathrm{I}}+K_{\mathrm{II}}\mathfrak{M}_{\mathrm{II}}+K_{\mathrm{III}}\mathfrak{M}_{\mathrm{III}}\ .\ .\ . \]

Thus, for any case of loading of the girders of a frame we can obtain the additional moments without determining the displacing forces and without solving a system of equations for each individual case.

It is easy to see that, under a horizontal load applied at the column joints, we can obtain the values of the additional moments directly, by proportionally increasing the moments \(m_{\mathrm{I}}, m_{\mathrm{II}}, m_{\mathrm{III}}, \ldots\), corresponding to the joint loads \(V_{\mathrm{I}}=1, V_{\mathrm{II}}=1, V_{\mathrm{III}}=1\), etc.

It follows from this that, in cases where, in addition to the loading of the girders, the horizontal joint loading of the columns must also be taken into account, the proposed method of using influence lines in the analysis of frames under a fixed load encounters no difficulties in its application.

The general formulas (23) and (24) for determining the coefficients \(K\) are considerably simplified for the case when the girder is taken as a beam of constant stiffness. In view of their practical interest, we shall give these formulas with numerical coefficients for cases frequently encountered in practice. The displacement \(\Delta x\) is taken equal to \(0.001\) m.

If a concentrated load \(P\) is located at a distance \(a\) from the left support \(A\), then:

\[ \begin{aligned} \text{for } a&=\frac{l}{4} & K&=P\beta l\left(328m^{A}+234m^{B}\right) \\[4pt] \text{for } a&=\frac{l}{3} & K&=P\beta l\left(370m^{A}+296m^{B}\right) \\[4pt] \text{for } a&=\frac{l}{2} & K&=375P\beta l\left(m^{A}+m^{B}\right) \end{aligned} \qquad\}\ .\ .\ .\ .\ .\ (25) \]

If the whole span of the girder is loaded by a uniformly distributed load \(p\ \dfrac{\text{tons}}{\text{lin. meter}}\),

\[ K=250p\beta l^{2}\left(m^{A}+m^{B}\right)\ .\ .\ .\ .\ .\ .\ .\ .\ (26) \]

In conclusion, it may be pointed out that, as is usual in the method of foci, the presence in the system of 2–3 tiers should be regarded as the limit of practical applicability of the procedure set forth. When the number of tiers is \(>3\), the computations become so cumbersome that use of the method of foci becomes practically disadvantageous, since analytic methods (for example, the method of fictitious hinges, Gehler’s method) give a more rapid and elegant solution.

Let us briefly summarize all that has been set forth in Chapter 2.

1) The construction of influence lines for fixing moments in frames with movable joints is divided into 2 independent computational parts: in the first part of the calculation fictitious supports are introduced and influence lines are constructed for the frame with immovable joints—that is, influence lines for the principal fixing moments. In the second part, influence lines are constructed for the additional moments arising as a result of displacement of the joints of the frame.

2) The method of constructing influence lines for additional moments, set forth in the present chapter, is applicable to all cases of frames without exception.

3) To construct influence lines for additional moments it is necessary first to construct influence lines for each independent displacement of the system (for example, in a 3-tier system it is necessary to construct 3 separate influents for the displacements of each tier). In the case of a one-tier frame, the influence line for the displacement may serve as a common influent for all fixing moments, since the influence lines of the moments differ from the influence line for the displacement only by the scale factor \(\dfrac{\mathfrak{M}}{\Delta x}\), known for each moment.

4) To construct the influence line for a displacement, one must apply to the frame, in the direction of this displacement, a displacing force \(V = 1\) and construct for this case the deflection line of the frame. To make such a construction possible, all independent displacements and fixing moments must be determined. If the load moves only over a definite group of members, the fixing moments are calculated only for these members.

5) To obtain from the influence lines for displacements the influence lines of additional moments, the ordinates of the former are multiplied by the coefficients

\[ \frac{\mathfrak{M}_{I}}{\Delta x}, \qquad \frac{\mathfrak{M}_{II}}{\Delta x}, \qquad \frac{\mathfrak{M}_{III}}{\Delta x}, \ldots \]

(which represent the corresponding fixing moments for displacement of the 1st, 2nd, etc. fictitious supports by \(\Delta x\), divided by the value \(\Delta x\) in meters) and are added algebraically.

6) In calculating frames with horizontal girders and vertical columns under a fixed load, application of the method set forth for constructing influence lines may provide a substantial simplification in the computation of additional moments from displacement, if it is possible to avoid constructing the influence lines themselves for displacement and, for the calculation of displacements, to use ready-made formulas. This condition is fulfilled when the load is absent within the limits of the columns and is present only on the girders and at the column joints.

III. Practical Application of the Method Set Forth for Constructing Influence Lines

§ 1. Use of the properties of angular foci.

The presence, in an elastically fixed-end member, of focal points for moments has as its direct consequence the existence of completely analogous focal points for the angles of rotation of the support joints of the given member.

Indeed, let some member \(AB\) (Fig. 18) of a frame structure (with immovable joints) be subjected to bending as a result of the rotation of joint \(B\), caused by the loading of the spans to the right of member \(AB\).

Let us denote the angles of rotation of joints \(A\) and \(B\) by \(\tau^A\) and \(\tau^B\), and let us consider these angles positive in the case of rotation of the joint in the clockwise direction.

If end \(B\) of member \(AB\), rigidly connected with joint \(B\), has rotated through the angle \(\tau^B\), then we must regard this rotation as the result of the action of the support moment \(M^B\). From the definition of the angular quantity \(\gamma^B\) (§ 1, Chap. I) there follows the equality \(\tau^B = M^B \gamma^B\). Further, reasoning exactly as in § 2, Chap. I, and using the usual sign rule for moments, from the scheme of action of the right-hand part of the structure on member \(AB\) (similarly to Fig. 9) we obtain the sign of this equality

\[ \tau^B = - M^B \gamma^B \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (27) \]

Next, the rotation of joint \(A\) through the angle \(\tau^A\) is the result of the action on the joint of the moment \(M^A\), opposite in direction to the fixing moment \(M^A\). From the definition of the angular quantity \(\varphi^A\) we have \(\tau^A = M^A \cdot \varphi^A\).

Representing to ourselves the scheme of interaction of the left-hand part of the structure and member \(AB\) (similarly to Fig. 9), we can assign to this last equality the necessary sign

\[ \tau^A = - M^A \varphi^A \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (28) \]

For our case the fixing moments are connected by the focal relation

\[ \frac{M^A}{M^B} = - K_a \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (29) \]

Dividing equality (28) by (27) and using (29), we obtain

\[ \frac{\tau^A}{\tau^B} = + \frac{M^A}{M^B} \cdot \frac{\varphi^A}{\gamma^B} = - K_a \cdot \frac{\varphi^A}{\gamma^B} = \mathrm{const} \ldots \ldots \ldots (30) \]

Thus, between the angles of rotation of the joints there exists a constant focal relation, completely analogous to the moment focal relation.

We shall denote the quantity \(K_a \dfrac{\varphi^A}{\gamma^B}\) by \(\chi_a\) and call it, by analogy with \(K_a\), the left angular focal relation. Then

\[ \frac{\tau^A}{\tau^B} = - \chi_a . \]

The equality

\[ \chi_a = K_a \frac{\varphi^A}{\gamma^B} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (31) \]

is the general formula for computing the left angular focal relation \(\chi_a\) from the left moment focal relation \(K_a\).

For the case when the loaded spans are located to the left of member \(AB\), by precisely the same reasoning we are convinced of the existence of the right angular focal ratio \(x_b\),

\[ \frac{\tau^B}{\tau^A}=-K_b\frac{\varphi^B}{\gamma^A}=-x_b=\mathrm{const}. \]

The equality

\[ x_b=K_b\frac{\varphi^B}{\gamma^A}\ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (32) \]

is the general formula for calculating the right angular focal ratio from the known right moment focal ratio.

The angular quantities \(\varphi\) and \(\gamma\) are known from the preliminary analysis of the frame. Therefore, the practical application of formulas (32) and (31) presents no difficulty. The formulas are extremely simple.

If desired, formulas (31) and (32) may be given another form. Namely, from the theory of foci it is known that

\[ K_a=\frac{\beta}{\alpha_1+\varphi^A},\quad \text{or}\quad \varphi^A=\frac{\beta-\alpha_1K_a}{K_a};\quad \gamma^B=\alpha_2-K_a\beta. \]

Substituting in formula (31), we obtain

\[ x_a=\frac{\beta-\alpha_1K_a}{\alpha_2-\beta K_a} \ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (31^{a}) \]

For the right angular focal ratio we analogously obtain

\[ x_b=\frac{\beta-\alpha_2K_b}{\alpha_1-\beta K_b} \ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (32^{a}) \]

For members with constant stiffness, when

\[ \alpha_1=\alpha_2=2\beta, \]

formulas \((31^a)\) and \((32^a)\) are simplified and take the form

\[ x_a=\frac{1-2K_a}{2-K_a} \ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (31^{b}) \]

\[ x_b=\frac{1-2K_b}{2-K_b} \ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (32^{b}) \]

Solving formulas \((31^b)\) and \((32^b)\) with respect to \(K_a\) and \(K_b\), we obtain

\[ K_a=\frac{1-2x_a}{2-x_a} \ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (33) \]

\[ K_b=\frac{1-2x_b}{2-x_b} \ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (34) \]

Let us give one more formula for calculating angular focal ratios for members with constant stiffness.

Substituting the expression for the moment focal ratio

\[ K_a=\frac{1}{2+\dfrac{\varphi^A}{\beta}} \]

in formula \((31^{b})\), after transformations we obtain

\[ \varkappa_{a}=\frac{1}{2+\dfrac{3\beta}{\varphi^{A}}}\ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (35) \]

In an entirely similar way we obtain for the right focus

\[ \varkappa_{b}=\frac{1}{2+\dfrac{3\beta}{\varphi^{B}}}\ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (36) \]

Repeating the usual course of reasoning of the theory of moment foci, we easily arrive at the conclusion that the existence of angular focal ratios \(\dfrac{\tau^{A}}{\tau^{B}}=-\varkappa_{a}\) and \(\dfrac{\tau^{B}}{\tau^{A}}=-\varkappa_{b}\) with the minus sign indicates that the angular focal points lie within the member \(AB\). The relations \(\varkappa_{a}=\dfrac{a'}{l-a'}\) and \(\varkappa_{b}=\dfrac{b'}{l-b'}\) (where \(a'\) and \(b'\) are the angular focal distances), which follow from this proposition and are analogous to the known formulas of the theory of moment foci: \(K_{a}=\dfrac{a}{l-a}\) and \(K_{b}=\dfrac{b}{l-b}\), make it possible to obtain from formulas (35) and (36) formulas for computing the angular focal distances

\[ a'=\frac{l}{3}\cdot\frac{1}{1+\dfrac{\beta}{\varphi^{A}}}; \qquad b'=\frac{l}{3}\cdot\frac{1}{1+\dfrac{\beta}{\varphi^{B}}} \ . \ . \ . \ (37) \]

With complete fixity of end \(A\), i.e. for \(\varphi^{A}=0\), \(a'=0\) and \(\varkappa_{a}'=0\); with hinged fastening \((\varphi^{A}=\infty)\), \(a'=\dfrac{l}{3}\) and \(\varkappa_{a}=\dfrac{1}{2}\). Thus, we obtain a picture opposite to that which occurs for moment foci.

With partial fixity (for example, the heel of the post rests on an elastic foundation) we have, for moment foci*):

\[ K=\frac{1}{2\nu} \qquad a=\frac{1}{1+2\nu}\cdot l, \quad \text{where} \quad \nu=1+\frac{3}{l} \quad \text{and } l \text{ is the length} \]

of the element (the height of the post) in meters.

From formulas \((31^{b})\) and \((32^{b})\), after substitution, we obtain

\[ \varkappa = \frac{2(\nu-1)}{4\nu-1} = \frac{\left(1+\dfrac{3}{l}-1\right)2}{4\left(1+\dfrac{3}{l}\right)-1} = \frac{2}{l+4} \ . \ . \ . \ (37) \]

Taking the mean value for the coefficient \(\nu=1.50\), we have

\[ \varkappa=\frac{1}{5} \qquad a'=\frac{l}{6}=0.167\,l. \]

*) See “Analysis of Frame Structures” by Prof. G. V. Ulyaninsky, 3rd and 4th eds.

With this we shall end the review of the general properties of angular foci and proceed to their practical application in constructing influence lines.

The existence of a focal relation between the angles of rotation of the joints of an unloaded member helps us simplify the construction of the influence line of the angles of rotation \(\tau\). Indeed, if the moving load \(P=1\) is located to the right of member \(AB\), then, in view of the relation

\[ \tau^{A}=-\chi_{a}\tau^{B}, \]

we are relieved of the need to construct (i.e., to compute the ordinates by the method of elastic lines—§ 4, Ch. I) the influence line for \(\tau^{A}\), if the influence line for \(\tau^{B}\) has been constructed for spans to the right of joint \(B\). It is enough to multiply the ordinates of the line \(\tau^{B}\) in these spans by the left angular focal ratio \((-\chi_{a})\), in order to obtain the ordinates for \(\tau^{A}\).

In exactly the same way, for spans to the left of member \(AB\), from the known ordinates of the influence line \(\tau^{A}\) we can obtain the ordinates of the influence line \(\tau^{B}\) by multiplying the former by \((-\chi_{b})\).

Applying similar reasoning to a consecutive series of joints, it is easy to arrive at the conclusion that the computation of the ordinates of the influence line of the angle of rotation of a certain joint, as the ordinates of the elastic line due to the action on the joint of a moment \(M=1\), need be carried out only on the portions of the members meeting at the given joint, since, for portions of the frame beyond these members (beyond the joints adjacent to the one under consideration), the influence line for the angle of rotation of the given joint can be obtained from the influence lines for the angles of rotation \(\tau\) of the adjacent joints, by multiplying the ordinates of the latter by the corresponding angular focal ratios.

The computation of the ordinates of elastic lines constitutes the most complicated part of the work in finding influence lines for the angles of rotation \(\tau\); therefore, the indicated use of angular focal ratios considerably facilitates the work.

It should be noted that the use of angular foci for the indicated purpose is possible only in the case when the rotation of one joint is transmitted to another only through one member, i.e., when the frame contains no closed sections. But we had the same restriction in Ch. I, when considering the new method of constructing influence lines in general in frames with fixed joints.

In the solution of the subsequent examples one can see the practical use of angular focal ratios.

From the considerations just presented concerning the construction of the influence lines \(\tau\), it may be concluded that the determination of the fixed-end moments due to the action on the frame of a moment \(M=1\) at the joint under consideration must be carried out only for the members meeting at this joint (in practice usually only for two members, i.e., a total of 4 fixed-end moments), since only along these members is the influence line constructed as an elastic line according to the directions of § 4, Ch. I of this article.

It is not difficult to see how much this consideration simplifies and shortens the computational work for distributing moments.

§ 2. Distribution coefficients for a moment applied externally to a joint.

In the described method of constructing influence lines, we practically encounter several times the operation of distributing a moment \(M=1\), applied to a joint externally, among the members meeting at this joint.

Let us have (Fig. 19) a joint \(A\), at which 4 elements meet. If an external moment \(M\) is applied to this joint and is to be distributed among members (1), (2), (3), and (4), then we can easily imagine,

that this moment is transmitted to the joint through a certain fictitious (5) member, under which assumption all the formulas of the general theory of foci for computing the distribution coefficients remain valid. That is, we shall have

\[ \mu_1=\mu_{5-1}= \frac{\dfrac{1}{\gamma_1^A}} {\dfrac{1}{\gamma_1^A}+\dfrac{1}{\gamma_2^A}+\dfrac{1}{\gamma_3^A}+\dfrac{1}{\gamma_4^A}} \]

\[ \mu_2=\mu_{5-2}= \frac{\dfrac{1}{\gamma_2^A}} {\dfrac{1}{\gamma_1^A}+\dfrac{1}{\gamma_2^A}+\dfrac{1}{\gamma_3^A}+\dfrac{1}{\gamma_4^A}}, \quad \text{or} \quad \mu_{5-2}=\mu_{5-1}\frac{\gamma_1^A}{\gamma_2^A} \]

\[ \mu_3=\mu_{5-3}= \frac{\dfrac{1}{\gamma_3^A}} {\dfrac{1}{\gamma_1^A}+\dfrac{1}{\gamma_2^A}+\dfrac{1}{\gamma_3^A}+\dfrac{1}{\gamma_4^A}}, \quad \text{or} \quad \mu_{5-3}=\mu_{5-1}\frac{\gamma_1^A}{\gamma_3^A} \]

\[ \mu_4=\mu_{5-4}= \frac{\dfrac{1}{\gamma_4^A}} {\dfrac{1}{\gamma_1^A}+\dfrac{1}{\gamma_2^A}+\dfrac{1}{\gamma_3^A}+\dfrac{1}{\gamma_4^A}}, \quad \text{or} \quad \mu_{5-4}=\mu_{5-1}\frac{\gamma_1^A}{\gamma_4^A} \]

Transforming the formula for \(\mu\), we have:

\[ \mu_1= \frac{\dfrac{1}{\gamma_1^A}} {\dfrac{1}{\gamma_1^A}+ \left( \dfrac{1}{\gamma_2^A}+ \dfrac{1}{\gamma_3^A}+ \dfrac{1}{\gamma_4^A} \right)}, \quad \text{but} \quad \dfrac{1}{\gamma_2^A}+\dfrac{1}{\gamma_3^A}+\dfrac{1}{\gamma_4^A} = \dfrac{1}{\varphi_1^A}, \]

\[ \text{therefore} \quad \mu_1= \frac{\dfrac{1}{\gamma_1^A}} {\dfrac{1}{\gamma_1^A}+\dfrac{1}{\varphi_1^A}}, \]

\[ \text{or} \quad \mu_1=\frac{\varphi_1^A}{\varphi_1^A+\gamma_1^A} \]

In exactly the same way,

\[ \left. \begin{aligned} \mu_2&=-\frac{\varphi_2^A}{\varphi_2^A+\gamma_2^A}\\[6pt] \mu_3&=-\frac{\varphi_3^A}{\varphi_3^A+\gamma_3^A}\\[6pt] \mu_4&=-\frac{\varphi_4^A}{\varphi_4^A+\gamma_4^A} \end{aligned} \right\} \qquad \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \tag{38} \]

The quantities \(\gamma_1^A\), \(\varphi_1^A\), and so on, are computed in the preliminary calculation; consequently, the use of formulas (38) greatly simplifies the computation of the distribution coefficients.

