PROCEEDINGS
of the Tomsk Technological Institute
of Emperor Nicholas II

Vol. 7. 1907. No. 4.

N. V. Nekrasov.

ON THE THEORY OF TRUSSES WITH RIGID JOINTS AT THE NODES.

1. An attempt at a comparative analysis of calculation methods. I–VIII, 1–182.

Contents

Literature.
Preface. Page

Chapter I. Investigation of the Deformation of a Rigid Truss.

1. Conditions of deformation of a rigid truss . . . . . . . . . . . . . . . . 1
2. General formulas for the bending of a beam with fixed ends, subjected to compression . . . . . . . . . . . . . . . . . . . . . .
3. Work of deformation of a compressed and bent beam with fixed ends . . . . . . . . . . . . . . . . . . . . . . . . .
4. Generalization of the derived formulas for the case of combined bending and tension . . . . . . . . . . . . . . . . . . . .
5. Investigation of the functions $\varphi$ and $\psi$ . . . . . . . . . . . . . . . . 17
6. Work of deformation of a rigid truss and its derivative. Necessary simplifications . . . . . . . . . . . . . . . . . . . . 22

Chapter II. Systematization of Methods for Calculating Stresses Due to the Rigidity of Joints.

1. Classification of existing methods . . . . . . . . . . . . . . . . . . 26
2. Generalized method . . . . . . . . . . . . . . . . . . . . . 26
3. Group of methods of Eng. Perederii . . . . . . . . . . . . . . . . 33
4. Methods of Manderla, Mohr, etc. . . . . . . . . . . . . . . . . 37
5. Methods of Engesser, Müller-Breslau, et al. . . . . . . . . . . . . . 40

Chapter III. Examples of Comparative Calculations of Rigidity Stresses.

1. On the selection of truss examples for comparative calculations . . . . 41
2. Example I. A two-lattice truss . . . . . . . . . . . . . . . . . 42
3. Derivation of the basic equations for calculating moments due to joint rigidity . . . . . . . . . . . . . . . . . . . . 49
4. Calculation of the coefficients of the equations . . . . . . . . . . . . 55
5. Abbreviated method for solving the equations . . . . . . . . . . . . 64
6. Example II. A two-panel truss . . . . . . . . . . . . . . . . . 81
7. Example III. A truss with a triangular web and additional posts . . . . . 100
8. Example IV. The truss of Example III with the stiffness of the chords reduced by half . . . . . . . . . . . . . . . . . . . . 109
9. Influence of the assumptions made by us (in substantiating the generalized method) . . . . . . . . . . . . . . . . . . . 112

1. Example V. A truss with triangular lattice and additional
   posts (bridge of the Moscow–Vindava–Rybinsk Railway) . . . 117

Chapter IV. Systematization of the data obtained by comparative calculations.

1. On the method of comparing the data obtained . . . . . . . . . . 142
2. Comparative accuracy of various methods of calculation . . . . . 153
3. Works of Engineers Paton and Perederiy . . . . . . . . . . . . . 160

Chapter V. Analysis of the diagrams constructed by Engineer Paton.

1. Relationship between the moments due to the rigidity of the joints and the areas
   of the cross-sections of individual members . . . . . . . . . . 164
2. Variations in the value of the coefficient $\dfrac{N}{m}$ . . . . . . 168
3. Diagrams of Engineer Paton . . . . . . . . . . . . . . . . . . 173
4. Critical review of later works . . . . . . . . . . . . . . . . 178

Literature *).

E. O. Paton: Calculation of openwork trusses with rigid joints. Moscow, 1901.

E. Yu. Pustolykors: Calculation of trusses with rigid joints on the basis of the principle of the work of constraints. St. Petersburg, 1903.

G. P. Perederii: The influence of the rigidity of the joints of metal trusses on the forces and stresses in their members. “Engineering Affairs,” 1904, books 1 and 2.

A. Fanc-de-Fleury: Bending of compressed and tensioned beams with fixed ends. St. Petersburg, 1904.

Winkler: Theorie der Brücken, II Heft. 1881.

Ritter und Tetmajer: Bericht über die Mönchensteiner Brückenkatastrophe. Schweiz. Bauzeit. 1891.

Th. Oppoltzer: Lehrbuch der Bahnbestimmung von Planeten und Kometen. II Band.

E. O. Paton: The so-called elastic forces. “Proceedings of the Assembly of Railway Engineers.” 1905.

E. V. Zotikov: Two-braced trusses and rigid joints. “Journal of the Ministry of Ways of Communication,” 1905, book 9.

E. O. Paton: On the question of two-braced trusses. “Engineer,” 1906, Nos. 9–10; 1907, Nos. 2–3.

G. V. Perederii: Learned equilibrists. “Engineering Affairs,” 1905, No. 3.

*) The list includes only those articles and monographs to which references are made in the present work.

In recent years Russian technical literature has been enriched by several very interesting investigations concerning additional stresses in trusses with rigid nodal connections. The impetus in this direction was given by the appearance of the book by Engineer Paton: “Calculation of Through Trusses with Rigid Joints”*) Moscow, 1901. In this book the author collected, and personally carried out, a considerable number of computations of stresses due to the rigidity of nodal connections, and arrived at certain conclusions whose importance went far beyond the limits of a purely theoretical investigation. Of particular interest was the computation of rigidity stresses for a two-plane truss, where Engineer Paton, applying Mohr’s method, came to very alarming conclusions regarding the strength of trusses of such a system.

Engineer Paton’s work gave rise also to the article by Engineer E. Pistol’kors, which appeared in 1903 and which, however, dealt chiefly with spatial trusses.

Finally, in 1904, the journal Engineering Affairs published an article by Engineer Perederii, who, when considering Mohr’s method as applied by Engineer Paton, found it inaccurate and, in some cases—including the above-mentioned two-plane truss—yielding greatly exaggerated results. Engineer Perederii, in turn, proposed a new method of calculation, based on assumptions different from those of Mohr’s method, applied it to certain trusses and, indeed, obtained a very considerable reduction in stresses due to the rigidity of the joints as compared with Engineer Paton’s data.

However, Engineer Perederii’s method is not entirely exact either; therefore, although it could not be denied that, in application to the above-mentioned two-plane truss, it should give far more accurate results than Mohr’s method, nevertheless for other cases doubts remained as to the limits of its applicability.

The present article aims to clarify, as far as possible, the degree of accuracy of the methods mentioned and to give certain indications for choosing one or another method of calculation in particular cases.

The course of the investigation will be as follows: first of all, we shall attempt to develop a method that is as accurate as possible and at the same time practicable—

*) In German, it was published in abridged form in Zeitschrift d. Arch. u. Ing. Vereins für Hannover for 1903.

the analytical method of calculation, proceeding from the expression for the work of deformation for a bar situated under the conditions of an element of a rigid truss*) and applying the theorem of least work of deformation to the entire truss with rigid connections at the joints.

Further on, we shall derive the former methods of calculation from the generalized one and show that all of them can be divided into three groups.

In order to form an idea of the degree of accuracy of the methods, we shall carry out a number of comparative computations for various trusses by several methods for each, and shall draw the corresponding conclusions.

In conclusion, we shall take up several propositions of Eng. Paton and shall try to illuminate them with the aid of the data obtained.

In our investigations, apart from those assumptions that are stipulated separately in each individual case, the following assumptions concerning the conditions of construction and operation of the trusses under consideration are constantly adopted without reservation:

a) the modulus of elasticity of the material of the trusses retains a constant value, and the stresses do not exceed the elastic limit;

b) defects in the assembly of trusses are not taken into consideration, and the centering of the elements is assumed to be perfect;

c) local strengthening of the cross-sections of elements by joint and splice plates is ignored;

d) the temperature is assumed to be constant and equal to the temperature during assembly of the truss;

e) riveted connections at the joints are considered to be completely rigid.

In its essential features the proposed work was completed in the spring of 1905, and in the autumn of the same year was presented to the civil-engineering department of the Tomsk Technological Institute. Owing to circumstances independent of the author, the printing of the article was greatly delayed, and it appears in print only in 1907.

Works on the same question that appeared during this interval, though in fact almost not touching upon the author’s conclusions, are noted in § 28.

The author considers it his duty to express his profound gratitude to Professors L. F. Nicolai and V. L. Kirpichev for their valuable comments on the present work, given by them at the request of the engineering-construction department of the Tomsk Technological Institute. Some of their remarks were taken into account by the author when the work was printed.

*) In investigating the elements of deformation of such a bar (§ 1), very useful indications were drawn by the author from the dissertation of A. A. van der Flit (see the list of literature).

1. Conditions of deformation of a rigid truss.

Let us consider the conditions of deformation of the truss \(O12\ldots\) with rigid riveted joints at the nodes. Suppose that, under the action of some external load, the truss deforms, while point \(O\) (the support point) remains in place, and the nodal points \(1, 2\ldots\) take new positions \(1', 2'\ldots\) (Fig. 1). In this case the new directions of the rectilinear axes of the members, i.e., of the chords, make with the basic ones certain angles \(\psi_{0-1}, \psi_{0-2}\ldots\) (we shall regard all angles as positive in the clockwise direction from the position corresponding to the basic line or to its parallel).

Fig. 1.

Meanwhile, the fundamental property of a rigid truss consists in the fact that the angles between the members remain constant, or, what is the same thing, all the ends of the members belonging to one node rotate through one and the same angle. The consequence of this is bending of the members, and constancy of the angles is preserved only between the tangents to the axes of the bent members at the nodal points.

Let us denote the angle of rotation of the ends of the members meeting at node \(O\) (the angle of rotation of node \(O\)) by \(\varphi_0\), at node 1 by \(\varphi_1\), etc.

It is clear from the drawing that the tangents to the curved axis of member \(O—1\) will, at points \(O\) and \(1'\), deviate from the chord of member \(O—1'\), respectively, by the angles \(\tau_0\) and \(\tau_1\).

Drawing through point \(1'\) a straight line parallel to \(O—1\), and recalling the sign convention, we may write

\[ \tau_0=\varphi_0-\psi_{0-1} \]

\[ \tau_1=\varphi_1-\psi_{0-1}. \]

A number of similar equalities may also be written for the other members.

As a consequence of the deformation considered, any member, for example \(O—1\), will not be subjected only to the action of an axial force, as is the case in a hinged truss, but will also be bent by certain moments \(M_0\) and \(M_1\), applied at the points \(O\) and 1, and by a certain pair of elastic forces, the necessity for whose presence will be clarified in the following paragraph.

Let us isolate the truss member \(O—1\) and suppose that all the forces acting on it are known to us, if the truss is in equilibrium under the action of the given external forces. Then the deformation of the truss member, expressed in the transition from the rectilinear bar \(O—1\) to the curved one \(O—1'\), may be reproduced in the following way.

We rotate the bar \(O—1\), together with the planes of restraint, into the position \(O—1''\); in doing so, obviously, no internal forces will arise in the bar and their work will be equal to zero. Then we apply to the bar the forces known to us, as a result of which it will assume the deformed position \(O—1'\); the work of deformation of the internal forces will in this case be equal to the work which the member absorbs during the deformation of the entire truss.

A similar operation may be performed with member \(1—2\), moving it into the position \(1'—2''\), fixing point \(1'\) and applying the known system of forces.

In general, the participation of a member of a rigid truss in its deformation may be divided into two parts: the displacement of the geometrical axis of the member (whereby the work of the internal forces is equal to zero) and its deformation proper.

2. General Formulas for the Bending of a Bar with Fixed Ends Subjected to Compression

We shall investigate the deformation and the stressed state of a straight bar with fixed ends, subjected to the action of a compressive load along its axis and bent by certain moments \(M_1\) and \(M_2\), applied to the end sections of the bar.

We shall determine the equation of the elastic line of the bar, find the expression for the bending moment in each section, investigate the dependence between the angles of rotation of the fixed-end sections \(\tau_1, \tau_2\), the moments \(M_1, M_2\), and the force \(S\), and, finally, write the expression for the total work of deformation for the bar under consideration.

As before, we shall regard as positive the moments rotating clockwise and, correspondingly, the positive angles \(\tau\) shall be measured clockwise from the chord to the position of the tangent to the axis of the bar after deformation.

We shall place the origin of coordinates at point \(A\) and direct the \(Y\)-axis vertically downward (Fig. 2). In addition to the force \(S\) and the moments \(M_1\) and \(M_2\),

Fig. 2.

for equilibrium of the bar it is necessary to apply to it also a pair of forces \(+P\) and \(-P\), as is evident from the equilibrium equation \(\Sigma M = O\).

\[ M_1 + M_2 - Pl = 0, \]

whence

\[ \tag{1} P=\frac{M_1+M_2}{l}. \]

We shall consider the forces \(P\) positive if they are directed downward.

The bending moment for any section of the member \(mn\), at a distance \(x\) from the left support, is written as

\[ \tag{2} M_x=M_1-Px+Sy. \]

The equation of the elastic line is

\[ EIy''=-(M_1-Px+Sy) \]

or

\[ \tag{3} EIy''+Sy=-M_1+Px. \]

This second-order differential equation is easily integrated in the following way:

\[ \tag{4} y=u+Y. \]

Here \(u\) is the general integral of the equation:

\[ u=C\Cos ax+D\Sin ax, \]

where

\[ a=\sqrt{\frac{S}{EI}}, \]

and \(Y\) is one of the particular solutions, to which we may give the form

\[ Y=mx+n; \]

the coefficients \(m\) and \(n\) are easily determined, taking into account that differentiation gives us

\[ Y'=m,\qquad Y''=0. \]

Substitution into equation (3) gives

\[ Smx+Sn=-M_1+Px, \]

and comparison of coefficients determines

\[ m=\frac{P}{S}, \]

\[ n=-\frac{M_1}{S}. \]

Thus the general integral of equation (3) is obtained in the following form

\[ y=C\Cos ax+D\Sin ax+\frac{P}{S}x-\frac{M_1}{S}, \tag{5} \]

where

\[ a=\sqrt{\frac{S}{EI}}. \]

The arbitrary constants \(C\) and \(D\) can be found from the conditions that

\[ \begin{aligned} &1)\quad \text{at } x=0 \qquad y=0,\\ &2)\quad \text{at } x=l \qquad y=0. \end{aligned} \]

The first of these conditions gives

\[ 0=C-\frac{M_1}{S}, \]

whence

\[ C=\frac{M_1}{S}, \]

and from the second we obtain

\[ 0=C\Cos al+D\Sin al+\frac{P}{S}l-\frac{M_1}{S}, \]

\[ 0=\frac{M_1}{S}\left(\Cos al-1\right)+D\Sin al+\frac{P}{S}l, \]

\[ D=\frac{M_1}{S\Sin al}\left(1-\Cos al\right)-\frac{Pl}{S\Sin al}. \]

Now we obtain the equation of the elastic line

\[ y=\frac{M_1}{S}\Cos ax+ \left[ \frac{M_1}{S\Sin al}\left(1-\Cos al\right) -\frac{Pl}{S\Sin al} \right]\Sin ax \]

\[ +\frac{P}{S}x-\frac{M_1}{S}, \]

or, replacing \(P\) by its expression from equation (1), after transformations we shall have, in final form,

\[ y=\frac{M_1}{S}\Cos ax+ \left[ -\frac{M_1\Cos al}{S\Sin al} -\frac{M_2}{S\Sin al} \right]\Sin ax +\frac{M_1+M_2}{Sl}x \]

\[ -\frac{M_1}{S}. \tag{6} \]

To determine the angles \(\tau_1\) and \(\tau_2\) as functions of \(M_1\), \(M_2\), and \(S\), we note that

\[ 1)\ \text{for } x=0 \qquad \left(\frac{dy}{dx}\right)_0 = \operatorname{Tg}\tau_1, \]

\[ 2)\ \text{for } x=l \qquad \left(\frac{dy}{dx}\right)_l = \operatorname{Tg}\tau_2, \]

or, owing to the smallness of the angles \(\tau_1\) and \(\tau_2\),

\[ \tau_1=\left(\frac{dy}{dx}\right)_0, \]

\[ \tau_2=\left(\frac{dy}{dx}\right)_l . \]

We differentiate \(y\) with respect to \(x\):

\[ y'=-\frac{aM_1}{S}\operatorname{Sin}ax +a\left[-\frac{M_1\operatorname{Cos}al}{S\operatorname{Sin}al} -\frac{M_2}{S\operatorname{Sin}al}\right]\operatorname{Cos}ax +\frac{M_1+M_2}{Sl}, \]

whence we obtain, respectively,

\[ \tau_1=-\frac{aM_1\operatorname{Cos}al}{S\operatorname{Sin}al} -\frac{aM_2}{S\operatorname{Sin}al} +\frac{M_1+M_2}{Sl}, \]

\[ \tau_2=-\frac{aM_1\operatorname{Sin}al}{S} -\frac{aM_1\operatorname{Cos}^2 al}{S\operatorname{Sin}al} -\frac{aM_2\operatorname{Cos}al}{S\operatorname{Sin}al} +\frac{M_1+M_2}{Sl}, \]

or, after transformations,

\[ \begin{aligned} \tau_1&=\frac{1}{Sl}\left[ M_1\left(1-\frac{al}{\operatorname{Tg}al}\right) -M_2\left(\frac{al}{\operatorname{Sin}al}-1\right) \right],\\[4pt] \tau_2&=\frac{1}{Sl}\left[ -M_1\left(\frac{al}{\operatorname{Sin}al}-1\right) +M_2\left(1-\frac{al}{\operatorname{Tg}al}\right) \right]. \end{aligned} \tag{7} \]

If in expressions (7) we put \(S=0\), i.e., ignore the influence of the longitudinal force on the deformation of the member, then these expressions assume an indeterminate form, since in this case

\[ \text{and } a=\sqrt{\frac{S}{EI}} \text{ becomes equal to zero.} \]

To ascertain the true value of expressions (7), we expand into series in powers of \(S\) the functions entering into them,

\[ \left(1-\frac{al}{\operatorname{Tg}al}\right) \quad \text{and} \quad \left(\frac{al}{\operatorname{Sin}al}-1\right). \]

Recalling the expansions

\[ \operatorname{Sin} al = al-\frac{a^3l^3}{3!}+\frac{a^5l^5}{5!}-\cdots, \]

\[ \operatorname{Tg} al = al+\frac{2}{3!}a^3l^3+\frac{2^4}{5!}a^5l^5+\frac{2^4\cdot 17}{7!}a^7l^7+\cdots, \]

we easily write the following new expansions:

\[ \frac{al}{\operatorname{Tg} al} =1-\frac{a^2l^2}{3}-\frac{a^4l^4}{45}-\frac{2a^6l^6}{945}-\cdots, \]

\[ \frac{al}{\operatorname{Sin} al} =1+\frac{a^2l^2}{6}+\frac{7}{360}a^4l^4+\frac{31}{15120}a^6l^6+\cdots. \]

Further, after reductions and transformations,

\[ \left(1-\frac{al}{\operatorname{Tg} al}\right) =\frac{a^2l^2}{3} \left(1+\frac{a^2l^2}{15}+\frac{2a^4l^4}{315}+\cdots\right), \]

\[ \left(\frac{al}{\operatorname{Sin} al}-1\right) =\frac{a^2l^2}{6} \left(1+\frac{7}{60}a^2l^2+\frac{31}{2520}a^4l^4+\cdots\right). \]

Replacing \(a^2\) by \(\dfrac{S}{EI}\) and substituting into expressions (7), we have

\[ \tau_1=\frac{l}{3EI} \left[ M_1\left(1+\frac{Sl^2}{15EI}+\frac{2S^2l^4}{315E^2I^2}+\cdots\right) -\frac{M_2}{2}\left(1+\frac{7}{60}\frac{Sl^2}{EI}+\frac{31S^2l^4}{2520E^2I^2}\cdots\right) \right], \]

\[ \tau_2=\frac{l}{3EI} \left[ -\frac{M_1}{2}\left(1+\frac{7Sl^2}{60EI}+\frac{31S^2l^4}{2520E^2I^2}+\cdots\right) +M_2\left(1+\frac{Sl^2}{15EI}+\frac{2S^2l^4}{315E^2I^2}+\cdots\right) \right]. \tag{8} \]

If in these expressions we put \(S=0\), we obtain the well-known dependence

\[ \tau_1=\frac{l}{3EI}\left(M_1-\frac{M_2}{2}\right), \]

\[ \tau_2=\frac{l}{3EI}\left(M_2-\frac{M_1}{2}\right). \]

On the other hand, from equations (7), solving them with respect to \(M_1\) and \(M_2\), we readily obtain these latter quantities as functions of \(\tau_1,\tau_2\) and \(S\).

\[ \begin{aligned} M_1 &= \frac{EI}{l} \frac{ \left[ \tau_1\left(1-\frac{al}{\operatorname{Tg}al}\right) + \tau_2\left(\frac{al}{\operatorname{Sin}al}-1\right) \right] }{ \left( \frac{\operatorname{Tg}\frac{al}{2}}{\left(\frac{al}{2}\right)}-1 \right) }, \\[1.2em] M_2 &= \frac{EI}{l} \frac{ \left[ \tau_2\left(1-\frac{al}{\operatorname{Tg}al}\right) + \tau_1\left(\frac{al}{\operatorname{Sin}al}-1\right) \right] }{ \left( \frac{\operatorname{Tg}\frac{al}{2}}{\left(\frac{al}{2}\right)}-1 \right) }. \end{aligned} \tag{9} \]

It is easy to verify by expansion in series that expressions (9), upon substituting \(S=0\), reduce to the well-known formulas

\[ M_1=\frac{2EI}{l}\left(2\tau_1+\tau_2\right), \]

\[ M_2=\frac{2EI}{l}\left(\tau_1+2\tau_2\right). \]

The bending moment for section \(mn\) is obtained by substituting \(y\), according to formula (6), into expression (2). After canceling like terms we obtain

\[ M_x = M_1\operatorname{Cos}ax - \frac{M_1\operatorname{Cos}al}{\operatorname{Sin}al}\operatorname{Sin}ax - M_2\frac{\operatorname{Sin}ax}{\operatorname{Sin}al}, \]

\[ M_x= \frac{M_1\operatorname{Sin}a(l-x)-M_2\operatorname{Sin}ax}{\operatorname{Sin}al}. \tag{10} \]

The vertical shearing force will be constant along the entire length of the beam and, in magnitude, equal to \(P\).

\[ V=P=\frac{M_1+M_2}{l}. \tag{11} \]

3. Work of deformation of a compressed and bending member with fixed ends.

Having determined the quantities \(M_x\) and \(V\), we can proceed to the construction of the formula for the total work of deformation of the member under consideration. Let us isolate, by two infinitely close sections, parallel to one another and perpendicular to the axis of the member, a certain element \(ABCD\) (Fig. 3), and consider the work performed by the forces acting on its surface. Suppose that one of the sections is fixed immovably, and let us investigate the displacements of the other section and the stresses arising in it.

Under the influence of the forces acting upon it, the section of the member \(CD\) will undergo the following displacements:

1) it will be displaced in the horizontal direction, parallel to itself, by the amount
\(CC_1=\Delta dx\);

2) it will rotate about the neutral axis through the angle
\(C_2OC_1=d\varphi\);

3) it will shift in the vertical direction by the amount
\(C_2C_3=du\).

Accordingly, we shall have three kinds of stress arising in the plane of the section \(CD\) under the influence of the external forces:

Fig. 3.

1) compressive normal stresses caused by the force \(S\) and equal to

\[ \sigma_1=\frac{S}{\omega} \]

2) compressive and tensile normal stresses due to bending, corresponding to the bending moment \(M_x\) and expressed by the formula

\[ \sigma_z=\frac{M_x z}{I}, \]

3) shearing stresses, determined by the vertical force \(V_x\).

Let us consider separately the work of deformation produced by each of these groups of stresses for the above-listed deformations of an element of a member.

1) The compressive stresses \(\sigma_1\), during displacement 1, produce the elementary work of deformation

\[ dA_1=\frac{1}{2}\int \sigma_1\,d\omega \cdot \Delta dx, \]

where the integration extends over the entire area of the section. Since

\[ \Delta dx=\frac{Sdx}{E\omega}, \]

we have

\[ \tag{12} dA_1=\frac{1}{2}\int \frac{S^2 dx}{E\omega^2}\,d\omega =\frac{S^2 dx}{2E\omega}. \]

When the section \(CD\) rotates, the element \(d\omega\) at a distance \(z\) from the neutral axis passes through the horizontal path

\[ zd\varphi=\frac{Mz}{EI}\,dx. \]

The corresponding work of the compressive stresses will be equal to

\[ \frac{1}{2}\int \frac{S}{\omega}\,d\omega\,\frac{Mz}{IE}\,dx = \frac{MS}{2E\omega I}\,dx\int z\,d\omega=0, \]

since

\[ \int z\,d\omega=0. \]

In shear deformation, the paths traversed by the points of the section are perpendicular to the direction of the compressive stresses and, consequently, the work of the latter is zero.

Thus all the work of the compressive stresses reduces to the quantity \(dA_1\).

2) The normal bending stresses perform work equal to zero under a deformation of compression. Indeed, in this case their work is equal to

\[ \int \frac{Mz}{I}\,d\omega \,\Delta dx = \frac{1}{2}\int \frac{Mz}{I}\,d\omega\,\frac{Sdx}{E\omega} = \frac{MSdx}{2E\omega I}\int z\,d\omega =0 \]

The work of the same stresses under rotation of the section \(CD\) will be expressed as follows:

\[ dA_2 = \frac{1}{2}\int \frac{Mz}{I}\,d\omega\cdot \frac{Mz}{EI}\,dx = \frac{1}{2}\frac{M^2dx}{EI^2}\int z^2d\omega, \]

or, since

\[ \int z^2d\omega=I, \]

\[ dA_2=\frac{M^2dx}{2EI}. \tag{13} \]

Finally, under a shear deformation the normal stresses do no work for the same reason as the compressive stresses.

3) Passing to the shearing stresses, we first observe that, for deformations of the first and second kind, the named stresses do no work, since the displacements of the points of the section are normal to the directions of the stresses.

As regards the shear deformation, in this case the elementary work of the shearing stresses is determined by

\[ dA_3=\frac{1}{2}Vdu, \]

or, taking into account that

\[ du=K\frac{Vdx}{G\omega}, \]

(where \(K\) is a certain coefficient, constant for the given section, and

\[ G=0.4E), \]

we obtain

\[ dA_3=K\frac{V^2dx}{0.8E\omega}. \tag{14} \]

Adding the obtained expressions \(dA_1, dA_2\), and \(dA_3\), we determine the elementary work of deformation of a bar element of length \(dx\)

\[ dA=\frac{S^2 dx}{2E\omega}+\frac{M^2 dx}{2EI}+K\frac{V^2 dx}{0.8\,E\omega}. \]

Integrating this expression over the entire bar, we obtain the total work of deformation of a bar subjected simultaneously to compression and bending, in the following form:

\[ \tag{15} A=\int_{0}^{l} dA =\int_{0}^{l}\frac{S^2 dx}{2E\omega} +\int_{0}^{l}\frac{M^2 dx}{2EI} +\int_{0}^{l}K\frac{V^2 dx}{0.8\,E\omega}. \]

Proceeding to compute the definite integrals in expression (15), we shall take the quantities \(\omega, I, K\) to be constant along the entire length of the bar, as is usually the case in the members of bridge trusses.

Therefore we rewrite expression (15) in the form

\[ A=\frac{S^2}{2E\omega}\int_{0}^{l} dx +\frac{1}{2EI}\int_{0}^{l} M^2 dx +\frac{K}{0.8\,E\omega}\int_{0}^{l} V^2 dx. \]

The first of the integrals obviously gives

\[ \tag{16} A=\frac{S^2 l}{2E\omega}. \]

To determine the second, we must substitute the value of \(M\) from formula (10):

\[ A_2=\frac{1}{2EI}\int_{0}^{l} \left[ \frac{M_1\Sin a(l-x)-M_2\Sin ax}{\Sin al} \right] dx. \]

Dividing into three integrals, we have

\[ \tag{17} A_2= \frac{1}{2EI\,\Sin^2 al} \left\{ \int_{0}^{l} M_1^2\Sin^2 a(l-x)\,dx +\int_{0}^{l} M_2^2\Sin^2 ax\,dx -\int_{0}^{l}2M_1M_2\Sin a(l-x)\Sin ax\,dx \right\}. \]

But, carrying out the integration, we have

\[ \int_{0}^{l}\operatorname{Sin}^{2} ax\,dx = \left|\frac{x}{2}-\frac{\operatorname{Sin} ax\,\operatorname{Cos} ax}{2a}\right|_{0}^{l} = \left[\frac{l}{2}-\frac{\operatorname{Sin} al\,\operatorname{Cos} al}{2a}\right], \]

\[ \int_{0}^{l}\operatorname{Sin}^{2} a(l-x)\,dx = \left|\frac{x}{2}+\frac{\operatorname{Sin} a(l-x)\,\operatorname{Cos} a(l-x)}{2a}\right|_{0}^{l} \]

\[ = \left[\frac{l}{2}-\frac{\operatorname{Sin} al\,\operatorname{Cos} al}{2a}\right], \]

\[ \int_{0}^{l}\operatorname{Sin} a(l-x)\,\operatorname{Sin} ax\,dx = \left|\operatorname{Sin} al\,\frac{\operatorname{Sin}^{2} ax}{2a} \right. \]

\[ \left. -\operatorname{Cos} al\left(\frac{x}{2}-\frac{\operatorname{Sin} ax\,\operatorname{Cos} ax}{2a}\right) \right|_{0}^{l} = \left[ \frac{\operatorname{Sin}^{3} al}{2a} -\operatorname{Cos} al\left(\frac{l}{2}-\frac{\operatorname{Sin} al\,\operatorname{Cos} al}{2a}\right) \right] = \left[ \frac{\operatorname{Sin} al}{2a}-\frac{l\,\operatorname{Cos} al}{2} \right]. \]

Substituting the expressions found into formula (17), we have

\[ A_{2} = \frac{1}{2EI} \left\{ \frac{M_{1}^{2}}{4a\,\operatorname{Sin}^{2} al}\,(2al-\operatorname{Sin}2al) + \frac{M_{2}^{2}}{4a\,\operatorname{Sin}^{2} al}\,(2al-\operatorname{Sin}2al) - \frac{M_{1}M_{2}}{a\,\operatorname{Sin}^{2} al}\, (\operatorname{Sin} al-al\,\operatorname{Cos} al) \right\}. \tag{28} \]

As for the third integral of formula (15), noting that

\[ V=\frac{M_{1}+M_{2}}{l}=\mathrm{Const.}, \]

we easily obtain

\[ A_{3}=K\,\frac{(M_{1}+M_{2})^{2}}{0{,}8\,E\omega\,l}. \tag{19} \]

Combining formulas (16), (18), and (19), we obtain the final expression for the work of deformation of the member

\[ \begin{aligned} A={}&\frac{S^{2}l}{2E\omega} +\frac{1}{2EI}\left\{ \frac{(M_1^2+M_2^2)}{4a\sin^{2}al}(2al-\sin 2al)\right.\\ &\left.-\frac{M_1M_2}{a\sin^{2}al}(\sin al-al\cos al) \right\} +K\frac{(M_1+M_2)^2}{0{,}8\,E\omega\,l}. \end{aligned} \tag{20} \]

It is easy to see that in the last formula the first and last terms do not change depending on whether or not the influence of the force \(S\) on the deformation of the member in bending is taken into account. In order to determine to what extent taking into account the force \(S\) affects the second term of expression (20), we shall use the method already applied above of expanding functions in series in powers of \(S\).

With the aid of the known expansions of \(\sin x\) and \(\cos x\), we can very easily write the expansions of the functions entering expression (20). Thus we obtain

\[ \sin^{2}al=\frac{1-\cos 2al}{2} =\frac{1}{2}\left[1-\left(1-\frac{1}{2}(2al)^2+\frac{1}{4!}(2al)^4-\cdots\right)\right] \]

\[ =a^{2}l^{2}\left(1-\frac{a^{2}l^{2}}{3}+\frac{2a^{4}l^{4}}{45}-\frac{a^{6}l^{6}}{315}+\cdots\right), \]

\[ (2al-\sin 2al)= \left[2al-\left(2al-\frac{(2al)^3}{3!}+\frac{(2al)^5}{5!}-\cdots\right)\right] \]

\[ =\frac{4a^{3}l^{3}}{3} \left[1-\frac{a^{2}l^{2}}{5}+\frac{2a^{4}l^{4}}{105}-\frac{a^{6}l^{6}}{945}+\cdots\right], \]

\[ (\sin al-al\cos al)= \left[\left(al-\frac{a^{3}l^{3}}{3!}+\frac{a^{5}l^{5}}{5!}-\frac{a^{7}l^{7}}{7!}+\cdots\right)\right. \]

\[ \left.-al\left(1-\frac{a^{2}l^{2}}{2}+\frac{a^{4}l^{4}}{4!}-\frac{a^{6}l^{6}}{6!}+\cdots\right)\right] \]

\[ =\frac{a^{3}l^{3}}{3} \left[1-\frac{a^{2}l^{2}}{10}+\frac{a^{4}l^{4}}{280}-\frac{a^{6}l^{6}}{15120}+\cdots\right]. \]

\[ \frac{2al-\sin 2al}{4a\sin^{2}al} =\frac{l}{3}\left(1+\frac{2}{15}a^{2}l^{2} +\frac{2}{105}a^{4}l^{4} +\frac{4}{1575}a^{6}l^{6}+\cdots\right) \]

\[ \frac{\sin al-al\cos al}{a\sin^{2}al} =\frac{l}{3}\left[1+\frac{7}{30}a^{2}l^{2} +\frac{31}{840}a^{4}l^{4} +\frac{127}{25200}a^{6}l^{6}\cdots\right]. \]

Replacing \(a^{2}\) by \(\dfrac{S}{EI}\) and introducing the expansions obtained into the formula

(20), we find the expression for the work of deformation in the following form:

\[ A=\frac{S^{2}l}{2E\omega}+\frac{l}{6EI} \left[ (M_{1}^{2}+M_{2}^{2})[1+\varphi(S)]-M_{1}M_{2}[1+\psi(S)] \right] \]

\[ +K\,\frac{(M_{1}+M_{2})^{2}}{0.8\,E\omega l}, \tag{21} \]

where by \(\varphi(S)\) and \(\psi(S)\) are denoted the series

\[ \varphi(S)=\frac{2}{15}\cdot\frac{Sl}{EI} +\frac{2}{105}\cdot\frac{S^{2}l^{4}}{E^{2}I^{2}} +\frac{4}{1575}\cdot\frac{S^{3}l^{6}}{E^{3}I^{3}}+\cdots, \]

\[ \psi(S)=\frac{7}{30}\cdot\frac{Sl^{2}}{EI} +\frac{31}{840}\cdot\frac{S^{2}l^{4}}{E^{2}I^{2}} +\frac{127}{25200}\cdot\frac{S^{3}l^{6}}{E^{3}I^{3}}+\cdots. \]

If in formula (21) \(S\) is put equal to zero, the well-known formula is obtained for a bar bent by two moments

\[ A'=\frac{l}{6EI}\left[M_{1}^{2}+M_{2}^{2}-M_{1}M_{2}\right] +K\,\frac{(M_{1}+M_{2})^{2}}{0.8\,E\omega l}. \]

4. Generalization of the Derived Formulas for the Case of Combined Bending and Tension.

Up to now we have carried out all derivations under the assumption of a compressive force \(S\); if we now wish to obtain the corresponding formulas for the case of a tensile force, there is no need to repeat all the computations. It is sufficient merely, in all eight formulas obtained by us, to replace \(S\) by \(-S\); accordingly, the quantities \(a, a^{2}, a^{3}, a^{4}\ldots\) will also change into the quantities \(ia,-a^{2},-ia^{3},a^{4}\ldots\).

In addition, all trigonometric circular functions of the arguments \(ax\) and \(al\) will have to be replaced by the corresponding hyperbolic functions according to the formulas

\[ \operatorname{Sin} ial = i\,\operatorname{Sinh} al \]

\[ \operatorname{Cos} ial = \operatorname{Cosh} al \quad \text{etc.} \]

We shall present the most important formulas from among those derived above, in transformed form for the case of a tensile member.

1) Equation of the elastic line:

\[ \tag{6^1} y=-\frac{M_1}{S}\operatorname{Cosh} ax+ \left\{ \frac{M_1}{S}\frac{\operatorname{Cosh} al}{\operatorname{Sinh} al} + \frac{M_1}{S\,\operatorname{Sinh} al} \right\}\operatorname{Sinh} ax -\frac{M_1+M_2}{Sl}\,x+\frac{M_1}{S}, \]

2) Angles \(\tau_1\) and \(\tau_2\):

\[ \tag{7^1} \tau_1=-\frac{1}{Sl}\left[ M_1\left(1-\frac{al}{\operatorname{Tgh} al}\right) - M_2\left(\frac{al}{\operatorname{Sinh} al}-1\right) \right], \]

\[ \tau_2=-\frac{1}{Sl}\left[ -M_1\left(\frac{al}{\operatorname{Sinh} al}-1\right) + M_2\left(1-\frac{al}{\operatorname{Tgh} al}\right) \right], \]

or, after expansion into series,

\[ \tag{8^1} \tau_1=\frac{l}{3EI}\left[ M_1\left(1-\frac{Sl^2}{15EI}+\frac{2S^2l^4}{315E^2I^2}-\cdots\right) -\frac{M_2}{2}\left(1-\frac{7Sl^2}{60EI}+\frac{31S^2l^4}{2520E^2I^2}-\cdots\right) \right], \]

\[ \tau_2=\frac{l}{3EI}\left[ -\frac{M_1}{2}\left(1-\frac{7Sl^2}{60EI}+\frac{31S^2l^4}{2520E^2I^2}-\cdots\right) + M_2\left(1-\frac{Sl^2}{15EI}+\frac{2S^2l^4}{315E^2I^2}-\cdots\right) \right]. \]

3) Bending moment:

\[ \tag{10^1} M_x=\frac{M_1\operatorname{Sinh} a(l-x)-M_2\operatorname{Sinh} ax}{\operatorname{Sinh} al}, \]

4) Work of deformation:

\[ \tag{20^1} A=\frac{S^2l}{2E\omega} + \frac{1}{2EI}\left\{ -\frac{M_1^2+M_2^2}{4a\operatorname{Sinh}^2 al}(2al-\operatorname{Sinh}2al) + \frac{M_1M_2}{a\operatorname{Sinh}^2 al} \left(\operatorname{Sinh} al-al\operatorname{Cosh} al\right) \right\} + K\frac{(M_1+M_2)^2}{0.8E\omega l}, \]

or, in a form transformed by means of expansion into series,

\[ A=\frac{S^2l}{2E\omega} +\frac{l}{6EI}\left[(M_1^2+M_2^2)[1+\varphi_1(S)]-M_1M_2[1+\psi_1(S)]\right] \]

\[ +K\frac{(M_1+M_2)^2}{0.8E\omega l}, \tag{21$^1$} \]

where

\[ \varphi_1(S)=-\frac{2}{15}\cdot\frac{Sl^2}{EI} +\frac{2}{105}\cdot\frac{S^2l^4}{E^2I^2} -\frac{4}{1575}\cdot\frac{S^3l^6}{E^3I^3} +\ldots, \]

\[ \psi_1(S)=-\frac{7}{30}\cdot\frac{Sl^2}{EI} +\frac{31}{840}\cdot\frac{S^2l^4}{E^2I^2} -\frac{127}{25200}\cdot\frac{S^3l^6}{E^3I^3} +\ldots . \]

As was to be expected, the imaginary expressions in these formulas vanish, and in this case as well the results are obtained as real quantities.

5. Investigation of the functions \(\varphi\) and \(\psi\).

Let us consider in somewhat greater detail the quantities \(\varphi(S)\), \(\psi(S)\), \(\varphi_1(S)\), and \(\psi_1(S)\); let us investigate how large the values of these coefficients may become for various changes in the magnitude of the longitudinal force \(S\).

For choosing a method of computing the quantities \(\varphi(S)\), etc., the guiding significance will be the value of the argument

\[ \frac{Sl^2}{EI}=a^2l^2. \]

For small values of this argument, especially those less than unity, the series by which the desired functions are expressed will be so rapidly convergent that, to attain quite sufficient accuracy, one may confine oneself to two or three terms of the expansion—and the computation is carried out very quickly. The matter is otherwise for large values of

\[ \frac{Sl^2}{EI}; \]

then it becomes necessary either to compute a considerable number of terms of the expansion, or to use, for the computation, directly the formulas from expressions (20) and (20$^1$).

For greater clarity, let us make use of the identical form of expressions (20), (20¹) and (21), (21¹); we may write

\[ \frac{1}{2EI}\cdot \frac{(M_1^2+M_2^2)}{4a\operatorname{Sin}^2 al} \left(2al-\operatorname{Sin}2al\right) = \frac{l}{6EI}(M_1^2+M_2^2)\left[1+\varphi(S)\right], \]

\[ \frac{1}{2EI}\cdot \frac{M_1\cdot M_2}{a\operatorname{Sin}^2 al} \left(\operatorname{Sin}al-al\operatorname{Cos}al\right) = \frac{l}{6EI}M_1M_2\left[1+\psi(S)\right], \]

whence, by canceling the corresponding common factors, we easily obtain the expressions \(\varphi(S)\), etc., in the form of trigonometric functions

\[ \begin{aligned} 1+\varphi(S)&=\frac{3(2al-\operatorname{Sin}^2 al)}{4al\,\operatorname{Sin}^2 al},\\ 1+\psi(S)&=\frac{3(\operatorname{Sin}al-al\operatorname{Cos}al)}{al\,\operatorname{Sin}^2 al}. \end{aligned} \tag{22} \]

Similarly, for the case of tension,

\[ \begin{aligned} 1+\varphi_1(S)&=\frac{3(2al-\operatorname{Sinh}2al)}{4al\,\operatorname{Sin}^2h\,al},\\ 1+\psi_1(S)&=\frac{3(\operatorname{Sinh}al-al\operatorname{Cosh}al)}{al\,\operatorname{Sin}^2h\,al}. \end{aligned} \tag{22¹} \]

Using these and the preceding forms of the expressions \(\varphi(S)\), etc., we shall make a brief survey of their variation depending on the values of the argument \(\dfrac{Sl^2}{EI}\).

As already noted above, when \(S=0\) (i.e. when \(\dfrac{Sl^2}{EI}\) is also equal to zero), all the functions \(\varphi(S)\), etc., also vanish. With the increase of the absolute value of \(\dfrac{Sl^2}{EI}\), the absolute values of the functions \(\varphi(S)\), etc., immediately begin to increase; moreover, in the case of compression these functions always remain positive, and in the case of tension—negative, i.e. the magnitudes of the coefficients \(\left[1+\varphi(S)\right]\) and \(\left[1+\psi(S)\right]\) are always greater, while the coefficients \(\left[1+\varphi_1(S)\right]\) and \(\left[1+\psi_1(S)\right]\) are always less than unity.

For the value

\[ \frac{Sl^2}{EI}=0.1 \]

these coefficients assume the following values:

\[ \begin{aligned} 1+\varphi(S)&=1.014, &\qquad 1+\psi(S)&=1.024,\\ 1+\varphi_1(S)&=0.987, & 1+\psi_1(S)&=0.977. \end{aligned} \]

For

\[ \frac{Sl^2}{EI}=1 \]

the coefficients will be equal to

\[ \begin{aligned} 1+\varphi(S)&=1.155, &\qquad 1+\psi(S)&=1.275,\\ 1+\varphi_1(S)&=0.883, & 1+\psi_1(S)&=0.799. \end{aligned} \]

Next, let us consider the case

\[ \frac{Sl^2}{EI}=a^2l^2=\frac{\pi^2}{4}, \]

whence

\[ al=\frac{\pi}{2}. \]

In this case, according to formulas (22) and (22′), we shall have

\[ \begin{aligned} 1+\varphi(S)&=1.500, &\qquad 1+\psi(S)&=1.910,\\ 1+\varphi_1(S)&=0.758, & 1+\psi_1(S)&=0.592, \end{aligned} \]

since

\[ \sinh al=\sinh \frac{\pi}{2}=2.3013, \]

\[ \cosh al=\cosh \frac{\pi}{2}=2.5092, \]

\[ \sinh 2al=\sinh \pi=11.5489. \]

Finally, for the value

\[ \frac{Sl^2}{EI}=\pi^2; \quad -al=\pi \]

we obtain, respectively, for compression

\[ 1+\varphi(S)=\infty, \qquad 1+\psi(S)=\infty. \]

This result is a consequence of an inaccurate derivation of the bending formula, since the radius of curvature was taken as equal to

\[ \frac{1}{y''}, \]

whereas its exact value is

\[ \rho=\frac{(1+y'^2)^{\frac{3}{2}}}{y''}. \]

It should be noted that the equality

\[ \frac{Sl^2}{EI}=\pi^2, \]

which may otherwise be written

\[ S=\pi^2\frac{l^2}{EI}, \]

is nothing other than the well-known Euler formula for the longitudinal bending of a straight bar with freely rotating ends.

For the case of tension of a bar, under the condition

\[ \frac{Sl^2}{EI}=\pi^2, \]

we obtain, from formula (22¹), the following values of the coefficients:

\[ 1+\varphi_1(S)=0{,}371,\qquad 1+\psi_1(S)=0{,}178, \]

since

\[ \operatorname{Sinh}2\pi=267{,}7278, \]

\[ \operatorname{Sinh}\pi=11{,}5489, \]

\[ \operatorname{Cosh}\pi=11{,}5921. \]

From the results we have obtained, the following conclusions may be drawn:

1) the influence of the longitudinal force on the work of deformation in bending increases rapidly with its increase;

2) for equal magnitudes of the tensile and compressive longitudinal force, the latter exerts a comparatively greater influence in the direction of increasing the work of deformation than the former does in the direction of decreasing it.

In order to clarify for ourselves to some extent the degree of influence of the participation of the longitudinal force, introduced by us, on the magnitude of the work of deformation for various ratios of the magnitudes of the bending moments, let us take a specific case.

Suppose we have a member with a fourfold safety factor according to Euler’s formula, i.e., for it

\[ \frac{S l^2}{EI} = \frac{\pi^2}{4}. \]

We saw above that in this case

\[ 1 + \varphi(S) = 1.500, \]

\[ 1 + \psi(S) = 1.910, \]

therefore the second term of (20) is

\[ A_2 = \frac{l}{6EI}\left[(M_1^2 + M_2^2)\,1.500 - 1.910\,M_1M_2\right]. \]

Let us first assume that both moments are equal and of the same sign,

\[ M_1 = M_2 . \]

Then the work of deformation in bending, if the influence of the compressive force is ignored, is:

\[ A'_2 = \frac{l}{6EI}\,M_1^2, \]

and, introducing the influence of the compressive force,

\[ A_2 = \frac{l}{6EI}\left[3M_1^2 - 1.910\,M_1^2\right] = 1.09\,\frac{l}{6EI}\,M_1^2 . \]

Now let one of the moments be equal to 0,

\[ M_2 = 0. \]

In this case

\[ A'_2 = \frac{l}{6EI}\,M_1^2, \]

\[ A_2 = 1.5\,\frac{l}{6EI}\,M_1^2 . \]

Finally, let both moments be equal, but opposite in sign:

\[ M_2 = -M_1 . \]

Then

\[ A_2'=\frac{l}{6EI}\,3M_1^2 \]

\[ A_2=\frac{l}{6EI}\cdot 4.91\,M_1^2 =1.637\,\frac{l}{6EI}\cdot 3M_1^2 . \]

As we see, taking the compressive force into account has the greatest influence when the moments are of different signs but close in absolute value, and the least influence when the moments have the same sign and are also close in magnitude.

6. Work of deformation of a rigid truss and the derivative from it. Necessary simplifications.

In composing the expression for the work of deformation of an entire truss with rigid nodal connections, we shall assume that the nodal connections do not perform work (an ideal connection without friction) and, consequently, the work of deformation of the truss can be obtained by simple summation of the amounts of work performed by the internal forces of each member separately. Since the members of the truss are subjected to the action of a longitudinal compressive or tensile force and of bending moments appearing as a result of the rigidity of the joints, the formulas derived in the preceding paragraphs are applicable to them.

Let the force in some member of the truss \(MN\) be denoted by \(S_{MN}\), and the moments due to the rigidity of the joints, acting respectively at the ends of the member \(M\) and \(N\), by

\[ M_{M-N}\ \text{and}\ M_{N-M}. \]

Then, on the basis of formula (21), we shall have the work of deformation for the named member

\[ \begin{aligned} A_{MN} &= \frac{S_{MN}^{2}}{2E\omega_{MN}}\,l_{MN} + \frac{l_{MN}}{6EI_{MN}} \left[ \left(M_{M-N}^{2}+M_{N-M}^{2}\right) \left[1+\varphi_{MN}(S)\right] \right.\\ &\qquad\left. - M_{M-N}\cdot M_{N-M} \left[1+\psi_{MN}(S)\right] \right] + K\frac{\left(M_{M-N}+M_{N-M}\right)^2}{0.8E\omega_{MN}l_{MN}} . \end{aligned} \]

For the whole truss, then,

\[ A=\sum A_{MN} \]

\[ \begin{aligned} &=\sum \frac{S_{MN}^{2} l_{MN}}{2E\omega_{MN}} +\sum \frac{l_{MN}}{6EI_{MN}} \Bigl\{ \left(M_{M-N}^{2}+M_{N-M}^{2}\right) +\left[1+\varphi_{MN}(S)\right] \\ &\qquad\qquad +M_{M-N}\cdot M_{N-M}\left[1+\psi_{MN}(S)\right] \Bigr\} +\sum K\frac{\left(\dfrac{M_{M-N}+M_{N-M}}{l_{MN}}\right)^{2}} {0.8E\omega_{MN}} \end{aligned} \tag{24} \]

Formula (24) is, under the assumptions indicated, a completely exact expression for the work of deformation of a rigid truss; but, as we shall now see, it is too complicated for our further investigations.

Any truss with rigid joints may be regarded as a statically indeterminate system with many unknowns, and for computing the latter we may apply the following method, based on the theorem of least work of deformation.

Let us take the unknown moments due to the rigidity of the joints as the statically indeterminate quantities, express the remaining quantities through them—in our case, the longitudinal forces \(S\)—and make use of the principle that in every body subjected to a given system of forces the stresses are distributed in such a way that the work of deformation of the system has the least value.

It follows from this that the derivatives of the expression for the work with respect to the statically indeterminate quantities must be equal to zero. It is evident that in this way we can obtain as many equations as there are unknown moments.

Suppose we have a truss consisting of a certain number of elements: \(0-1\), \(0-2\), \(0-3\), \(\ldots\), \(1-2\), \(1-3\), \(\ldots\), \(M-N\), etc. Then the force in any element of the rigid truss may be expressed as a function of the unknown moments in the linear form

\[ S_{MN}=S^{0}_{MN}+\alpha_{MN}M_{0-1}+\beta_{MN}M_{1-0}+\gamma_{MN}M_{1-2}+\cdots\cdots\cdots \]

\[ +\mu_{MN}M_{M-N}+\cdots\cdots\cdots, \tag{25} \]

where \(S^{0}_{MN}\) denotes the force in the element \(M-N\) under the action of the external load alone in the absence of moments, i.e.

if the joints are taken to be hinged; the quantities \(\alpha_{MN}, \beta_{MN}\ldots \mu_{MN}\) represent the forces, likewise in the member \(MN\) of the hinged truss, but in the absence of external forces and under the action of a load consisting of the moments \(M_{0-1}, M_{1-0}\ldots\) equal to unity.

Let us now differentiate expression (24) with respect to some unknown \(M_{K-L}\); here it should be noted that the quantities \(M\) are not variables fully independent of one another, but are connected in groups by the condition that the sum of the moments applied at each joint must be equal to zero.

Denoting by \(M_{M-N}, M_{M-P}, \ldots, M_{M-Q}\) the moments acting at joint \(M\), we shall have the condition

\[ \tag{26} M_{M-N}+M_{M-P}+\ldots+M_{M-Q}=0, \]

or, for the derivatives,

\[ 1+\frac{\partial M_{M-P}}{\partial M_{M-N}}+\ldots+ \frac{\partial M_{M-Q}}{\partial M_{M-N}}=0, \]

\[ \tag{27} \frac{\partial M_{M-N}}{\partial M_{M-P}}+1+\ldots+ \frac{\partial M_{M-Q}}{\partial M_{M-P}}=0, \]

\[ \ldots \]

\[ \frac{\partial M_{M-N}}{\partial M_{M-Q}}+ \frac{\partial M_{M-P}}{\partial M_{M-Q}}+\ldots+1=0. \]

Taking these conditions into consideration, we have

\[ \begin{aligned} \frac{\partial A}{\partial M_{K-L}} ={}& \sum \frac{S_{MN} l_{MN}}{E\omega_{MN}}\cdot \frac{\partial S_{MN}}{\partial M_{K-L}} + \sum_{K} \frac{l_{K-S}}{6EI_{K-S}} \Biggl\{ 2M_{K-S}\cdot \frac{\partial M_{K-S}}{\partial M_{K-L}} \left[1+\varphi_{KS}(S)\right] \\ &\qquad - M_{S-K}\cdot \frac{\partial M_{K-S}}{\partial M_{K-L}} \left[1+\psi_{KS}(S)\right] \Biggr\} \\ &+ \sum \frac{l_{MN}}{6EI_{MN}} \Biggl\{ \left(M_{M-N}^{2}+M_{N-M}^{2}\right) \frac{\partial \varphi_{MN}(S)}{\partial S} \cdot \frac{\partial S_{MN}}{\partial M_{K-L}} \\ &\qquad - M_{M-N}\cdot M_{N-M}\cdot \frac{\partial \psi_{MN}(S)}{\partial S_{MN}} \cdot \frac{\partial S_{MN}}{\partial M_{K-L}} \Biggr\} \\ &+ \sum_{K} K\, \frac{2\left(M_{K-S}+M_{S-K}\right)} {0.8\,E\omega_{KS}l_{KS}}\, \frac{\partial M_{K-S}}{\partial M_{K-L}} =0 . \end{aligned} \tag{28} \]

As already mentioned above, we shall obtain as many such equations as there are unknown moments, which would thus be determined from such a system.