Practically it proves more convenient to compute, by formulas (38), only one of the distribution coefficients, for example \(\mu_1\), and to obtain the rest from the relations

\[ \mu_2=\mu_1\cdot\frac{\gamma_1^\Delta}{\gamma_2^\Delta}; \qquad \mu_3=\mu_1\cdot\frac{\gamma_1^\Delta}{\gamma_3^\Delta}; \qquad \mu_4=\mu_1\cdot\frac{\gamma_1^\Delta}{\gamma_4^\Delta}\ldots\ldots\ldots\ldots \tag{39} \]

It is appropriate to mention here a practical device for computing the angular quantities \(\varphi\) from the angular quantities \(\gamma\). As is known, the quantities \(\varphi\) are obtained from the equations

\[ \frac{1}{\varphi}=\frac{1}{\gamma_1}+\frac{1}{\gamma_2}+\frac{1}{\gamma_3}\ldots \]

and their calculation constitutes one of the most cumbersome operations in the preliminary computation of the basic quantities. To facilitate these calculations, the following graphical device may be proposed, consisting in the construction of a scale for determining the quantities \(\varphi\). If only three members meet at a joint, then

\[ \frac{1}{\varphi_{1-2}}=\frac{1}{\gamma_1}+\frac{1}{\gamma_2}. \]

Draw an angle \(=120^\circ\) and draw its bisector (Fig. 20). If on the sides of the angle the segments \(\overline{OA}=\gamma_1\) and \(\overline{OB}=\gamma_2\) are set off (on a known scale), and the secant \(AB\) is drawn, then on the bisector we obtain the segment \(\overline{OX}\), which will be equal to \(\varphi_{1-2}\), on the same scale as \(\gamma_1\) and \(\gamma_2\). The proof is extremely simple and may be based on elementary propositions of analytic geometry: taking the sides of the angle as the axes \(Ox\) and \(Oy\) of oblique coordinates, we have the equation of the straight line \(AB\) “in intercepts”:

\[ \frac{x}{\gamma_2}+\frac{y}{\gamma_1}=1, \]

the equation of the bisector \(OX\) is

\[ x=y. \]

Moreover, for a coordinate angle of \(120^\circ\), the radius vector \(\overline{OX}=x_0=y_0\), as may be seen from the equilateral \(\triangle OXN\), in which all angles are \(60^\circ\).

For the point \(X\), the intersection of the straight line \(AB\) with the bisector \(OX\), we have

\[ \frac{\overline{OX}}{\gamma_2}+\frac{\overline{OX}}{\gamma_1}=1 \quad\text{or}\quad \frac{1}{\gamma_2}+\frac{1}{\gamma_1}=\frac{1}{\overline{OX}}, \quad\text{and since}\quad \frac{1}{\gamma_2}+\frac{1}{\gamma_1}=\frac{1}{\varphi_{1-2}}, \]

then \(\overline{OX}=\varphi_{1-2}\), as was required to be proved.

Thus, by drawing an angle \(=120^\circ\) and drawing its bisector, we shall have a scale for direct measurement of the quantity \(\varphi_{1-2}\) for known \(\gamma_1\) and \(\gamma_2\). If not 3 members but more meet at a joint, the computations should be carried out in parts, grouping the quantities \(\gamma\) two at a time.

Namely, if

\[ \frac{1}{\varphi}=\frac{1}{\gamma_1}+\frac{1}{\gamma_2}+\frac{1}{\gamma_3}+\frac{1}{\gamma_4}, \]

then, having separately found

\[ \frac{1}{\gamma_1}+\frac{1}{\gamma_2}=\frac{1}{\varphi_{1-2}} \quad\text{and}\quad \frac{1}{\gamma_3}+\frac{1}{\gamma_4}=\frac{1}{\varphi_{3-4}} \]

by means of the scale, we finally obtain

\[ \frac{1}{\varphi}=\frac{1}{\varphi_{1-2}}+\frac{1}{\varphi_{3-4}}, \]

using the same scale.

The construction of the scale is as elementary as its use; therefore, with frequent use of the method of foci for the analysis of frames

its use may prove of substantial benefit, if not for the computations themselves, then for checking them.

§ 3. Simplifications in the computation of ordinates of elastic lines.

The general formula for computing the ordinates of an elastic line (Ch. I, § 4)

\[ y=\beta l\left(M^A\eta_A+M^B\eta_B\right)\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (11) \]

for the case of frames with fixed joints and continuous beams, when there is a focal relation between the moments \(M^A\) and \(M^B\),

\[ \frac{M^B}{M^A}=-K_b \]

takes the form

\[ y=\beta l M^A\left(\eta_A-K_b\eta_B\right)\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (12) \]

In this form the expression for \(y\) is most convenient for computations.

Having computed for each point of the span the factor

\[ (\eta_A-K_b\eta_B), \]

we, multiplying it by \(\beta l M_A\), obtain the ordinates \(y\).

In exactly the same way, if the focal relation is

\[ \frac{M^A}{M^B}=-K_a \]

(for the span to the left of the joint whose influence line \(\tau\) we determine),

\[ y=\beta l M^A\left(\eta_B-K_a\eta_A\right),\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (12\text{-}a) \]

that is, it is necessary to compute the factor

\[ (\eta_B-K_a\eta_A). \]

For the particular case of members with constant stiffness—the case corresponding to the usual methods of design practice—the construction of a graph may be recommended for quickly obtaining the values of the factor \((\eta_A-K_b\eta_B)\).

Assigning to the individual points of the member (for example, at intervals of \(\frac{1}{12}\) of the span length) \(\eta_A\) and \(\eta_B\) equal to the corresponding numerical values from the tables, we easily see that the magnitude of the factor \((\eta_A-K_b\eta_B)\) will in each case be connected with the focal ratio \(K_b\) by a linear dependence, which on the graph gives a straight line.

Plotting the quantities \(K_b\) on the axis of abscissas, and the values of the factor \((\eta_A-K_b\eta_B)\) on the axis of ordinates, we shall evidently obtain a series of straight lines, in which each line corresponds to a definite point of the member.

Use of the graph will considerably shorten the computation of the ordinates of elastic lines.

§ 4. Determination of influence lines for support moments in continuous beams.

Application of the described method of constructing influence lines to continuous beams—the simplest type of frame structures—gives a number of essential simplifications. These should be mentioned.

Let us have (Fig. 21) a multi-span continuous beam. Let us find the influence line for the moment \(M_c\) at support \(C\).

The distribution coefficients for an external moment, according to formulas (38), will be

\[ \text{will be }\quad \psi_n=\frac{\varphi_n^c}{\varphi_n^c+\gamma_n^c} =\frac{\gamma_{n+1}^c}{\gamma_{n+1}^c+\gamma_n^c}; \quad M_{n+1}= \frac{\varphi_{n+1}^c}{\varphi_{n+1}^c+\gamma_{n+1}^c} =\frac{\gamma_n^c}{\gamma_n^c+\gamma_{n+1}^c}, \]

Since, obviously, \(\varphi_n^c=\gamma_{n+1}^c\) and \(\varphi_{n+1}^c=\gamma_n^c\). Taking the sign into account, from the external moment \(M=1\) we obtain the support moments

\[ M_n^c=-\frac{\gamma_{n+1}^c}{\gamma_{n+1}^c+\gamma_n^c} \qquad M_{n+1}^c=+\frac{\gamma_n^c}{\gamma_{n+1}^c+\gamma_n^c}. \]

The ordinates of the influence line for the angle of rotation \(\tau^c\) of the support section \(C\) are obtained from formulas (12):

\[ \text{for the } n\text{-th span}\quad \tau^c=M_n^c \beta_n l_n(\eta_B-K_{an}\eta_A) =-\frac{\gamma_{n+1}^c}{\gamma_{n+1}^c+\gamma_n^c}\,\beta_n l_n(\eta_B-K_{an}\eta_A) \]

\[ \text{for the }(n+1)\text{-st span}\quad \tau^c=M_{n+1}^c \beta_{n+1}l_{n+1}(\eta_A-K_{b(n+1)}\eta_B)= \]

\[ =+\frac{\gamma_n^c}{\gamma_{n+1}^c+\gamma_n^c}\,\beta_{n+1}l_{n+1}(\eta_A-K_{b(n+1)}\eta_B) \]

To pass from the influence line for \(\tau^c\) to the influence line for \(M_c\), we have the scale numbers (with their signs):

\[ \text{for the } n\text{ span}\quad \left(+\frac{1}{\gamma_{n+1}^c}\right)\ \text{or}\ \left(+\frac{1}{\varphi_n^c}\right); \quad \text{for the }(n+1)\text{-st span} \]

\[ \left(-\frac{1}{\varphi_{n+1}^c}\right)\ \text{or}\ \left(-\frac{1}{\gamma_n^c}\right). \]

Thus, the ordinates of the influence line for the support moment \(M_c\) will be, in the \(n\)-th span,

\[ y=\tau_c\cdot \frac{1}{\gamma_{n+1}^c}; \]

in the \((n+1)\)-st,

\[ y=-\tau_c\cdot \frac{1}{\gamma_n^c}. \]

Substituting the expressions derived for \(\tau_c\), we obtain the formulas for the ordinates of the influence line of the support moment \(M_c\):

\[ \left. \begin{aligned} \text{for the } n\text{-th span}\quad y&=-\frac{\beta_n}{\gamma_{n+1}^c+\gamma_n^c}\cdot l_n(\eta_B-K_{an}\eta_A)\\ \text{for the }(n+1)\text{-st span}\quad y&=-\frac{\beta_{n+1}}{\gamma_{n+1}^c+\gamma_n^c}\cdot l_{n+1}(\eta_A-K_{b(n+1)}\eta_B) \end{aligned} \right\} \qquad \ldots (40) \]

In these formulas the coefficients \(\eta_A\) and \(\eta_B\) must be taken separately for each span.

Formulas (40) serve for the direct calculation of the ordinates of the influence lines of any support moment for the left and right spans adjacent to the support under consideration.

For spans not adjacent to the support under consideration, the ordinates of the influence line are obtained from the ordinates of the influence lines of the neighboring support moments, by multiplying these by the focal ratios \(K_{bn}\) and \(K_{a(n+1)}\) for the moments in the \(n\)-th and \((n+1)\)-st spans.

Use of the graph for the multipliers \((\eta_A-K_b\eta_B)\) (see § 3 of this chapter) can considerably shorten the computations.

The general formula for the area of a segment of an influence line (formula 17) takes the form:

\[ \left. \begin{aligned} \text{for the } n\text{-th span}\quad \Omega_{(n)} &=-\frac{\beta_n l_n^2(1-K_{an})}{4(\gamma_{n+1}^c+\gamma_n^c)} \\ \text{for the }(n+1)\text{-st span}\quad \Omega_{(n+1)} &=-\frac{\beta_{n+1} l_{n+1}^2(1-K_{b(n+1)})}{4(\gamma_{n+1}^c+\gamma_n^c)} \end{aligned} \right\} \qquad \ldots\ldots (41) \]

§ 5. Construction of influence lines of bending moments for sections in the spans of members.

For sections in the spans of members, the determination of influence lines is carried out by the well-known formula

\[ M_x = M_{ox} + \frac{M^A(l-x)}{l} + \frac{M^B x}{l}\ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (42) \]

The quantities \(M_{ox}\)—the ordinates of the influence line of the bending moment in the section \((x)\) of a simple beam on two supports—can be easily calculated. Their values may be found in many handbooks.

\(M^A\) and \(M^B\) are the ordinates of the influence lines of the support moments.

It is convenient to reduce the calculation of the ordinates \(M_x\) to tables.

The ordinates of the influence lines of shearing forces are calculated by the formula

\[ Q_x = Q_{ox} + \frac{M^B - M^A}{l}\ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (43) \]

The values of the ordinates \(Q_{ox}\) are easily calculated or are taken ready-made from handbooks. The calculations of \(Q_x\) are likewise more conveniently reduced to tables.

When calculating the ordinates \(M_x\) and \(Q_x\) simultaneously, the areas of individual portions of the influence lines may be determined in the same tables, since the values of these areas for the moments \(M^A\) and \(M^B\) will be known; as regards the corresponding areas for \(M_{ox}\) and \(Q_{ox}\), in view of the elementary form of their influence lines, their calculation is very easy.

§ 6. Practical remarks of a general character.

1. In order not to have to deal with very small quantities when calculating the ordinates of influence lines for the angles \(\tau\) and displacements \(\delta_l\), it is more convenient to find the influence lines for the angles \(E\tau\) and displacements \(E\delta_l\), in the same way as, in practical calculations by the method of foci, the angular quantities \(E\beta\), \(E\gamma\), \(E\varphi\), etc., are used.

In passing from the influence lines \(E\tau\) and \(E\delta_l\) to the influence lines of moments, we use the formulas

\[ M = \pm \frac{\tau}{\gamma} \quad \text{or} \quad M = \pm \frac{\tau}{\varphi} \]

and

\[ M = \delta_l^{I}\cdot \frac{\mathfrak{M}_{I}}{\Delta x} + \delta_l^{II}\frac{\mathfrak{M}_{II}}{\Delta x} + \delta_l^{III}\frac{\mathfrak{M}_{III}}{\Delta x} + \ldots , \]

which will then have the form:

\[ M = \pm \frac{E\tau}{E\gamma} \quad \text{or} \quad M = \pm \frac{E\tau}{E\varphi} \quad \text{(the factor \(E\) cancels)} \]

\[ M = E\delta_l^{I}\cdot \frac{\mathfrak{M}_{I}}{E\Delta x} + E\delta_l^{II}\cdot \frac{\mathfrak{M}_{II}}{E\Delta x} + E\delta_l^{III}\frac{\mathfrak{M}_{III}}{E\Delta x} + \ldots \ . \ . \ . \ . \ . \ . \ (44) \]

In the latter case the displacements will have to be multiplied by the moments \(\mathfrak{M}\), reduced \(E\Delta x\) times (for \(\Delta x = 0.001\) m and \(E = 21 \cdot 10^5\ \frac{\mathrm{t}}{\mathrm{m}^2}\), \(E\Delta x = 2100\ \frac{\mathrm{t}}{\mathrm{m}}\)).

1. All calculations of ordinates are reduced to tables. For the practice of computation, a rational choice of the type of tables is of known importance. The sequence of the individual operations for obtaining the final result must be chosen as successfully as possible.

The general formula for computing the ordinates (11):

\[ y=\beta l\left(M^{A}\eta_{A}+M^{B}\eta_{B}\right), \]

if there is no graph for determining the multipliers \((\eta_{A}-K_{b}\eta_{B})\) or \((\eta_{B}-K_{a}\eta_{A})\) (see § 3 of this chapter), is more conveniently represented for practical use in the form

\[ y=M^{A}\beta l\eta_{A}+M^{B}\beta l\eta_{B}=A\cdot\eta_{A}+B\eta_{B}\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (45) \]

where \(A=M^{A}\beta l\) and \(B=M^{B}\beta l\) are multipliers common to all values of \(\eta_{A}\) and \(\eta_{B}\) on the segment of an individual member. They are computed in a special column of the table, after which the coefficients \(\eta_{A}\) and \(\eta_{B}\) are multiplied by them. The final operation will be the addition of the products \(A\eta_{A}\) and \(B\eta_{B}\).

With this order of computations the determination of the areas of influence lines is also facilitated. Namely, formula (16) may be represented in the form

\[ \Omega=M^{A}\beta l^{2}\nabla_{A}+M^{B}\beta l^{2}\nabla_{B}=Al\nabla_{A}+Bl\nabla_{B}\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (46) \]

For symmetrical members we obtain the entirely simple formula

\[ \Omega=\frac{1}{4}\beta l^{2}\left(M^{A}+M^{B}\right)=\frac{A+B}{4}\cdot l\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (47) \]

Examples of Calculations.

Example I. Find the influence lines of the fixed-end moments and of the bending moment in the middle of the 2nd span for a reinforced-concrete frame of a four-span trestle (Fig. 22).

Basic data.

Length of girders: \(l_{1}=6.00\) m, \(l_{2}=l_{3}=8.00\) m, \(l_{4}=6.00\) m.

Height of columns: \(h_{5}=9.00\) m, \(h_{6}=8.00\) m, \(h_{7}=7.00\) m, \(h_{8}=6.00\) m.

At the bases of columns (8), (7), and (6) there is partial fixity (the column shoes rest on an elastic foundation). At joint (0) there is a movable support; at joint (V), a hinged restraint.

Moments of inertia: for the girders in the span,

\[ J_{1}=J_{2}=J_{3}=J_{4}=0.12\ \text{m}^{4}, \]

moreover:

for the 1st span—the moment of inertia at the support is \(J_{01}=0.60\ \text{m}^{4}\), the length of the haunch is

\[ 1.50\ \text{m},\quad \lambda=\frac{1.50}{6.00}=\frac{1}{4};\quad \frac{J_{1}}{J_{01}}=\frac{0.12}{0.60}=0.20; \]

for the 2nd span—the moment of inertia at the support is \(J_{02}=1.20\ \text{m}^{4}\), the length of the haunch is

\[ 1.60\ \text{m},\quad \lambda=\frac{1.60}{8.00}=\frac{1}{5};\quad \frac{J_{2}}{J_{02}}=\frac{0.12}{1.20}=0.10; \]

for the 3rd span—the same quantities as for the 2nd span;

for the 4th span—the moment of inertia at the support is \(J_{04}=1.20\ \text{m}^{4}\). The length of the haunch is

\[ 1.50\ \text{m},\quad \lambda=\frac{1.50}{6.00}=\frac{1}{4};\quad \frac{J_{4}}{J_{04}}=\frac{0.12}{1.20}=0.10. \]

The moment of inertia of the columns is \(J_{8}=J_{7}=0.03\ \text{m}^{4}\); \(J_{6}=J_{5}=0.04\ \text{m}^{4}\), with the columns being taken as members of constant stiffness.

Determination of the principal quantities and focal ratios.

Formulas for calculation:

\[ a_n=\frac{l_n\cdot E\beta_n}{E\alpha_n^{A}+E\varphi_n^{A}}; \qquad b_n=\frac{l_n\cdot E\beta_n}{E\alpha_n^{B}+E\varphi_n^{B}}; \]

\[ E\alpha_n^{A}=\frac{l_n}{2J_n}\cdot C_1; \quad E\alpha_n^{B}=\frac{l_n}{2J_n}\cdot C_2; \quad E\beta_n=\frac{l_n}{6J_n}\cdot C_3, \]

where \(C_1\), \(C_2\), and \(C_3\) are coefficients from Suter’s tables.

\[ E\gamma_n^{B}=E\alpha_n^{B}-(1+K_{an})E\beta_n; \quad E\gamma_n^{A}=E\alpha_n^{A}-(1+K_{bn})E\beta_n; \]

\[ K_{an}=\frac{a_n}{l_n-a_n}; \quad K\beta_n=\frac{b_n}{l_n-b_n}; \quad E\varphi=\frac{1}{\sum \frac{1}{E\gamma}}. \]

A. Determination of the left focal ratios:

1st member: \(a_1=0,\ K_{a1}=0\)

\[ E\beta_1=\frac{l_1}{6J_1}\cdot C_3 =\frac{6.00}{6\cdot 0.12}\cdot 0.933=7.77 \]

\[ E\alpha_1^{B}=\frac{l_1}{2J_1}\cdot C_1 =\frac{6.00}{2\cdot 0.12}\cdot 0.758=18.89. \]

Here \(C_3\) and \(C_1\) are coefficients from Suter’s tables for a left-sided straight haunch with

\[ \lambda=\frac{1}{4} \quad\text{and}\quad n=\frac{J}{J_0}=0.20. \]

\[ E\gamma_1^{B} = E\alpha_1^{B}-E\beta_1(1+K_{a1}) = E\alpha_1^{B}-E\beta_1 = 11.12. \]

8th member: for the case of partial fixity we have (see “Calculation of Frame Structures,” Prof. G. V. Ul’yaninskii, 4th ed., pp. 188–189)

\[ a_8=\frac{1}{1+2\nu_8}\cdot l_8; \quad \text{where } \nu_8=1+\frac{3}{l_8}=1+\frac{3}{6.00}=1.50; \quad K_{a8}=\frac{1}{2\nu_8}; \]

substituting the numerical values, we obtain

\[ a_8=\frac{1}{1+3}\cdot 6.00=1.50\ \text{m.}; \qquad K_{a8}=\frac{1}{3}=0.333; \]

\[ E\beta_8=\frac{l_8}{6J_8} = \frac{6.00}{6\cdot 0.03} = 33.33; \]

\[ E\gamma_8^{B} = E\beta_8(2-K_{a8}) = 33.33(2-0.333) = 55.6 \]

(column 8 is a member of constant stiffness);

\[ E\varphi_2^{\,l} = \frac{E\gamma_8^{B}\cdot E\gamma_1^{B}} {E\gamma_8^{B}+E\gamma_1^{B}} = \frac{55.6\cdot 11.12}{55.6+11.12} = 9.26. \]

2nd member. \(E\beta_2=\dfrac{l_2}{6J_2}\cdot C_2=\dfrac{8.00}{6\cdot0.12}\cdot0.885=9.84;\quad E\alpha_2^{\mathrm{B}}=\)

\[ =E\alpha_2^{\mathrm{A}}=\frac{l_2}{2J_2}\cdot C_1 =\frac{8.00}{2\cdot0.12}\cdot0.734=24.48. \]

Here \(C_1\) and \(C_2\) are coefficients from the tables for the case of double-sided guys, with \(\lambda=\dfrac{1}{5}\) and \(\dfrac{J}{J_0}=0.10\).

\[ a_2=l_2\frac{E\beta_2}{E\alpha_2^{\mathrm{A}}+E\varphi_2^{\mathrm{I}}} =8.00\frac{9.84}{24.48+9.26} =\frac{8\cdot9.84}{33.74}=2.33\ \text{m.}; \]

\[ K_{a2}=\frac{2.33}{8.00-2.33}=0.411; \]

\[ E\gamma_2^{\mathrm{B}}=E\alpha_2^{\mathrm{B}}-(1+K_{a2})E\beta_2 =24.48-1.411\cdot9.84=24.48-13.91=10.57. \]

7th member. By analogy with member (8)—partial restraint—we have:

\[ \nu_7=1+\frac{3}{l_7}=1+\frac{3}{7.00}=1.43, \]

\[ a_7=\frac{1}{1+2\nu_7}l_7 =\frac{1}{1+2.86}\cdot7.00 =\frac{7.00}{3.86}=1.814\ \text{m.}; \quad K_{a7}=\frac{1}{2\nu_7}=\frac{1}{2.86}=0.35; \]

\[ E\beta_7=\frac{l_7}{6J_7} =\frac{7.00}{6\cdot0.03}=38.90; \quad E\gamma_7^{\mathrm{B}}=E\beta_7(2-K_{a7})=38.9(2-0.35)=64.10. \]

\[ E\varphi_3^{\mathrm{II}} =\frac{E\gamma_7^{\mathrm{B}}\cdot E\gamma_2^{\mathrm{B}}} {E\gamma_7^{\mathrm{B}}+E\gamma_2^{\mathrm{B}}} =\frac{64.10\cdot10.57}{64.10+10.57} =\frac{64.10\cdot10.57}{74.67}=9.06. \]

3rd member is identical in dimensions to member (2):

\[ E\beta_3=E\beta_2=9.84;\quad E\alpha_3=E\alpha_3^{\mathrm{A}}=24.48; \]

\[ a_3=l_3\frac{E\beta_3}{E\alpha_3^{\mathrm{A}}+E\varphi_3^{\mathrm{II}}} =8.00\frac{9.84}{24.48+9.06} =\frac{8.00\cdot9.84}{33.54}=2.35\ \text{m.} \]

\[ K_{a3}=\frac{a_3}{l_3-a_3}=\frac{2.35}{8-2.35}=0.416; \]

\[ E\gamma_3^{\mathrm{B}}=E_3\alpha^{\mathrm{B}}-(1+K_{a3})E\beta_3 =24.48-1.416\cdot9.84=24.48-13.94=10.54. \]

6th member. By analogy with posts (7) and (8) we have

\[ \nu_6=1+\frac{3}{l_6}=1+\frac{3}{8.00}=1.375; \quad a_6=\frac{l_6}{1+2.75}=\frac{8.00}{3.75}=2.135\ \text{m.} \]

\[ K_{a6}=\frac{1}{2\nu_6}=\frac{1}{2.75}=0.366; \quad E\beta_6=\frac{l_6}{6J_6}=\frac{8.00}{6\cdot0.04}=33.35; \]

\[ E\gamma_6^{\mathrm{B}}=E\beta_6(2-K_{a6}) =33.35(2-0.366)=33.35\cdot1.634=54.40; \]

\[ E\varphi_4^{\mathrm{III}} =\frac{E\gamma_6^{\mathrm{B}}\cdot E\gamma_3^{\mathrm{B}}} {E\gamma_6^{\mathrm{B}}+E\gamma_3^{\mathrm{B}}} =\frac{54.40\cdot10.54}{54.40+10.54}=8.82 \]

4th member.

\[ E\beta_4=\frac{l_4}{6J_4}\cdot C_2=\frac{6.00}{6\cdot 0.12}\,0.827=6.89 \]

\[ E\alpha_4^{A}=E\alpha_4^{B}=\frac{l_4}{2J_4}\cdot C_1=\frac{6.00}{2\cdot 0.12}\cdot 0.669=16.72. \]

Here \(C_1\) and \(C_2\) are coefficients from the tables for double-sided brackets at

\[ \lambda=\frac{1}{4}\quad \text{and}\quad \frac{J}{J_0}=0.10. \]

\[ a_4=l_4\,\frac{E\beta_4}{E\alpha_4^{A}+E\varphi_4^{III}} =6.00\cdot \frac{6.89}{16.72+8.82} =\frac{6.00\cdot 6.89}{25.54}=1.62\ \text{m}. \]

\[ K_{a4}=\frac{a_4}{l_4-a_4}=\frac{1.62}{6.00-1.62}=0.369; \]

\[ E\gamma_4^{B}=E\alpha_4^{B}-E\beta_4(1+K_{a4}) =16.72-6.89\cdot 1.369=7.30. \]

5th member. With the adopted principal direction for the posts—from bottom to top—the lower focal distance \(a_5=0\), \(K_{a5}=0\), since we have a hinged fastening of the fifth post (5).

B. Determination of the right focal distances:

5th member.

\[ E\beta_5=\frac{l_5}{6J_5}=\frac{9.00}{6\cdot 0.04}=37.5; \qquad E\gamma_5^{B}=2E\beta_5=75.0; \]

\[ b_5=\frac{l_5}{3+\dfrac{E\varphi_5^{IV}}{E\beta_5}} =\frac{9.00}{3+\dfrac{7.30}{37.50}} =\frac{9.00}{3+0.1947} =\frac{9.00}{3.1947}=2.82\ \text{m}, \]

since

\[ E\varphi_5^{IV}=E\gamma_4^{B}=7.30; \]

\[ K_{b5}=\frac{b_5}{l_5-b_5}=\frac{2.82}{9.00-2.82}=0.456. \qquad E\gamma_5^{A}=E\beta_5(2-K_{b5}) \]

4th member.

\[ E\varphi_4^{IV}=E\gamma_5^{B}=2E\beta_5=75.0; \]

\[ b_4=l_4\,\frac{E\beta_4}{E\alpha_4^{B}+E\varphi_4^{IV}} =6.00\,\frac{6.89}{16.72+75.0} =\frac{6.00\cdot 6.89}{91.72}=0.45\ \text{m}; \]

\[ K_{b4}=\frac{0.45}{6.00-0.45}=\frac{0.45}{5.55}=0.081; \]

\[ E\gamma_4^{A}=E\alpha_4^{A}-(1+K_{b4})E\beta_4 =16.72-1.081\cdot 6.89 =16.72-7.44=9.28. \]

6th member.

\[ E\varphi_6^{III} =\frac{E\gamma_4^{A}\cdot E\gamma_3^{B}}{E\gamma_4^{A}+E\gamma_3^{B}} =\frac{9.28\cdot 10.54}{9.28+10.54} =\frac{9.28\cdot 10.54}{19.82}=4.92; \]

\[ b_6=\frac{l_6}{3+\dfrac{E\varphi_6^{III}}{E\beta_6}} =\frac{8.00}{3+\dfrac{4.92}{33.35}} =\frac{8.00}{3+0.148}=2.54\ \text{m}; \]

\[ K_{b6}=\frac{1}{2+0.148}=0.466 \]

3rd member.

\[ E\varphi_3^{\mathrm{III}} = \frac{E\gamma_4^{A}\cdot E\gamma_6^{B}}{E\gamma_4^{A}+E\gamma_6^{B}} = \frac{9{,}28\cdot 54{,}40}{9{,}28+54{,}40} = \frac{9{,}28\cdot 54{,}40}{63{,}68} = 7{,}91; \]

\[ b_3 = l_3\frac{E\beta_3}{E\alpha_3^{B}+E\varphi_3^{\mathrm{III}}} = 8{,}00\frac{9{,}84}{24{,}48+7{,}91} = \frac{8{,}00\cdot 9{,}84}{32{,}39} = 2{,}428\ \text{m}; \]

\[ K_{b3} = \frac{2{,}428}{8{,}00-2{,}428} = 0{,}436 \]

\[ E\gamma_3^{A} = E\alpha_3^{A}-(1+K_{b3})E\beta_3 = 24{,}48-1{,}436\cdot 9{,}84 = 24{,}48-14{,}13 = 10{,}35. \]

7th member.

\[ E\varphi_7^{\mathrm{II}} = \frac{E\gamma_3^{A}\cdot E\gamma_2^{B}}{E\gamma_3^{A}+E\gamma_2^{B}} = \frac{10{,}35\cdot 10{,}57}{10{,}35+10{,}57} = 5{,}23; \]

\[ b_7 = \frac{l_7}{3+\dfrac{E\varphi_7^{\mathrm{II}}}{E\beta_7}} = \frac{7{,}00}{3+\dfrac{5{,}23}{38{,}90}} = \frac{7{,}00}{3+0{,}1343} = \frac{7{,}00}{3{,}134} = 2{,}238\ \text{m}; \]

\[ K_{b7} = \frac{2{,}238}{7{,}00-2{,}238} = 0{,}468. \]

2nd member.

\[ E\varphi_2^{\mathrm{II}} = \frac{E\gamma_3^{A}\cdot E\gamma_7^{B}}{E\gamma_3^{A}+E\gamma_7^{B}} = \frac{10{,}35\cdot 64{,}10}{10{,}35+64{,}10} = \frac{10{,}35\cdot 64{,}10}{74{,}45} = 8{,}91; \]

\[ b_2 = l_2\frac{E\beta_2}{E\alpha_2^{B}+E\varphi_2^{\mathrm{II}}} = 8{,}00\frac{9{,}84}{24{,}48+8{,}91} = \frac{8\cdot 9{,}84}{33{,}59} = 2{,}36\ \text{m}; \]

\[ K_{b2} = \frac{2{,}36}{8{,}00-2{,}36} = 0{,}418; \]

\[ E\gamma_2^{A} = E\alpha_2^{A}-(1+K_{b2})E\beta_2 = 24{,}48-1{,}418\cdot 9{,}84 = 24{,}48-13{,}96 = 10{,}52. \]

8th member.

\[ E\varphi_8^{\mathrm{II}} = \frac{E\gamma_2^{A}\cdot E\gamma_1^{B}}{E\gamma_2^{A}+E\gamma_1^{B}} = \frac{10{,}52\cdot 11{,}12}{10{,}52+11{,}12} = \frac{10{,}52\cdot 11{,}12}{21{,}64} = 5{,}40; \]

\[ b_8 = \frac{l_8}{3+\dfrac{E\varphi_8^{\mathrm{II}}}{E\beta_8}} = \frac{6{,}00}{3+\dfrac{5{,}40}{33{,}33}} = \frac{6{,}00}{3+0{,}162} = 1{,}897\ \text{m}; \]

\[ K_{b8} = \frac{1}{2+0{,}162} = 0{,}463. \]

1st member.

\[ E\varphi_1^{\mathrm{I}} = \frac{E\gamma_2^{A}\cdot E\gamma_8^{B}}{E\gamma_2^{A}+E\gamma_8^{B}} = \frac{10{,}52\cdot 55{,}6}{10{,}52+55{,}6} = \frac{10{,}52\cdot 55{,}6}{66{,}12} = 8{,}836; \]

\[ b_1 = l_1\frac{E\beta_1}{E\alpha_1^{B}+E\varphi_1^{\mathrm{I}}} = 6{,}00\frac{7{,}77}{18{,}90+8{,}836} = \frac{6\cdot 7{,}77}{27{,}736} = 1{,}682\ \text{m}; \]

\[ K_{b1} = \frac{1{,}682}{6{,}00-1{,}682} = 0{,}389. \]

Summary table of preliminary data.