In formula (28), under the sign \(\sum\), the summation must extend over all elements of the truss, while under the sign \(\sum_K\) it must extend only over the elements meeting at node \(K\), where the moment \(M_{K-L}\) is applied.

Let us now consider whether a practical solution of the problem is feasible in the form outlined above, and, if not, what changes must be made in order for it to become possible.

In this respect the difficulty is presented by the functions \(\varphi(S)\) and \(\psi(S)\); as we saw above, they may be represented either in transcendental form or in the form of infinite series. In either case, retaining them in the equation makes the solution of the problem impossible. Therefore, in what follows, we shall take \(\varphi(S)\) and \(\psi(S)\) to be equal to zero; in other words, we shall ignore the influence of the longitudinal force on the bending of the element, although we have seen above that this influence can reach rather large magnitudes.

At the end of Chapter III we shall try to estimate approximately the effect of this assumption and shall see that, for the cases considered by us, it is insignificant, which by no means can be said in general.

As for the last term of formula (28), which relates to the work of the shearing stresses, its introduction into the computations presents no difficulty.

But since the influence of this factor is less than that of the term with \(\varphi(S)\) and \(\psi(S)\), which must necessarily be discarded, and since, on the other hand, the authors of the methods that we are to examine neglected the influence of the work of the shearing stresses, we too shall not introduce it into the computations.

After these reductions, the basic formula for the work of deformation of a rigid truss assumes the following form:

\[ A = \sum \frac{S_{MN}^{2} l_{MN}}{2 E \omega_{MN}} + \sum \frac{l_{MN}}{6 E I_{MN}} \left( M_{M-N}^{2} + M_{N-M}^{2} - M_{M-N}\cdot M_{N-M} \right). \tag{29} \]

This is the formula from which we shall proceed in the subsequent exposition.

The transformed derivative of the work with respect to \(M_{K-L}\)

\[ \frac{\partial A}{\partial M_{K-L}} = \sum \frac{S_{MN}\cdot l_{MN}}{E\omega_{MN}}\cdot \frac{\partial S_{MN}}{\partial M_{K-L}} + \sum_{K}\frac{l_{KS}}{6EI_{KS}} \left( 2M_{K-S} - M_{S-K} \right) \frac{\partial M_{K-S}}{\partial M_{K-L}}, \tag{30} \]

where the summation symbols retain their previous meaning.

7. Grouping of existing methods.

In proceeding to characterize the existing methods for calculating bending moments and stresses due to the rigidity of the joints, we must distinguish the following four groups of methods:

1) The generalized method, based directly on formulas (29) and (30); the results of calculations by this method will serve us as a criterion for comparing the other methods;

2) the group of methods proposed by Eng. Perederii,

3) the methods of Mohr, Müller-Breslau, Manderla, and others,

4) the methods of Engesser, Müller-Breslau, etc.

We shall now proceed to a sequential exposition of these methods.

8. The generalized method.

We could make use, for determining the unknown moments, directly of equations of the form (30). If the truss under consideration has \(n\) joints, then for a statically determinate truss the number of members will be \(2n-3\), and the number of unknowns \(2(2n-3)\); for a statically indeterminate system the number of members will be \((2n-3)+m\), and the number of unknowns \(2(2n-3)+2m\), where \(m\) is the number of statically indeterminate quantities*) (redundant members).

*) Of course, trusses statically indeterminate only with respect to support reactions are not included here.

It would therefore be necessary to solve the corresponding number of linear equations. However, the number of the latter can be considerably reduced if one uses the condition that, for each joint, the sum of the moments is equal to zero.

Let us have some joint \(0\), in which \(n\) members meet: \(0—1, 0—2, \ldots, 0—(n-1), 0—n\); let us denote the moments acting on these members at joint \(0\) (except for \(0—n\)), respectively, by

\[ M_{0-1}=M_1; \]

\[ M_{0-2}=M_2, \]

\[ \ldots \]

\[ M_{0-(n-1)}=M_{n-1}. \]

Then, according to the condition

\[ \Sigma M=0, \]

\[ M_{0-n}=-(M_1+M_2+\cdots+M_{n-1}). \]

We shall now understand by \(M_1\) the aggregate of the moment \(+M_1\), acting on the member \(0—1\), and \(-M_1\), acting on the member \(0—n\); and likewise \(M_2\), etc. Thus the number of unknown elements at this joint will be smaller by one. Repeating this manipulation for all joints, we reduce the number of unknowns for a statically determinate truss to \((3n-6)\), and for a statically indeterminate one to \((3n-6)+2m\).

For statically determinate trusses the introduction of balanced moments is advantageous also in the respect that, in this case, many coefficients \(\alpha,\beta\), etc., from formulas of the form

\[ S=S^0+\alpha M_1+\beta M_2+\ldots, \]

turn into \(0\), thereby facilitating the calculations.

It is very important to note here certain properties of the systems of equations to which the application of this method leads—properties on which the abbreviated method, set forth by us below, for solving the said systems is based.

Let us have a certain truss \(01234\ldots\) (Fig. 4), for which we wish to determine the moments due to the rigidity of the joints.

Let us introduce, as indicated above, the notation

Fig. 4.

\[ \begin{aligned} M_{0-1} &= M_1, &\qquad M_{1-0} &= -(M_2+M_3),\\ M_{0-2} &= -M_1, & M_{2-0} &= -(M_4+M_5),\\ M_{1-4} &= M_3, & M_{4-1} &= M_7,\\ M_{1-3} &= M_2, & M_{3-1} &= M_6,\\ M_{2-4} &= M_5, & M_{4-2} &= M_8,\\ M_{2-3} &= M_4, & M_{3-2} &= M_9,\\ &\ldots, & &\ldots . \end{aligned} \]

The work of deformation of the individual members of the truss will be expressed, on the basis of (29), by the following formulas:

\[ A_{0-1} = \frac{S_{0-1}^{2}}{2E\omega_{0-1}}\,l_{0-1} + \frac{l_{0-1}}{6EI_{0-1}} \left[ M_1^{2}+(M_2+M_3)^{2}+M_1(M_2+M_3) \right], \]

\[ A_{0-2} = \frac{S_{0-1}^{2}}{2E\omega_{0-2}}\,l_{0-2} + \frac{l_{0-2}}{6EI_{0-2}} \left[ M_1^{2}+(M_4+M_5)^{2}-M_1(M_4+M_5) \right], \]

\[ A_{1-3} = \frac{S_{1-3}^{2}}{2E\omega_{1-3}}\,l_{1-3} + \frac{l_{1-3}}{6EI_{1-3}} \left[ M_2^{2}+M_6^{2}-M_6M_2 \right], \]

\[ A_{1-4} = \frac{S_{1-4}^{2}}{2E\omega_{1-4}}\,l_{1-4} + \frac{l_{1-4}}{6EI_{1-4}} \left[ M_3^{2}+M_7^{2}-M_3M_7 \right], \]

\[ \ldots\ldots\ldots\ldots\ldots\ldots\ldots \]

We differentiate the total work of the whole truss,

\[ A=A_{0-1}+A_{0-2}+A_{1-3}+A_{1-4}+\ldots\ldots, \]

successively with respect to \(M_1, M_2\), etc.; at the same time we note that

\[ S_{0-1}=S_{0-1}^{0}+\alpha_{0-1}M_1+\beta_{0-1}M_2+\gamma_{0-1}M_3+\ldots, \]

\[ S_{0-2}=S_{0-2}^{0}+\alpha_{0-2}M_1+\beta_{0-2}M_2+\gamma_{0-2}M_3+\ldots, \]

\[ \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots \]

and consequently

\[ \frac{\partial S_{0-1}}{\partial M_1}=\alpha_{0-1}; \qquad \frac{\partial S_{0-1}}{\partial M_2}=\beta_{0-1}; \qquad \ldots \]

\[ \frac{\partial S_{0-2}}{\partial M_1}=\alpha_{0-2}; \qquad \frac{\partial S_{0-2}}{\partial M_2}=\beta_{0-2}; \qquad \ldots \]

\[ \ldots \]

Assuming, moreover,

\[ \frac{l_{0-1}}{6EI_{0-1}}=\Delta\gamma_{0-1}; \qquad \frac{l_{0-2}}{6EI_{0-2}}=\Delta\gamma_{0-2}; \qquad \ldots, \]

\[ \frac{l_{0-1}}{E\omega_{0-1}}=\Delta\lambda_{0-1}; \qquad \frac{l_{0-2}}{E\omega_{0-2}}=\Delta\lambda_{0-2}; \qquad \ldots, \]

we obtain a system of equations for determining the unknown moments

\[ \begin{aligned} \frac{\partial A}{\partial M_1} &= \sum S\Delta\lambda\cdot \alpha +\Delta\gamma_{0-1}(2M_1+M_2+M_3) \\ &\quad +\Delta\gamma_{0-2}(2M_1-M_4-M_5)=0, \\[1ex] \frac{\partial A}{\partial M_2} &= \sum S\Delta\lambda\cdot \beta +\Delta\gamma_{0-1}(2M_2+2M_3+M_1) \\ &\quad +\Delta\gamma_{1-3}(2M_2-M_6)=0, \\[1ex] \frac{\partial A}{\partial M_3} &= \sum S\Delta\lambda\cdot \gamma +\Delta\gamma_{0-1}(2M_2+2M_3+M_1) \\ &\quad +\Delta\gamma_{1-4}(2M_3-M_7)=0, \end{aligned} \tag{31} \]

\[ \ldots \]

Let us consider the procedure for composing the sums entering into these equations,

\[ \sum S\Delta\lambda\cdot \alpha = \Delta\lambda_{0-1} \left[ S^{0}_{0-1}\alpha_{0-1} +\alpha_{0-1}\alpha_{0-1}M_1 +\alpha_{0-1}\beta_{0-1}M_2 \right. \]

\[ \left. +\alpha_{0-1}\gamma_{0-1}M_3+\ldots \right] \]

$$ +\Delta\lambda_{0-2}\left[ S^0_{0-2}\alpha_{0-2} +\alpha_{0-2}\alpha_{0-2}M_1 +\alpha_{0-2}\beta_{0-2}M_2 \right. $$

$$ \left. +\alpha_{0-2}\gamma_{0-2}M_3+\cdots \right] $$

$$ +\Delta\lambda_{1-3}\left[ S^0_{1-3}\alpha_{1-3} +\alpha_{1-3}\alpha_{1-3}M_1 +\alpha_{1-3}\beta_{1-3}M_2 \right. $$

$$ \left. +\alpha_{1-3}\gamma_{1-3}M_3+\cdots \right] $$

$$ +\cdots, $$

$$ \sum S\Delta\lambda\cdot\beta = \Delta\lambda_{0-1}\left[ S^0_{0-1}\beta_{0-1} +\alpha_{0-1}\beta_{0-1}M_1 +\beta_{0-1}\beta_{0-1}M_2 \right. $$

$$ \left. +\beta_{0-1}\gamma_{0-1}M_3+\cdots \right] $$

$$ +\Delta\lambda_{0-2}\left[ S^0_{0-2}\beta_{0-2} +\alpha_{0-2}\beta_{0-2}M_1 +\beta_{0-2}\beta_{0-2}M_2 \right. $$

$$ \left. +\beta_{0-2}\gamma_{0-2}M_3+\cdots \right] $$

$$ +\cdots, $$

$$ \sum S\Delta\lambda\cdot\gamma = \Delta\lambda_{0-1}\left[ S^0_{0-1}\gamma_{0-1} +\alpha_{0-1}\gamma_{0-1}M_1 +\beta_{0-1}\gamma_{0-1}M_2 \right. $$

$$ \left. +\gamma_{0-1}\gamma_{0-1}M_3+\cdots \right] $$

$$ +\Delta\lambda_{0-2}\left[ S^0_{0-2}\gamma_{0-2} +\alpha_{0-2}\gamma_{0-2}M_1 +\beta_{0-2}\gamma_{0-2}M_2 \right. $$

$$ \left. +\gamma_{0-2}\gamma_{0-2}M_3+\cdots \right] $$

$$ +\cdots. $$

It is easy to see that these sums may be rewritten with symbolic notations in the following manner:

$$ \begin{aligned} \sum S\Delta\lambda\cdot\alpha &=(S^0\alpha)+(\alpha\alpha)M_1+(\alpha\beta)M_2+(\alpha\gamma)M_3+\cdots,\\ \sum S\Delta\lambda\cdot\beta &=(S^0\beta)+(\beta\alpha)M_1+(\beta\beta)M_2+(\beta\gamma)M_3+\cdots,\\ \sum S\Delta\lambda\cdot\gamma &=(S^0\gamma)+(\gamma\alpha)M_1+(\gamma\beta)M_2+(\gamma\gamma)M_3+\cdots, \end{aligned} \tag{32} $$

$$ \cdots $$

Here, the symbols \((S^{0}\alpha)\), \((\alpha\alpha)\), etc. denote the sums

\[ \left(S^{0}\alpha\right)= \left(S^{0}_{0-1}\Delta\lambda_{0-1}\alpha_{0-1} +S^{0}_{0-2}\Delta\lambda_{0-2}\alpha_{0-2} +\cdots\right), \]

\[ \left(S^{0}\beta\right)= \left(S^{0}_{0-1}\Delta\lambda_{0-1}\beta_{0-1} +S^{0}_{0-2}\Delta\lambda_{0-2}\beta_{0-2} +\cdots\right), \]

\[ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots \]

\[ (\alpha\alpha)= \left(\Delta\lambda_{0-1}\alpha^{2}_{0-1} +\Delta\lambda_{0-2}\alpha^{2}_{0-2} +\cdots\right), \]

\[ (\beta\beta)= \Delta\lambda_{0-1}\beta^{2}_{0-1} +\Delta\lambda_{0-2}\beta^{2}_{0-2} +\cdots, \]

\[ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots \]

\[ (\alpha\beta)= \Delta\lambda_{0-1}\alpha_{0-1}\beta_{0-1} +\Delta\lambda_{0-2}\alpha_{0-2}\beta_{0-2} +\cdots, \]

\[ (\beta\alpha)= \Delta\lambda_{0-1}\beta_{0-1}\alpha_{0-1} +\Delta\lambda_{0-2}\beta_{0-2}\alpha_{0-2} +\cdots, \]

\[ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots \]

Directly from the rule by which these sums are formed it follows, obviously,

\[ (\alpha\beta)=(\beta\alpha);\quad (\alpha\gamma)=(\gamma\alpha);\quad \text{etc.} \]

Thus, if we compile a table of the coefficients of \(M_{1}\) obtained from the sums under consideration, we shall see that they are arranged symmetrically with respect to the diagonal on which the double symbols of the form \((\alpha\alpha)\), \((\beta\beta)\), etc. are located.

\[ \begin{array}{ccccc} (\alpha\alpha) & (\alpha\beta) & (\alpha\gamma) & (\alpha\delta) & \cdots\\ (\alpha\beta) & (\beta\beta) & (\beta\gamma) & (\beta\delta) & \cdots\\ (\alpha\gamma) & (\beta\gamma) & (\gamma\gamma) & (\gamma\delta) & \cdots\\ (\alpha\delta) & (\beta\delta) & (\gamma\delta) & (\delta\delta) & \cdots \end{array} \]

It is also of interest to note that sums of the form \((\alpha\alpha)\), \((\beta\beta)\), etc. are always positive.

If we now turn to the remaining terms of equations (31) and arrange the coefficients occurring in them likewise in a table, we shall again find the characteristic symmetry with respect to the diagonal, which is quite evident from the table

\[ \begin{array}{ccccc} (2\Delta\lambda'_{0-1}+2\Delta\lambda'_{0-2}) & \Delta\lambda'_{0-1} & \Delta\lambda'_{0-1} & -\Delta\lambda'_{0-2} & -\Delta\lambda'_{0-2} \\ \Delta\lambda'_{0-1} & (2\Delta\lambda'_{0-1}+2\Delta\lambda'_{1-3}) & 2\Delta\lambda'_{0-1} & 0 & 0 \\ \Delta\lambda'_{0-1} & 2\Delta\lambda'_{0-1} & (2\Delta\lambda'_{0-1}+2\Delta\lambda'_{1-4}) & 0 & 0 \end{array} \tag{33} \]

It is obvious that, when the corresponding coefficients for \(M\) are added into the general table, symmetry will be preserved; moreover, it is easy to see that, from the row of coefficients for each unknown, the largest will be located on the diagonal.

In its final form the system of equations (31) may be represented in the following table (34):

Coefficients at \(M_1\)	Coefficients at \(M_2\)	Coefficients at \(M_3\)	. . .	Free term
\((\alpha\alpha)+2\left(\Delta\gamma'_{0-1}+\Delta\gamma'_{0-2}\right)\)	\((\alpha\beta)+\Delta\gamma'_{0-1}\)	\((\alpha\gamma)+\Delta\gamma'_{0-1}\)	. . .	\((S^0_\alpha)\)
\((\alpha\beta)+\Delta\gamma'_{0-1}\)	\((\beta\beta)+2\left(\Delta\gamma'_{0-1}+\Delta\gamma'_{1-3}\right)\)	\((\beta\gamma)+2\Delta\gamma'_{0-1}\)	. . .	\((S^0_\beta)\)
\((\alpha\gamma)+\Delta\gamma'_{0-1}\)	\((\beta\gamma)+2\Delta\gamma'_{0-1}\)	\((\gamma\gamma)+2\left(\Delta\gamma'_{0-1}+\Delta\gamma'_{1-4}\right)\)	. . .	\((S^0_\gamma)\)
\((\alpha\delta)-\Delta\gamma'_{0-2}\)	\((\beta\delta)\)	. . . . . .	. . .	\((S^0_\delta)\)
\((\alpha\varepsilon)-\Delta\gamma'_{0-5}\)	\((\beta\varepsilon)\)	. . . . . .	. . .	\((S^0_\varepsilon)\)
. . . . . .	. . . . . .	. . . . . .	. . .	. . .

\[ \tag{34} \]

Engineer Pistol’kors arrives at the method just set forth by a somewhat different method, based on the “principle of the work of constraints.” However, he does not apply this method of calculation at all to plane systems, in view of the large number of equations; for spatial systems he applies it with many assumptions. Since in the present article we are concerned exclusively with plane systems, we shall not dwell in detail on this work.

In engineer Perederii’s article the author also mentions this method in general terms, but does not apply it in practice, considering it too complicated. Indeed, for two of the trusses examined by engineer Perederii, the application of this method is practically impossible. From our calculations we have become convinced that, even when using an abbreviated method for solving the equations, an increase in the number of equations in the system beyond 14–15 leads to extremely intricate computations.

Our attempt—to apply the method set forth to the investigation—is therefore the first in the literature.

9. Group of Methods of Eng. Perederii.

As already mentioned above, the generalized method of calculation cannot be applied to trusses with any appreciable number of members.

In order to reduce the number of unknown quantities, it is possible in some cases, with an accuracy sufficient for practical purposes, to assume that rigid nodal connections exist only between parts of the chords, while the web of the truss is attached by hinges. This assumption is equivalent to saying that all moments due to the rigidity of the joints acting on the web members are set equal to zero.

In such a case, at each joint of the truss there remain only two moments equal in magnitude and opposite in sign; consequently, the number of unknown quantities falls to the number of joints \(n\).

Of course, such a simplification will not significantly affect the accuracy of the results only in the case where the bending moments in the web are small in comparison with the moments in the chord members.

For the truss \(01234\ldots\) considered by us above, the equations required for determining the unknowns are easily obtained by setting, in equations (31) and formula (32), the corresponding moments equal to zero:

\[ M_{1-4}=M_{3}=0,\qquad M_{4-1}=M_{7}=0, \]

\[ M_{2-3}=M_{4}=0,\qquad M_{3-2}=M_{9}=0. \]

We obtain

\[ \sum S\Delta\lambda\cdot\alpha+\Delta\gamma_{0-1}(2M_{1}+M_{2})+\Delta\gamma_{0-2}(2M_{1}-M_{5})=0, \]

\[ \sum S\Delta\lambda\cdot\beta+\Delta\gamma_{0-1}(2M_{2}+M_{1})+\Delta\gamma_{1-3}(2M_{2}-M_{6})=0, \tag{31'} \]

\[ \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots, \]

where this time

\[ \sum S \Delta \lambda \cdot \alpha = (S^0\alpha) + (\alpha\alpha)M_1 + (\alpha\beta)M_2 + (\alpha\varepsilon)M_3 + \cdots, \]

\[ \tag{32′} \sum S \Delta \lambda \cdot \beta = S^0\beta + (\alpha\beta)M_1 + (\beta\beta)M_2 + (\beta\varepsilon)M_3 + \cdots, \]

\[ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ , \]

Engineer Perederii has proposed in his article several methods of calculation, different in form but, in their internal content, homogeneous with the one just set forth by us, which he likewise schematically designates by the name “method of least work.”

Of the other methods described by him, the “method of changes of angles” consists in the following.

Two expressions are compared for the change of the angle between two chord members meeting at some node; on the one hand, this change can be expressed through the moments acting at the given node—\(M_m\)—and at the two adjacent nodes—\(M_{m-1}\) and \(M_{m+1}\):

\[ \tag{35} \Delta \Theta_m = - \frac{l_m}{6EI_m}\left(2M_m + M_{m-1}\right) - \frac{l_{m+1}}{6EI_{m+1}}\left(2M_m + M_{m+1}\right) \]

(Here \(l_m\) and \(l_{m+1}\) denote the lengths of the members \((m-1)-m\) and \((m+1)-m\).)

In determining the sign of the moment, it is necessary mentally to place oneself inside the rigid contour of the truss; then, for the right end of each member, the moment rotating clockwise will be considered positive, and for the left end—the reverse. Applying formula (35), for example, to node \(O\) of our truss 01234... (Fig. 4), and paying attention to the signs of the moments, we obtain

\[ \tag{36} \Delta \Theta_0 = - \frac{l_{0-1}}{6EI_{0-1}}\left(2M_1 + M_2\right) - \frac{l_{0-2}}{6EI_{0-2}}\left(2M_1 - M_5\right). \]

On the other hand, \(\Delta \Theta_m\) can be expressed as a function of the unknown moments, either by the well-known Mohr–Breslau formula, or on the basis of the principle of virtual displacements:

\[ 1.\quad \Delta \Theta_m = \sum \Delta l \cdot S_0, \]

where \(\Delta l\) denotes the elongations of the truss members under the actual load of the truss.

\[ \Delta l=\frac{Sl}{E\omega}=\left(S^0+\alpha M_1+\beta M_2+\ldots\right)\frac{l}{E\omega}, \]

where \(S_0\) are the forces in the members of the hinged truss due to the action of the moment \(M=1\).

It is easy to see that \(\Delta\Theta_m=\sum S_0\Delta l\) is identical with the quantity

\[ \sum S\Delta l\cdot\alpha; \]

substituting the latter into formula (36), we obtain the first of equations \((31')\); the remaining ones are found in the same way.

The “method of possible displacements” is also reduced to that given by us at the beginning of this paragraph.

Somewhat different from the others is the “method of displacements of the joints for trusses with parallel chords.” Whereas in the methods set forth above the rigid contour of the truss is assumed to be closed, i.e., for trusses with parallel chords the connection of the support posts with the chords is rigid, in this latter method the author assumes that the bending moments for the support posts are equal to zero. Consequently both chords are separated and rest freely on the supports, whereby the number of unknowns is reduced to \(n-4\) for a statically determinate truss and to \(n-4+2m\) for a statically indeterminate truss.

In addition, this method differs from the preceding ones in that the unknowns adopted are not bending moments, but forces \(P\) (see Chapter I), called by Eng. Perederii forces of elasticity. Nevertheless the basis of the method remains the same, as is easily seen from the following considerations.

To obtain the equations from which the forces \(P\) are determined, Eng. Perederii equates to one another two expressions for the deflections of the truss joints; the first of them is obtained as a function of the forces \(P\), as the deflections of the joints of a rigid truss under the action of the given load, or, what is the same, of a hinged truss under the action of the given load plus the forces of elasticity; the second expression for the deflections is obtained by considering each chord as a beam on two supports, loaded by a system of forces inverse to the forces of elasticity. Consequently, equations of the form must hold:

\[ y_i=-f_i^*). \]

*) Eng. Perederii writes \(y_i+f_i=0\) and, in the second case, loads the beam directly by the forces of elasticity.

The quantity \(f_i\) can be obtained by differentiating the deformation work of the chord, bent by the elastic forces, with respect to these unknown forces.

But the work just mentioned is nothing other than the work of the chord members in bending under the action of moments due to the rigidity of the joints; this is easily verified by substituting, in the formulas, their expressions in terms of the moments in place of the elastic forces. Thus

\[ f_i=\frac{\partial A_{\text{bend. chord}}}{\partial P_i}; \]

on the other hand

\[ y_i=\sum \sigma_i \cdot S\,\frac{l}{E\omega}, \]

where \(S=(S^0+\sigma_1 P_1+\sigma_2 P_2+\cdots \sigma_i P_i+\cdots)\) is the force in a member of the rigid truss under the action of the given load, and \(\sigma_1,\sigma_2,\ldots \sigma_i,\ldots\) are the forces in the parts of the hinged truss from the loading by the forces \(P_1, P_2,\ldots P_i\ldots\)

It is clear that

\[ \sigma_i=\frac{\partial S}{\partial P_i}, \]

and consequently

\[ y_i=\frac{\partial \sum S^2\frac{l}{2E\omega}}{\partial P_i} \]

is the derivative of the deformation work of the longitudinal forces of the truss, taken with respect to the force \(P_i\).

From comparison of the quantities \(f_i\) and \(y_i\) it is evident that the equation

\[ y_i+f_i=0 \]

is nothing other than the derivative of the total deformation work, set equal to zero, while the work of the web in bending has not been taken into account.

Since it is entirely immaterial whether the moments or the elastic forces are determined first, this method also turns out to be of the same kind as the preceding ones*).

*) Only, as mentioned, the support posts are assumed to be pin-connected with the chords.

10. The Methods of Manderla, Mohr, etc.

If it is assumed that in the expression

\[ S = S^0 + \alpha M_1 + \beta M_2 + \gamma M_3 \ldots \]

the terms \(\alpha M_1\), \(\beta M_2\), etc. are insignificant in comparison with the magnitude \(S^0\), then their sum may be put equal to zero and it may be considered that \(S = S^0\), i.e., that the longitudinal forces in the members of a rigid truss are equal to the corresponding forces in the hinged truss under the same loading.

Methods for calculating rigidity stresses based on such an assumption are the most numerous and, up to the present time, have been applied most often.

Under the stated assumption, equations (31) take the following form:

\[ \Delta \gamma_{0-1}(2M_1 + M_2 + M_3) + \Delta \gamma_{0-2}(2M_1 - M_4 - M_5) = -\sum S^0 \Delta \gamma \cdot \alpha \]

\[ \Delta \gamma_{0-1}(2M_2 + 2M_3 + M_1) + \Delta \gamma_{1-3}(2M_2 - M_6) = -\sum S^0 \Delta \gamma \cdot \beta \]

\[ \Delta \gamma_{0-1}(2M_2 + 2M_3 + M_1) + \Delta \gamma_{1-4}(2M_3 - M_7) = -\sum S^0 \Delta \gamma \cdot \gamma \]

\[ \ldots \tag{37} \]

In these equations the sums appearing on the right-hand sides evidently represent the changes in the angles between the members of the hinged truss when it is deformed under the action of the given load. In this form the equations are given by engineer Pistol’kors.

From them other forms of equations, used in various methods of calculation, are easily obtained.

Thus, for example, recalling that

\[ M_1 = M_{0-1} = \frac{2EI_{0-1}}{l_{0-1}} (2\tau_{0-1} + \tau_{1-0}) = \frac{1}{3\Delta \gamma_{1-0}} (2\tau_{0-1} + \tau_{1-0}), \]

\[ (M_2 + M_3) = -M_{1-0} = -\frac{1}{3\Delta \gamma_{0-1}} (2\tau_{1-0} + \tau_{0-1}), \]

\[ M_1=-M_{0-2}=-\frac{1}{3\Delta\gamma_{0-2}}\left(2\tau_{0-2}+\tau_{2-0}\right), \]

\[ (M_4+M_8)=-M_{2-0}=-\frac{1}{3\Delta\gamma_{0-2}}\left(2\tau_{2-0}+\tau_{0-2}\right),\quad \text{etc.,} \]

we obtain Manderla’s basic equations:

\[ \tau_{0-1}-\tau_{0-2}=-\Delta\Theta_{102}, \]

\[ -\tau_{1-0}+\tau_{1-3}=-\Delta\Theta_{310}, \]

\[ \ldots\ldots\ldots\ldots \]

Since most other methods in this group differ only in the details of solving the equations, we shall discuss in detail only Mohr’s method and its application to calculations by Eng. Paton. As is known, the main distinction of Mohr’s method consists in the fact that, instead of directly calculating the moments \(M\) or the angles \(\tau\), the so-called angles of rotation of the nodes \(\varphi\) and the angles of rotation of the members \(\psi\) are first determined (see § 1). The relation between \(M\), \(\varphi\), and \(\psi\), whose derivation it is unnecessary to give, is as follows:

\[ M_{M-N}=\frac{2EI_{MN}}{l_{MN}}\left(2\varphi_M+\varphi_N-3\psi_{MN}\right) \]

Using the condition \(\sum M=0\), for any node \(M\) one obtains the equation

\[ \tag{38} 2\varphi_M\sum N_{MX}+\sum_X\cdot N_{MX}=3\sum N_{MX}\cdot \psi_{MX}, \]

where it is denoted that

\[ N_{MX}=\frac{2EI_{MX}}{l_{MX}}=\frac{1}{3\Delta\gamma_{MX}}; \]

to form the sums, one must successively substitute for \(X\) the numbers of all nodes directly connected with \(M\). Equation (38) can also be obtained directly from equations (37).

From the system of equations of type (38), the angles \(\varphi\) are determined, while the quantities \(\psi\) are calculated beforehand. For this purpose Mohr uses the principle of virtual displacements, applying to the member \(MN\) a pair of forces perpendicular to the axis of the member and equal—

TRUSSES WITH RIGID JOINTS.

... equal to \(\dfrac{1}{l_{MN}}\) (i.e., the moment of the couple is equal to unity). Then the virtual work of this couple must be equal to the virtual work of the deformation of all elements of the truss:

\[ 1 \cdot \psi_{MN}=\sum \sigma_{MN}\cdot \Delta l, \]

where \(\Delta l=\dfrac{Sl}{E\omega}\) is the change in the length of an element under the actual load, and \(\sigma_{MN}\) are the forces due to the action of the couple. In this way the angles \(\psi\) are determined for all elements.

In applying Mohr’s method in his calculations, Eng. Paton, intentionally or inadvertently, allowed an inaccuracy at this point. Instead of a couple of forces perpendicular to the axis of the member, he applies to all elements except the posts couples of forces directed vertically. For horizontal chords this makes no difference, but for diagonals and inclined chords a rather significant error is obtained. Indeed, let a couple of vertical forces, whose moment is equal to 1, be applied to some element \(MN\), inclined at an angle \(\alpha\) to the horizontal (Fig. 5).

Fig. 5.

The magnitude of the forces is

\[ P=\frac{1}{l_{MN}\cdot \operatorname{Cos}\alpha}. \]

Resolve each of the forces into two components: one normal to the axis of the element,

\[ R=\frac{1}{l_{MN}\operatorname{Cos}\alpha}\cdot \operatorname{Cos}\alpha=\frac{1}{l_{MN}} \]

and one along the axis of the element,

\[ Q=\frac{1}{l_{MN}\operatorname{Cos}\alpha}\cdot \operatorname{Sin}\alpha=\frac{1}{l_{MN}}\operatorname{Tg}\alpha . \]

Then the virtual work of the couple of forces \(R\) when the element is rotated through the angle \(\psi_{MN}\) will be

\[ \frac{1}{l_{MN}} \cdot \psi_{MN} \cdot l_{MN} = 1 \cdot \psi_{MN}. \]

But, in addition, the forces \(Q\) will perform further work in the shortening or elongation of the member,

\[ \pm Q \cdot \Delta l = \pm \frac{\Delta l_{MN}}{l_{MN}} \operatorname{Tg}\alpha . \]

Thus the equation of work (38) for a vertical couple of forces is not applicable. The error in this case may, in some cases, turn out to be quite considerable; we shall return to this again in Chapter IV.

11. The Methods of Engesser, Müller-Breslau, and Others

As we have seen above, the methods of the second group are based on the assumption that, owing to the small stiffness of the web members as compared with the chords, the web may be considered as attached by hinges and, consequently, its bending moments taken equal to zero. On the other hand, the methods of the third group consider it possible to ignore the difference between the forces in hinged and rigid trusses and take as the basis of the calculation, for determining the stresses due to the stiffness of the joints, the increased deformation, generally speaking, of the hinged truss.

The methods of the fourth group combine both assumptions; thus the equations for this case are easily obtained from formula (37), by taking the bending moments for the web equal to zero.

\[ \Delta \gamma_{0-1}(2M_1 + M_2) + \Delta \gamma_{0-2}(2M_1 - M_5) = - \sum S^0 \Delta \lambda \cdot \alpha, \]

\[ \Delta \gamma_{0-1}(2M_2 + M_1) + \Delta \gamma_{1-3}(2M_2 - M_6) = - \sum S^0 \Delta \lambda \cdot \beta, \]

\[ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots \]

We shall not examine these methods in detail, since a priori it may be said that the results of their application cannot be

considered reliable. In fact, it is clear that the two assumptions contradict one another: if one can assume the lattice to be hinged, in view of the great rigidity of the chords, then precisely this rigidity does not allow one to neglect the work of the deformation of the chords in bending. We shall see from examples that this a priori conclusion is justified by the results of comparative computations.

From a review of all the calculation methods considered, it is evident that, in order to compare their accuracy, it is sufficient to choose one method from each group, since the others, differing in form and details, are in essence based on common principles and must therefore give identical results for each group.

12. On the choice of examples of trusses for comparative computations.

In choosing trusses as objects for carrying out comparative computations, we had to be guided by various considerations. In the first place stood the possibility of applying to the given truss the generalized method of calculation, which necessarily limited the number of members, as well as the span of the trusses.

As for the system of trusses, here it was desirable, of course, to choose precisely those for which previous computations had given the most divergent results. The chief place in this respect was occupied by two-panel and two-lattice trusses, for which the computations of engineers Paton and Perederiy led to such sharply different conclusions. Since, in the opinion of Eng. Perederiy, similar results could be expected in general in trusses with ∞-shaped bending of the chords, for the third example a truss of a triangular system with additional posts was taken.

As for the sections of the truss members, experience has shown that the most divergent results are obtained with sections having comparatively large moments of inertia. Therefore the sections were selected with a considerable reserve against the load customary for bridge trusses. Accordingly, the design load was also chosen to be considerably greater than the prescribed one. Since the matter concerns comparative computations, such a discrepancy between the assignment and practical conditions cannot be of great significance.

A large part of the cross-sections was borrowed from the trusses analyzed in Engineer Paton’s book, in order to achieve a certain homogeneity of the material; thus, for example, the cross-sections borrowed from a two-lattice truss were introduced precisely into a truss of the same system, etc.

For each of the three trusses we selected, parallel calculations were carried out by the following four methods:

1) the generalized method;
2) the principal method of the 2nd group;
3) Mohr’s method;
4) the method of the 4th group.

For truss No. 3, moreover, the same calculations were performed under the assumption that the moments of inertia of the chords were reduced by half, while all other data remained unchanged.

13. Example 1. Two-lattice truss.

The two-lattice truss shown in Fig. 6, the principal dimensions and cross-sections of whose members are given in Table 1, is a statically indeterminate system with one redundant unknown. Indeed, the number of joints of the truss is \(n = 10\); consequently, the possible number of equations for determining the forces in the members of a hinged truss is

\[ k = 2n - 3 = 17; \]

whereas the number of unknown forces in the members is

\[ m = 18. \]

We shall therefore apply, for determining the forces, the theorem of least work of deformation. Assuming that there is no friction in the supports and that the work of the support reactions is equal to zero, and also recalling that hinged connections, according to our fundamental premise, perform no work at all, we may write the work of deformation of the hinged truss as

\[ A = \sum \frac{S^{2}l}{2E\omega}, \]

where the summation must be extended over all members of the truss. Let us take the force in member \(0\text{--}3\) as the unknown and denote it by \(x\).

Fig. 6.

\[ \alpha = 38^\circ 40', \qquad \beta = 51^\circ 20' \]

\[ \operatorname{Sin}\alpha=\operatorname{Cos}\beta=0.6247 \]

\[ \operatorname{Cos}\alpha=\operatorname{Sin}\beta=0.7809 \]

\[ \operatorname{Tg}\alpha=\operatorname{Cotg}\beta=0.8000 \]

Table I.

Name of element	No. No. of elements	\(\omega\) brutto, cm\(^2\)	\(I\) brutto, cm\(^4\)	\(e_1\), cm	\(e_2\), cm	\(l\), cm	\(N=\dfrac{2EI}{l}\), 10 tn. m.	\(\Delta\lambda=\dfrac{l}{6EI}\), \(\dfrac{1}{10^5\ \mathrm{tn}\cdot \mathrm{m}}\)	\(\dfrac{l}{\omega}\), \(\dfrac{1}{\mathrm{cm}}\)
Upper chord.	1—3	80	3700	7,2	15,8	400	37,0	90,09	5,0000
Upper chord.	3—5	100	4500	5,3	18,7	400	45,0	74,07	4,0000
Lower chord.	0—2	80	3700	7,2	15,8	400	37,0	90,09	5,0000
Lower chord.	2—4	100	4500	5,3	18,7	400	45,0	74,07	4,0000
Diagonals.	1—2	36	2100	13	13	640	13,1	254,45	17,7778
Diagonals.	3—4	28	900	10	10	640	5,6	595,24	22,8571
Diagonals.	0—3	106	2500	12	12	640	15,6	213,68	6,0377
Diagonals.	2—5	96	1700	10	10	640	10,6	314,47	6,6667
Post.	0—1	96	1400	8	8	500	11,2	297,62	5,2083

Then the force in any element can be represented in the following form:

\[ S=S^0+ax, \tag{a} \]

where \(S^0\) are the forces from the external load in the principal statically deter-

of the divisible truss (Fig. 7), obtained from the given one by removing member \(0—3\); \(a\) is the force in the member of the same truss under the action of the single force \(x = 1^{tn}\).

Differentiating the expression for the work with respect to \(x\) and noting that

\[ \frac{\partial S}{\partial x}=a, \]

we obtain, by setting the derivative equal to zero, the equation for determining \(x\)

\[ \tag{b} \frac{\partial A}{\partial x} = \sum \frac{Sl}{E\omega}\cdot a = \sum \frac{S^{0}l}{E\omega}\cdot a + \sum \frac{l}{E\omega}\cdot a^{2}x = 0 \]

Hence

\[ \tag{c} x = - \frac{\displaystyle \sum \frac{S^{0}l}{\omega}\cdot a} {\displaystyle \sum \frac{l}{\omega}\cdot a^{2}}. \]

For the purposes of our calculation, we shall determine the forces in the following cases of unit loading:

1) for a load \(P = 1^{tn}\), applied successively at joints 1, 2, 3, 4, 5;

2) for a pair of forces whose moment is equal to unity, and whose components are applied successively normal to the axes of members \(0—1\), \(0—3\), \(2—5\), \(1—2\), \(3—4\) at their end points.

The results of the computations are grouped in tables.

Table II gives the determination of the quantity \(\sum a^{2}\frac{l}{\omega}\), which remains constant for all loading cases; Tables III and IV contain the quantities \(S^{0}\) and \(S^{0}\frac{al}{\omega}\) for the aforementioned loading cases, and in the two lower rows are placed the sums \(S^{0}\frac{al}{\omega}\) and the values \(x\) obtained from them.

Table V presents the final values of the forces for the loading cases considered. In addition, from these forces, by the appropriate addition, the forces due to the action of pairs of forces \(= 1\) ton-m on the chord members have been determined and placed in the first four columns of Table VI.

For the investigation of stresses due to the rigidity of the joints, the loading has been chosen symmetrical but nonuniform, in order to obtain significant

magnitudes of the stresses mentioned: loads of \(50^{tn}\) are assumed at the nodes \(3\) and \(3'\). For this load, the forces, stresses, and changes in the lengths of the members of the hinged truss have been calculated and are placed in the last three columns of Table VI.

Fig. 7.

Table II.

Names of members	\(\dfrac{l}{\omega}\left(\dfrac{1}{\mathrm{cm}.}\right)\)	\(\alpha\ \mathrm{tn}.\)	\(\alpha\cdot\dfrac{l}{\omega}\left(\dfrac{\mathrm{tn}}{\mathrm{cm}.}\right)\)	\(\alpha^{2}\dfrac{l}{\omega}\left(\dfrac{\mathrm{tn}^{2}}{\mathrm{cm}.}\right)\)
\(0—2\)	5,0000	\(-0,625\)	\(-3,125\)	\(+1,953\)
\(2—4\)	4,0000	\(+0,625\)	\(+2.500\)	\(+1,563\)
\(4—2'\)	4,0000	\(-0,625\)	\(-2,500\)	\(+1,563\)
\(2'—0'\)	5,0000	\(+0,625\)	\(+3,125\)	\(+1,953\)
\(1—3\)	5,0000	\(-0,625\)	\(-3,125\)	\(+1,953\)
\(3—5\)	4,0000	\(+0,625\)	\(+2,500\)	\(+1,563\)
\(5—3'\)	4,0000	\(-0,625\)	\(-2,500\)	\(+1,563\)
\(3'—1'\)	5,0000	\(+0,625\)	\(+3,125\)	\(+1,953\)
\(0—3\)	6,0377	\(+1,000\)	\(+6,038\)	\(+6,038\)
\(2—5\)	6,6667	\(-1,000\)	\(-6,667\)	\(+6,667\)
\(5—2'\)	6,6667	\(+1,000\)	\(+6,667\)	\(+6,667\)
\(3'—0'\)	6,0377	\(-1,000\)	\(-6,038\)	\(+6,038\)
\(1—2\)	17,7778	\(+1,000\)	\(+17,773\)	\(+17,778\)
\(3—4\)	22,8571	\(-1,000\)	\(-22,857\)	\(+22,857\)
\(4—3'\)	22,8571	\(+1,000\)	\(+22,857\)	\(+22,857\)
\(2'—1'\)	17,7778	\(-1,000\)	\(-17,778\)	\(+17,778\)
\(0—1\)	5,2083	\(-0,781\)	\(-4,068\)	\(+3,177\)
\(0'—1'\)	5,2083	\(+0,781\)	\(+4,068\)	\(+3,177\)

\[ \sum \alpha^{2}\frac{l}{\omega}=+127.098 \]

Table III

Elements	Group \(P=1\) tn. at node 1, \(S^\circ\)	Group \(P=1\) tn. at node 1, \(S^\circ \dfrac{al}{\omega}\)	Group \(P=1\) tn. at node 2, \(S^\circ\)	Group \(P=1\) tn. at node 2, \(S^\circ \dfrac{al}{\omega}\)	Group \(P=1\) tn. at node 3, \(S^\circ\)	Group \(P=1\) tn. at node 3, \(S^\circ \dfrac{al}{\omega}\)	Group \(P=1\) tn. at node 4, \(S^\circ\)	Group \(P=1\) tn. at node 4, \(S^\circ \dfrac{al}{\omega}\)	Group \(P=1\) tn. at node 5, \(S^\circ\)	Group \(P=1\) tn. at node 5, \(S^\circ \dfrac{al}{\omega}\)
\(0-2\)	0	0	0	0	0	0	0	0	0	0
\(2-4\)	0	0	\(+0,400\)	\(+1,000\)	\(+1,200\)	\(+3,000\)	\(+0,800\)	\(+2,000\)	\(+0,800\)	\(+2,000\)
\(4-2'\)	0	0	\(+0,400\)	\(-1,000\)	\(-0,400\)	\(+1,000\)	0	0	\(+0,800\)	\(-2,000\)
\(2'-0'\)	0	0	0	0	\(+0,800\)	\(+2,500\)	\(+0,800\)	\(+2,500\)	0	0
\(1-3\)	0	0	\(-0,600\)	\(+1,875\)	\(-0,600\)	\(+1,875\)	\(-0,400\)	\(+1,250\)	\(-0,400\)	\(+1,250\)
\(3-5\)	0	0	\(-0,600\)	\(-1,500\)	\(+0,200\)	\(+0,500\)	\(-0,400\)	\(-1,000\)	\(-0,400\)	\(-1,000\)
\(5-3'\)	0	0	\(-0,200\)	\(+0,500\)	\(-1,000\)	\(+2,500\)	\(-1,200\)	\(+3,000\)	\(-0,400\)	\(+1,000\)
\(3'-1'\)	0	0	\(-0,200\)	\(-0,625\)	\(+0,600\)	\(+1,875\)	\(+0,400\)	\(+1,250\)	\(-0,400\)	\(-1,250\)
\(0-3\)	0	0	0	0	0	0	0	0	0	0
\(2-5\)	0	0	\(+0,320\)	\(-2,133\)	\(-0,960\)	\(+6,400\)	\(-0,640\)	\(+4,267\)	\(-0,640\)	\(+4,267\)
\(5-2'\)	0	0	\(-0,320\)	\(-2,133\)	\(+0,960\)	\(+6,400\)	\(+0,640\)	\(+4,267\)	\(-0,640\)	\(-4,267\)
\(3'-0'\)	0	0	0	0	\(-1,280\)	\(+7,729\)	\(-1,280\)	\(+7,729\)	0	0
\(1-2\)	0	0	\(+0,960\)	\(+17,067\)	\(+0,960\)	\(+17,067\)	\(+0,640\)	\(+11,378\)	\(+0,640\)	\(+11,378\)
\(3-4\)	0	0	0	0	\(-1,280\)	\(+29,257\)	0	0	0	0
\(4-3'\)	0	0	0	0	\(+1,280\)	\(+29,257\)	\(+1,280\)	\(+29,257\)	0	0
\(2'-1'\)	0	0	\(+0,320\)	\(-5,689\)	\(-0,960\)	\(+5,689\)	\(-0,640\)	\(+11,378\)	\(+0,640\)	\(-11,378\)
\(0-1\)	\(-1,000\)	\(+4,068\)	\(-0,750\)	\(+3,051\)	\(-0,750\)	\(+3,051\)	\(-0,500\)	\(+2,034\)	\(-0,500\)	\(+2,034\)
\(0-1'\)	0	0	\(-0,250\)	\(-1,017\)	\(+0,750\)	\(-3,051\)	\(+0,500\)	\(+2,034\)	\(-0,500\)	\(-2,034\)
\(\sum \dfrac{S^\circ al}{\omega}\)		\(+4,068\)		\(+9,913\)		\(+115,049\)		\(+81,344\)		0
\(x\)		\(-0,032\)		\(-0,078\)		\(-0,905\)		\(-0,640\)		0

Table IV.

Elements	Pair of forces, applied to element 0–1: $S^0$	Pair of forces, applied to element 0–1: $S^0 \dfrac{al}{\omega}$	Pair of forces, applied to element 0–3: $S^0$	Pair of forces, applied to element 0–3: $S^0 \dfrac{al}{\omega}$	Pair of forces, applied to element 2–5: $S^0$	Pair of forces, applied to element 2–5: $S^0 \dfrac{al}{\omega}$	Pair of forces, applied to element 1–2: $S^0$	Pair of forces, applied to element 1–2: $S^0 \dfrac{al}{\omega}$	Pair of forces, applied to element 1–4: $S^0$	Pair of forces, applied to element 1–4: $S^0 \dfrac{al}{\omega}$
0–2	+2.000	−6.250	+1.220	−3.813	+2.000	0	0	0	0	0
2–4	+1.000	+2.500	+1.781	+4.453	+1.000	+2.500	+1.000	+2.500	−1.000	−2.500
4–2′	+1.000	−2.500	+0.219	−0.548	+1.000	−2.500	+1.000	−2.500	+1.000	−2.500
2′–0′	0	0	+0.781	+2.441	0	0	0	0	0	0
1–3	−1.500	+4.688	−0.281	+0.878	+0.500	−1.563	−1.500	+4.688	+0.500	+1.563
3–5	−1.500	−3.750	−0.720	−1.800	+0.500	+1.250	−1.500	−3.750	−1.500	−3.750
5–3′	−0.500	+1.250	−1.281	+3.203	−0.500	+1.250	−0.500	+1.250	−0.500	+1.250
3′–1′	−0.500	−1.563	+0.281	+0.878	−0.500	−1.563	−0.500	−1.563	−0.500	−1.563
0–3	0	0	0	0	0	0	0	0	0	0
2–5	+0.800	−5.333	−0.449	−2.993	−0.449	+2.993	+0.800	−5.333	+0.800	−5.333
5–2′	−0.800	−5.333	−0.449	+2.993	−0.800	−5.333	−0.800	−5.333	−0.800	−5.333
3′–0′	0	0	−1.250	+7.548	0	0	0	0	0	0
1–2	−0.800	−14.222	+0.449	+7.982	−0.800	−14.222	+0.449	+7.982	−0.800	−14.222
3–4	0	0	−1.250	−28.571	0	0	0	0	+1.250	−28.571
4–3′	0	0	+1.250	+28.571	0	0	0	0	0	0
2′–1	+0.800	−14.222	−0.449	+7.982	+0.800	−14.222	+0.800	−14.222	+0.800	−14.222
0–1	+0.625	−2.543	−0.351	+1.428	+0.625	−2.543	+0.625	−2.543	+0.625	−2.543
0′–1′	−0.625	−2.543	+0.351	+1.428	−0.625	−2.543	−0.625	−2.543	−0.625	−2.543
$\sum \dfrac{S^0 al}{\omega}\,x$		−49.821		+95.188		−36.496		−21.367		−83.393
		+0.392		−0.749		+0.287		+0.168		+0.656

Table V.

Elements	Load \(P=1\) tm at nodes: 1	Load \(P=1\) tm at nodes: 2	Load \(P=1\) tm at nodes: 3	Load \(P=1\) tm at nodes: 4	Load \(P=1\) tm at nodes: 5	Couple \(Pl=1\) tm-m at nodes: 0–1	Couple \(Pl=1\) tm-m at nodes: 0–3	Couple \(Pl=1\) tm-m at nodes: 2–5	Couple \(Pl=1\) tm-m at nodes: 1–2	Couple \(Pl=1\) tm-m at nodes: 3–4
0–2	+0,020	+0,049	+0,566	+0,400	0	+1,755	+1,688	−0,179	−0,105	−0,410
2–4	−0,020	+0,351	+0,634	+0,400	+0,800	+1,245	+1,312	+1,179	+1,105	−0,590
4–2′	+0,020	+0,449	+0,166	−0,400	+0,800	+0,755	+0,688	+0,821	+0,895	+0,590
2′–0′	−0,020	−0,049	+0,234	+0,400	0	+0,245	+0,312	+0,179	+0,105	+0,410
1–3	+0,020	−0,551	−0,034	0	−0,400	−1,745	+0,188	+0,321	−1,605	+0,090
3–5	−0,020	−0,649	−0,366	−0,800	−0,400	−1,255	−1,188	+0,679	−1,395	−1,090
5–3′	+0,020	−0,151	−0,434	−0,800	−0,400	−0,745	−0,812	−0,679	−0,605	−0,910
3′–1′	−0,020	−0,249	+0,034	0	−0,400	−0,255	−0,188	−0,321	−0,395	−0,090
0–3	−0,032	−0,078	−0,905	−0,640	0	+0,392	−0,749	+0,287	+0,168	+0,656
2–5	+0,032	+0,398	−0,055	0	−0,640	+0,408	+0,300	−0,736	+0,632	+0,144
5–2′	−0,032	−0,398	+0,055	0	−0,640	−0,408	−0,300	−0,513	−0,632	−0,144
3′–0′	+0,032	+0,078	−0,375	−0,640	0	−0,392	−0,500	−0,287	−0,168	−0,656
1–2	−0,032	−0,882	+0,055	0	+0,640	−0,408	−0,300	−0,513	+0,617	−0,144
3–4	+0,032	+0,078	−0,375	+0,640	0	−0,392	−0,500	−0,287	−0,168	+0,594
4–3′	−0,032	−0,078	+0,375	+0,640	0	+0,392	+0,500	+0,287	+0,168	+0,656
2′–1′	+0,032	+0,398	−0,055	0	+0,640	+0,408	+0,300	+0,513	+0,632	+0,144
0–1	−0,975	−0,689	−0,043	0	−0,500	+0,319	+0,234	+0,401	+0,494	+0,112
0′–1′	−0,025	−0,311	+0,043	0	−0,500	−0,319	−0,234	−0,401	−0,494	−0,112

Table VI.

Elem.	Forces from the pair of forces \(P'l = 1\), applied at the nodes: 0—2	Forces from the pair of forces \(P'l = 1\), applied at the nodes: 2—4	Forces from the pair of forces \(P'l = 1\), applied at the nodes: 1—3	Forces from the pair of forces \(P'l = 1\), applied at the nodes: 3—5	Forces from loads \(P = 1\) tn, applied at 3 and \(3'\).	Forces from load \(P = 50\) tn, applied at nodes 3 and \(3'\): Forces, tn.	Forces from load \(P = 50\) tn, applied at nodes 3 and \(3'\): Stresses, kg./cm².	Forces from load \(P = 50\) tn, applied at nodes 3 and \(3'\): \(\lambda = \dfrac{Sl}{E\omega}\cdot 10^2\), cm.
0—2	\(+0{,}122\)	\(+0{,}878\)	\(+1{,}365\)	\(-1{,}415\)	\(+0{,}800\)	\(+40{,}00\)	\(+500\)	\(+10{,}000\)
2—4	\(+0{,}878\)	\(+0{,}122\)	\(+1{,}635\)	\(+0{,}415\)	\(+0{,}800\)	\(+40{,}00\)	\(+400\)	\(+8{,}000\)
4—\(2'\)	\(+1{,}122\)	\(-0{,}122\)	\(+0{,}365\)	\(+1{,}585\)	\(+0{,}800\)	\(+40{,}00\)	\(+400\)	\(+8{,}000\)
\(2'\)—\(0'\)	\(-0{,}122\)	\(+1{,}122\)	\(+0{,}635\)	\(-0{,}585\)	\(+0{,}800\)	\(+40{,}00\)	\(+500\)	\(+10{,}000\)
1—3	\(-1{,}378\)	\(+1{,}378\)	\(-0{,}135\)	\(-0{,}915\)	\(0\)	\(0\)	\(0\)	\(0\)
3—5	\(-1{,}622\)	\(-0{,}378\)	\(-0{,}865\)	\(-0{,}085\)	\(-0{,}800\)	\(-40{,}00\)	\(-400\)	\(-8{,}000\)
5—\(3'\)	\(-0{,}378\)	\(-0{,}622\)	\(-1{,}135\)	\(-0{,}085\)	\(-0{,}800\)	\(-40{,}00\)	\(-400\)	\(-8{,}000\)
\(3'\)—\(1'\)	\(-0{,}622\)	\(+0{,}622\)	\(+0{,}135\)	\(-1{,}085\)	\(0\)	\(0\)	\(0\)	\(0\)
0—3	\(-0{,}195\)	\(-1{,}405\)	\(-2{,}183\)	\(+2{,}263\)	\(-1{,}280\)	\(-64{,}00\)	\(-604\)	\(-19{,}321\)
2—5	\(+0{,}995\)	\(-0{,}995\)	\(-0{,}217\)	\(-1{,}463\)	\(0\)	\(0\)	\(0\)	\(0\)
5—\(2'\)	\(-0{,}995\)	\(+0{,}995\)	\(+0{,}217\)	\(-1{,}737\)	\(0\)	\(0\)	\(0\)	\(0\)
\(3'\)—\(0'\)	\(+0{,}195\)	\(-1{,}795\)	\(-1{,}017\)	\(+0{,}937\)	\(-1{,}280\)	\(-64{,}00\)	\(-604\)	\(-19{,}321\)
1—2	\(+2{,}205\)	\(-2{,}205\)	\(+0{,}217\)	\(+1{,}463\)	\(0\)	\(0\)	\(0\)	\(0\)
3—4	\(+0{,}195\)	\(+1{,}405\)	\(-1{,}017\)	\(+0{,}937\)	\(0\)	\(0\)	\(0\)	\(0\)
4—\(3'\)	\(-0{,}195\)	\(+1{,}795\)	\(+1{,}017\)	\(-0{,}937\)	\(0\)	\(0\)	\(0\)	\(0\)
\(2'\)—\(1'\)	\(+0{,}995\)	\(-0{,}995\)	\(-0{,}217\)	\(+1{,}737\)	\(0\)	\(0\)	\(0\)	\(0\)
0—1	\(-1{,}722\)	\(+1{,}722\)	\(+2{,}330\)	\(-1{,}143\)	\(0\)	\(0\)	\(0\)	\(0\)
\(0'\)—\(1'\)	\(-0{,}778\)	\(+0{,}778\)	\(+0{,}170\)	\(-1{,}357\)	\(0\)	\(0\)	\(0\)	\(0\)

14. Derivation of the basic equations for calculating the moments due to the rigidity of the nodes.

Having the principal forces in the hinged truss, we can now proceed to the study of the truss under the assumption of rigid nodal connections. As we saw in the second chapter, the influence of rigid nodes on the longitudinal forces in the members of the truss can be expressed by formulas of the form

\[ S = S^0 + \alpha M_1 + \beta M_2 + \gamma M_3 + \cdots . \tag{d} \]

By the symbols \(M_1\), \(M_2\), etc., we meant the aggregate of mutually equilibrating moments, for example \(+M_{0-2}\) and \(-M_{0-1}\), etc.

In the case now under consideration, the symmetry of the truss and of the load makes it possible to assign to the quantities \(M_1\), \(M_2\), . . . . . a somewhat different meaning. The point is that in this case the absolute magnitudes of the moments acting on symmetrically situated members will be equal, i.e., we shall have

\[ \left| M_{0-2} \right| = \left| M_{0'-2'} \right|, \]

\[ \ldots\ldots\ldots \]

\[ \left| M_{0-1} \right| = \left| M_{0'-1'} \right|,\quad \text{etc.} \]

On this basis we may understand by \(M_1\) the aggregate of four mutually equilibrating moments: \(+M_{0-2}\), \(-M_{0-1}\); \(+M_{0'-2'}\), \(-M_{0'-1'}\).

In the appended table, for clarity, the loads for each of the 14 cases \(M_1\), \(M_2\), etc. are shown schematically (Fig. 8).

Here the moments acting on the ends of the members are replaced by equivalent couples of forces.

With the aid of the diagram it is easy to write the following table of moments:

\[ M_{0-1}=-M_1, \qquad M_{1-0}=M_3, \]

\[ M_{0-2}=M_1+M_2, \qquad M_{2-0}=-(M_5+M_6+M_7), \]

\[ M_{0-3}=-M_2, \qquad M_{3-0}=M_8, \]

\[ M_{1-2}=-M_4, \qquad M_{2-1}=M_5, \]

\[ M_{1-3}=M_3+M_4, \qquad M_{3-1}=-(M_8+M_9+M_{10}), \]

\[ M_{2-4}=M_7, \qquad M_{4-2}=-M_{11}, \]

\[ M_{2-5}=M_6,\qquad M_{5-2}=-M_{12}, \]

\[ M_{3-4}=M_9,\qquad M_{4-3}=-M_{13}, \]

\[ M_{3-5}=M_{10},\qquad M_{5-3}=-M_{14}. \]

The work of deformation of the individual elements will be expressed, according to these notations, as

\[ A_{0-1}=A'_{0-1}+\Delta\gamma_{0-1}\left[M_1^2+M_3^2-M_1M_3\right], \]

\[ A_{0-2}=A'_{0-2}+\Delta\gamma_{0-2}\left[(M_1+M_2)^2+(M_5+M_9+M_7)^2\right. \]

\[ \left.+(M_1+M_2)(M_5+M_6+M_7)\right], \]

\[ A_{0-3}=A'_{0-3}+\Delta\gamma_{0-3}\left[M_2^2+M_8^2+M_2M_8\right], \]

\[ A_{1-2}=A'_{1-2}+\Delta\gamma_{1-2}\left[M_4^2+M_5^2+M_4M_5\right], \]

\[ A_{1-3}=A'_{1-3}+\Delta\gamma_{1-3}\left[(M_3+M_4)^2+(M_8+M_9+M_{10})^2\right. \]

\[ \left.+(M_3+M_4)(M_8+M_9+M_{10})\right], \]

\[ A_{2-4}=A'_{2-4}+\Delta\gamma_{2-4}\left[M_7^2+M_{11}^2+M_7M_{11}\right], \]

\[ A_{2-5}=A'_{2-5}+\Delta\gamma_{2-5}\left[M_6^2+M_{12}^2+M_6M_{12}\right], \]

\[ A_{3-4}=A'_{3-4}+\Delta\gamma_{3-4}\left[M_9^2+M_{13}^2+M_9M_{13}\right], \]

\[ A_{3-5}=A'_{3-5}+\Delta\gamma_{3-5}\left[M_{10}^2+M_{14}^2+M_{10}M_{14}\right], \]

N. V. Nekrasov.

Fig. 8.

Here, by \(A'_{0--1}\), \(A'_{0--2}\), etc., are denoted expressions of the form

\[ A'_{0--1}=\frac{S^{2}_{0--1}l_{0--1}}{2E\omega_{0--1}}, \]

i.e., the work of the internal forces of the member under the action of the longitudinal force.

Noting that the work of deformation for the whole truss is equal to the sum of the works of all members,

\[ A=(A_{0--1}+A_{0--2}+A_{0--3}+\cdots +A_{2--4}+A_{3--5})\times 2 \]

and differentiating \(A\) successively with respect to the unknowns \(M\), we obtain 14 equations

\[ \frac{\partial A}{\partial M_1} = \sum \frac{\partial A'}{\partial M_1} +\Delta\gamma_{0--1}(2M_1-M_3) +\Delta\gamma_{0--2}(2M_1+2M_2 \]

\[ +M_5+M_6+M_7)=0, \]

\[ \frac{\partial A}{\partial M_2} = \sum \frac{\partial A'}{\partial M_2} +\Delta\gamma_{0--2}(2M_1+2M_2+M_5+M_6+M_7) \]

\[ +\Delta\gamma_{0--3}(2M_2+M_8)=0, \]

\[ \frac{\partial A}{\partial M_3} = \sum \frac{\partial A'}{\partial M_3} +\Delta\gamma_{0--1}(2M_3-M_1) +\Delta\gamma_{1--3}(2M_3+2M_4 \tag{e} \]

\[ +M_8+M_9+M_{10})=0, \]

\[ \frac{\partial A}{\partial M_4} = \sum \frac{\partial A'}{\partial M_4} +\Delta\gamma_{1--2}(2M_4+M_5) +\Delta\gamma_{1--3}(2M_3+2M_4 \]

\[ +M_8+M_9+M_{10})=0, \]

\[ \frac{\partial A}{\partial M_5} = \sum \frac{\partial A'}{\partial M_5} +\Delta\gamma_{1--2}(2M_5+M_4) +\Delta\gamma_{0--2}(2M_5+2M_6 \]

\[ +2M_7+M_1+M_2)=0, \]

\[ \begin{aligned} \frac{\partial A}{\partial M_6} &= \sum \frac{\partial A'}{\partial M_6} + \Delta\gamma_{0-2}\left(2M_5+2M_6+2M_7+M_1+M_2\right) \\ &\quad + \Delta\gamma_{2-5}\left(2M_6+M_{12}\right)=0, \\[1.2em] \frac{\partial A}{\partial M_7} &= \sum \frac{\partial A'}{\partial M_7} + \Delta\gamma_{0-2}\left(2M_5+2M_6+2M_7+M_1+M_2\right) \\ &\quad + \Delta\gamma_{2-4}\left(2M_7+2M_{11}\right)=0, \\[1.2em] \frac{\partial A}{\partial M_8} &= \sum \frac{\partial A'}{\partial M_8} + \Delta\gamma_{1-3}\left(2M_8+2M_9+2M_{10}+M_3+M_4\right) \\ &\quad + \Delta\gamma_{0-3}\left(2M_8+M_2\right)=0, \\[1.2em] \frac{\partial A}{\partial M_9} &= \sum \frac{\partial A'}{\partial M_9} + \Delta\gamma_{1-3}\left(2M_8+2M_9+2M_{10}+M_3+M_4\right) \\ &\quad + \Delta\gamma_{3-4}\left(2M_9+M_{13}\right)=0, \tag{e} \\[1.2em] \frac{\partial A}{\partial M_{10}} &= \sum \frac{\partial A'}{\partial M_{10}} + \Delta\gamma_{1-3}\left(2M_8+2M_9+2M_{10}+M_3+M_4\right) \\ &\quad + \Delta\gamma_{3-5}\left(2M_{10}+M_{14}\right)=0, \\[1.2em] \frac{\partial A}{\partial M_{11}} &= \sum \frac{\partial A'}{\partial M_{11}} + \Delta\gamma_{2-4}\left(2M_{11}+M_7\right)=0, \\[1.2em] \frac{\partial A}{\partial M_{12}} &= \sum \frac{\partial A'}{\partial M_{12}} + \Delta\gamma_{2-5}\left(2M_{12}+M_6\right)=0, \\[1.2em] \frac{\partial A}{\partial M_{13}} &= \sum \frac{\partial A'}{\partial M_{13}} + \Delta\gamma_{3-4}\left(2M_{13}+M_9\right)=0, \\[1.2em] \frac{\partial A}{\partial M_{14}} &= \sum \frac{\partial A'}{\partial M_{14}} + \Delta\gamma_{3-5}\left(2M_{14}+M_{10}\right)=0. \end{aligned} \]

15. Computation of the coefficients of the equations.

We shall now take up the finding of the sums entering into the left-hand sides of equations \((e)\).

In the second chapter we saw that these sums may be written:

\[ \sum \frac{\partial A'}{\partial M_1} = (S^0\alpha)+(\alpha\alpha)M_1+(\alpha\beta)M_2+\cdots+(\alpha\rho)M_{14}, \]

\[ \sum \frac{\partial A'}{\partial M_2} = (S^0\beta)+(\alpha\beta)M_1+(\beta\beta)M_2+\cdots+(\beta\rho)M_{14}, \]

\[ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots \]

The meanings of the symbols \((S^0\alpha)\), \((\alpha\alpha)\), etc., retain the definitions of the second chapter.

Let us note that, thanks to the property of symmetry of the symbols pointed out by us, we shall have to compute all 15 coefficients only for the first sum; then, for each subsequent sum, one coefficient fewer will have to be computed, so that for the last it will be necessary only to determine the quantities \((\rho\rho)\) and \((S^0\rho)\).

We first find the quantities \(\alpha,\beta,\ldots,\rho\); with the aid of the data from Tables V and VI we first compile Table VII of the forces in the members of the whole truss due to symmetrically applied moments \(=1\) (for example, to members \(0—1\) and \(0'—1'\), etc.), and then Table VIII from symmetrical loading by the moments \(M_1=1;\ M_2=1\), etc.; these will be the required \(\alpha,\beta,\ldots,\rho\).

N. V. NEKRASOV.

Table VII.

Forces from symmetrical loading of the truss; equal elements.

Elements	0–2 0′–2′	2–4 4–2′	1–3 1′–3′	3–5 5′–3′	0–3 0′–3′	2–5 5–2′	1–2 1′–2′	3–4 3′–4′	0–1 0′–1′
0–2	0	+0,200	+0,200	−0,200	+0,200	0	0	0	+0,200
2–4	+0,200	0	+0,200	+0,200	+0,200	+0,200	+0,200	0	+0,200
1–3	−0,200	+0,200	0	−0,200	0	0	−0,200	0	−0,200
3–5	−0,200	−0,200	−0,200	0	−0,200	0	−0,200	−0,200	−0,200
0–3	0	−0,320	−0,320	+0,320	−0,125	0	0	0	0
2–5	0	0	0	−0,320	0	−0,125	0	0	0
1–2	+0,320	−0,320	0	+0,320	0	0	+0,125	0	0
3–4	0	+0,320	0	0	0	0	0	+0,125	0
0–1	−0,250	+0,250	+0,250	−0,250	0	0	0	0	0

Table VIII and IX.

Elem.	$\alpha$	$\beta$	$\gamma$	$\delta$	$\varepsilon$	$\eta$	$\vartheta$	$x$	$\lambda$	$\mu$	$\nu$	$o$	$\tau$	$\rho$
0–2	−0,200	−0,200	0	+0,200	0	0	+0,200	0	−0,200	−0,400	−0,200	0	0	+0,200
2–4	0	0	0	0	0	0	−0,200	0	−0,200	0	0	−0,200	0	−0,200
1–3	0	−0,200	+0,200	+0,200	0	+0,200	+0,400	0	0	−0,200	−0,200	0	0	+0,200
3–5	0	0	0	0	0	+0,200	0	0	0	+0,200	+0,200	+0	+0,200	0
0–3	0	+0,125	−0,320	−0,320	0	0	−0,320	+0,195	+0,320	+0,640	+0,320	0	0	−0,320
2–5	0	0	0	0	0	−0,125	0	0	0	−0,320	0	+0,125	0	+0,320
1–2	+0,320	+0,320	0	−0,125	−0,195	−0,320	−0,640	0	0	+0,320	+0,320	0	−0,125	−0,320
3–4	0	0	0	0	0	0	+0,320	0	+0,125	0	−0,320	0	−0,125	0
0–1	−0,250	−0,250	+0,250	+0,250	+0,250	+0,250	+0,500	−0,250	−0,250	−0,500	−0,250	0	0	+0,250

Products of the quantities $\Delta_i=\dfrac{l}{E\omega}$ by the coefficients.

Elem.	$\alpha$	$\beta$	$\gamma$	$\delta$	$\varepsilon$	$\eta$	$\vartheta$	$x$	$\lambda$	$\mu$	$\nu$	$o$	$\tau$	$\rho$
0–2	−0,500	−0,500	0	−0,500	0	0	−0,500	0	−0,500	−1,000	−0,500	0	0	+0,500
2–4	0	0	0	0	0	0	−0,400	0	−0,400	0	0	−0,400	0	−0,400
1–3	0	−0,500	+0,500	+0,500	0	+0,500	+1,000	0	0	−0,500	−0,500	0	0	+0,500
3–5	0	0	0	0	0	+0,400	0	0	0	+0,400	+0,400	0	+0,400	0
0–3	0	+0,377	−0,966	−0,966	0	0	−0,966	+0,589	+0,966	+1,932	+0,966	0	0	−0,966
2–5	0	0	0	0	0	−0,417	0	0	0	−1,067	0	+0,417	0	+1,067
1–2	+2,844	+2,844	0	−1,111	−1,733	−2,844	−5,689	0	0	+2,844	+2,844	0	0	−2,844
3–4	0	0	0	0	0	0	+3,657	0	+1,429	0	−3,657	0	−1,429	0
0–1	−0,651	−0,651	+0,651	+0,651	+0,651	+0,651	+1,302	−0,651	−0,651	−1,302	−0,651	0	0	+0,651

Table X.

elem.	coeff.	$(\alpha\chi)$	$(\alpha\beta)$	$(\alpha\gamma)$	$(\alpha\delta)$	$(\alpha\varepsilon)$	$(\alpha\eta)$	$(\alpha\xi)$	$(\alpha\lambda)$	$(\alpha\mu)$	$(\alpha\nu)$	$(\alpha o)$	$(\alpha\pi)$	$(\alpha\rho)$
0—2		+0,100	+0,100	0	−0,100	0	−0,100	0	+0,100	+0,200	+0,100	0	0	−0,100
2—4		0	0	0	0	0	0	0	0	0	0	0	0	0
1—3		0	0	0	0	0	0	0	0	0	0	0	0	0
3—5		0	0	0	0	0	0	0	0	0	0	0	0	0
0—3		0	0	0	0	0	0	0	0	0	0	0	0	0
2—5		0	0	0	0	0	0	0	0	0	0	0	0	0
1—2		+0,910	+0,910	0	−0,356	−0,555	−0,910	−1,820	0	+0,910	+0,910	0	0	−0,910
3—4		0	0	0	0	0	0	0	0	0	0	0	0	0
0—1		+0,163	+0,163	−0,163	−0,163	−0,163	−0,163	−0,326	+0,163	+0,326	+0,163	0	0	−0,163
$\Sigma$		+1,173	+1,173	−0,163	−0,619	−0,718	−1,073	−2,246	+0,263	+1,436	+1,173	0	0	−1,173

elem.	coeff.	$(\beta\chi)$	$(\beta\beta)$	$(\beta\gamma)$	$(\beta\delta)$	$(\beta\varepsilon)$	$(\beta\eta)$	$(\beta\xi)$	$(\beta\lambda)$	$(\beta\mu)$	$(\beta\nu)$	$(\beta o)$	$(\beta\pi)$	$(\beta\rho)$
0—2		+0,100	+0,100	0	−0,100	0	−0,100	0	+0,100	+0,200	+0,100	0	0	−0,100
2—4		0	0	0	0	0	0	0	0	0	0	0	0	0
1—3		+0,100	−0,100	−0,100	−0,100	0	−0,100	−0,200	0	+0,100	+0,100	0	0	−0,100
3—5		0	0	0	0	0	0	0	0	0	0	0	0	0
0—3		+0,047	−0,121	−0,121	−0,121	0	−0,121	0	+0,121	+0,242	+0,121	0	0	−0,121
2—5		0	0	0	0	0	0	0	0	0	0	0	0	0
1—2		+0,910	0	0	−0,356	−0,555	−0,910	−1,820	0	+0,910	+0,910	0	0	−0,910
3—4		0	0	0	0	0	0	0	0	0	0	0	0	0
0—1		+0,163	−0,163	−0,163	−0,163	−0,163	−0,163	−0,326	+0,163	+0,326	+0,173	0	0	−0,163
$\Sigma$		+1,320	−0,384	−0,384	−0,840	−0,718	−1,173	−2,567	+0,384	+1,778	+1,394	0	0	−1,394

Table XI.

\(M_1\)	\(M_2\)	\(M_3\)	\(M_4\)	\(M_5\)	\(M_6\)	\(M_7\)	\(M_8\)	\(M_9\)	\(M_{10}\)	\(M_{11}\)	\(M_{12}\)	\(M_{13}\)	\(M_{14}\)
\(+1,17\)	\(+1,17\)	\(-0,16\)	\(-0,62\)	\(-0,72\)	\(-1,07\)	\(-2,25\)	\(+0,16\)	\(+0,26\)	\(+1,44\)	\(+1,17\)	\(0\)	\(0\)	\(-1,17\)
\(+1,17\)	\(+1,32\)	\(-0,38\)	\(-0,84\)	\(-0,72\)	\(-1,17\)	\(-2,57\)	\(+0,24\)	\(+0,38\)	\(+1,78\)	\(+1,39\)	\(0\)	\(0\)	\(-1,39\)
\(-0,16\)	\(-0,38\)	\(+0,57\)	\(+0,57\)	\(+0,16\)	\(+0,26\)	\(+0,84\)	\(-0,35\)	\(-0,47\)	\(-1,04\)	\(-0,57\)	\(0\)	\(0\)	\(+0,57\)
\(-0,62\)	\(-0,84\)	\(+0,57\)	\(+0,81\)	\(+0,38\)	\(+0,62\)	\(+1,65\)	\(-0,35\)	\(-0,57\)	\(-1,60\)	\(-1,03\)	\(0\)	\(0\)	\(+1,03\)
\(-0,72\)	\(-0,72\)	\(+0,16\)	\(+0,38\)	\(+0,50\)	\(+0,72\)	\(+1,44\)	\(-0,16\)	\(-0,16\)	\(-0,88\)	\(-0,72\)	\(0\)	\(0\)	\(+0,72\)
\(-1,07\)	\(-1,17\)	\(+0,26\)	\(+0,62\)	\(+0,72\)	\(+1,31\)	\(+2,35\)	\(-0,16\)	\(-0,16\)	\(-1,12\)	\(-1,09\)	\(-0,05\)	\(+0,08\)	\(+1,04\)
\(-2,25\)	\(-2,57\)	\(+0,84\)	\(+1,65\)	\(+1,44\)	\(+2,35\)	\(+6,35\)	\(-0,51\)	\(-0,20\)	\(-3,49\)	\(-3,93\)	\(+0,08\)	\(-0,46\)	\(+2,84\)
\(+0,16\)	\(+0,24\)	\(-0,35\)	\(-0,35\)	\(-0,16\)	\(-0,16\)	\(-0,51\)	\(+0,29\)	\(+0,35\)	\(+0,70\)	\(+0,35\)	\(0\)	\(0\)	\(-0,35\)
\(+0,26\)	\(+0,38\)	\(-0,47\)	\(-0,57\)	\(-0,16\)	\(-0,16\)	\(-0,20\)	\(+0,35\)	\(+0,83\)	\(+1,14\)	\(+0,12\)	\(+0,08\)	\(-0,18\)	\(-0,49\)
\(+1,44\)	\(+1,78\)	\(-1,04\)	\(-1,60\)	\(-0,88\)	\(-1,12\)	\(-3,49\)	\(+0,70\)	\(+1,14\)	\(+3,72\)	\(+2,23\)	\(-0,13\)	\(+0,08\)	\(-2,50\)
\(+1,17\)	\(+1,39\)	\(-0,57\)	\(-1,03\)	\(-0,72\)	\(-1,09\)	\(-3,93\)	\(+0,35\)	\(+0,12\)	\(+2,83\)	\(+1,58\)	\(0\)	\(+0,54\)	\(-1,58\)
\(0\)	\(0\)	\(0\)	\(0\)	\(0\)	\(-0,05\)	\(+0,08\)	\(0\)	\(+0,08\)	\(-0,13\)	\(0\)	\(+0,13\)	\(+0,26\)	\(+0,21\)
\(0\)	\(0\)	\(0\)	\(0\)	\(0\)	\(+0,08\)	\(-0,46\)	\(0\)	\(-0,18\)	\(+0,08\)	\(+0,54\)	\(+0,26\)	\(0\)	\(0\)
\(-1,17\)	\(-1,39\)	\(+0,57\)	\(+1,03\)	\(+0,72\)	\(+1,04\)	\(+2,84\)	\(-0,35\)	\(-0,49\)	\(-2,50\)	\(-1,58\)	\(+0,21\)	\(0\)	\(+2,00\)

Table XII.

$M_1$	$M_2$	$M_3$	$M_4$	$M_5$	$M_6$	$M_7$	$M_8$	$M_9$	$M_{10}$	$M_{11}$	$M_{12}$	$M_{13}$	$M_{14}$
$-775,42$	$+180,18$	$-297,62$	$0$	$+90,09$	$+90,09$	$+90,09$	$0$	$0$	$0$	$0$	$0$	$0$	$0$
$+180,18$	$+607,54$	$0$	$0$	$+90,09$	$+90,09$	$+90,09$	$+213,68$	$0$	$0$	$0$	$0$	$0$	$0$
$-297,62$	$0$	$+775,42$	$+180,18$	$0$	$0$	$0$	$+90,09$	$+90,09$	$+90,09$	$0$	$0$	$0$	$0$
$0$	$0$	$+180,18$	$+689,08$	$+254,45$	$0$	$0$	$+90,09$	$+90,09$	$+90,09$	$0$	$0$	$0$	$0$
$+90,09$	$+90,09$	$0$	$+254,45$	$+689,08$	$+180,18$	$+180,18$	$0$	$0$	$0$	$0$	$0$	$0$	$0$
$+90,09$	$+90,09$	$0$	$0$	$+180,18$	$-809,12$	$+180,18$	$0$	$0$	$0$	$0$	$+314,47$	$0$	$0$
$+90,09$	$+90,09$	$0$	$0$	$+180,18$	$+180,18$	$+328,32$	$0$	$0$	$0$	$+74,07$	$0$	$0$	$0$
$0$	$+213,68$	$+90,09$	$+90,09$	$0$	$0$	$0$	$+607,54$	$+180,18$	$+180,18$	$0$	$0$	$0$	$0$
$0$	$0$	$+90,09$	$+90,09$	$0$	$0$	$0$	$+180,18$	$+1370,66$	$+180,18$	$0$	$0$	$+595,24$	$0$
$0$	$0$	$+90,09$	$+90,09$	$0$	$0$	$0$	$+180,18$	$+180,18$	$+328,32$	$0$	$0$	$0$	$+74,07$
$0$	$0$	$0$	$0$	$0$	$0$	$+74,07$	$0$	$0$	$0$	$+148,14$	$0$	$0$	$0$
$0$	$0$	$0$	$0$	$0$	$+314,47$	$0$	$0$	$0$	$0$	$0$	$+628,94$	$0$	$0$
$0$	$0$	$0$	$0$	$0$	$0$	$0$	$0$	$+595,24$	$0$	$0$	$0$	$+1190,48$	$0$
$0$	$0$	$0$	$0$	$0$	$0$	$0$	$0$	$0$	$+74,07$	$0$	$0$	$0$	$+148,14$

Table XIII.

Element	\((S^0_\alpha)\)	\((S^0_\beta)\)	\((S^0_\gamma)\)	\((S^0_\delta)\)	\((S^0_\varepsilon)\)	\((S^0_\eta)\)	\((S^0_\vartheta)\)	\(\lambda \times 10^4\) m.
0—2	−20,00	−20,00	0	+20,00	0	0	+20,00	+10,00
2—4	0	0	0	0	0	0	−16,00	+8,000
1—3	0	0	0	0	0	0	0	0
3—5	0	0	0	0	0	−16,00	0	−8,000
0—3	0	−24,15	+61,88	+61,83	0	0	+61,83	−19,321
2—5	0	0	0	0	0	0	0	0
1—2	0	0	0	0	0	0	0	0
3—4	0	0	0	0	0	0	0	0
0—1	0	0	0	0	0	0	0	0
\(\Sigma\)	−20,00	−44,15	+61,83	+81,83	0	−16,00	+65,83

Element	\((S^0_\kappa)\)	\((S^0_\lambda)\)	\((S^0_\mu)\)	\((S^0_\nu)\)	\((S^0_o)\)	\((S^0_\pi)\)	\((S^0_\rho)\)	\(\lambda \times 10^4\) m.
0—2	0	−20,00	−40,00	−20,00	0	0	+20,00	+10,000
2—4	0	−16,00	0	0	−16,00	0	−16,00	+8,000
1—3	0	0	0	0	0	0	0	0
3—5	0	0	−16,00	−16,00	0	−16,00	0	−8,000
0—3	−37,68	−61,83	−123,65	−61,83	0	0	+61,83	−19,321
2—5	0	0	0	0	0	0	0	0
1—2	0	0	0	0	0	0	0	0
3—4	0	0	0	0	0	0	0	0
0—1	0	0	0	0	0	0	0	0
\(\Sigma\)	−37,68	−97,83	−179,65	−97,83	−16,00	−16,00	+65,83

Multiplying the coefficients of Table VIII by the corresponding quantities

\[ \Delta \lambda = \frac{l}{E\omega} \times 10^5, \]

expressed in \(\dfrac{\mathrm{m}}{\mathrm{tn}}\), we shall compile Table IX of quantities of the form \(a\,\Delta \lambda \times 10^5\), etc.

Now the sums denoted by symbols are easily obtained by multiplying Table IX successively by the vertical column of Table VIII corresponding to each moment, and by summing the products. For clarity, in Table X the course of the computations is shown in detail for the first two sums, while for the others only the final values are given (expressed in \(\frac{1}{\text{tn.}\times\text{m.}}\) and increased \(10^3\) times) in Table XI.

The latter, therefore, contains the coefficients of the unknown moments, representing the result of expanding the sums of equations \((e)\) denoted by symbols.

Then in Table XII are placed the coefficients of the unknowns obtained by expanding the parentheses with the multipliers \(\Delta\gamma=\dfrac{l}{6EI}\times 10^3\);

(they are likewise expressed in \(\frac{1}{\text{tn.}\times\text{m.}}\) and increased \(10^3\) times).

It remains further to compute the quantities denoted by the symbols \((S^0\alpha)\), \((S^0\beta)\), etc.

Let us recall their meaning:

\[ \left(S^0\alpha\right)=\sum S^0\Delta\lambda\cdot\alpha =\sum \frac{S^0l}{E\omega}\cdot\alpha =\sum \lambda\cdot\alpha\quad\text{etc.} \]

We take the values \(\alpha, \beta\ldots\) from Table VIII, and \(\lambda\) from the last column of Table VI.

The products \(\alpha\lambda, \beta\lambda\ldots\) and their sums are given in Table XIII, increased \(10^3\) times. We now have all the quantities entering into equations \((e)\). We add the coefficients corresponding to identical \(M\)’s and place them in Table XVI.

In the column denoted by \(A\), we place the quantities \((S^0\alpha)\), \((S^0\beta)\), etc., with opposite signs, since, for solving, the free terms are transferred to the right-hand sides of equations \((e)\). The penultimate column of Table XVI contains (it is marked by the letter \(\Sigma\)) the so-called control sums, serving to verify the correctness of the computations during the solution of the equations.

To obtain these sums, the quantities of all coefficients in the given equation, including the free term, are added algebraically.

The significance of these sums will become clear when we set forth the shortened method of solving equations that we have applied, to which we shall now proceed.

Table XIV.

$M_1$	$M_2$	$M_3$	$M_4$	$M_5$	$M_6$	$M_7$	$M_8$	$M_9$
+ 776,59	+ 181,35	− 297,78	− 0,62	+ 89,37	+ 89,02	+ 87,84	+ 0,16	+ 0,26
+ 181,35	+ 608,86	− 0,38	− 0,84	+ 89,37	+ 88,92	+ 87,52	+ 213,92	+ 0,38
− 297,78	− 0,38	+ 775,99	+ 180,75	+ 0,16	+ 0,26	+ 0,84	+ 89,74	+ 89,62
− 0,62	− 0,84	+ 180,75	+ 689,89	+ 254,83	+ 0,62	+ 1,65	+ 89,74	+ 89,52
+ 89,37	+ 89,37	+ 0,16	+ 254,83	+ 689,58	+ 180,90	+ 181,62	− 0,16	− 0,16
+ 89,02	+ 88,92	+ 0,26	+ 0,62	+ 180,90	+ 810,43	+ 182,53	− 0,16	− 0,16
+ 87,84	+ 87,52	+ 0,84	+ 1,65	+ 181,62	+ 182,53	+ 334,67	− 0,51	− 0,20
+ 0,16	+ 213,92	+ 89,74	+ 89,74	− 0,16	− 0,16	− 0,51	+ 607,83	+ 180,53
+ 0,26	+ 0,38	+ 89,62	+ 89,52	− 0,16	− 0,16	− 0,20	+ 180,53	+ 1371,49
+ 1,44	+ 1,78	+ 89,05	+ 88,49	− 0,88	− 1,12	− 3,49	+ 180,88	+ 181,32
+ 1,17	+ 1,39	− 0,57	− 1,03	− 0,72	− 1,09	+ 70,14	+ 0,35	+ 0,12
0	0	0	0	0	+ 314,42	+ 0,08	0	+ 0,08
0	0	0	0	0	+ 0,08	− 0,46	0	+ 595,06
− 1,17	− 1,39	+ 0,57	+ 1,03	+ 0,72	+ 1,04	+ 2,84	− 0,35	− 0,49

$M_{10}$	$M_{11}$	$M_1$	$M_{13}$	$M_{14}$	$A$	$\Sigma$
+ 1,44	+ 1,17	0	0	− 1,17	+ 20,00	+ 947,63	1
+ 1,78	+ 1,39	0	0	− 1,39	+ 44,15	+ 1315,03	2
+ 89,05	− 0,57	0	0	+ 0,57	− 61,83	+ 866,42	3
+ 88,49	− 1,03	0	0	+ 1,03	− 81,83	+ 1312,20	4
− 0,88	− 0,72	0	0	+ 0,72	0	+ 1484,63	5
− 1,12	− 1,09	+ 314,42	+ 0,08	+ 1,04	+ 16,00	+ 1681,69	6
− 3,49	+ 70,14	+ 0,08	− 0,46	+ 2,84	− 65,83	+ 879,24	7
+ 180,88	+ 0,35	0	0	− 0,35	+ 37,68	+ 1399,65	8
+ 181,32	+ 0,12	+ 0,08	+ 595,06	− 0,49	+ 97,83	+ 2605,20	9
+ 332,04	+ 2,23	− 0,13	+ 0,08	+ 71,57	+ 179,65	+ 1122,91	10
+ 2,23	+ 150,97	0	+ 0,54	− 1,58	+ 97,83	+ 319,75	11
− 0,13	0	+ 629,07	0	+ 0,21	+ 16,00	+ 959,73	12
+ 0,08	+ 0,54	0	+ 1190,74	0	+ 16,00	+ 1802,04	13
+ 71,57	− 1,58	+ 0,21	0	+ 150,14	− 65,83	+ 157,31	14

16. An Abbreviated Method for Solving Equations*)

Let us first examine the question in an abstract form, with literal coefficients. For definiteness of exposition, let us take a system of equations with six unknowns; for a greater or smaller number of unknowns it is easy to expand or shorten the proposed scheme of solution.

We shall write the equations with symbolic notation for the coefficients:

\[ \begin{aligned} &(aa)M_1+(ab)M_2+(ac)M_3+(ad)M_4+(ae)M_5\\ &\qquad +(af)M_6=(am) \qquad (an),\\[6pt] &(ab)M_1+(bb)M_2+(bc)M_3+(bd)M_4+(be)M_5\\ &\qquad +(bf)M_6=(bm) \qquad (bn),\\[6pt] &(ac)M_1+(bc)M_2+(cc)M_3+(cd)M_4+(ce)M_5\\ &\qquad +(cf)M_6=(cm) \qquad (cn),\\[6pt] &(ad)M_1+(bd)M_2+(cd)M_3+(dd)M_4+(de)M_5\\ &\qquad +(df)M_6=(dm) \qquad (dn),\\[6pt] &(ae)M_1+(be)M_2+(ce)M_3+(de)M_4+(ee)M_5\\ &\qquad +(ef)M_6=(em) \qquad (en),\\[6pt] &(af)M_1+(bf)M_2+(cf)M_3+(df)M_4+(ef)M_5\\ &\qquad +(ff)M_6=(fm) \qquad (fn). \end{aligned} \tag{f} \]

By the symbols \((an), (bn)\ldots\) are denoted the check sums.

*) The idea of this method belongs to Gauss; in its exposition we follow Oppolzer’s excellent book (see the list of literature).

We shall begin solving the equations by eliminating the unknown \(M_1\); as we indicated in Chapter II, the largest coefficient at \(M_1\) will be on the diagonal, i.e., in the first equation, whence we shall determine it, so as to reduce the error as much as possible; in all computations we shall carry out the same operations with the control sums as with the other coefficients.

\(M_1\) is determined as follows:

\[ M_1 = -\frac{(ab)}{(aa)}M_2 -\frac{(ac)}{(aa)}M_3 -\frac{(ad)}{(aa)}M_4 -\frac{(ae)}{(aa)}M_5 -\frac{(af)}{(aa)}M_6 \]

\[ +\frac{(am)}{(aa)},\qquad -\frac{(an)}{(aa)}. \]

Substitute this value \(M_1\) into the remaining five equations and collect the coefficients of like \(M\). We obtain

\[ \left[(bb)-\frac{(ab)}{(aa)}(ab)\right]M_2 + \left[(bc)-\frac{(ab)}{(aa)}(ac)\right]M_3 + \left[(bd)-\frac{(ab)}{(aa)}(ad)\right]M_4 \]

\[ + \left[(be)-\frac{(ab)}{(aa)}(ae)\right]M_5 + \left[(bf)-\frac{(ab)}{(aa)}(af)\right]M_6 = \left[(bm)-\frac{(ab)}{(aa)}(am)\right], \]

\[ \left[(bn)-\frac{(ab)}{(aa)}(an)\right], \]

\[ \left[(bc)-\frac{(ac)}{(aa)}(ab)\right]M_2 + \left[(cc)-\frac{(ac)}{(aa)}(ac)\right]M_3 + \left[(cd)-\frac{(ac)}{(aa)}(ad)\right]M_4 \]

\[ + \left[(ce)-\frac{(ac)}{(aa)}(ae)\right]M_5 + \left[(cf)-\frac{(ac)}{(aa)}(af)\right]M_6 = \left[(cm)-\frac{(ac)}{(aa)}(am)\right], \]

\[ \left[(cn)-\frac{(ac)}{(aa)}(an)\right], \]

\[ \left[(bd)-\frac{(ad)}{(aa)}(ab)\right]M_2+ \left[(cd)-\frac{(ad)}{(aa)}(ac)\right]M_3+ \left[(dd)-\frac{(ad)}{(aa)}(ad)\right]M_4 \]

\[ +\left[(de)-\frac{(ad)}{(aa)}(ae)\right]M_5+ \left[(df)-\frac{(ad)}{(aa)}(af)\right]M_6 = \left[(dm)-\frac{(ad)}{(aa)}(am)\right], \]

\[ \left[(dn)-\frac{(ad)}{(aa)}(an)\right], \]

\[ \left[(be)-\frac{(ae)}{(aa)}(ab)\right]M_2+ \left[(ce)-\frac{(ae)}{(aa)}(ac)\right]M_3+ \left[(de)-\frac{(ae)}{(aa)}(ad)\right]M_4 \]

\[ +\left[(ee)-\frac{(ae)}{(aa)}(ae)\right]M_5+ \left[(ef)-\frac{(ae)}{(aa)}(af)\right]M_6 = \left[(em)-\frac{(ae)}{(aa)}(am)\right], \]

\[ \left[(en)-\frac{(ae)}{(aa)}(an)\right], \]

\[ \left[(bf)-\frac{(af)}{(aa)}(ab)\right]M_2+ \left[(cf)-\frac{(af)}{(aa)}(ac)\right]M_3+ \left[(df)-\frac{(af)}{(aa)}(ad)\right]M_4 \]

\[ +\left[(ef)-\frac{(af)}{(aa)}(ae)\right]M_5+ \left[(ff)-\frac{(af)}{(aa)}(af)\right]M_6 = \left[(fm)-\frac{(af)}{(aa)}(am)\right], \]

\[ \left[(fn)-\frac{(af)}{(aa)}(an)\right]. \]

From consideration of these new equations it is easy to derive two consequences:

1) the new coefficients of the unknowns are likewise arranged symmetrically with respect to the diagonal;

2) the sum of all the coefficients of each new equation is equal to the quantity obtained by transforming the check sum of the old equation in parallel with the other coefficients.

Indeed, take, for example, the value of the transformed sum of the first of the new equations and substitute into it, in place of \((an)\) and \((bn)\), their expressions:

\[ \left[(bn)-\frac{(ab)}{(aa)}(an)\right] = \left[(ab)+(bb)+(bc)+(bd)+(be)+(bf)\right. \]

\[ \left. +(bn)-\frac{(ab)}{(aa)} \left\{(aa)+(ab)+(ac)+(ad)+(ae)+(af)+(an)\right\} \right]; \]

obviously, this quantity expresses the sum of all the coefficients of the new equation.

Let us introduce new symbolic designations

\[ \left[(bb)-\frac{(ab)}{(aa)}(ab)\right]=(bb_1);\quad \left[(bc)-\frac{(ab)}{(aa)}(bc)\right]=(bc_1) \quad \text{and so on,} \]

and rewrite the transformed system in the following form:

\[ \begin{aligned} &(bb_1)M_2+(bc_1)M_3+(bd_1)M_4+(be_1)M_5+(bf_1)M_6=(bm_1),\ (bn_1),\\ &(bc_1)M_2+(cc_1)M_3+(cd_1)M_4+(ce_1)M_5+(cf_1)M_6=(cm_1),\ (cn_1),\\ &(bd_1)M_2+(cd_1)M_3+(dd_1)M_4+(de_1)M_5+(df_1)M_6=(dm_1),\ (dn_1),\\ &(be_1)M_2+(ce_1)M_3+(de_1)M_4+(ee_1)M_5+(ef_1)M_6=(em_1),\ (en_1),\\ &(bf_1)M_2+(cf_1)M_3+(df_1)M_4+(ef_1)M_5+(ff_1)M_6=(fm_1),\ (fn_1). \end{aligned} \tag{g} \]

From the first equation of the preceding system we determine

\[ M_2=-\frac{(bc_1)}{(bb_1)}M_3-\frac{(bd_1)}{(bb_1)}M_4-\frac{(be_1)}{(bb_1)}M_5 -\frac{(bf_1)}{(bb_1)}M_6+\frac{(bm_1)}{(bb_1)},\quad -\frac{(bn_1)}{(bb_1)}. \]

Substituting into the remaining equations and introducing the new symbols \((cc_2)\), \((cd_2)\ldots\), we obtain

\[ \begin{aligned} &(cc_2)M_3+(cd_2)M_4+(ce_2)M_5+(cf_2)M_6=(cm_2),\ (cn_2),\\ &(cd_2)M_3+(dd_2)M_4+(de_2)M_5+(df_2)M_6=(dm_2),\ (dn_2),\\ &(ce_2)M_3+(de_2)M_4+(ee_2)M_5+(ef_2)M_6=(em_2),\ (en_2),\\ &(cf_2)M_3+(df_2)M_4+(ef_2)M_5+(ff_2)M_6=(fm_2),\ (fn_2). \end{aligned} \tag{h} \]

We determine

\[ M_3=-\frac{(cd_2)}{(cc_2)}M_4-\frac{(ce_2)}{(cc_2)}M_5-\frac{(cf_2)}{(cc_2)}M_6+\frac{(cm_2)}{(cc_2)},\ -\frac{(cn_2)}{(cc_2)}, \]

and, after substitution into the equations, we have

\[ \tag{i} \begin{aligned} (dd_3)M_4+(de_3)M_5+(df_3)M_6&=(dm_3),\ (dn_3),\\ (de_3)M_4+(ee_3)M_5+(ef_3)M_6&=(em_3),\ (en_3),\\ (df_3)M_4+(ef_3)M_5+(ff_3)M_6&=(fm_3),\ (fn_3). \end{aligned} \]

Further,

\[ M_4=-\frac{(de_3)}{(dd_3)}M_5-\frac{(df_3)}{(dd_3)}M_6+\frac{(dm_3)}{(dd_3)},\ -\frac{(dn_3)}{(dd_3)}. \]

Hence

\[ \tag{k} \begin{aligned} (ee_4)M_5+(ef_4)M_6&=(em_4),\ (en_4),\\ (ef_4)M_5+(ff_4)M_6&=(fm_4),\ (fn_4). \end{aligned} \]

From the first equation of system \((k)\) we find

\[ M_5=-\frac{(ef_4)}{(ee_4)}M_6+\frac{(em_4)}{(ee_4)},\ -\frac{(en_4)}{(ee_4)}, \]

and after substitution we have the equation

\[ \tag{l} (ff_5)M_6=(fm_5),\ (fn_5), \]

from which, finally, one obtains

\[ M_6=+\frac{(fm_5)}{(ff_5)}. \]

After this the remaining unknowns are easily found by back substitution.

In solving such a system with numerical coefficients, an enormous reduction in the work is achieved by using the logarithmic scheme presented in Table XV.

In the upper row of the table all coefficients of the first equation are written down, and beneath them their logarithms; then the coefficients of the second equation are entered below, except for the first coefficient, which is the same as the second coefficient of the first equation. In the first column on the left we place \(\lg \dfrac{(ab)}{(aa)}\)—the logarithm of the quantity by which, as we saw above, all coefficients (beginning with the second) of the first equation must be multiplied, and the products subtracted from the corresponding coefficients of the second equation.

Therefore we write \(\lg \dfrac{(ab)}{(aa)}\) on the lower edge of a paper strip and, holding it successively above each logarithm of the second row, obtain by addition the logarithms of the aforementioned products.

By entering the values of these products beneath the corresponding coefficients of the second equation and carrying out the subtraction, we find the coefficients of a new equation from which \(M_1\) has been eliminated; beneath them we again write their logarithms.

Then we write out the coefficients (except for the first two) of the third equation, with which we perform the operation just described twice, and so on. The course of the computation is clear from the scheme; let us note that the final equations obtained from each principal one and marked by us on the right with the letter \(E\) may be called elimination equations (Eliminationsgleichungen); with their aid the back-substitution of the unknowns is carried out. Verification of the computations is easily achieved by calculating the check sums; it is only necessary to compute with particular care the logarithms placed in the first column, since an error in them easily escapes the check.

With some practice in computation, for checking one may confine oneself merely to calculating the sums marked by \(!\).

The scheme for computing the unknowns after completion of the elimination process is shown in Table XVI and requires no explanation.

Experience has shown that, for computations with entirely sufficient accuracy, five-place logarithms may be used. Tables XVII and XVIII give a numerical example of solving a system of six equations by means of the described scheme. In the last column of Table XVII the check sums are computed, by which one may judge the degree of accuracy of the check.

Table XV.

$M_1$	$M_2$	$M_3$	$M_4$	$M_5$	$M_6$	$A$	$\Sigma$
$(aa)$ $\lg(aa)$	$(ab)$ $\lg(ab)$	$(ac)$ $\lg(ac)$	$(ad)$ $\lg(ad)$	$ae$ $\lg(ac)$	$af$ $\lg(af)$	$am$ $\lg(am)$	$an$ $\lg(an)$	E
$\lg\dfrac{(ab)}{(aa)}$	$(bb)$ $\dfrac{(ab)}{(aa)}(ab)$	$(bc)$ $\dfrac{(ab)}{(aa)}(ac)$	$(bd)$ $\dfrac{(ab)}{(aa)}(ad)$	$(be)$ $\dfrac{(ab)}{(aa)}(ae)$	$(bf)$ $\dfrac{(ab)}{(aa)}(af)$	$(bm)$ $\dfrac{(ab)}{(aa)}(am)$	$(bn)$ $\dfrac{(ab)}{(aa)}(an)$
$\lg\dfrac{(ab)}{(aa)}$	$(bb_1)$ $\lg(bb_1)$	$(bc_1)$ $\lg(bc_1)$	$(bd_1)$ $\lg(bd_1)$	$(be_1)$ $\lg(be_1)$	$(bf_1)$ $\lg(bf_1)$	$(bm_1)$ $\lg(bm_1)$	$(bn_1)$ $\lg(bn_1)$	$1\ \mathrm{E}!$
$\lg\dfrac{(ac)}{(aa)}$		$(cc)$ $\dfrac{(ac)}{(aa)}(ac)$	$(cd)$ $\dfrac{(ac)}{(aa)}(ad)$	$(ce)$ $\dfrac{(ac)}{(aa)}(ae)$	$(cf)$ $\dfrac{(ac)}{(aa)}(af)$	$(cm)$ $\dfrac{(ac)}{(aa)}(am)$	$(cn)$ $\dfrac{(ac)}{(aa)}(an)$
$\dfrac{(bc_1)}{(bb_1)}$		$(cc_1)$ $\dfrac{(bc_1)}{(bb_1)}(bc_1)$	$(cd_1)$ $\dfrac{(bc_1)}{(bb_1)}(bd_1)$	$(ce_1)$ $\dfrac{(bc_1)}{(bb_1)}(be_1)$	$(cf_1)$ $\dfrac{(bc_1)}{(bb_1)}(bf_1)$	$(cm_1)$ $\dfrac{(bc_1)}{(bb_1)}(bm_1)$	$(cn_1)$ $\dfrac{(bc_1)}{(bb_1)}(bn_1)$	2
		$(cc_2)$ $\lg(cc_2)$	$(cd_2)$ $\lg(cd_2)$	$(ce_2)$ $\lg(be_2)$	$(cf_2)$ $\lg(cf_2)$	$(cm_2)$ $\lg(cm_2)$	$(cn_2)$ $\lg(cn_2)$	$3\ \mathrm{E}!$
$\lg\dfrac{(ad)}{(aa)}$			$(dd)$ $\dfrac{(ad)}{(aa)}(ad)$	$(de)$ $\dfrac{(ad)}{(aa)}(ae)$	$(df)$ $\dfrac{(ad)}{(aa)}(af)$	$(dm)$ $\dfrac{(ad)}{(aa)}(am)$	$(dn)$ $\dfrac{(ad)}{(aa)}(an)$
$\lg\dfrac{(bd_1)}{(bb_1)}$			$(dd_1)$ $\dfrac{(bd_1)}{(bb_1)}(bd_1)$	$(de_1)$ $\dfrac{(bd_1)}{(bb_1)}(be_1)$	$(df_1)$ $\dfrac{(bd_1)}{(bb_1)}(bf_1)$	$(dm_1)$ $\dfrac{(bd_1)}{(bb_1)}(bm_1)$	$(dn_1)$ $\dfrac{(bd_1)}{(bb_1)}(bn_1)$	4
$\lg\dfrac{(cd_2)}{(cc_2)}$			$(dd_2)$ $\dfrac{(cd_2)}{(cc_2)}(cd_2)$	$(de_2)$ $\dfrac{(cd_2)}{(cc_2)}(ce_2)$	$(df_2)$ $\dfrac{(cd_2)}{(cc_2)}(cf_2)$	$(dm_2)$ $\dfrac{(cd_2)}{(cc_2)}(cm_2)$	$(dn_2)$ $\dfrac{(cd_2)}{(cc_2)}(cn_2)$	5

Table XV (continued).

$M_1$	$M_2$	$M_3$	$M_4$	$M_5$	$M_6$	$A$	$\Sigma$
			$(dd_3)$ $\lg(dd_3)$	$(de_3)$ $\lg(de_3)$	$(df_3)$ $\lg(df_3)$	$(dm_3)$ $\lg(dm_3)$	$(dn_3)$ $\lg(dn_3)$	$6\,E!$
$\lg \dfrac{(ae)}{(aa)}$				$(ee)$ $\dfrac{(ae)}{(aa)}(ae)$	$(ef)$ $\dfrac{(ae)}{(aa)}(af)$	$(em)$ $\dfrac{(ae)}{(aa)}(am)$	$(en)$ $\dfrac{(ae)}{(aa)}(an)$
$\lg \dfrac{(be_1)}{(bb_1)}$				$ee_1$ $\dfrac{(be_1)}{(bb_1)}(be_1)$	$ef_1$ $\dfrac{(be_1)}{(bb_1)}(bf_1)$	$em_1$ $\dfrac{(be_1)}{(bb_1)}(bm_1)$	$en_1$ $\dfrac{(be_1)}{(bb_1)}(bn_1)$	$7$
$\lg \dfrac{(ce_2)}{(cc_2)}$				$(ee_2)$ $\dfrac{(ce_2)}{(cc_2)}(ce_2)$	$(ef_2)$ $\dfrac{(ce_2)}{(cc_2)}(cf_2)$	$(em_2)$ $\dfrac{(ce_2)}{(cc_2)}(cm_2)$	$(en_2)$ $\dfrac{(ce_2)}{(cc_2)}(cn_2)$	$8$
$\lg \dfrac{(de_3)}{(dd_3)}$				$(ee_3)$ $\dfrac{(de_3)}{(dd_3)}(de_3)$	$(ef_3)$ $\dfrac{(de_3)}{(dd_3)}(df_3)$	$(em_3)$ $\dfrac{(de_3)}{(dd_3)}(dm_3)$	$(en_3)$ $\dfrac{(de_3)}{(dd_3)}(dn_3)$	$9$
				$(ee_4)$ $\lg(ee_4)$	$(ef_4)$ $\lg(ef_4)$	$(em_4)$ $\lg(em_4)$	$(en_4)$ $\lg(en_4)$	$10\,E!$
$\lg \dfrac{(af)}{(aa)}$					$(ff)$ $\dfrac{(af)}{(aa)}(af)$	$(fm)$ $\dfrac{(af)}{(aa)}(am)$	$(fn)$ $\dfrac{(af)}{(aa)}(an)$
$\lg \dfrac{(bf_1)}{(bb_1)}$					$(ff_1)$ $\dfrac{(bf_1)}{(bb_1)}(bf_1)$	$(fm_1)$ $\dfrac{(bf_1)}{(bb_1)}(bm_1)$	$(fn_1)$ $\dfrac{(bf_1)}{(bb_1)}bn_1)$	$11$
$\lg \dfrac{(cf_2)}{(cc_2)}$					$(ff_2)$ $\dfrac{(cf_2)}{(cc_2)}(cf_2)$	$(fm_2)$ $\dfrac{(cf_2)}{(cc_2)}(cm_2)$	$(fn_2)$ $\dfrac{(cf_2)}{(cc_2)}cn_2)$	$12$

Table XV (continued).