No. of member	\(l\)	\(E\beta\)	\(a\)	\(K_a\)	\(b\)	\(K_b\)	\(E\gamma^A\)	\(E\gamma^B\)	\(E\varphi^A\)	\(E\varphi^B\)	Distribution coefficients	Distribution coefficients	Distribution coefficients	Distribution coefficients
1	6.0	7.77	0	0	1.682	0.389	—	\(\gamma_1^{\mathrm I}=11.2\)	\(\infty\)	\(\varphi_1^{\mathrm I}=8.836\)	\(\mu_{1-2}=+0.84\)	\(\mu_{1-8}=-0.16\)	—	—
2	8.0	9.84	2.33	0.411	2.36	0.418	\(\gamma_2^{\mathrm I}=10.52\)	\(\gamma_2^{\mathrm I}=10.57\)	\(\varphi_2^{\mathrm I}=9.26\)	\(\varphi_2^{\mathrm{II}}=8.91\)	\(\mu_{2-1}=+0.833\)	\(\mu_{2-3}=+0.861\)	\(\mu_{2-7}=-0.139\)	\(\mu_{2-8}=+0.167\)
3	8.0	9.84	2.35	0.416	2.428	0.436	\(\gamma_3^{\mathrm{II}}=10.35\)	\(\gamma_3^{\mathrm{II}}=10.54\)	\(\varphi_3^{\mathrm{II}}=9.06\)	\(\varphi_3^{\mathrm{III}}=7.91\)	\(\mu_{3-7}=+0.14\)	\(\mu_{3-2}=+0.86\)	\(\mu_{3-4}=+0.864\)	\(\mu_{3-6}=-0.145\)
4	6.0	6.89	1.62	0.369	0.45	0.081	\(\gamma_4^{\mathrm{III}}=9.28\)	\(\gamma_4^{\mathrm{IV}}=7.30\)	\(\varphi_4^{\mathrm{III}}=8.82\)	\(\varphi_4^{\mathrm{IV}}=75.0\)	\(\mu_{4-6}=+0.163\)	\(\mu_{4-3}=+0.837\)	\(\mu_{4-5}=-1\)	—
5	9.0	37.50	0	0.456	2.82	0.456	—	\(\gamma_5^{\mathrm V}=75.00\)	—	\(\varphi_5^{\mathrm V}=7.30\)	\(\mu_{5-4}=-1\)	—	—	—
6	8.0	33.35	2.135	0.366	2.54	0.466	—	\(\gamma_6^{\mathrm{III}}=54.40\)	—	\(\varphi_6^{\mathrm{III}}=4.92\)	\(\mu_{6-4}=+0.533\)	\(\mu_{6-3}=-0.467\)	—	—
7	7.0	38.90	1.814	0.350	2.238	0.468	—	\(\gamma_7^{\mathrm{II}}=64.10\)	—	\(\varphi_7^{\mathrm{II}}=5.23\)	\(\mu_{7-3}=+0.505\)	\(\mu_{7-2}=-0.495\)	—	—
8	6.0	33.33	1.50	0.333	1.897	0.463	—	\(\gamma_8^{\mathrm I}=55.60\)	—	\(\varphi_8^{\mathrm I}=5.40\)	\(\mu_{8-2}=+0.515\)	\(\mu_{8-1}=-0.485\)	—	—

The distribution coefficients, which we shall need later only in finding the epure of moments from displacement, are computed by the formulas

\[ \psi_{1-2}=\frac{\varphi_1^{\mathrm I}}{\gamma_2^{\mathrm I}},\qquad \psi_{2-1}=\frac{\varphi_2^{\mathrm I}}{\gamma_1^{\mathrm I}} \quad \text{and so on.} \]

These computations have been carried out in the table of preliminary data, and, for convenience, the proper signs have been assigned to the distribution coefficients. (See p. 35.)

Finding the epure of moments from displacement of the upper joints of the frame to the right by \(\Delta x=0.001\) m.

With a displacement to the right by \(\Delta x=0.001\) m, we shall have the angles of rotation of the columns

\[ \vartheta_8=\frac{0.001}{6.00}=1.667\cdot 10^{-4};\qquad \vartheta_7=\frac{0.001}{7.00}=1.428\cdot 10^{-4}; \]

\[ \vartheta_6=\frac{0.001}{8.00}=1.25\cdot 10^{-4};\qquad \vartheta_5=\frac{0.001}{9.00}=1.111\cdot 10^{-4}. \]

The moments in the columns due to displacement, for \(E=21\cdot 10^5\ \dfrac{\mathrm t}{\mathrm m^2}\) (for reinforced concrete),

\[ M_8^{\mathrm{VIII}} = -\frac{E\vartheta_8}{E\beta_8}\cdot \frac{a_8}{l_8-a_8-b_8} = \]

\[ = -\frac{21\cdot 10^5\cdot 1.667\cdot 10^{-4}}{33.33} \cdot \frac{1.50}{6.00-1.50-1.897} = -10.51\,\frac{1.50}{2.603} = \]

\[ =-6.05\ \mathrm{tm}. \]

\[ M_8^{\mathrm I} = -(-6.05)\frac{1.897}{1.50} = +7.65\ \mathrm{tm}. \]

\[ M_7^{\mathrm{VII}} = -\frac{E\vartheta_7}{E\beta_7}\cdot \frac{a_7}{l_7-a_7-b_7} = \]

\[ = -\frac{21\cdot 10^5\cdot 1.428\cdot 10^{-4}}{38.90} \cdot \frac{1.814}{7.00-1.814-2.238} = -7.70\,\frac{1.814}{2.947} = \]

\[ =-4.74\ \mathrm{tm}. \]

\[ M_7^{\mathrm{II}} = -(-4.74)\frac{2.238}{1.814} = +5.84\ \mathrm{tm}. \]

\[ M_6^{\mathrm{VI}} = -\frac{21\cdot 10^5\cdot 1.25\cdot 10^{-4}}{33.35} \cdot \frac{2.135}{8.00-2.135-2.54} = \]

\[ = -7.87\,\frac{2.135}{3.325} = -5.05\ \mathrm{tm}. \]

\[ M_6^{\mathrm{III}} = -(-5.05)\frac{2.54}{2.135} = +6.01\ \mathrm{tm}. \]

\[ M_5^{\mathrm V}=0 \]

\[ M_5^{\mathrm{IV}} = +\frac{21\cdot 10^5\cdot 1.111\cdot 10^{-4}}{37.50} \cdot \frac{2.82}{9-2.82} = +6.22\ \mathrm{tm}. \]

We carry out the distribution and summation of the moments in the table.

Moments	Support moment from rotation: Column (5)	Support moment from rotation: Column (6)	Support moment from rotation: Column (7)	Support moment from rotation: Column (8)	Sum \(\mathfrak{M}\)	\(m=\dfrac{\mathfrak{M}}{Z}\)
\(M_1^{\mathrm{I}}\)	\(+0,2355\)	\(-0,344\)	\(+0,987\)	\(-3,710\)	\(-2,83\)	\(-0,468\)
\(M_2^{\mathrm{I}}\)	\(+0,283\)	\(-0,413\)	\(+1,188\)	\(+3,940\)	\(+5,00\)	\(+0,827\)
\(M_2^{\mathrm{II}}\)	\(-0,689\)	\(+1,005\)	\(-2,890\)	\(-1,647\)	\(-4,22\)	\(-0,698\)
\(M_3^{\mathrm{II}}\)	\(-0,800\)	\(+1,170\)	\(+2,950\)	\(-1,420\)	\(+1,90\)	\(+0,314\)
\(M_3^{\mathrm{III}}\)	\(+1,925\)	\(-2,810\)	\(-1,285\)	\(+0,618\)	\(-1,55\)	\(-0,256\)
\(M_4^{\mathrm{III}}\)	\(+2,300\)	\(+3,200\)	\(-1,098\)	\(+0,528\)	\(+4,93\)	\(+0,815\)
\(M_4^{\mathrm{IV}}\)	\(-6,220\)	\(-0,259\)	\(+0,089\)	\(-0,043\)	\(-6,43\)	\(-1,062\)
\(M_5^{\mathrm{IV}}\)	\(+6,220\)	\(+0,259\)	\(-0,089\)	\(+0,043\)	\(+6,43\)	\(+1,062\)
\(M_6^{\mathrm{III}}\)	\(+0,375\)	\(+6,010\)	\(+0,187\)	\(-0,090\)	\(+6,48\)	\(+1,070\)
\(M_6^{\mathrm{VI}}\)	\(-0,137\)	\(-5,050\)	\(-0,069\)	\(+0,033\)	\(-5,22\)	\(-0,864\)
\(M_7^{\mathrm{II}}\)	\(-0,111\)	\(+0,165\)	\(+5,84\)	\(+0,227\)	\(+6,12\)	\(+1,012\)
\(M_7^{\mathrm{VII}}\)	\(+0,039\)	\(-0,058\)	\(-4,74\)	\(-0,079\)	\(-4,84\)	\(-0,800\)
\(M_8^{\mathrm{I}}\)	\(+0,047\)	\(-0,069\)	\(+0,199\)	\(+7,650\)	\(+7,83\)	\(+1,295\)
\(M_8^{\mathrm{VIII}}\)	\(-0,016\)	\(+0,023\)	\(-0,066\)	\(-6,050\)	\(-6,11\)	\(-1,010\)

Calculation of the restraining force \(Z\) for a displacement \(\Delta x=0,001\ \mathrm{m}\).

Reaction of the support of the 8th member.

\[ B_8^{\mathrm{I}} = \frac{\mathfrak{M}_8^{\mathrm{I}}-\mathfrak{M}_8^{\mathrm{VIII}}}{l_8} = \frac{7,83+6,11}{6,00} = +2,32\ \mathrm{t}. \]

The same for the 7th member.

\[ B_7^{\mathrm{II}} = \frac{\mathfrak{M}_7^{\mathrm{II}}-\mathfrak{M}_7^{\mathrm{VII}}}{l_7} = \frac{6,12+4,84}{7,00} = +1,56\ \mathrm{t}. \]

The same for the 6th member.

\[ B_6^{\mathrm{III}} = \frac{\mathfrak{M}_6^{\mathrm{III}}-\mathfrak{M}_6^{\mathrm{VI}}}{l_6} = \frac{6,48+5,22}{8,00} = +1,46\ \mathrm{t}. \]

The same for the 5th member.

\[ B_5^{\mathrm{IV}} = \frac{\mathfrak{M}_5^{\mathrm{IV}}}{l_5} = \frac{6,43}{9,00} = +0,714\ \mathrm{t}. \]

\[ \text{Restraining force } Z = B_8^{\mathrm{I}}+B_7^{\mathrm{II}}+B_6^{\mathrm{III}}+B_5^{\mathrm{IV}} = \]

\[ =2,32+1,56+1,46+0,714=+6,05\ \mathrm{t}. \]

The restraining force has come out positive, i.e., it is directed from left to right. The moments \(m\) correspond to the displacing force \(V=1\ \mathrm{t}\).

All the calculations and computations carried out so far are equally necessary both in the former and in the new method of constructing influence lines.

In the former method, the first operation leading to the determination of influence lines would have been the construction of the basic diagram.

In the proposed method, the preliminary operation is the determination of the influence lines for the angles of rotation of the nodes \(E\tau\) and the influence line for the displacement \(E\delta_l\).

Determination of influence lines.

1. Construction of the influence line for the angle of rotation \(E\tau^{\mathrm{I}}\) of node 1.

We divide each member (girder) into 6 equal parts, so that in each span we shall have 5 points. The total number of points along the entire length of the frame, including the nodal points, will be \(20+5=25\).

We assume that at joint IV a fictitious support has been introduced, preventing displacement of the joints.

We apply from outside to joint I a moment \(M=1\) tm (clockwise).

The distribution coefficients (formulas 38 and 39) will be

\[ \mu_1=\frac{E\varphi_1^{\mathrm I}}{E\varphi_1^{\mathrm I}+E\gamma_1^{\mathrm I}} =\frac{8.836}{8.836+11.12}=0.443; \]

\[ \mu_2=\mu_1\cdot\frac{E\gamma_1^{\mathrm I}}{E\gamma_2^{\mathrm I}} =0.443\,\frac{11.12}{10.52}=0.468. \]

Taking the signs into account, we obtain the moments:

\[ M_1^{\mathrm I}=-0.443\ \text{tm};\qquad M_1^{0}=-K_{a1}\cdot M_1^{\mathrm I}=0 \]

\[ M_2^{\mathrm I}=+0.468\ \text{tm};\qquad M_2^{\mathrm{II}}=-K_{b2}\cdot M_2^{\mathrm I} =-0.418\cdot0.468=-0.196\ \text{tm}. \]

We carry out the calculation of the ordinates of the influence line \(E\tau^{\mathrm I}\) in Table No. 1.

The table is constructed for the application of formulas (45), (46), and (47). The coefficients \(\eta_A\) and \(\eta_B\) for the 1st member are taken from Suter’s tables for the case of a one-sided straight haunch at \(\lambda=\frac14\) and \(\frac{J}{J_0}=0.20\).

The coefficient \(\nabla_B\) of the area is also taken from these tables, but for the case of a uniformly distributed load (in “Analysis of Frame Structures,” 4th ed., this table is given on p. 184).

It is taken into account here that the table is compiled for a left-hand haunch, whereas in our case the haunch is on the right.

For the 2nd span, \(\eta_A\) and \(\eta_B\) are taken from Suter’s table for symmetrical members with straight haunches at \(\lambda=\frac15\) and \(\frac{J}{J_0}=0.10\).

For the 3rd and 4th members, calculation of the ordinates in a similar manner becomes unnecessary, since when the moving load is located to the right of joint II there is the focal relation

\[ E\tau^{\mathrm I}=-x_{a2}\cdot E\tau^{\mathrm{II}}, \]

where

\[ x_{a2}=K_{a2}\frac{E\varphi_2^{\mathrm I}}{E\gamma_2^{\mathrm{II}}} =0.411\,\frac{9.26}{10.57}=0.361 \]

is the left angular focal ratio*).

Therefore the ordinates \(E\tau^{\mathrm I}\) are obtained from the ordinates \(E\tau^{\mathrm{II}}\) and from Table No. 2 by multiplying the latter by the factor \((-0.361)\).

Everything else is clear from Table No. 1.

1. Construction of the influence line for the angle of rotation \(E\tau^{\mathrm{II}}\) of joint II.

We apply from outside to joint II a moment \(M=1\) tm.

*) The angular focal ratios are calculated by formulas (31) and (32), Ch. III.

Distribution coefficients:

\[ \mu_2=\frac{E\varphi_2^{II}}{E\varphi_2^{II}+E\gamma_2^{II}} =\frac{8.91}{8.91+10.57}=0.458; \]

\[ \mu_3=\mu_2\frac{E\gamma_2^{II}}{E\gamma_3^{II}} =0.458\frac{10.57}{10.35}=0.467. \]

Moments:

\[ M_2^{II}=-0.458\ \text{tm}; \qquad M_2^{I}=-K_{a2}\cdot M_2^{II}=+0.411\cdot 0.458=+0.188\ \text{tm}. \]

\[ M_3^{II}=+0.467\ \text{tm}; \qquad M_3^{III}=-K_{b3}\cdot M_3^{II}=-0.436\cdot 0.467=-0.204\ \text{tm}. \]

All computations of the ordinates of the influence line \(E\tau^{II}\) have been carried out in Table No. 2.

The coefficients \(\eta_A\) and \(\eta_B\) for the 2nd and 3rd spans are identical.
On the segments of the 1st and 4th members the computations are unnecessary, since we can use angular focal ratios.

For the 1st span

\[ E\tau^{II}=-x_{b2}\cdot E\tau^{I}, \quad \text{where}\quad x_{b2}=K_{b2}\frac{E\varphi_2^{II}}{E\gamma_2^{I}}= \]

\[ =-0.418\frac{8.91}{10.52}=0.354; \]

therefore the ordinates \(E\tau^{II}\) are obtained by multiplying the ordinates \(E\tau^{I}\) from Table No. 1 by the factor \((-0.354)\).

For the 4th span we have:

\[ E\tau^{II}=-x_{a3}\cdot E\tau^{III}, \quad \text{where}\quad x_{a3}=K_{a3}\frac{E\varphi_3^{II}}{E\gamma_3^{III}}= \]

\[ =0.416\frac{9.06}{10.54}=0.358. \]

Therefore the ordinates \(E\tau^{II}\) are obtained by multiplying the ordinates \(E\tau^{III}\) from Table No. 3 by the factor \((-0.358)\).

3. Construction of the influence line of the angle of rotation \(E\tau^{III}\) of joint III.

We apply an external moment \(M=1\ \text{tm}\) to joint III.