$\lg$	$M_6$	$A$	$\Sigma$
$\lg \dfrac{(df_3)}{(dd_3)}$	$(ff_3)$ $\dfrac{(df_3)}{(dd_3)}\, (df_3)$	$(fm_3)$ $\dfrac{(df_3)}{(dd_3)}\, (dm_3)$	$(fn_3)$ $\dfrac{(df_3)}{(dd_3)}\, (dn_3)$	$13$
$\lg \dfrac{(ef_4)}{(ee_4)}$	$(ff_4)$ $\dfrac{(ef_4)}{(ee_4)}\, (ef_4)$	$(fm_4)$ $\dfrac{(ef_4)}{(ee_4)}\, (em_4)$	$(fn_4)$ $\dfrac{(ef_4)}{(ee_4)}\, (en_4)$	$14$
	$(ff_5)$ $\lg (ff_5)$	$(fm_5)$ $\lg (fm_5)$	$(fn_5)$	$15\ E!$

Table XVI.

$M_6$	$M_5$	$M_4$	$M_3$	$M_2$	$M_1$
$(fm_5)$	$(em_4)$ $-\,(ef_4)\,M_6$	$(dm_3)$ $-\,(df_3)\,M_6$ $-\,(de_3)\,M_5$	$(cm_2)$ $-\,(cf_2)\,M_6$ $-\,(ce_2)\,M_5$ $-\,(cd_2)\,M_4$	$(bm_1)$ $-\,(bf_1)\,M_6$ $-\,(be_1)\,M_5$ $-\,(bd_1)\,M_4$ $-\,(bc_1)\,M_3$	$(am)$ $-\,(af)\,M_6$ $-\,(ae)\,M_5$ $-\,(ad)\,M_4$ $-\,(ac)\,M_3$ $-\,(ab)\,M_2$	$A$ $M_6$ $M_5$ $M_4$ $M_3$ $M_2$
$(fm_5)$ $\lg (fm_5)$ $\lg (ff_5)$	$\Sigma (ex_4)$ $\lg \Sigma (ex_4)$ $\lg (ee_4)$	$\Sigma (dx_3)$ $\lg \Sigma (dx_3)$ $\lg (dd_3)$	$\Sigma (cx_2)$ $\lg \Sigma (cx_2)$ $\lg (cc_2)$	$\Sigma (bx_1)$ $\lg \Sigma (bx_1)$ $\lg (bb_1)$	$\Sigma (ax)$ $\lg \Sigma (ax)$ $\lg (aa)$	$\Sigma$ $\lg \Sigma$ $\lg D$
$\lg M_6$ $M_6$	$\lg M_5$ $M_5$	$\lg M_4$ $M_4$	$\lg M_3$ $M_3$	$\lg M_2$ $M_2$	$\lg M_1$ $M_1$	$\lg M$ $M$

Check sums.

\[ (bb_1) + (bc_1) + (bd_1) + (be_1) + (bf_1) + (bm_1) = (bn_1) \qquad 1! \]

\[ (bc_1) + (cc_1) + (cd_1) + (ce_1) + (cf_1) + (cm_1) = (cn_1) \qquad 2 \]

\[ (cc_2) + (cd_2) + (ce_2) + (cf_2) + (cm_2) = (cn_2) \qquad 3! \]

\[ (bd_1) + (cd_1) + (dd_1) + (de_1) + (df_1) + (dm_1) = (dn_1) \qquad 4 \]

\[ (cd_2) + (dd_2) + (de_2) + (df_2) + (dm_2) = (dn_2) \qquad 5 \]

\[ (dd_3) + (de_3) + (df_3) + (dm_3) = (dn_3) \qquad 6! \]

\[ (be_1) + (ce_1) + (de_1) + (ee_1) + (ef_1) + (em_1) = (en_1) \qquad 7 \]

\[ (ce_2) + (de_2) + (ee_2) + (ef_2) + (em_2) = (en_2) \]

\[ (de_3) + (ee_3) + (ef_3) + (em_3) = (en_3) \qquad 9 \]

\[ (ee_4) + (ef_4) + (em_4) = (en_4) \qquad 10! \]

\[ (bf_1) + (cf_1) + (df_1) + (ef_1) + (ff_1) + (fm_1) = (fn_1) \qquad 11 \]

\[ (cf_2) + (df_2) + (ef_2) + (ff_2) + (fm_2) = (fn_2) \qquad 12 \]

\[ (df_3) + (ef_3) + (ff_3) + (fm_3) = (fn_3) \qquad 13 \]

\[ (ef_4) + (ff_4) + (fm_4) = (fn_4) \qquad 14 \]

\[ (ff_5) + (fm_5) = (fn_5) \qquad 15! \]

Applying the method just described to the system of equations whose coefficients are given in Table XIV, we determine the values of the unknowns

\[ \begin{aligned} M_1&=-0.0392, & M_8&=-0.2344,\\ M_2&=+0.1975, & M_9&=-0.0063,\\ M_3&=-0.1158, & M_{10}&=+0.9437,\\ M_4&=-0.2466, & M_{11}&=+0.8862,\\ M_5&=+0.1968, & M_{12}&=-0.0234,\\ M_6&=+0.0988, & M_{13}&=+0.0159,\\ M_7&=-0.5662, & M_{14}&=-0.8666. \end{aligned} \]

In order to obtain from the equations of the generalized method the equations for the second group of methods, it is sufficient, as we saw in Chapter II, to set equal to zero the bending moments for the elements of the lattice, i.e., in our case to put

\[ M_2=M_4=M_5=M_6=M_8=M_9=M_{12}=M_{13}=0. \]

Under this condition, discarding the corresponding terms of Table XIV, we obtain a new system

\[ \begin{aligned} &+776.59\,M_1-297.78\,M_3+87.84\,M_7+1.44\,M_{10}+1.17\,M_{11}\\ &\qquad -1.17\,M_{14}=20.00, \end{aligned} \]

\[ \begin{aligned} &-297.78\,M_1+775.99\,M_3+0.84\,M_7+89.05\,M_{10}-0.57\,M_{11}\\ &\qquad +0.57\,M_{14}=-61.83, \end{aligned} \]

\[ \begin{aligned} &+87.84\,M_1+0.84\,M_3+334.67\,M_7-3.49\,M_{10}+70.14\,M_{11}\\ &\qquad +2.84\,M_{14}=-65.83, \end{aligned} \]

\[ \begin{aligned} &+1.44\,M_1+89.05\,M_3-3.49\,M_7+332.04\,M_{10}+2.23\,M_{11}\\ &\qquad +71.57\,M_{14}=179.65, \end{aligned} \]

\[ -\,1.17\,M_1 - 0.57\,M_3 + 70.14\,M_7 + 2.23\,M_{10} + 150.97\,M_{11} \]
\[ -\,1.58\,M_{14} = 97.83, \]

\[ -\,1.17\,M_1 + 0.57\,M_3 + 2.84\,M_7 + 71.57\,M_{10} - 1.58\,M_{11} \]
\[ +\,150.14\,M_{14} = -65.83. \]

The solution of this system has just been given in Tables XVII and XVIII. We have

\[ \begin{aligned} M_1 &= -0.0017\ \text{tn.-m}, &\qquad M_{10} &= +0.7437,\\ M_3 &= -0.1641, & M_{11} &= +0.7895,\\ M_7 &= -0.3470, & M_{14} &= -0.7775. \end{aligned} \]

Let us now pass to the calculation by Mohr’s method. The magnitudes of the angles of rotation of the members are determined by formulas of the form

\[ \psi_{0-1}=\sum \lambda \cdot \sigma_{0-1}\ \text{and so forth}, \]

where the summation extends over all the members of the truss. The quantities \(\lambda\) are taken from Table VI, and \(\sigma\) from Tables V and VI.

Having the angles \(\psi\), it is easy to determine the angles of rotation of the joints \(\varphi\)—from a system of equations of the form (see Chapter II)

\[ 2\varphi_N \sum N_{NX} + \sum N_{NX}\varphi_X = 3\sum N_{NX}\psi_{NX} \]

The sums \(\sum N_{NX}\) and \(\sum N_{NX}\psi_{NX}\) entering into this equation are placed in Table XX for joints 0, 1, 2, 3; in view of the symmetry of the load, the angles of rotation of joints 4 and 5 are equal to zero

\[ \varphi_4=\varphi_5=0. \]

Table XVI. Frame with rigid joints.

\(M_1\)	\(M_2\)	\(M_3\)	\(M_4\)	\(M_5\)	\(M_6\)	\(A\)	\(S\)	\(E\)	\(v\)
\(776.59\) \(2.89019\)	\(297.78\) \(2_{\varphi}\ 47390\)	\(87.84\) \(1.94369\)	\(1.44\) \(0.15336\)	\(1.17\) \(0.06819\)	\(-1.17\) \(0.06819\)	\(20.00\) \(1.30103\)	\(588.09\) \(2.76944\)
\(9_{\varphi}\ 58370\)	\(775.990\) \(114.182\)	\(0.840\) \(-33.682\)	\(89.050\) \(-0.552\)	\(0.570\) \(-0.449\)	\(-0.570\) \(-0.449\)	\(-61.830\) \(-7.669\)	\(-506.270\) \(-225.501\)
	\(661.808\) \(2.82073\)	\(-34.522\) \(1.53810\)	\(-89.602\) \(1.95232\)	\(-0.121\) \(9_{\psi}\ 08279\)	\(0.121\) \(9_{\psi}\ 08279\)	\(54.161\) \(1_{\psi}\ 73369\)	\(731.771\) \(2.86438\)	\(1E'\)	\(731.771\)
\(9.05350\)		\(334.670\) \(1.936\)	\(3.490\) \(0.163\)	\(70.140\) \(-0.132\)	\(-2.840\) \(-0.132\)	\(-65.830\) \(-2.262\)	\(-427.010\) \(66.519\)
\(8.71737\)		\(324.734\) \(1.801\)	\(3.663\) \(4.674\)	\(70.008\) \(-0.006\)	\(-2.972\) \(-0.006\)	\(-68.092\) \(-2.825\)	\(360.491\) \(-38.172\)	\(2\)	\(360.491\)
		\(322.933\) \(2.50911\)	\(3.327\) \(0_{\psi}\ 92049\)	\(70.014\) \(1.84518\)	\(-2.966\) \(0.47212\)	\(-65.267\) \(1_{\psi}\ 81469\)	\(322.319\) \(2.50829\)	\(3E'\)	\(322.319\)
\(7.26817\)			\(332.040\) \(1.0063\)	\(2.230\) \(-0.002\)	\(71.570\) \(0.002\)	\(179.650\) \(-0.037\)	\(672.490\) \(1.090\)
\(9.13159\)			\(332.037\) \(1.2131\)	\(2.228\) \(0.016\)	\(71.570\) \(0.016\)	\(179.613\) \(1.293\)	\(671.400\) \(9.007\)	\(4\)	\(671.399\)
\(8_{\varphi}\ 44138\)			\(319.906\) \(0.215\)	\(2.244\) \(1.805\)	\(71.556\) \(-0.076\)	\(186.946\) \(1.683\)	\(572.323\) \(8.311\)	\(5\)	\(572.325\)
			\(319.691\) \(2.50473\)	\(4.049\) \(0.60735\)	\(71.632\) \(1.85511\)	\(185.263\) \(2.26779\)	\(580.634\) \(2.76390\)	\(6E'\)	\(580.635\)
\(7.17800\)				\(150.970\) \(-0.002\)	\(-1.580\) \(-0.002\)	\(97.830\) \(0.030\)	\(320.190\) \(0.886\)
\(6_{\varphi}\ 26206\)				\(150.968\) \(0\)	\(-1.578\) \(0\)	\(97.800\) \(0.010\)	\(319.304\) \(-0.134\)	\(7\)	\(319.305\)
\(9.33607\)				\(150.968\) \(1.179\)	\(-1.578\) \(0.643\)	\(97.790\) \(14.150\)	\(319.438\) \(6.982\)	\(8\)	\(319.438\)
\(8.10262\)				\(135.789\) \(0.051\)	\(2.221\) \(0.307\)	\(111.940\) \(-2.347\)	\(249.557\) \(7.354\)	\(9\)	\(249.557\)
				\(135.738\) \(2.13270\)	\(2.228\) \(0_{\psi}\ 49527\)	\(109.593\) \(2.03978\)	\(242.203\) \(2.38418\)	\(10E'\)	\(242.203\)
\(7_{\psi}\ 17800\)					\(150.140\) \(0.002\)	\(-65.830\) \(-0.030\)	\(-155.610\) \(-0.886\)
\(6.26206\)					\(150.138\) \(0\)	\(-65.800\) \(-0.010\)	\(156.496\) \(0.134\)	\(11\)	\(157.425\)
\(7.96306\)					\(150.138\) \(-0.027\)	\(-65.790\) \(-0.599\)	\(157.292\) \(2.960\)	\(12\)	\(157.292\)
\(9.35038\)					\(150.111\) \(1.6050\)	\(-65.191\) \(41.512\)	\(154.332\) \(130.101\)	\(13\)	\(154.331\)
\(8_{\psi}\ 36257\)					\(134.061\) \(0.072\)	\(-106.703\) \(-2.526\)	\(-24.231\) \(-5.562\)	\(14\)	\(-24.230\)
					\(133.989\) \(2.12707\)	\(-104.177\) \(2_{\varphi}\ 01777\)	\(-29.813\)	\(15E'\)	\(29.512\)

[[unclear: the table continues/contains faint ruled entries at the lower edge of the visible page; several blank cells and fine dividing lines are present in the source image but contain no legible values.]]

N. V. NEKRASOV.

Table XVIII.

$M_6$	$M_5$	$M_4$	$M_3$	$M_2$	$M_1$	$A$
$-104{,}177$	$+109{,}593$	$+185{,}263$	$-65{,}267$	$-54{,}161$	$+20{,}000$	$A$
	$-2{,}432$	$+55{,}693$	$+2{,}306$	$+0{,}094$	$-0{,}910$	$M_6$
		$-3{,}197$	$-55{,}274$	$+0{,}096$	$-0{,}924$	$M_5$
			$+6{,}193$	$-66{,}640$	$-1{,}071$	$M_4$
				$+11{,}978$	$+30{,}476$	$M_3$
					$-48{,}880$	$M_2$
						$M_1$
$-104{,}177$	$+107{,}161$	$+237{,}759$	$-112{,}042$	$-108{,}633$	$-1{,}309$	$\Sigma$
$2_n\,01777$	$2{,}03004$	$2{,}37614$	$2_n\,04938$	$2_n\,03596$	$0_n\,11694$	$\lg \Sigma$
$2{,}12707$	$2{,}13270$	$2{,}50473$	$2{,}50911$	$2{,}82073$	$2{,}89019$	$\lg D$
$9_n\,89070$	$9{,}89734$	$9{,}87141$	$9_n\,54027$	$9_n\,21523$	$7_n\,22675$	$\lg M$
$-0{,}7775$	$+0{,}7895$	$0{,}7437$	$-0{,}3470$	$-0{,}1641$	$-0{,}0017$	$M$

Table XIX.

Angles $\psi$ are increased $10^5$ times.

Elements	$0—2$	$2—4$	$1—3$	$3—5$	$0—3$	$2—5$	$1—2$	$3—4$	$0—1$	$\lambda \times 10^2$ cm.
$0—2$	$0$	$+20{,}00$	$+20{,}00$	$-20{,}00$	$+20{,}00$	$0$	$0$	$0$	$+20{,}00$	$+10{,}000$
$0'—2'$	$0$	$+20{,}00$	$+20{,}00$	$-20{,}00$	$+20{,}00$	$0$	$0$	$0$	$+20{,}00$	$+10{,}000$
$2—4$	$+16{,}00$	$0$	$+16{,}00$	$+16{,}00$	$+16{,}00$	$+16{,}00$	$+16{,}00$	$0$	$+16{,}00$	$+8{,}000$
$4—2'$	$+16{,}00$	$0$	$+16{,}00$	$+16{,}00$	$+16{,}00$	$+16{,}00$	$+16{,}00$	$0$	$+16{,}00$	$+8{,}000$
$1—3$	$0$	$0$	$0$	$0$	$0$	$0$	$0$	$0$	$0$	$0$
$1'—3'$	$0$	$0$	$0$	$0$	$0$	$0$	$0$	$0$	$0$	$0$
$3—5$	$+16{,}00$	$+16{,}00$	$+16{,}00$	$0$	$+16{,}00$	$0$	$+16{,}00$	$+16{,}00$	$+16{,}00$	$+8{,}000$
$5—3'$	$+16{,}00$	$+16{,}00$	$+16{,}00$	$0$	$+16{,}00$	$0$	$+16{,}00$	$+16{,}00$	$+16{,}00$	$+8{,}000$
$0—3$	$0$	$+61{,}83$	$+61{,}83$	$-61{,}83$	$+24{,}15$	$0$	$0$	$0$	$0$	$-19{,}321$
$0'—3'$	$0$	$+61{,}83$	$+61{,}83$	$-61{,}83$	$+24{,}15$	$0$	$0$	$0$	$0$	$-19{,}321$
$\psi$	$+32{,}00$	$+97{,}83$	$+113{,}83$	$-65{,}83$	$+76{,}15$	$+16{,}00$	$+32{,}10$	$+16{,}00$	$+52{,}00$

Table XX.

Node	Element	\(N=\dfrac{2EI}{l}\) 10 tn.-m.	\(\Sigma N\)	\(\psi \times 10^4\)	\(\dfrac{\psi N \times 10^4}{10\ \mathrm{tn.-m.}}\)	\(\Sigma \psi N\)	\(3\Sigma \psi N\)
0	0—1	11,2	63,8	\(+\,5,200\)	\(+\,58,24\)	\(+\,295,43\)	\(+\,886,29\)
0	0—2	37,0	63,8	\(+\,3,200\)	\(+\,118,40\)	\(+\,295,43\)	\(+\,886,29\)
0	0—3	15,6	63,8	\(+\,7,615\)	\(+\,118,79\)	\(+\,295,43\)	\(+\,886,29\)
1	1—0	11,2	61,3	\(+\,5,200\)	\(+\,58,24\)	\(+\,521,33\)	\(+\,1563,99\)
1	1—2	13,1	61,3	\(+\,3,200\)	\(+\,41,92\)	\(+\,521,33\)	\(+\,1563,99\)
1	1—3	37,0	61,3	\(+\,11,383\)	\(+\,421,17\)	\(+\,521,33\)	\(+\,1563,99\)
2	2—0	37,0	105,7	\(+\,3,200\)	\(+\,118,40\)	\(+\,617,52\)	\(+\,1852,56\)
2	2—1	13,1	105,7	\(+\,3,200\)	\(+\,41,92\)	\(+\,617,52\)	\(+\,1852,56\)
2	2—4	45,0	105,7	\(+\,9,783\)	\(+\,440,24\)	\(+\,617,52\)	\(+\,1852,56\)
2	2—5	10,6	105,7	\(+\,1,600\)	\(+\,16,96\)	\(+\,617,52\)	\(+\,1852,56\)
3	3—0	15,6	103,2	\(+\,7,615\)	\(+\,118,79\)	\(+\,252,68\)	\(+\,758,04\)
3	3—1	37,0	103,2	\(+\,11,383\)	\(+\,421,17\)	\(+\,252,68\)	\(+\,758,04\)
3	3—4	5,6	103,2	\(+\,1,600\)	\(+\,8,96\)	\(+\,252,68\)	\(+\,758,04\)
3	3—5	45,0	103,2	\(-\,6,583\)	\(-\,296,24\)	\(+\,252,68\)	\(+\,758,04\)

Substituting the quantities found in Table XX into the equations, we obtain the system

\[ \begin{aligned} 127,6\varphi_0+11,2\varphi_1+37,0\varphi_2+15,6\varphi_3&=886,29,\\ 11,2\varphi_0+122,6\varphi_1+13,1\varphi_2+37,0\varphi_3&=1563,99,\\ 37,0\varphi_0+13,1\varphi_1+211,4\varphi_2&=1852,56,\\ 15,6\varphi_0+37,0\varphi_1+206,4\varphi_3&=758,04. \end{aligned} \]

Solving it, we find the values

\[ \begin{aligned} \varphi_0&=3,6369, & \varphi_2&=7,4321,\\ \varphi_1&=11,2117, & \varphi_3&=1,3879. \end{aligned} \]

Then, by the formula

\[ M_{MN}=N_{MN}\left(2\varphi_M+\varphi_X-3\psi_{MN}\right) \]

we determine the bending moments, as well as the stresses and the ratios of the total stresses to the basic ones. All these quantities are placed in Table XXI.

Table XXI.

Elements	Angles	Magnitude of angles	Resultant angles	Moments, kg.-cm	Max. stress, kg./cm.$^2$	$\dfrac{N}{n}$
0—1	$\varphi_0$	$+\,3,6369$	$+\,2,8855$	$+\,3232$	$\pm\,67$	—
0—1	$\varphi_1$	$+\,11,2117$	$+\,10,4603$	$+\,11716$	$\pm\,67$	—
0—1	$\psi_{0-1}$	$+\,5,200$			$\pm\,67$	—
0—2	$\varphi_0$	$+\,3,6369$	$+\,5,1059$	$+\,18892$	$+\,142$	$1,28$
0—2	$\varphi_2$	$+\,7,4321$	$+\,8,9011$	$+\,32934$	$+\,142$	$1,28$
0—2	$\psi_{0-2}$	$+\,3,200$			$+\,142$	$1,28$
0—3	$\varphi_0$	$+\,3,6369$	$-\,14,1833$	$-\,22126$	$\pm\,123$	$1,20$
0—3	$\varphi_3$	$+\,1,3879$	$-\,16,4323$	$-\,25634$	$\pm\,123$	$1,20$
0—3	$\psi_{0-3}$	$+\,7,615$			$\pm\,123$	$1,20$
1—2	$\varphi_1$	$+\,11,2117$	$+\,20,2555$	$+\,26535$	$\pm\,164$	—
1—2	$\varphi_2$	$+\,7,4321$	$+\,16,4759$	$+\,21583$	$\pm\,164$	—
1—2	$\psi_{1-2}$	$+\,3,200$			$\pm\,164$	—
1—3	$\varphi_1$	$+\,11,2117$	$-\,10,3377$	$-\,38249$	$-\,164$	—
1—3	$\varphi_3$	$+\,1,3879$	$-\,20,1615$	$-\,74598$	$+\,318$	—
1—3	$\psi_{1-3}$	$+\,11,383$				—
2—4	$\varphi_2$	$+\,7,4321$	$-\,14,4848$	$-\,65182$	$+\,270$	$1,68$
2—4	$\varphi_4$	$0$	$-\,21,9169$	$-\,98626$	$+\,270$	$1,68$
2—4	$\psi_{2-4}$	$+\,9,783$			$+\,270$	$1,68$
2—5	$\varphi_2$	$+\,7,4321$	$+\,10,0642$	$+\,10668$	$\pm\,63$	—
2—5	$\varphi_5$	$0$	$+\,2,6321$	$+\,2790$	$\pm\,63$	—
2—5	$\psi_{2-5}$	$+\,1,600$			$\pm\,63$	—
3—4	$\varphi_3$	$+\,1,3879$	$-\,2,0242$	$-\,1134$	$\pm\,21$	—
3—4	$\varphi_4$	$0$	$-\,3,4121$	$-\,1911$	$\pm\,21$	—
3—4	$\psi_{3-4}$	$+\,1,600$			$\pm\,21$	—
3—5	$\varphi_3$	$+\,1,3879$	$+\,22,5248$	$+\,101362$	$-\,376$	$1,94$
3—5	$\varphi_5$	$0$	$+\,21,1369$	$+\,95116$	$-\,376$	$1,94$
3—5	$\psi_{3-5}$	$-\,6,583$			$-\,376$	$1,94$

Finally, for the calculation by the fourth method it is necessary to ignore the influence of the rigidity of the joints on the magnitude of the longitudinal forces and, moreover, to suppose the lattice to be hinged at its supports.

In accordance with these assumptions, we obtain the necessary equations from Tables XII and XIII, again setting

\[ M_2=M_4=M_5=M_6=M_8=M_9=M_{12}=M_{13}=0 \]

and discarding the equations corresponding to these moments.

We have the system

\[ \begin{aligned} +775{,}42 M_1-297{,}62 M_3+90{,}09 M_7&=+20{,}00,\\ -297{,}62 M_1+775{,}42 M_3+90{,}09 M_{10}&=-61{,}83,\\ +90{,}09 M_1+328{,}32 M_7+74{,}07 M_{11}&=-65{,}83,\\ +90{,}09 M_3+328{,}32 M_{10}+74{,}07 M_{14}&=+179{,}65,\\ +74{,}07 M_7+148{,}14 M_{11}&=+97{,}83,\\ +74{,}07 M_{10}+148{,}14 M_{14}&=-66{,}83. \end{aligned} \]

Solving it, we find

\[ \begin{aligned} M_1&=+0{,}0075\ \text{tn.-m.}, & M_{10}&=+0{,}7816,\\ M_3&=-0{,}1677, & M_{11}&=+0{,}8585,\\ M_7&=-0{,}3962, & M_{14}&=-0{,}8352. \end{aligned} \]

17. Example II. Two-lattice truss.

The two-lattice truss of Example II is a statically indeterminate system, since the number of its joints is \(n=10\), and the number of members is equal to 19, which gives two statically indeterminate quantities. Therefore the calculation of the forces in the members of the hinged truss must be carried out according to the theorem of least work of deformation. In doing this we shall not consider the influence of temperature changes on the forces and shall neglect the work of the support reactions. Then, in order to compose the expression for the work of deformation, it will be necessary to take the sum of the works of the individual members

\[ A=\sum \frac{S^2 l}{2E\omega} \tag{a} \]

Let us take the forces in the members \(3—2'\) and \(2—3'\) as the unknown quantities and denote them by

\[ \begin{aligned} S_{3—2'}&=x,\\ S_{2—3'}&=y. \end{aligned} \]

Fig. 9.

\[ \begin{gathered} \alpha = 38^\circ 40' \qquad \sin \alpha = \cos \beta = 0.6247 \qquad \gamma = 58^\circ \qquad \sin \gamma = \cos \delta = 0.8480,\\ \beta = 51^\circ 20' \qquad \cos \alpha = \sin \beta = 0.7809 \qquad \delta = 32^\circ \qquad \sin \delta = \cos \gamma = 0.5299,\\ \tan \alpha = \cot \beta = 0.8000 \qquad\qquad \tan \gamma = \cot \delta = 1.6000. \end{gathered} \]

Table XXII.

Name of member	Member Nos.	\(\omega\) brutto, cm²	\(\omega\) netto, cm²	\(I\) brutto, cm⁴	\(I\) netto, cm⁴	\(e_1\)	\(e_2\)	\(l\)
Lower chord	0—2	94	80	10127	8266	10	25	400
Lower chord	2—4	94	80	10127	8266	10	25	400
Upper chord	1—3	142	121	17124	16932	7	32	400
Upper chord	3—5	200	172	37397	36867	8	42	400
Diagonals	1—2	87	81	8542	8098	17	17	640
Diagonals	1—4	91	81	1877	1471	10	10	943
Diagonals	3—2′	37	32.5	1286	1220	13	13	943
Posts	0—1	131	120	9775	9091	17.5	17.5	500
Posts	2—3	79	70	2890	2151	13	13	500
Posts	4—5	64	56	1066	864	9	9	500

Then the force \(S\) in some member of a statically indeterminate truss is expressed by the formula

\[ \tag{b} S = S^0 + ax + by. \]

Trusses with Rigid Joints at the Nodes

Here \(S^0\) is the force in the corresponding member of the principal statically determinate truss (obtained from the given one by removing members \(3—2'\) and \(2—3'\)) under the action of the prescribed load; \(a\) and \(b\) are, respectively, the forces in the same member from the action of forces \(x=1\) and \(y=1\) on the principal truss (Fig. 10). Differentiating \(A\) \((a)\) successively with respect to \(x\) and to \(y\), and setting the derivatives equal to zero, we obtain

\[ \frac{\partial A}{\partial x} = \sum \frac{Sl}{E\omega}\cdot \frac{\partial S}{\partial x} =0, \]

\[ \frac{\partial A}{\partial y} = \sum \frac{Sl}{E\omega}\cdot \frac{\partial S}{\partial y} =0, \]

or

\[ \sum \frac{S^0l}{E\omega}\cdot a + \sum \frac{l}{E\omega}\cdot a^2x + \sum \frac{l}{E\omega}\cdot aby =0, \]

\[ \sum \frac{S^0l}{E\omega}\cdot b + \sum \frac{l}{E\omega}\cdot abx + \sum \frac{l}{E\omega}\cdot b^2y =0. \]

Canceling \(E\) and putting

\[ \sum a\cdot \frac{S^0l}{\omega}=C,\qquad \sum b\cdot \frac{S^0l}{\omega}=D,\qquad \sum ab\cdot \frac{l}{\omega}=B, \]

and, owing to the symmetry of the truss,

\[ \sum a^2 \frac{l}{\omega} = \sum b^2 \frac{l}{\omega} = A, \]

we obtain the equations

\[ Ax+By+C=0, \]
\[ Bx+Ay+D=0. \]

Whence

\[ x=\frac{BD-AC}{A^2-B^2}; \qquad y=\frac{BC-AD}{A^2-B^2}. \]

Substitution of the computed \(x\) and \(y\) into formulas of the form \((b)\) gives the forces in the statically indeterminate hinged truss.

The computation was carried out for a load \(P=1^{tn}\), placed successively at nodes 2, 3, 4, 5. When the load is placed at node 1, evidently, only \(S_{0-1}=-1^{tn}\), while the remaining forces are equal to zero.

Since in the subsequent calculation it is necessary to use the forces from horizontal and inclined forces, in addition to the calculations for the load \(P\), determinations were made of the forces from the action of a horizontal force \(P'=1^{tn}\), directed from left to right and applied successively at all the nodes of the truss. In this case, when the force \(P'=1\) is placed at node \(2'\), only \(S^{1'-2'}=-1^{tn}\), the remaining...

Fig. 10.

Table XXIII.

Elements	$a$	$b$	$\dfrac{l}{\omega}\left(\dfrac{1}{\mathrm{cm}}\right)$	$\dfrac{al}{\omega}\left(\dfrac{1}{\mathrm{cm}}\right)$	$\dfrac{bl}{\omega}\left(\dfrac{1}{\mathrm{cm}}\right)$	$\dfrac{a^{2}l}{\omega}\left(\dfrac{1}{\mathrm{cm}}\right)$	$\dfrac{abl}{\omega}\left(\dfrac{1}{\mathrm{cm}}\right)$
$0—2$	0	0	4,2553	0	0	0	0
$2—4$	$+0,424$	$-1,272$	4,2553	$+1,804$	$-5,413$	$+0,765$	$-2,295$
$4—2'$	$-1,272$	$+0,424$	4,2553	$-5,413$	$+1,804$	$+6,885$	$-2,295$
$2'—0'$	0	0	4,2553	0	0	0	0
$1—3$	$+0,424$	$-0,424$	2,8169	$+1,194$	$-1,194$	$+0,506$	$-0,506$
$3—5$	$-0,424$	$-0,424$	2,0000	$-0,848$	$-0,848$	$+0,360$	$+0,360$
$5—3'$	$-0,424$	$-0,424$	2,0000	$-0,848$	$-0,848$	$+0,360$	$+0,360$
$3'—1'$	$-0,424$	$+0,424$	2,8169	$-1,194$	$+1,194$	$+0,506$	$-0,506$
$0—1$	0	0	3,8168	0	0	0	0
$2—3$	$-0,530$	0	6,3291	$-3,354$	0	$+1,778$	0
$4—5$	0	0	7,8125	0	0	0	0
$3'—2'$	0	$-0,530$	6,3291	0	$-3,354$	0	0
$0'—1'$	0	0	3,8168	0	0	0	0
$1—2$	$+0,679$	$-0,679$	7,3553	$+4,995$	$-4,995$	$+3,392$	$-3,392$
$1—4$	$-1,000$	$+1,000$	10,3626	$-10,363$	$-10,363$	$+10,363$	$-10,363$
$2—3'$	$+1,000$	0	25,4855	$+25,487$	0	$+25,487$	0
$3—2'$	0	$+1,000$	25,4865	0	$+25,487$	0	0
$4—1'$	$+1,000$	$-1,000$	10,3626	$+10,363$	$-10,363$	$+10,363$	$-10,363$
$2—1'$	$-0,679$	$+0,679$	7,3553	$-4,995$	$+4,995$	$+3,392$	$-3,392$
$\sum =$						$+64,157$	$-32,392$

Table XXIV.

Elements	Load \(P = 1^{tn}\) at node 2	Load \(P = 1^{tn}\) at node 3	Load \(P = 1^{tn}\) at node 4	Load \(P = 1^{tn}\) at node 5	Horizontal force \(P = 1^{tn}\) at node 0	Horizontal force \(P = 1^{tn}\) at node 1	Horizontal force \(P = 1^{tn}\) at node 2	Horizontal force \(P = 1^{tn}\) at node 3	Horizontal force \(P = 1^{tn}\) at node 4	Horizontal force \(P = 1^{tn}\) at node 5	Horizontal force \(P = 1^{tn}\) at node \(3'\)	Horizontal force \(P = 1^{tn}\) at node \(1'\)
\(0-2\)	0	0	0	0	\(-1{,}000\)	0	0	0	0	0	0	0
\(2-4\)	\(+0{,}800\)	\(+0{,}800\)	0	0	\(-1{,}000\)	0	\(-1{,}000\)	0	0	0	0	0
\(4-2'\)	0	0	0	0	\(-1{,}000\)	\(-1{,}000\)	\(-1{,}000\)	\(-1{,}000\)	\(-1{,}000\)	\(-1{,}000\)	\(-1{,}000\)	\(-1{,}000\)
\(2'-0'\)	0	0	0	0	\(-1{,}000\)	\(-1{,}000\)	\(-1{,}000\)	\(-1{,}000\)	\(-1{,}000\)	\(-1{,}000\)	\(-1{,}000\)	\(-1{,}000\)
\(1-3\)	\(-0{,}400\)	\(-0{,}400\)	\(-0{,}800\)	\(-0{,}800\)	0	\(-0{,}500\)	0	\(+0{,}500\)	0	\(+0{,}500\)	\(+0{,}500\)	\(+0{,}500\)
\(3-5\)	\(-0{,}400\)	\(-0{,}400\)	\(-0{,}800\)	\(-0{,}800\)	0	\(-0{,}500\)	0	\(-0{,}500\)	0	\(+0{,}500\)	\(+0{,}500\)	\(+0{,}500\)
\(5-3'\)	\(-0{,}400\)	\(-0{,}400\)	\(-0{,}800\)	\(-0{,}800\)	0	\(-0{,}500\)	0	\(-0{,}500\)	0	\(-0{,}500\)	\(+0{,}500\)	\(-0{,}500\)
\(3'-1'\)	\(-0{,}400\)	\(-0{,}400\)	\(-0{,}800\)	\(-0{,}800\)	0	\(-0{,}500\)	0	\(-0{,}500\)	0	\(-0{,}500\)	\(-0{,}500\)	\(+0{,}500\)
\(0-1\)	\(-0{,}750\)	\(-0{,}750\)	\(-0{,}500\)	\(-0{,}500\)	0	\(+0{,}313\)	0	\(+0{,}313\)	0	\(+0{,}313\)	\(+0{,}313\)	\(+0{,}313\)
\(2-3\)	0	\(-1{,}000\)	0	0	0	0	0	0	0	0	0	0
\(4-5\)	0	0	\(-1{,}000\)	0	0	0	0	0	0	0	0	0
\(2'-3'\)	0	0	0	0	0	0	0	0	0	0	0	0
\(0'-1'\)	\(-0{,}250\)	\(-0{,}250\)	\(-0{,}500\)	\(-0{,}500\)	0	\(-0{,}313\)	0	\(-0{,}313\)	0	\(-0{,}313\)	\(-0{,}313\)	\(-0{,}313\)
\(1-2\)	\(+1{,}280\)	\(+1{,}280\)	0	0	0	0	0	0	0	0	0	0
\(1-4\)	\(-0{,}472\)	\(-0{,}472\)	\(+0{,}943\)	\(+0{,}943\)	0	\(-0{,}590\)	0	\(-0{,}590\)	0	\(-0{,}590\)	\(-0{,}590\)	\(-0{,}590\)
\(2'-3'\)	0	0	0	0	0	0	0	0	0	0	0	0
\(3-2'\)	0	0	0	0	0	0	0	0	0	0	0	0
\(4-1'\)	\(+0{,}472\)	\(+0{,}472\)	\(+0{,}943\)	\(+0{,}943\)	0	\(+0{,}590\)	0	\(+0{,}590\)	0	\(+0{,}590\)	\(+0{,}590\)	\(+0{,}590\)
\(2'-1'\)	0	0	0	0	0	0	0	0	0	0	0	0

Table XXXV.

Load	Element	Load \(P=1\) tn at node 2: \(S^0 \dfrac{al}{\omega}\), tn/cm	Load \(P=1\) tn at node 2: \(S^0 \dfrac{bl}{\omega}\), tn/cm	Load \(P=1\) tn at node 3: \(S^0 \dfrac{al}{\omega}\)	Load \(P=1\) tn at node 3: \(S^0 \dfrac{bl}{\omega}\)	Load \(P=1\) tn at node 4: \(S^0 \dfrac{al}{\omega}\)	Load \(P=1\) tn at node 4: \(S^0 \dfrac{bl}{\omega}\)	Load \(P=1\) tn at node 5: \(S^0 \dfrac{al}{\omega}\)	Load \(P=1\) tn at node 5: \(S^0 \dfrac{bl}{\omega}\)	Force \(P'=1\) tn at node 0: \(S^0 \dfrac{al}{\omega}\)	Force \(P'=1\) tn at node 0: \(S^0 \dfrac{bl}{\omega}\)	Force \(P'=1\) tn at node 1: \(S^0 \dfrac{al}{\omega}\)	Force \(P'=1\) tn at node 1: \(S^0 \dfrac{bl}{\omega}\)
Load	\(0-2\)	0	0	0	0	0	0	0	0	0	0	0	0
Load	\(2-4\)	\(+1{,}443\)	\(-4{,}330\)	\(+1{,}443\)	\(-4{,}330\)	0	0	0	0	\(-1{,}804\)	\(+5{,}413\)	\(+5{,}413\)	\(-1{,}804\)
Load	\(4-2'\)	0	0	0	0	0	0	0	0	0	0	0	0
Load	\(2'-0'\)	0	0	0	0	0	0	0	0	0	0	0	0
Load	\(1-3\)	\(-0{,}478\)	\(+0{,}478\)	\(-0{,}478\)	\(+0{,}478\)	\(-0{,}955\)	\(+0{,}955\)	\(-0{,}955\)	\(+0{,}955\)	0	0	\(-0{,}597\)	\(+0{,}597\)
Load	\(3-5\)	\(+0{,}339\)	\(+0{,}339\)	\(+0{,}339\)	\(+0{,}339\)	\(+0{,}678\)	\(+0{,}678\)	\(+0{,}678\)	\(+0{,}678\)	0	0	\(+0{,}424\)	\(+0{,}424\)
Load	\(5-3'\)	\(+0{,}339\)	\(+0{,}339\)	\(+0{,}339\)	\(+0{,}339\)	\(+0{,}678\)	\(+0{,}678\)	\(+0{,}678\)	\(+0{,}678\)	0	0	\(+0{,}424\)	\(+0{,}424\)
Load	\(3'-1'\)	\(+0{,}478\)	\(-0{,}478\)	\(+0{,}478\)	\(-0{,}478\)	\(+0{,}955\)	\(-0{,}955\)	\(+0{,}955\)	\(-0{,}955\)	0	0	\(+0{,}597\)	\(-0{,}597\)
Load	\(0-1\)	0	0	0	0	0	0	0	0	0	0	0	0
Load	\(2-3\)	0	0	\(+3{,}354\)	0	0	0	0	0	0	0	0	0
Load	\(4-5\)	0	0	0	0	0	0	0	0	0	0	0	0
Load	\(2'-3'\)	0	0	0	0	0	0	0	0	0	0	0	0
Load	\(0'-1'\)	0	0	0	0	0	0	0	0	0	0	0	0
Load	\(1-2\)	\(+6{,}394\)	\(-6{,}394\)	\(+6{,}394\)	\(-6{,}394\)	\(-9{,}772\)	\(+9{,}772\)	\(-9{,}772\)	\(+9{,}772\)	0	0	\(+6{,}114\)	\(-6{,}114\)
Load	\(1-4\)	\(+4{,}891\)	\(-4{,}891\)	\(+4{,}891\)	\(-4{,}891\)	0	0	0	0	0	0	0	0
Load	\(2-3'\)	0	0	0	0	0	0	0	0	0	0	0	0
Load	\(3-2'\)	0	0	0	0	0	0	0	0	0	0	\(+6{,}114\)	\(-6{,}114\)
Load	\(4-1'\)	\(+4{,}891\)	\(-4{,}891\)	\(+4{,}891\)	\(-4{,}891\)	\(+9{,}772\)	\(-9{,}772\)	\(+9{,}772\)	\(-9{,}772\)	0	0	0	0
Load	\(2'-1'\)	0	0	0	0	0	0	0	0	0	0	0	0
\(\Sigma\)		\(+18{,}297\)	\(-19{,}828\)	\(+21{,}651\)	\(-19{,}828\)	\(+1{,}356\)	\(+1{,}356\)	\(+1{,}356\)	\(+1{,}356\)	\(+3{,}609\)	\(+3{,}609\)	\(+18{,}489\)	\(-13{,}184\)

Table XVI.

Load	Element	Force \(P'=1\) ton at node 2: \(S^0 \dfrac{al}{\omega}\), cm	Force \(P'=1\) ton at node 2: \(S^0 \dfrac{bl}{\omega}\), cm	Force \(P'=1\) ton at node 3: \(S^0 \dfrac{al}{\omega}\)	Force \(P'=1\) ton at node 3: \(S^0 \dfrac{bl}{\omega}\)	Force \(P'=1\) ton at node 4: \(S^0 \dfrac{al}{\omega}\)	Force \(P'=1\) ton at node 4: \(S^0 \dfrac{bl}{\omega}\)	Force \(P'=1\) ton at node 5: \(S^0 \dfrac{al}{\omega}\)	Force \(P'=1\) ton at node 5: \(S^0 \dfrac{bl}{\omega}\)	Force \(P'=1\) ton at node \(3'\): \(S^0 \dfrac{al}{\omega}\)	Force \(P'=1\) ton at node \(3'\): \(S^0 \dfrac{bl}{\omega}\)	Force \(P'=1\) ton at node \(1'\): \(S^0 \dfrac{al}{\omega}\)	Force \(P'=1\) ton at node \(1'\): \(S^0 \dfrac{bl}{\omega}\)
	\(0-2\)	0	0	0	0	0	0	0	0	0	0	0	0
	\(2-4\)	\(-1{,}804\)	\(+5{,}413\)	0	0	0	0	0	0	0	0	0	0
	\(4-2'\)	\(+5{,}413\)	\(-1{,}804\)	\(+5{,}413\)	\(-1{,}804\)	\(+5{,}413\)	\(-1{,}804\)	\(+5{,}413\)	\(-1{,}804\)	\(+5{,}413\)	\(-1{,}804\)	\(+5{,}413\)	\(-1{,}804\)
	\(2'-0'\)	0	0	0	0	0	0	0	0	0	0	0	0
	\(1-3\)	0	0	\(+0{,}597\)	\(-0{,}597\)	0	0	\(+0{,}597\)	\(-0{,}597\)	\(+0{,}597\)	\(-0{,}597\)	\(+0{,}597\)	\(-0{,}597\)
	\(3-5\)	0	0	\(+0{,}424\)	\(+0{,}424\)	0	0	\(-0{,}424\)	\(-0{,}424\)	\(-0{,}424\)	\(-0{,}424\)	\(-0{,}424\)	\(-0{,}424\)
	\(5-3'\)	0	0	\(+0{,}424\)	\(+0{,}424\)	0	0	\(+0{,}424\)	\(+0{,}424\)	\(-0{,}424\)	\(-0{,}424\)	\(-0{,}424\)	\(-0{,}424\)
	\(3'-1'\)	0	0	\(+0{,}597\)	\(-0{,}597\)	0	0	\(+0{,}597\)	\(-0{,}597\)	\(+0{,}597\)	\(-0{,}597\)	\(-0{,}597\)	\(+0{,}597\)
	\(0-1\)	0	0	0	0	0	0	0	0	0	0	0	0
	\(2-3\)	0	0	0	0	0	0	0	0	0	0	0	0
	\(4-5\)	0	0	0	0	0	0	0	0	0	0	0	0
	\(2'-3'\)	0	0	0	0	0	0	0	0	0	0	0	0
	\(0'-1'\)	0	0	0	0	0	0	0	0	0	0	0	0
	\(1-2\)	0	0	\(+6{,}114\)	\(-6{,}114\)	0	0	\(+6{,}114\)	\(-6{,}114\)	\(+6{,}114\)	\(-6{,}114\)	\(+6{,}114\)	\(-6{,}114\)
	\(1-4\)	0	0	0	0	0	0	0	0	0	0	0	0
	\(2-3'\)	0	0	0	0	0	0	0	0	0	0	0	0
	\(3-2'\)	0	0	0	0	0	0	0	0	0	0	0	0
	\(4-1'\)	0	0	\(+6{,}114\)	\(-6{,}114\)	0	0	\(+6{,}114\)	\(-6{,}114\)	\(+6{,}114\)	\(-6{,}114\)	\(+6{,}114\)	\(-6{,}114\)
	\(2'-1'\)	0	0	0	0	0	0	0	0	0	0	0	0
\(\sum\)		\(+3{,}609\)	\(+3{,}609\)	\(+19{,}683\)	\(-14{,}378\)	\(+5{,}413\)	\(-1{,}804\)	\(+18{,}835\)	\(-15{,}226\)	\(+17{,}987\)	\(-16{,}074\)	\(+16{,}793\)	\(-14{,}880\)

Table XXVII.

Designation of quantities	Load \(P=1^{tn.}\), applied at joint: 2	Load \(P=1^{tn.}\), applied at joint: 3	Load \(P=1^{tn.}\), applied at joint: 4	Load \(P=1^{tn.}\), applied at joint: 5	Force \(P'=1^{tn.}\) at joint: 0	Force \(P'=1^{tn.}\) at joint: 1
\(C\)	\(+\ 18,297\)	\(+\ 21,651\)	\(+\ 1,356\)	\(+\ 1,356\)	\(+\ 3,609\)	\(+\ 18,489\)
\(D\)	\(-\ 19,828\)	\(-\ 19,828\)	\(+\ 1,356\)	\(+\ 1,356\)	\(+\ 3,609\)	\(-\ 13,184\)
\(BD\)	\(+\ 642,269\)	\(+\ 642,269\)	\(-\ 43,924\)	\(-\ 43,924\)	\(-\ 116,903\)	\(+\ 427,056\)
\(AC\)	\(+1173,881\)	\(+1389,063\)	\(+\ 86,997\)	\(+\ 86,997\)	\(+\ 231,543\)	\(+1186,199\)
\(BC\)	\(-\ 592,676\)	\(-\ 701,319\)	\(-\ 43,924\)	\(-\ 43,924\)	\(-\ 116,903\)	\(-\ 598,896\)
\(AD\)	\(-1272,105\)	\(-1272,105\)	\(+\ 86,997\)	\(+\ 86,997\)	\(+\ 231,543\)	\(-\ 845,846\)
\(x\)	\(-\ 0,1733\)	\(-\ 0,2435\)	\(-\ 0,0427\)	\(-\ 0,0427\)	\(-\ 0,1136\)	\(-\ 0,2475\)
\(y\)	\(+\ 0,2215\)	\(+\ 0,1861\)	\(-\ 0,0427\)	\(-\ 0,0427\)	\(-\ 0,1136\)	\(+\ 0,0805\)

Designation of quantities	Force \(P'=1^{tn.}\) applied at joint: 2	Force \(P'=1^{tn.}\) applied at joint: 3	Force \(P'=1^{tn.}\) applied at joint: 4	Force \(P'=1^{tn.}\) applied at joint: 5	Force \(P'=1^{tn.}\) applied at joint: \(3^1\)	Force \(P'=1^{tn.}\) applied at joint: \(1^1\)
\(C\)	\(+\ 3,609\)	\(+\ 19,683\)	\(+\ 5,413\)	\(+\ 18,835\)	\(+\ 17,987\)	\(+\ 16,793\)
\(D\)	\(+\ 3,609\)	\(-\ 14,378\)	\(-\ 1,804\)	\(-\ 15,226\)	\(-\ 16,074\)	\(-\ 14,880\)
\(BD\)	\(-\ 116,903\)	\(+\ 465,732\)	\(+\ 58,435\)	\(+\ 493,201\)	\(+\ 520,669\)	\(+\ 481,993\)
\(AC\)	\(+\ 231,543\)	\(+1262,802\)	\(+\ 347,282\)	\(+1208,397\)	\(+\ 1153,992\)	\(+1077,389\)
\(BC\)	\(-\ 116,903\)	\(-\ 637,572\)	\(-\ 175,338\)	\(-\ 610,103\)	\(-\ 582,635\)	\(-\ 543,959\)
\(AD\)	\(+\ 231,543\)	\(-\ 922,449\)	\(-\ 115,739\)	\(-\ 976,854\)	\(-\ 1031,260\)	\(-\ 954,656\)
\(x\)	\(-\ 0,1136\)	\(-\ 0,2599\)	\(-\ 0,0942\)	\(-\ 0,2332\)	\(-\ 0,2065\)	\(-\ 0,1941\)
\(y\)	\(-\ 0,1136\)	\(+\ 0,0929\)	\(-\ 0,0194\)	\(+\ 0,1196\)	\(+\ 0,1463\)	\(+\ 0,1339\)

the same forces are equal to zero. In Table XXIII the forces \(a\) and \(b\) are given for the elements of the principal statically determinate truss; next, the products

\[ \frac{al}{\omega},\quad \frac{bl}{\omega},\quad \frac{a^{2}l}{\omega},\quad \frac{abl}{\omega} \]

were computed, and the sums of the last two products were calculated for all the elements. Table XXIV contains the forces in the elements of the principal truss for the above-mentioned cases of unit load.

Table XXXVIII.

Elements	Load \(P' = 1\) tn. at nodes: 2	Load \(P' = 1\) tn. at nodes: 3	Load \(P' = 1\) tn. at nodes: 4	Load \(P' = 1\) tn. at nodes: 5	Horizontal force \(P' = 1\) tn. at nodes: 0	Horizontal force \(P' = 1\) tn. at nodes: 1	Horizontal force \(P' = 1\) tn. at nodes: 2	Horizontal force \(P' = 1\) tn. at nodes: 3	Horizontal force \(P' = 1\) tn. at nodes: 4	Horizontal force \(P' = 1\) tn. at nodes: 5	Horizontal force \(P' = 1\) tn. at nodes: \(3'\)	Horizontal force \(P' = 1\) tn. at nodes: \(1'\)
\(0\!-\!2\)	0	0	0	0	\(-1,000\)	0	0	0	0	0	0	0
\(2\!-\!4\)	\(+0,445\)	\(+0,460\)	\(+0,037\)	\(+0,037\)	\(-0,903\)	\(-0,208\)	\(-0,903\)	\(-0,228\)	\(-0,016\)	\(-0,252\)	\(-0,274\)	\(-0,252\)
\(4\!-\!2'\)	\(+0,314\)	\(+0,389\)	\(+0,037\)	\(+0,037\)	\(-0,903\)	\(-0,651\)	\(-0,903\)	\(-0,630\)	\(-0,888\)	\(-0,653\)	\(-0,675\)	\(-0,696\)
\(2'\!-\!0'\)	0	0	0	0	\(-1,000\)	\(-1,000\)	\(-1,000\)	\(-1,000\)	\(-1,000\)	\(-1,000\)	\(-1,000\)	\(-1,000\)
\(1\!-\!3\)	\(-0,567\)	\(-0,582\)	\(-0,800\)	\(-0,800\)	0	\(-0,639\)	0	\(+0,351\)	\(-0,032\)	\(+0,350\)	\(+0,350\)	\(+0,361\)
\(3\!-\!5\)	\(-0,421\)	\(-0,376\)	\(-0,764\)	\(-0,764\)	\(+0,097\)	\(-0,429\)	\(+0,097\)	\(-0,429\)	\(+0,048\)	\(+0,548\)	\(+0,526\)	\(+0,525\)
\(5\!-\!3'\)	\(-0,421\)	\(-0,376\)	\(-0,764\)	\(-0,764\)	\(+0,097\)	\(-0,429\)	\(+0,097\)	\(-0,429\)	\(+0,048\)	\(-0,452\)	\(+0,526\)	\(+0,525\)
\(3'\!-\!1'\)	\(-0,233\)	\(-0,218\)	\(-0,800\)	\(-0,800\)	0	\(-0,361\)	0	\(-0,351\)	\(+0,032\)	\(-0,350\)	\(-0,350\)	\(+0,639\)
\(0\!-\!1\)	\(-0,750\)	\(-0,750\)	\(-0,500\)	\(-0,500\)	0	\(+0,313\)	0	\(+0,313\)	0	\(+0,313\)	\(+0,313\)	\(+0,313\)
\(2\!-\!3\)	\(+0,092\)	\(-0,871\)	\(+0,023\)	\(+0,023\)	\(+0,060\)	\(+0,131\)	\(+0,060\)	\(+0,138\)	\(+0,050\)	\(+0,123\)	\(+0,110\)	\(+0,103\)
\(4\!-\!5\)	0	0	0	\(-1,000\)	0	0	0	0	0	0	0	0
\(2'\!-\!3'\)	\(-0,118\)	\(-0,099\)	\(+0,023\)	\(+0,023\)	\(+0,060\)	\(-0,043\)	\(+0,060\)	\(-0,049\)	\(+0,010\)	\(-0,064\)	\(-0,077\)	\(-0,071\)
\(0'\!-\!1'\)	\(-0,250\)	\(-0,250\)	\(-0,500\)	\(-0,500\)	0	\(-0,313\)	0	\(-0,313\)	0	\(-0,313\)	\(-0,313\)	\(-0,313\)
\(1\!-\!2\)	\(+1,012\)	\(+0,988\)	0	0	0	\(-0,223\)	0	\(-0,240\)	\(-0,051\)	\(-0,239\)	\(-0,240\)	\(-0,223\)
\(1\!-\!4\)	\(-0,077\)	\(-0,042\)	\(+0,943\)	\(+0,943\)	0	\(-0,261\)	0	\(-0,237\)	\(+0,075\)	\(-0,237\)	\(-0,237\)	\(-0,262\)
\(2\!-\!3'\)	\(+0,222\)	\(+0,186\)	\(-0,043\)	\(-0,043\)	\(-0,114\)	\(+0,081\)	\(-0,114\)	\(+0,093\)	\(-0,019\)	\(+0,120\)	\(+0,146\)	\(+0,134\)
\(3\!-\!2'\)	\(-0,173\)	\(-0,244\)	\(-0,043\)	\(-0,043\)	\(-0,114\)	\(-0,248\)	\(-0,114\)	\(-0,260\)	\(-0,094\)	\(-0,233\)	\(-0,207\)	\(-0,194\)
\(1'\!-\!4'\)	\(+0,077\)	\(+0,042\)	\(+0,943\)	\(+0,943\)	0	\(+0,261\)	0	\(+0,237\)	\(-0,075\)	\(+0,237\)	\(+0,237\)	\(+0,262\)
\(1'\!-\!2'\)	\(+0,268\)	\(+0,292\)	0	0	0	\(+0,223\)	0	\(+0,240\)	\(+0,051\)	\(+0,239\)	\(+0,240\)	\(+0,223\)

Table XXXIX.

Forces are expressed in \( \frac{1}{10m} \).

Forces in the elements under the action of pairs of forces \(1\,t^*\), applied to the elements:

Element	0–2	2–4	1–3	3–5	0–1	2–3	1–2	1–4	3–2′	2–3′
0–2	0	0	0	0	+2,000	0	0	0	0	0
2–4	+1,113	−1,020	+1,150	−1,058	+1,390	+1,350	+1,282	−0,075	−0,257	+0,303
4–2′	+0,785	−0,693	+0,973	−0,880	+0,504	+0,546	+0,613	+0,166	−0,304	+0,257
2′–0′	0	0	0	0	0	0	0	0	0	0
1–3	−1,418	−0,582	−1,455	−0,545	−1,278	+0,702	−1,333	−1,060	+0,511	−0,511
3–5	−1,053	−0,858	−0,940	−0,970	−1,052	−1,052	−1,053	−0,955	−0,281	+0,281
5–3′	−1,053	−0,858	−0,940	−0,970	−1,052	−1,052	−1,053	−0,955	−0,281	+0,281
3′–1′	−0,582	−1,418	−0,545	−1,455	−0,722	−0,702	−0,667	−0,940	−0,511	+0,511
0–1	−1,875	+0,625	+0,625	+0,625	+0,625	+0,625	+0,625	+0,625	+0,625	+0,625
2–3	+0,230	−0,173	−2,178	+2,235	+0,142	+0,156	+0,177	+0,067	+0,755	−0,144
4–5	0	0	0	−2,500	0	0	0	0	0	0
2′–3′	−0,295	+0,353	−0,248	+0,305	−0,206	−0,218	−0,241	−0,009	+0,144	−0,755
0′–1′	−0,625	−0,625	−0,625	−0,625	−0,625	−0,625	−0,625	−0,625	−0,625	−0,625
1–2	+2,530	−2,530	+2,470	−2,470	−0,446	−0,480	+0,716	−0,097	−0,782	−0,782
1–4	−0,193	+2,550	−0,105	+2,463	−0,522	−0,474	−0,393	+0,659	−0,026	−0,026
3–2′	−0,433	+0,325	−0,610	+0,503	−0,268	−0,292	−0,332	−0,126	+0,271	+0,271
2–3′	+0,555	−0,663	+0,465	−0,573	+0,390	+0,414	+0,455	+0,017	−0,271	−0,271
4–1′	+0,193	+2,165	−0,105	+2,253	+0,522	+0,474	+0,393	+1,037	+0,026	+0,026
2′–1′	+0,670	−0,670	+0,730	−0,730	+0,446	+0,480	+0,534	+0,097	−0,782	+0,782

Further, Tables XXV and XXVI contain the quantities \(S^0\dfrac{al}{\omega}\) and \(S^0\dfrac{bl}{\omega}\), and their sums are also computed.

Table XXVII gives the scheme for computing the statically indeterminate quantities \(x\) and \(y\), while Table XXVIII contains the forces in the members of the statically indeterminate pin-jointed truss for 12 cases of unit loading. From these data it is also easy to compute the forces due to the loading of the various members by unit pairs of forces normal to the axes of the members; these forces are given in Table XXIX.

The further calculation of the stresses due to the rigidity of the joints is carried out in exactly the same way as in the first example. We shall therefore not dwell on it in detail, but shall confine ourselves to the explanations necessary for the tables.

The load on the truss is chosen, in this case as well, to be symmetrical, but non-uniform, so that one system of diagonals is loaded considerably more heavily; hence one may expect considerable stresses due to the rigidity of the joints.

At joints 2 and \(2'\) loads of \(50^{tn}\) each are placed, while at joint 4 there is only \(20^{tn}\).

The forces, stresses, and changes in the lengths of the members resulting from such loading, as well as some other quantities needed for the calculation, are given in Table XXX.

We choose 14 unknowns \(M_1 \ldots M_{14}\), connected with the moments due to the rigidity of the joints in the following manner:

\[ \begin{aligned} M_{0-1}&=-M_1, & M_{1-0}&=+M_2,\\ M_{0-2}&=+M_1, & M_{2-0}&=-(M_5+M_6+M_7+M_8),\\ M_{1-2}&=+M_3, & M_{2-1}&=+M_5,\\ M_{1-3}&=-(M_2+M_3+M_4), & M_{3-1}&=+M_9,\\ M_{1-4}&=+M_4, & M_{4-1}&=+M_{13},\\ M_{2-3}&=+M_6, & M_{3-2}&=+M_{10},\\ M_{2-4}&=+M_8, & M_{4-2}&=-M_{12},\\ M_{2-3'}&=+M_7, & M_{3'-2}&=-M_{11},\\ M_{3-5}&=-(M_9+M_{10}+M_{11}), & M_{5-3}&=-M_{14},\\ M_{3-2'}&=+M_{11}, & M_{2'-3}&=-M_7. \end{aligned} \]

Here, by the quantities \(M_1 \ldots M_{14}\) we shall again mean the entire set of the symmetrical loading of the truss by four moments equal in absolute value and pairwise equilibrated.

The magnitudes of the forces \(\alpha, \beta, \gamma \ldots \rho\) in the members of the truss from the action of the corresponding loads \(M_1 \ldots M_{14}\), equal to unity, are given in Table XXXI.

Load scheme.

Fig. 11.

Table XXX.

Element	Forces from \(P = 1\ \mathrm{tn.}\) at joints 2 and \(2'\)	Forces from \(P = 1\ \mathrm{tn.}\) at joint 4	Forces under loading according to the scheme	\(\lambda = \dfrac{Sl}{E\omega}\cdot 10^2\), m.	\(\Delta \gamma \times 10^5\), \(\dfrac{1}{\mathrm{tn.-m.}}\)	\(\Delta_\lambda = \dfrac{l}{E\omega}+10^5\), \(\dfrac{\mathrm{m.}}{\mathrm{tn.}}\)	Stress, kg./cm²
0—2	0	0	0	0	30,62	1,9792	0
2—4	\(+0,759\)	\(+0,037\)	\(+38,69\)	\(+7,658\)	30,62	1,9792	\(+484\)
1—3	\(-0,800\)	\(-0,800\)	\(-56,00\)	\(-7,337\)	18,11	1,3102	\(-463\)
3—5	\(-0,842\)	\(-0,764\)	\(-57,38\)	\(-5,338\)	8,29	0,9302	\(-334\)
0—1	\(-1,000\)	\(-0,500\)	\(-60,00\)	\(-10,652\)	39,65	1,7753	\(-500\)
2—3	\(-0,026\)	\(+0,023\)	\(-0,84\)	\(-0,247\)	134,11	2,9438	\(-12\)
1—2	\(+1,280\)	0	\(+64,00\)	\(+21,898\)	58,08	3,4215	\(+790\)
1—4	0	\(+0,943\)	\(+18,86\)	\(+9,090\)	389,46	4,8198	\(+233\)
\(2—3'\)	\(+0,049\)	\(-0,043\)	\(+1,59\)	\(+1,885\)	568,44	11,8542	\(+49\)

Table XXXI.

Elem.	$\alpha$	$\beta$	$\gamma$	$\delta$	$\varepsilon$	$\eta$	$\theta$	$\iota$	$\lambda$	$\mu$	$\nu$	$o$	$\pi$	$\rho$
0–2	−0.200	+0.200	0	0	0	0	0	0	0	0	0	0	0	0
2–4	0	−0.023	−0.023	0.203	0	0	−0.134	−0.361	+0.406	+0.383	+0.138	+0.171	+0.009	−0.194
1–3	0	0	0	0	0	+0.200	+0.200	0	0	+0.200	+0.200	+0.200	−0.200	−0.200
3–5	0	−0.023	−0.023	−0.003	0	0	+0.267	+0.039	+0.006	−0.016	+0.138	+0.171	−0.191	−0.194
0–1	−0.250	0	0	0	+0.250	+0.250	0	+0.250	0	0	0	0	0	0
2–3	0	+0.236	+0.236	+0.248	0	0	−0.083	+0.025	−0.496	−0.270	−0.164	−0.018	+0.006	+0.254
1–2	+0.320	−0.320	−0.195	−0.320	−0.195	−0.320	−0.320	−0.640	+0.640	+0.320	+0.320	+0.320	0	−0.320
1–4	0	0	0	+0.170	0	0	0	+0.472	−0.472	−0.472	−0.472	−0.472	+0.170	+0.472
2–3′	0	+0.027	+0.027	+0.004	0	0	−0.012	−0.046	−0.007	+0.019	+0.007	+0.034	−0.101	−0.007

Table XXXII.

Equation	$M_1$	$M_2$	$M_3$	$M_4$	$M_5$	$M_6$	$M_7$	$M_8$	$M_9$	$M_{10}$	$M_{11}$	$M_{12}$	$M_{13}$	$M_{14}$
1	$+0,54$	$-0,43$	$-0,21$	$-0,35$	$-0,33$	$-0,46$	$0,46$	$-0,81$	$+0,70$	$+0,35$	$+0,35$	$+0,35$	$0$	$-0,35$
2	$-0,43$	$+0,60$	$+0,39$	$+0,53$	$+0,21$	$+0,35$	$+0,29$	$+0,72$	$-1,07$	$-0,54$	$-0,47$	$-0,36$	$0$	$+0,54$
3	$-0,21$	$+0,39$	$+0,31$	$+0,40$	$+0,13$	$+0,21$	$+0,15$	$+0,44$	$-0,79$	$-0,40$	$-0,33$	$-0,23$	$0$	$+0,40$
4	$-0,35$	$+0,53$	$+0,40$	$+0,75$	$+0,21$	$+0,35$	$+0,34$	$+1,25$	$-1,61$	$-1,08$	$-0,91$	$-0,82$	$+0,14$	$+1,00$
5	$-0,33$	$+0,21$	$+0,13$	$+0,21$	$+0,24$	$+0,33$	$+0,33$	$+0,54$	$-0,43$	$-0,21$	$-0,21$	$-0,21$	$0$	$+0,21$
6	$-0,46$	$+0,35$	$+0,21$	$+0,35$	$+0,33$	$+0,51$	$+0,51$	$+0,81$	$-0,70$	$-0,30$	$-0,30$	$-0,30$	$-0,05$	$+0,30$
7	$-0,46$	$+0,29$	$+0,15$	$+0,34$	$+0,32$	$+0,51$	$+0,64$	$+0,92$	$-0,68$	$-0,34$	$-0,26$	$-0,30$	$-0,10$	$+0,24$
8	$-0,81$	$+0,72$	$+0,44$	$+1,25$	$+0,54$	$+0,81$	$+0,92$	$+2,87$	$-2,80$	$-2,08$	$-1,88$	$-1,91$	$+0,38$	$+1,93$
9	$+0,70$	$-1,07$	$-0,79$	$-1,61$	$-0,43$	$-0,70$	$-0,68$	$-2,80$	$+3,53$	$+2,46$	$+2,13$	$+1,94$	$-0,39$	$-2,30$
10	$+0,35$	$-0,54$	$-0,40$	$-1,08$	$-0,21$	$-0,30$	$-0,34$	$-2,08$	$+2,46$	$+1,97$	$+1,70$	$+1,62$	$-0,44$	$-1,82$
11	$+0,35$	$-0,47$	$-0,33$	$-0,91$	$-0,21$	$-0,30$	$-0,26$	$-1,88$	$+2,13$	$+1,70$	$+1,61$	$+1,56$	$-0,47$	$-1,68$
12	$+0,35$	$-0,36$	$-0,23$	$-0,82$	$-0,21$	$-0,30$	$-0,30$	$-1,91$	$+1,94$	$+1,62$	$+1,56$	$+1,58$	$-0,47$	$-1,59$
13	$0$	$0$	$0$	$+0,14$	$0$	$-0,05$	$-0,10$	$+0,38$	$-0,39$	$-0,44$	$-0,47$	$-0,47$	$+0,23$	$+0,47$
14	$-0,35$	$+0,54$	$+0,40$	$+1,00$	$+0,21$	$+0,30$	$+0,24$	$+1,93$	$-2,30$	$-1,82$	$-1,68$	$-1,59$	$+0,47$	$+1,78$

Table XXXIII.

Eqn.	$M_1$	$M_2$	$M_3$	$M_4$	$M_5$	$M_6$	$M_7$	$M_8$	$M_9$	$M_{10}$	$M_{11}$	$M_{12}$	$M_{13}$	$M_{14}$
1	+140,54	+39,65	0	0	+30,62	+30,62	+30,62	+30,62	0	0	0	0	0	0
2	+39,65	+115,52	+36,22	+36,22	0	0	0	0	+18,11	0	0	0	0	0
3	0	+36,22	+152,38	+36,22	−58,08	0	0	0	+18,11	0	0	0	0	0
4	0	+36,22	+36,22	+815,14	0	0	0	0	+18,11	0	0	0	−389,46	0
5	+30,62	0	−58,08	0	+177,40	+61,24	+61,24	+61,24	0	0	0	0	0	0
6	+30,62	0	0	0	+61,24	+329,46	+61,24	+61,24	0	−134,11	0	0	0	0
7	+30,62	0	0	0	+61,24	+61,24	+1198,12	+61,24	0	0	+568,44	0	0	0
8	+30,62	0	0	0	+61,24	+61,24	+61,24	+122,48	0	0	0	+30,62	0	0
9	0	+18,11	+18,11	+18,11	0	0	0	0	+52,80	+16,58	+16,58	0	0	+8,29
10	0	0	0	0	0	−134,11	0	0	+16,58	+284,80	+16,58	0	0	+8,29
11	0	0	0	0	0	0	+568,44	0	+16,58	+16,58	+1153,46	0	0	+8,29
12	0	0	0	0	0	0	0	+30,62	0	0	0	+61,24	0	0
13	0	0	0	−389,46	0	0	0	0	0	0	0	0	+778,92	0
14	0	0	0	0	0	0	0	0	+8,29	+8,29	+8,29	0	0	+16,58

N. V. Nekrasov.

Table XXXIV.

$M_1$	$M_2$	$M_3$	$M_4$	$M_5$	$M_6$	$M_7$	$M_8$
+ 141,08	+ 39,22	− 0,21	− 0,35	+ 30,29	+ 30,16	+ 30,16	+ 29,81
+ 39,22	+ 116,12	+ 36,61	+ 36,75	+ 0,21	+ 0,35	+ 0,29	+ 0,72
− 0,21	+ 36,61	+ 152,69	+ 36,62	− 57,95	+ 0,21	+ 0,15	+ 0,44
− 0,35	+ 36,75	+ 36,62	+ 815,89	+ 0,21	+ 0,35	+ 0,34	+ 1,25
+ 30,29	+ 0,21	− 57,95	+ 0,21	+ 177,64	+ 61,57	+ 61,57	+ 61,78
+ 30,16	+ 0,35	+ 0,21	+ 0,35	+ 61,57	+ 329,97	+ 61,75	+ 62,05
+ 30,16	+ 0,29	+ 0,15	+ 0,34	+ 61,57	+ 61,75	+ 1198,76	+ 62,16
+ 29,81	+ 0,72	+ 0,44	+ 1,25	+ 61,78	+ 62,05	+ 62,16	+ 125,35
+ 0,70	+ 17,04	+ 17,32	+ 0,56	− 0,43	− 0,70	− 0,68	− 2,80
+ 0,35	− 0,54	− 0,40	− 1,08	− 0,21	− 134,41	− 0,34	− 2,08
+ 0,35	− 0,47	− 0,34	− 0,91	− 0,21	− 0,30	+ 568,18	− 1,88
+ 0,35	− 0,36	− 0,23	− 0,82	− 0,21	− 0,30	− 0,30	+ 28,71
0	0	0	− 389,32	0	− 0,05	− 0,10	+ 0,38
− 0,35	+ 0,54	+ 0,40	+ 1,00	+ 0,21	+ 0,30	+ 0,24	+ 1,93

$M_9$	$M_{10}$	$M_{11}$	$M_{12}$	$M_{13}$	$M_{14}$	$A$	$\sum$
+ 0,70	+ 0,35	+ 0,35	+ 0,35	0	− 0,35	− 96,74	+ 204,82	1
+ 17,04	− 0,54	− 0,47	− 0,36	0	+ 0,54	+ 70,70	+ 317,18	2
+ 17,32	− 0,40	− 0,34	− 0,23	0	+ 0,40	+ 43,31	+ 228,62	3
+ 16,50	− 1,08	− 0,91	− 0,82	− 389,32	+ 1,00	+ 70,60	+ 587,03	4
− 0,43	− 0,21	− 0,21	− 0,21	0	+ 0,21	+ 69,35	+ 403,82	5
− 0,70	− 134,41	− 0,30	− 0,30	− 0,05	+ 0,30	+ 111,40	+ 522,35	6
− 0,68	− 0,34	+ 568,18	− 0,30	− 0,10	+ 0,24	+ 135,88	+ 2118,06	7
− 2,80	− 2,08	− 1,88	+ 28,71	+ 0,38	+ 1,93	+ 154,58	+ 522,40	8
+ 56,33	+ 19,04	+ 18,71	+ 1,94	− 0,39	+ 5,99	− 129,14	+ 19,43	9
+ 19,04	+ 286,77	+ 18,28	+ 1,62	− 0,44	+ 6,47	− 43,78	+ 149,25	10
+ 18,71	+ 18,28	+ 1155,07	+ 1,56	− 0,47	+ 6,61	− 16,26	+ 1747,92	11
+ 1,94	+ 1,62	+ 1,56	+ 62,82	− 0,47	− 1,59	− 17,18	+ 75,54	12
− 0,39	− 0,44	− 0,47	− 0,47	+ 779,15	+ 0,47	− 40,77	+ 347,99	13
+ 5,99	+ 6,47	+ 6,61	− 1,59	+ 0,47	+ 18,36	+ 17,78	+ 58,36	14

Similarly to how we proceeded in the first example, it is easy to form the expression for the work of deformation of the truss (in view of the symmetry of the loading—for half of the truss) as a function of the unknowns \(M\), then, differentiating this expression successively with respect to each of the unknowns \(M\) and equating the derivatives to zero, to obtain a system of 14 equations for determining all the \(M\)’s.

In Table XXXII we have placed the quantities denoted by us above by the symbols \((\alpha\alpha)\), \((\alpha\beta)\), etc.; in Table XXXIII—the coefficients of the form \(\Delta_\gamma'\) and \(2\Delta_\gamma'\); finally, Table XXXIV gives the final system of equations.

Solving it by the method indicated in Example I, we obtain the values of the unknowns

\[ \begin{aligned} M_1&=-1{,}4450\ \text{tn.-m.}, & M_8&=+1{,}4807,\\ M_2&=+1{,}3931, & M_9&=-2{,}9041,\\ M_3&=+0{,}3024, & M_{10}&=+0{,}1170,\\ M_4&=+0{,}0482, & M_{11}&=+0{,}0004,\\ M_5&=+0{,}1225, & M_{12}&=-0{,}8047,\\ M_6&=+0{,}1959, & M_{13}&=-0{,}0318,\\ M_7&=+0{,}0538, & M_{14}&=+1{,}5663. \end{aligned} \]

From the equations of Table XXXIV it is easy to obtain the simplified equations of the second method, by setting

\[ M_3=M_4=M_5=M_6=M_7=M_{10}=M_{11}=M_{13}=0. \]

This gives the system

\[ \begin{aligned} &+141{,}08M_1+39{,}22M_2+29{,}81M_8+0{,}70M_9+0{,}35M_{12}\\ &\qquad -0{,}35M_{14}=-96{,}74,\\[4pt] &+39{,}22M_1+116{,}12M_2+0{,}72M_8+17{,}04M_9-0{,}36M_{12}\\ &\qquad +0{,}54M_{14}=+70{,}70,\\[4pt] &+29{,}81M_1+0{,}72M_2+125{,}35M_8-2{,}80M_9+28{,}71M_{12}\\ &\qquad +1{,}93M_{14}=+154{,}58,\\[4pt] &+0{,}70M_1+17{,}04M_2-2{,}80M_8+56{,}33M_9+1{,}94M_{12}\\ &\qquad +5{,}99M_{14}=-129{,}14,\\[4pt] &+0{,}35M_1-0{,}36M_2+28{,}71M_8+1{,}94M_9+62{,}82M_{12}\\ &\qquad -1{,}59M_{14}=-17{,}18,\\[4pt] &-0{,}35M_1+0{,}54M_2+1{,}93M_8+5{,}90M_9-1{,}59M_{12}\\ &\qquad +18{,}36M_{14}=+17{,}78. \end{aligned} \]

Resolving it, we find

\[ \begin{aligned} M_1&=-1{,}4380\ \text{tn.-m.}, & M_9&=-2{.}7718,\\ M_2&=+1{,}4793, & M_{12}&=-0{,}9031,\\ M_8&=+1{,}6867, & M_{14}&=+1{,}5463. \end{aligned} \]

The computational data for Mohr’s method are given in Tables XXXV and XXXVI, entirely analogous to Tables XX and XXI of Example I, and therefore requiring no explanation.

To determine the angles of rotation of the joints, the equations obtained are

\[ 386{,}0\,\varphi_0 + 84{,}1\,\varphi_1 + 108{,}9\,\varphi_2 = 5514{,}72, \]

\[ 84{,}1\,\varphi_0 + 668{,}4\,\varphi_1 + 57{,}4\,\varphi_2 + 184{,}1\,\varphi_3 = 8453{,}82, \]

\[ 108{,}9\,\varphi_0 + 57{,}4\,\varphi_1 + 612{,}0\,\varphi_2 + 19{,}0\,\varphi_3 = 5296{,}35, \]

\[ 184{,}1\,\varphi_1 + 19{,}0\,\varphi_2 + 1233{,}8\,\varphi_3 = 4198{,}20. \]

Whence we obtain

\[ \begin{aligned} \varphi_0&=10{,}4004, & \varphi_2&=5{,}7773,\\ \varphi_1&=10{,}3560, & \varphi_3&=1{,}7684. \end{aligned} \]

After this, the moments and stresses are easily determined in the usual way, as has been done in Table XXXVI.

Table XXXV.

Joints	Elem.	\(N=\dfrac{2EIbr}{l}\) 10 tn.-m.	\(\Sigma N\) 10 tn.-m.	\(\psi \times 10^4\)	\(\psi N \times 10^4\) 10 tn.-m.	\(\Sigma \psi N\)	\(3\Sigma \psi N\)
0	0—1	84,1	193,0	+ 4,066	+ 341,95	+ 1838,24	+ 5514,72
0	0—2	108,9	193,0	+ 13,740	+ 1496,29	+ 1838,24	+ 5514,72
1	1—0	84,1	334,2	+ 4,066	+ 341,95	+ 2817,94	+ 8453,82
1	1—2	57,4	334,2	+ 6,805	+ 390,61	+ 2817,94	+ 8453,82
1	1—3	184,1	334,2	+ 11,137	+ 2050,32	+ 2817,94	+ 8453,82
1	1—4	8,6	334,2	+ 4,077	+ 35,06	+ 2817,94	+ 8453,82
2	2—0	108,9	306,0	+ 13,740	+ 1496,29	+ 1765,45	+ 5296,35
2	2—1	57,4	306,0	+ 6,805	+ 390,61	+ 1765,45	+ 5296,35
2	2—3	24,9	306,0	+ 2,600	+ 64,74	+ 1765,45	+ 5296,35
2	2—4	108,9	306,0	− 1,718	− 187,09	+ 1765,45	+ 5296,35
2	2—3′	5,9	306,0	+ 0,152	+ 0,90	+ 1765,45	+ 5296,35
3	3—1	184,1	616,9	+ 11,137	+ 2050,32	+ 1399,40	+ 4198,20
3	3—2	24,9	616,9	+ 2,600	+ 64,74	+ 1399,40	+ 4198,20
3	3—5	402,0	616,9	− 1,778	− 714,76	+ 1399,40	+ 4198,20
3	[[unclear: element label]]	5,9	616,9	− 0,152	− 0,90	+ 1399,40	+ 4198,20

Table XXXVI.

Elements	Angles	Magnitude of angles	Resultant angles	Moments, kg.-cm.	Stresses, kg./cm.^2	\(N/n\)
0—1	\(\varphi_0\)	\(+10,4004\)	\(+18,9588\)	\(+159444\)	\(\pm 307\)	1,61
0—1	\(\varphi_1\)	\(+10,3560\)	\(+18,9144\)	\(+159070\)	\(\pm 307\)	1,61
0—1	\(\psi_{0-1}\)	\(+4,066\)			\(\pm 307\)	1,61
0—2	\(\varphi_0\)	\(+10,4004\)	\(-14,6419\)	\(-159450\)	\(+482\) \(-193\) \(-635\) \(+234\)	—
0—2	\(\varphi_2\)	\(+5,7773\)	\(-19,2650\)	\(-20979\)	\(+482\) \(-193\) \(-635\) \(+234\)	—
0—2	\(\psi_{0-2}\)	\(+13,740\)			\(+482\) \(-193\) \(-635\) \(+234\)	—
1—2	\(\varphi_1\)	\(+10,3560\)	\(+6,0743\)	\(+34866\)	\(\pm 65\)	1,08
1—2	\(\varphi_2\)	\(+5,7773\)	\(+1,4956\)	\(+8585\)	\(\pm 65\)	1,08
1—2	\(\psi_{1-2}\)	\(+6,805\)			\(\pm 65\)	1,08
1—3	\(\varphi_1\)	\(+10,3560\)	\(-10,9306\)	\(-201232\)	\(+83\) \(-380\) \(-149\) \(+679\)	1,82
1—3	\(\varphi_3\)	\(+1,7684\)	\(-19,5182\)	\(-359330\)	\(+83\) \(-380\) \(-149\) \(+679\)	1,82
1—3	\(\psi_{1-3}\)	\(+11,137\)			\(+83\) \(-380\) \(-149\) \(+679\)	1,82
1—4	\(\varphi_1\)	\(+10,3560\)	\(+8,4810\)	\(+7294\)	\(\pm 50\)	1,21
1—4	\(\varphi_4\)	\(0\)	\(-1,8750\)	\(-1613\)	\(\pm 50\)	1,21
1—4	\(\psi_{1-4}\)	\(+4,077\)			\(\pm 50\)	1,21
2—3	\(\varphi_2\)	\(+5,7773\)	\(+5,5230\)	\(+13752\)	\(\pm 83\)	7,92
2—3	\(\varphi_3\)	\(+1,7684\)	\(+1,5141\)	\(+3770\)	\(\pm 83\)	7,92
2—3	\(\psi_{2-3}\)	\(+2,600\)			\(\pm 83\)	7,92
2—4	\(\varphi_2\)	\(+5,7773\)	\(+16,7086\)	\(+181957\)	\(-552\) \(+220\) \(+360\) \(-144\)	1,74
2—4	\(\varphi_4\)	\(0\)	\(+10,9313\)	\(+119042\)	\(-552\) \(+220\) \(+360\) \(-144\)	1,74
2—4	\(\psi_{2-4}\)	\(-1,718\)			\(-552\) \(+220\) \(+360\) \(-144\)	1,74
2—3′	\(\varphi_2\)	\(+5,7773\)	\(+9,3302\)	\(+5505\)	\(\pm 59\)	2,20
2—3′	\(\varphi_{3′}\)	\(-1,7684\)	\(+1,7845\)	\(+1053\)	\(\pm 59\)	2,20
2—3′	\(\psi_{2-3′}\)	\(+0,152\)			\(\pm 59\)	2,20
3—5	\(\varphi_3\)	\(+1,7684\)	\(+8,8708\)	\(+356606\)	\(-77\) \(+407\) \(+62\) \(-325\)	1,97
3—5	\(\varphi_5\)	\(0\)	\(+7,1022\)	\(+285508\)	\(-77\) \(+407\) \(+62\) \(-325\)	1,97
3—5	\(\psi_{3-5}\)	\(-1,778\)			\(-77\) \(+407\) \(+62\) \(-325\)	1,97

Finally, for the fourth method, as before, with the aid of Tables XXXIII and XXXIV, we write the equations

\[ \begin{aligned} 140{,}54 M_1 + 39{,}65 M_2 + 30{,}62 M_8 &= -96{,}74,\\ 39{,}65 M_1 + 115{,}52 M_2 + 18{,}11 M_9 &= +70{,}70.\\ 30{,}62 M_1 + 122{,}48 M_8 + 30{,}62 M_{12} &= +154{,}58,\\ 18{,}11 M_2 + 52{,}80 M_9 + 8{,}29 M_{14} &= -129{,}14,\\ 30{,}62 M_8 + 61{,}24 M_{12} &= -17{,}18,\\ 8{,}29 M_9 + 16{,}58 M_{14} &= +17{,}18. \end{aligned} \]

Solving the system, we obtain

\[ \begin{array}{ll} M_1=-1{,}6016\ \text{tn.-m.}, & M_9=-3{,}4726,\\ M_2=+1{,}7064, & M_{12}=-1{,}2710,\\ M_8=+1{,}9805, & M_{14}=+2{,}8084. \end{array} \]

18. Example III. A truss with a triangular web and additional posts.

The determination of the forces in the truss (Fig. 12), as a statically determinate system, presents no difficulty. In Table XXXVII the principal dimensions of the sections and the forces from the adopted load shown in the drawing are given. Likewise, Table XXXVIII contains the forces, necessary for the further calculation, due to a load by pairs of forces. For the calculation of the bending moments due to the rigidity of the joints, this time we shall have to determine 12 unknowns, through which all the moments are expressed in the following manner:

\[ \begin{array}{ll} M_{0-1}=-M_1, & M_{1-0}=+M_2,\\ M_{0-2}=+M_1, & M_{2-0}=-(M_5+M_6),\\ M_{1-2}=+M_3, & M_{2-1}=+M_5,\\ M_{1-3}=+M_4, & M_{3-1}=+M_7,\\ M_{1-4}=-(M_2+M_3+M_4), & M_{4-1}=+M_{10},\\ M_{2-3}=+M_6, & M_{3-2}=-(M_7+M+M_9),\\ M_{3-4}=+M_8, & M_{4-3}=-M_{11},\\ M_{3-5}=+M_9, & M_{5-3}=-M. \end{array} \]

Fig. 12.

\[ \alpha = 38^\circ 40' \qquad \sin \alpha = \cos \beta = 0.6247 \]

\[ \beta = 51^\circ 20' \qquad \cos \alpha = \sin \beta = 0.7809 \]

\[ \tan \alpha = \cot \beta = 0.8000 \]

Table XXXVII.

Elem.	$\omega$ br. cm.$^2$	$I$ br. cm.$^4$	$e_1$ cm.	$e_2$ cm.	$l$ cm.	$N=\dfrac{2EI}{l}$ 10 tn.-m.	$\Delta\gamma=\dfrac{l}{6EI}$ $\dfrac{1}{10^5\ \text{tn.-m.}}$	Forces, $S$ tn.	Stress kg./cm.$^2$	$\lambda=\dfrac{Sl}{E\omega}\times 10^2$ cm.
0—2	106	14700	9,0	31,8	400	147,0	22,68	+ 44,00	+ 415	+ 8,302
2—3	106	14700	9,0	31,8	400	147,0	22,68	+ 44,00	+ 415	+ 8,302
3—5	138	16850	7,6	34,0	400	168,5	19,78	+ 68,00	+ 493	+ 9,855
1—4	138	16850	7,6	34,0	800	84,2	39,56	− 56,00	− 406	− 16,232
0—1	145	5080	16	16	640	31,7	104,99	− 70.40	− 486	− 15,537
1—3	92	3000	12,5	12,5	640	18,7	177,78	+ 19.20	+ 209	+ 6,678
3—4	92	3000	12,5	12,5	640	18,7	177,78	− 19.20	− 209	− 6,678
1—2	80	2890	13	13	500	23,1	144,18	+ 40,00	+ 500	+ 12,500
4—5	80	2890	13	13	500	23,1	144,18	+ 30.00	+ 375	+ 9.375

Accordingly, let us determine the magnitudes of the forces \(\alpha, \beta \ldots \theta\) in the members of the truss under the action of symmetric loads \(M_1 \ldots M_{12}\).

In Table XXXIX these forces are given, and in Table XL their products by

\[ \Delta\lambda=\frac{l}{E\omega}. \]

To determine the unknowns we proceed exactly as in the preceding examples; Tables XLII and XLI give us the data for forming a system of equations with 12 unknowns, and the final values of the coefficients are combined in Table XLIII.

The solution of this system will give us the values of the unknowns

\[ \begin{aligned} M_1&=-0.4209\ \text{tn.-m.}, & M_7&=+0.0081,\\ M_2&=+0.0663, & M_8&=+0.0572,\\ M_3&=+0.0833, & M_9&=-1.1586,\\ M_4&=+0.0414, & M_{10}&=-0.8147,\\ M_5&=+0.1289, & M_{11}&=+0.0483,\\ M_7&=+1.6439, & M_{12}&=+2.1070. \end{aligned} \]

Next, the equations of the second method, putting

\[ M_3=M_4=M_5=M_7=M_8=M_{11}=0, \]

are obtained from Table XLIII in the following form:

\[ +255.62M_1+105.04M_2+22.15M_6+0.20M_9=-61.55, \]

\[ +105.04M_1+289.49M_2-0.42M_6-0.06M_9+39.41M_{10} \]
\[ +0.18M_{12}=-47.49, \]

\[ +22.15M_1-0.42M_2+92.23M_6+21.93M_9 \]
\[ +0.18M_{10}=124.06, \]

\[ +0.20M_1-0.06M_2+21.93M_6+86.72M_9-0.06M_{10} \]
\[ +18.64M_{12}=-22.22, \]

\[ +39.41M_2+0.18M_6-0.06M_9+79.39M_{10} \]
\[ -0.24M_{12}=-57.31, \]

\[ +0.18M_2+18.64M_9-0.24M_{10}+40.76M_{12}=+64.52. \]

Table XXXVIII.

(Forces are expressed in \(\frac{1}{10\ \mathrm{m}}\).)

Elem.	0—2	2—3	3—5	1—4	0—1	1—3	3—4	1—2	4—5
	Forces in the elements of the truss due to the action of a pair of forces \(+1\) tn·m, applied normally to the elements:	Forces in the elements of the truss due to the action of a pair of forces \(+1\) tn·m, applied normally to the elements:	Forces in the elements of the truss due to the action of a pair of forces \(+1\) tn·m, applied normally to the elements:	Forces in the elements of the truss due to the action of a pair of forces \(+1\) tn·m, applied normally to the elements:	Forces in the elements of the truss due to the action of a pair of forces \(+1\) tn·m, applied normally to the elements:	Forces in the elements of the truss due to the action of a pair of forces \(+1\) tn·m, applied normally to the elements:	Forces in the elements of the truss due to the action of a pair of forces \(+1\) tn·m, applied normally to the elements:	Forces in the elements of the truss due to the action of a pair of forces \(+1\) tn·m, applied normally to the elements:
0—2	+1.667	−0.333	−0.333	−0.333	+1.667	−0.333	−0.333	−0.333
2—3	+1.667	−0.333	−0.333	−0.333	+1.667	−0.333	−0.333	+1.667
3—5	+1.000	+1.000	+1.000	+1.000	+1.000	+1.000	+1.000	+1.000
5—3′	+1.000	+1.000	+1.000	+1.000	+1.000	+1.000	+1.000	+1.000	In view of the symmetry of the loading, member 4—5 is not subjected to bending.
3′—2′	+0.333	+0.333	+0.333	+0.333	+0.333	+0.333	+0.333	+0.333	In view of the symmetry of the loading, member 4—5 is not subjected to bending.
2′—0′	+0.333	+0.333	+0.333	+0.333	+0.333	+0.333	+0.333	+0.333	In view of the symmetry of the loading, member 4—5 is not subjected to bending.
1—4	−1.333	−1.333	+0.667	−0.333	−1.333	−1.333	+0.667	−1.333	In view of the symmetry of the loading, member 4—5 is not subjected to bending.
4—1′	−0.667	−0.667	−0.667	−0.667	−0.667	−0.667	−0.667	−0.667	In view of the symmetry of the loading, member 4—5 is not subjected to bending.
0—1	−2.667	+0.533	+0.533	+0.533	−0.717	+0.533	+0.533	+0.533	In view of the symmetry of the loading, member 4—5 is not subjected to bending.
1—3	−0.533	−2.667	−0.533	+1.067	−0.533	+0.717	−0.533	−0.533	In view of the symmetry of the loading, member 4—5 is not subjected to bending.
3—4	+0.533	+0.533	−2.667	−1.067	+0.533	+0.533	−0.717	+0.533	In view of the symmetry of the loading, member 4—5 is not subjected to bending.
4—3′	−0.533	−0.533	−0.533	−0.533	−0.533	−0.533	−0.533	−0.533	In view of the symmetry of the loading, member 4—5 is not subjected to bending.
3′—1′	+0.533	+0.533	+0.533	+0.533	+0.533	+0.533	+0.533	+0.533	In view of the symmetry of the loading, member 4—5 is not subjected to bending.
1′—0′	−0.533	−0.533	−0.533	−0.533	−0.533	−0.533	−0.533	−0.533	In view of the symmetry of the loading, member 4—5 is not subjected to bending.
1—2	+2.500	−2.500	0	0	0	0	0	0
4—5	0	0	+2.500	0	0	0	0	0
1′—2′	0	0	0	0	0	0	0	0

Table XXXIX.

Elements	$\alpha$	$\beta$	$\gamma$	$\delta$	$\varepsilon$	$\eta$	$\vartheta$	$x$	$\lambda$	$\mu$	$\nu$	$o$
0—2	0	+0,200	0	0	−0,200	−0,200	0	0	0	0	0	0
2—3	0	+0,200	+0,200	0	0	−0,200	0	0	0	0	0	0
3—5	0	0	0	0	0	0	0	0	0	0	0	0
1—4	0	−0,100	−0,100	−0,100	0	0	0	+0,200	+0,200	+0,200	−0,200	−0,200
0—1	−0,195	−0,125	0	0	+0,320	+0,320	0	0	0	−0,100	0	0
1—3	0	−0,160	−0,160	−0,035	0	+0,320	−0,195	−0,320	−0,320	+0,160	0	0
3—4	0	+0,160	+0,160	+0,160	0	0	0	−0,125	−0,320	−0,160	+0,125	+0,320
1—2	+0,250	0	0	0	−0,250	−0,500	+0,250	+0,250	+0,250	0	0	0
4—5	0	0	0	0	0	0	0	0	+0,500	0	0	−0,500

Table XL.

Produced $\Delta\lambda$, for unit external force:

elem.	$\alpha$	$\beta$	$\gamma$	$\delta$	$\varepsilon$	$\eta$	$\vartheta$	$x$	$\lambda$	$\mu$	$\nu$	$o$	$\Delta_{\lambda} = \dfrac{l}{E\omega}\,10^3$ m, tn.
0—2	0	+0,377	0	0	−0,377	−0,377	0	0	0	0	0	0	1,8868
2—3	0	+0,377	+0,377	0	0	−0,377	0	0	0	0	0	0	1,8868
3—5	0	0	0	0	0	0	0	0	0	0	0	0	1,4493
1—4	0	−0,290	−0,290	−0,290	0	0	0	+0,580	+0,580	+0,290	−0,290	−0,290	2,8986
0—1	−0,430	−0,276	0	0	+0,706	+0,706	0	0	0	0	0	0	2,2069
1—3	0	−0,557	−0,557	−0,122	0	+1,113	−0,678	−1,113	−1,113	+0,557	0	0	3,4783
3—4	0	+0,557	+0,557	+0,557	0	0	0	−0,435	−1,113	−0,557	+0,435	+1,113	3,4783
1—2	+0,781	0	0	0	−0,781	−1,562	+0,781	+0,781	+0,781	0	0	0	3,1250
4—5	0	0	0	0	0	0	0	0	+1,562	0	0	−1,562	3,1250

Table XII.

No.	$M_1$	$M_2$	$M_3$	$M_4$	$M_5$	$M_6$	$M_7$	$M_8$	$M_9$	$M_{10}$	$M_{11}$	$M_{12}$	$A$
1	+255.34	+104.99	0	0	+22.68	+22.68	0	0	0	0	0	0	−61.55
2	+104.99	+289.10	+79.12	+79.12	0	0	0	0	0	+39.56	0	0	−47.49
3	0	+79.12	+367.48	+79.12	−144.18	0	0	0	0	+39.56	0	0	−11.47
4	0	+79.12	+79.12	+434.68	0	0	−177.78	0	0	+39.56	0	0	−3.21
5	+22.68	0	−144.18	0	+333.72	−45.36	0	0	0	0	0	0	+97.57
6	+22.68	0	0	0	+45.36	+90.72	+22.68	+22.68	+22.68	0	0	0	+124.06
7	0	0	0	−177.78	0	+22.68	+400.92	+45.36	+45.36	0	0	0	−18.23
8	0	0	0	0	0	+22.68	+45.36	+400.92	+45.36	0	+177.78	0	+14.24
9	0	0	0	0	0	+22.68	+45.36	+45.36	+84.92	0	0	−19.78	−22.22
10	0	+39.56	+39.56	+39.56	0	0	0	0	0	+79.12	0	0	−57.31
11	0	0	0	0	0	0	0	+177.78	0	0	+355.56	0	+28.06
12	0	0	0	0	0	0	0	0	+19.78	0	0	+39.56	+64.52

Table XLII.

\(M_1\)	\(M_2\)	\(M_3\)	\(M_4\)	\(M_5\)	\(M_6\)	\(M_7\)	\(M_8\)	\(M_9\)	\(M_{10}\)	\(M_{11}\)	\(M_{12}\)
\(+0.28\)	\(+0.05\)	\(0\)	\(0\)	\(-0.33\)	\(-0.53\)	\(+0.20\)	\(+0.20\)	\(+0.20\)	\(0\)	\(0\)	\(0\)	1
\(+0.05\)	\(+0.39\)	\(+0.28\)	\(+0.14\)	\(-0.16\)	\(-0.42\)	\(+0.11\)	\(+0.05\)	\(-0.06\)	\(-0.15\)	\(+0.07\)	\(+0.18\)	2
\(0\)	\(+0.28\)	\(+0.28\)	\(+0.14\)	\(0\)	\(-0.25\)	\(+0.11\)	\(+0.05\)	\(-0.06\)	\(-0.15\)	\(+0.07\)	\(+0.18\)	3
\(0\)	\(+0.14\)	\(+0.14\)	\(+0.12\)	\(0\)	\(-0.04\)	\(+0.02\)	\(-0.09\)	\(-0.20\)	\(-0.08\)	\(+0.07\)	\(+0.18\)	4
\(-0.33\)	\(-0.16\)	\(0\)	\(0\)	\(+0.50\)	\(+0.69\)	\(-0.20\)	\(-0.20\)	\(-0.20\)	\(0\)	\(0\)	\(0\)	5
\(-0.53\)	\(-0.42\)	\(-0.25\)	\(-0.04\)	\(+0.69\)	\(+1.51\)	\(-0.61\)	\(-0.75\)	\(-0.75\)	\(+0.18\)	\(0\)	\(0\)	6
\(+0.20\)	\(+0.11\)	\(+0.11\)	\(+0.02\)	\(-0.20\)	\(-0.61\)	\(+0.33\)	\(+0.41\)	\(+0.41\)	\(-0.11\)	\(0\)	\(0\)	7
\(+0.20\)	\(+0.05\)	\(+0.05\)	\(-0.09\)	\(-0.20\)	\(-0.75\)	\(+0.41\)	\(+0.72\)	\(+0.81\)	\(-0.17\)	\(-0.05\)	\(-0.14\)	8
\(+0.20\)	\(-0.06\)	\(-0.06\)	\(-0.20\)	\(-0.20\)	\(-0.75\)	\(+0.41\)	\(+0.81\)	\(+1.80\)	\(-0.06\)	\(-0.14\)	\(-1.14\)	9
\(0\)	\(-0.15\)	\(-0.15\)	\(-0.08\)	\(0\)	\(+0.18\)	\(-0.11\)	\(-0.17\)	\(-0.06\)	\(+0.27\)	\(-0.13\)	\(-0.24\)	10
\(0\)	\(+0.07\)	\(+0.07\)	\(+0.07\)	\(0\)	\(0\)	\(0\)	\(-0.05\)	\(-0.14\)	\(-0.13\)	\(+0.11\)	\(+0.20\)	11
\(0\)	\(+0.18\)	\(+0.18\)	\(+0.18\)	\(0\)	\(0\)	\(0\)	\(-0.14\)	\(-1.14\)	\(-0.24\)	\(+0.20\)	\(+1.20\)	12

Hence we obtain

\[ \begin{aligned} M_1&=-0.4302\ \text{tn.-m.}, &\qquad M_9&=-1.1895,\\ M_2&=+0.0976, & M_{10}&=-0.7686,\\ M_6&=+1.7332, & M_{12}&=+2.1004. \end{aligned} \]

In the calculation by Mohr’s method, after computing the angles \(\psi\), we shall have the following system of equations for determining the angles \(\varphi\):

\[ 357.4\varphi_0 + 31.7\varphi_1 + 147.0\varphi_2 = 8332.71, \]

\[ 31.7\varphi_0 + 315.4\varphi_1 + 23.1\varphi_2 + 18.7\varphi_3 = 3260.46, \]

\[ 147.0\varphi_0 + 23.1\varphi_1 + 634.2\varphi_2 + 147.0\varphi_3 = 9677.67, \]

\[ 18.7\varphi_1 + 147.0\varphi_2 + 705.8\varphi_3 = 5623.38. \]

Solving this system, we find its roots

\[ \begin{aligned} \varphi_0&=+18.8452, &\qquad \varphi_2&=+9.2674,\\ \varphi_1&=+7.4182, & \varphi_3&=+5.8408. \end{aligned} \]

After this, by the usual procedure, we determine the moments and stresses, which has also been done in Table XLIV.

In view of the fact that the difference between the three calculations carried out for this case is insignificant, no calculation by the fourth method was made for this truss.

Table XLIII.

No.	$M_1$	$M_2$	$M_3$	$M_4$	$M_5$	$M_6$	$M_7$	$M_8$	$M_9$	$M_{10}$	$M_{11}$	$M_{12}$	$A$	$\sum$
1	−255,62	+105,04	0	0	+22,35	+22,15	+0,20	+0,20	+0,20	0	0	0	−61,55	+344,21
2	+105,04	+289,49	+79,40	+79,26	−0,16	−0,42	+0,11	+0,05	−0,06	+39,41	+0,07	+0,18	−47,49	+544,88
3	0	−79,40	+367,76	+79,26	−144,18	−0,25	+0,11	+0,05	−0,06	+39,41	+0,07	+0,18	−11,47	+410,28
4	0	+79,26	+79,26	+434,80	0	−0,04	−177,76	−0,09	−0,20	+39,48	+0,07	+0,18	−3,21	+451,75
5	+22,35	−0,16	−144,18	0	+334,22	+46,05	−0,20	−0,20	−0,20	0	0	0	+97,57	+355,25
6	+22,15	−0,42	−0,25	−0,04	+46,05	+92,23	+22,07	+21,93	+21,93	+0,18	0	0	+124,06	+349,89
7	+0,20	+0,11	+0,11	−177,76	−0,20	+22,07	+401,25	+45,77	+45,77	−0,11	0	0	−18,23	+318,98
8	+0,20	+0,05	+0,05	−0,09	−0,20	+21,93	+45,77	+401,64	+46,17	−0,17	+177,73	+0,14	+14,24	+707,18
9	+0,20	−0,06	−0,06	−0,20	−0,20	+21,93	+45,77	+46,17	+86,72	−0,06	+0,14	+18,64	−22,22	+196,49
10	0	+39,41	+39,41	+39,48	0	+0,18	−0,11	−0,17	−0,06	+79,39	−0,13	−0,24	−57,31	+139,85
11	0	+0,07	+0,07	+0,07	0	0	0	+177,73	−0,14	−0,13	+355,67	+0,20	+28,06	+561,60
12	0	+0,18	+0,18	+0,18	0	0	0	−0,14	+18,64	−0,24	+0,20	+40,76	+64,52	+124,28

Table XLIV.

Elem.	Angles	Angular quantities	Result. angles	Moments, kg.-cm.	Stresses, kg./cm²	$\dfrac{N}{n}$
0—1	$\varphi_0$	$+\,18{,}8452$	$+\,13{,}6686$	$+\,43329$	$\pm\,137$	$1{,}28$
0—1	$\varphi_1$	$+\,7{,}4182$	$+\,2{,}2416$	$+\,7106$	$\pm\,137$	$1{,}28$
0—1	$\psi_{0-1}$	$+\,10{,}480$			$\pm\,137$	$1{,}28$
0—2	$\varphi_0$	$+\,18{,}8452$	$-\,2{,}9472$	$-\,43324$	$+\,94$	$1{,}23$
0—2	$\varphi_2$	$+\,9{,}2674$	$-\,12{,}5250$	$-\,184118$	$+\,94$	$1{,}23$
0—2	$\psi_{0-2}$	$+\,16{,}635$			$+\,94$	$1{,}23$
1—2	$\varphi_1$	$+\,7{,}4182$	$+\,3{,}4698$	$+\,8015$	$\pm\,55$	$1{,}11$
1—2	$\varphi_2$	$+\,9{,}2674$	$+\,5{,}3190$	$+\,12287$	$\pm\,55$	$1{,}11$
1—2	$\psi_{1-2}$	$+\,6{,}878$			$\pm\,55$	$1{,}11$
1—3	$\varphi_1$	$+\,7{,}4182$	$+\,2{,}5212$	$+\,4715$	$\pm\,20$	$1{,}10$
1—3	$\varphi_3$	$+\,5{,}8408$	$+\,0{,}9438$	$+\,1765$	$\pm\,20$	$1{,}10$
1—3	$\psi_{1-3}$	$+\,6{,}052$			$\pm\,20$	$1{,}10$
1—4	$\varphi_1$	$+\,7{,}4182$	$-\,2{,}3566$	$-\,19843$	$+\;9$ $-\;40$	$1{,}10$
1—4	$\varphi_4$	$0$	$-\,9{,}7748$	$-\,82304$	$+\,166$ $-\;37$	$1{,}10$
1—4	$\psi_{1-4}$	$+\,5{,}731$				$1{,}10$
2—3	$\varphi_2$	$+\,9{,}2674$	$+\,11{,}6886$	$+\,171822$	$+\,263$	$1{,}63$
2—3	$\varphi_3$	$+\,5{,}8408$	$+\,8{,}2620$	$+\,121451$	$+\,263$	$1{,}63$
2—3	$\psi_{2-3}$	$+\,4{,}229$			$+\,263$	$1{,}63$
3—4	$\varphi_3$	$+\,5{,}8408$	$+\,3{,}2636$	$+\,6103$	$\pm\,25$	$1{,}12$
3—4	$\varphi_4$	$0$	$-\,2{,}5772$	$-\,4819$	$\pm\,25$	$1{,}12$
3—4	$\psi_{3-4}$	$+\,2{,}806$			$\pm\,25$	$1{,}12$
3—5	$\varphi_3$	$+\,5{,}8408$	$-\,7{,}6744$	$-\,129314$	$+\,261$	$1{,}53$
3—5	$\varphi_5$	$0$	$-\,13{,}5152$	$-\,227731$	$+\,261$	$1{,}53$
3—5	$\psi_{3-5}$	$+\,6{,}452$			$+\,261$	$1{,}53$

19. Example IV. The truss of Example III with the stiffness of the chords reduced by half.

For a better clarification of the significance that the stiffness of the truss chords has when comparing calculations of stiffness stresses by various methods, let us compute the aforementioned stresses for truss III under the assumption that the moments of inertia and the section moduli of all elements of the truss chords are reduced by half, while the other data remain unchanged.

In the calculation by the generalized method, such a change will be reflected first of all in Table XLIII and will be expressed in the fact that the values \(\Delta'\), corresponding to the chord elements, will be doubled.