Distribution coefficients:

\[ \mu_3=\frac{E\varphi_3^{III}}{E\varphi_3^{III}+E\gamma_3^{III}} =\frac{7.91}{7.91+10.54}=0.429; \]

\[ \mu_4=\mu_3\frac{E\gamma_3^{III}}{E\gamma_4^{III}} =0.429\frac{10.54}{9.28}=0.487. \]

Moments:

\[ M_3^{III}=-0.429\ \text{tm}; \qquad M_3^{II}=-K_{a3}\cdot M_3^{III}=+0.416\cdot 0.429=+0.178\ \text{t. m.} \]

\[ M_4^{III}=+0.487\ \text{tm}; \qquad M_4^{IV}=-K_{b4}\cdot M_4^{III}=-0.081\cdot 0.487=-0.039\ \text{t. m.} \]

The computation of ordinates has been carried out in Table No. 3.

The coefficients $\eta_A$ and $\eta_B$ for the 4th member are taken from Suter’s tables for members with two-sided rectilinear haunches, for the case

\[ \lambda=\frac{1}{4}\quad \text{and}\quad \frac{J}{J_0}=0.10. \]

The calculation of ordinates is not needed for the 1st and 2nd spans, since we can use the angular focal relation:

\[ E\tau^{\mathrm{III}}=-x_{b3}\cdot E\tau^{\mathrm{II}}, \quad \text{where}\quad x_{b3}=K_{b3}\frac{E\varphi_3^{\mathrm{III}}}{E\gamma_3^{\mathrm{II}}} =0.436\,\frac{7.91}{10.35}=0.333 \]

We take the ordinates $E\tau^{\mathrm{II}}$ from Table No. 2 and multiply by the factor $(-0.333)$.

4. Construction of the influence line for the angle of rotation $E\tau^{\mathrm{IV}}$ of joint IV.

We apply an external moment $M=1$ tm to joint IV.
The distribution coefficient is

\[ \mu_4=\frac{E\varphi_4^{\mathrm{IV}}}{E\varphi_4^{\mathrm{IV}}+E\gamma_4^{\mathrm{IV}}} =\frac{75.00}{75.00+7.30}=0.911 \]

Moments:

\[ M_4^{\mathrm{IV}}=-0.911\ \text{tm.}; \qquad M_4^{\mathrm{III}}=-K_{a4}\cdot M_4^{\mathrm{IV}} =+0.369\cdot 0.911=+0.336. \]

The calculation of the ordinates has been carried out in Table No. 4.

For the 1st, 2nd, and 3rd members we do not perform calculations, since we have the possibility of using the angular focal relation:

\[ E\tau^{\mathrm{IV}}=-x_{b4}E\tau^{\mathrm{III}}, \quad \text{where}\quad x_{b4}=K_{b4}\frac{E\varphi_4^{\mathrm{IV}}}{E\gamma_4^{\mathrm{III}}} =0.081\,\frac{75.00}{9.28}=0.655. \]

Therefore we take the ordinates $E\tau^{\mathrm{III}}$ from Table No. 3 and multiply them by the factor $(-0.655)$.

The graphical construction of the influence lines for the angles of rotation of the joints $E\tau^{\mathrm{I}}$, $E\tau^{\mathrm{II}}$, $E\tau^{\mathrm{III}}$, $E\tau^{\mathrm{IV}}$ is unnecessary, since for further calculations we need to know only the numerical values of the ordinates of these influence lines.

5. Construction of the influence line for the displacement $E\delta_l$.

The calculation of the ordinates has been carried out in Table No. 5. The moments $m^A$ and $m^B$ corresponding to the displacing force $V=1$ t are taken from the table of moments due to displacement of the joints.

The coefficients $\eta_A$ and $\eta_B$ for all members were established when finding the influence lines for the angles of rotation.

The influence line for the displacement $E\delta_l$ can be used to determine all additional moments due to displacement (see Chapter II).

Fig. 22-a shows the influence line $E\delta_l$.

After calculating the ordinates of the influence lines for the angles of rotation and for the displacement, we can compute the ordinates for the influence lines of all fixed-end moments.

6. Finding the influence lines for the fixed-end moments.

The calculation of the ordinates has been carried out in Table No. 6.

In this table, for each moment the ordinates corresponding to the principal moment are first determined; for this purpose the ordinates of the original line for the angle of rotation of the joint are multiplied by the

scale numbers corresponding to the given moment. The formula, in a form convenient for computation, is given in each upper line. The ordinates \(E_t\) are taken from the appropriate table Nos. 1–4.

To the ordinates of the principal moment are added the ordinates of the additional moment due to displacement, computed from the ordinates of the influence line for \(E\delta_l\) according to formula (44):

\[ M' = E\delta_l \cdot \frac{\mathfrak{M}}{E\Delta x}, \]

which, for the modulus of reinforced concrete \(E = 21 \cdot 10^5 \dfrac{\mathrm{t}}{\mathrm{m}^2}\) and \(\Delta x = 0.001\ \mathrm{m}\), takes the form

\[ M' = E\delta_l \cdot \frac{\mathfrak{M}}{2100} = E\delta_l \left(4.76 \cdot 10^{-4}\mathfrak{M}\right). \]

Here \(\mathfrak{M}\)—the support moment due to the displacement of the upper joints by \(\Delta x = -0.001\ \mathrm{m}\)—must be taken from the table of moments due to displacement of the joints.

Together with the ordinates, in Table No. 6 the areas of the portions of the influence lines have been computed; moreover, with the areas taken from Tables Nos. 1–4 and No. 5, exactly the same operations are performed as with the ordinates (multiplication by the same factors and final addition).

As a check on the computations, one may use a verification of the equilibrium of moments at the joint for each position of the moving load. For example, for joint I, taking into account the signs of the support moments, the condition must be satisfied:

\[ M_1^{\mathrm{I}} + M_8^{\mathrm{I}} - M_2^{\mathrm{I}} = 0. \]

For joint II it must be:

\[ M_2^{\mathrm{II}} + M_7^{\mathrm{II}} - M_3^{\mathrm{II}} = 0. \]

For joint III it must be:

\[ M_3^{\mathrm{II}} + M_6^{\mathrm{III}} - M_4^{\mathrm{III}} = 0. \]

By summing the ordinates (with the corresponding signs) for one and the same point of load position, one should obtain confirmation of these equalities.

In view of the fact that the solution of the present example is intended only to acquaint the reader with the practical application of the proposed method of constructing influence lines, the influence lines have been found in Table No. 6 (for reasons of saving space) not for all fixed-end moments, but only for the moments

\[ M_1^{\mathrm{I}},\ M_2^{\mathrm{I}},\ M_8^{\mathrm{I}},\ M_8^{\mathrm{VIII}},\ M_2^{\mathrm{II}},\ M_3^{\mathrm{II}},\ M_7^{\mathrm{II}}. \]

If desired, Table No. 6 can easily be supplemented by computing the ordinates for all the remaining moments, preserving the order of computation set forth above.

7. Finding the influence line for the bending moment at the middle of the 2nd span is carried out by the well-known formula for the moment in the span

\[ M_x = M_{0x} + \frac{M^{\mathrm{A}}(l-x)}{l} + \frac{M^{\mathrm{B}}x}{l}. \]

The computation of the ordinates has been carried out in Table No. 7.

No. 1. Table for computing the influence ordinate \(E^{\mathrm I}\) of node I.

No. of story and No. of point	Story 1	Story 1	Story 1	Story 1	Story 1	Story 2	Story 2	Story 2	Story 2	Story 2	Story 3	Story 3	Story 3	Story 3	Story 3	Story 4	Story 4	Story 4	Story 4	Story 4
	0	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20
\(\eta_A=\)	0	0.393	0.396	0.310	0.165	0.237	0.372	0.385	0.306	0.165	0
\(\eta_B=\)	0	0.294	0.366	0.349	0.216	0.165	0.306	0.385	0.372	0.237	0
\(A=-M^A\cdot E\beta\cdot l=\)				0			\(M_1^1\cdot E\beta_1\cdot l_1=0\)							Computation is omitted.
\(B=-M^B\cdot E\beta\cdot l=\)		\(-0.443\cdot 77.6\cdot 0\)	\(=-20.65\)				\(M_1^1\cdot E\beta_1\cdot l_1=-0.196\cdot 9.984\cdot 8=-15.45\)							In view of the existence of the focal dependence
\(A\cdot\eta_A=\)		0	0	0	0	0	\(+8.74\)	\(+13.72\)	\(+14.20\)	\(+11.28\)	\(+6.08\)	0				\(E^{\mathrm I}_1=-\chi_{a2}\cdot E^{\mathrm I}_1=-0.361\cdot E^{\mathrm I}_1\),
\(B\cdot\eta_B=\)		\(-3.34\)	\(-6.06\)	\(-7.55\)	\(-7.20\)	\(-4.46\)	\(-2.55\)	\(-4.72\)	\(-5.94\)	\(-5.74\)	\(-3.66\)	0				the ordinates \(E^{\mathrm I}\) are obtained by taking from table No. 2 the quantities of the ordinates \(E^{\mathrm{II}}\) and multiplying them by \((-0.361)\).
\(E^{\mathrm I}=A\eta_A+B\eta_B=\)		\(-3.34\)	\(-6.06\)	\(-7.55\)	\(-7.20\)	\(-4.46\)	\(+6.19\)	\(+9.00\)	\(+8.26\)	\(+5.54\)	\(+2.42\)	0	\(-2.20\)	\(-3.17\)	\(-2.89\)	\(-1.91\)	\(-0.81\)	0	\(+0.56\)	\(+0.90\)

Participation-area ordinates of the influence line \(\Omega\):

\[ \Omega_{(1)}=-E\beta_1 l_1^2 \nabla_B M_1^1 =77.6^2\cdot 0.237(-0.443)=-29.30 \]

\[ \Omega_{(2)}=\frac{A+B}{4}\cdot l_2 =\frac{36.85-15.45}{4}\cdot 8 =+42.80 \]

\[ \Omega_{(3)}=-0.361\cdot 41.40=-14.95 \]

\[ \Omega_{(4)}=-0.361\cdot(-9.95)=+3.59 \]

No. 2. Table for computing the influence ordinate \(E^{\mathrm{II}}\) of node II.

No. of story and No. of point	Story 1	Story 1	Story 1	Story 1	Story 1	Story 2	Story 2	Story 2	Story 2	Story 2	Story 3	Story 3	Story 3	Story 3	Story 3	Story 4	Story 4	Story 4	Story 4	Story 4
	0	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20
\(\eta_A\)	Computation is omitted, since \(E^{\mathrm{II}}=-\chi_{b1}\cdot E^{\mathrm I}_1=-0.354\cdot E^{\mathrm I}\). Taking the ordinates \(E^{\mathrm I}\) from table No. 1, multiplied by \((-0.354)\).					0.237	0.372	0.385	0.306	0.165	0.237	0.372	0.385	0.306	0.165	0
\(\eta_B\)						0.165	0.306	0.385	0.372	0.237	0.165	0.306	0.385	0.372	0.237	0
\(A=-M^A\cdot E\beta\cdot l=\)						\(M_2^1\cdot E\beta_2\cdot l_2=0.188\cdot 9.984\cdot 8=14.82\)					\(M_3^{\mathrm{II}}\cdot E'_3\cdot l_3=0.467\cdot 9.984\cdot 8=36.80\)					Computation is omitted, since \(E^{\mathrm{III}}=-\chi_{a3}\cdot E^{\mathrm{II}}=-0.358E^{\mathrm{II}}\)
\(B=-M^B\cdot E\beta\cdot l=\)						\(M_2^1\cdot E\beta_2\cdot l_2=-0.458\cdot 9.984\cdot 8=-36.10\)					\(M_3^{\mathrm{III}}\cdot E\beta'_3 l_3=-0.204\cdot 9.984\cdot 8=-16.05\)					Taking the ordinates \(E^{\mathrm{III}}\) from table No. 3, multiplied by \((-0.358)\).
\(A\cdot\eta_A=\)		\(+1.18\)	\(+2.15\)	\(+2.67\)	\(+2.55\)	\(+1.58\)	\(+3.52\)	\(+5.51\)	\(+5.70\)	\(+4.54\)	\(+2.45\)	\(+8.73\)	\(+13.70\)	\(+14.18\)	\(+11.27\)	\(+6.07\)	0
\(B\cdot\eta_B=\)		\(-1.18\)	\(-2.15\)	\(-2.67\)	\(-2.55\)	\(-1.58\)	\(-2.44\)	\(-5.96\)	\(-11.05\)	\(-13.90\)	\(-13.45\)	\(-8.55\)	\(-2.65\)	\(-4.91\)	\(-6.18\)	\(-5.98\)	\(-3.81\)	0
\(E^{\mathrm{II}}=A\eta_A+B\eta_B=\)		0	0	0	0	0	\(+1.08\)	\(+2.44\)	\(-0.45\)	\(-8.20\)	\(-8.91\)	\(-6.10\)	\(+6.08\)	\(+8.79\)	\(+5.29\)	\(+2.26\)	0	\(-1.55\)	\(-2.50\)	\(-2.58\)

Participation-area ordinates of the influence line \(\Omega\):

\[ \Omega_{(1)}=-0.354(-29.30)=+10.35 \]

\[ \Omega_{(2)}=\frac{A+B}{4}\cdot l_2 =\frac{14.82-36.10}{4}\cdot 8 =-42.50 \]

\[ \Omega_{(3)}=\frac{A+B}{4}\cdot l_3 =\frac{36.80-16.05}{4}\cdot 8 =+41.40 \]

\[ \Omega_{(4)}=-0.358\cdot 27.80=-9.95 \]

No. 3. Table for Calculating the Influence of Rotation \(E^{-\mathrm{III}}\) of Node III.

No. of rod and No. of point	0	1	2	3	4	5	6	7	8	9	10	II	11	12	13	14	15	III	16	17	18	19	20	IV
\(\eta_A=\)							Deflections drop out.	Deflections drop out.	Deflections drop out.	Deflections drop out.	Deflections drop out.	0	0.237	0.372	0.385	0.306	0.165	0	0.229	0.372	0.390	0.309	0.165	0
\(\eta_B=\)							In view of the focal dependence	In view of the focal dependence	In view of the focal dependence	In view of the focal dependence	In view of the focal dependence	0	0.165	0.306	0.385	0.372	0.237	0	0.165	0.309	0.390	0.372	0.229	0
\(A=M^A\cdot E\beta\cdot l=\)							\(E^{-\mathrm{III}}=-x_b\cdot E^{-\mathrm{II}}\),	\(E^{-\mathrm{III}}=-x_b\cdot E^{-\mathrm{II}}\),	\(E^{-\mathrm{III}}=-x_b\cdot E^{-\mathrm{II}}\),	\(E^{-\mathrm{III}}=-x_b\cdot E^{-\mathrm{II}}\),	\(E^{-\mathrm{III}}=-x_b\cdot E^{-\mathrm{II}}\),	\(M_3^{\mathrm{II}}\cdot E\beta_3 l_3=+0.178\cdot9.984\cdot8=+14.02\)	\(M_3^{\mathrm{II}}\cdot E\beta_3 l_3=+0.178\cdot9.984\cdot8=+14.02\)	\(M_3^{\mathrm{II}}\cdot E\beta_3 l_3=+0.178\cdot9.984\cdot8=+14.02\)	\(M_3^{\mathrm{II}}\cdot E\beta_3 l_3=+0.178\cdot9.984\cdot8=+14.02\)	\(M_3^{\mathrm{II}}\cdot E\beta_3 l_3=+0.178\cdot9.984\cdot8=+14.02\)	\(M_3^{\mathrm{II}}\cdot E\beta_3 l_3=+0.178\cdot9.984\cdot8=+14.02\)	\(M_4^{\mathrm{III}}\cdot E\beta_4 l_4=+0.487\cdot6.89\cdot6=-20.12\)	\(M_4^{\mathrm{III}}\cdot E\beta_4 l_4=+0.487\cdot6.89\cdot6=-20.12\)	\(M_4^{\mathrm{III}}\cdot E\beta_4 l_4=+0.487\cdot6.89\cdot6=-20.12\)	\(M_4^{\mathrm{III}}\cdot E\beta_4 l_4=+0.487\cdot6.89\cdot6=-20.12\)	\(M_4^{\mathrm{III}}\cdot E\beta_4 l_4=+0.487\cdot6.89\cdot6=-20.12\)	\(M_4^{\mathrm{III}}\cdot E\beta_4 l_4=+0.487\cdot6.89\cdot6=-20.12\)
\(B=M^B\cdot E\beta\cdot l=\)							we take the ordinates \(E^{-\mathrm{II}}\) from table No. 2 and multiply them by the multiplier \((-0.333)\).	we take the ordinates \(E^{-\mathrm{II}}\) from table No. 2 and multiply them by the multiplier \((-0.333)\).	we take the ordinates \(E^{-\mathrm{II}}\) from table No. 2 and multiply them by the multiplier \((-0.333)\).	we take the ordinates \(E^{-\mathrm{II}}\) from table No. 2 and multiply them by the multiplier \((-0.333)\).	we take the ordinates \(E^{-\mathrm{II}}\) from table No. 2 and multiply them by the multiplier \((-0.333)\).	\(M_3^{\mathrm{III}}\cdot E\beta_3 l_3=-0.429\cdot9.984\cdot8=-33.74\)	\(M_3^{\mathrm{III}}\cdot E\beta_3 l_3=-0.429\cdot9.984\cdot8=-33.74\)	\(M_3^{\mathrm{III}}\cdot E\beta_3 l_3=-0.429\cdot9.984\cdot8=-33.74\)	\(M_3^{\mathrm{III}}\cdot E\beta_3 l_3=-0.429\cdot9.984\cdot8=-33.74\)	\(M_3^{\mathrm{III}}\cdot E\beta_3 l_3=-0.429\cdot9.984\cdot8=-33.74\)	\(M_3^{\mathrm{III}}\cdot E\beta_3 l_3=-0.429\cdot9.984\cdot8=-33.74\)	\(M_4^{\mathrm{IV}}\cdot E\beta_4 l_4=-0.039\cdot6.89\cdot6=-1.61\)	\(M_4^{\mathrm{IV}}\cdot E\beta_4 l_4=-0.039\cdot6.89\cdot6=-1.61\)	\(M_4^{\mathrm{IV}}\cdot E\beta_4 l_4=-0.039\cdot6.89\cdot6=-1.61\)	\(M_4^{\mathrm{IV}}\cdot E\beta_4 l_4=-0.039\cdot6.89\cdot6=-1.61\)	\(M_4^{\mathrm{IV}}\cdot E\beta_4 l_4=-0.039\cdot6.89\cdot6=-1.61\)	\(M_4^{\mathrm{IV}}\cdot E\beta_4 l_4=-0.039\cdot6.89\cdot6=-1.61\)
\(A\cdot\eta_A=\)												0	+3.32	+5.21	+5.40	+4.29	+2.31	0	+4.61	+7.48	+7.85	+6.21	+3.32	0
\(B\cdot\eta_B=\)												0	−5.57	−10.32	−13.60	−12.55	−7.99	0	−0.27	−0.50	−0.63	−0.60	−0.37	0
\(E^{-\mathrm{III}}=A\eta_A+B\eta_B\)	0	−0.39	−0.72	−0.89	−0.85	+0.81	+1.85	+2.74	+2.97	+2.03	0	0	−2.25	−5.11	−7.60	−8.26	−5.68	0	+4.34	+6.98	+7.22	+5.61	+2.95	0