A new Table XLIII′ is obtained; changing accordingly the general form of the equations, we arrive at a new system, shown in Table XLIV′, from which

\[ \begin{aligned} M_1&=-0{,}3546\ \text{tn.-m.}, & M_7&=+0{,}0040,\\ M_2&=+0{,}0190, & M_8&=+0{,}0557,\\ M_3&=+0{,}0703, & M_9&=-0{,}6262,\\ M_4&=+0{,}0319, & M_{10}&=-0{,}4213,\\ M_5&=+0{,}1227, & M_{11}&=+0{,}0500,\\ M_6&=+0{,}8424; & M_{12}&=+1{,}1013. \end{aligned} \]

For the second method we shall have the system

\[ 300{,}98M_1+105{,}04M_2+44{,}83M_6+0{,}20M_9=-61{,}55, \]

\[ 105{,}04M_1+368{,}61M_2-0{,}42M_6-0{,}06M_9+78{,}97M_{10} +0{,}18M_{12}=-47{,}49, \]

\[ 44{,}83M_1-0{,}42M_2+182{,}95M_6+44{,}61M_9+0{,}18M_{10}=+124{,}06, \]

\[ 0{,}20M_1-0{,}06M_2+44{,}61M_6+171{,}64M_9-0{,}06M_{10} +38{,}42M_{12}=-22{,}22, \]

\[ +78{,}97M_2+0{,}18M_6-0{,}06M_9+158{,}51M_{10}-0{,}24M_{12}=-57{,}31, \]

\[ +0{,}18M_2+38{,}42M_9-0{,}24M_{10}+80{,}32M_{12}=+64{,}52. \]

Table XLIII.

	$M_1$	$M_2$	$M_3$	$M_4$	$M_5$	$M_6$	$M_7$	$M_8$	$M_9$	$M_{10}$	$M_{11}$	$M_{12}$
1	+300,70	+104,99	0	0	+45,36	+45,36	0	0	0	0	0	0
2	+104,99	+368,22	+158,24	+158,24	0	0	0	0	0	+79,12	0	0
3	0	+158,24	+446,60	+158,24	−144,18	0	0	0	0	+79,12	0	0
4	0	+158,24	+158,24	+513,80	0	0	−177,78	0	0	+79,12	0	0
5	+45,36	0	−144,18	0	+379,08	+90,72	0	0	0	0	0	0
6	+45,36	0	0	0	+90,72	+181,44	+45,36	+45,36	+45,36	0	0	0
7	0	0	0	−177,78	0	+45,36	+446,28	+90,72	+90,72	0	+177,78	0
8	0	0	0	0	0	+45,36	+90,72	+446,28	+90,72	0	+177,78	0
9	0	0	0	0	0	+45,36	+90,72	+90,72	+169,84	0	0	+39,56
10	0	+79,12	+79,12	+79,12	0	0	0	0	0	+158,24	0	0
11	0	0	0	0	0	0	0	+177,78	0	0	+355,56	0
12	0	0	0	0	0	0	0	0	+39,56	0	0	+79,12

Table XLIV.

\(M_1\)	\(M_2\)	\(M_3\)	\(M_4\)	\(M_5\)	\(M_6\)	\(M_7\)	\(M_8\)	\(M_9\)	\(M_{10}\)	\(M_{11}\)	\(M_{12}\)	\(A\)	\(\Sigma\)
\(+200,98\)	\(+105,04\)	\(0\)	\(0\)	\(+45,03\)	\(+44,83\)	\(+0,20\)	\(+0,20\)	\(+0,20\)	\(0\)	\(0\)	\(0\)	\(-61,55\)	\(+434,93\)	1
\(+105,04\)	\(+368,61\)	\(+158,52\)	\(+158,38\)	\(-0,16\)	\(-0,42\)	\(+0,11\)	\(+0,05\)	\(-0,06\)	\(+78,97\)	\(+0,07\)	\(+0,18\)	\(-47,49\)	\(+821,80\)	2
\(0\)	\(+158,52\)	\(+446,88\)	\(+158,38\)	\(-144,18\)	\(-0,25\)	\(+0,11\)	\(+0,05\)	\(-0,06\)	\(+78,97\)	\(+0,07\)	\(+0,18\)	\(-11,47\)	\(+687,20\)	3
\(0\)	\(+158,38\)	\(+158,38\)	\(+513,92\)	\(0\)	\(-0,04\)	\(-177,76\)	\(-0,09\)	\(-0,20\)	\(+79,04\)	\(+0,07\)	\(+0,18\)	\(-3,21\)	\(+728,67\)	4
\(+45,03\)	\(-0,16\)	\(-144,14\)	\(0\)	\(+379,58\)	\(+91,41\)	\(+0,20\)	\(-0,20\)	\(-0,20\)	\(0\)	\(0\)	\(0\)	\(+97,57\)	\(+468,65\)	5
\(+44,83\)	\(-0,42\)	\(-0,25\)	\(-0,04\)	\(+91,41\)	\(+182,95\)	\(+44,75\)	\(+44,61\)	\(+44,61\)	\(+0,18\)	\(0\)	\(0\)	\(+124,06\)	\(+576,69\)	6
\(+0,20\)	\(+0,11\)	\(+0,11\)	\(-177,76\)	\(+0,20\)	\(+44,75\)	\(+446,61\)	\(+91,13\)	\(+91,13\)	\(-0,11\)	\(0\)	\(0\)	\(-18,23\)	\(+477,74\)	7
\(+0,20\)	\(+0,05\)	\(+0,05\)	\(-0,09\)	\(-0,20\)	\(+44,61\)	\(+91,13\)	\(+447,00\)	\(+91,53\)	\(-0,17\)	\(+177,73\)	\(-0,14\)	\(+14,24\)	\(+865,94\)	8
\(+0,20\)	\(-0,06\)	\(-0,06\)	\(-0,20\)	\(-0,20\)	\(+44,61\)	\(+91,13\)	\(+91,53\)	\(+171,64\)	\(-0,06\)	\(-0,14\)	\(+38,42\)	\(-22,22\)	\(+414,59\)	9
\(0\)	\(+78,97\)	\(+78,97\)	\(+79,04\)	\(0\)	\(+0,18\)	\(-0,11\)	\(-0,17\)	\(-0,06\)	\(+158,51\)	\(-0,13\)	\(-0,24\)	\(-57,31\)	\(+337,65\)	10
\(0\)	\(+0,07\)	\(+0,07\)	\(+0,07\)	\(0\)	\(0\)	\(0\)	\(+177,73\)	\(-0,14\)	\(-0,13\)	\(+355,67\)	\(+0,20\)	\(+28,06\)	\(+561,60\)	11
\(0\)	\(+0,18\)	\(+0,18\)	\(+0,18\)	\(0\)	\(0\)	\(0\)	\(-0,14\)	\(+38,42\)	\(-0,24\)	\(+0,20\)	\(+80,32\)	\(+64,52\)	\(+183,62\)	12

Hence

\[ \begin{aligned} M_1&=-0.3608\ \text{tn.-m.},& M_9&=-0.6124,\\ M_2&=+0.0579,& M_{10}&=-0.3900,\\ M_6&=+0.9164,& M_{12}&=+1.0949. \end{aligned} \]

For Mohr’s method the previously computed angles \(\psi\) do not change, while \(\varphi\) are computed from the equations

\[ \begin{aligned} 210.4\varphi_0+31.7\varphi_1+73.5\varphi_2&=4664.67,\\ 31.7\varphi_0+231.2\varphi_1+23.1\varphi_2+18.7\varphi_3&=2536.65,\\ 73.5\varphi_0+23.1\varphi_1+340.2\varphi_2+73.5\varphi_3&=5077.14,\\ 18.7\varphi_1+73.5\varphi_2+390.2\varphi_3&=3059.19. \end{aligned} \]

We find

\[ \begin{aligned} \varphi_0&=+17.8308,& \varphi_2&=+9.3480,\\ \varphi_1&=+7.1289,& \varphi_3&=+5.7374. \end{aligned} \]

For the moments we obtain the values

\[ \begin{aligned} M_{0-1}&=+35981\ \text{kg.-cm.},& M_{1-0}&=+2056\ \text{kg.-cm.},\\ M_{0-2}&=-35981,& M_{2-0}&=-98330,\\ M_{1-2}&=+6865,& M_{2-1}&=+11991,\\ M_{0-3}&=+3439,& M_{3-1}&=+837,\\ M_{1-4}&=-12357,& M_{4-1}&=-42370,\\ M_{2-3}&=+86336,& M_{3-2}&=+59798,\\ M_{3-4}&=+5716,& M_{4-3}&=-5013,\\ M_{3-5}&=-66360,& M_{5-3}&=-114669. \end{aligned} \]

20. The Influence of the Assumptions We Have Made (in Explaining the General Method).

In concluding our calculations, we shall try to determine at least approximately to what degree the final results of the calculations are affected by those assumptions which had to be introduced (as we saw in Chapter I) so that the problem of determining the moments from the rigidity of the joints could be solved without the aid of several approximations—from a single calculation.

The most important assumption consisted in the fact that, when composing the expression for the work of deformation of the truss members in bending, we ignored the influence of the longitudinal force in the member on its bending.

Some idea of the magnitude of the error that we thereby make can be formed in the following way. As we saw (formula [21], Ch. 1), the work of deformation of a member in bending is expressed exactly as

\[ A=\frac{l}{6EI}\left\{(M_1^2+M_2^2)\,[1+\varphi(S)]-M_1M_2\,[1+\psi(S)]\right\}. \]

In all our calculations we took \(\varphi(S)\) and \(\psi(S)\) to be equal to zero.

Greater accuracy in the calculation can be obtained by assuming

\[ \varphi(S)=\alpha=\mathrm{Const.} \qquad \text{and} \qquad \psi(S)=\beta=\mathrm{Const.} \]

Thus we suppose that the functions \(\varphi(S)\) and \(\psi(S)\) do not depend on the moments \(M\), and we compute the values of these functions by substituting into the corresponding formulas or series, in place of \(S\), the quantity \(S^\circ\)—the force in the member of the hinged truss.

Then, substituting the quantities \(\alpha\) and \(\beta\) computed for each member into the formulas for the work of deformation and differentiating, we obtain for all the members a series of expressions of the form

\[ \frac{\partial A_{0-1}}{\partial M_i} = \frac{\partial A'_{0-1}}{\partial M_i} + \Delta\gamma_{0-1} \left[ \frac{\partial \sum M^2}{\partial M_i}\,(1+\alpha_{0-1}) + \frac{\partial \sum MM}{\partial M_i}\,(1+\beta_{0-1}) \right], \]

which can be rewritten otherwise as

\[ \frac{\partial A_{0-1}}{\partial M_i} = \frac{\partial A'_{0-1}}{\partial M_i} + \Delta\gamma_{0-1}(1+\alpha_{0-1}) \frac{\partial \sum M^2}{\partial M_i} + \Delta\gamma_{0-1}(1+\beta_{0-1}) \frac{\partial \sum MM}{\partial M_i}, \]

Summing these expressions for each \(M_i\) and setting the sums equal to zero, we obtain equations entirely similar to the equations of the generalized method set forth above.

Solving them, we shall find the moments due to the rigidity of the joints, and then also the forces in the members of the rigid truss.

Comparing the latter with the forces in the hinged truss, we shall see approximately how significant an error we made in determining \(\alpha\) and \(\beta\).

Let us apply the method set forth to the truss of Example III. To calculate \(\alpha\) and \(\beta\), for small values of

\[ \alpha^2 l^2 = -\frac{S l^2}{EI} \]

we shall use the series expansions (21) and (21′), and, for considerable values, formulas (22) and (22′).

The results of calculating \(\alpha\), \(\beta\), and the other quantities are given in Table XLIV.

Proceeding now to the compilation of the ordinary equations, we note that the data of Table XLII retain their value for the new calculation as well; Table XLI, however, will be replaced by Table XLI bis, placed below.

From a comparison of Tables XLI bis and XLII we obtain Table XLIII bis, representing the equations we need for determining the quantities \(M\).

\[ \begin{aligned} M_1 &= -0{,}3275\;(-0{,}4209), & M_7 &= +0{,}0030\;(+0{,}0081),\\ M_2 &= +0{,}1214\;(+0{,}0663), & M_8 &= +0{,}0466\;(+0{,}0572),\\ M_3 &= +0{,}0712\;(+0{,}0833), & M_9 &= -1{,}1422\;(-1{,}1586),\\ M_4 &= +0{,}0388\;(+0{,}0414), & M_{10} &= -0{,}8116\;(-0{,}8147),\\ M_5 &= +0{,}1342\;(+0{,}1289), & M_{11} &= +0{,}0368\;(+0{,}0483),\\ M_6 &= +1{,}6527\;(+1{,}6439), & M_{12} &= +2{,}1456\;(+2{,}1070). \end{aligned} \]

For comparison, the quantities calculated by us above according to the first method are given in parentheses; they agree very closely with the newly obtained

Table XLIV.

Elements	Forces in the hinged truss	\(\alpha^2 l^2 = -\dfrac{S l^2}{EI}\)	\((1+\beta)\)	\((1+\alpha)\)	\(\Delta \gamma(1+\beta)\)	\(2\Delta \gamma(1+\alpha)\)	Forces in the rigid truss
0—2	\(+44{,}00\)	\(-0{,}24\)	0,946	0,969	21,46	43,95	\(+43{,}67\)
2—3	\(+44{,}00\)	\(-0{,}24\)	0,946	0,969	21,46	43,95	\(+43{,}71\)
3—5	\(+68{,}00\)	\(-0{,}32\)	0,929	0,959	18,38	37,94	\(+67{,}40\)
1—4	\(-56{,}00\)	\(+0{,}27\)	1,066	1,038	42,17	82,13	\(-56{,}16\)
0—1	\(-70{,}40\)	\(+2{,}84\)	2,139	1,622	224,57	340,59	\(-69{,}77\)
1—3	\(+19{,}20\)	\(-1{,}31\)	0,747	0,852	132,80	302,94	\(+19{,}92\)
3—4	\(-19{,}20\)	\(+1{,}31\)	1,381	1,214	245,51	431,65	\(-17{,}99\)
1—2	\(+40{,}00\)	\(-1{,}73\)	0,684	0,816	98,62	235,30	\(+40{,}00\)
4—5	\(+30{,}00\)	\(-1{,}30\)	0,749	0,853	107,99	245,97	\(+28{,}37\)

Table XLI bis.

No.	$M_1$	$M_2$	$M_3$	$M_4$	$M_5$	$M_6$	$M_7$	$M_8$	$M_9$	$M_{10}$	$M_{11}$	$M_{12}$
1	+384,54	+224,57	0	0	+21,46	+21,46	0	0	0	0	0	0
2	+224,57	+422,72	+82,13	+82,13	0	0	0	0	0	+42,17	0	0
3	0	+82,13	+317,43	+82,13	−98,62	0	0	0	0	+42,17	0	0
4	0	+82,13	+82,13	+385,07	0	0	−132,80	0	0	+42,17	0	0
5	+21,46	0	−98,62	0	+279,25	+43,95	0	0	0	0	0	0
6	+21,46	0	0	0	+43,95	+87,90	+21,46	+21,46	+21,46	0	0	0
7	0	0	0	−132,80	0	+21,46	+346,89	+43,95	+43,95	0	0	0
8	0	0	0	0	0	+21,46	+43,95	+475,60	+43,95	0	+245,51	0
9	0	0	0	0	0	+21,46	+43,95	+43,95	+81,89	0	0	+18,38
10	0	+42,17	+42,17	+42,17	0	0	0	0	0	+82,13	0	0
11	0	0	0	0	0	0	0	+245,51	0	0	+431,65	0
12	0	0	0	0	0	0	0	0	+18,38	0	0	+37,94

Table XLIII bis.

No.	$M_1$	$M_2$	$M_3$	$M_4$	$M_5$	$M_6$	$M_7$	$M_8$	$M_9$	$M_{10}$	$M_{11}$	$M_{12}$	$A$	$\sum$
1	+384,82	+224,62	0	0	+21,13	+20,93	+0,20	+0,20	+0,20	0	0	0	−61,55	+590,55
2	+224,62	+423,11	+82,41	+82,27	−0,16	−0,42	+0,11	+0,05	−0,05	+42,02	+0,07	+0,18	−47,49	+806,71
3	0	+82,41	+317,71	+82,27	−98,62	0	+0,11	+0,05	−0,06	+42,02	+0,07	+0,18	−11,47	+414,42
4	0	+82,27	+82,27	+385,19	0	−0,04	−132,78	−0,09	−0,20	+42,09	+0,07	+0,18	−3,21	+455,75
5	+21,13	−0,16	−98,62	0	+279,75	+44,64	−0,20	−0,20	−0,20	0	0	0	+97,57	+343,71
6	+20,93	−0,42	−0,25	−0,04	+44,64	+89,41	+20,85	+20,71	+20,71	+0,18	0	0	+124,06	+340,78
7	+0,20	+0,11	+0,11	−132,78	−0,20	+20,85	+347,22	+44,36	+44,36	−0,11	0	0	−18,23	+305,89
8	+0,20	+0,05	+0,05	−0,09	−0,20	+20,71	+44,36	+476,32	+44,76	−0,17	+245,46	+0,14	+14,24	+845,55
9	+0,20	−0,06	−0,06	−0,20	−0,20	+20,71	+44,36	+44,76	+83,69	+0,06	−0,14	+17,24	−22,22	+188,02
10	0	+42,02	+42,02	+42,09	0	+0,18	−0,11	−0,17	−0,06	+82,40	−0,13	−0,24	−57,31	+150,69
11	0	+0,07	+0,07	+0,07	0	0	0	+245,46	+0,14	−0,13	+431,76	+0,20	+28,06	+705,42
12	0	+0,18	+0,18	+0,18	0	0	0	+0,14	+17,24	−0,24	+0,20	+39,14	+64,52	+121,76

valuable. A certain deviation is represented only by the moments \(M_1\) and \(M_2\) occurring at the ends of member 0—1, on which taking the longitudinal force into account had the strongest effect.

In general, the accuracy of our method may therefore be regarded as sufficient for the trusses under consideration.

As can be seen from the following comparative table, the forces calculated in the rigid truss differ somewhat from the forces in the hinged truss (the latter are placed in parentheses).

\[ \begin{aligned} S_{0-2}&=+43.67\,(44.00), & S_{1-4}&=-56.16\,(-56.00),\\ S_{0-1}&=-69.77\,(-70.40), & S_{3-4}&=-17.99\,(-19.20),\\ S_{2-3}&=+43.71\,(44.00), & S_{3-5}&=+67.40\,(+68.00),\\ S_{1-3}&=+19.92\,(+19.20), & S_{1-2}&=+38.78\,(+40.00). \end{aligned} \]

21. Example V. A truss with triangular bracing and additional posts. (Bridge of the Moscow–Vindavo–Rybinsk Railway.)

All the examples of trusses considered by us up to now were specially designed by us for the definite purpose of comparative calculations. In striving to obtain the most clear-cut results of comparison, we deliberately sacrificed the full correspondence of our trusses to the requirements of practical applicability. This circumstance by no means deprives our calculations of practical significance, since comparisons with analogous calculations by other authors for practical cases give coinciding results.

Nevertheless, for completeness of the picture we shall consider one more example, choosing it from among structures used in practice.

The bridge truss shown in Fig. \(12^{\mathrm{bis}}\) belongs to the number of those which, according to the well-known circular of the Railway Administration of Sept. 29, 1900, could be used without special permission from the Ministry of Ways of Communication. The bridge, with a span of 20 sazhens, was first designed in 1898 for the Moscow–Vindava line of the Company of the Moscow–Vindavo–Rybinsk Railway.

The only change made by us consists in the fact that all panels have been taken of equal length, whereas according to the design the two end panels are \(4''\) longer than the others. The principal dimensions of the truss are shown—

are shown in Fig. 12 bis, and the lengths of the elements, the dimensions of their cross-sections, and certain other data necessary for our calculations are given in Table XLV.

The cross-sections of the chords have an I-shaped form and are composed of two vertical plates measuring \(20'' \times 7/16''\), one or two horizontal plates \(20'' \times 3/8''\), and two to four angles \(3 \tfrac{1}{2}'' \times 3 \tfrac{1}{2}'' \times 3/8''\).

The distance between the vertical plates is \(12''\).

The inclined chord member 1—3 is framed along the lower edge of the vertical plates with angles \(3 \tfrac{1}{2}'' \times 3 \tfrac{1}{2}'' \times 3/8''\).

The diagonals are of built-up double-tee section, composed of four angles \(6'' \times 4'' \times 7/16''\) and one plate \(12'' \times 5/16''\) (diagonal 3—4), or of two plates \(12 \tfrac{1}{2}'' \times 7/16''\) and four angles \(6'' \times 4'' \times 3/8''\) (diagonals 4—7 and 7—8).

The posts, made of four angles measuring \(3'' \times 3 \tfrac{1}{2}'' \times 3/8''\), are riveted to the outside of the chord plates.

The load on the truss is taken as twofold:

a. permanent: 5 tons at each joint of the upper chord and 8 tons at a joint of the lower chord;

b. live, in the form of two locomotives with tenders and one car, the locomotives being turned with their tenders toward one another.

Fig. 12 bis.

\[ \alpha = 53^\circ,\qquad \sin \alpha = 0.7986 \sim 0.80,\qquad \cos \alpha = 0.6018 \sim 0.60 \]

The live load is evidently arranged symmetrically on the span, and moreover in such a way as to cause, as far as possible, the greatest forces in the posts.

The arrangement of the live load is shown in Fig. 12bis for the left half of the span; on the right side another possible variant of the load arrangement is shown (locomotives with their tubes facing one another). With the selected position of the train, the live load is distributed over the nodes of the lower chord as follows:

\[ \text{Node } 2:\quad Q_2=\frac{7.5}{5.5}(4.8+2.3+3.6+4.9)=21^{tn}27 \]

\[ \ldots\;4:\quad Q_4=\frac{7.5}{5.5}(0.6+1.9+3.2)+\frac{6.25}{5.5}(0.1+1.7+3.3)=13^{tn}57 \]

\[ \ldots\;6:\quad Q_6=\frac{6.25}{5.5}(2.2+3.8+5.4)+\frac{5\times1.9}{5.5}=14^{tn}68 \]

\[ \ldots\;8:\quad Q_8=\frac{5\times3.6}{5.5}\times2=6^{tn}55 \]

\[ \ldots\;1:\quad Q_1=\frac{7.5\times0.7}{5.5}=0^{tn}96 \]

To determine the forces in the members of the truss, we first calculate the forces from a unit load; since we shall have to deal exclusively with a symmetrical load, Table XLVI gives the forces in the members of the left half of the truss from the successive loading, by forces equal to \(1^{tn}\), of the nodes in pairs: 2 and \(2'\), 3 and \(3'\), etc.; the same table gives the forces from loading by horizontal forces also equal to \(1^{tn}\), in pairs at the nodes 2 and \(2'\), ... 7 and \(7'\).

Then Table XLVII gives the calculation of the total forces in the truss members due to the action of the specified permanent and live load.

The total forces are placed in the last column of this table, while the corresponding stresses may be found in the last column of Table XLV.

Using Table XLVI, it is easy to compile a table of forces in the members under the action of pairs of forces equal to \(1^{tn\cdot cm}\) and applied in pairs to symmetrically situated members. The forces composing a pair, as always, are directed perpendicular to the axis of the member. The corresponding forces in tons, increased 100 times, are presented in Table XLVIII.

Proceeding to the investigation of the stresses due to the rigidity of the joints by the first (generalized) method, we must introduce 21 unknown quantities, each of which represents the absolute value of four equal mutually balancing pairs.

V. N. NEKRASOV.

Table XLV.

Name of element	Length $l$, cm	Area $\omega_1$, dm.$^2$	Area $\omega$, cm.$^2$	Moment of inertia $I$, dm.$^4$	Moment of inertia $I$, cm.$^4$	Dist. c. of grav. $e_1$, cm	Dist. c. of grav. $e_2$, cm	$\dfrac{l}{\omega}$, $\dfrac{1}{\mathrm{cm}}$	$\Delta_\gamma=\dfrac{l}{6I}10^4$, $\dfrac{1}{\mathrm{cm}^3}$	$N=\dfrac{2I}{l}$, cm.$^3$	Stress, kg./cm.$^2$
1—2	550	29,996	193,5	1275,7	53100	35,6	16,1	2,84	17,26	193,1	+ 381
2—4	550	29,996	193,5	1275,7	53100	35,6	16,1	2,84	17,26	193,1	+ 381
4—6	550	34,992	225,8	1391,0	57900	37,5	14,3	2,44	15,83	210,5	+ 664
6—8	550	34,992	225,8	1391,0	57900	37,5	14,3	2,44	15,83	210,5	+ 664
1—3	916	34,992	225,8	2014,6	83850	20,8	30,9	4,06	18,21	183,1	— 544
3—5	550	29,996	193,5	1275,7	53100	16,1	35,6	2,84	17,26	193,1	— 629
5—7	550	29,996	193,5	1275,7	53100	16,1	35,6	2,84	17,26	193,1	— 629
7—9	550	37,496	241,9	1535,8	63920	13,7	39,0	2,27	14.34	232,4	— 650
3—4	916	20,522	132,4	135,4	5640	15,6	15,6	6,92	270,69	12.3	+ 604
4—7	916	25,402	163,9	258,5	10760	15,6	15,6	5.59	141,88	23,5	— 286
7—8	916	25,402	163,9	258,5	10760	15,6	15,6	5.59	141,88	23,5	+ 75
2—3	732	9,268	59,9	24,6	1020	9,3	9,3	12,22	1196,08	2,8	+ 489
4—5	732	9.268	59,9	24,6	1020	9,3	9,3	12,22	1196,08	2,8	— 83
6—7	732	9,268	59,9	24,6	1020	9,3	9,3	12,22	1196,08	2,8	+ 379
8—9	732	9,268	58,9	24,6	1020	9,3	9,3	12,22	1196,08	2,8	— 83

The bending moments acting on the end sections of the elements as a result of the rigidity of the joints are expressed in terms of the above-mentioned unknowns as follows:

\[ \begin{aligned} M_{1-2}&=+M_1,\\ M_{1-3}&=-M_1,\\ M_{2-1}&=-(M'_2+M_3),\\ M_{2-3}&=+M'_2,\\ M_{2-4}&=+M_3,\\ M_{3-1}&=-(M'_4+M'_5+M_6),\\ M_{3-2}&=+M'_4,\\ M_{3-4}&=+M'_5,\\ M_{3-5}&=+M_6,\\ M_{4-2}&=-(M'_7+M'_8\\ &\qquad +M'_9+M_{10}), \end{aligned} \qquad \begin{aligned} M_{4-3}&=+M'_7,\\ M_{4-5}&=+M'_8,\\ M_{4-7}&=+M'_9,\\ M_{4-6}&=+M_{10},\\ M_{5-3}&=-(M'_{11}+M_{12}),\\ M_{5-4}&=+M'_{11},\\ M_{5-7}&=+M_{12},\\ M_{6-4}&=-(M'_{13}+M_{14}),\\ M_{6-7}&=+M'_{13},\\ M_{6-8}&=+M_{14}, \end{aligned} \]

Table XLVI.

Forces in the members of a hinged truss under the action of load:

Members	Vertical forces = 1 tn, applied successively at nodes: 2 and 2′	Vertical forces = 1 tn, applied successively at nodes: 3 and 3′	Vertical forces = 1 tn, applied successively at nodes: 4 and 4′	Vertical forces = 1 tn, applied successively at nodes: 5 and 5′	Vertical forces = 1 tn, applied successively at nodes: 6 and 6′	Vertical forces = 1 tn, applied successively at nodes: 7 and 7′	At nodes: 8 (2 tn)	At nodes: 9 (2 tn)	Horizontal forces = 1 tn, applied successively at nodes: 2 and 2′	Horizontal forces = 1 tn, applied successively at nodes: 3 and 3′	Horizontal forces = 1 tn, applied successively at nodes: 4 and 4′	Horizontal forces = 1 tn, applied successively at nodes: 5 and 5′	Horizontal forces = 1 tn, applied successively at nodes: 6 and 6′	Horizontal forces = 1 tn, applied successively at nodes: 7 and 7′
1—2	+0,75	+0,75	+0,75	+0,75	+0,75	+0,75	+0,75	+0,75	0	0	0	0	0	0
2—4	+0,75	+0,75	+0,75	+0,75	+0,75	+0,75	+0,75	+0,75	0	0	0	0	0	0
4—6	+0,75	+0,75	+1,50	+1,50	+2,25	+2,25	+2,25	+2,25	0	0	0	0	0	0
6—8	+0,75	+0,75	+1,50	+1,50	+2,25	+2,25	+2,25	+2,25	0	0	+1,00	0	0	0
3—5	−0,75	−0,75	−1,50	−1,50	−1,50	−1,50	−1,50	−1,50	0	0	0	+1,00	0	0
5—7	−0,75	−0,75	−1,50	−1,50	−1,50	−1,50	−1,50	−1,50	0	0	0	+1,00	0	0
7—9	−0,75	−0,75	−1,50	−1,50	−2,25	−2,25	−3,00	−3,00	0	0	0	+1,00	0	0
1—3	−1,25	−1,25	−1,25	−1,25	−1,25	−1,25	−1,25	−1,25	0	0	0	0	0	0
3—4	0	0	+1,25	+1,25	+1,25	+1,25	+1,25	+1,25	0	0	0	0	0	0
4—7	0	0	0	0	−1,25	−1,25	−1,25	−1,25	0	0	0	0	0	0
7—8	0	0	0	0	0	0	+1,25	+1,25	0	0	0	0	0	0
2—3	+1,00	0	0	0	0	0	0	0	+1,00	0	0	0	0	0
4—5	0	0	0	−1,00	0	0	0	0	0	0	−1,00	0	0	0
6—7	0	0	0	0	+1,00	0	0	0	0	0	0	0	+1,00	0
—9	0	0	0	0	0	0	0	−2,00	0	0	0	0	0	0

Table XLVII.

Forces in the members of a pin-jointed truss under the action of loads:

Members	5 tn. at the nodes of the upper chord	8 tn. at the nodes of the lower chord	21 tn.,27 at nodes 2 and 2′	13 tn.,57 at nodes 4 and 4′	14 tn.,68 at nodes 6 and 6′	6 tn.,55 at node 8	full load of the truss
1—2	+ 13,125	+ 21,000	+ 15,953	+ 10,178	+ 11,010	+ 2,456	+ 73,722
2—4	+ 13,125	+ 21,000	+ 15,953	+ 10,178	+ 11,010	+ 2,456	+ 73,722
4—6	+ 28,125	+ 45,000	+ 15,953	+ 20,355	+ 33,030	+ 7,369	+ 149,832
6—8	+ 28,125	+ 45,000	+ 15,953	+ 20,355	+ 33,030	+ 7,369	+ 149,832
3—5	− 22,500	− 36,000	− 15,953	− 20,355	− 22,020	− 4,913	− 121,741
5—7	− 22,500	− 36,000	− 15,953	− 20,355	− 22,020	− 4,913	− 121,741
7—9	− 30,000	− 48,000	− 15,953	− 20,355	− 33,030	− 9,825	− 157,163
1—3	− 21,875	− 35,000	− 26,588	− 16,963	− 18,350	− 4,094	− 122,870
3—4	+ 15,625	+ 25,000	0	+ 16,963	+ 18,350	+ 4,094	+ 80,032
4—7	− 9,375	− 15,000	0	0	− 18,350	− 4,094	− 46,819
7—8	+ 3,125	+ 5,000	0	0	0	+ 4,094	+ 12,219
2—3	0	+ 8,000	+ 21,270	0	0	0	+ 29,270
4—5	− 5,000	0	0	0	0	0	− 5,000
6—7	0	+ 8,000	0	0	+ 14,680	0	+ 22,680
8—9	− 5,000	0	0	0	0	0	− 5,000

\[ \begin{aligned} M_{7-5}&=-(M'_{15}+M'_{16}\\ &\quad+M'_{17}+M_{18}),\\ M_{7-4}&=+M'_{15},\\ M_{7-6}&=+M'_{16},\\ M_{7-8}&=+M'_{17}, \end{aligned} \qquad \begin{aligned} M_{7-9}&=+M_{18},\\ M_{8-6}&=-M_{19},\\ M_{8-7}&=-M'_{20},\\ M_{9-7}&=-M_{21},\\ M_{8-9}&=M_{9-8}=0. \end{aligned} \]

From Table XLV it is easy to see the considerable difference in the values of the quantities \(\Delta\gamma\) for the elements of the truss contour and its lattice; as a consequence of this, the coefficients of the quantities \(M\) in the equations for their calculation will be very different in magnitude, which adversely affects the accuracy of the solution of the equations.

To avoid this, let us agree to express the moments for the lattice in tenths of ton-meters, and for the contour, as before, in ton-meters. Then we shall obtain the expressions for the deformation work of the individual elements

\[ A_{1-2}=A'_{1-2}+\frac{\Delta\gamma'_{1-2}}{E} \left[ M_1^2+\left(\frac{M_2}{10}+M_3\right)^2 +M_1\left(\frac{M_2}{10}+M_3\right) \right], \]

\[ \begin{aligned} A_{2-4}=A'_{2-4}+\frac{\Delta\gamma'_{2-4}}{E} \Bigg[ &M_3^2+\left(\frac{M_7+M_8+M_9}{10}+M_{10}\right)^2\\ &+M_3\left(\frac{M_7+M_8+M_9}{10}+M_{10}\right) \Bigg], \end{aligned} \]

\[ A_{4-6}=A'_{4-6}+\frac{\Delta\gamma'}{E} \left[ M_{10}^2+\left(\frac{M_{13}}{10}+M_{14}\right)^2 +M_{10}\left(\frac{M_{13}}{10}+M_{14}\right) \right], \]

\[ A_{6-8}=A'_{6-8}+\frac{\Delta\gamma'_{6-8}}{E} \left[ M_{14}^2+M_{19}^2+M_{14}\cdot M_{19} \right], \]

\[ A_{3-5}=A'_{3-5}+\frac{\Delta\gamma'_{3-5}}{E} \left[ M_6^2+\left(\frac{M_{11}}{10}+M_{12}\right)^2 +M_6\left(\frac{M_{11}}{10}+M_{12}\right) \right], \]

\[ \begin{aligned} A_{5-7}=A'_{5-7}+\frac{\Delta\gamma'_{5-7}}{E} \Bigg[ &M_{12}^2+\left(\frac{M_{15}+M_{16}+M_{17}}{10}+M_{18}\right)^2\\ &+M_{12}\left(\frac{M_{15}+M_{16}+M_{17}}{10}+M_{18}\right) \Bigg], \end{aligned} \]

\[ A_{7-9}=A'_{7-9}+\frac{\Delta\gamma'_{7-9}}{E} \left[ M_{18}^2+M_{21}^2+M_{18}\cdot M_{21} \right], \]

Table XLVIII.
Forces in members, increased 100 times.

Forces in the members of a hinged truss at \(1\,t_m=-1\ \text{t-m}\), by which the members are loaded.

Member	\(1\!-\!2\) and \(1'\!-\!2'\)	\(2\!-\!4\) and \(2'\!-\!4'\)	\(4\!-\!6\) and \(4'\!-\!6'\)	\(6\!-\!8\) and \(6'\!-\!8'\)	\(3\!-\!5\) and \(3'\!-\!5'\)	\(5\!-\!7\) and \(5'\!-\!7'\)	\(7\!-\!9\) and \(7'\!-\!9'\)	\(1\!-\!3\) and \(1'\!-\!3'\)	\(3\!-\!4\) and \(3'\!-\!4'\)	\(4\!-\!7\) and \(4'\!-\!7'\)	\(7\!-\!8\) and \(7'\!-\!8'\)	\(2\!-\!3\) and \(2'\!-\!3'\)	\(4\!-\!5\) and \(4'\!-\!5'\)	\(6\!-\!7\) and \(6'\!-\!7'\)
\(1\!-\!2\)	\(+0,136\)	0	0	0	0	0	0	\(+0,136\)	0	0	0	0	0	0
\(2\!-\!4\)	\(+0,136\)	\(+0,136\)	0	0	0	0	0	\(+0,136\)	0	0	0	0	0	0
\(4\!-\!6\)	\(+0,136\)	\(+0,136\)	\(+0,136\)	0	\(+0,136\)	\(+0,136\)	0	\(+0,136\)	\(+0,136\)	0	0	\(+0,136\)	\(+0,136\)	0
\(6\!-\!8\)	\(+0,136\)	\(+0,136\)	\(+0,136\)	0	\(+0,136\)	\(+0,136\)	0	\(+0,136\)	\(+0,136\)	0	0	\(+0,136\)	\(+0,136\)	\(+0,136\)
\(3\!-\!5\)	\(-0,136\)	\(-0,136\)	0	0	\(-0,136\)	0	0	\(-0,136\)	\(-0,136\)	0	0	0	\(-0,136\)	0
\(5\!-\!7\)	\(-0,136\)	\(-0,136\)	0	0	\(-0,136\)	0	0	\(-0,136\)	\(-0,136\)	0	0	0	\(-0,136\)	0
\(7\!-\!9\)	\(-0,136\)	\(-0,136\)	\(-0,136\)	\(-0,136\)	\(-0,136\)	\(-0,136\)	\(-0,136\)	\(-0,136\)	\(-0,136\)	\(-0,136\)	\(-0,136\)	\(-0,136\)	\(-0,136\)	\(-0,136\)
\(1\!-\!3\)	\(-0,227\)	0	0	0	0	0	0	\(-0,082\)	0	0	0	0	0	0
\(3\!-\!4\)	0	\(+0,227\)	0	0	\(+0,227\)	0	0	0	\(+0,082\)	0	0	0	0	0
\(4\!-\!7\)	0	0	\(-0,227\)	0	0	\(-0,227\)	0	0	0	\(-0,082\)	0	0	0	0
\(7\!-\!8\)	0	0	0	\(+0,227\)	0	0	\(+0,227\)	0	0	0	\(+0,082\)	0	0	0
\(2\!-\!3\)	\(+0,182\)	\(-0,182\)	0	0	0	0	0	0	0	0	0	0	0	0
\(4\!-\!5\)	0	0	\(+0,182\)	\(-0,182\)	\(-0,182\)	\(+0,182\)	0	0	0	0	0	0	0	0
\(6\!-\!7\)	0	0	0	0	0	0	\(+0,182\)	0	0	\(-0,182\)	0	0	0	0
\(8\!-\!9\)	0	0	0	0	0	0	\(-0,364\)	0	0	0	0	0	0	0

$$ A_{1-3}=A'_{1-3}+\frac{\Delta\gamma_{1-3}}{E}\left[ M_1^2+\left(\frac{M_4+M_5}{10}+M_6\right)^2 - M_1\left(\frac{M_4+M_5}{10}+M_6\right) \right], $$

$$ A_{3-4}=A'_{3-4}+\frac{\Delta\gamma_{3-4}}{100E}\left[ M_5^2+M_7^2-M_5\cdot M_7 \right], $$

$$ A_{4-7}=A'_{4-7}+\frac{\Delta\gamma_{4-7}}{100E}\left[ M_9^2+M_{15}^2-M_9\cdot M_{15} \right], $$

$$ A_{7-8}=A'_{7-8}+\frac{\Delta\gamma_{7-8}}{100E}\left[ M_{17}^2+M_{20}^2+M_{17}\cdot M_{20} \right], $$

$$ A_{2-3}=A'_{2-3}+\frac{\Delta\gamma_{2-3}}{100E}\left[ M_2^2+M_4^2-M_2\cdot M_4 \right], $$

$$ A_{4-5}=A'_{4-5}+\frac{\Delta\gamma_{4-5}}{100E}\left[ M_8^2+M_{11}^2-M_8\cdot M_{11} \right], $$

$$ A_{6-7}=A'_{6-7}+\frac{\Delta\gamma_{6-7}}{100E}\left[ M_{13}^2+M_{16}^2-M_{13}\cdot M_{16} \right]. $$

In these formulas, \(A'\) as before denotes the amount of work of the members under simple tension or compression; in addition, it is assumed that

$$ \begin{aligned} M_2&=10M'_2; & M_4&=10M'_4; & M_5&=10M'_5; & M_7&=10M'_7;\\ M_8&=10M'_8; & M_9&=10M'_9; & M_{11}&=10M'_{11}; & M_{13}&=10M'_{13};\\ M_{15}&=10M'_{15}; & M_{16}&=10M'_{16}; & M_{17}&=10M'_{17}; & M_{20}&=10M'_{20}. \end{aligned} $$

The quantities \(M'_2, M'_4, \ldots, M'_{20}\) are expressed in \(\dfrac{\mathrm{tn.-m.}}{10}\).

We shall now compute, on the basis of the notation introduced, with the aid of Table XLVIII, the magnitudes of the forces in the members of the truss under the action of a unit load by groups of forces \(M_1=1;\ M_2=1;\ \ldots\ M_{21}=1\).

Table II.

Forces increased 100-fold

Elem.	\(M_1\)	\(M_2\)	\(M_3\)	\(M_4\)	\(M_5\)	\(M_6\)	\(M_7\)	\(M_8\)	\(M_9\)	\(M_{10}\)
1—2	0	−0.136	−0.136	−0.136	−0.136	−0.136	0	0	0	0
2—4	0	0	0.136	0	−0.136	−0.136	0	0	0	0
4—6	0	0	0	0	0	0	0	0	0	0
6—8	0	0	0	0	0	0	0	0	0	0
3—5	0	0	0	0	0	0	0	+0.136	+0.136	+0.136
5—7	0	0	0	0	0	0	0	0	+0.136	+0.136
7—9	0	0	0	0	0	0	0	0	0	0
1—3	−0.145	+0.227	+0.227	+0.082	+0.082	+0.082	0	0	0	0
3—4	0	0	+0.227	0	+0.082	+0.227	−0.145	−0.227	−0.227	−0.227
4—7	0	0	0	0	0	0	0	0	−0.082	−0.227
7—8	0	0	0	0	0	0	0	0	0	0
2—3	+0.182	−0.182	−0.364	0	0	0	+0.182	+0.182	+0.182	+0.182
4—5	0	0	0	0	0	−0.182	0	0	0	0
6—7	0	0	0	0	0	0	0	0	0	+0.182
8—9	0	0	0	0	0	0	0	0	0	0

Table II. (Continuation).

Forces increased 100 times.

Forces in the members of the hinged truss under the action of groups of forces equal to 1 ton-cm.

$M_{11}$	$M_{12}$	$M_{13}$	$M_{14}$	$M_{15}$	$M_{16}$	$M_{17}$	$M_{18}$	$M_{19}$	$M_{20}$	$M_{21}$
0	0	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	0	0	0
0	0	−0,136	−0,136	0	−0,136	−0,136	−0,136	0	0	0
0	0	0	−0,136	0	0	−0,136	−0,136	0	0	0
+0,136	+0,136	0	0	0	0	0	0	0	0	0
0	+0,136	0	0	0	0	0	0	0	0	0
0	0	0	0	0	0	0	0	+0,136	+0,136	+0,136
0	0	0	0	0	0	0	0	0	0	0
−0,227	−0,227	0	0	0	0	0	0	0	0	0
0	−0,227	+0,227	+0,227	+0,145	+0,227	+0,227	+0,227	0	0	0
0	0	0	+0,227	0	0	+0,082	+0,227	−0,227	−0,082	−0,227
0	0	0	0	0	0	0	0	0	0	0
+0,182	+0,364	0	0	−0,182	−0,182	−0,182	−0,182	0	0	0
0	0	−0,182	−0,364	0	0	0	0	+0,182	0	0
0	0	0	0	0	0	0	−0,364	0	0	+0,346

It is easy to see that, in order to obtain the forces from \(M_1=1\), it is sufficient to subtract from the values in the 1st column of Table XLVIII the corresponding values of the 8th column; for \(M_2=1\), to subtract the 1st from the 12th column, and so on.

All these values of the forces are brought together in Table IL. (The forces have been increased 100-fold.)

Then, by multiplying the numbers of Table IL by the corresponding value of \(\dfrac{l}{\omega}\) for each element (see Table XLV), Table L was obtained.

It should not be forgotten that we agreed to express the bending moments for the elements of the truss in tenths of a tonne-meter, whereas Tables IL and L are compiled for values of \(M\) equal to 1 tonne-centimeter, but increased 100-fold, i.e., they in fact correspond to 1 tonne-meter. In other words, the values in Table IL represent the coefficients \(\alpha, \beta,\ldots \mu,\nu\) of the following expression for the forces in the elements of the rigid truss:

\[ S = S^0 + \alpha M_1 + \beta M_2' + \ldots + \mu M_{20}' + \nu M_{21}; \]

whereas, passing to the values \(M_2, M_4,\ldots M_{20}\), expressed in tenths of tn.-m., we obtained the formula

\[ S = S^0 + \alpha M_1 + \frac{\beta}{10} M_2 + \ldots + \frac{\mu}{10} M_{20} + \nu M_{21}, \]

i.e., the values in Tables IL and L that correspond to the moments bending the truss must be reduced by a factor of 10.

Taking this circumstance into account, we proceed to compute the coefficients at \(M\) in the basic equations; for this, as is known, we must successively multiply the numbers of Table L by the numbers in the vertical rows of Table IL corresponding to each \(M\), and in the 21 tables thus obtained sum the vertical columns.

Without presenting the intermediate calculations, we give in Table LI the final results of calculating these coefficients. In the last column are placed the values of the constant term for the equations; the latter are obtained by successive multiplication of the complete forces of the pin-jointed truss by the vertical columns of Table L, with subsequent summation.

Then Table LII contains the values of the coefficients for \(M\) of type \(\dfrac{\Delta r}{E}\); finally, Table LIII gives the final form of the basic equations of the generalized method for calculating the quantities \(M\).

The calculation itself presents no particular features in comparison with the preceding examples, apart from the extremely large number of computations caused by the considerable number of unknowns.

A count carried out according to the usual abbreviated logarithmic scheme determines the values of the unknowns

\[ \begin{aligned} M_1^{*} &= -3.2070, & M_{12}^{*} &= +2.8820,\\ M_2 &= +0.9934, & M_{13} &= +0.2666,\\ M_3^{*} &= +3.3565, & M_{14}^{*} &= +4.3942,\\ M_4 &= +0.8401, & M_{15} &= +0.3735,\\ M_5 &= +1.4860, & M_{16} &= +0.2355,\\ M_6^{*} &= -2.1770, & M_{17} &= +1.8910,\\ M_7 &= +0.5659, & M_{18}^{*} &= +0.4845,\\ M_8 &= +0.7849, & M_{19}^{*} &= -1.5787,\\ M_9 &= +4.0064, & M_{20} &= +0.9815,\\ M_{10}^{*} &= -1.3913, & M_{21}^{*} &= +2.3672,\\ M_{11} &= +0.8759. & \end{aligned} \]

The moments marked with \(*\) are expressed in ton-meters; the remaining ones—in tenths of a ton-meter. In order to pass from the generalized method to the methods of the second group, we must, as is known from the preceding material, set equal to zero the moments bending the lattice, i.e., put

\[ M_2 = M_4 = M_5 = M_7 = M_8 = M_9 = M_{11} = M_{13} = M_{15} = M_{16} = \]
\[ = M_{17} = M_{20} = 0. \]

Table L

Products of influence numbers \(M\) by \(\omega\)
(products are expressed in ton·cm)

Elements	\(l\omega\)	\(M_1\)	\(M_2\)	\(M_3\)	\(M_4\)	\(M_5\)	\(M_6\)	\(M_7\)	\(M_8\)	\(M_9\)	\(M_{10}\)	\(M_{11}\)	\(M_{12}\)	\(M_{13}\)	\(M_{14}\)	\(M_{15}\)	\(M_{16}\)	\(M_{17}\)	\(M_{18}\)	\(M_{19}\)	\(M_{20}\)
1–2	2,84	0	−0,386	−0,386	−0,386	−0,386	−0,386	0	0	0	0	0	0	0	0	0	0	0	0	0	0
2–4	2,84	0	0	−0,386	0	−0,386	−0,386	0	0	0	0	0	0	0	0	0	0	0	0	0	0
4–6	2,44	0	0	0	0	0	0	0	0	0	0	0	0	−0,332	−0,332	0	−0,332	−0,332	−0,332	0	0
6–8	2,44	0	0	0	0	0	0	0	0	0	0	0	0	0	−0,332	0	0	−0,332	−0,332	0	0
3–5	2,84	0	0	0	0	0	0	0	−0,386	−0,386	+0,386	+0,386	+0,386	0	0	0	0	0	0	0	0
5–7	2,84	0	0	0	0	0	0	0	0	−0,386	+0,386	0	+0,386	0	0	0	0	0	0	0	0
7–9	2,27	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	+0,309	+0,309
1–3	4,06	−0,559	+0,922	+0,922	+0,333	+0,333	+0,333	0	0	0	0	0	0	0	0	0	0	0	0	0	0
3–4	6,92	0	+1,571	+1,571	0	+0,567	+1,571	−1,003	−1,571	−1,571	−1,571	−1,571	−1,571	0	0	0	0	0	0	0	0
4–7	5,59	0	0	0	0	0	0	0	−0,458	−1,269	−1,269	0	−1,269	−1,269	−0,811	−1,269	+1,269	+1,269	−1,269	0	0
7–8	5,59	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	+0,458	−1,269	−1,269	−1,269
2–3	12,22	−2,224	−2,224	−4,448	0	0	0	+2,224	+2,224	+2,224	+2,224	0	+4,448	0	0	0	0	0	0	0	0
4–5	12,22	0	0	0	0	−2,224	0	0	0	+2,224	+4,448	+4,448	0	0	0	−2,224	−2,224	−0,224	−2,224	0	0
6–7	12,22	0	0	0	0	0	0	0	0	0	0	0	−2,224	−2,224	−4,448	0	0	0	0	+2,224	0
8–9	12,22	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	−4,448	0	−4,448

Table LI.