Span participation	Formula
\(Q_{(1)}\)	\(=-0.333\cdot10.33=-3.44\)
\(Q_{(2)}\)	\(=-0.333\cdot(-42.50)=+14.16\)
\(Q_{(3)}\)	\(=\dfrac{A+B}{4}\cdot l_3=\dfrac{14.02-33.74}{4}\cdot8=-39.50\)
\(Q_{(4)}\)	\(=\dfrac{A+B}{4}\cdot l_4=\dfrac{20.12-1.61}{4}\cdot6=+27.80\)

No. 4. Table for Calculating the Influence of Rotation \(E^{-\mathrm{IV}}\) of Node IV.

No. of rod and No. of point	0	1	2	3	4	5	6	7	8	9	10	II	11	12	13	14	15	III	16	17	18	19	20	IV
\(\eta_A=\)							Deflections drop out.	Deflections drop out.	Deflections drop out.	Deflections drop out.	Deflections drop out.							0	0.229	0.372	0.390	0.309	0.165	0
\(\eta_B=\)							In view of the existing focal dependence	In view of the existing focal dependence	In view of the existing focal dependence	In view of the existing focal dependence	In view of the existing focal dependence							0	0.165	0.309	0.390	0.372	0.229	0
\(A=M^A\cdot E\beta\cdot l=\)							\(E^{-\mathrm{IV}}=-x_b\cdot E^{-\mathrm{III}}=-0.655E^{-\mathrm{III}},\)	\(E^{-\mathrm{IV}}=-x_b\cdot E^{-\mathrm{III}}=-0.655E^{-\mathrm{III}},\)	\(E^{-\mathrm{IV}}=-x_b\cdot E^{-\mathrm{III}}=-0.655E^{-\mathrm{III}},\)	\(E^{-\mathrm{IV}}=-x_b\cdot E^{-\mathrm{III}}=-0.655E^{-\mathrm{III}},\)	\(E^{-\mathrm{IV}}=-x_b\cdot E^{-\mathrm{III}}=-0.655E^{-\mathrm{III}},\)							\(M_4^{\mathrm{III}}\cdot E\beta_4 l_4=+0.336\cdot6.89\cdot6=+13.90\)	\(M_4^{\mathrm{III}}\cdot E\beta_4 l_4=+0.336\cdot6.89\cdot6=+13.90\)	\(M_4^{\mathrm{III}}\cdot E\beta_4 l_4=+0.336\cdot6.89\cdot6=+13.90\)	\(M_4^{\mathrm{III}}\cdot E\beta_4 l_4=+0.336\cdot6.89\cdot6=+13.90\)	\(M_4^{\mathrm{III}}\cdot E\beta_4 l_4=+0.336\cdot6.89\cdot6=+13.90\)	\(M_4^{\mathrm{III}}\cdot E\beta_4 l_4=+0.336\cdot6.89\cdot6=+13.90\)
\(B=M^B\cdot E\beta\cdot l=\)							we take the ordinates for \(E^{-\mathrm{III}}\) from table No. 3, multiplied by the multiplier \((-0.655)\).	we take the ordinates for \(E^{-\mathrm{III}}\) from table No. 3, multiplied by the multiplier \((-0.655)\).	we take the ordinates for \(E^{-\mathrm{III}}\) from table No. 3, multiplied by the multiplier \((-0.655)\).	we take the ordinates for \(E^{-\mathrm{III}}\) from table No. 3, multiplied by the multiplier \((-0.655)\).	we take the ordinates for \(E^{-\mathrm{III}}\) from table No. 3, multiplied by the multiplier \((-0.655)\).							\(M_4^{\mathrm{IV}}\cdot E\beta_4 l_4=-0.911\cdot6.89\cdot6=-37.70\)	\(M_4^{\mathrm{IV}}\cdot E\beta_4 l_4=-0.911\cdot6.89\cdot6=-37.70\)	\(M_4^{\mathrm{IV}}\cdot E\beta_4 l_4=-0.911\cdot6.89\cdot6=-37.70\)	\(M_4^{\mathrm{IV}}\cdot E\beta_4 l_4=-0.911\cdot6.89\cdot6=-37.70\)	\(M_4^{\mathrm{IV}}\cdot E\beta_4 l_4=-0.911\cdot6.89\cdot6=-37.70\)	\(M_4^{\mathrm{IV}}\cdot E\beta_4 l_4=-0.911\cdot6.89\cdot6=-37.70\)
\(A\cdot\eta_A\)																		0	+3.18	+5.16	+5.42	+4.29	+2.29	0
\(B\cdot\eta_B\)																		0	−6.22	−11.65	−14.70	−14.04	−8.64	0
\(E^{-\mathrm{IV}}=A\eta_A+B\eta_B\)	0	+0.26	+0.47	+0.58	+0.56	+0.35	0	−0.53	−1.21	−1.80	−1.94	−1.33	0	+1.47	+3.35	+4.97	+5.42	+3.72	0	−3.04	−6.49	−9.28	−9.75	−6.35

Span participation	Formula
\(Q_{(1)}\)	\(=-0.655\cdot(-3.44)=+2.27\)
\(Q_{(2)}\)	\(=-0.655\cdot14.16=-9.28\)
\(Q_{(3)}\)	\(=-0.655\cdot(-39.50)=+25.90\)
\(Q_{(4)}\)	\(=\dfrac{A+B}{4}\cdot l_4=\dfrac{13.90-37.70}{4}\cdot6=-36.62\)

Table No. 5. Calculation of the influence lines for the displacements \(E\delta_i\).

No. of member and point nos.	0	1	2	3	4	5	6	7	8	9	10	II	11	12	13	14	15	III	16	17	18	19	20	IV
\(\eta_A=\)		0	0,271	0,383	0,396	0,310	0,165	0,237	0,372	0,385	0,306	0,165	0	0,237	0,372	0,385	0,306	0,165	0	0,229	0,372	0,390	0,309	0,165
\(\eta_B=\)		0	0,162	0,294	0,366	0,349	0,216	0,165	0,306	0,385	0,385	0,306	0,237	0	0,165	0,306	0,372	0,385	0	0,165	0,309	0,390	0,372	0,229
\(A=m^1\cdot E\beta\cdot l=\)					\(m^1_1\cdot E\beta_1\cdot l_1=0\)				\(m^1_2\cdot E\beta_2\cdot l_2=+0,827\cdot 9,84\cdot 8=+65,20\)				\(m^1_3\cdot E\beta_3\cdot l_3=+0,314\cdot 9,84\cdot 8=+24,72\)				\(m^1_4\cdot E\beta_4\cdot l_4=+0,815\cdot 6,89\cdot 6=+33,68\)
\(B=m^B\cdot E\beta\cdot l=\)					\(m^B_1\cdot E\beta_1\cdot l_1=-0,468\cdot 8,77\cdot 6=-21,85\)				\(m^B_2\cdot E\beta_2\cdot l_2=-0,698\cdot 9,84\cdot 8=-55,00\)				\(m^B_3\cdot E\beta_3\cdot l_3=-0,256\cdot 9,84\cdot 8=-20,18\)				\(m^B_4\cdot E\beta_4\cdot l_4=-1,062\cdot 6,89\cdot 6=-43,90\)
\(A\cdot\eta_A=\)		0	0	0	0	0	0	+24,28	+25,10	+19,95	+10,77	0	+5,86	+9,19	+9,51	+7,56	+4,08	0	+7,71	+12,52	+13,15	+10,42	+5,56	0
\(B\cdot\eta_B=\)		0	−3,54	−6,42	−7,98	−7,62	−4,72	−9,08	−16,83	−21,20	−21,20	−16,83	−13,33	−6,17	−7,77	−7,50	−4,78	0	−7,25	−13,58	−17,15	−16,35	−10,07	0
\(E\delta_j=A\eta_A+B\eta_B=\)		0	−3,54	−6,42	−7,98	−7,62	−4,72	+15,20	+8,27	−1,25	−10,43	−16,83	−7,47	+3,02	+1,74	+0,06	−0,70	0	+0,46	−1,06	−4,00	−5,93	−4,51	0

Area of a section of the influence line:

\[ \Omega_{(1)}=E\beta_1\cdot l_1^{2}\cdot\nabla_B m^1 =-7,77\cdot 6^2\cdot 0,237\cdot 0,468=-30,96 \]

\[ \Omega_{(2)}=\frac{A+B}{4}\cdot l_2\cdot 8 =\frac{65,20-55,00}{4}\cdot 8=+20,30 \]

\[ \Omega_{(3)}=\frac{A+B}{4}\cdot l_3\cdot 8 =\frac{24,72-20,18}{4}\cdot 8=+9,12 \]

\[ \Omega_{(4)}=\frac{A+B}{4}\cdot l_4 =\frac{33,68-43,90}{4}\cdot 6=-15,32 \]

Table No. 6. Calculation of the influence lines of the fixed-end moments.

	No. of member and point nos.	0	1	2	3	4	5	I	6	7	8	9	10	II	11	12	13	14	15	III	16	17	18	19	20	IV
\(M_1^1\)	Ordinate of the basic moment		\(M_1^1=E\tau^1\cdot\frac{1}{E\varphi_1^1}=E\tau^1\cdot\frac{1}{8,836}=0,1132\,E\tau^1\)																			\(M_1^1=E\tau^1\left(-\frac{1}{E\gamma_1^1}\right)=E\tau^1\left(-\frac{1}{11,12}\right)=-0,0899\,E\tau^1\)
\(M_1^1\)	Ordinate of the basic moment	0	−0,378	−0,685	−0,854	−0,814	−0,592	0	−0,557	−0,809	−0,744	−0,498	−0,218	0	+0,198	+0,286	+0,260	+0,172	+0,073	0	−0,050	−0,081	−0,084	−0,065	−0,034	0
	Ordinate of the additional moment		\((M_1^1)=E\delta_i(4,76\cdot10^{-4}\mathfrak{M}_1^1)=E\delta_i(4,76\cdot10^{-4}\cdot2,83)=-1,35\cdot10^{-3}\cdot E\delta_i\)
\(M_1^1\)	Ordinate of the additional moment	0	+0,005	+0,009	+0,011	+0,010	+0,005	0	−0,009	−0,010	−0,005	+0,001	+0,003	0	+0,001	−0,004	−0,002	0,000	+0,001	0	−0,001	+0,001	+0,005	+0,008	+0,006	0
\(M_1^1\)	Full values	0	−0,373	−0,676	−0,843	−0,804	−0,496	0	−0,566	−0,819	−0,749	−0,497	−0,215	0	+0,196	+0,282	+0,258	+0,172	+0,074	0	−0,051	−0,080	−0,079	−0,057	−0,028	0

Areas of the sections of the influence line:

\[ \Omega_{(1)} =\frac{-29,30\cdot0,1132+30,96\cdot1,35\cdot10^{-3}}{} =-3,32+0,042=-3,28 \]

\[ \Omega_{(2)} =\frac{-42,80\cdot0,0899-20,30\cdot1,35\cdot10^{-3}}{} =-3,85-0,027=-3,88 \]

\[ \Omega_{(3)} =\frac{+14,95\cdot0,0899-9,12\cdot1,35\cdot10^{-3}}{} =+1,35-0,012=+1,34 \]

\[ \Omega_{(4)} =\frac{-3,59\cdot0,0899+15,32\cdot1,35\cdot10^{-3}}{} =-0,323+0,021=-0,302 \]

No. 6. (Continuation)

		Stringer 1-I	Stringer 1-I	Stringer 1-I	Stringer 1-I	Stringer 1-I	Stringer 2-I	Stringer 2-I	Stringer 2-I	Stringer 2-I	Stringer 2-I	Stringer 3-I	Stringer 3-I	Stringer 3-I	Stringer 3-I	Stringer 3-I	Stringer 4-I	Stringer 4-I	Stringer 4-I	Stringer 4-I	Stringer 4-I
	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20
\(M_2^{\mathrm{I}}\)	Ordinates of the principal moment	0	\(-0.317\)	\(-0.576\)	\(-0.718\)	\(-0.685\)	\(-0.424\)	0	\(-0.668\)	\(-0.970\)	\(-0.891\)	\(-0.598\)	\(-0.261\)	0	\(+0.238\)	\(+0.342\)	\(+0.312\)	\(+0.206\)	\(+0.087\)	0	\(-0.060\)
\(M_2^{\mathrm{I}}\)	Ordinates of the additional moment	0	\(-0.008\)	\(-0.015\)	\(-0.019\)	\(-0.018\)	\(-0.011\)	0	\(+0.015\)	\(+0.018\)	\(+0.009\)	\(-0.001\)	\(-0.005\)	0	\(+0.006\)	\(+0.007\)	\(+0.004\)	0.000	\(-0.002\)	0	\(+0.001\)
\(M_2^{\mathrm{I}}\)	Total	0	\(-0.325\)	\(-0.591\)	\(-0.737\)	\(-0.703\)	\(-0.435\)	0	\(-0.653\)	\(-0.952\)	\(-0.882\)	\(-0.599\)	\(-0.266\)	0	\(+0.244\)	\(+0.349\)	\(+0.316\)	\(+0.206\)	\(+0.085\)	0	\(-0.059\)
\(M_2^{\mathrm{I}}\)	Influence-area ordinates						\(\Omega_{(1)}=-29.30\cdot0.195+\dfrac{30.96\cdot2.38\cdot10^{-3}}{2.780-0.074}=-2.85\)					\(\Omega_{(2)}=-42.80\cdot0.108+\dfrac{20.30\cdot2.38\cdot10^{-3}}{4.620+0.048}=4.57\)					\(\Omega_{(3)}=+14.95\cdot0.108+\dfrac{9.12\cdot2.38\cdot10^{-3}}{1.610+0.022}=+1.63\)
\(M_8^{\mathrm{I}}\)	Ordinates of the principal moment	0	\(+0.060\)	\(+0.109\)	\(+0.136\)	\(+0.136\)	\(+0.080\)	0	\(-0.111\)	\(-0.162\)	\(-0.148\)	\(-0.100\)	\(-0.043\)	0	\(+0.057\)	\(+0.052\)	\(+0.034\)	\(+0.015\)	0	\(-0.010\)	\(-0.016\)
\(M_8^{\mathrm{I}}\)	Ordinates of the additional moment	0	\(-0.013\)	\(-0.024\)	\(-0.030\)	\(-0.028\)	\(-0.018\)	0	\(+0.024\)	\(+0.028\)	\(+0.015\)	\(-0.002\)	\(-0.008\)	0	\(+0.011\)	\(+0.006\)	0.000	\(-0.003\)	\(+0.002\)	0	\(-0.004\)
\(M_8^{\mathrm{I}}\)	Total	0	\(+0.047\)	\(+0.085\)	\(+0.106\)	\(+0.102\)	\(+0.062\)	0	\(-0.087\)	\(-0.134\)	\(-0.133\)	\(-0.102\)	\(-0.051\)	0	\(+0.068\)	\(+0.058\)	\(+0.034\)	\(+0.012\)	\(+0.008\)	\(-0.020\)	\(-0.032\)
\(M_8^{\mathrm{I}}\)	Influence-area ordinates						\(\Omega_{(1)}=+29.30\cdot0.018+\dfrac{30.96\cdot3.72\cdot10^{-3}}{0.526-0.115}=+0.411\)					\(\Omega_{(2)}=-42.80\cdot0.018+\dfrac{20.30\cdot3.72\cdot10^{-3}}{-0.77+0.08}=-0.69\)					\(\Omega_{(3)}=+14.95\cdot0.018+\dfrac{9.12\cdot3.72\cdot10^{-3}}{0.269+0.034}=+0.303\)
\(M_8^{\mathrm{VIII}}\)	Ordinates of the principal moment	0	\(-0.020\)	\(-0.036\)	\(-0.045\)	\(-0.043\)	\(-0.027\)	0	\(+0.037\)	\(+0.054\)	\(+0.049\)	\(+0.033\)	\(+0.014\)	0	\(-0.019\)	\(-0.017\)	\(-0.011\)	\(-0.005\)	0	\(+0.003\)	\(+0.004\)
\(M_8^{\mathrm{VIII}}\)	Ordinates of the additional moment	0	\(+0.010\)	\(+0.019\)	\(+0.023\)	\(+0.022\)	\(+0.014\)	0	\(-0.019\)	\(-0.022\)	\(-0.011\)	\(+0.001\)	\(+0.005\)	0	\(-0.007\)	\(-0.009\)	\(-0.005\)	0.000	\(+0.002\)	0	\(-0.001\)
\(M_8^{\mathrm{VIII}}\)	Total	0	\(-0.010\)	\(-0.017\)	\(-0.022\)	\(-0.021\)	\(-0.013\)	0	\(+0.018\)	\(+0.032\)	\(+0.038\)	\(+0.034\)	\(+0.020\)	0	\(-0.022\)	\(-0.028\)	\(-0.011\)	\(-0.005\)	0	\(+0.002\)	\(+0.008\)
\(M_8^{\mathrm{VIII}}\)	Influence-area ordinates						\(\Omega_{(1)}=-0.333\cdot0.526+\dfrac{30.96\cdot2.91\cdot10^{-3}}{-0.197+0.090}=-0.107\)					\(\Omega_{(2)}=+0.333\cdot0.77-\dfrac{20.30\cdot2.91\cdot10^{-3}}{0.257-0.059}=+0.198\)					\(\Omega_{(3)}=-0.333\cdot0.269-\dfrac{9.12\cdot2.91\cdot10^{-3}}{-0.090-0.027}=-0.117\)