No. of equation	\(M_1\)	\(M_2\)	\(M_3\)	\(M_4\)	\(M_5\)	\(M_6\)	\(M_7\)	\(M_8\)	\(M_9\)	\(M_{10}\)	\(M_{11}\)	\(M_{12}\)	\(M_{13}\)	\(M_{14}\)	\(M_{15}\)	\(M_{16}\)	\(M_{17}\)	\(M_{18}\)	\(M_{19}\)	\(M_{20}\)	\(M_{21}\)	Free term, \(A\)
1	\(+0.49\)	\(-0.05\)	\(-0.94\)	0	0	\(-0.05\)	\(+0.04\)	\(+0.04\)	\(+0.04\)	\(+0.41\)	0	0	0	0	0	0	0	0	0	0	0	\(+137.47\)
2	\(-0.05\)	\(+0.01\)	\(+0.11\)	0	0	\(+0.01\)	0	0	0	\(+0.04\)	0	0	0	0	0	0	0	0	0	0	0	\(-20.68\)
3	\(-0.94\)	\(+0.11\)	\(+2.29\)	\(+0.01\)	\(+0.03\)	\(+0.54\)	\(-0.10\)	\(-0.12\)	\(-0.12\)	\(-1.17\)	\(-0.36\)	\(-0.04\)	0	0	0	0	0	0	0	0	0	\(-174.66\)
4	0	0	\(+0.01\)	0	0	\(+0.01\)	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	\(-6.94\)
5	0	0	\(+0.03\)	0	0	\(+0.03\)	0	0	0	\(-0.01\)	0	\(-0.01\)	0	0	0	0	0	0	0	0	0	\(-5.25\)
6	\(-0.05\)	\(+0.01\)	\(+0.54\)	\(+0.01\)	\(+0.03\)	\(+0.89\)	\(-0.02\)	\(-0.04\)	\(-0.04\)	\(-0.36\)	\(-1.17\)	\(-0.08\)	0	0	0	\(+0.04\)	\(+0.04\)	\(+0.04\)	\(+0.41\)	0	0	\(+39.02\)
7	\(+0.04\)	0	\(-0.10\)	0	0	\(-0.02\)	\(+0.01\)	\(+0.01\)	\(+0.01\)	\(+0.06\)	0	\(+0.02\)	0	0	0	0	0	0	0	0	0	\(-1.52\)
8	\(+0.04\)	0	\(-0.12\)	0	0	\(-0.04\)	\(+0.01\)	\(+0.01\)	\(+0.01\)	\(+0.08\)	0	\(+0.04\)	0	0	0	0	0	0	0	0	0	\(-10.76\)
9	\(+0.04\)	0	\(-0.12\)	0	0	\(-0.04\)	\(+0.01\)	\(+0.01\)	\(+0.01\)	\(+0.10\)	0	\(+0.06\)	0	0	0	0	0	0	0	0	0	\(-13.32\)
10	\(+0.41\)	\(+0.04\)	\(-1.17\)	0	\(-0.01\)	\(-0.36\)	\(+0.06\)	\(+0.08\)	\(+0.10\)	\(+1.56\)	\(+0.75\)	\(+0.04\)	\(-0.07\)	\(-0.01\)	0	\(-0.03\)	\(-0.02\)	\(+1.10\)	\(-0.29\)	\(+0.41\)	0	\(-44.77\)
11	0	0	\(-0.04\)	0	0	\(-0.08\)	0	0	0	\(+0.04\)	\(+0.10\)	0	0	0	0	0	0	0	\(-0.04\)	0	0	\(-18.38\)
12	0	0	\(-0.36\)	0	\(-0.01\)	\(-1.17\)	\(+0.02\)	\(+0.04\)	\(+0.06\)	\(+0.75\)	\(+2.37\)	\(+0.12\)	\(-0.03\)	\(-0.11\)	\(-0.10\)	\(-0.11\)	\(-0.10\)	\(-0.29\)	\(+0.67\)	\(-1.10\)	0	\(-182.54\)
13	0	0	0	0	0	0	0	0	0	\(-0.07\)	\(-0.03\)	0	\(+0.01\)	\(+0.01\)	0	0	0	0	\(-0.03\)	\(-0.04\)	0	\(-15.96\)
14	0	0	0	0	0	0	0	0	\(-0.01\)	\(-1.10\)	\(-0.29\)	0	\(+0.11\)	\(+2.29\)	\(+0.02\)	\(+0.03\)	\(+0.05\)	\(+0.03\)	\(+0.67\)	\(-1.10\)	0	\(-244.28\)
15	0	0	0	0	0	\(+0.04\)	0	0	0	\(-0.02\)	\(-0.10\)	0	0	\(+0.02\)	\(+0.01\)	\(+0.01\)	\(+0.01\)	\(-0.07\)	\(-0.29\)	\(+0.01\)	0	\(-2.69\)
16	0	0	0	0	0	\(+0.04\)	0	0	0	\(-0.03\)	\(-0.11\)	0	0	\(+0.03\)	\(+0.01\)	\(+0.01\)	\(+0.01\)	\(+0.06\)	\(-0.01\)	\(+0.01\)	0	\(-9.80\)
17	0	0	0	0	0	\(+0.04\)	0	0	0	\(-0.03\)	\(-0.11\)	0	0	\(+0.03\)	\(+0.01\)	\(+0.01\)	\(+0.01\)	\(+0.07\)	\(-0.09\)	\(-0.01\)	0	\(-14.22\)
18	0	0	0	0	0	\(+0.41\)	0	0	0	\(-0.29\)	\(-1.10\)	\(-0.04\)	\(+0.03\)	\(+1.10\)	\(+0.07\)	\(+0.09\)	\(+0.06\)	\(+1.89\)	\(-0.29\)	\(+0.29\)	\(-1.10\)	\(-121.16\)
19	0	0	0	0	0	0	0	0	0	\(+0.41\)	0	0	\(-0.04\)	0	0	\(-1.10\)	0	0	\(+0.74\)	\(+0.33\)	0	\(-136.63\)
20	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	\(-0.01\)	0	0	\(+0.01\)	\(+0.01\)	0	\(-5.42\)
21	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	\(-0.29\)	0	0	\(-0.33\)	\(+1.14\)	0	\(-75.19\)

Table LII

Eq. No.	\(M_1\)	\(M_2\)	\(M_3\)	\(M_4\)	\(M_5\)	\(M_6\)	\(M_7\)	\(M_8\)	\(M_9\)	\(M_{10}\)	\(M_{11}\)	\(M_{12}\)	\(M_{13}\)	\(M_{14}\)	\(M_{15}\)	\(M_{16}\)	\(M_{17}\)	\(M_{18}\)	\(M_{19}\)	\(M_{20}\)	\(M_{21}\)
1	\(+70.94\)	\(+1.73\)	\(+17.26\)	\(-1.82\)	\(-1.82\)	\(-18.21\)	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
2	\(+1.73\)	\(+24.27\)	\(+3.45\)	\(-11.96\)	0	0	0	\(+1.73\)	\(+0.35\)	\(+3.45\)	0	\(+17.26\)	0	0	0	0	0	0	0	0	0
3	\(+17.26\)	\(+3.45\)	\(+69.04\)	0	0	0	\(+1.73\)	\(+0.35\)	\(+0.35\)	\(+3.45\)	\(+17.26\)	\(+69.04\)	0	0	0	\(+1.73\)	\(+1.73\)	\(+17.26\)	0	0	0
4	\(-1.82\)	\(-11.96\)	0	\(+24.28\)	\(+0.36\)	\(+3.64\)	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
5	\(-1.82\)	0	0	\(+0.36\)	\(+5.76\)	\(+3.64\)	\(-2.70\)	0	0	0	0	0	0	0	0	0	0	0	0	0	0
6	\(-18.21\)	0	0	\(+3.64\)	\(+3.64\)	\(+70.94\)	0	0	0	0	\(+1.73\)	\(+17.26\)	0	0	0	0	0	0	0	0	0
7	0	0	\(+1.73\)	0	\(-2.70\)	0	\(+15.75\)	\(+0.35\)	\(+0.35\)	\(+3.45\)	0	0	0	0	0	0	0	0	0	0	0
8	0	0	\(+1.73\)	0	0	0	\(+0.35\)	\(+24.27\)	\(+0.35\)	\(+3.45\)	\(-11.96\)	0	0	0	0	0	0	0	0	0	0
9	0	0	\(+1.73\)	0	0	0	\(+0.35\)	\(+0.35\)	\(+3.19\)	\(+3.45\)	0	0	0	\(-1.42\)	0	0	0	0	0	0	0
10	0	0	\(+17.26\)	0	0	0	\(+3.45\)	\(+3.45\)	\(+3.45\)	\(+66.18\)	0	0	\(+1.58\)	\(+15.83\)	0	0	0	0	0	0	0
11	0	0	0	0	0	\(+1.73\)	0	\(-11.96\)	0	0	\(+24.27\)	\(+3.45\)	0	0	0	0	0	0	0	0	0
12	0	0	0	0	0	\(+17.26\)	0	0	0	0	\(+3.45\)	\(+69.04\)	0	0	\(+1.73\)	\(+1.73\)	\(+17.26\)	\(+17.26\)	0	0	0
13	0	0	0	0	0	0	0	0	0	\(+1.58\)	0	0	\(+24.24\)	\(+3.17\)	0	\(-11.96\)	0	0	0	0	0
14	0	0	0	0	0	0	0	0	0	\(+15.83\)	0	0	\(+3.17\)	\(+63.32\)	0	0	0	0	\(+15.83\)	0	0
15	0	0	0	0	0	0	0	0	\(-1.42\)	0	0	\(+1.73\)	0	0	\(+3.19\)	\(+0.35\)	\(+0.35\)	\(+3.45\)	0	0	0
16	0	0	0	0	0	0	0	0	0	0	0	\(+1.73\)	\(-11.96\)	0	\(+0.35\)	\(+24.27\)	\(+0.35\)	\(+3.45\)	0	0	0
17	0	0	0	0	0	0	0	0	0	0	0	\(+1.73\)	0	0	\(+0.35\)	\(+0.35\)	\(+3.19\)	\(+3.45\)	0	\(+1.42\)	0
18	0	0	0	0	0	0	0	0	0	0	0	\(+17.26\)	0	0	\(+3.45\)	\(+3.45\)	\(+3.45\)	\(+63.20\)	0	0	\(+14.34\)
19	0	0	0	0	0	0	0	0	0	0	0	0	0	\(+15.83\)	0	0	0	0	\(+31.66\)	0	0
20	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	\(+1.42\)	0	0	\(+2.84\)	0
21	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	\(+14.34\)	0	0	\(+28.68\)

Table LIII. Truss with rigid joints.

No. of face	\(M_1\)	\(M_2\)	\(M_3\)	\(M_4\)	\(M_5\)	\(M_6\)	\(M_7\)	\(M_8\)	\(M_9\)	\(M_{10}\)	\(M_{11}\)	\(M_1\)	\(M_2\)	\(M_3\)	\(M_4\)	\(M_5\)	\(M_6\)	\(M_7\)	\(M_8\)	\(M_9\)	\(M_{10}\)	\(M_{11}\)	Free term \(A\)	Check sums
1	\(+71,43\)	\(+1,68\)	\(+16,32\)	\(+1,82\)	\(-1,82\)	\(+18,26\)	\(+0,04\)	\(+0,04\)	\(+0,04\)	\(+0,04\)	\(+0,41\)	0	0	0	0	0	0	0	0	0	0	0	\(+137,47\)	\(205,53\)
2	\(+1,68\)	\(+24,28\)	\(+3,56\)	\(-11,96\)	0	\(+0,01\)	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	\(-20,68\)	\(3,15\)
3	\(+16,32\)	\(+3,56\)	\(+71,33\)	\(-0,01\)	\(+0,03\)	\(+0,54\)	\(+1,63\)	\(+1,61\)	\(+1,61\)	\(+16,09\)	\(+0,04\)	\(-0,36\)	0	0	0	0	0	0	0	0	0	0	\(-174,66\)	\(-62,33\)
4	\(-1,82\)	\(-11,96\)	\(+0,01\)	\(+24,28\)	\(+3,65\)	\(-0,36\)	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	\(-6,94\)	\(7,58\)
5	\(-1,82\)	0	\(+0,03\)	\(+0,36\)	\(+3,67\)	\(+5,76\)	\(-2,70\)	0	0	\(-0,01\)	0	\(-0,01\)	0	0	0	0	0	0	0	0	0	0	\(-5,25\)	\(0,03\)
6	\(-18,26\)	\(+0,01\)	\(+0,54\)	\(+3,65\)	\(+3,67\)	\(+71,83\)	\(-0,02\)	\(-0,04\)	\(-0,04\)	\(-0,36\)	\(+1,65\)	\(+16,09\)	0	0	\(+0,04\)	\(+0,04\)	\(+0,04\)	\(+0,04\)	\(+0,41\)	0	0	0	\(+39,02\)	\(118,27\)
7	\(+0,04\)	0	\(+1,63\)	0	\(-2,70\)	\(-0,02\)	\(+5,76\)	\(+0,36\)	\(+0,36\)	\(+3,51\)	0	\(+0,02\)	0	0	0	0	0	0	0	0	0	0	\(-1,52\)	\(7,44\)
8	\(+0,04\)	0	\(+1,61\)	0	0	\(+0,04\)	\(+0,36\)	\(+24,28\)	\(+0,36\)	\(+3,53\)	\(-11,96\)	\(+0,04\)	0	0	0	0	0	0	0	0	0	0	\(-10,76\)	\(7,46\)
9	\(+0,04\)	0	\(+1,61\)	0	0	\(+0,04\)	\(+0,36\)	\(+3,20\)	\(+3,53\)	\(+3,55\)	0	\(+0,06\)	0	0	\(-0,01\)	\(-1,42\)	0	0	\(+0,01\)	0	0	0	\(-13,32\)	\(5,62\)
10	\(+0,41\)	\(+0,04\)	\(+16,09\)	0	\(-0,01\)	\(-0,36\)	\(-3,51\)	\(-3,53\)	\(-3,55\)	\(+67,74\)	\(+6,04\)	\(+0,75\)	\(+1,51\)	\(+14,73\)	\(+0,02\)	\(-0,02\)	\(+0,03\)	\(-0,03\)	\(+0,29\)	\(-1,41\)	0	0	\(-44,77\)	\(66,72\)
11	0	0	\(-0,04\)	0	\(+1,65\)	0	\(-11,96\)	0	0	\(+0,04\)	\(+24,28\)	\(+3,57\)	\(+24,25\)	\(+0,03\)	\(+0,29\)	\(+1,63\)	\(+1,62\)	\(+1,62\)	\(+16,16\)	\(+1,41\)	0	0	\(-18,38\)	\(0,88\)
12	0	0	\(-0,36\)	0	\(+16,09\)	\(-0,01\)	\(+0,02\)	\(+0,04\)	\(+0,04\)	\(+0,06\)	\(+0,75\)	\(+71,41\)	\(-0,03\)	\(+3,28\)	\(+65,61\)	\(+0,02\)	\(+0,03\)	\(+0,05\)	\(+0,67\)	\(+0,29\)	0	0	\(-182,54\)	\(70,26\)
13	0	0	0	0	0	0	0	0	0	\(+1,51\)	0	\(+0,03\)	\(+24,25\)	\(+3,28\)	\(+65,61\)	\(+3,20\)	\(+0,36\)	\(+0,36\)	\(+3,51\)	\(+0,67\)	\(+0,29\)	0	\(-15,96\)	\(1,08\)
14	0	0	0	0	0	0	0	0	0	\(+14,73\)	0	\(+0,29\)	\(+4,25\)	\(+3,28\)	\(+0,02\)	\(+3,20\)	\(+0,36\)	\(+0,36\)	\(+3,51\)	\(+0,67\)	\(+0,29\)	\(-0,01\)	\(-244,20\)	\(145,76\)
15	0	0	0	0	\(+0,04\)	0	0	0	0	\(-0,02\)	0	\(+1,63\)	0	0	\(+0,02\)	\(+3,20\)	\(+0,36\)	\(+0,36\)	\(+3,51\)	0	0	0	\(-2,69\)	\(4,99\)
16	0	0	0	0	\(+0,04\)	0	0	0	0	\(-0,03\)	0	\(+1,62\)	\(-11,96\)	\(+0,03\)	\(+0,03\)	\(+0,36\)	\(+24,28\)	\(+0,36\)	\(+3,52\)	0	0	0	\(-9,80\)	\(8,42\)
17	0	0	0	0	\(+0,04\)	0	0	0	0	\(-0,03\)	0	\(+1,62\)	0	\(-11,96\)	\(+0,05\)	\(+0,36\)	\(+0,36\)	\(+3,20\)	\(+3,54\)	\(+0,01\)	\(+1,42\)	\(-0,01\)	\(-14,22\)	\(3,68\)
18	0	0	0	0	\(+0,41\)	0	0	0	0	\(-0,29\)	\(+0,04\)	\(+16,16\)	\(+0,03\)	\(+24,25\)	\(+0,67\)	\(+3,51\)	\(+3,52\)	\(+3,54\)	\(+65,08\)	\(+0,29\)	\(+0,01\)	\(+13,24\)	\(-121,16\)	\(15,64\)
19	0	0	0	0	0	0	0	0	0	\(+0,41\)	0	0	0	\(+0,04\)	\(+14,73\)	0	0	\(-0,01\)	\(+0,29\)	\(-32,40\)	\(+0,01\)	\(+0,33\)	\(-13,63\)	\(33,91\)
20	0	0	0	0	0	0	0	0	0	0	0	0	0	0	\(-0,01\)	0	0	\(+1,42\)	\(-0,01\)	\(+0,01\)	\(+2,84\)	\(+0,01\)	\(-5,42\)	\(1,16\)
21	0	0	0	0	0	0	0	0	0	0	0	0	0	0	\(+0,29\)	0	0	\(-0,01\)	\(+13,24\)	\(+0,33\)	\(+0,01\)	\(-29,82\)	\(-75,19\)	\(32,09\)

Under this condition we obtain a new system of 9 equations with the same number of unknowns, whose coefficients are presented in Table LIV.

Solving this system, we obtain new values for the nine moments, expressed in tn.-m.,

\[ \begin{aligned} M_1&=-3{,}2369, & M_{14}&=+4{,}3404,\\ M_3&=+3{,}4720, & M_{18}&=+0{,}6230,\\ M_6&=-2{,}0572, & M_{19}&=-1{,}5566,\\ M_{10}&=-1{,}1193, & M_{21}&=+2{,}3043.\\ M_{12}&=+2{,}9261, \end{aligned} \]

It remains to give the calculation of the moments by Mohr’s method, which for this example has been carried out somewhat differently than for the preceding ones. In the equation for calculating the angles of rotation of the joints

\[ 2\varphi'_N \sum N'_{NX}+\sum N'_{NX}\cdot\varphi'_X=3\sum N'_{NX}\cdot\psi'_{NX} \]

we denote by \(N'_{NX}\) not the quantity \(\dfrac{2EI}{l}\), as before, but only

\[ \frac{2I}{l}; \]

correspondingly, the quantities \(\varphi'\) and \(\psi'\) have the meanings

\[ \varphi'=\varphi\cdot E,\qquad \psi'=\psi\cdot E, \]

where \(\varphi\) and \(\psi\) are the known deformation elements of the hinged truss.

The above-mentioned equation may, consequently, be rewritten in our notation as

\[ 2\varphi_N\cdot E\sum N'_{NX}-\sum N'_{NX}\varphi_X\cdot E =3\sum \psi_{NX}\cdot N'_{NX}\cdot E. \]

The calculation of the quantities entering equations of this type is shown in Table LV; the quantities \(\psi E\) are computed by the formula

\[ \psi_{NX}\cdot E=\sum S^0\cdot \frac{l}{\omega}\cdot a_{NX}, \]

where \(S^0\) denotes the forces in the members of the pin-jointed truss under the given load, and \(a_{NX}\) the forces due to loading of the member \(N—X\) by a pair of forces \(=\pm 1\), while the summation extends over all members. We obtain the system of equations

\[ 75.20\varphi'_1 + 19.30\varphi'_2 + 18.30\varphi'_3 = 462.96, \]

\[ 19.30\varphi'_1 + 77.80\varphi'_2 + 0.30\varphi'_3 + 19.30\varphi'_4 = 454.06, \]

\[ 18.30\varphi'_1 + 0.30\varphi'_2 + 78.20\varphi'_3 + 1.20\varphi'_4 + 19.30\varphi'_5 = 418.77, \]

\[ 19.50\varphi'_2 + 1.20\varphi'_3 + 88.60\varphi'_4 + 0.30\varphi'_5 + 21.10\varphi'_6 + 2.4\varphi'_7 = 362.82, \]

\[ 19.30\varphi'_3 + 0.30\varphi'_4 + 77.80\varphi'_5 + 19.30\varphi'_7 = 334.80, \]

\[ 21.10\varphi'_4 + 85.00\varphi'_6 + 0.30\varphi'_7 = 172.75, \]

\[ 2.40\varphi'_4 + 19.30\varphi'_5 + 0.30\varphi'_6 + 95.20\varphi'_7 = 182.99. \]

Solving the system gives, in \(\dfrac{\mathrm{tn.}}{10\ \mathrm{cm.^2}}\),

\[ \begin{aligned} \varphi'_1 &= +4.2512, & \varphi'_5 &= +3.1143,\\ \varphi'_2 &= +4.0736, & \varphi'_6 &= +1.3333,\\ \varphi'_3 &= +3.5330, & \varphi'_7 &= +1.2161,\\ \varphi'_4 &= +2.7989. & \end{aligned} \]

Furthermore, by the symmetry of the truss and of the load,

\[ \varphi_8 = 0, \qquad \varphi_9 = 0. \]

Having the values of \(\varphi'\) and \(\psi'\), it is easy to determine all moments due to the rigidity of the joints by the formula

\[ M_{MN} = N'_{MN}\left(2\varphi'_M + \varphi'_N - 3\psi'_{MN}\right). \]

The results of the calculation are given in the last column of Table LV.

Table LIV.

No. of equation	$M_1$	$M_3$	$M_6$	$M_{10}$	$M_{12}$	$M_{14}$	$M_{18}$	$M_{19}$	$M_{21}$	Free term $A$	Control sum
1	$+71,43$	$+16,32$	$+18,26$	$+0,41$	$0$	$0$	$0$	$0$	$0$	$+137,45$	$+207,35$
2	$+16,32$	$+71,33$	$+0,54$	$+16,09$	$-0,36$	$0$	$0$	$0$	$0$	$-174,66$	$-70,74$
3	$-18,26$	$+0,54$	$+71,83$	$-0,36$	$+16,09$	$0$	$+0,41$	$0$	$0$	$+39,02$	$+109,27$
4	$+0,41$	$+16,09$	$-0,36$	$+67,74$	$+0,75$	$+14,73$	$-0,29$	$+0,41$	$0$	$-44,77$	$+54,71$
5	$0$	$-0,36$	$+16,09$	$+0,75$	$+71,41$	$-0,29$	$+16,16$	$0$	$0$	$-182,54$	$-78,78$
6	$0$	$0$	$0$	$+1,73$	$-0,29$	$+65,61$	$+0,67$	$+14,73$	$-0,29$	$-244,28$	$-149,12$
7	$0$	$0$	$+0,41$	$-0,29$	$+16,16$	$+0,67$	$+65,08$	$-0,29$	$+13,24$	$-121,16$	$-26,18$
8	$0$	$0$	$0$	$+0,41$	$0$	$+14,73$	$-0,29$	$+32,40$	$+0,33$	$-13,63$	$+33,95$
9	$0$	$0$	$0$	$0$	$0$	$-0,29$	$+13,24$	$+0,33$	$+29,82$	$-75,19$	$-32,09$

Table LV.

No. of node	Designation of elements	\(N' \cdot \dfrac{2I}{l}\), in \(10\ \mathrm{cm}^3\)	\(\sum N'\), in \(10\ \mathrm{cm}^3\)	\(\psi E\), in \(\dfrac{\mathrm{tn.}}{10\ \mathrm{cm}^2}\)	\(N'\psi E\), in \(\mathrm{tn.}\!-\!\mathrm{cm.}\)	\(\sum N'\psi E\), \(\mathrm{tn.}\!-\!\mathrm{cm.}\)	\(3\sum N'\psi E\), \(\mathrm{tn.}\!-\!\mathrm{cm.}\)	Bending moment, in \(\mathrm{tn.}\!-\!\mathrm{cm.}\)
1	1—2	19,3	37,6	+47,73	+921,19	+1543,21	+4629,63	−336,41
1	1—3	18,3	37,6	+33,99	+622,02	+1543,21	+4629,63	+336,41
2	2—1	19,3	38,9	+47,73	+921,19	+1513,52	+4540,56	−370,67
2	2—3	0,3	38,9	+27,05	+8,12	+1513,52	+4540,56	+10,70
2	2—4	19,3	38,9	+30,27	+584,21	+1513,52	+4540,56	+359,97
3	3—1	18,3	39,1	+33,99	+622,02	+1395,91	+4187,73	+205,02
3	3—2	0,3	39,1	+27,05	+8,12	+1395,91	+4187,73	+9,07
3	3—4	1,2	39,1	+28,74	+34,49	+1395,91	+4187,73	+14,92
3	3—5	19,3	39,1	+37,89	+731,28	+1395,91	+4187,73	−229,01
4	4—2	19,3	44,3	+30,27	+584,21	+1209,40	+3628,20	+113,91
4	4—3	1,2	44,3	+28,74	+34,49	+1209,40	+3628,20	+6,11
4	4—5	0,3	44,3	+19,50	+5,85	+1209,40	+3628,20	+8,59
4	4—6	21,1	44,3	+25,79	+544,17	+1209,40	+3628,20	−170,09
4	4—7	2,4	44,3	+16,95	+40,68	+1209,40	+3628,20	+41,48
5	5—3	19,3	38,9	+37,89	+731,28	+1115,99	+3347,97	−309,82
5	5—4	0,3	38,9	+19,50	+5,85	+1115,99	+3347,97	+9,53
5	5—7	19,3	38,9	+19,63	+378,86	+1115,99	+3347,97	+300,28
6	6—4	21,1	42,5	+25,79	+544,17	+575,82	+1727,46	−479,32
6	6—7	0,3	42,5	+9,83	+2,95	+575,82	+1727,46	+2,80
6	6—8	21,1	42,5	+1,36	+28,70	+575,82	+1727,46	+476,52
7	7—4	2,4	47,6	+16,95	+40,68	+609,96	+1829,88	+3,51
7	7—5	19,3	47,6	+19,63	+378,86	+609,96	+1829,88	−66,14
7	7—6	0,3	47,6	+9,83	+2,95	+609,96	+1829,88	+2,45
7	7—8	2,4	47,6	+5,42	+13,01	+609,96	+1829,88	+19,34
7	7—9	23,2	47,6	+7,52	+174,46	+609,96	+1829,88	+40,85
8	8—6	21,1	47,3	+1,36	+28,70	0	0	+195,24
8	8—7	2,4	47,3	+5,42	+13,01	0	0	−9,84
8	8—9	0,3	47,3	0	0	0	0	0
8	8—7′	2,4	47,3	−5,42	−13,01	0	0	+9,84
8	8—6′	21,1	47,3	−1,36	−28,70	0	0	−195,24
9	9—7	23,2	46,7	+7,52	+174,46	0	0	−218,06
9	9—8	0,3	46,7	0	0	0	0	0
9	9—7′	23,2	46,7	−7,52	−174,46	0	0	+218,06

22. On the Method of Comparing the Data Obtained.

In undertaking a comparative evaluation of the results of the calculations presented in the preceding chapter, we must first of all dwell on the question of the method of comparison.

The point is that one may compare directly the computed values of the moments due to the rigidity of the joints, the stresses due to rigidity, and the total stresses, or else make use of certain auxiliary quantities, such as the coefficient most often employed,

\[ \frac{N}{n} \]

(the ratio of the greatest total stress in a member of the rigid truss to the corresponding stress in a member of the hinged truss).

In our opinion, which we shall try to substantiate in Chapter V, the single coefficient

\[ \frac{N}{n} \]

is in general insufficiently graphic for determining the relation of a given truss to the stresses due to rigidity.

Therefore, in the tables placed below, the following quantities, computed above by the various methods, are compared:

1) the moments due to the rigidity of the joints and, for the approximate methods, the ratios of these moments to those obtained by the generalized method;

2) the forces in the members of the rigid truss;

3) the stresses from longitudinal forces (primary stresses) and from bending;

4) the greatest total stresses and the coefficients

\[ \frac{N}{n}. \]

With regard to item 2, the following must also be noted: the basis of Mohr’s method and of methods similar to it is the assumption that the longitudinal forces in the members of the truss do not depend on the rigid fastenings of the joints; in other words, that the forces in the rigid and hinged trusses are identical. Consequently, strictly speaking, the determination of the forces in the rigid truss by Mohr’s method is not consistent with its foundations. However, if this is disregarded and the forces in the rigid truss are computed using the values of the moments found by Mohr’s method, then, as the data given below show, the quantities obtained in this way come much closer to the exact ones than do the forces in the hinged truss. For those cases in which Mohr’s method itself is sufficiently accurate, the named quantities almost coincide with the exact ones, as, of course, was to be expected. For this reason the corresponding data have been included by us in the appended tables.

Further, we have already mentioned that the fourth method of calculation (see Chapters II and III) contains a contradiction in its very foundation; therefore we do not enter into a detailed examination of it and merely present in the tables certain data indicating its extreme inaccuracy.

Example I. A.

Elements	Moments from rigid joints, calculated by methods: I	Moments from rigid joints, calculated by methods: II	Moments from rigid joints, calculated by methods: III	Moments from rigid joints, calculated by methods: IV	Ratios to the data of method I: II	Ratios to the data of method I: III	Ratios to the data of method I: IV	Forces in members of the rigid truss, calculated by methods: I	Forces in members of the rigid truss, calculated by methods: II	Forces in members of the rigid truss, calculated by methods: III
0—2	+0,159	−0,002	+0,189	+0,008	—	1,19	—	+39,08	+39,32	+38,99
2—0	+0,270	+0,347	+0,329	+0,396	1,28	1,22	1,47	+39,08	+39,32	+38,99
1—3	−0,363	−0,164	−0,382	−0,168	0,45	1,05	0,46	−0,86	−0,63	−0,95
3—1	−0,704	−0,744	−0,746	−0,782	1,06	1,06	1,11	−0,86	−0,63	−0,95
2—4	−0,566	−0,347	−0,652	−0,396	0,53	1,15	0,70	+40,29	+40,22	+40,33
4—2	−0,886	−0,790	−0,986	−0,858	0,89	1,11	0,97	+40,29	+40,22	+40,33
3—5	+0,944	+0,744	+1,014	+0,782	0,79	1,07	0,83	−39,61	−39,69	−39,57
5—3	+0,867	+0,778	+0,951	+0,835	0,90	1,10	0,96	−39,61	−39,69	−39,57
0—1	+0,039	+0,002	+0,032	−0,008	—	0,82	—	−1,19	−0,98	−1,31
1—0	+0,116	+0,164	+0,117	+0,168	1,41	1,01	1,45	−1,19	−0,98	−1,31
0—3	−0,198		−0,221			1,12		−62,56	−63,10	−62,42
3—0	−0,234		−0,256			1,09		−62,56	−63,10	−62,42
1—2	+0,247		+0,265			1,07		+1,23	+0,96	+1,38
2—1	+0,197		+0,216			1,10		+1,23	+0,96	+1,38
2—5	+0,099		+0,107			1,08		−0,59	−0,49	−0,65
5—2	+0,023		+0,028			1,22		−0,59	−0,49	−0,65
3—4	−0,006		−0,011			1,83		−0,47	−0,36	−0,53
4—3	−0,016		−0,019			1,19		−0,47	−0,36	−0,53

N. V. NEKRASOV.

Example 1. B.

Elements	In the hinged truss	Stresses from longitudinal forces in a rigid truss, computed by method I	Stresses from longitudinal forces in a rigid truss, computed by method II	Stresses from longitudinal forces in a rigid truss, computed by method III	Stresses from rigid joints, computed by method I	Stresses from rigid joints, computed by method II	Stresses from rigid joints, computed by method III	Full greatest stress and coefficient \(N/n\), computed by method I	Full greatest stress and coefficient \(N/n\), computed by method II	Full greatest stress and coefficient \(N/n\), computed by method III	Full greatest stress and coefficient \(N/n\), computed by method III*)
0—2	\(+500\)	\(+489\)	\(+492\)	\(+487\)	\(-68\)	\(+1\)	\(-81\)	\(+604\)	\(+640\)	\(+628\)	\(+641\)
0—2	\(+500\)	\(+489\)	\(+492\)	\(+487\)	\(+31\)	\(0\)	\(+37\)	1.21	1.28	1.26	1.28
2—0	\(+500\)	\(+489\)	\(+492\)	\(+487\)	\(+115\)	\(+148\)	\(+141\)
2—0	\(+500\)	\(+489\)	\(+492\)	\(+487\)	\(-52\)	\(-67\)	\(-64\)
1—3	\(0\)	\(-11\)	\(-8\)	\(-12\)	\(+71\)	\(+32\)	\(+74\)	\(+290\)	\(+310\)	\(+306\)	\(+318\)
1—3	\(0\)	\(-11\)	\(-8\)	\(-12\)	\(-155\)	\(-70\)	\(-163\)	\(-166\)	\(-153\)	\(-175\)	\(-163\)
3—1	\(0\)	\(-11\)	\(-8\)	\(-12\)	\(-137\)	\(-145\)	\(-145\)	—	—	—	—
3—1	\(0\)	\(-11\)	\(-8\)	\(-12\)	\(+301\)	\(+318\)	\(+318\)
2—4	\(+400\)	\(+403\)	\(+402\)	\(+403\)	\(+238\)	\(+145\)	\(+271\)	\(+641\)	\(+547\)	\(+674\)	\(+671\)
2—4	\(+400\)	\(+403\)	\(+402\)	\(+403\)	\(-67\)	\(-41\)	\(-77\)	1.60	1.37	1.68	1.68
4—2	\(+400\)	\(+403\)	\(+402\)	\(+403\)	\(-368\)	\(-329\)	\(-411\)
4—2	\(+400\)	\(+403\)	\(+402\)	\(+403\)	\(+104\)	\(+93\)	\(+116\)
3—5	\(-400\)	\(-396\)	\(-397\)	\(-396\)	\(-111\)	\(-88\)	\(-119\)	\(-757\)	\(-721\)	\(-792\)	\(-796\)
3—5	\(-400\)	\(-396\)	\(-397\)	\(-396\)	\(+393\)	\(+310\)	\(+423\)	1.89	1.80	1.98	1.99
5—3	\(-400\)	\(-396\)	\(-397\)	\(-396\)	\(+102\)	\(+92\)	\(+112\)
5—3	\(-400\)	\(-396\)	\(-397\)	\(-396\)	\(-361\)	\(-324\)	\(-396\)
0—1	\(0\)	\(-12\)	\(-10\)	\(-14\)	\(\pm22\)	\(\pm1\)	\(\pm18\)	\(-78\)	\(-104\)	\(-81\)	\(\pm67\)
0—1	\(0\)	\(-12\)	\(-10\)	\(-14\)				—	—	—	—
1—0	\(0\)	\(-12\)	\(-10\)	\(-14\)	\(\pm66\)	\(\pm94\)	\(\pm67\)
0—3	\(-604\)	\(-590\)	\(-595\)	\(-589\)	\(\pm95\)		\(\pm106\)	\(-703\)		\(-712\)	\(-727\)
0—3	\(-604\)	\(-590\)	\(-595\)	\(-589\)	\(\pm113\)		\(\pm123\)	1.16		1.18	1.20
3—0	\(-604\)	\(-590\)	\(-595\)	\(-589\)
1—2	\(0\)	\(+34\)	\(+27\)	\(+38\)	\(\pm154\)		\(\pm165\)	\(+188\)		\(+203\)	\(\pm165\)
1—2	\(0\)	\(+34\)	\(+27\)	\(+38\)	\(\pm122\)		\(\pm134\)	—		—	—
2—1	\(0\)	\(+34\)	\(+27\)	\(+38\)
2—5	\(0\)	\(-6\)	\(-5\)	\(-7\)	\(\pm58\)		\(\pm63\)	\(-64\)		\(-70\)	\(\pm58\)
2—5	\(0\)	\(-6\)	\(-5\)	\(-7\)	\(\pm14\)		\(\pm16\)	—		—	—
5—2	\(0\)	\(-6\)	\(-5\)	\(-7\)
3—4	\(0\)	\(-17\)	\(-13\)	\(-19\)	\(\pm7\)		\(\pm12\)	\(-35\)		\(-40\)	\(\pm21\)
3—4	\(0\)	\(-17\)	\(-13\)	\(-19\)	\(\pm18\)		\(\pm21\)	—		—	—
4—3	\(0\)	\(-17\)	\(-13\)	\(-19\)

*) The longitudinal forces for the hinged truss are taken to be the same as for the rigid one.

Example II A.

Elements	Moments from rigid joints, computed by method: I	Moments from rigid joints, computed by method: II	Moments from rigid joints, computed by method: III	Moments from rigid joints, computed by method: IV	Ratios to the data of method I: II	Ratios to the data of method I: III	Ratios to the data of method I: IV	Forces in rigid trusses, computed by method: I	Forces in rigid trusses, computed by method: II	Forces in rigid trusses, computed by method: III
0—2	−1,445	−1,438	−1,594	−1,602	1,00	1,10	1,11	+0,57	+0,58	+0,64
2—0	−1,854	−1,687	−2,098	−1,981	0,91	1,13	1,07
1—3	−1,743	−1,479	−2,012	−1,706	0,85	1,15	0,98	−56,39	−56,49	−56,76
3—1	−2,904	−2,772	−3,593	−3,473	0,95	1,24	1,19
2—4	+1,481	+1,687	+1,820	+1,981	1,14	1,23	1,34	+36,51	+36,47	+35,76
4—2	+0,805	+0,903	+1,190	+1,271	1,12	1,48	1,58
3—5	+2,787	+2,772	+3,566	+3,473	0,99	1,28	1,24	−57,80	−57,82	−58,10
5—3	+1,566	+1,546	+2,855	+2,808	0,99	1,82	1,80
0—1	+1,445	+1,438	+1,594	+1,602	1,00	1,10	1,11	−59,18	−59,22	−59,08
1—0	+1,393	+1,479	+1,591	+1,706	1,06	1,14	1,22
1—2	+0,302		+0,349			1,15		+59,39	+59,43	+58,05
2—1	+0,123		+0,086			0,70
1—4	+0,048		+0,073			1,52		+15,73	+15,60	+14,42
4—1	−0,032		−0,016			0,50
2—3	+0,196		+0,138			0,70		+1,43	+1,33	+2,19
3—2	+0,117		+0,038			0,33
2—3′	+0,054		+0,055			1,02		+1,56	+1,53	+1,57
3—2′	0		+0,011			—

Example II B.

Elements	Stresses from longitudinal forces in the members: hinged truss	Stresses from longitudinal forces in the members: rigid truss, calculated by method I	Stresses from longitudinal forces in the members: rigid truss, calculated by method II	Stresses from longitudinal forces in the members: rigid truss, calculated by method III	Stresses from rigid joints, calculated by method I (kg/cm²)	Stresses from rigid joints, calculated by method II (kg/cm²)	Stresses from rigid joints, calculated by method III (kg/cm²)	Complete greatest stresses and coefficient $N/n$, I	Complete greatest stresses and coefficient $N/n$, II	Complete greatest stresses and coefficient $N/n$, III	Complete greatest stresses and coefficient $N/n$, III*)
0—2	0	+7	+7	+8	+437 −175	+434 −174	+482 −193	+444 −553	+441 −503	+490 −627	+482 −635
2—0	0	+7	+7	+8	−560 +224	−510 +204	−635 +254	—	—	—	—
1—3	−463	−466	−467	−469	+72 −329	+61 −279	+83 −380	−795	−746	−849	−843
3—1	−463	−466	−467	−469	−120 +548	−115 +524	−148 +678	1.71	1.61	1.83	1.82
2—4	+484	+456	+456	+447	−447 +179	−509 +204	−550 +220	+699	+729	+807	+844
4—2	+484	+456	+456	+447	+243 −97	+273 −109	+360 −144	1.44	1.51	1.67	1.74
3—5	−334	−336	−336	−338	−60 +316	−60 +315	−77 +406	−514	−512	−663	−659
5—3	−334	−336	−336	−338	+34 −178	+33 −176	+62 −325	1.54	1.53	1.98	1.97
0—1	−500	−493	−494	−492	+278	+276	+307	−771	−778	−799	−807
1—0	−500	−493	−494	−492	±268	±284	±306	1.54	1.56	1.60	1.61
1—2	+790	+733	+734	+717	±64		±73	+797		+790	+863
2—1	+790	+733	+734	+717	±26		±18	1.01		1.00	1.08
1—4	+233	+194	+193	+178	±33		±50	+227		+228	+283
4—1	+233	+194	+193	+178	±22		±11	0.97		0.98	1.21
2—3	−12	+20	+19	+31	±119		±84	+139		+115	−96
3—2	−12	+20	+19	+31	±71		±23	—		—	—
2—3′	+49	+48	+47	+48	±57		±59	+105		+107	+108
3′—2	+49	+48	+47	+48	0		±12	2.14		2.18	2.20

*) See the note to the table of Example I B.

Example III A.

Elements	Moments from rigid joints, computed by method: I	Moments from rigid joints, computed by method: II	Moments from rigid joints, computed by method: III	Moments from rigid joints, computed by method: V*)	Ratios to the data of the 1st method: II	Ratios to the data of the 1st method: III	Ratios to the data of the 1st method: V*)	Forces in the rigid truss, computed by method: I	Forces in the rigid truss, computed by method: II	Forces in the rigid truss, computed by method: III	Forces in the rigid truss, computed by method: V*)
0—2	−0.421	−0.430	−0.433	−0.328	1.02	1.03	0.80	+43.66	+43.67	+43.65	+43.67
2—0	−1.773	−1.733	−1.841	−1.787	0.98	1.04	1.01
2—3	+1.644	+1.733	+1.718	+1.653	1.05	1.04	1.00	+43.70	+43.67	+43.69	+43.71
3—2	+1.094	+1.190	+1.215	+1.092	1.09	1.11	1.00
1—4	−0.190	−0.098	−0.198	−0.232	0.52	1.04	1.22	−56.16	−56.17	−56.18	−56.16
4—1	−0.815	−0.769	−0.823	−0.812	0.94	1.01	1.00
3—5	−1.159	−1.190	−1.293	−1.142	1.03	1.12	0.99	+67.41	+67.43	+67.37	+67.40
5—3	−2.107	−2.100	−2.277	−2.146	1.00	1.08	1.02
0—1	+0.421	+0.430	+0.433	+0.328	1.02	1.03	0.80	−69.76	−69.78	−69.74	−69.77
1—0	+0.066	+0.098	+0.071	+0.121	1.49	1.18	1.84
1—3	+0.041		+0.047	+0.039		1.15	0.95	+19.92	+19.84	+19.99	+19.92
3—1	+0.008		+0.018	+0.003		2.25	0.38
3—4	+0.057		+0.061	+0.047		1.07	0.82	−17.99	−18.01	−17.89	−17.99
4—3	−0.048		−0.048	−0.037		1.00	0.77
1—2	+0.083		+0.080	+0.071		1.07	0.86	+38.77	+38.73	+38.70	+38.78
2—1	+0.129		+0.123	+0.134		1.00	1.04

*) The data were obtained by the calculation performed in § 20.

Example III B.

Elements	In hinged truss	Longitudinal-force stresses in a rigid truss, method I	Longitudinal-force stresses in a rigid truss, method II	Longitudinal-force stresses in a rigid truss, method III	Bending stresses, method I	Bending stresses, method II	Bending stresses, method III	Total greatest stresses and coefficient \(N/n\), method I	Total greatest stresses and coefficient \(N/n\), method II	Total greatest stresses and coefficient \(N/n\), method III	Total greatest stresses and coefficient \(N/n\), method III*)
0—2	+415	+412	+412	+412	+91 −26	+93 −26	+94 −27	+521	+518	+525	+528
2—0	+415	+412	+412	+412	−384 +109	−375 +106	−398 +113	1,25	1,25	1,26	1,27
2—3	+415	+412	+412	+412	−356 +101	−376 +106	−372 +105	+649	+670	+675	+678
3—2	+415	+412	+412	+412	+237 −67	+258 −73	+263 −74	1,56	1,61	1,63	1,64
1—4	−406	−407	−407	−407	+9 −38	+4 −20	+9 −40	−445	−442	−447	−446
4—1	−406	−407	−407	−407	−37 +165	−35 +155	−37 +166	1,10	1,09	1,10	1,10
3—5	+493	+488	+489	+488	+234 −52	+240 −54	+261 −58	+722	+729	+749	+754
5—3	+493	+488	+489	+488	−425 +95	−424 +95	−459 +103	1,46	1,48	1,52	1,53
0—1	−486	−481	−481	−481	+217	+222	+223	−698	−703	−704	−709
1—0	−486	−481	−481	−481	±34	±51	±37	1,44	1,45	1,45	1,46
1—3	+209	+217	+216	+217	±17		±20	+234		+237	+229
3—1	+209	+217	+216	+217	±3		±8	1,12		1,13	1,09
3—4	−209	−196	−196	−194	±24		±52	−220		−219	−234
4—3	−209	−196	−196	−194	±20		±20	1,05		1,05	1,12
1—2	+500	+485	+484	+484	±39		±37	+545		+541	+557
2—1	+500	+485	+484	+484	±60		±57	1,09		1,08	1,12

*) See the note to Table B in Example I.

Example IV A.

Elements	Moments due to the rigidity of the joints, calculated by method: I	Moments due to the rigidity of the joints, calculated by method: II	Moments due to the rigidity of the joints, calculated by method: III	Ratios to those given by method I: II	Ratios to those given by method I: III	Longitudinal forces in rigid trusses, calculated by method: I	Longitudinal forces in rigid trusses, calculated by method: II	Longitudinal forces in rigid trusses, calculated by method: III
0—2	−0,355	−0,361	−0,360	1,02	1,01	+43,81	+43,83	+43,81
2—0	−0,965	−0,916	−0,983	0,95	1,02
2—3	+0,842	+0,916	+0,863	1,09	1,01
3—2	+0,566	+0,612	+0,598	1,08	1,06	+43,85	+43,83	+43,85
1—4	−0,120	−0,058	−0,124	0,48	1,03
4—1	−0,421	−0,390	−0,424	0,93	1,01	−56,08	−56,09	−56,09
3—5	−0,626	−0,612	−0,664	0,98	1,06
5—3	−1,101	−1,095	−1,147	0,99	1,04	+67,69	+67,70	+67,68
0—1	+0,355	+0,361	+0,360	1,02	1,01
1—0	+0,019	+0,058	+0,021	3,05	1,10	−70,02	−70,05	−70,02
1—3	+0,032		+0,034		1,06
3—1	+0,004		+0,008		2,00	+19,57	+19,62	+19,61
3—4	+0,056		+0,057		1,02
4—3	−0,050		−0,050		1,00	−18,56	−18,58	−18,57
1—2	+0,070		+0,069		0,99
2—1	+0,123		+0,120		0,98	+39,32	+39,30	+39,30

Example IV B.

Elements	In a hinged truss	Longitudinal-force stresses in a rigid truss, method I	Longitudinal-force stresses in a rigid truss, method II	Longitudinal-force stresses in a rigid truss, method III	Bending stresses, method I	Bending stresses, method II	Bending stresses, method III	Complete greatest stress and coefficient \(N/n\), method I	Complete greatest stress and coefficient \(N/n\), method II	Complete greatest stress and coefficient \(N/n\), method III	Complete greatest stress and coefficient \(N/n\), method III*)
0—2		\(+415\)	\(+413\)	\(+414\)	\(+154\) \(-44\)	\(+156\) \(-44\)	\(+156\) \(-44\)	\(+467\)	\(+470\)	\(+459\)	\(+471\)
2—0		\(+415\)	\(+413\)	\(+414\)	\(-417\) \(+118\)	\(-396\) \(+112\)	\(-425\) \(+120\)	1,12	1,13	1,13	1,13
2—3		\(+415\)	\(+414\)	\(+414\)	\(-364\) \(+103\)	\(-396\) \(+112\)	\(-373\) \(+106\)	\(+659\)	\(+679\)	\(+673\)	\(+674\)
3—2		\(+415\)	\(+414\)	\(+414\)	\(+245\) \(-69\)	\(+265\) \(-75\)	\(+259\) \(-73\)	1,59	1,64	1,62	1,62
1—4		\(-406\)	\(-406\)	\(-406\)	\(+11\) \(-48\)	\(+5\) \(-23\)	\(+11\) \(-50\)	\(-454\)	\(-441\)	\(-456\)	\(-456\)
4—1		\(-406\)	\(-406\)	\(-406\)	\(-38\) \(+170\)	\(-35\) \(+157\)	\(-38\) \(+171\)	1,12	1,08	1,12	1,12
3—5		\(+493\)	\(+490\)	\(+491\)	\(+252\) \(-57\)	\(+246\) \(-55\)	\(+267\) \(-60\)	\(+742\)	\(+737\)	\(+757\)	\(+760\)
5—3		\(+493\)	\(+490\)	\(+491\)	\(-444\) \(+99\)	\(-441\) \(+99\)	\(-462\) \(+103\)	1,51	1,49	1,53	1,54
0—1		\(-486\)	\(-483\)	\(-483\)	\(+183\) \(+9\)	\(+186\) \(+36\)	\(+186\) \(+11\)	\(-666\)	\(-669\)	\(-669\)	\(-672\)
1—0		\(-486\)	\(-483\)	\(-483\)	\(+13\) \(+2\)		\(+14\) \(+3\)	1,37	1,38	1,38	1,38
1—3		\(+209\)	\(+213\)	\(+213\)	\(+13\) \(+2\)		\(+14\) \(+3\)	\(+226\) 1,08		\(+227\) 1,08	\(+223\) 1,07
3—1		\(+209\)	\(+213\)	\(+213\)	\(+23\) \(+21\)		\(+24\) \(+21\)	\(-225\) 1,08		\(-226\) 1,08	\(-233\) 1,11
3—4		\(-209\)	\(-202\)	\(-202\)	\(+23\) \(+21\)		\(+24\) \(+21\)	\(-225\) 1,08		\(-226\) 1,08	\(-233\) 1,11
4—3		\(-209\)	\(-202\)	\(-202\)	\(+33\)		\(+33\)	\(+549\)		\(+548\)	\(+557\)
1—2		\(+500\)	\(+491\)	\(+491\)	\(+33\) \(+58\)		\(+33\) \(+57\)	\(+549\) 1,09		\(+548\) 1,09	\(+557\) 1,12
2—1		\(+500\)	\(+491\)	\(+491\)	\(+58\)		\(+57\)	1,09		1,09	1,12

Example V A.

Elements	Moments from rigid joints, computed by method I	Moments from rigid joints, computed by method II	Moments from rigid joints, computed by method III	Ratio to the given generalized method II	Ratio to the given generalized method III	Forces in members of the rigid truss, computed by method I	Forces in members of the rigid truss, computed by method II	Forces in members of the rigid truss, computed by method III
1—2	− 3,207	− 3,237	− 3,364	1,01	1,05	+ 73,52	+ 73,53	+ 73,50
2—1	− 3,456	− 3,472	− 3,707	1,01	1,07
2—4	+ 3,357	+ 3,472	+ 3,600	1,03	1,07	+ 73,54	+ 73,53	+ 73,52
4—2	+ 0,856	+ 1,119	+ 1,139	1,31	1,33
4—6	− 1,391	− 1,119	− 1,701	0,80	1,22	+ 149,14	+ 149,16	+ 149,10
6—4	− 4,421	− 4,340	− 4,793	0,98	1,08
6—8	+ 4,394	+ 4,340	+ 4,765	0,99	1,08	+ 149,14	+ 149,16	+ 149,10
8—6	+ 1,579	+ 1,557	+ 1,952	0,99	1,24
3—5	− 2,177	− 2,057	− 2,290	0,95	1,05	− 121,46	− 121,50	− 121,48
5—3	− 2,970	− 2,926	− 3,098	0,99	1,04
5—7	+ 2,882	+ 2,926	+ 3,003	1,02	1,04	− 121,49	− 121,50	− 121,51
7—5	− 0,735	− 0,623	− 0,661	0,85	0,90
7—9	+ 0,485	+ 0,623	+ 0,409	1,28	0,84	− 157,04	− 157,06	− 157,12
9—7	+ 2,367	− 2,304	− 2,181	0,97	0,92
1—3	+ 3,207	+ 3,237	+ 3,364	1,01	1,05	− 121,78	− 121,78	− 121,71
3—1	+ 1,944	+ 2,057	+ 2,050	1,06	1,06
3—4	+ 0,1486		+ 0,1492		1,00	+ 79,84	+ 79,94	+ 79,90
4—3	+ 0,0566		+ 0,0611		1,08
4—7	+ 0,4006		+ 0,4148		1,04	− 46,02	− 46,10	− 45,91
7—4	+ 0,0374		+ 0,0351		0,94
7—8	+ 0,1891		+ 0,1934		1,02	+ 13,16	+ 13,18	+ 13,35
8—7	− 0,0982		− 0,0984		1,00
2—3	+ 0,0993		+ 0,1070		1,08	+ 27,29	+ 27,21	+ 27,12
3—2	+ 0,0340		+ 0,0907		1,08
4—5	+ 0,0785		+ 0,0859		1,09	− 3,67	− 3,67	− 3,59
5—4	+ 0,0876		+ 0,0953		1,08
6—7	+ 0,0267		+ 0,0280		1,05	+ 20,54	+ 20,61	+ 20,28
7—6	+ 0,0236		+ 0,0245		1,04
8—9	0		0		1,00	− 4,32	− 4,39	− 4,35
9—8	0		0		1,00

Example V B.

Designation of moments	In the hinged truss	From longitudinal forces; in the rigid truss, computed by method I	From longitudinal forces; in the rigid truss, computed by method II	From longitudinal forces; in the rigid truss, computed by method III	Stresses from bending, computed by method I	Stresses from bending, computed by method II	Stresses from bending, computed by method III	Total greatest stresses and coefficients \(N/n\), computed by method I	Total greatest stresses and coefficients \(N/n\), computed by method II	Total greatest stresses and coefficients \(N/n\), computed by method III	Total greatest stresses and coefficients \(N/n\), computed by method III *)
1—2 2—1	+381	+380	+380	+380	+215 −97 −232 +105	+216 −98 −232 +105	+225 −102 −248 +112	+595 1.56	+596 1.56	+606 1.59	+605 1.59
2—4 4—2	+381	+380	+380	+380	−225 +102 +57 −26	−232 +105 +75 −34	−241 +109 +76 −34	+482 1.26	+485 1.27	+490 1.29	+489 1.28
4—6 6—4	+664	+660	+661	+660	+90 −34 −286 +109	+73 −27 −280 +107	+110 −42 −310 +118	+769 1.16	+768 1.16	+782 1.18	+778 1.17
6—8 8—6	+664	+660	+661	+660	−284 +108 +102 −39	−280 +107 +101 −38	−308 +118 +126 −48	+766 1.15	+761 1.15	+790 1.19	+786 1.18
3—5 5—3	−629	−628	−628	−628	+66 −146 −90 +199	+62 −138 −89 +196	+69 −154 −94 +207	−774 1.23	−766 1.22	−783 1.25	−782 1.24
5—7 7—5	−629	−628	−628	−628	−88 +193 −22 +49	−89 +196 −17 +38	−91 +201 −18 +43	−716 1.14	−717 1.14	−720 1.14	−719 1.14
7—9 9—7	−650	−649	−649	−650	+30 −10 +144 −51	+38 −13 +141 −50	+25 −9 +136 −47	−700 1.08	−699 1.08	−697 1.07	−697 1.07
1—3 3—1	−544	−539	−539	−539	−80 +118 −48 −72	−80 +119 +51 −76	−84 +121 +51 −76	−619 1.14	−619 1.14	−628 1.15	−623 1.14

Example V b. (Continuation).

Moment designation	Stress from longitudinal forces: in the hinged truss	Stress from longitudinal forces: in the rigid truss, computed by method I	Stress from longitudinal forces: in the rigid truss, computed by method II	Stress from longitudinal forces: in the rigid truss, computed by method III	Bending stress, computed by method I	Bending stress, computed by method II	Bending stress, computed by method III	Total greatest stress and coefficients \(N/n\), computed by method I	Total greatest stress and coefficients \(N/n\), computed by method II	Total greatest stress and coefficients \(N/n\), computed by method III	Total greatest stress and coefficients \(N/n\), computed by method III
3—4					\(\pm 41\)		\(\pm 41\)	\(+644\)		\(+645\)	\(+644\)
4—3	\(+604\)	\(+603\)	\(+604\)	\(+603\)	\(\pm 17\)		\(\pm 17\)	1.07		1.07	1.07
4—7					\(\pm 58\)		\(\pm 60\)	\(-339\)		\(-346\)	\(-340\)
7—4	\(-286\)	\(-281\)	\(-281\)	\(-280\)	\(\pm 5\)		\(\pm 5\)	1.19		1.21	1.19
7—8					\(\pm 27\)		\(\pm 28\)	\(+107\)		\(+103\)	\(+109\)
8—7	\(+75\)	\(+80\)	\(+80\)	\(+81\)	\(\pm 14\)		\(\pm 14\)	1.43		1.37	1.45
2—3					\(\pm 91\)		\(\pm 97\)	\(+547\)		\(+586\)	\(+550\)
3—2	\(+489\)	\(+456\)	\(+454\)	\(+453\)	\(\pm 77\)		\(\pm 83\)	1.12		1.20	1.12
4—5					\(\pm 72\)		\(\pm 78\)	\(-141\)		\(-169\)	\(-146\)
5—4	\(-83\)	\(-61\)	\(-61\)	\(-60\)	\(\pm 80\)		\(\pm 86\)	1.70		2.04	1.76
6—7					\(\pm 24\)		\(\pm 26\)	\(+367\)		\(+405\)	\(+365\)
7—6	\(+379\)	\(+343\)	\(+344\)	\(+339\)	\(\pm 21\)		\(\pm 22\)	0.97		1.07	0.96
8—9					\(0\)		\(0\)	\(-72\)		\(-83\)	\(-73\)
9—8	\(-83\)	\(-72\)	\(-73\)	\(-73\)	\(0\)		\(0\)	0.87		1.00	0.88

23. Comparative accuracy of the various methods of calculation.

Considering in Table A of Example I the relative values of the moments (columns 6, 7, 8), we shall easily see that the quantities most ...

close to the data of the generalized method are obtained by method III, i.e. Mohr’s. Indeed, even if the moments \(M_{0-1}\) and \(M_{0-2}\) are excluded (for which methods II and IV give entirely incorrect values), then the greatest error by the two methods reaches 55% and 54% (\(M_{1-3}\)), whereas by Mohr’s method the greatest error is only 22% (for \(M_{2-0}\); of course only the moments on the rigid contour are considered). The same result is obtained also for the mean values of the errors: 28% and 26% for methods II and IV, and only 10% for method III. Turning next to the values of the longitudinal forces in the rigid truss, here too we notice a difference in favor of method III; this is especially convenient to see from the stresses in Table B.

Let us finally consider the values of the total stresses and the coefficients \(\dfrac{N}{n}\); here, too, the advantage of Mohr’s method is quite clear; for example, for member 2–4 the second method gives an error of 23%, whereas Mohr’s method gives only 8%, and so on.

Consequently, all methods of comparison lead us to the conclusion that, for the truss of Example I, method III (Mohr’s) gives better results than the other methods. Quite different conclusions are obtained when examining the calculations for the truss of Example II.

Already from the first columns of Table A it is evident that, for this truss, Mohr’s-method data differ considerably from the data of the generalized method, whereas the values of method II are close to the true ones. Consideration of the relative values of the moments fully confirms this conclusion: the greatest error by Mohr’s method reaches 82% (\(M_{5-3}\)), while by method II it is only 15%; the mean errors will be, respectively, 27% and 6%.

All the other methods of comparison likewise give a very considerable difference in favor of method II; the values of the overstress coefficients given in Table B are especially characteristic.

Let us also note that method IV, in this case opposite to the first, gives results just as far from the correct ones; thus the a priori conclusion we gave in Chapter II concerning its unsuitability is fully confirmed.

The truss of Example III occupies, with respect to the accuracy of the methods, a middle position between the two preceding ones.

In this case it is difficult to give preference, in respect of accuracy, to any one of the approximate methods of calculation; although the relative values of the moments (Table A) give indications in favor of Mohr’s method (greatest error by method II—49%, by method III—18%; mean error 13% and 7%), the remaining methods of comparison do not show any substantial difference between methods II and III.

On the other hand, for this truss we have carried out above a computation taking into account (approximately) the longitudinal forces in the deformation of the truss members in bending (§ 20).

It is evident from the table that, precisely in the magnitudes of the relative values of the moments, the indicated calculation gives a rather large difference (84% maximum and 15% average for the rigid contour) in comparison with method I (the generalized method *).

If, in a calculation with the greatest accuracy possible for us, we can make an error close to the difference between the exact and approximate methods, then from the standpoint of accuracy, of course, it is quite immaterial which of the approximate methods is chosen.

The picture changes somewhat if the stiffness of the chords is reduced by half, i.e., if we pass to example IV; in this case the errors according to method II increase considerably, while according to method III they decrease, so that it is already possible to give decisive preference to Mohr’s method as the more accurate one.

Conversely, if we turn to the last example considered by us—example V—then we arrive at the conclusion that method II gives better results than Mohr’s method. This is clearly seen both from the average magnitudes of the errors in computing the moments and from the other results of the computations given in Tables A and B.

It is, of course, of the greatest interest and desirability to establish some convenient criterion that would make it possible to foresee in advance, for each given truss, which method of calculating the stresses due to the stiffness of the joints will give the best results. The calculations we have performed do not, naturally, give the right to state any definitive conclusions on this matter; for this purpose it would have been necessary to carry out a still considerable number of separate calculations, studying perhaps each truss system separately. The execution of this task, given the mass of purely mechanical work, is hardly within the power of a single person.

Nevertheless, the results we have obtained make it possible to establish certain general propositions, which will not constitute sharply delineated rules, but will give a sufficiently clear idea of the indisputable limits of applicability of the various methods and of their comparative merits.

As a criterion for classifying trusses, engineer Perederii proposed using the so-called stiffness of the members, comparing the latter by means of a coefficient of stiff—

*) As we have already said at the end of Chapter III, this difference does not exert a substantial influence on the overall accuracy of the adopted method; in the present case, however, it has decisive importance, since the only difference between methods II and III consists in the magnitudes of the (relative) moments.

stiffness, the magnitude of which is proportional to the moment of inertia of the section and inversely proportional to the third power of the length of the element.

Proceeding from purely theoretical considerations, it is, of course, generally difficult to justify the establishment of any criterion, since it is not possible to establish the theoretical dependence exactly. In favor of the criterion proposed by Eng. Perederii in particular, it is hardly possible to cite any theoretical considerations. On the other hand, however, it cannot be denied that the classification of trusses according to this criterion gives satisfactory results.

Below we shall give the values, computed by us for the five trusses examined, of the stiffness coefficients

\[ \eta=\frac{I_{\text{av.}}}{l_{\text{av.}}^{3}}\cdot 10^{6}. \]

Here, the quantities \(I_{\text{av.}}\) and \(l_{\text{av.}}\) mean the average values of the moment of inertia of the section and of the length of the group of elements under consideration. For each truss the calculations are carried out separately for the elements of the rigid contour of the truss and of the web.

If we agree to designate the results of approximate calculations with an average error (in comparison with the data of the generalized method) not exceeding 10% as good, with an error up to 20% on average as sufficient, and with an error greater than 20% as insufficient, then the following table may be compiled, classifying the trusses according to the stiffness of their elements in connection with the accuracy of calculations by approximate methods.

Table C.

Example No.	Avg. moment of inertia, \(I_{\text{av.}}\), cm	Avg. length, \(l_{\text{av.}}\), cm	Stiffness coeff. \(\eta\), for r.c.	Stiffness coeff. \(\eta\), for web	Ratio \(\chi=\dfrac{\eta_{\text{web}}}{\eta_{\text{r.c.}}}\)	Accuracy of calculation results by Mohr’s method	Accuracy of calculation results by method II
II	16910 3362	420 728	227.8	8.7	0.038	insufficient	good
V	67994 4463	681 811	215.7	8.4	0.039	sufficient	good
III	13636 2945	528 570	92.5	15.9	0.172	good	sufficient
I	3560 1800	420 640	48.0	5.9	0.144	sufficient	insufficient
IV	6818 2945	528 570	46.3	15.9	0.344	good	insufficient

Table D.

Joint No.	Sum designation	Truss I. $\Sigma$	Truss I. $\chi$	Truss II. $\Sigma$	Truss II. $\chi$	Truss III. $\Sigma$	Truss III. $\chi$	Truss IV. $\Sigma$	Truss IV. $\chi$	Truss V. $\Sigma$	Truss V. $\chi$
0	$\Sigma_k$	48,2	0,32	193,0	0	188,7	0	94,4	0
0	$\Sigma_p$	15,6	0,32	0	0	0	0	0	0
1	$\Sigma_k$	48,2	0,27	268,2	0,25	115,9	0,36	57,9	0,72	376,2	0
1	$\Sigma_p$	13,1	0,27	66,0	0,25	41,8	0,36	41,8	0,72	0	0
2	$\Sigma_k$	82,0	0,29	217,8	0,40	294,0	0,08	147,0	0,16	386,2	0,01
2	$\Sigma_p$	23,7	0,29	88,2	0,40	23,1	0,08	23,1	0,16	2,8	0,01
3	$\Sigma_k$	82,0	0,26	586,1	0,05	315,5	0,12	157,7	0,24	376,2	0,04
3	$\Sigma_p$	21,2	0,26	30,8	0,05	37,4	0,12	37,4	0,24	15,1	0,04
4	$\Sigma_k$	90,0	0,12	217,8	0,08	168,4	0,36	84,2	0,72	403,6	0,10
4	$\Sigma_p$	11,2	0,12	17,2	0,08	60,5	0,36	60,5	0,72	38,6	0,10
5	$\Sigma_k$	90,0	0,24	804,0	0	337,0	0,07	168,5	0,14	386,2	0,01
5	$\Sigma_p$	21,2	0,24	0	0	23,1	0,07	23,1	0,14	2,8	0,01
6	$\Sigma_k$									425,0	0,01
6	$\Sigma_p$									2,8	0,01
7	$\Sigma_k$									425,5	0,12
7	$\Sigma_p$									49,8	0,12
8	$\Sigma_k$									421,0	0,12
8	$\Sigma_p$									49,8	0,12
9	$\Sigma_k$									464,8	0,01
9	$\Sigma_p$									2,8	0,01
Average $\chi$.			0,25		0,13		0,17		0,33		0,05

From examination of the table it is clearly seen that, as the rigidity of the contour increases, the inaccuracy of Mohr’s method increases, while method II gives ever better results, since, in parallel with the increase in the rigidity of the contour, there is a decrease in the ratio between the rigidity of the lattice and that of the contour. The table does not give a complete coincidence of the sequence, mainly, probably, in view of the difference between the systems of trusses. For our part, we would consider it possible, proceeding from approximate theoretical considerations, to give another system of classification, which also gives a sufficiently close agreement of the results, especially with respect to method II.

Considering the equilibrium between the moments appearing in one and the same joint of a truss, we shall arrive at the conclusion that the error in determining the moments in the rigid contour by method II will be the greater, the more significant are the moments bending the lattice. But approximately we may assume that the magnitudes of the moments are proportional to the values

\[ N=\frac{2EI}{l}; \]

in order to judge the magnitude of the error, we may calculate, for each joint, \(\sum N\) separately for the lattice and for the rigid contour, and take the ratio of these two sums.

The results of such a calculation are given in Table D; taking the mean values of \(\chi\) for each truss, we may, using our terminology, again draw up a comparative table of the dependence of the accuracy of the second method on the value of \(\chi\).

Trusses	\(\chi\)	Accuracy of method II
IV	0.33	insufficient
I	0.25	insufficient
III	0.17	sufficient
II	0.13	good
V	0.05	good

From this small table it is seen that the value \(\chi\) gives a better classification than the coefficient \(\eta\).

But the coefficient proposed by us is also of interest in that it apparently makes it possible to trace the accuracy of the determination of moments in individual joints of trusses as a function of the values of the coefficient \(\chi\) for these joints.

This dependence can be traced most distinctly on the truss of example V.

If we compare, in a small table, the values of $\chi$ for the various joints with the mean values of the errors in determining the moments in the rigid contour by the second method, we shall see an almost parallel variation of these two quantities.

Joints	$\chi$	Mean err. %	Joints	$\chi$	Mean err. %
1	0	1	6	0.01	1.5
2	0.01	2	7	0.12	21.5
3	0.04	5.5	8	0.12	1
4	0.10	25.5	9	0.01	3
5	0.01	1.5

The sole exception is joint 8, where, with a relatively large value of $\chi$, the mean error is only 1%; this, however, is quite understandable because of the symmetry of the load, owing to which the influence of the two web members is mutually canceled.

The same picture is presented also by truss example III, with the corresponding reservation concerning joints 4 and 5. For this truss we have the table:

Joints	$\chi$	Mean err. %	Joints	$\chi$	Mean err. %
0	0	2	3	0.62	6
1	0.36	48.5	4	0.36	6
2	0.08	4	5	1.07	0

We observe the same correspondence for truss IV as well, the data for which we do not give because of their complete identity with the preceding ones.

The indicated dependence is seen very clearly also in truss example II; for example I, however, the picture is obscured by the fact that at all joints the coefficient $\chi$ has a comparatively large value, which makes it difficult to investigate the variation of the mean errors.

In any case, from the calculations we have made it follows that, when the value of $\chi$ is above 0.25, neglecting the stiffness of the web appreciably affects the accuracy of the computation.

In assessing the applicability, for any given truss, of approximate methods of calculation, one should, in our opinion, be guided by a twofold criterion. For assessing the suitability of Mohr’s method

the best results are apparently given by the stiffness criterion proposed by Engineer Perederii*).

The system of trusses plays, as is evident from Table C, a considerable role in determining the applicability of Mohr’s method; therefore, on the basis of the available data it is difficult to draw a strict boundary in this respect. Nevertheless, with sufficient probability one may assert that, for values of the contour stiffness coefficient exceeding 200, Mohr’s method should be avoided.

With regard to the second method (Engineer Perederii), one should be guided by the coefficient \(\chi\) given by us; for average values of this coefficient for the truss under consideration greater than 0.20, this method begins to give incorrect results.

It should, however, be borne in mind that sometimes the average value of \(\chi\) increases considerably under the influence of one or two joints, where web members with large values of the quantity \(N\) meet.

In such cases the method can be made more applicable by taking into account the rigid attachment of the corresponding web members, which, of course, introduces into the equations two additional unknowns for each such member.

In choosing one method or another, in addition to accuracy, one must also be guided by the comparative complexity of the calculations. Engineer Perederii believes that computation by his method (our second) requires almost as much work as by Mohr’s method.

We cannot fully agree with this: both methods lead to the solution of a system of first-degree equations, the number of whose unknowns is equal to the number of joints of the truss under consideration.

At the same time, however, in Mohr’s method a significant part of the coefficients at the unknowns are equal to zero, which greatly simplifies the solution; whereas in Engineer Perederii’s method, generally speaking, all the coefficients turn out to be significant. With a large number of unknowns, the saving of labor by Mohr’s method becomes substantial.

24. Works of Engineers Paton and Perederii.

Let us now turn to the works of Engineers Paton and Perederii and compare their conclusions, guided by the data obtained by—

*) There are some theoretical grounds in favor of choosing, as a criterion, the quantity \(\dfrac{I}{l^2}\), but our calculations have shown that, in the sense of practical agreement with the calculation data, this criterion is inferior to that proposed by Engineer Perederii.

with comparative calculation. The two named authors obtained sharply different results in calculating the stiffness stresses for the two-diagonal truss shown in Fig. 13.

Fig. 13.

Engineer Paton calculated the stiffness stresses for this truss by Mohr’s method*), arranging the asymmetrical load in such a way that the system of diagonals indicated by the dotted line was loaded more than the second system. In doing so he obtained very considerable overstresses in the chord members, reaching as much as 293%, so that the strength of trusses of this system appeared doubtful.

Engineer Perederiy determined, for the same truss, the stiffness stresses by the method he proposed and found that the overstresses reached only 74%. In his calculations the load does not fully coincide with Engineer Paton’s problem, since, in order to shorten the exposition, he assumed it to be symmetrical.

As can be seen, the results obtained by Engineer Perederiy are of a very reassuring character. However, his method, like Mohr’s method, is not entirely exact, as we saw in Chapter II. The question therefore arises as to how close these reassuring conclusions are to reality.

In this connection Engineer Perederiy points to the very small magnitude of the work of deformation in bending of the web system in comparison with the work of deformation in bending of the chords, and asserts that the first of these works may be neglected. There is no objection to this, especially as applied to the truss under consideration.

However, by taking the web system as hinged, we not only neglect the work of bending of the web system, but also alter the distribution of the work of bending in the chords, as is easily seen from Tables A given above. The question therefore remains disputable.

*) However, when determining the centers of rotation of the diagonals, forces constituting a couple equal to 1 were applied vertically (see above, Chapter II).

To clarify the matter, let us turn to our Example II; it is a truss of the same type as the truss in Fig. 13, and even the greater part of the cross-sections entering into it were deliberately taken by us from the truss of Eng. Paton.

For this truss, as we have convinced ourselves above, decisive preference should be given to Eng. Perederiy’s method. The error in determining the value of the overstress coefficient \(\frac{N}{n}\) reaches, in this case, only \(10\%\) (by Mohr’s method, \(44\%\)).

Let us compare the values of the stiffness coefficients: the average stiffness of the chords

\[ \begin{aligned} &\text{for our truss—}227.8,\\ &\text{for the truss in Fig. 13—}428; \end{aligned} \]

the stiffness of the chords of the second truss is almost twice as great as that of the first; consequently, Mohr’s method, when applied to it, is still less accurate than for our truss.

Let us compare, on the other hand, the ratios of the average stiffness of the web to the average stiffness of the chords:

\[ \text{for our truss: } \frac{8.7}{227.8}=0.038, \]

\[ \text{for Paton’s truss: }\qquad =0.019. \; *) \]

Obviously, with respect to Eng. Perederiy’s method, the second truss is in still better conditions than the first, for which Perederiy’s method gave only a \(10\%\) greatest error in comparison with the exact method.

It may therefore be said with confidence that the results of Eng. Perederiy’s computations for the given truss are very close to the exact ones, i.e., there is indeed no basis for any particular apprehension regarding the strength of trusses of the two-lattice system. **)

However, Eng. Perederiy does not confine himself in his article to the truss just cited, but also applies his method to a two-lattice truss which was likewise analyzed by Eng. Paton (Fig. 14). In doing so he obtains a reduction of the greatest additional stresses from \(177\%\) to \(115\%\).

*) Using our criterion, we have

\[ \begin{aligned} &\text{for Paton’s truss } \chi = 0.008,\\ &\text{for our truss } \chi = 0.013. \end{aligned} \]

**) In addition to what has been said, cf. also the polemic between Paton and Perederiy in § 28.

This result cannot be considered more accurate than Paton’s data.

Indeed, for a truss of the same type (Example 1) we obtained above an error in the overstress coefficient, in comparison with the generalized method, of 23% by Pérderie’s method and 9% by Mohr’s method, the former being directed toward a reduction of stresses.

Comparing, for these two trusses, the stiffness coefficients of the contours, we have

for our truss—48.0,
for Paton’s truss—31.8;

the ratios of the stiffness of the web to the stiffness of the chords (contour)

Fig. 14.

for our truss—0.146,
for Paton’s truss—0.107. *)

From these numbers it is evident that Paton’s truss (the stiffness stresses for it were computed by Winkler) should relate to the methods of Mohr and Pérderie approximately in the same way as the one we have analyzed, i.e., Mohr’s method should give results considerably closer to the exact ones.

This is true to an even greater degree for the third example of Eng. Pérderie (Fig. 15). Here the mean stiffness of the contour (or, what is the same for this case, of the chords) is 136.3, and that of the web is 33.8; i.e., the ratio of these two quantities is 0.248.

It is clear that in this case neglecting the rigid attachment of the web must lead to erroneous conclusions. **)

Thus, the reduction of stresses obtained by Eng. Pérderie in the three cases he considered is only in one—

*) The coefficient $\gamma$ gives, for our truss—0.25,
for Paton’s truss—0.20.

**) Generally speaking, no conclusions can be drawn from Eng. Pérderie’s computation for the truss of Fig. 15, since an unfortunate error has crept into the calculation: on p. 24, 8 lines from the bottom, instead of $-53.20 M_3$ there should stand $+53.20 M_3$, which, of course, changes the subsequent conclusions.

case is due to the greater accuracy of the method applied by him. In general, however, whereas Mohr’s method gives excessively large stresses, Perderii’s method has the property, in certain cases, of understating them.

Fig. 15.

Summarizing the data of the comparative analysis we have carried out, we arrive at the conclusion that both the second and the third groups of methods are not universal, but within certain limits may be applied with quite sufficient accuracy; moreover, for the second group these limits are apparently narrower than is assumed in his work by Engineer Perderii.

25. Dependence between the moments due to the rigidity of the joints and the sectional areas of the individual members.

In his book, Engineer Paton, among other things, constructs diagrams of the relationship between the quantities $\frac{l}{e}$ and $\frac{N}{n}$; here $l$ is the length of the member; $e$ is the distance from the neutral axis of the member section to the most highly stressed fiber; $\frac{N}{n}$ is the overstress coefficient. These diagrams deserve attention as an attempt to apply the theoretical data on rigidity stresses in practice and to give the designer indications as to which sections are most advantageous with respect to the stresses from the rigidity of the nodal connections. Engineer Paton has even included these data in the course on bridges now being published by him. It is therefore of interest to investigate how far these diagrams reflect the actual state of affairs. Let us first note one property of the rigidity stresses that follows from calculating them by Mohr’s method, and therefore

and existing only when the named method can be applied with sufficient accuracy. Only within these limits shall we make use of the mentioned property; moreover, to simplify the derivations, we shall prove it only for statically determinate trusses, since it is only to such trusses that we shall have to apply it.

Let us recall the equations which we obtained in Chapter II for group III of the methods of calculation:

\[ \Delta\gamma'_{0-1}(2M_1+M_2+M_3)+\Delta\gamma'_{0-2}(2M_1-M_4-M_5) = -\sum S^0\Delta\lambda.\alpha, \]

\[ \Delta\gamma'_{0-1}(2M_2+2M_3+M_1)+\Delta\gamma'_{1-3}(2M_2-M_6) = -\sum S^0\Delta\lambda.\beta, \tag{37} \]

\[ \Delta\gamma'_{0-1}(2M_2+2M_3+M_1)+\Delta\gamma'_{1-4}(2M_3-M_7) = -\sum S^0\Delta\lambda.\gamma, \]

\[ \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot \]

where

\[ \Delta\gamma'_{0-1}=\frac{l_{0-1}}{6EI_{0-1}}; \quad \text{etc.,} \]

and the sign \(\sum\) denotes that the summation extends over all elements of the truss; \(\alpha,\ \beta,\ \gamma \ldots\) are the magnitudes of the forces in the elements of the truss under the action of the corresponding loads \(M_1=1;\ M_2=1;\ M_3=1;\) etc.

Finally,

\[ \Delta\lambda'_{0-1}=\frac{l}{E\omega}. \]

Solving equations (37), we may express the quantities \(M_1, M_2,\ldots\) by means of determinants:

\[ M'_1=\frac{\Delta'_1}{\Delta}, \]

\[ M'_2=\frac{\Delta'_2}{\Delta}, \tag{a} \]

\[ \cdot\quad \cdot\quad \cdot \]

Let us now reduce the cross-sectional area of some element of the truss, say, for definiteness, \(0-2\), by \(n\) times, while leaving unchanged the moment of inertia and the section modulus of this section and preserving all other conditions unchanged.

The new cross-sectional area is

\[ (\omega_1)_{0-2}=\frac{(\omega_0)_{0-2}}{n}. \]

Consequently,

\[ (\Delta\lambda_1)_{0-2} = \frac{l_{0-2}\cdot n}{E(\omega_0)_{0-2}} = \frac{l_{0-2}}{E(\omega_0)_{0-2}} + \frac{(n-1)l_{0-2}}{E(\omega_0)_{0-2}}. \]

Substituting this quantity into the sums (37), we obtain their expressions for the truss modified (by reducing the cross section of the member \(0-2\)):

\[ \sum S^0\Delta\lambda_1\alpha = \sum S^0\Delta\lambda\cdot\alpha + \frac{(n-1)l_{0-2}}{E(\omega_0)_{0-2}}\cdot\alpha_{0-2}, \]

\[ \sum S^0\Delta\lambda_1\beta = \sum S^0\Delta\lambda\cdot\beta + \frac{(n-1)l_{0-2}}{E(\omega_0)_{0-2}}\cdot\beta_{0-2}, \]

\[ \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot \]

To determine \(M\), we therefore obtain a new system of equations

\[ \Delta'_{0-1}(2M_1+M_2+M_3) + \Delta'_{0-2}(2M_1-M_4-M_5) = \]

\[ -\sum S^0\Delta\lambda\cdot\alpha - (n-1)\Delta\lambda_{0-2}\alpha_{0-2}, \]

\[ \tag{b} \Delta'_{0-1}(2M_2+2M_3+M_1) + \Delta'_{1-3}(2M_2-M_6) = \]

\[ -\sum S^0\Delta\lambda\cdot\beta - (n-1)\Delta\lambda_{0-2}\beta_{0-2}, \]

\[ \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot \]

As is known from the theory of determinants, a system of equations of this kind can be solved in the following way.

We decompose the given system into two, of which the first will be identical with system (37), and its roots will be represented in the form of the expressions (a), while the second will have the following form:

\[ \Delta r'_{0-1}(2M_1+M_2+M_3)+\Delta r'_{0-2}(2M_1-M_4-M_5)= \]
\[ =-(n-1)\Delta\lambda_{0-2}\alpha_{0-2}, \]
\[ \Delta r'_{0-1}(2M_2+2M_3+M_1)+\Delta r'_{1-3}(2M_2-M_6)= \tag{c} \]
\[ =-(n-1)\Delta\lambda_{0-2}\beta_{0-2}, \]
\[ \Delta r'_{0-1}(2M_2+2M_3+M_1)+\Delta r'_{1-4}(2M_3-M_7)= \]
\[ =-(n-1)\Delta\lambda_{0-2}\gamma_{0-2}, \]
\[ \cdots \]

According to the theory of determinants, the solutions of this system may be written

\[ M''_1=(n-1)\Delta\lambda_{0-2}\cdot\frac{\Delta''_1}{\Delta}, \]
\[ M''_2=(n-1)\Delta\lambda_{0-2}\cdot\frac{\Delta''_2}{\Delta}, \tag{d} \]
\[ \cdots \]

Now the values of the roots of system (d) are obtained by simple addition of the corresponding expressions (a) and (d)

\[ M_1=\frac{\Delta'_1}{\Delta}+(n-1)\Delta\lambda_{0-2}\cdot\frac{\Delta''_1}{\Delta}, \]
\[ M_2=\frac{\Delta'_2}{\Delta}+(n-1)\Delta\lambda_{0-2}\cdot\frac{\Delta''_2}{\Delta}, \tag{e} \]
\[ \cdots \]

We substitute, in place of \(n\), its value

\[ n=\frac{(\omega_0)_{0-2}}{(\omega_1)_{0-2}} \]

and write the variable cross-sectional area without a sign

\[ M_1=\frac{\Delta'_1}{\Delta}+\frac{l_{0-2}}{E\omega_{0-2}}\cdot\frac{\Delta''_1}{\Delta} -\frac{l_{0-2}}{E(\omega_0)_{0-2}}\cdot\frac{\Delta''_1}{\Delta}, \]
\[ M_2=\frac{\Delta'_2}{\Delta}+\frac{l_{0-2}}{E\omega_{0-2}}\cdot\frac{\Delta''_2}{\Delta} -\frac{l_{0-2}}{E(\omega_0)_{0-2}}\cdot\frac{\Delta''_2}{\Delta}, \tag{f} \]
\[ \cdots, \]

or, substituting, in place of quantities independent of the change in area, the abbreviated designations, according to (g)

\[ \begin{aligned} \frac{\Delta_1'}{\Delta} - \frac{l_{0-2}}{E(\omega_0)_{0-2}}\cdot\frac{\Delta_1''}{\Delta} &= A_1, & \frac{l_{0-2}}{E}\cdot\frac{\Delta_1''}{\Delta} &= \beta_1, \\ \frac{\Delta_2'}{\Delta} - \frac{l_{0-2}}{E(\omega_0)_{0-2}}\cdot\frac{\Delta_2''}{\Delta} &= A_2, & \frac{l_{0-2}}{E}\cdot\frac{\Delta_2''}{\Delta} &= \beta_2, \end{aligned} \tag{g} \]

. . . . . . . . . . . . . . . . .

we find

\[ \begin{aligned} M_1 &= A_1 + \frac{\beta_1}{\omega_{0-2}},\\ M_2 &= A_2 + \frac{\beta_2}{\omega_{0-2}}, \end{aligned} \tag{h} \]

. . . . . . . . . . . . . . . . .

Formulas (h) express the property of the moments and stresses due to the rigidity of the joints which we wished to note: if, in a statically determinate rigid truss subjected to a given load, one changes the cross-sectional area of some member, while keeping unchanged both the moments of inertia and the section moduli of that cross section, as well as all other conditions, then the dependence between the indicated change in area and the changes in all the moments and rigidity stresses will be expressed by the equation of a hyperbola.

26. Fluctuations in the magnitude of the coefficient \(\dfrac{N}{n}\).

Before proceeding to a critique of Eng. Paton’s diagrams themselves, let us consider for a while the coefficient \(\dfrac{N}{n}\) entering into them and its changes when the cross-sectional area of some member is increased or decreased.

For definiteness of judgment, let us take as an example the truss placed in Eng. Paton’s book under No. 5.

The schematic form of this truss and its loading are shown in Fig. 16, and the principal dimensions are given in Table E.

We determine, by Mohr’s method, the moments and stresses due to the rigidity of the joints in the following three cases:

1) for the basic truss;

2) on the assumption that the cross-sectional areas of members 1—3 and 5—7 are reduced by half;

3) on the assumption that the cross-sectional areas of members 3—4 and 4—5 are reduced by half.

Drawing 16.

Table E.

Element	Element No.	$\omega$ gross, cm²	$I$ gross, cm⁴	$l$, cm	$e_1 = e_2$, cm	Stress $n$, kg/cm²	Sections
Upper chord	2—4	60	1000	500	9.5	−567	4 angles
Upper chord	4—6	60	1000	500	9.5	−467	90 × 90 × 9
Lower chord	1—3	50	750	500	8.5	+360	4 angles
Lower chord	5—7	50	750	500	8.5	+300	80 × 80 × 9
Lower chord	3—5	70	1125	500	8.5	+457	4 ang. 80 × 80 × 12
Ascending braces	1—2	50	790	350	8.5	−520	4 angles
Ascending braces	6—7	50	790	350	8.5	−420	80 × 80 × 9
Ascending braces	3—4	30	440	350	8.5	+100	4 angles
Ascending braces	4—5	30	440	350	8.5	−200	80 × 40 × 7
Descending braces	2—3	40	610	350	8	+575	4 angles
Descending braces	5—6	40	610	350	8	+450	75 × 75 × 8

Table F.

Elements	Nodes	Main truss: Moment, kg·cm	Main truss: Stress, kg/cm²	Main truss: \(N/n\)	Sections 3–4 and 4–5 reduced by half: Moment, kg·cm	Sections 3–4 and 4–5 reduced by half: Stress, kg/cm²	Sections 3–4 and 4–5 reduced by half: \(N/n\)	Sections 1–3 and 5–7 reduced by half: Moment, kg·cm	Sections 1–3 and 5–7 reduced by half: Stress, kg/cm²	Sections 1–3 and 5–7 reduced by half: \(N/n\)
1–3	1	− 3052	101	1,28	− 3275	108	1,30	− 1546	110	1,15
1–3	3	− 8883	101	1,28	− 9571	108	1,30	− 9698	110	1,15
3–5	3	+ 9160	69	1,15	+ 11129	84	1,18	+ 8931	68	1,15
3–5	5	− 4310	69	1,15	− 2465	84	1,18	− 4924	68	1,15
5–7	5	+ 5438	62	1,21	+ 4649	53	1,18	+ 5855	66	1,11
5–7	7	+ 2089	62	1,21	+ 1912	53	1,18	+ 880	66	1,11
2–4	2	+ 1287	86	1,15	+ 1676	89	1,16	+ 3373	76	1,13
2–4	4	− 9097	86	1,15	− 9394	89	1,16	− 7974	76	1,13
4–6	4	+ 8761	83	1,18	+ 9419	89	1,19	+ 7662	73	1,16
4–6	6	+ 150	83	1,18	+ 1331	89	1,19	− 2207	73	1,16
1–2	1	+ 3052	33	1,06	+ 3275	35	1,07	+ 1546	51	1,10
1–2	2	− 820	33	1,06	− 738	35	1,07	− 4697	51	1,10
3–4	3	+ 3902	75	1,80	+ 3527	68	1,36	+ 3992	77	1,82
3–4	4	+ 126	75	1,80	− 368	68	1,36	+ 221	77	1,82
4–5	4	+ 208	67	1,35	+ 353	63	1,16	+ 90	73	1,39
4–5	5	− 3455	67	1,35	− 3259	63	1,16	− 3789	73	1,39
6–7	6	− 135	22	1,05	− 403	21	1,05	+ 2474	27	1,07
6–7	7	− 2089	22	1,05	− 1918	21	1,05	− 880	27	1,07
2–3	2	− 468	55	1,10	− 940	67	1,12	+ 1323	42	1,08
2–3	3	− 4176	55	1,10	− 5075	67	1,12	− 3221	42	1,08
5–6	5	+ 2324	31	1,07	+ 1076	14	1,03	+ 2856	37	1,08
5–6	6	− 14	31	1,07	− 928	14	1,03	− 266	37	1,08

It should be noted that such changes in the cross-sectional area, while keeping the other quantities constant, may prove difficult—

Drawing 17.

Diagram No. 1 of the relation between $\dfrac{L}{e}$ and $\dfrac{N}{n}$, by Eng. Paton, for chords near supports.

Trusses with parallel chords and a lattice:

• simple diagonal lattice . . . examples 1, 2
• compound diagonal lattice . . . examples 13, XIII
• triangular lattice with additional posts . . . examples 9, 10
• triangular lattice without additional posts . . . examples 5, 6, 7, 8

$\circ$ — denotes double bending of a simple [[unclear: type]].

*) The sign *) marks cases in which the critical fiber for the upper chord is the lower one, and for the lower chord, the upper fiber.

—while, for small cross-sectional dimensions, as in the present case, they may even be structurally impossible owing to the insufficient variety of available rolled sections.

But we are considering the question only from the fundamental, theoretical side; if necessary, however, the hyperbolic formula—

…dependence will always enable us to pass to a practically feasible modification.

Fig. 19. Diagram for the lower chord of Paton No. 1 after excluding members 9, 10, 13, XIII.

The results of all three calculations are brought together in Table F, from which it is evident that the value of the coefficient \(\frac{N}{n}\) for one and the same member is subject to considerable fluctuations depending on the change in the cross-sectional area of this member. Moreover, for the given—

...of the truss, the stresses due to the stiffness of the joints are comparatively very small, so that in other cases one may expect still sharper fluctuations. Meanwhile, the magnitudes of the principal stresses obtained when changing the sections do not represent anything special: 720 and 600 \(\frac{\mathrm{kg}}{\mathrm{cm}^2}\) for the chords; 200 and 400 \(\frac{\mathrm{kg}}{\mathrm{cm}^2}\) for the diagonals.

It is clear from this that the coefficient \(\frac{N}{n}\), very convenient in its simplicity, is in itself insufficient for characterizing the danger of stiffness stresses for a given truss; it may assume very large values for individual members of the truss and yet the strength of the truss will not be in doubt.

27. Diagrams of Engineer Paton.

Let us now turn to the diagrams of Engineer Paton; there are four of them in all—two each for the chords and for the diagonals, with the members located at the supports separated into one group, and those nearer the middle of the span into another. For our purposes it will suffice to examine in detail one of the diagrams for the chords, namely the first, which we reproduce in Fig. 17.

[Figure: truss diagram with panels and cross-bracing; panel length \(4^m\), height \(4^m\).]

Examples 13 and XIII represent one and the same diagonal truss (Fig. 19), for which the stresses due to the stiffness of the joints were calculated under two assumptions:

1) the diagonals are riveted at the intersections;
2) the diagonals are not riveted.

The chords in this truss are designed to be very stiff: of a vertical plate \(45 \times 1.5\) cm, two angles \(11.8 \times 11.8 \times 1.3\) cm, and from one to three horizontal plates \(48 \times 1.2\) cm. With the comparatively small panel length (\(4^m\)), our conventional coefficient of average stiffness of the contour comes out above 500, i.e., according to the data we have obtained, for it one should certainly have applied the method of Engineer Perederii (all the more so since the lattice possesses...

with comparatively small rigidity—on average about 20); Mora’s method, or one of like nature, must likewise yield, at the very least, doubtful results.

But even leaving aside the question of the method of calculation, one cannot fail to notice that the points on the upper branch of the curve which are most important for the diagram, from this example, correspond to very small principal stresses \((150\) and \(200\ \frac{\mathrm{kg}}{\mathrm{cm}^{2}})\), whereas the stresses of all the other members exceed them by \(2.5\)–\(4\) times. Above we saw that precisely in such a case disproportionately large values may be obtained for the coefficient \(\frac{N}{n}\).

Finally, one cannot but draw attention also to the circumstance that Eng. Paton included in the diagram only this one truss of a complicated diagonal system (No. 12).

It was necessary either to exclude this system altogether, which would have been the most correct course, or else to introduce into the diagram all the examples available for the given system. Otherwise the diagram loses its generality and becomes a collection of artificially selected examples.

Further, apart from example 13 and XIII, examples 9 and 10 served Eng. Paton for determining the vertical part of the curve; let us examine them in more detail.

Both of these examples relate to trusses of a triangular system with additional posts—a system of which Eng. Paton himself says (p. 47): “the additional posts, being subjected to elastic changes in their length, may be the cause of more or less considerable deformations of the chords,” and further (p. 48): “examples 9 and 10 clearly show that these stresses *) are the greater, the more considerable are the changes in length of the posts.”

Despite the fact that Eng. Paton thus himself identifies the true cause of the considerable additional stresses in these trusses (a cause entirely independent of the magnitude of the ratio \(\frac{l}{e}\)), he nevertheless includes this system of trusses in his diagram.

It is interesting to show how greatly exaggerated values of the coefficient \(\frac{N}{n}\) entered the diagram not because of the low value of the ratio \(\frac{l}{e}\), but

*) From the rigidity of the nodal connections.

Fig. 20.

Table G.

Member group	Nos. of members	Basic stress according to the data of engineer Paton	\(N/n\) according to the data of engineer Paton	Basic \(H/n\) according to our calculation	\(N/n\) according to our calculation, in the main truss	\(N/n\) with enlarged struts	\(l/e\)	Sections
Lower chord	\(0—1\)	\(+280\)	1,54	\(+280\)	1,55	1,38	39	1 vertical plate \(40 \times 1\). 1 horizontal plate \(40 \times 0{,}8\). 2 angles \(9 \times 9 \times 1\).
Lower chord	\(0'—1'\)	\(+160\)	1,19	\(+160\)	1,27	1,25	39	1 vertical plate \(40 \times 1\). 1 horizontal plate \(40 \times 0{,}8\). 2 angles \(9 \times 9 \times 1\).
Lower chord	\(1—3\)	\(+280\)	2,43	\(+280\)	2,34	1,46	11	1 vertical plate \(40 \times 1\). 1 horizontal plate \(40 \times 0{,}8\). 2 angles \(9 \times 9 \times 1\).
Lower chord	\(1'—3'\)	\(+160\)	1,22	\(+160\)	1,13	1,06	39	1 vertical plate \(40 \times 1\). 1 horizontal plate \(40 \times 0{,}8\). 2 angles \(9 \times 9 \times 1\).
Lower chord	\(3—5\)	\(+480\)	1,83	\(+500\)	1,81	1,36	10,3	1 vertical plate \(40 \times 1\). 2 horizontal plates \(40 \times 0{,}8\). 2 angles \(9 \times 9 \times 1\).
Lower chord	\(3'—5'\)	\(+350\)	1,11	\(+340\)	1,16	1,04	46	1 vertical plate \(40 \times 1\). 2 horizontal plates \(40 \times 0{,}8\). 2 angles \(9 \times 9 \times 1\).
Lower chord	\(5—6\)	\(+480\)	1,87	\(+500\)	1,79	1,28	10,3	1 vertical plate \(40 \times 1\). 2 horizontal plates \(40 \times 0{,}8\). 2 angles \(9 \times 9 \times 1\).
Lower chord	\(5'—6'\)	\(+350\)	1,11	\(+340\)	1,08	1,03	46	1 vertical plate \(40 \times 1\). 2 horizontal plates \(40 \times 0{,}8\). 2 angles \(9 \times 9 \times 1\).
Lower chord	\(6—8\)	\(+450\)	2,11	\(+460\)	1,86	1,34	9,8	1 vertical plate \(40 \times 1\). 2 horizontal plates \(40 \times 0{,}8\). 1 horizontal plate \(40 \times 1\). 2 angles \(9 \times 9 \times 1\).
Lower chord	\(6'—8'\)	\(+400\)	1,10	\(+390\)	1,15	1,03	52	1 vertical plate \(40 \times 1\). 2 horizontal plates \(40 \times 0{,}8\). 1 horizontal plate \(40 \times 1\). 2 angles \(9 \times 9 \times 1\).
Lower chord	\(8—9\)	\(+450\)	1,45	\(+460\)	1,42	1,17	9,8	1 vertical plate \(40 \times 1\). 2 horizontal plates \(40 \times 0{,}8\). 1 horizontal plate \(40 \times 1\). 2 angles \(9 \times 9 \times 1\).
Lower chord	\(8'—9'\)	\(+400\)	1,13	\(+390\)	1,47	1,04	9,8	1 vertical plate \(40 \times 1\). 2 horizontal plates \(40 \times 0{,}8\). 1 horizontal plate \(40 \times 1\). 2 angles \(9 \times 9 \times 1\).
Upper chord	\(2—4\)	\(-480\)	1,21	\(-490\)	1,18	1,15	22	As \(0—1\) and \(1—3\).
Upper chord	\(2'—4'\)	\(-300\)	1,13	\(-310\)	1,17	1,16	78	As \(0—1\) and \(1—3\).
Upper chord	\(4—7\)	\(-480\)	1,08	\(-490\)	1,08	1,08	101	1 vertical plate \(40 \times 1\). 1 horizontal plate \(40 \times 0{,}8\). 1 horizontal plate \(40 \times 1{,}4\). 2 angles \(9 \times 9 \times 1\).
Upper chord	\(4'—7'\)	\(-350\)	1,09	\(-370\)	1,02	1,02	101	1 vertical plate \(40 \times 1\). 1 horizontal plate \(40 \times 0{,}8\). 1 horizontal plate \(40 \times 1{,}4\). 2 angles \(9 \times 9 \times 1\).
Upper chord	\(7—7'\)	\(-400\)	1,10	\(-430\)	1,08	1,08	99	As preceding \(+1\) horizontal plate \(20 \times 1\).

exclusively because of the unsuccessful selection of the sections of the additional struts.

We carried out, for the truss of Example 10, two parallel calculations under one and the same load and otherwise identical conditions—only in the second case were the sectional areas of all the additional struts doubled.

The load adopted is that indicated in the primary source—the report of Prof. Ritter and Tetmajer on the causes of the collapse of this truss (the bridge over the River Birs at Mönchenstein in Switzerland).

The scheme of the truss is given in Fig. 20, and the results of the calculation in Table G.

Fig. 21.

Table K.

Element Nos.	Length cm	$\omega$ cm$^2$	$I$ cm$^4$	$e_1=e_2$ cm	$n$ kg./cm$^2$	$\dfrac{N}{n}$
1—2	500	50	750	8.5	350	1.12
1—3	300	50	750	8.5	210	1.50
2—3	400	40	600	8.0	250	1.06
3—4	300	50	750	8.5	210	1.45
2—4	500	50	750	8.5	150	1.22

The table also gives the data of Eng. Paton, since they differ somewhat from the results obtained by us*), although not so much as to change the essence of the matter.

As is seen from the table, the initially very considerable values of the coefficients $\dfrac{N}{n}$ decrease extremely strongly when

*) Unfortunately, for lack of data it was not possible to establish precisely the cause of the discrepancy. Apparently the matter is explained by a change in the load in Eng. Paton’s work as compared with Ritter’s data, and by an error in determining the length of the diagonals (712 cm in Eng. Paton instead of 695 cm = $\sqrt{600^2+350^2}$, as it should be).

increases in the cross-sections of the posts, and instead of the greatest value 2.34 (according to Eng. Paton, 2.43) reach only 1.46 at the very insignificant value

\[ \frac{l}{e}=11. \]

But such values of the coefficient can easily be obtained in trusses of the system under consideration even for much larger values of

\[ \frac{l}{e}. \]

This is readily seen from consideration of the example of a simple truss shown in Fig. 21. As is evident from the table, the stresses in the chords and posts are almost equal; the ratio \(\frac{l}{e}\) is sufficiently large \(\left(\frac{l}{e}=35.3\right)\), yet nevertheless for members 1—3 and 3—4 considerable values of \(\frac{N}{n}\) were obtained: 1.50 and 1.45.

It is clear that trusses of a triangular system with additional posts, with respect to the stresses due to the rigidity of the joints, are in quite special conditions as compared with trusses of other systems. Therefore it seems impossible to us to base a diagram having general significance on data obtained from calculations of trusses of the indicated system.

Among the other examples included in the diagram, example 2 attracts attention: whereas for the other trusses the calculation was performed by Mohr’s method or by another method equivalent to it, example 2 was computed by a method belonging to group 4, which, as we have seen, gives very inaccurate results.

Without insisting on this last circumstance, however, we consider it, on the basis of the considerations set forth above, certainly necessary to exclude from the diagram cases 9, 10, 13, and XIII, as being entirely heterogeneous with the others.

As a result, the diagram takes the form shown in Fig. 18; it is clear that the most important part of the curve thereby proves to be wholly unfounded, and Eng. Paton’s position—that these diagrams can be used as an aid in determining allowable stresses—falls away of itself.

Of course, everything said with respect to the first diagram also applies to the remaining ones.

In general, the phenomenon of additional bending under the influence of rigid nodal connections is so complex and depends on such a variety of...

for various reasons, and the material collected to date is so heterogeneous that in any case it is premature to draw any general conclusions.

28. Critical review of the latest works.

In proceeding to examine the most recent investigations on the question of additional stresses due to the rigidity of joints, we must first of all note the article by E. O. Paton, “The So-Called Elastic Forces,” which appeared in Izv. Sobr. Inzh. P. S. for the year 1905.

This article is a response to the critical remarks of G. P. Perederii (in the work cited by us more than once) concerning E. O. Paton’s dissertation.

Analyzing the method of calculation proposed by Engineer Perederii, Engineer Paton asserts, first of all, that priority in indicating the role of “elastic forces” does not belong to Engineer Perederii, since as early as 1898 an article by A. Franke appeared in the Zeitschrift für Bauwesen, containing indications of the substantial influence, in certain cases, of “transverse forces” and outlining a scheme of calculation with these forces taken into account.

In Mr. Paton’s opinion, the methods of calculation of Franke and Perederii are essentially the same, but their justification is presented more elegantly and simply by the German scholar. Further, Engineer Paton finds that the assumption made by Engineer Perederii of hinged fastening of the web is, for the case under consideration of a two-diagonal truss, too crude an approximation.

Although German scholars, such as Müller-Breslau, Engesser, and others, accepted this assumption, they did so chiefly for trusses with curved chords, where the web has considerably less rigidity than in trusses with parallel chords.

Giving examples of the computation of bending moments from the rigidity of joints in the members of trusses with curved and parallel chords, Engineer Paton comes to the conclusion that, for trusses with parallel chords, the assumption made by Engineer Perederii is too inaccurate.

Then Engineer Paton, in opposition to Engineer Perederii’s method, proposes to use, for the accurate computation of bending moments, the “method of successive approximation,” i.e., starting from Mohr’s method, to calculate, according to the bending moments found...

“elastic forces,” then to recompute the forces in the members from the given load plus the elastic forces, again to compute, by Mohr’s method, the deformations of the truss and the moments due to the rigidity of the joints, and so on. According to Engineer Paton, already the third or fourth approximation gives sufficient accuracy.

As regards the calculation, carried out by Engineer Perederiy, of the additional stresses due to the rigidity of the joints for a two-panel truss (see above, § 24), E. O. Paton notes a significant error, in his opinion, in the computation of the “elastic forces”; until this error is corrected, it is impossible to say whether it will serve to the benefit or to the detriment of the two-panel truss.

With respect to his own calculations, Mr. Paton observes that he knew their conditional character and at the time pointed out, as one of the reasons for this conditional character, the neglect of the transverse component of the forces in the members—in other words, of the “elastic forces.” On the other hand, Mr. Paton seeks to rehabilitate Mohr’s method, asserting that the inaccuracy of the results acknowledged by him is due not to the application of Mohr’s method, but to the introduction of exaggerated longitudinal forces in the members of the pin-jointed truss as a consequence of the use of an improved method for calculating these forces.

Further, Engineer Paton seeks to justify the application of this improved method to multi-panel and multi-lattice trusses, contrasting it with the former method of calculating such trusses by decomposing them into simple systems, and proving that the substantial structural deficiencies of such trusses compel one to treat with caution the former method, which presupposes the perfectly serviceable action of the structure.

Without disputing the obvious fact of the successful existence of trusses of multi-panel and multi-lattice systems, Engineer Paton nevertheless finds that Engineer Perederiy’s calculations do not rehabilitate the indicated systems from the charge, brought against them in E. O. Paton’s dissertation, of being less suitable in comparison with simple systems.

The considerations just set forth by Engineer Paton did not remain unanswered by Engineer Perederiy, who responded to them in the article “Learned Equilibrists,” published in The Engineer’s Business for 1905.

Leaving aside the sharply polemical character of this article, it must be acknowledged that some of Engineer Perederiy’s remarks introduce a substantial correction to Engineer Paton’s assertions.

First of all, Engineer Perederiy correctly notes the rearrangement

subject of polemic in Mr. Paton’s new article; the dispute arose not over the general structural merits or shortcomings of double-lattice trusses, but over the special question of the magnitude of the additional stresses due to the rigidity of the joints and of the application of Mohr’s method to the calculation of these stresses in a double-lattice truss. The method again proposed by Mr. Paton for the exact determination of the indicated stresses by successive approximation does not achieve its aim, contrary to Engineer Paton’s assumptions: the calculations performed by Mr. Perederii prove this.

On the other hand, the error noted by Mr. Paton in Perederii’s calculation of the “elastic forces” turns out to be far from so significant—only 1.5%.

Then Engineer Perederii, with full justification, returns the reproach made against him of ignoring the actual conditions of operation of multiple-lattice trusses, since, after all, Engineer Paton in his work proceeded from the same theoretical considerations.

It seems to us, on the other hand, that Engineer Paton goes too far in his effort to rehabilitate Mohr’s method; of course, the reason for the inaccuracy of Engineer Paton’s calculations was the exaggeration of the longitudinal forces in the elements of the lattice, which in a hinged truss are more highly stressed than in a rigid one; however, it is precisely the approximate assumption of sufficient accuracy of the forces in the hinged truss that is taken as the basis of Mohr’s method and of methods similar to it.

This does not mean that we reject Mohr’s method altogether—in the preceding exposition we gave due credit to this highly elegant and convenient method—it means only that Mohr’s method is applicable under certain conditions, the observance of which must be monitored and which were not fulfilled in Engineer Paton’s calculations for the double-lattice truss.

On the cardinal point of disagreement between Messrs. Paton and Perederii—concerning the comparative accuracy of the calculations of the one author and the other—we have already expressed ourselves above (in § 24); now we shall note only that Engineer Perederii’s calculation does not everywhere give the same accuracy, and namely, nearer the supports its accuracy is lower than at midspan. This will be clear from the small table given below, where the values of the quantities \(\Sigma_k\), \(\Sigma_p\), and \(\chi\) are collected for all joints of Paton’s double-lattice truss (Fig. 13).

Joints	\(\Sigma_k\)	\(\Sigma_p\)	\(\chi = \dfrac{\Sigma_p}{\Sigma_k}\)
0	22.6	0	0
1	29.3	11.7	0.40

2	19.6	9.7	0.49
3	52.6	7.0	0.13
4	36.5	8.7	0.24
5	82.8	4.4	0.05
6	68.1	4.9	0.07
7	95.6	2.5	0.03
8	123.8	2.9	0.02
9	97.8	1.4	0.01
10	164.8	2.2	0.01
11	97.8	0.6	0.01

From consideration of this table one may, among other things, draw the conclusion that almost complete accuracy of the calculation would have been achieved if, in addition to the moments in the rigid contour, the rigid attachment of diagonals 1—2 and 1—4 had been taken into account in the calculation.

In addition to the articles by Messrs. Paton and Perederii that we have considered, in the area of interest to us it is necessary to point to the article by Eng. E. V. Zotikov: “Double-diagonal trusses and rigid joints” (Journal of the Ministry of Railways, 1905) and the reply by E. O. Paton prompted by it: “On the question of double-diagonal trusses” (The Engineer, 1906 and 1907).

This polemic is of interest to us not in its entirety, since both authors widened the scope of the dispute, one being an ardent defender and the other an equally ardent opponent of the double-diagonal system of trusses; we shall therefore cite only considerations directly related to the question occupying us.

E. V. Zotikov sets as the task of his investigation the same aim as that set forth in our work: “to verify the conclusions obtained by them (Paton and Perederii) with the aim of clarifying on whose side the truth lies, whose conclusions are closer to the truth.” His attention is directed above all to the excessively large absolute magnitude of the stresses calculated by E. O. Paton; including the various additional stresses, the total magnitude of the stress in the material of the trusses would have to reach 3—4000 kg/cm², which contradicts the available data on the service of such bridges.

The esteemed author sees the cause of such a contradiction quite rightly in the neglect of the rigidity of the chords, and seeks to prove the exaggeration of the stresses by means of an approximate method which, unfortunately, as Eng. Paton showed in his reply, does not fully achieve the aim.

E. V. Zotikov removes diagonals 7—10 and 7′—10 (Fig. 13) at joint 10 of the lower chord, which, of course, places the segment 8—8′ of the lower chord in worse conditions compared with the actual ones;

Moreover, he assumes the ends 8 and \(8'\) of the chord to be horizontally fixed, which also constitutes the most unfavorable assumption.

Next, E. V. Zotikoff considers the bending of two chords, \(8—8'\) and \(9—9'\), connected in the middle by the post \(10—11\) and loaded at point 10 by the design load; under these most unfavorable assumptions, the stresses turn out to be half as large as those obtained by Eng. Paton; from this one might a fortiori conclude that Paton’s data are in general greatly exaggerated. E. O. Paton’s calculations showed, however, that the desired result is obtained only for the middle panel; in other cases the stresses by E. V. Zotikoff’s method are much greater—even greater than those obtained by Paton. In any case this method, based on a separate consideration of only part of the truss, could not give any sufficiently accurate representation of the actual distribution of stresses.

Further, E. V. Zotikoff criticizes Mohr’s method and points out a number of contradictions to which this method leads if its application is not limited only to trusses with chords of small rigidity. In reply to this, E. O. Paton again emphasizes the acknowledged conditional character of his calculations.

Finally, E. V. Zotikoff warmly defends double-diagonal and double-lattice systems, proving that the additional stresses in trusses with a simple lattice system may reach a very large magnitude. As proof he cites the truss of the bridge over the river Birs, which we analyzed in § 27; as we saw, this example cannot be regarded as a conclusive argument against simple systems, in view of the role played by the additional posts in the deformation of the chords and in view of the inaccuracy of Mohr’s method for this case as well.

On the other hand, E. V. Zotikoff cites a number of experimental data which, in his opinion, confirm that the actual stresses in double-diagonal and double-lattice trusses are less than those computed theoretically, which demonstrates the beneficial influence of rigid chords. Without entering into a detailed critique of these data, we shall merely point out that Eng. Paton disputes the significance of these investigations very vigorously and, in our opinion, quite justifiably.

From the survey we have made of the most recent works, the numerous gaps in the study of the question of additional stresses due to the rigidity of nodal connections are sufficiently clear. It may therefore be that the present work will prove useful in elucidating certain details of this highly complex question.