\[ M_2^{\mathrm{I}}=E\tau^{\mathrm{I}}\cdot\frac{1}{E\gamma_2^{\mathrm{I}}} =E\tau^{\mathrm{I}}\left(-\frac{1}{9.26}\right) =E\tau^{\mathrm{I}}(-0.108). \]

\[ (M_2^{\mathrm{I}})=E\delta_l(4.76\cdot10^{-4}\cdot5.00) =-2.38\cdot10^{-3}E\delta_l. \]

\[ M_8^{\mathrm{I}}=E\tau^{\mathrm{I}}\left(-\frac{1}{E\gamma_8^{\mathrm{I}}}\right) =E\tau^{\mathrm{I}}\left(-\frac{1}{55.6}\right) =-0.0180\,E\tau^{\mathrm{I}}. \]

\[ (M_8^{\mathrm{I}})=E\delta_l(4.76\cdot10^{-4}\cdot7.83) =+3.72\cdot10^{-3}E\delta_l. \]

\[ M_8^{\mathrm{VIII}}=-K_{a,8}M_8^{\mathrm{I}}=-0.333\,M_8^{\mathrm{I}} \quad\text{(here \(K_{a,8}\) is the left focus ratio)} \]

\[ (M_8^{\mathrm{VIII}})=-E\delta_l(4.76\cdot10^{-4}\cdot6.11) =-2.91\cdot10^{-3}E\delta_l. \]

No. 6. (Continuation of the table)

Quantity	Frames 1st	Frames 2nd	Frames 3rd	Frames 4th
	0–5	6–10	11–15	16–20, IV
\(M_2^{\mathrm{II}}\)
Main-moment ordinates	\(M_2^{\mathrm{II}}=E\varphi_2^{\mathrm{II}}=E r^{\mathrm{II}}\dfrac1{8.91}=+0.1124\,E r^{\mathrm{II}}\)		\(M_2^{\mathrm{II}}=E r^{\mathrm{II}}\left(-\dfrac1{E\gamma_2^{\mathrm{II}}}\right)=E r^{\mathrm{II}}\left(-\dfrac1{10.57}\right)=-0.0945\,E r^{\mathrm{II}}\)
Main-moment ordinates	\(0,\ +0.132,\ +0.242,\ +0.300,\ +0.286,\ +0.177,\ 0\)	\(-0.274,\ -0.622,\ -0.922,\ -1.001,\ -0.685,\ 0\)	\(-0.575,\ -0.830,\ -0.756,\ -0.500,\ -0.214,\ 0\)	\(+0.147,\ +0.237,\ +0.244,\ +0.190,\ +0.100,\ 0\)
Additional-moment ordinates	\((M_2^{\mathrm{II}})=E\delta_1(4.76\cdot10^{-4};\,4.22)=-2.01\cdot10^{-3}\,E\delta_1\)
Additional-moment ordinates	\(0,\ +0.007,\ +0.013,\ +0.016,\ +0.015,\ +0.009,\ 0\)	\(-0.013,\ -0.015,\ -0.008,\ +0.001,\ +0.005,\ 0\)	\(-0.005,\ -0.006,\ -0.003,\ 0,\ +0.001,\ 0\)	\(-0.001,\ +0.002,\ +0.008,\ +0.012,\ +0.009,\ 0\)
Total	\(0,\ +0.139,\ +0.255,\ +0.316,\ +0.301,\ +0.186,\ 0\)	\(-0.287,\ -0.637,\ -0.930,\ -1.000,\ -0.680,\ 0\)	\(-0.580,\ -0.836,\ -0.759,\ -0.500,\ -0.213,\ 0\)	\(+0.146,\ +0.239,\ +0.252,\ +0.202,\ +0.109,\ 0\)
Areas of influence sectors	\(Q_{(1)}=+10.35\cdot0.1124+\dfrac{30.96\cdot2.01\cdot10^{-3}}{1.062}+\dfrac{0.062}{1.062}=+1.124\)	\(Q_{(2)}=-42.50\cdot0.1124-\dfrac{20.30\cdot2.01\cdot10^{-3}}{4.780}-\dfrac{0.041}{4.780}=-4.821\)	\(Q_{(3)}=-41.40\cdot0.0945-\dfrac{9.12\cdot2.01\cdot10^{-3}}{3.920}-\dfrac{0.019}{3.920}=-3.94\)	\(Q_{(4)}=+9.95\cdot0.0945+\dfrac{15.32\cdot2.01\cdot10^{-3}}{0.940}+\dfrac{0.031}{0.940}=+0.97\)
\(M_3^{\mathrm{II}}\)
Main-moment ordinates	\(M_3^{\mathrm{II}}=E\gamma_3^{\mathrm{II}}=E r^{\mathrm{II}}\dfrac1{10.35}=+0.0965\,E r^{\mathrm{II}}\)		\(M_3^{\mathrm{II}}=E r^{\mathrm{II}}\left(-\dfrac1{\varphi_3^{\mathrm{II}}}\right)=E r^{\mathrm{II}}\left(-\dfrac1{9.06}\right)=-0.1105\,E r^{\mathrm{II}}\)
Main-moment ordinates	\(0,\ +0.114,\ +0.208,\ +0.258,\ +0.246,\ +0.153,\ 0\)	\(-0.236,\ -0.535,\ -0.791,\ -0.860,\ -0.589,\ 0\)	\(-0.672,\ -0.970,\ -0.885,\ -0.535,\ -0.252,\ 0\)	\(+0.171,\ +0.277,\ +0.285,\ +0.222,\ +0.117,\ 0\)
Additional-moment ordinates	\((M_3^{\mathrm{II}})=E\delta_1(4.76\cdot10^{-4};\,1.90)=+0.904\cdot10^{-3}\,E\delta_1\)
Additional-moment ordinates	\(0,\ -0.003,\ -0.006,\ -0.007,\ -0.007,\ -0.004,\ 0\)	\(+0.006,\ +0.007,\ +0.004,\ 0,\ -0.002,\ 0\)	\(+0.002,\ +0.003,\ +0.002,\ 0,\ -0.001,\ 0\)	\(+0.000,\ -0.001,\ -0.004,\ -0.006,\ -0.004,\ 0\)
Total	\(0,\ +0.111,\ +0.202,\ +0.251,\ +0.239,\ +0.149,\ 0\)	\(-0.230,\ -0.528,\ -0.787,\ -0.860,\ -0.591,\ 0\)	\(-0.670,\ -0.967,\ -0.883,\ -0.535,\ -0.253,\ 0\)	\(+0.171,\ +0.276,\ +0.281,\ +0.216,\ +0.113,\ 0\)
Areas of influence sectors	\(Q_{(1)}=+10.35\cdot0.0965-\dfrac{30.96\cdot0.904\cdot10^{-3}}{1.000}+\dfrac{0.028}{1.000}=+0.972\)	\(Q_{(2)}=-42.50\cdot0.0965+\dfrac{20.30\cdot0.904\cdot10^{-3}}{4.11}+\dfrac{0.018}{4.11}=-4.092\)	\(Q_{(3)}=-41.40\cdot0.1105+\dfrac{9.12\cdot0.904\cdot10^{-3}}{4.58}+\dfrac{0.008}{4.58}=-4.572\)	\(Q_{(4)}=-9.95\cdot0.1105-\dfrac{15.32\cdot0.904\cdot10^{-3}}{1.098}-\dfrac{0.014}{1.098}=-1.084\)
\(M_7^{\mathrm{II}}\)
Main-moment ordinates	\(M_7^{\mathrm{II}}=E\varphi_7^{\mathrm{II}}=E r^{\mathrm{II}}\left(-\dfrac1{E\gamma_7^{\mathrm{II}}}\right)=E r^{\mathrm{II}}\left(-\dfrac1{64.10}\right)=-0.0156\,E r^{\mathrm{II}}\)
Main-moment ordinates	\(0,\ -0.018,\ -0.034,\ -0.042,\ -0.040,\ -0.025,\ 0\)	\(+0.038,\ +0.087,\ +0.128,\ +0.139,\ +0.095,\ 0\)	\(-0.137,\ -0.125,\ -0.082,\ -0.035,\ 0,\ +0.024\)	\(+0.039,\ +0.040,\ +0.031,\ +0.017,\ 0\)
Additional-moment ordinates	\((M_7^{\mathrm{II}})=E\delta_1(4.76\cdot10^{-4};\,6.12)=+2.92\cdot10^{-3}\,E\delta_1\)
Additional-moment ordinates	\(0,\ -0.010,\ -0.019,\ -0.023,\ -0.022,\ -0.014,\ 0\)	\(+0.019,\ +0.022,\ +0.011,\ -0.001,\ -0.005,\ 0\)	\(+0.005,\ 0,\ -0.002,\ 0,\ +0.001,\ 0\)	\(-0.003,\ -0.012,\ -0.017,\ -0.013,\ 0\)
Total	\(0,\ -0.028,\ -0.053,\ -0.065,\ -0.062,\ -0.039,\ 0\)	\(+0.057,\ +0.109,\ +0.139,\ +0.138,\ +0.088,\ 0\)	\(-0.128,\ -0.120,\ -0.082,\ -0.037,\ 0,\ +0.024\)	\(+0.036,\ +0.028,\ +0.014,\ +0.004,\ 0\)
Areas of influence sectors	\(Q_{(1)}=-10.35\cdot0.156-\dfrac{30.96\cdot2.92\cdot10^{-3}}{0.161}-\dfrac{0.090}{0.161}=-0.251\)	\(Q_{(2)}=+42.50\cdot0.0156+\dfrac{20.30\cdot2.92\cdot10^{-3}}{0.833}+\dfrac{0.059}{0.833}=+0.722\)	\(Q_{(3)}=-41.40\cdot0.0156+\dfrac{9.12\cdot2.92\cdot10^{-3}}{-0.646}-\dfrac{0.027}{-0.646}=-0.673\)	\(Q_{(4)}=+9.95\cdot0.0156-\dfrac{15.32\cdot2.92\cdot10^{-3}}{0.155}-\dfrac{0.045}{0.155}=+0.110\)

No. 7. Table for computing the ordinates of the influence line of the bending moment at the middle of the 2nd span.

\[ M_x=M_{0x}+\frac{M^A(l-x)}{l}+\frac{M^B x}{l}, \quad \text{which for } x=\frac{l}{2} \text{ gives} \]

\[ M_x=M_0+\frac{M^A+M^B}{2}; \qquad M^A=M_2^I;\quad M^B=M_2^{II};\quad l_2=8.0\ \text{m}. \]

Span	Point Nos.	\(M_{0x}\)	\(M_2^I\)	\(M_2^{II}\)	\(\dfrac{M_2^I+M_2^{II}}{2}\)	\(M_x\)	Areas of portions of the i.l.
1st span	0	0	0	0	0	0
1st span	1	0	−0.325	+0.139	−0.096	−0.096	\(\displaystyle \Omega_{(1)}=\frac{-2.85+1.124}{2}=\)
1st span	2	0	−0.591	+0.255	−0.168	−0.168	\(\displaystyle =-0.863\)
1st span	3	0	−0.737	+0.316	−0.210	−0.210
1st span	4	0	−0.703	+0.301	−0.200	−0.200
1st span	5	0	−0.435	+0.186	−0.125	−0.125
1st span	I	0	0	0	0	0
2nd span	6	0.667	−0.653	−0.287	−0.470	+0.197	\(\displaystyle \Omega_{(2)}=\frac{2.8}{2}-\)
2nd span	7	1.333	−0.952	−0.637	−0.795	+0.538	\(\displaystyle -\frac{4.57+4.821}{2}=\)
2nd span	8	2.00	−0.882	−0.930	−0.906	+1.094	\(\displaystyle =8-4.695=+3.305\)
2nd span	9	1.333	−0.599	−1.000	−0.800	+0.533
2nd span	10	0.667	−0.266	−0.680	−0.473	+0.194
2nd span	II	0	0	0	0	0
3rd span	11	0	+0.244	−0.580	−0.168	−0.168	\(\displaystyle \Omega_{(3)}=\frac{+1.63-3.94}{2}=\)
3rd span	12	0	+0.349	−0.836	−0.243	−0.243	\(\displaystyle =-1.155\)
3rd span	13	0	+0.316	−0.759	−0.221	−0.221
3rd span	14	0	+0.206	−0.500	−0.147	−0.147
3rd span	15	0	+0.085	−0.213	−0.064	−0.064
3rd span	III	0	0	0	0	0
4th span	16	0	−0.059	+0.146	+0.043	+0.043	\(\displaystyle \Omega_{(4)}=\frac{-0.42+0.97}{2}=\)
4th span	17	0	−0.100	+0.239	+0.069	+0.069	\(\displaystyle =+0.275\)
4th span	18	0	−0.110	+0.252	+0.071	+0.071
4th span	19	0	−0.092	+0.202	+0.055	+0.055
4th span	20	0	−0.052	+0.109	+0.028	+0.028
4th span	IV	0	0	0	0	0

From the ordinates computed in Fig. 22ᵇ, the influence line has been constructed. With this we finish Example 1. Owing to the detailed explanations accompanying each table, the treatment of this example has grown very large in volume. In practical work, however, when such explanations are superfluous, the solution takes on a quite compact form, and the entire work is reduced to the successive filling-in of tables.

Example II. A reinforced-concrete three-span frame viaduct (Fig. 23) is loaded by a concentrated force \(P_1 = 20\) t in the first span, a concentrated force \(P_2 = 10\) t in the 2nd span, and a continuous uniformly distributed load along the length of the 3rd span.

From the preliminary computations that have been made, the angular quantities are known:

\[ E\beta_2 = E\beta_4 = 8.94; \qquad E\beta_3 = 10.40 \]

(all elements of the frame are assumed to be members of constant stiffness). In addition, the moment diagram has been constructed for the displacement of the upper nodes of the frame to the right by \(\Delta x = 0.001\) m, and the moments \(m\) corresponding to the displacing force \(V = 1\) t have been computed.

Namely, for the support moments of the cross-girders the following values were obtained:

from displacement by \(\Delta x = 0.001\) m:

\[ \mathfrak{M}_2^{\mathrm{I}} = +8.447 \ \text{t m.} \qquad \mathfrak{M}_3^{\mathrm{II}} = +1.935 \ \text{t m.} \qquad \mathfrak{M}_4^{\mathrm{III}} = +5.325 \ \text{t m.} \]

\[ \mathfrak{M}_2^{\mathrm{II}} = -5.325 \quad „ \qquad \mathfrak{M}_3^{\mathrm{III}} = -1.935 \quad „ \qquad \mathfrak{M}_4^{\mathrm{IV}} = -8.447 \quad „ \]

from the displacing force \(V = 1\) t:

\[ m_2^{\mathrm{I}} = +0.846 \ \text{t m.} \qquad m_3^{\mathrm{II}} = +0.194 \ \text{t m.} \qquad m_4^{\mathrm{III}} = +0.534 \ \text{t m.} \]

\[ m_2^{\mathrm{II}} = -0.534 \quad „ \qquad m_3^{\mathrm{III}} = -0.194 \quad „ \qquad m_4^{\mathrm{IV}} = -0.846 \quad „ \]

It is required, for the loading case shown in Fig. 23, to find the displacement of the upper nodes of the frame without determining the displacing forces. In addition, it is required to compute the additional support moments for the cross-girders corresponding to the displacement found.

Solution.

The problem reduces to applying formulas (25) and (26) to compute the displacements (the displacement coefficients \(K\)).

In the 1st span we have (load at midspan):

\[ \text{displacement} \qquad \delta_l' = K\Delta x = 0.375 P_1 \beta l \left(m^{\mathrm{A}} + m^{\mathrm{B}}\right) = \frac{0.375 P_1 E\beta_2 l_2\left(m_2^{\mathrm{I}} + m_2^{\mathrm{II}}\right)}{E}, \]

or

\[ \delta_l' = \frac{0.375 \cdot 20 \cdot 8.94 \cdot 7.50(0.846 - 0.534)} {21 \cdot 10^5} = \frac{0.375 \cdot 20 \cdot 8.94 \cdot 7.50 \cdot 0.312} {21 \cdot 10^5} = \]

\[ = +7.46 \cdot 10^{-5}\ \text{m.} \]

In the 2nd span (load at \(\dfrac{1}{3}\) of the span from the left support: \(\dfrac{2.92}{8.75} = \dfrac{1}{3}\)):

\[ \text{displacement} \qquad \delta_l'' = K\Delta x = P_2 \beta l \left(0.370 m^{\mathrm{A}} + 0.296 m^{\mathrm{B}}\right) = \]

\[ = \frac{P_2 E\beta_3 l_3 \left(0.370 m_3^{\mathrm{II}} + 0.296 m_3^{\mathrm{III}}\right)} {E}, \]

or

\[ \delta_l'' = \frac{10 \cdot 10.40 \cdot 8.75(0.370 \cdot 0.194 - 0.296 \cdot 0.194)} {21 \cdot 10^5} = \]

\[ = \frac{10 \cdot 10.40 \cdot 8.75 \cdot 0.194 \cdot 0.074} {21 \cdot 10^5} = +0.62 \cdot 10^{-5}\ \text{m.} \]

In the 3rd span (uniformly distributed continuous load): displacement

\[ \delta_l'''=K\Delta x=0.250\,q\,l^2\,\beta\,(m^{A}+m^{B})= \]

\[ =\frac{0.250\,q\,l_4^{\,2}\,E\beta_4\,(m_4^{\mathrm{III}}+m_4^{\mathrm{IV}})}{E}, \]

or

\[ \delta_l'''=\frac{0.250\cdot20\cdot7.50^2\cdot8.94\,(0.534-0.846)} {21\cdot10^5}= \]

\[ =\frac{0.250\cdot20\cdot7.5^2\cdot8.94\,(-0.312)} {21\cdot10^5} =-37.7\cdot10^{-5}\ \text{m}. \]

Total displacement:

\[ \delta_l=\delta_l'+\delta_l''+\delta_l''' =7.46\cdot10^{-5}+0.62\cdot10^{-5}-37.7\cdot10^{-5} =-29.62\cdot10^{-5}\ \text{m} \]

\[ =-0.296\ \text{m/m}. \]

The sign \((-)\) indicates that the displacement will occur to the left.
The additional fixing moments for the girders are obtained from formula (44):

\[ M'=\delta_l\cdot\frac{\mathfrak{M}}{\Delta x} =\mathfrak{M}\cdot\frac{\delta_l}{\Delta x} =\mathfrak{M}\cdot\left(-\frac{29.62\cdot10^{-5}}{0.001}\right) =-0.296\,\mathfrak{M}. \]

\[ (M_2^{\mathrm{I}})'=-0.296\,\mathfrak{M}_2^{\mathrm{I}} =-0.296\cdot8.447=-2.498\ \text{t·m}. \]

\[ (M_2^{\mathrm{II}})'=-0.296\,\mathfrak{M}_2^{\mathrm{II}} =+0.296\cdot5.325=+1.584\ \text{''} \]

\[ (M_3^{\mathrm{II}})'=-0.296\,\mathfrak{M}_3^{\mathrm{II}} =-0.296\cdot1.935=-0.573\ \text{''} \]

\[ (M_3^{\mathrm{III}})'=-0.296\,\mathfrak{M}_3^{\mathrm{III}} =+0.296\cdot1.935=+0.573\ \text{''} \]

\[ (M_4^{\mathrm{III}})'=-0.296\,\mathfrak{M}_4^{\mathrm{III}} =-0.296\cdot5.325=-1.584\ \text{''} \]

\[ (M_4^{\mathrm{IV}})'=-0.296\,\mathfrak{M}_4^{\mathrm{IV}} =+0.296\cdot8.447=+2.498\ \text{''} \]

Conclusion

The fundamental principles of the new method for constructing influence lines are given at the end of Chapter I and Chapter II.

Let us draw final conclusions from everything set forth in the present article.

1. The new method of finding influence lines for restraining moments in frames with immovable joints is applicable to all cases of frames, provided that they do not contain closed sections.

In particular, the new method is applicable to the analysis of continuous beams.

If a lesser accuracy of calculation is admitted, use of the method presented is also possible in the case of frames with closed sections, if the device is employed of cutting a complex multi-story system into elementary single-story frames,* which of course gives approximate results, but sometimes with an accuracy sufficient for practical purposes.

1. In finding influence lines for frames with displacing joints, the calculation is divided into two entirely independent parts: 1) finding the influence lines for the principal restraining moments, on the assumption that the joints of the frame are immovable; 2) finding the influence lines for additional moments arising as a result of displacement of the joints. Summation of the ordinates obtained for the first and second parts of the calculation gives the influence lines of the total restraining moments.

The proposed method for finding the influence lines of additional moments is valid for all types of frames, including frames with closed sections.

1. Certain conclusions and formulas following from the theory of the new method of constructing influence lines for additional moments may be applied to the analysis of frames under a fixed load, creating simplification in the computation of displacements and the determination of additional moments in frames with horizontal girders and vertical columns for cases of loads on the girders and at the column joints (if the load is absent within the columns).

The advantages of the new method of constructing influence lines may include the following:

1) In finding the influence lines of the principal moments, the need to construct the principal diagram disappears; i.e., the tedious work connected with the distribution of moments is eliminated.

1. In finding the influence lines of the additional moments, it is not necessary to compute the displacing forces for each position of the moving load; i.e., the most cumbersome part of the computational work in determining displacements and additional moments by the old method disappears.

2. In the new method, all the work of computing the ordinates of the influence lines is reduced to the mechanical multiplication of a series of coefficients by multipliers common to the entire series,** and subsequent addition of the results.

The mechanical nature and simplicity of the computations make the work untiring and create the possibility of rapid checking, whereby great accuracy is guaranteed. All computations of ordinates may be placed in tables.

* See on this, Analysis of Frame Structures, 4th ed., pp. 273–274.
** When using a slide rule, such an arithmetical operation may be performed with one position of the rule’s slider. When using an arithmometer—with the setting of one and the same multiplier on the drum of the arithmometer.

1. Increasing the accuracy of constructing influence lines, i.e., increasing the number of points in the span of a member for which ordinates are determined, has almost no effect on the amount and time of computational work, whereas in the former method the number of separate positions of the moving load entirely determined the volume of computations of the ordinates of influence lines.

2. With the new method, the possibility is retained of making direct use of Suter’s tables for members with variable stiffness, which gives the method of foci, in general, a special practical value.

3. The construction of a special graph (§ 3, Ch. III) for the case of members with constant stiffness can facilitate the computation of the ordinates of influence lines and make the computational work still more elementary.

4. The transfer of the new principles for constructing influence lines to the field of calculating frames under a fixed load brings, in known cases, practical simplifications in determining additional moments and makes it possible to compute displacements without determining the displacing forces.

5. A substantial advantage of the new method is the possibility of an extremely easy auxiliary computation of the areas of portions of influence lines, the computation being carried out with complete mathematical accuracy. This eliminates the need to perform this work subsequently, after the construction of the influence lines, by the approximate trapezoidal rule (or Simpson’s rule)—as is usually done.

6. The theoretical merits of the new method consist in the fact that the interpretation of an influence line as the elastic line of a member loaded by support moments makes it possible to obtain a clearer and fuller understanding of the form of the influence line in frame systems.

In conclusion, let us say a few words about the method of foci in general.

The discovery of the existence and remarkable properties of the focal points (Festpunkte) of an elastically fixed member is undoubtedly a major piece of positive knowledge about the behavior of a system of such members, i.e., of a frame structure. Therefore the method of foci, in addition to purely practical conveniences in solving certain categories of frame problems, has great theoretical significance, revealing the essence of the behavior of a frame structure in general.

The existing literature on frame structures sufficiently illuminates the question of the practical applicability of the method of foci. The practice of computations has established fairly clear limits to the cases of frame problems in which the specific advantages of the method of foci—simplicity and visual clarity—recede into the background before the cumbersomeness of the computational operations (frames with closed panels, high-rise multi-story frames), while the availability of the latest, more convenient analytical techniques compels one either to abandon the use of the method of foci “pur sang” in favor of a combined method, or else to turn to purely analytical methods of solution.

As for single-story systems, especially unsymmetrical and multi-bay ones, here the method of foci has great practical advantages. Still greater practical value is recognized for the method of foci in constructing influence lines in frames and continuous beams.

The present article is devoted precisely to this latter field of application of the method of foci.

Tomsk, May 1929.

Plate I. For the article by N. E. Strobenko.

Fig. 1.

Fig. 2.

Fig. 3.

Fig. 4.

Fig. 5.

Fig. 6.

Plate II.

For the article by Ya. E. Strebeiko.

Fig. 7

Labels and notations visible in the figure:

• \(A\), \(B\)
• \(+M^A\), \(-M^A\)
• \(+M^B\), \(-M^B\)

Fig. 8

Labels and notations visible in the frame diagram:

• Joints and supports: \(0\), I, II, III, IV, V, VI, VII, VIII, IX, X, XI, XII, XIII
• Members: \((1)\), \((2)\), \((3)\), \((4)\), \((5)\), \((6)\), \((7)\), \((8)\), \((9)\), \((10)\), \((11)\), \((12)\), \((13)\)
• Loads and points:
• \(P_n = 1\)
• \(P_m = 1\)
• \(n\), \(m\)
• \(Z_1\)

Fig. 9

Load to the right of joint IV.

Action of the right-hand part on the left-hand part

Labels and notations visible in the figure:

• IV
• \((4)\), \((8)\), \((12)\)
• \(+M^{IV}_{12}\)
• \(-M^{IV}_{4}\)
• \(-M^{IV}_{8}\)
• \(\tau_{IV}\), \(\tau'_{IV}\), \(\tau''_{IV}\)

Action of the left-hand part on the right-hand part

Labels and notations visible in the figure:

• IV
• \((5)\)
• \(-M^{IV}_{5}\)
• \(\tau_{IV}\)

Plate III

To the article by N. E. Strel’bicko.

Fig. 10
Member to the left of joint IV

Action of the right part on the left

\[ +M^{\mathrm{IV}}_{4} \]

\[ \tau^{\mathrm{IV}} \]

Action of the left part on the right

\[ +M^{\mathrm{IV}}_{12} \]

\[ -M^{\mathrm{IV}}_{8} \]

\[ +M^{\mathrm{IV}}_{5} \]

Fig. 11

Influence line of the angle \(\tau^{\mathrm{III}}\)

Influence line of moment \(M^{\mathrm{III}}_{3}\)

\[ M^{\mathrm{III}}_{3} = \tau^{\mathrm{III}}\!\cdot\!\left(+\frac{1}{\varphi^{\mathrm{III}}_{3}}\right) \]

\[ M^{\mathrm{III}}_{3} = \tau^{\mathrm{III}}\!\cdot\!\left(-\frac{1}{\varphi^{\mathrm{III}}_{3}}\right) \]

Influence line of moment \(M^{\mathrm{III}}_{4}\)

\[ M^{\mathrm{III}}_{4} = \tau^{\mathrm{III}}\!\cdot\!\left(+\frac{1}{\varphi^{\mathrm{III}}_{4}}\right) \]

\[ M^{\mathrm{III}}_{4} = \tau^{\mathrm{III}}\!\cdot\!\left(-\frac{1}{\varphi^{\mathrm{III}}_{4}}\right) \]

Influence line of moment \(M^{\mathrm{III}}_{8}\)

\[ M^{\mathrm{III}}_{8} = \tau^{\mathrm{III}}\!\cdot\!\left(-\frac{1}{\delta^{\mathrm{III}}_{8}}\right) \]

Influence line of moment \(M^{\mathrm{III}}_{8}\)

\[ M^{\mathrm{III}}_{8} = \tau^{\mathrm{III}}\!\cdot\!\left(+\frac{K_{89}}{\delta^{\mathrm{III}}_{8}}\right) \]

Plate IV

To the article by N. E. Streboiko.

Fig. 12.

Fig. 13.

Fig. 14.

Fig. 15.

Fig. 16.

horizontal line of displacement \(\delta_\ell^{\mathrm{II}}\)

\[ \left[\delta_\ell^{\mathrm{II}} = K_{\mathrm{II}}\cdot \Delta x\right] \]

Plate V

To the article by N. E. Streboyko.

Fig. 17

Visible labels in the diagram: \(A\), \(B\), \(A'\), \(B'\), \(\delta_n\), \(\delta_B\), \(m^A\), \(m^B\).

Fig. 18

Visible labels in the diagram: \(C^A\), \(M^A\), \(M^B\), \(\tau^A\), \(\tau^B\).

Fig. 19

Visible labels in the diagram: \(A\), \(M\), \((1)\), \((2)\), \((3)\), \((4)\), \((5)\).

Fig. 20

Scale for calculating the quantities \(\varphi\).

Visible labels in the diagram: \(A\), \(O\), \(B\), \(X\), \(N\), \(\delta_1\), \(\delta_2\), \(\varphi_{1-2}\), \(x_0\), \(y_0\), \(60^\circ\).

\[ OA=\delta_1;\qquad OB=\delta_2, \]

\[ OX=\varphi_{1-2}. \]

\[ \frac{1}{\varphi_{1-2}}=\frac{1}{\delta_1}+\frac{1}{\delta_2}. \]

Fig. 21

Visible labels in the diagram: \(B\), \(C\), \(A\), \((n)\), \((n+1)\), \(M=1\), \(l_n\), \(J_n\), \(l_{n+1}\), \(J_{n+1}\).

Plate VI

To the article by N. E. Strobeiko.

Fig. 22

For numerical example I

Frame diagram with spans:

• \(\ell_1 = 6.00\)
• \(\ell_2 = 8.00\)
• \(\ell_3 = 8.00\)
• \(\ell_4 = 6.00\)

Column heights:

• \(h_8 = 6.00\)
• \(h_7 = 7.00\)
• \(h_6 = 8.00\)
• \(h_5 = 6.00\)

Visible joint and member labels include:

• Supports / bases: \(I\), \(II\), \(III\), \(IV\), \(V\), \(VI\), \(VII\), \(VIII\)
• Top points: \(b_1\), \(a_2\), \(b_2\), \(a_3\), \(b_3\), \(a_4\), \(b_4\), \(b_5\)
• Member numbers: \((1)\), \((2)\), \((3)\), \((4)\), \((5)\), \((6)\), \((7)\), \((8)\)
• Column labels: \(\delta_8\), \(d_8\); \(\delta_7\), \(d_7\); \(\delta_6\), \(d_6\); \(\delta_5\)

Fig. 22a

Influence line \(E\delta_2\)

Visible station marks along the baseline:
\(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\)

Marked points on the line:

• \(I\) at station \(5\)
• \(II\) at station \(10\)
• \(III\) at station \(16\)
• \(IV\) at station \(20\)

Visible ordinate values:

• \(-7.98\)
• \(+7.45\)
• \(-2.28\)
• \(+3.02\)
• \(-0.70\)
• \(+0.46\)
• \(-5.93\)

Sign convention shown: “\(-\)” above the baseline, “\(+\)” below the baseline.

Plate VII

To the article by N. E. Streletsky.

Fig. 22

Line of the bending moment in the section of the 2nd span \((M_2^{(8)})\)

Line of influence of \(M_{\max}\)

Span II

\[ \Omega_1 = -0.863 \]

\[ \Omega_2 = +3.305 \]

\[ \Omega_3 = -1.155 \]

\[ \Omega_4 = +0.071 \]

Fig. 23

Span II

\[ q = 20\ \mathrm{t/m} \]

\[ P_1 = 20\ \mathrm{t} \]

\[ P_2 = 10\ \mathrm{t} \]

\[ J_2 = J_3 = J_4 = 0.14\ \mathrm{m}^4 \]

\[ J_1 = J_7 = J_5 = 0.09\ \mathrm{m}^4 \]